Calculus 9e Purcell-Varberg-Rigdon (Solution).

16. ( 5− 3) = ( 5) − 2( 5)( 3) + ( 3)17.18.19.20.21.22.23.24.25.26.2 2 212 4 227.2x + 2x + x + x+2= 5− 2 15+ 3= 8−2 1512 4( x + 2) 2x= + +xx ( + 2) xx ( + 2) xx ( + 2)2(3x− 4)( x+ 1) = 3x + 3x−4x−412 + 4x+ 8 + 2x 6x+202= == 3x−x−4xx ( + 2) xx ( + 2)2(3x+ 10)2(2x− 3) = (2x−3)(2x−3)=xx ( + 2)2= 4x −6x− 6x+922 y= 4x− 12x+928. +6y− 2 29y−12(3 x– 9)(2x+ 1) = 6x + 3 x–18 x– 92y= +2= 6 x –15 x–92(3y− 1) (3y+ 1)(3y−1)2(3y+1) 2y= +2(4x−11)(3x− 7) = 12x −28x− 33x+772(3y+ 1)(3 y− 1) 2(3y+ 1)(3y−1)2= 12x− 61x+776y+ 2+2y8y+ 2==2(3y+ 1)(3 y−1)2(3y+ 1)(3y−1)2 2 2 2(3t − t+ 1) = (3t − t+ 1)(3t − t+1)2(4y+ 1) 4y+1= =4 3 2 3 2 2= 9t − 3t + 3t − 3t + t − t+ 3t − t+12(3y+ 1)(3y− 1) (3y+ 1)(3y−1)4 3 2= 9t − 6t + 7t − 2t+1029. a. 00 ⋅ = 0 b.30(2t+ 3) = (2t+ 3)(2t+ 3)(2t+3)2= (4t + 12t+ 9)(2t+3)0c. 03 2 217 = d. 30= 8t + 12t + 24t + 36t+ 18t+273 2= 8t + 36t + 54t+2750e. 0 = 0 f. 17 = 1x 2 –4 ( x–2)( x+2)= = x + 2 , x ≠ 230. If 0 x–2 x–202x −x−6 ( x− 3)( x+2)= = x + 2 , x ≠ 3single value satisfies 0 = a .x−3 ( x−3)02t –4 t –21 ( t+3)( t –7) 31. .083= = t – 7 , t ≠− 3t+ 3 t+312 1.0009622x−2x 2 x(1 −x)=403 2 2x − 2 x + x x( x − 2x+1)36−2 xx ( −1)=4xx ( −1)( x−1)2=−x − 1is undefined.is undefined.= a , then 0= 0⋅ a , but this is meaninglessbecause a could be any real number. No2 Section 0.1 Instructor’s Resource Manual

32. .2857147 2.000000146056403550491073028233. .14285721 3.0000002190846042180168120105150147334. .294117...17 5.000000... → 0.2941176470588235341601537068201730171301191135. 3.63 11.092018236. .84615313 11.00000010 4605280782013706550391137. x = 0.123123123...1000x= 123.123123...x = 0.123123...999x= 123123 41x = =999 33338. x = 0.217171717…1000x= 217.171717...10x= 2.171717...990x= 215215 43x = =990 19839. x = 2.56565656...100x= 256.565656...x = 2.565656...99x= 254254x =9940. x = 3.929292…100x= 392.929292...x = 3.929292...99x= 389389x =99Instructor’s Resource Manual Section 0.1 3

62.2 2 3= (8.004) (270) − (8) (270) ≈ 54.3 ft.V π π63. a. If I stay home from work today then itrains. If I do not stay home from work,then it does not rain.b. If the candidate will be hired then shemeets all the qualifications. If thecandidate will not be hired then she doesnot meet all the qualifications.64. a. If I pass the course, then I got an A on thefinal exam. If I did not pass the course,thn I did not get an A on the final exam.b. If I take off next week, then I finished myresearch paper. If I do not take off nextweek, then I did not finish my researchpaper.65. a. If a triangle is a right triangle, then2 2 2 a + b = c . If a triangle is not a right2 2 2triangle, then a + b ≠ c .b. If the measure of angle ABC is greater than0 o and less than 90 o , it is acute. If themeasure of angle ABC is less than 0 o orgreater than 90 o , then it is not acute.66. a. If angle ABC is an acute angle, then itsmeasure is 45 o . If angle ABC is not anacute angle, then its measure is not 45 o .2 2b. If a < b then a < b.If aa ≥ b.67. a. The statement, converse, andcontrapositive are all true.b. The statement, converse, andcontrapositive are all true.2 2≥ b then68. a. The statement and contrapositive are true.The converse is false.b. The statement, converse, andcontrapositive are all false.69. a. Some isosceles triangles are notequilateral. The negation is true.b. All real numbers are integers. The originalstatement is true.c. Some natural number is larger than itssquare. The original statement is true.b. Every circle has area less than or equal to9π. The original statement is true.c. Some real number is less than or equal toits square. The negation is true.71. a. True; If x is positive, thenb. False; Take x =− 2 . Thenx < 0 .c. False; Take2x is positive.2x > 0 but12 1x = . Then x = < x24d. True; Let x be any number. Take2y = x + 1 . Then y > x .e. True; Let y be any positive number. Takeyx = . Then 0 < x < y .272. a. True; x ( x) x ( x)+ − < + 1 + − :0 .xxd. True; 1/n can be made arbitrarily closeto 0.e. True; 1/ 2 n can be made arbitrarily closeto 0.73. a. If n is odd, then there is an integer k suchthat n = 2k+ 1. Then2 2 2n = (2k + 1) = 4k + 4k+12= 2(2k+ 2 k) + 1b. Prove the contrapositive. Suppose n iseven. Then there is an integer k such that2 2 2 2n = 2 k.Then n = (2 k) = 4k = 2(2 k ) .2Thus n is even.74. Parts (a) and (b) prove that n is odd if and2only if n is odd.75. a. 243= 33333 ⋅ ⋅ ⋅ ⋅b.2124 = 4⋅ 31 = 2⋅2⋅31 or 2 ⋅ 31270. a. Some natural number is not rational. Theoriginal statement is true.Instructor’s Resource Manual Section 0.1 5

c.5100 = 2⋅ 2550 = 2⋅2⋅1275= 2⋅2⋅3⋅ 425 = 2⋅2⋅3⋅5⋅8576. For example, let77.2 2 4 62 2= 2235517 ⋅ ⋅ ⋅ ⋅ ⋅ or 2 ⋅35 ⋅ ⋅17A = b⋅c 2 ⋅ d3 ; thenA = b ⋅c ⋅ d , so the square of the numberis the product of primes which occur an evennumber of times.2p p 2 22 ; 2 ; 2 q p ;q 2q= = = Since the primefactors of p 2 must occur an even number oftimes, 2q 2 pwould not be valid and 2q =must be irrational.22 278. 3 = p p; 3 ; 3 q p ;q= 2q= Since the prime2factors of p must occur an even number of2ptimes, 3q would not be valid and 3q =must be irrational.79. Let a, b, p, and q be natural numbers, so a band p a p aq + bpare rational. + = Thisq b q bqsum is the quotient of natural numbers, so it isalso rational.p80. Assume a is irrational, ≠ 0 is rational, andqp rq⋅ra ⋅ = is rational. Then a = isq sp⋅srational, which is a contradiction.81. a. – 9 = –3; rational82. a. –2 b. –2c. x = 2.4444...;10x= 24.4444...x = 2.4444...9x= 2222x =9d. 1e. n = 1: x = 0, n = 2:f. 25n = 4: x =4The upper bound is 3 .23x = , n = 3:22x = – ,383. a. Answers will vary. Possible answer: Anexample2is S = { x: x < 5, x a rational number}.Here the least upper bound is 5, which isreal but irrational.b. True0.2 Concepts Review1. [ −1,5);( −∞, − 2]2. b > 0; b < 03. (b) and (c)4. −1≤ x ≤ 5Problem Set 0.21. a.b.30.375 = ; rational8b.c. (3 2)(5 2) = 15 4 = 30; rationald.2(1+ 3) = 1+ 2 3 + 3 = 4 + 2 3;irrationalc.d.6 Section 0.2 Instructor’s Resource Manual

e.f.9.–3 < 1– 6x≤ 4–4 < –6x≤32 1 ⎡ 1 2⎞> x ≥ – ; – ,3 2 ⎢ ⎟ ⎣ 2 3 ⎠2. a. (2,7) b. [ − 3,4)c. ( −∞, − 2] d. [ − 1,3]3. x− 7< 2x−5− 2 < x;( −2, ∞)10. 4< 5− 3x< 7− 1< − 3x< 21 2 ⎛ 2 1⎞> x >− ; ⎜ − , ⎟3 3 ⎝ 3 3⎠4.5.3x− 5< 4x−6( )1 < x; 1, ∞7 x–2≤ 9x+3–5 ≤ 2x5 ⎡ 5 ⎞x ≥ – ; – ,2 ⎢ ∞ ⎟⎣ 2 ⎠11. x 2 + 2x – 12 < 0;2–2 ± (2) – 4(1)(–12) –2±52x = =2(1) 2= –1±13⎡x– (–1+ 13 )⎤⎡x– ( –1– 13)⎤ < 0;⎣ ⎦⎣ ⎦(–1– 13, –1+13)6. 5x− 3> 6x−41 > x;( −∞,1)12.2x −5x− 6>0( x+ 1)( x− 6) > 0;( −∞, −1) ∪(6, ∞ )7. − 4< 3x+ 2 0; (2x – 1)(x + 3) > 0;⎛1⎞( −∞, −3) ∪⎜, ∞⎟⎝2⎠8. − 3< 4x− 9

16.17.3x− 2 ⎛ 2⎤≥0; ⎜ −∞, ∪(1, ∞)x −1 3⎥⎝ ⎦2< 5x2− 5 < 0x2−5x< 0;x⎛2⎞(– ∞, 0) ∪⎜, ∞⎟⎝5⎠20.3> 2x + 53− 2>0x + 53− 2( x + 5)> 0x + 5−2x−7 ⎛ 7⎞> 0; ⎜−5,− ⎟x + 5 ⎝ 2⎠21. ( x+ 2)( x−1)( x− 3) > 0; ( −2,1) ∪ (3,8)18.7≤ 74x22.⎛ 3⎞ ⎛1⎞(2x+ 3)(3x−1)( x− 2) < 0; ⎜−∞, − ⎟∪⎜ ,2⎟⎝ 2⎠ ⎝3⎠7−7≤04x7−28x≤ 0;4x⎡1⎞( −∞,0 ) ∪ ⎢ , ∞ ⎟⎣4⎠23.2(2 -3)( -1) ( -3) 0;3⎤2⎥⎦x x x ≥ – ∞, ∪[ 3, ∞)⎛⎜⎝19.1≤ 43x− 224.2(2x− 3)( x−1) ( x− 3) > 0;⎛ 3 ⎞⎜ ⎟⎝ 2 ⎠( −∞,1) ∪ 1, ∪ ( 3, ∞)1−4≤03x− 21−4(3x−2)≤ 03x− 29−12x⎛ 2⎞ ⎡3⎞≤0; ⎜−∞, ⎟∪ , ∞⎟3x− 2 3 ⎢⎝ ⎠ ⎣4⎠25.3 2x – 5x – 6x02( x −1)( x− 1) > 02( x+ 1)( x− 1) > 0;( −1,1) ∪(1, ∞ )27. a. False. b. True.c. False.8 Section 0.2 Instructor’s Resource Manual

28. a. True. b. True.c. False.2229. a. ⇒ Let a < b, so ab 0 , we can divide by 2b to getb− a > 0 .b. We can divide or multiply an inequality byany positive number.a 1 1a 1 and 2x + 1 < 33x > –6 and 2x < 2x > –2 and x < 1; (–2, 1)b. 3x + 7 > 1 and 2x + 1 > –43x > –6 and 2x > –55x > –2 and – ; −2,∞2x > ( )c. 3x + 7 > 1 and 2x + 1 < –45x > –2 and x < – ; ∅232. a. 2x− 7 > 1 or 2x+ 1 8 or 2x 4 or x

36. x + 2 < 1;–1< x + 2 6;x1 1− 36xx1 1+ 3< 0 or − 9>0x x1+ 3x1−9x< 0 or > 0;x x⎛ 1 ⎞ ⎛ 1⎞⎜−,0⎟∪⎜0,⎟⎝ 3 ⎠ ⎝ 9⎠38. 2 x –1 > 2;2x – 1 < –2 or 2x – 1 > 22x < –1 or 2x > 3;1 3 ⎛ 1⎞ ⎛3⎞x< – or x > , ⎜– ∞, – ⎟∪⎜ , ∞⎟2 2 ⎝ 2⎠ ⎝2⎠39.2x−5 ≥772x2x−5≤ −7 or −5≥77 72x2x≤−2 or ≥127 7x≤−7 or x≥42;( −∞, −7] ∪[42, ∞ )x40. + 1 < 14x− 1< + 1 1;5x− 6< −1 or 5x− 6>15x< 5 or 5x> 77 ⎛7⎞x< 1 or x > ;( −∞,1) ∪⎜, ∞⎟5 ⎝5⎠42. 2 x –7 > 3;2x – 7 < –3 or 2x – 7 > 32x < 4 or 2x > 10x < 2 or x > 5; ( −∞,2) ∪ (5, ∞ )44.45.46.52+ > 1;x5 52 + < –1 or 2+ > 1xx5 53+ < 0 or 1+ > 0x x3x+ 5 x+5< 0 or > 0;x x⎛ 5 ⎞(– ∞, – 5) ∪⎜– , 0 ⎟∪(0, ∞)⎝ 3 ⎠x2− 3x−4≥0;23 ± (–3) – 4(1)(–4) 3±5x = = = –1, 42(1) 2( x+ 1)( x− 4) = 0;( −∞, −1] ∪[4, ∞ )2 4 ± ( −4) −4(1)(4)x − 4x+ 4≤ 0; x = = 22(1)( x−2)( x−2) ≤ 0; x = 247. 3x 2 + 17x – 6 > 0;48.2–17 ± (17) – 4(3)(–6) –17 ± 19 1x = = = –6,2(3) 6 3⎛1⎞(3x – 1)(x + 6) > 0; (– ∞, – 6) ∪⎜, ∞⎟⎝3⎠214x+ 11x−15 ≤ 0;2− 11 ± (11) −4(14)( −15) − 11±31x = =2(14) 283 5x =− ,2 7⎛ 3 5 3 5⎜x+ ⎞⎛ ⎟⎜x− ⎞ ⎟≤0; ⎡ − ,⎤2 7 ⎢ 2 7⎥⎝ ⎠⎝ ⎠ ⎣ ⎦249. x− 3 < 0.5 ⇒5 x− 3 < 5(0.5) ⇒ 5x− 15 < 2.550. x+ 2 < 0.3⇒ 4 x+ 2 < 4(0.3) ⇒ 4x+ 18 < 1.210 Section 0.2 Instructor’s Resource Manual

ε51. x− 2 < ⇒6 x− 2 < ε ⇒ 6x− 12 < ε6ε52. x+ 4 < ⇒ 2 x+ 4 < ε ⇒ 2x+ 8 < ε253. 3x− 15 < ε ⇒ 3( x− 5) < ε⇒3 x − 5 < εε ε⇒ x − 5 < ; δ =3 354. 4x− 8 < ε ⇒ 4( x− 2) < ε⇒ 4 x − 2 < εε ε⇒ x − 2 < ; δ =4 455. 6x+ 36 < ε ⇒ 6( x+ 6) < ε⇒ 6 x + 6 < εε ε⇒ x + 6 < ; δ =6 656. 5x+ 25 < ε ⇒ 5( x+ 5) < ε⇒ 5 x + 5 < εε ε⇒ x + 5 < ; δ =5 557. C = π dC –10 ≤ 0.02πd–10 ≤0.02⎛ 10 ⎞π⎜d– ⎟ ≤0.02⎝ π ⎠10 0.02d – ≤ ≈0.0064π πWe must measure the diameter to an accuracyof 0.0064 in.5−50 ≤1.5, −32 −50 ≤ 1.5;95( F −32)−90 ≤1.59F −122 ≤ 2.758. C( F )We are allowed an error of 2.7 F.59.60.61.62.x–1 < 2 x–3x–1 < 2 x–62 2( x–1) < (2 x–6)2 2– –2 –x 2x+ 1< 4x 24x+363x22x+ 35>0(3 x– 7)( x– 5) > 0;⎛ 7 ⎞⎜– ∞, ⎟∪(5, ∞)⎝ 3 ⎠2x−1 ≥ x+12( x ) 2(2x−1) ≥ + 12 24x − 4x+ 1≥ x + 2x+123x−6x≥03 xx ( −2) ≥0( −∞,0] ∪[2, ∞ )22x− 3< x+104x− 6 < x+102 2(4x− 6) < ( x+10)2 216x − 48x+ 36 < x + 20x+100215x− 68x− 64 < 0(5x+ 4)(3x− 16) < 0;⎛ 4 16⎞⎜– , ⎟⎝ 5 3 ⎠2 2(3x− 1) < (2x+12)2 29x − 6x+ 1 < 4x + 48x+14425x−54x− 143

63.x < y ⇒ x x ≤ x y and x y < y y Order property: x< y ⇔ xz < yz when z is positive.2 2⇒ x < yTransitivity2( )2 2 2⇒ x < y x = xConversely,( x y )( x y )( )2 2 2 2 2 2x < y ⇒ x < y x = x2 2 2⇒ x – y < 0 Subtract y from each side.⇒ – + < 0 Factor the difference of two squares.⇒ x – y < 0 This is the only factor that can be negative.⇒ x < y Add y to each side.0 < < ⇒ = 2 = 2 , so64. a b a ( a ) and b ( b )( a ) ( b )2 2,< and, by Problem 63,a < b ⇒ a < b .65. a. a– b = a+ (– b) ≤ a + – b = a + bb. a– b ≥ a – b ≥ a – b Use Property 4of absolute values.c. a+ b+ c = ( a+ b)+ c ≤ a+ b + c≤ a + b + c67.x– 2 x+(–2)=2 2x + 9 x + 9x– 2 x –2≤ +2 2 2x + 9 x + 9 x + 9x –2 x 2 x + 2≤ + =2 2 2 2x + 9 x + 9 x + 9 x + 92 1 1Since x + 9≥9,≤2x + 9 9x + 2 x + 2≤2x + 9 9x –2 x + 2≤2x + 9 966.1 1 1 ⎛ 1 ⎞− = + −2 2 2x + 3 x + x + 3⎜ x 2⎟⎝ + ⎠1 1≤ + −2x + 3 x + 21 1= +2x + 3 x + 21 1= +2x + 3 x + 2by the Triangular Inequality, and since21 1x + 3>0, x + 2 > 0 ⇒ 0, 0.2x + 3> x + 2>2x + 3≥ 3 and x + 2≥ 2, so1 1 1 1≤ and ≤ , thus,2x + 3 3 x + 2 21 + 1 ≤ 1 +12x + 3 x + 2 3 268.69.2 2x ≤ 2⇒ x + 2x+ 7 ≤ x + 2x+ 7≤ 4+ 4+ 7=1521and x + 1 ≥ 1 so ≤ 1.2x + 1Thus,x2+ 2x+7 1x2= x + 2x+72 2+ 1 x + 1≤15⋅ 1 = 154 1 3 1 2 1 1x + x + x + x+2 4 8 164 1 3 1 2 1 1≤ x + x + x + x +2 4 8 161 1 1 1≤ 1 + + + + since x ≤1.2 4 8 164 1 3 1 2 1 1So x + x + x + x+ ≤ 1.9375 < 2.2 4 8 1612 Section 0.2 Instructor’s Resource Manual

70. a.b.x 122x < x2x − x

Problem Set 0.31.5.2 2d 1 = (5 + 2) + (3 – 4) = 49 + 1 = 50d2= (5 − 10)2+ (3 − 8)2= 25 + 25 = 50d3= ( −2− 10)2+ (4−8)2= 144 + 16 = 160d1= d2so the triangle is isosceles.2.2 2d = (3 –1) + (1–1) = 4 = 26.2 2a = (2 − 4) + ( −4 − 0) = 4 + 16 = 202 2b = (4 − 8) + (0 + 2) = 16 + 4 = 202 2c = (2 − 8) + ( − 4 + 2) = 36 + 4 = 402 2a + b = c2 , so the triangle is a right triangle.7. (–1, –1), (–1, 3); (7, –1), (7, 3); (1, 1), (5, 1)8.2 2 2 2( x− 3) + (0 − 1) = ( x− 6) + (0 −4) ;2 2x − 6x+ 10= x − 12x+526x= 42( )x = 7⇒7,03.2 2d = ( −3− 2) + (5 + 2) = 74 ≈ 8.609.⎛–2+ 4 –2+3⎞ ⎛ 1⎞⎜ , ⎟=⎜1, ⎟;⎝ 2 2 ⎠ ⎝ 2⎠122 ⎛ ⎞ 25d = (1 + 2) + ⎜ – 3⎟= 9 + ≈3.91⎝2 ⎠ 44.2 2d = (4 – 5) + (5 + 8) = 170 ≈ 13.04⎛1+ 2 3+6⎞ ⎛3 9⎞10. midpoint of AB = ⎜ , ⎟=⎜ , ⎟⎝ 2 2 ⎠ ⎝2 2⎠⎛4+ 3 7+4⎞ ⎛7 11⎞midpoint of CD = ⎜ , ⎟=⎜ , ⎟⎝ 2 2 ⎠ ⎝2 2 ⎠2 2⎛3 7⎞ ⎛9 11⎞d = ⎜ − ⎟ + ⎜ − ⎟⎝2 2⎠ ⎝2 2 ⎠= 4+ 1= 5 ≈ 2.2411 (x – 1) 2 + (y – 1) 2 = 112.2 2 2( x+ 2) + ( y− 3) = 42 2( x+ 2) + ( y− 3) = 1613.2 2 2( x − 2) + ( y+ 1) = r2 2 2(5 − 2) + (3 + 1) = r2r = 9+ 16=252 2( x− 2) + ( y+ 1) = 252 2d = ( −1− 6) + (5− 3) = 49+ 4 = 53≈ 7.2814 Section 0.3 Instructor’s Resource Manual

14.15.2 2 2( x − 4) + ( y− 3) = r2 2 2(6 − 4) + (2 − 3) = r2r = 4+ 1=52 2( x− 4) + ( y− 3) = 5⎛1+ 3 3+7⎞center = ⎜ , ⎟=(2, 5)⎝ 2 2 ⎠1 2 2 1radius = (1– 3) + (3 – 7) = 4 + 162 21= 20 = 522 2( x–2) + ( y–5) = 516. Since the circle is tangent to the x-axis, r = 4.2 2( x− 3) + ( y− 4) = 1617.18.2 2x x y y2 2x x y y2 2x x y y2 2x y+ 2 + 10 + –6 –10=0+ 2 + –6 = 0( + 2 + 1) + ( –6 + 9) = 1+9( + 1) + ( –3) = 10center = (–1, 3); radius = 102 2x y y2 2x y y2 2x y+ − 6 = 16+ ( − 6 + 9) = 16+9+ ( − 3) = 25center = (0, 3); radius = 521.22.23.25.2 2x x y y4 + 16 + 15+ 4 + 6 = 02 ⎛ 2 3 9 ⎞94( x + 4x+ 4) + 4⎜y + y+ ⎟=− 15 + 16 +⎝ 2 16⎠42 ⎛ 3⎞134( x+ 2) + 4⎜y+ ⎟ =⎝ 4⎠42 ⎛ 3⎞13( x+ 2) + ⎜y+ ⎟ =⎝ 4⎠16⎛ 3 ⎞center = ⎜−2, − ⎟;radius =⎝ 4 ⎠221342 105 24x + 16x+ + 4y + 3y= 0162 ⎛ 2 3 9 ⎞4( x + 4x+ 4) + 4⎜y + y+⎟⎝ 4 64⎠105 9=− + 16 +16 1622 ⎛ 3 ⎞4( x+ 2) + 4⎜y+ ⎟ = 10⎝ 8 ⎠22 ⎛ 3⎞5( x+ 2) + ⎜y+ ⎟ =⎝ 8⎠2⎛ 3⎞5 10center = ⎜−2, − ⎟; radius = =⎝ 8⎠2 22–1 12–1 = 24. 7− 5 = 24−3–6–3 9= 26.–5–2 7− 6+ 4 = 10−219.2x + y 2 –12x+ 35 = 02 2x –12 x+ y = –352 2( x –12x+ 36) + y = –35 + 362 2( x–6) + y = 1center = (6, 0); radius = 127.5–0 5= – 28.0–3 329. y− 2=−1( x−2)y− 2= − x+2x+ y− 4=06− 0 = 10+620.2 2x y x y2 2x x y y2 2x y+ − 10 + 10 = 0( − 10 + 25) + ( + 10 + 25) = 25 + 25( − 5) + ( + 5) = 50center = ( 5, − 5 ); radius = 50 = 5 230. y− 4=−1( x−3)y− 4= − x+3x+ y− 7=031. y = 2x+ 32 x– y+ 3=032. y = 0x+ 50x+ y− 5=0Instructor’s Resource Manual Section 0.3 15

33.34.8–3 5m = = ;4–2 25y– 3 = ( x–2)22 y–6=5 x–105 x–2 y–4=02−1 1m = = ;8−4 41y− 1 = ( x−4)44y− 4= x−4x− 4y+ 0=02 135. 3y = –2x + 1; y = – x+;3 31y -intercept =336. − 4y= 5x−65 3y =− x+4 25 3slope =− ; y-intercept=4 237. 6–2y= 10 x–2–2y= 10 x–8y = –5x+4;slope = –5; y-intercept = 438. 4x+ 5y= −205y=−4x−204y =− x−454slope =− ; -intercept = 45 y−39. a. m = 2;y+ 3=2( x–3)y = 2 x–9b.1m = – ;21y+ 3 = – ( x–3)21 3y = – x–2 22slope = – ;3c. 2x+ 3y= 63 y = –2x+62y = – x+2;32m = – ;32y+ 3 = – ( x–3)32y = – x–13d.e.3m = ;23y+ 3 = ( x–3)23 15y = x–2 2–1–2 3m = = – ;3+1 43y+ 3 = – ( x–3)43 3y = – x–4 4f. x = 3 g. y = –340. a. 3x+ cy = 53(3) + c(1) = 5c = −4b. c = 0c. 2x+ y =−1y = −2x−1m = − 2;3x+ cy = 5cy = − 3x+ 53 5y = − x+c c3− 2 =−c3c =2d. c must be the same as the coefficient of x,so c = 3.16 Section 0.3 Instructor’s Resource Manual

41.e. y− 2= 3( x+3);1perpendicular slope =− ;31 3− =−3 cc = 93m = ;23y+ 1 = ( x+2)23y = x+2242. a. m = 2;kx − 3y= 10− 3y= − kx+10k 10y = x−3 3k= 2; k = 63b.1m =− ;2k 1=−3 23k =−2c. 2x+ 3y= 63y=− 2x+62y =− x+2;33 3 9m = ; k = ; k =2 3 2 243. y = 3(3) – 1 = 8; (3, 9) is above the line.44.b−0 b( a,0),(0, b);m = = −0 − a ab bx x yy =− x+ b; + y = b; + = 1a a a b45. 2x+ 3y= 4–3x+ y = 52x+ 3y= 49 x–3 y = –1511 x = –11x = –1–3(–1) + y = 5y = 2Point of intersection: (–1, 2)3 y = –2x+42 4y = – x+3 33m =23y− 2 = ( x+1)23 7y = x+2 246. 4x− 5y= 82x+ y =−104x− 5y= 8−4x− 2y= 20− 7y= 28y = −44x− 5( − 4) = 84x= −12x = −3Point of intersection: ( −3, − 4 );4x− 5y= 8− 5y=− 4x+84 8y = x−5 55m = −45y+ 4 =− ( x+3)45 31y =− x−4 4Instructor’s Resource Manual Section 0.3 17

47. 3 x–4y= 52x+ 3y= 99 x–12y= 158x+ 12y= 3617 x = 51x = 33(3) – 4 y = 5–4 y = –4y = 1Point of intersection: (3, 1); 3x – 4y = 5;–4 y = –3x+53 5y = x–4 44m = – 34y–1 = – ( x–3)34y = – x+5348. 5 x–2y= 52x+ 3y= 615 x– 6y= 154x+ 6y= 1219 x = 2727x =19⎛27⎞2⎜⎟ + 3y= 6⎝19⎠603y=1920y =19⎛27 20 ⎞Point of intersection: ⎜ , ⎟;⎝19 19 ⎠5x− 2y= 5–2 y = –5x+55 5y = x–2 22m = – 520 2 ⎛ 27 ⎞y– = – ⎜x−⎟19 5 ⎝ 19 ⎠2 54 20y = – x+ + 5 95 192 154y =− x+5 9518 Section 0.3 Instructor’s Resource Manual49.⎛2+ 6 –1+3⎞ center: ⎜ , ⎟ = (4, 1)⎝ 2 2 ⎠⎛2+ 6 3+3⎞midpoint = ⎜ , ⎟=(4, 3)⎝ 2 2 ⎠2 2inscribed circle: radius = (4 – 4) + (1– 3)= 4 = 22 2( x–4) + ( y–1) = 4circumscribed circle:2 2radius = (4 – 2) + (1– 3) = 82 2( x–4) + ( y–1) = 850. The radius of each circle is 16 = 4. The centersare ( 1, −2 ) and ( − 9,10 ).The length of the belt isthe sum of half the circumference of the firstcircle, half the circumference of the second circle,and twice the distance between their centers.1 12 2L = ⋅ 2 π(4) + ⋅ 2 π(4) + 2 (1+ 9) + ( −2−10)2 2= 8π+ 2 100 + 144≈ 56.3751. Put the vertex of the right angle at the originwith the other vertices at (a, 0) and (0, b). The⎛a b⎞midpoint of the hypotenuse is ⎜ , ⎟.⎝2 2⎠ Thedistances from the vertices are2 2 2 2⎛ a⎞ ⎛ b⎞a b⎜a– ⎟ + ⎜0–⎟ = +⎝ 2⎠ ⎝ 2⎠4 41 2 2= a + b ,22 2 2 2⎛ a⎞ ⎛ b⎞a b⎜0– ⎟ + ⎜b– ⎟ = +⎝ 2⎠ ⎝ 2⎠4 41 2 2= a + b , and22 2 2 2⎛ a⎞ ⎛ b⎞a b⎜0– ⎟ + ⎜0–⎟ = +⎝ 2⎠ ⎝ 2⎠4 41 2 2= a + b ,2which are all the same.52. From Problem 51, the midpoint of thehypotenuse, ( 4,3, ), is equidistant from thevertices. This is the center of the circle. Theradius is 16 + 9 = 5. The equation of thecircle is2 2( x− 4) + ( y− 3) = 25.

53.54.x2 + y2 – 4 x–2 y–11=02 2( x –4x+ 4) + ( y –2y+ 1) = 11+ 4+12 2( x– 2) + ( y–1) = 162 2x + y + 20 x–12y+ 72 = 02 2( x + 20x+ 100) + ( y –12y+36)= –72 + 100 + 362 2( x+ 10) + ( y– 6) = 64center of first circle: (2, 1)center of second circle: (–10, 6)2 2d = (2 + 10) + (1– 6) = 144 + 25= 169 = 13However, the radii only sum to 4 + 8 = 12, sothe circles must not intersect if the distancebetween their centers is 13.2 2x + ax+ y + by+ c = 0⎛ 2 22 a ⎞ ⎛2 b ⎞x + ax+ + y + by+⎜ 4 ⎟ ⎜ 4 ⎟⎝ ⎠ ⎝ ⎠2 2a b=− c + +4 42 2 2 2⎛ a⎞ ⎛ b⎞a + b −4c⎜x+ ⎟ + ⎜y+ ⎟ =⎝ 2⎠ ⎝ 2⎠42 2a + b −4c2 2> 0⇒ a + b > 4c455. Label the points C, P, Q, and R as shown in thefigure below. Let d = OP , h = OR , anda = PR . Triangles Δ OPR and Δ CQR aresimilar because each contains a right angle andthey share angle ∠ QRC . For an angle of30 d 3,h = 2and a 1h 2ah = 2⇒ = . Using aproperty of similar triangles, QC / RC = 3/2,2 3 4= → a = 2 +a − 2 2 3By the Pythagorean Theorem, we have2 2d = h − a = 3a= 2 3 + 4 ≈ 7.46456. The equations of the two circles are2 2 2( x − R) + ( y− R)= R2 2 2( x − r) + ( y− r)= rLet ( aa , ) denote the point where the twocircles touch. This point must satisfy2 2 2( a− R) + ( a− R)= R22 R( a− R)=2⎛ 2 ⎞a = ⎜1±R2 ⎟⎝ ⎠⎛ 2 ⎞Since a < R,a = ⎜1 −R.2 ⎟⎝ ⎠At the same time, the point where the twocircles touch must satisfy2 2 2( a− r) + ( a− r)= r⎛ 2 ⎞a = ⎜1±r2 ⎟⎝ ⎠⎛ 2 ⎞Since a > r,a = ⎜1 +r.2 ⎟⎝ ⎠Equating the two expressions for a yields⎛ 2 ⎞ ⎛ 2 ⎞⎜1− R = 1+r2 ⎟ ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠2⎛22 ⎞1− ⎜1−22 ⎟r = R =⎝ ⎠R2 ⎛ 2 ⎞⎛ 2 ⎞1+ 1 12 ⎜+ −2 ⎟⎜ 2 ⎟⎝ ⎠⎝ ⎠11− 2+r =2R11−2r = (3 −2 2) R ≈ 0.1716RInstructor’s Resource Manual Section 0.3 19

57. Refer to figure 15 in the text. Given ine l 1 withslope m, draw58. ABC with vertical andhorizontal sides m, 1.Line l 2 is obtained from l 1 by rotating itaround the point A by 90° counter-clockwise.Triangle ABC is rotated into triangle AED .We read offslope of l2= 1 1m=− .− m2 2 2 22 ( x− 1) + ( y− 1) = ( x− 3) + ( y−4)2 24( x − 2x+ 1+ y − 2y+1)2 2= x − 6x+ 9 + y − 8y+162 23x − 2x+ 3y= 9+ 16−4−4;2 2 2 2 2 173x − 2x+ 3y = 17; x − x+ y = ;3 3⎛ 2 2 1⎞2 17 1⎜x − x+ ⎟+ y = +⎝ 3 9⎠3 92⎛ 1⎞2 52⎜x− ⎟ + y =⎝ 3⎠9⎛1 ⎞ ⎛ 52 ⎞center: ⎜ ,0 ⎟; radius:⎝3 ⎠⎜ 3 ⎟⎝ ⎠59. Let a, b, and c be the lengths of the sides of theright triangle, with c the length of thehypotenuse. Then the Pythagorean Theorem2 2 2says that a + b = c2 2 2πa πb πc+ = or8 8 82 2 21 ⎛ a ⎞ 1 ⎛ b ⎞ 1 ⎛ c ⎞π ⎜ ⎟ + π ⎜ ⎟ = π⎜ ⎟2 ⎝2⎠ 2 ⎝2⎠ 2 ⎝2⎠Thus,60. See the figure below. The angle at T is a rightangle, so the Pythagorean Theorem gives2 2 2( PM + r) = ( PT ) + r2 2 2 2⇔ ( PM ) + 2 rPM + r = ( PT ) + r2⇔ PM ( PM + 2 r) = ( PT )PM + 2 r = PN so this gives ( PM )( PN) = ( PT )61. The lengths A, B, and C are the same as thecorresponding distances between the centers ofthe circles:A = (–2)2+ (8)2= 68 ≈8.2B = (6)2+ (8)2= 100 = 102 2C = (8) + (0) = 64 = 8Each circle has radius 2, so the part of the beltaround the wheels is2(2π − a −π) + 2(2π −b −π) + 2(2π −c − π)= 2[3π - ( a+ b+ c)] = 2(2 π) = 4πSince a + b + c = π , the sum of the angles of atriangle.The length of the belt is ≈ 8.2 + 10 + 8 + 4π≈ 38.8 units.221 ⎛ x ⎞π ⎜ ⎟2 ⎝2⎠is the area of a semicircle withdiameter x, so the circles on the legs of thetriangle have total area equal to the area of thesemicircle on the hypotenuse.From2 2a + b = c2 ,3 3 3a + b = c4 4 42 2 23 2x is the area of an equilateral triangle4with sides of length x, so the equilateraltriangles on the legs of the right triangle havetotal area equal to the area of the equilateraltriangle on the hypotenuse of the right triangle.62 As in Problems 50 and 61, the curved portionsof the belt have total length 2 π r.The lengthsof the straight portions will be the same as thelengths of the sides. The belt will have length2 π r+ d1+ d2+ … + d n .20 Section 0.3 Instructor’s Resource Manual

63. A = 3, B = 4, C = –63(–3) + 4(2) + (–6) 7d = =2 2(3) + (4) 564. A= 2, B = − 2, C = 42(4) −2( − 1) + 4) 14 7 2d = = =2 2(2) + (2) 8 265. A = 12, B = –5, C = 112(–2) – 5(–1) + 1 18d = =2 2(12) + (–5) 1366.A= 2, B = − 1, C =−52(3) −1( −1) −5 2 2 5d = = =2 2(2) + ( −1)5 567. 2x+ 4(0) = 55x =225( 2 ) + 4(0)–7 2 5d = = =2 2(2) + (4) 20 568. 7(0) − 5y=−11y =5⎛1⎞7(0) −5⎜⎟−6⎝5 ⎠ 7 7 74d = = =2 2(7) + ( −5)74 7469.2 3 5m− −3= =− ; m = ; passes through1+2 3 5⎛− 2+1 3−2⎞ ⎛ 1 1⎞⎜ , ⎟= ⎜−, ⎟⎝ 2 2 ⎠ ⎝ 2 2⎠1 3⎛1⎞y− = ⎜x+⎟2 5⎝2⎠3 4y = x+5 570.0–4 1m= = –2; m= ; passes through2–0 2⎛0+ 2 4+0⎞ ⎜ , ⎟ = (1, 2)⎝ 2 2 ⎠1y– 2 = ( x–1)21 3y = x+2 26–0 1m = = 3; m= – ; passes through4–2 3⎛2+ 4 0+6⎞ ⎜ , ⎟ = (3, 3)⎝ 2 2 ⎠1y– 3 = – ( x–3)31y = – x+431 3 1x+ = – x+42 2 35 5x =6 2x = 31 3y = (3) + = 32 2center = (3, 3)71. Let the origin be at the vertex as shown in thefigure below. The center of the circle is then4 − rr , , so it has equation( )2 2 2( x −(4 − r)) + ( y− r) = r . Along the side oflength 5, the y-coordinate is always 3 4 timesthe x-coordinate. Thus, we need to find thevalue of r for which there is exactly one x-22 32solution to ( x 4 r) ⎛ ⎞− + + ⎜ x− r⎟= r .⎝4⎠Solving for x in this equation gives16 ⎛2 ⎞x = ⎜ 16 − r± 24 ( − r + 7 r−6 ) ⎟ . There is25 ⎝⎠2exactly one solution when − r + 7r− 6=0,that is, when r = 1 or r = 6 . The root r = 6 isextraneous. Thus, the largest circle that can beinscribed in this triangle has radius r = 1.Instructor’s Resource Manual Section 0.3 21

72. The line tangent to the circle at ( ab , ) will beperpendicular to the line through ( ab , ) and thecenter of the circle, which is ( 0,0 ). The linethrough ( ab , ) and ( 0,0)has slope0 − b b 2 a rm = = ; ax+ by = r ⇒ y =− x+0 − a a b b2aso ax + by = r has slope − and isbperpendicular to the line through ( ab , ) and( 0,0 ), so it is tangent to the circle at ( ab , ).73. 12a + 0b = 36a = 32 23 + b = 36b =± 3 33 x–3 3y= 36x– 3y= 123x+ 3 3y= 36x+ 3y= 1274. Use the formula given for problems 63-66, for( xy , ) = ( 0,0).A= m, B =− 1, C = B−b;(0,0)m(0) − 1(0) + B−b B−bd = =2 2 2m + ( − 1) m + 175. The midpoint of the side from (0, 0) to (a, 0) is⎛0+ a 0+0⎞ ⎛a⎞⎜ , ⎟=⎜ , 0⎟⎝ 2 2 ⎠ ⎝2⎠The midpoint of the side from (0, 0) to (b, c) is⎛0+ b 0+c ⎞ ⎛ b c ⎞⎜ , ⎟=⎜ , ⎟⎝ 2 2 ⎠ ⎝2 2⎠c–0cm1= = b – a b – ac –02 cm2 = = ; mb1 = m2– a b–a2 2277.The slope of PS is1[ y1+ y4 − ( y1+y2)]2y4 − y2= . The slope of1[ x ]4 21+ x4 − ( x1+x2)x − x21[ y3 + y4 − ( y2 + y3)]4 2QR is2y − y= . Thus1[ x ]4 23 + x4 − ( x2 + x3)x − x2PS and QR are parallel. The slopes of SR andy3 − y1PQ are both , so PQRS is ax3 − x1parallelogram.2 2x ( y– 6) 25;+ = passes through (3, 2)tangent line: 3x – 4y = 1The dirt hits the wall at y = 8.0.4 Concepts Review1. y-axis 3. 8; –2, 1, 42. ( 4, − 2)4. line; parabolaProblem Set 0.41. y = –x 2 + 1; y-intercept = 1; y = (1 + x)(1 – x);x-intercepts = –1, 1Symmetric with respect to the y-axis76. See the figure below. The midpoints of thesides are⎛ x1+ x2 y, 1+yP2 ⎞ ⎛ x2 + x3 y2 + y3⎞⎜,2 2⎟ Q⎜, ,⎝⎠ 2 2⎟⎝⎠⎛ x3 + x4 y3 + y4⎞R ⎜ , ,2 2⎟ and⎝⎠⎛ x1+ x4 y1+y4⎞S ⎜ , .2 2⎟⎝⎠22 Section 0.4 Instructor’s Resource Manual

2.2x =− y + 1; y -intercepts =− 1,1;x -intercept = 1 .Symmetric with respect to the x-axis.5. x 2 + y = 0; y = –x 2x-intercept = 0, y-intercept = 0Symmetric with respect to the y-axis3. x = –4y 2 – 1; x-intercept = –1Symmetric with respect to the x-axis6.2y = x − 2 x; y-intercept = 0y = x(2 − x); x-intercepts = 0,24.2y = 4x−1;y -intercept =− 11 1y = (2x+ 1)(2 x− 1); x-intercepts =− ,2 2Symmetric with respect to the y-axis.7. 7x 2 + 3y = 0; 3y = –7x 2 ;y =7– 32xx-intercept = 0, y-intercept = 0Symmetric with respect to the y-axis8.2y = 3x − 2x+ 2; y -intercept = 2Instructor’s Resource Manual Section 0.4 23

9. x 2 + y 2 = 4x-intercepts = -2, 2; y-intercepts = -2, 2Symmetric with respect to the x-axis, y-axis,and origin12.2 24x+ 3y= 12; y -intercepts =− 2, 2x-intercepts =− 3, 3Symmetric with respect to the x-axis, y-axis,and origin10.2 23x4y12;+ = y-intercepts =− 3, 3x -intercepts =− 2,2Symmetric with respect to the x-axis, y-axis,and origin13. x 2 – y 2 = 4x-intercept = -2, 2Symmetric with respect to the x-axis, y-axis,and origin11. y = –x 2 – 2x + 2: y-intercept = 22± 4+ 8 2±2 3x-intercepts = = = –1±3–2 –214.2 2x + ( y− 1) = 9; y-intercepts = − 2,4x-intercepts = − 2 2,2 2Symmetric with respect to the y-axis24 Section 0.4 Instructor’s Resource Manual

15. 4(x – 1) 2 + y 2 = 36;y-intercepts =± 32 =± 4 2x-intercepts = –2, 4Symmetric with respect to the x-axis18.4 4x + y = 1; y-intercepts =− 1,1x-intercepts = − 1,1Symmetric with respect to the x-axis, y-axis,and origin16.2 2x − 4x+ 3y=− 2x-intercepts = 2±2Symmetric with respect to the x-axis19. x 4 + y 4 = 16; y-intercepts =− 2, 2x-intercepts = − 2, 2Symmetric with respect to the y-axis, x-axisand origin17. x 2 + 9(y + 2) 2 = 36; y-intercepts = –4, 0x-intercept = 0Symmetric with respect to the y-axis20. y = x 3 – x; y-intercepts = 0;y = x(x 2 – 1) = x(x + 1)(x – 1);x-intercepts = –1, 0, 1Symmetric with respect to the originInstructor’s Resource Manual Section 0.4 25

21.1y = ; y-intercept = 12x + 1Symmetric with respect to the y-axis24. ( ) 2 24 x− 5 + 9( y+ 2) = 36; x-intercept = 522.xy = ; y-intercept = 02x + 1x-intercept = 0Symmetric with respect to the origin25. y = (x – 1)(x – 2)(x – 3); y-intercept = –6x-intercepts = 1, 2, 326. y = x 2 (x – 1)(x – 2); y-intercept = 0x-intercepts = 0, 1, 223.2 22x – 4x+ 3y + 12 y = –22 22( x – 2x+ 1) + 3( y + 4y+ 4) = –2 + 2 + 122 22( x–1) + 3( y+ 2) = 1230y-intercepts = –2 ±3x-intercept = 127.2 2y x ( x 1) ;= − y-intercept = 0x-intercepts = 0, 126 Section 0.4 Instructor’s Resource Manual

28.4 4 4y = x ( x− 1) ( x+ 1) ; y-intercept = 0x-intercepts =− 1,0,1Symmetric with respect to the y-axisIntersection points: (0, 1) and (–3, 4)29. x + y = 1; y-intercepts = –1, 1;x-intercepts = –1, 1Symmetric with respect to the x-axis, y-axisand origin32.22x+ 3 =−( x−1)22x + 3=− x + 2x−12x + 4=0No points of intersection30. x + y = 4; y-intercepts = –4, 4;x-intercepts = –4, 4Symmetric with respect to the x-axis, y-axisand origin33.2− 2x+ 3=−2( x−4)2− 2x+ 3=− 2x + 16x−3222x−18x+ 35=018 ± 324 – 280 18 ± 2 11 9 ± 11x = = = ;4 4 2⎛9– 11 ⎞Intersection points:⎜ , – 6 + 11 ,2⎟⎝⎠⎛9+11 ⎞⎜ ,–6– 112⎟⎝⎠31.2− x+ 1 = ( x+1)2− x + 1= x + 2x+12x + 3x= 0xx ( + 3) = 0x = 0, −3Instructor’s Resource Manual Section 0.4 27

34.− 2x+ 3= 3x − 3x+1223x− x+ 9=0No points of intersection237.y = 3x+12 2x + 2 x+ (3x+ 1) = 152 2x + 2x+ 9x + 6x+ 1=15210x+ 8x− 14 = 022(5x+ 4x− 7) = 0− 2±39x = ≈− 1.65, 0.855Intersection points:⎛−2−39 −1−3 39 ⎞,and⎜ 5 5 ⎟⎝⎠⎛− 2+ 39 − 1+3 39 ⎞,⎜ 5 5 ⎟⎝⎠[ or roughly (–1.65, –3.95) and (0.85, 3.55) ]35.2 2x + x = 42x = 2x =±2Intersection points: (– 2, – 2 ),( 2, 2 )36.2 22x+ 3( x− 1) = 122 22x + 3x − 6x+ 3=1225x−6x− 9=06 ± 36 + 180 6 ± 6 6 3 ± 3 6x = = =10 10 5Intersection points:⎛3−3 6 −2− 3 6 ⎞ ⎛3+ 3 6 − 2+3 6 ⎞⎜, , ,5 5 ⎟ ⎜ 5 5 ⎟⎝ ⎠ ⎝ ⎠38.2 2x + (4x+ 3) = 812 2x + 16x + 24x+ 9 = 81217x+ 24x− 72 = 0− 12 ± 38x = ≈ − 2.88, 1.4717Intersection points:⎛−12 − 38 3 −24 38 ⎞,and⎜ 17 17 ⎟⎝⎠⎛− 12 + 38 3 + 24 38 ⎞,⎜ 17 17 ⎟⎝⎠− − ][ or roughly ( 2.88, 8.52 ),( 1.47,8.88)28 Section 0.4 Instructor’s Resource Manual

39. a.b.c.d.y = x2 ; (2)3 2ax + bx + cx + d, with a > 0: (1)3 2ax + bx + cx + d, with a < 0: (3)y = ax3 , with a > 0 : (4)( )22d6= (2 − 2) + ⎡1+ 13 − 1−13 ⎤⎣⎦( ) ( )2 2= 0+ 13+ 13 = 2 13= 2 13 ≈7.21Four such distances ( d2 = d4 and d1 = d5).40.2 2x + y = 13;( −2, −3), ( −2,3), (2, − 3), (2,3)d1=2 2(2 + 2) + ( − 3+ 3) = 4d2=2 2(2 + 2) + ( −3− 3) = 52 = 2 13d3=2 2(2 − 2) + (3+ 3) = 6Three such distances.41. x 2 + 2x + y 2 – 2y = 20; (–2, 1+21 ),( –2, 1 – 21 ), ( 2, 1+13 ), ( 2, 1 – 13)( )( )22d1= (–2–2) + ⎡1+ 21– 1+13 ⎤⎣⎦2= 16 + 21 – 13= 50 – 2 273 ≈4.12d2=22(–2–2) + ⎡1+21– ( 1– 13)⎤⎣⎦= 16 + 21 + 132( )= 50 + 2 273 ≈9.112d3= ( − 2+ 2) + ⎡1+ 21− 1−21 ⎤⎦2⎣ ( )2 2( ) ( )= 0+ 21+ 21 = 2 21= 2 21≈9.1722d4= ( −2− 2) + ⎡1 21 (1 13) ⎤⎣− − +⎦2= 16 + − 21 − 13( )= 50 + 2 273 ≈9.11d5=22( −2− 2) + ⎡1− 21−( 1−13)⎤⎣⎦= 16 + 13 −221( )= 50 −2 273 ≈ 4.120.5 Concepts Review1. domain; range2.2 2 2f (2 u) = 3(2 u) = 12 u ; f( x+ h) = 3( x+h)3. asymptote4. even; odd; y-axis; originProblem Set 0.51. a.b.c.d.e.f.2. a.2f (1) = 1 – 1 = 02f (–2) = 1– (–2) = –32f (0) = 1– 0 = 12f ( k) = 1– k2f (–5) = 1– (–5) = –242f ⎛ 1 1 1 15⎜ ⎞ ⎟= 1– ⎛ ⎜ ⎞⎟ = 1– =⎝4⎠ ⎝4⎠16 16f 1+h = 1−1+h = −2h− hg. ( ) ( )22 2h. f ( 1+ h) − f ( 1) = −2h−h − 0= −2h−hi. f ( h) f ( ) ( h) 2b.c.2+ − 2 = 1− 2+ + 33F (1) = 1 + 3 ⋅ 1 = 42=−4h−h3F ( 2) = ( 2) + 3( 2) = 2 2+3 2= 5 23F ⎛ 1 1 1 1 3 49⎜ ⎞ ⎟= ⎛ ⎜ ⎞ ⎟ + 3⎛ ⎜ ⎞⎟= + =⎝4⎠ ⎝4⎠ ⎝4⎠64 4 642Instructor’s Resource Manual Section 0.5 29

3. a.d. F( 1+ h) = ( 1+ h) 3+ 3( 1+h)2 3= 1+ 3h+ 3h + h + 3+3h2 3= 4+ 6h+ 3h + h2 3e. F ( 1+ h) − 1= 3+ 6h+ 3h + hf. F( 2+ h) −F( 2)= ( 2+ h) 3 + 3( 2+ h) −⎡2 3 −3( 2)b.c.d.e.f.⎤⎣ ⎦2 3= 8+ 12h+ 6h + h + 6+ 3h−142 3= 15h+ 6h + h1G (0) = = –10–11G (0.999) = = –10000.999 –11G (1.01) = = 1001.01–12Gy ( ) =y12–11 1G(– x) = = –– x –1 x + 12⎛ 1 ⎞ 1 xG ⎜ 2 ⎟ = =x1 –1 2⎝ ⎠ 2 1– xxf.5. a.b.c.2 2 22 ( x + x) + ( x + x)Φ ( x + x)=2x + x4 3 2x + 2x + 2x + x=2x + x1 1f (0.25) = = is not0.25 − 3 − 2.75defined1f( x) = ≈ 2.658π − 31 1f (3 + 2) = =3+ 2−3 2−0.25= 2 ≈0.841(0.79) 2 + 96. a. f(0.79) = ≈− 3.2930.79 – 3(12.26) + 9b. f(12.26) = ≈ 1.19912.26 – 3c.2( 3) + 9f ( 3) = ; undefined3– 37. a. x 2 + y 2 = 1y 2 = 1– x 2 2y =± 1– x ; not a function24. a.b.c.d.e.21+1Φ (1) = = 212 2– t+(– t) t – tΦ (– t)= =– t – t( ) 2⎛1⎞1 +2 1 2 34Φ ⎜ ⎟= = ≈1.06⎝2⎠ 1 12 22 2( u+ 1) + ( u+ 1) u + 3u+2Φ ( u + 1) = =u+ 1 u+12 2 2 2 42 ( x ) + ( x ) x + xΦ ( x ) = =2x xb. xy + y + x = 1y(x + 1) = 1 – x1– x 1– xy = ; f( x)=x + 1 x + 1c. x = 2y+ 1x 2 = 2y + 12 2x –1 x –1y = ; f( x)=2 2d. x = yy + 1xy + x = yx = y – xyx = y(1 – x)x xy = ; f( x)=1– x 1– x30 Section 0.5 Instructor’s Resource Manual

8. The graphs on the left are not graphs offunctions, the graphs on the right are graphs offunctions.9.10.11.2 2f( a+ h) – f( a) [2( a+h) – 1] – (2 a – 1)=hh24ah+ 2h= = 4a+2hh3 3F( a+ h)– F( a) 4( a+h) –4a=hh3 2 2 3 34a + 12a h+ 12ah + 4 h –4a=h2 2 312a h+ 12ah + 4h=h2 2= 12a + 12ah+4h3 3gx ( + h)– gx ( ) –x+h–2 x–2=hh3x−6−3x− 3h+62=x − 4x+ hx− 2h+4h−3h=2hx ( − 4x+ hx− 2h+4)3= –2x – 4 x+ hx–2h+4c.d.14. a.b.2( x) x –9ψ =x 2 –9≥ 0; x 2 ≥ 9; x ≥ 3Domain: { x∈: x ≥3}4H ( y) = – 625– y4 4625 – y ≥0; 625 ≥ y ; y ≤ 5Domain: { y∈: y ≤5}2 24– x 4– xf( x)= =2x – x–6( x–3)( x+2)Domain: { x∈: x ≠ −2, 3}Gy ( ) = ( y+1)–11≥ 0; y > –1y + 1Domain: { y∈ : y > −1}c. φ ( u) = 2u+3Domain: (all real numbers)d.2/3Ft () = t –4Domain: (all real numbers)15. f(x) = –4; f(–x) = –4; even function12.a+h( )– ( ) –aGa+h Ga a+ h+ 4 a+4=hh2 2a + 4a+ ah+ 4h−a −ah−4a2=a + 8a+ ah+ 4h+16h4h=2ha ( + 8a+ ah+ 4h+16)4=2a + 8a+ ah+ 4h+1616. f(x) = 3x; f(–x) = –3x; odd function13. a. F( z) = 2z+32z + 3 ≥ 0; z ≥ – 3 2Domain:⎧3⎫⎨z∈: z ≥ − ⎬⎩2⎭b.1gv () =4 v –14v – 1 = 0; v = 1 4Domain:⎧ 1 ⎫⎨v∈: v ≠ ⎬⎩ 4⎭Instructor’s Resource Manual Section 0.5 31

17. F(x) = 2x + 1; F(–x) = –2x + 1; neither20.3 3uugu ( ) ; g(– u) – ;8 8= = odd function18. F( x) = 3 x– 2; F(– x) = –3 x– 2; neither21.x– xgx ( ) = ; g(– x) = ; odd2 2x –1 x –119.2 2g( x) = 3x + 2 x–1; g(– x) = 3 x –2 x–1;neither22.2z1 –2z1φ( z) = + ; φ(– z) = + ; neitherz–1 – z–132 Section 0.5 Instructor’s Resource Manual

23. f( w) = w–1; f(– w) = – w–1;neither26. Ft () = – t+ 3; F(–) t = –– t+ 3; neither24.2 2hx ( ) = x + 4; h(– x) = x + 4; evenfunctionx x 27. gx ( ) = ; g( − x) = − ;2 2 neither28. Gx x G x x ( ) = 2 −1 ; ( − ) = − 2 + 1 ; neither25. f ( x) = 2 x ; f(– x) = –2x = 2 x ; evenfunction29.⎧1 if t ≤ 0⎪gt () = ⎨t+ 1 if 0< t

30.⎧ 2x + x≤⎪– 4 if 1hx ( ) = ⎨⎪⎩ 3xif x > 1neither35. Let y denote the length of the other leg. Then2 2 2x + y = h2 2 2y = h −x2 2y = h −x( )2 2L x = h −x36. The area isA x = 1 base× height = 1 x h − x2 2( )2 237. a. E(x) = 24 + 0.40x31. T(x) = 5000 + 805xDomain: { x∈integers: 0 ≤ x≤100}T( x) 5000ux ( ) = 805x= x+Domain: { x∈ integers: 0 < x≤100}32. a. Px ( ) = 6 x–(400+5 xx ( –4))= 6 x–400–5 x( x–4)b. 120 = 24 + 0.40x0.40x = 96; x = 240 mi2 ,38. The volume of the cylinder is π r h where h isthe height of the cylinder. From the figure,r 2 + h 2⎛⎜⎞⎟ = (2r) 2 ; h2 ⎝ 2⎠4 = 3r 2 ;2h = 12r = 2r3.2 3V() r =π r (2r 3) = 2πr 3b. P(200) 190≈− ; ( )P 1000 ≈ 610c. ABC breaks even when P(x) = 0;6 x– 400 – 5 x( x– 4) = 0; x ≈ 39033.Ex ( ) = x–x2y0.5−0.5120.51 xexceeds its square by the maximum amount.p34. Each side has length . The height of the33 ptriangle is .621⎛ p⎞⎛ 3p⎞3pAp ( ) = ⎜ ⎜=2⎝ 3⎟⎜ 6 ⎟⎠⎝ ⎠ 362πd39. The area of the two semicircular ends is .4The length of each parallel side is 1– πd.22 2 2π 1– –( )d ⎛ π d ⎞ πAd dd d π= + ⎜ ⎟= +d4 ⎝ 2 ⎠ 4 222 d – πd=4Since the track is one mile long, π d < 1, so1 ⎧1 ⎫d < . Domain: ⎨d∈ : 0 < d < ⎬π ⎩π⎭34 Section 0.5 Instructor’s Resource Manual

40. a.1 3A (1) = 1(1) + (1)(2 − 1) =2 242. a. f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y)b.2 2 2f ( x+ y) = ( x+ y) = x + 2xy+y≠ f ( x) + f( y)c. f(x + y) = 2(x + y) + 1 = 2x + 2y + 1≠ f(x) + f(y)d. f(x + y) = –3(x + y) = –3x – 3y = f(x) + f(y)b.1A (2) = 2(1) + (2)(3− 1) = 42c. A ( 0) = 0d.e.1 1 2A() c = c(1) + ()( c c+ 1− 1) = c + c2 243. For any x, x + 0 = x, sof(x) = f(x + 0) = f(x) + f(0), hence f(0) = 0.Let m be the value of f(1). For p in N,p = p⋅ 1= 1+ 1 + ... + 1, sof(p) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1)= pf(1) = pm.⎛ 1 ⎞ 1 1 11 = p ⎜ ⎟= + + ... + , so⎝ p ⎠ p p p⎛ 1 1 1 ⎞m = f(1) = f ⎜ + + ... + ⎟⎝ p p p⎠⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞= f ⎜ ⎟+ f ⎜ ⎟+ ... + f ⎜ ⎟=pf ⎜ ⎟,⎝ p⎠ ⎝ p⎠ ⎝ p⎠ ⎝ p⎠⎛ 1 ⎞ mhence f ⎜ ⎟ = . Any rational number can⎝ p⎠pbe written as p qwith p, q in N.f. Domain: { c∈: c≥0}Range: { y∈: y ≥0}41. a. B (0) = 0b.⎛1⎞ 1 1 1 1B⎜⎟ = B(1)= ⋅ =⎝2⎠2 2 6 12p 1 1 1 1p⎛ ⎞= ⎜ ⎟= + + ... + ,q ⎝q⎠q q q⎛ p ⎞ ⎛1 1 1⎞so f ⎜ ⎟= f ⎜ + + ... + ⎟⎝ q ⎠ ⎝q q q⎠⎛1⎞ ⎛1⎞ ⎛1⎞= f ⎜ ⎟+ f ⎜ ⎟+ ... + f ⎜ ⎟⎝q⎠ ⎝q⎠ ⎝q⎠⎛1 ⎞ ⎛m⎞ ⎛ p⎞= pf ⎜ ⎟= p⎜ ⎟=m⎜ ⎟⎝q⎠ ⎝ q ⎠ ⎝ q ⎠c.Instructor’s Resource Manual Section 0.5 35

44. The player has run 10t feet after t seconds. Hereaches first base when t = 9, second base whent = 18, third base when t = 27, and home platewhen t = 36. The player is 10t – 90 feet fromfirst base when 9 ≤ t ≤ 18, hence2 290 + (10t− 90) feet from home plate. Theplayer is 10t – 180 feet from second base when18 ≤ t ≤ 27, thus he is90 – (10t – 180) = 270 – 10t feet from third base2 2and 90 + (270 − 10 t)feet from home plate.The player is 10t – 270 feet from third basewhen 27 ≤t≤ 36, thus he is90 – (10t – 270) = 360 – 10t feet from homeplate.a.⎧10tif 0 ≤t≤9⎪2 2⎪90 + (10t− 90) if 9 < t ≤18s = ⎨⎪ 2 290 + (270 − 10 t) if 18 < t ≤ 27⎪⎪⎩ 360 –10tif 27 < t ≤3646. a. f(1.38) ≈ –76.8204f(4.12) ≈ 6.7508b. x f(x)–4 –6.1902–3 0.4118–2 13.7651–1 9.95790 01 –7.33692 –17.73883 –0.45214 4.4378b.⎧180 −180 −10tif 0 ≤t≤9⎪⎪or 27 < t ≤36⎪ 2 2s = ⎨ 90 + (10t− 90) if 9 < t ≤18⎪2 2⎪ 90 + (270 − 10 t) if 18 < t ≤ 27⎪⎪⎩47.45. a. f(1.38) ≈ 0.2994f(4.12) ≈ 3.6852b. x f(x)–4 –4.05–3 –3.1538–2 –2.375–1 –1.80 –1.251 –0.22 1.1253 2.38464 3.5548.a. Range: {y ∈ R: –22 ≤ y ≤ 13}b. f(x) = 0 when x ≈ –1.1, 1.7, 4.3f(x) ≥ 0 on [–1.1, 1.7] ∪ [4.3, 5]a. f(x) = g(x) at x ≈ –0.6, 3.0, 4.636 Section 0.5 Instructor’s Resource Manual

49.b. f(x) ≥ g(x) on [-0.6, 3.0] ∪ [4.6, 5]c. f ( x)– g( x )3 2 2= x – 5x + x+ 8–2x + 8x+13 2= x – 7x + 9x+9Largest value f(–2) – g (–2) = 45b. On ⎡⎣ −6, −3), g increases from13g ( − 6)= ≈ 4.3333 to ∞ . On (2,6⎤3⎦ , gdecreased from ∞ to 26 2.88899 ≈ . On( − 3, 2)the maximum occurs aroundx = 0.1451 with value 0.6748 . Thus, therange is ( −∞,0.6748⎤∪⎡2.8889,∞)⎦ ⎣ .c.2x + x– 6= 0; (x + 3)(x – 2) = 0Vertical asymptotes at x = –3, x = 2d. Horizontal asymptote at y = 30.6 Concepts Review1.2 3( x + 1)50.a. x-intercept: 3x – 4 = 0; x = 4 330–4 ⋅ 2y-intercept: =20 + 0–6 3b. c.2x x– 6 0;+ = (x + 3)(x – 2) = 0Vertical asymptotes at x = –3, x = 2d. Horizontal asymptote at y = 02. f(g(x))3. 2; left4. a quotient of two polynomial functionsProblem Set 0.61. a.b.( f + g)(2) = (2+ 3) + 2 = 9( f ⋅ g)(0) = (0+ 3)(0 ) = 022c.3 9 3( g f )(3) = = =3+3 6 22d. ( f g)(1)= f (1 ) = 1+3 = 4e. ( g f )(1) = g(1+ 3) = 4 1622 =2 =f. ( g f )(–8) = g(–8+ 3) = (–5) 25a. x-intercepts:2 4 2 33 x –4= 0; x = ± = ±3 3y-intercept: 2 32 2 2 282. a. ( f – g )(2) = (2 + 2)– = 6– =2+3 5 51 + 1 2b. ( f g)(1)= = = 4221+3242 ⎡ 2 ⎤ ⎛ 1 ⎞ 1c.g (3) = ⎢ = ⎜ ⎟ =3 3⎥⎣ + ⎦ ⎝ 3 ⎠ 922Instructor’s Resource Manual Section 0.6 37

⎛ 2 ⎞ ⎛ 1 ⎞ 1 3( f g)(1)= f+ =d.⎜ ⎟ = ⎜ ⎟⎝1+3 ⎠ ⎝ 2 ⎠ 2 42 2 2e. ( g f )(1) = g(1+ 1) = =2 + 3 5f.3. a.⎛ 2 ⎞( g g)(3)= g⎜⎟ =⎝ 3 + 3 ⎠3 1( Φ+Ψ )( t) = t + 1+t1322=+ 32103=35⎛ 1 ⎞ ⎛ 1 ⎞ 1b. ( Φ Ψ)(r)= Φ⎜⎟ = ⎜ ⎟ + 1 = + 13⎝ r ⎠ ⎝ r ⎠ rc.( Ψ Φ)(r)= Ψ(rd. Φ 3 (z) = (z 3 + 1) 333+ 1) =re. (Φ – Ψ)(5t) = [(5t) 3 +1] – 1 5t= 125t 3 + 1– 1 5t31+ 1⎛1⎞f. (( Φ – Ψ) Ψ)(t)= ( Φ – Ψ)⎜ ⎟⎝ t ⎠⎛= ⎜ 1⎞3⎟ + 1– 1 ⎝ t⎠1 = 1t 3 + 1–tt4. a.22 x –1( f ⋅ g)( x)=xDomain: (–∞, –1]∪ [1, ∞)5. ( ) ( )f g ( x) = f 1+ x = 1+ x −42= x + 2 x–3( ) =⎛4⎞⎜ − ⎟= 1+ −4⎝ ⎠2 2( gf ) x g x x2= 1 + x –46. g 3 (x) = (x 2 +1) 3 = (x 4 + 2x 2 + 1)(x 2 + 1)= x 6 + 3x 4 + 3x 2 + 1( g g g)(x)= ( g g)(x 2 + 1)= g[(x 2 + 1) 2 + 1] = g( x 4 + 2x 2 + 2)= ( x 4 + 2x 2 + 2) 2 + 1= x 8 + 4x 6 + 8x 4 + 8x 2 + 57. g(3.141) ≈ 1.1888. g(2.03) ≈ 0.0002052 1/3 21/3( π) − ( π) = ⎡ 11−7π −11−7π⎤⎣ ⎦ ⎢⎣ ⎥⎦≈ 4.7899. ⎡gg ⎤ ( )10.3 1/3 3 1/3[ g ( π)– g( π )] = [(6 π–11) –(6 π –11)]≈ 7.80711. a. gx ( ) = x, f( x) = x+712. a.b.b. g(x) = x 15 , f (x) = x 2 + xf( x)f( x)2= ,3x22g( x) = x + x+11= , g(x) = x 3 + 3xxb.4 44 4 2 ⎛2⎞f ( x) + g ( x) =⎛x –1⎞⎜ ⎟ + ⎜ ⎟⎝ ⎠ ⎝ x ⎠2 2 16= ( x –1) +4xDomain: (–∞, 0)∪ (0, ∞)13. p = f g h if f(x) =1/ x , g( x) = x,2hx ( ) = x + 1p = f g h if f ( x) = 1/ x , g(x) = x + 1,hx ( )= x2⎛ 2 ⎞ ⎛ 2 ⎞ 4c. ( f g)(x)= f ⎜ ⎟ = ⎜ ⎟ –1 = – 12⎝ x ⎠ ⎝ x ⎠ xDomain: [–2, 0) ∪ (0, 2]214. p = f g h l if f( x) = 1/ x , g( x) = x,h(x) = x + 1, l( x) = x 2d.2 2( g f )( x)= g⎛x –1⎞⎜ ⎟ =⎝ ⎠ 2x –1Domain: (– ∞ , –1) ∪ (1, ∞ )38 Section 0.6 Instructor’s Resource Manual

15. Translate the graph of g ( x)= x to the right 2units and down 3 units.17. Translate the graph of y = x 2 to the right 2units and down 4 units.18. Translate the graph of y = x 3 to the left 1 unitand down 3 units.16. Translate the graph of hx ( )units and down 4 units.= x to the left 319.x –3( f + g)( x)= + x220. ( f + g)( x)= x+xInstructor’s Resource Manual Section 0.6 39

21.Ft () =t – tt24. a. F(x) – F(–x) is odd becauseF(–x) – F(x) = –[F(x) – F(–x)]b. F(x) + F(–x) is even becauseF(–x) + F(–(–x)) = F(–x) + F(x)= F(x) + F(–x)c.F( x)– F(– x)F( x) + F(– x)is odd andis22even.F( x) −F( − x) F( x) + F( −x) 2 F( x)+ = = F( x)2 2 222. Gt () = t−t25. Not every polynomial of even degree is aneven function. For example f ( x)= x2 + x isneither even nor odd. Not every polynomial ofodd degree is an odd function. For example3 2g ( x)= x + x is neither even nor odd.26. a. Neitherb. PFc. RFd. PFe. RF23. a. Even;(f + g)(–x) = f(–x) + g(–x) = f(x) + g(x)= (f + g)(x) if f and g are both evenfunctions.b. Odd;(f + g)(–x) = f(–x) + g(–x) = –f(x) – g(x)= –(f + g)(x) if f and g are both oddfunctions.c. Even;( f ⋅g)( − x) = [ f( −x)][ g( −x)]= [ f( x)][ g( x)] = ( f ⋅ g)( x)if f and g are both even functions.d. Even;( f ⋅ g)(−x)= [ f ( −x)][g(−x)]= [ −f( x)][ − gx ( )] = [ f( x)][ gx ( )]= ( f ⋅ g)( x)if f and g are both odd functions.e. Odd;( f ⋅ g)(−x)= [ f ( −x)][g(−x)]= [ f( x)][ − g( x)] = −[ f( x)][ g( x)]=−( f ⋅g)( x)if f is an even function and g is an oddfunction.27. a.f. Neither2P = 29 – 3(2 + t) + (2 + t)= t+ t + 27b. When t = 15, P = 15 + 15 + 27 ≈ 6.77328. R(t) = (120 + 2t + 3t 2 )(6000 + 700t)= 2100t 3 + 19, 400t 2 + 96, 000t + 720, 00029.⎧⎪400tif 0 < t < 1Dt () = ⎨2 2⎪⎩ (400 t) + [300( t−1)] if t ≥1t⎧⎪400 if 0 < < 1() = ⎨2⎪⎩ 250,000t − 180,000t+ 90,000 if t ≥1Dt30. D(2.5) ≈ 1097 mit40 Section 0.6 Instructor’s Resource Manual

31.32.33. a.ax+ ba( ) + bcx–aax+bc( )–a⎛ ax + b ⎞f ( f ( x))= f ⎜ ⎟ =⎝ cx – a ⎠cx–a2 2a x+ ab+ bcx– ab x( a + bc)= = = x2 2acx + bc – acx + a a + bcIf a 2 + bc = 0 , f(f(x)) is undefined, while ifx = a , f(x) is undefined.cx–3⎛ ⎛ x –3⎞⎞⎛ – 3 ⎞x+1f( f( f( x)))= f ⎜ f ⎜ ⎟⎟=f ⎜ ⎟x 1 x–3⎝ ⎝ + ⎠⎠ ⎜ + 1 ⎟⎝ x+1 ⎠⎛ x– 3–3 x–3 ⎞ ⎛–2 x–6 ⎞ ⎛– x–3⎞= f ⎜ ⎟= f ⎜ ⎟=f ⎜ ⎟⎝ x– 3+ x+1 ⎠ ⎝ 2 x–2 ⎠ ⎝ x–1⎠– x –3– 3x –1 – x – 3 – 3x+ 3 – 4x= == = x– x –3+ 1 – x – 3 + x –1 – 4x –1If x = –1, f(x) is undefined, while if x = 1,f(f(x)) is undefined.b.c.34. a.⎛ 1 ⎞f ⎜ ⎟ =⎝ x ⎠1x1x 1=– 1 1 – x⎛ x ⎞f ( f ( x))= f ⎜ ⎟ =⎝ x – 1⎠= xxx– x+1=⎛ 1 ⎞ ⎛ x – 1⎞f⎜ = f ⎜ ⎟ =f ( x)⎟⎝ ⎠ ⎝ x ⎠= 1 – xf ( 1/ x)=1/ x=1/ x −1xx –1xx –1– 1x –1xx –1x1x − xx – 1=– 1 x – 1 – xb. x/( x −1)f( f( x)) = f( x/( x − 1)) =x−1x −1x=x( x − 1) + 1−x35. ( f1( f2 f3))( x) = f1(( f2f3)( x))= f1( f2( f3( x)))(( f1 f2) f3)( x) = ( f1f2)( f3( x))= f1( f2( f3( x)))= ( f1( f2 f3))( x)36. f1( f1( x)) = x;1f1( f2( x)) = ;xf ( f ( x)) = 1 −x;1 31f1( f4( x)) = ;1 − xx −1f1( f5( x)) = ;xxf1( f6( x)) = ;x − 11f2( f1( x)) = ;x1f2( f2( x)) = = x;1x1f2( f3( x)) = ;1 − x1f2( f4( x)) = = 1 −x;11−x1 xf2(f5(x))= ;x −=1x –1xf ( f26( x))=1xx−1x –1= ;xf3( f1( x)) = 1 −x;1 x −1f3( f2( x)) = 1 − = ;x xf3( f3( x)) = 1–(1– x) = x;1 xf3( f4( x)) = 1– = ;1– x x–1–1 1f3( f5( x)) = 1– x = ;x xx 1f3( f6( x)) = 1– = ;x –1 1– x1f4( f1( x)) = ;1 − x1 xf4( f2( x)) = = ;1 − 1 x − 1 x1 1f4( f3( x)) = = ;1–(1– x)x1 1 − x x–1f4( f4( x)) = = = ;1– 1 1−x−1x1– x1 xf4( f5 ( x)) = ;1– x –1= xx−( x−1)=x1 x −1f4(f6(x))= = = 1 – x;x1 – x −1−xx –1Instructor’s Resource Manual Section 0.6 41

x −1f5( f1( x)) = ;x1 −1xf5( f2( x)) = = 1 −x;1x1– x –1 xf5( f3( x)) = = ;1– x x–11– 11– x 1−(1 − x)f5(f4(x))= = = x;11f ( f555f ( f6( x))=( x))=1– xx –1xx –1xxx –1xx –1–1 x −1−x 1= = ;x −11 – x–1 x − ( x −1)1= = ;x xxf6( f1( x)) = ;x –11x 1f6( f2( x)) = = ;1–1 1– xx1– x x–1f6( f3( x)) = = ;1– x –1 xf ( f64( x))=11– x 1= =1–1 1−(1 − x)1– xx –11;xx x −1f6(f5(x))= = = 1 – x;x –1–1 x −1−xf ( f ( x))=66xxx –1xx –1x= = x–1 x − ( x −1)37.38.a. f 3 f 3 f 3 f 3 f 3= (((( f 3 f 3 ) f 3 ) f 3 ) f 3 )= ((( f f ) f ) f )1 3 3 3= (( f f ) f )3 3 31 f 3 f 3= f =b. f1 f 2 f 3 f 4 f 5 f 6= ((((( f1 f 2 ) f 3 ) f 4 ) f 5 ) f 6 )= (((( f 2 f 3 ) f 4 ) f 5 ) f 6 )= ( f f ) ( f f )4 4 5 6= f f = f5 2 3c. If F f 6 = f1, then F = f 6 .d. If G f 3 f 6 = f1,then G f 4 = f1soG = f 5 .e. If f 2 f 5 H = f 5 , then f 6 H = f 5 soH = f 3 . f 1 f 2 f 3 f 4 f 5 f 6f 1 f 1 f 2 f 3 f 4 f 5 f 6f 2 f 2 f 1 f 4 f 3 f 6 f 539.f 3 f 3 f 5 f 1 f 6 f 2 f 4f 4 f 4 f 6 f 2 f 5 f 1 f 3f 5 f 5 f 3 f 6 f 1 f 4 f 2f 6 f 6 f 4 f 5 f 2 f 3 f 142 Section 0.6 Instructor’s Resource Manual

40.Problem Set 0.71. a.⎛ π ⎞ π30⎜⎟ =⎝ 180 ⎠ 6b.⎛ π ⎞ π45⎜⎟ =⎝ 180 ⎠ 441. a.⎛ π ⎞ πc. –60 ⎜ ⎟ = –⎝180 ⎠ 3d.e.⎛ π ⎞ 4π240⎜⎟ =⎝ 180 ⎠ 3⎛ π ⎞ 37π–370 ⎜ ⎟ = –⎝180 ⎠ 18b.c.f.⎛ π ⎞ π10⎜⎟ =⎝ 180 ⎠ 182. a.76 π ⎛⎜180 ⎟⎞= 210°⎝ π ⎠b.34 π ⎛⎜180 ⎟⎞= 135°⎝ π ⎠c. – 1 3 π ⎛⎜180 ⎟⎞= –60°⎝ π ⎠d.43 π ⎛⎜180 ⎟⎞= 240°⎝ π ⎠e. – 3518 π ⎛⎜180 ⎟⎞= –350°⎝ π ⎠42.f.3 1803018 π ⎜⎞ ⎟ = °⎝ π ⎠3. a.⎛ π ⎞33.3⎜⎟ ≈ 0.5812⎝180⎠b.⎛ π ⎞46⎜⎟ ≈ 0.8029⎝180⎠c.⎛ π ⎞–66.6 ⎜ ⎟ ≈ –1.1624⎝180⎠0.7 Concepts Review1. (– ∞ , ∞ ); [–1, 1]2. 2π ; 2π ; π3. odd; evenx4. r = (–4) + 3 = 5; cos θ = = –r 52 2 4d.⎛ π ⎞240.11⎜⎟ ≈ 4.1907⎝180⎠e.⎛ π ⎞–369 ⎜ ⎟ ≈ –6.4403⎝180⎠f.⎛ π ⎞11⎜⎟ ≈ 0.1920⎝180⎠Instructor’s Resource Manual Section 0.7 43

⎛4. a. 3.141⎜180 ⎟⎞≈ 180°⎝ π ⎠Thus⎛b. 6. 28⎜180 ⎟⎞≈ 359. 8°⎝ π ⎠⎛c. 5. 00⎜180 ⎟⎞≈ 286.5°⎝ π ⎠⎛d. 0. 001⎜180 ⎟⎞≈ 0.057°⎝ π ⎠5. a.⎛e. –0.1⎜180 ⎟⎞≈ –5.73°⎝ π ⎠⎛f. 36. 0⎜180 ⎟⎞≈ 2062.6°⎝ π ⎠56. 4 tan34. 2°sin 34.1°≈ 68.37π 3cos = 6 2. The resultsπ π 2sin = cos = were derived in the text.4 4 2If the angle is π / 3 then the triangle in theπ 1figure below is equilateral. Thus cos = 3 2b.5.34 tan 21.3°sin 3.1°+cot 23.5° ≈ 0.8845and by the Pythagorean Identity,π 3sin = . 3 2c. tan (0.452) ≈ 0.4855d. sin (–0.361) ≈ –0.35326. a.b.234.1sin(1.56)cos(0.34)≈ 248.32sin (2.51) + cos(0.51) ≈ 1.28287. a.b.56. 3tan34. 2°sin 56.1°⎛ sin 35°⎜⎞⎟⎝ sin 26°+cos 26° ⎠≈ 46.0973≈ 0. 07898. Referring to Figure 2, it is clear thatsin 0 = 0 and cos 0 = 1. If the angle is π / 6 ,then the triangle in the figure below is1 1equilateral. Thus, PQ = OP = . This2 2π 1implies that sin = . By the Pythagorean6 22π 2π ⎛1⎞3Identity, cos = 1− sin = 1− ⎜ ⎟ = .6 6 ⎝2⎠42πReferring to Figure 2, it is clear that sin = 12πand cos = 0 . The rest of the values are2obtained using the same kind of reasoning inthe second quadrant.9. a.b.⎛π⎞sin ⎜ ⎟tan ⎛π⎞ ⎝ 6 ⎠ 3⎜ ⎟= =⎝ 6 ⎠ ⎛π⎞ 3cos⎜⎟⎝ 6 ⎠1sec( π ) = = –1cos( π)c.d.⎛3π⎞ 1sec ⎜ ⎟ = = – 2⎝ 4 ⎠ cosπ3( )⎛π⎞ 1csc⎜⎟ = = 1⎝2 ⎠ sin π( )2444 Section 0.7 Instructor’s Resource Manual

e.f.10. a.b.c.π( 4 )π( )⎛π⎞ coscot ⎜ ⎟ = = 1⎝4 ⎠ sin4π( 4 )π( )⎛ π ⎞ sin –tan ⎜– ⎟ = = –1⎝ 4 ⎠ cos –π( 3 )π( )⎛π⎞ sintan ⎜ ⎟ = = 3⎝3 ⎠ cos⎛π⎞ 1sec⎜⎟ = = 2⎝3 ⎠ cos π3( )3π( 3 )π( )⎛π⎞ cos 3cot ⎜ ⎟ = =⎝3⎠sin 334b. cos3t = cos(2 t+ t) = cos 2tcos t – sin 2tsint2 2= (2cos t –1)cos t – 2sin tcost3 2= 2cos t – cos t – 2(1– cos t)cost3 3= 2cos t – cos t – 2cost+2cos t3= 4cos t –3costc. sin 4x = sin[2(2 x)] = 2sin 2xcos 2x2= 2(2sin xcos x)(2cos x–1)d.13. a.3= 2(4sin x cos x–2sin xcos x)3= 8sin x cos x– 4sin xcosx2 2(1+ cos θ )(1− cos θ) = 1− cos θ = sin θsin u cosu2 2+ = sin u+ cos u = 1cscusecud.e.f.11. a.⎛π⎞ 1csc⎜⎟ = = 2⎝4 ⎠ sin π( )4π( 6 )π( )⎛ π ⎞ sin – 3tan ⎜– ⎟ = = –⎝ 6⎠cos – 3⎛ π ⎞ 1cos ⎜– ⎟ =⎝ 3 ⎠ 22z z z2 1z2z(1+ sin )(1 – sin ) = 1 – sin= cos =sec6b.c.d.2 2 2 2(1 − cos x)(1 + cot x) = (sin x)(csc x)= sin 2 ⎛ 1 ⎞x⎜⎝ sin 2 ⎟ = 1x ⎠⎛ 1 ⎞sin t(csc t – sin t) = sin t⎜ – sin t⎟⎝sint ⎠= 1– sin 2 t = cos 2 tcos2t2 21–csc t cot t sin2t= – = –2 2 1csc t csc tsin 2t2 1= –cos t = –sec2tb.c.d.12. a.2 2t t t t(sec –1)(sec + 1) = sec –1 = tan21 sin tsec t – sin ttan t = –costcost2 2= 1–sin t cos tcostcost= cost=sin2t2 2sec t –1 tan t cos2t= = =2 2 1sec t sec tcos 2t2sin t2 1 2 2sin v+ = sin v+ cos v = 12sec v14. a. y = sin 2xInstructor’s Resource Manual Section 0.7 45

. y = 2 sin tb. y = 2 cos t⎛ π ⎞c. y = cos⎜x−⎟⎝ 4 ⎠c. y = cos3td. y = sect⎛ π ⎞d. y = cos⎜t+⎟⎝ 3 ⎠15. a. y = csc t16. y = 3cos x 2Period = 4π , amplitude = 346 Section 0.7 Instructor’s Resource Manual

17. y = 2 sin 2xPeriod = π , amplitude = 221. y = 21 + 7 sin(2x+ 3)Period = π , amplitude = 7, shift: 21 units up,3 units left218. y = tan xPeriod = π⎛ π ⎞22. y = 3cos ⎜x– ⎟–1⎝ 2 ⎠Period = 2π , amplitude = 3, shifts: π 2 unitsright and 1 unit down.119. y = 2 + cot(2x)6πPeriod = , shift: 2 units up2⎛23. y = tan⎜2x – π ⎟⎞⎝ 3⎠Period = π 2 , shift: π units right620. y = 3+ sec( x−π )Period = 2π , shift: 3 units up, π units rightInstructor’s Resource Manual Section 0.7 47

⎛ π ⎞24. a. and g.: y = sin ⎜x+ ⎟= cos x = – cos( π– x)⎝ 2 ⎠⎛ π ⎞b. and e.: y = cos⎜x+ ⎟= sin( x+π)⎝ 2 ⎠=−sin( π−x)⎛ π ⎞c. and f.: y = cos⎜x− ⎟=sin x⎝ 2 ⎠=− sin( x +π)⎛ π ⎞d. and h.: y = sin ⎜x− ⎟= cos( x+π)⎝ 2 ⎠= cos( x −π)25. a. –t sin (–t) = t sin t; evenb.2 2sin (– t) sin t;c. csc(–t) == even1= –csc t; oddsin(–t)d. sin( − t) = – sin t = sin t ; evene. sin(cos(–t)) = sin(cos t); evenf. –x + sin(–x) = –x – sin x = –(x + sin x); odd26. a. cot(–t) + sin(–t) = –cot t – sin t= –(cot t + sin t); odd27.28.b.3 3sin (– t) – sin t;c. sec(–t) =d.= odd1= sec t; evencos(–t )4 4sin (– t) = sin t;evene. cos(sin(–t)) = cos(–sin t) = cos(sin t); evenf. (–x ) 2 + sin(–x ) = x 2 – sin x; neither2 22 π ⎛ 1 1cos cosπ ⎞ ⎛ ⎞= ⎜ ⎟ = ⎜ ⎟ =3 ⎝ 3 ⎠ ⎝2⎠42 22 π ⎛ 1 1sin sinπ ⎞ ⎛ ⎞= ⎜ ⎟ = ⎜ ⎟ =6 ⎝ 6 ⎠ ⎝2⎠430.31.ππ 3( )2 π 1 + cos2 12 1 + cos 6cos 1 += = = 212 2 2 2=2+34ππ 2( )2 π 1–cos2 8 1–cos 4 1– 2sin = = =8 2 2 2=2– 2432. a. sin(x – y) = sin x cos(–y) + cos x sin(–y)= sin x cos y – cos x sin y33.b. cos(x – y) = cos x cos(–y) – sin x sin (–y)= cos x cos y + sin x sin yc.tan x + tan(– y)tan( x– y)=1–tanxtan(– y)tan x – tan y=1 + tanxtan ytan t+ tan π tan t+0tan( t +π ) = =1– tan ttan π 1– (tan t)(0)= tan t34. cos( x− π ) = cos xcos( −π) −sin xsin( − π)= –cos x – 0 · sin x = –cos x35. s = rt = (2.5 ft)( 2π rad) = 5π ft, so the tiregoes 5π feet per revolution, or 15π revolutionsper foot.⎛ 1 rev⎞⎛ mi⎞⎛ 1 hr ⎞⎛ ft ⎞⎜ ⎟⎜60 ⎟⎜ ⎟⎜5280⎟⎝5πft ⎠⎝ hr ⎠⎝60 min⎠⎝ mi⎠≈ 336 rev/min36. s = rt = (2 ft)(150 rev)( 2π rad/rev) ≈ 1885 ft37. rt 11 = r2t2; 6(2 π ) t1= 8(2 π )(21)t 1 = 28 rev/sec38.Δy = sin α and Δx = cos αm = ΔyΔx = sin αcos α = tanα29.3 33 π ⎛ π⎞ ⎛1⎞1sin = ⎜sin⎟ = ⎜ ⎟ =6 ⎝ 6⎠ ⎝2⎠848 Section 0.7 Instructor’s Resource Manual

39. a. tanα = 344. Divide the polygon into n isosceles triangles byα = π drawing lines from the center of the circle to3the corners of the polygon. If the base of eachtriangle is on the perimeter of the polygon, thenb. 3x+ 3y= 6the angle opposite each base has measure 2 π.n3 y = – 3x+6Bisect this angle to divide the triangle into two3 3right triangles (See figure).y = – x+ 2; m = –3 33tan α = – 35πα =640. m 1 = tan θ 1 and m 2 = tanθ 2tanθ2 + tan( −θ1)tanθ = tan( θ2 − θ1)=1 − tanθ2 tan( − θ1)tanθ2 −tanθ1 m2 −m1= =1+ tanθ2 tanθ1 1+mm 1 2sin π bh= so b = 2rsin π and cosπ = so41. a. tan θ = 3 – 21+ 3(2) = 1 n 2r n n r7h = r cos π n .θ ≈ 0.1419πP = nb = 2rnsin–1– 1 nb. tan θ = 2= –31 + 1⎛1⎞ 2 π π( 2 )(–1)A = n⎜bh⎟=nr cos sin⎝2⎠ n nθ ≈ 1.892545. The base of the triangle is the side opposite thec. 2x – 6y = 12 2x + y = 0t–6y = –2x + 12y = –2xangle t. Then the base has length 2rsiny = 1 23 x –2(similar to Problem 44). The radius of them 1 = 1 3 , m t2 = –2semicircle is r sin and the height of the2–2 – 1 ttan θ = 3triangle is r cos .= –7; θ ≈1.712721 + 1( 3 )(–2)21 ⎛ t ⎞⎛ t ⎞ π ⎛ t ⎞A= ⎜ 2 r sin ⎟⎜r cos ⎟+⎜rsin ⎟22⎝ 2⎠⎝ 2⎠ 2⎝ 2⎠42. Recall that the area of the circle is π r . The2measure of the vertex angle of the circle is 2π .2 t t πr 2 t= r sin cos + sinObserve that the ratios of the vertex angles2 2 2 2must equal the ratios of the areas. Thus,t A=2π 2π r, so1 2A = r t.243. A = 1 2 (2)(5)2 = 25cm 2Instructor’s Resource Manual Section 0.7 49

x x x x46. cos cos cos cos2 4 8 16= 1 ⎡2 cos 3 4 x + cos 1 ⎣ ⎢4 x ⎤ 1 ⎡⎦ ⎥ 2 cos 316 x + cos 1⎣ ⎢16 x ⎤⎥ ⎦1⎡ 3 1 ⎤⎡ 3 1 ⎤= cos x cos x cos x cos x4⎢ +4 4 ⎥⎢ +16 16 ⎥⎣ ⎦⎣ ⎦1⎡3 3 3 1= cos cos cos cos4⎢x x+x x⎣ 4 16 4 16+ cos 1 4 x cos 316 x + cos 1 4 x cos 116 x⎤ ⎦ ⎥1 ⎡1 ⎛ 15 9 ⎞ 1 ⎛ 13 11 ⎞= cos cos x cos x cos x4⎢ ⎜ + ⎟+ ⎜ + ⎟⎣2⎝ 16 16 ⎠ 2⎝ 16 16 ⎠1 7 1 1 5 3+ ⎛ ⎜cos x + cos x ⎞ ⎟+ ⎛ ⎜cos x+cos x ⎞⎤2 16 16 2 16 16⎟⎥⎝ ⎠ ⎝ ⎠⎦= 1 ⎡ 15 13 11cos x + cos x + cos8 16 16 16 x + cos 9⎣ ⎢16 x+ 7 5 3 1cos x cos cos cos16 + x x x ⎤16 + 16 + 16 ⎥ ⎦47. The temperature function is⎛ 2π⎛ 7 ⎞⎞T ( t)= 80 + 25sin⎜ ⎜t− ⎟⎟⎝ 12 ⎝ 2 .⎠⎠The normal high temperature for November15 th is then T ( 10.5) = 67. 5 °F.49. As t increases, the point on the rim of thewheel will move around the circle of radius 2.a. x ( 2) ≈ 1. 902y ( 2) ≈ 0.618x ( 6) ≈ −1.176y ( 6) ≈ −1.618x ( 10) = 0y ( 10) = 2x ( 0) = 0y ( 0) = 2⎛ π ⎞ ⎛ π ⎞b. x(t)= −2 sin⎜t ⎟,y(t)= 2 cos⎜t⎟ ⎝ 5 ⎠ ⎝ 5 ⎠π πc. The point is at (2, 0) when t = ; that5 25is , when t = .22π50. Both functions have frequency . When 10you add functions that have the samefrequency, the sum has the same frequency.a. yt ( ) = 3sin( πt/ 5) − 5cos( πt/ 5)+ 2sin(( πt/5) −3)48. The water level function is⎛ 2π⎞F( t)= 8.5 + 3.5sin⎜( t − 9)⎟ .⎝ 12 ⎠The water level at 5:30 P.M. is thenF(17.5) ≈ 5.12 ft .b. yt ( ) = 3cos( πt/ 5 − 2) + cos( πt/ 5)+ cos(( πt/ 5) −3)50 Section 0.7 Instructor’s Resource Manual

51. a. Csin( ωt+ φ) = ( Ccos φ)sin ωt+ ( Csin φ)cos ωt.Thus A= C⋅cosφand B = C⋅ sinφ.b. 2 2 2 2 2 2 2 2 2A + B = ( Ccos ) + ( Csin ) = C (cos ) + C (sin ) = CB C⋅sinφAlso, = = tanφA C⋅cosφφ φ φ φc.A sin( ωt+ φ ) + A sin( ωt+ φ ) + A (sin ωt+ φ )1= A (sin ωtcosφ+ cosωtsin φ )1+ A (sin ωtcosφ+ cosωtsin φ )2+ A (sin ωtcosφ+ cosωtsin φ )3= ( A cosφ+ A cosφ+ A cosφ) sin ωt1+ ( A sin φ + A sin φ + A sin φ ) cosωt111= C +1sin ( ωtφ)221232where C and φ can be computed fromA= A cosφ + A cosφ + A cosφ1 1 2 2 3 3B = A sinφ + A sinφ + A sinφ1 1 2 2 3 3as in part (b).223321233333d. Written response. Answers will vary.52. ( a.), (b.), and (c.) all look similar to this:53. a.d.b.e.c.The windows in (a)-(c) are not helpful becausethe function oscillates too much over thedomain plotted. Plots in (d) or (e) show thebehavior of the function.The plot in (a) shows the long term behavior ofthe function, but not the short term behavior,whereas the plot in (c) shows the short termbehavior, but not the long term behavior. Theplot in (b) shows a little of each.Instructor’s Resource Manual Section 0.7 51

54. a. = ( )hx ( ) f g ( x)3 cos(100 x ) + 2=1002⎛ 1 ⎞ 2⎜ ⎟ cos (100 x) + 1⎝100⎠1 ⎛ 3x+ 2⎞jx ( ) = ( gf)( x) = cos⎜100100 2 ⎟⎝ x + 1 ⎠b.⎧ ⎪ ⎡ ⎢ ⎤4 4⎥⎨ ⎣ ⎦⎪⎩0.0625, otherwisewhere n is an integer.56. f ( x) = ( x−2 n) 2 , x∈ 2 n− 1 ,2n+10.50.25y−2−1 1 2 xc.0.8 Chapter ReviewConcepts Test1. False: p and q must be integers.⎧ ⎡ 1 ⎞⎪ 4( x− x )+ 1 : x∈ ⎢n,n+⎟⎪ ⎣ 4 ⎠= ⎨⎪ 4 7 ⎡ 1 ⎞− ( x − x)+ : x ∈ n + , n + 1⎪ ⎟3 3 ⎢⎩⎣ 4 ⎠where n is an integer.55. f ( x)21yp2. True:1 p2 p1q2 − p2q− = 1;sinceq q q q1 2 1 21, 1, 2, and 2are pq 1 2−p2q1and qq 1 2.p q p q are integers, so3. False: If the numbers are opposites(–π and π ) then the sum is 0,which is rational.4. True: Between any two distinct realnumbers there are both a rationaland an irrational number.5. False: 0.999... is equal to 1.n6. True: ( ) ( )m nmmna = a = a−11 x7. False: ( a* b)* c = a ; a*( b* c)= abcbc8. True: Since x ≤ y ≤ z and x≥ z,x = y = zx9. True: If x was not 0, then ε = would2be a positive number less than x .52 Section 0.8 Instructor’s Resource Manual

10. True: y− x =−( x− y)so( x − y)( y− x) = ( x− y)( −1)( x−y)2= ( −1)( x−y) .2( x y) 0− ≥ for all x and y, so2−( x− y) ≤ 0.11. True: a 1; 0 so ≤ ;1+r 1– ralso, –1 < r < 1.If –1 < r < 0, then r = – r and1– r = 1 + r , so1 1 1= ≤ .1+r 1– r 1– rIf 0 < r < 1, then r = r and1– r = 1– r , so20. True: If r > 1, then 1− r < 0. Thus,1 1since 1+ r ≥1 − r , ≤ .1− r 1+rIf r > 1, r = r,and 1− r = 1 − r , so1 1 1= ≤ .1− r 1− r 1+rIf r 0, y < 0) orx – y = – x–y = x+y(x < 0, y > 0). In either casex – y = x+y .If either x = 0 or y = 0, theinequality is easily seen to be true.22. True: If y is positive, then x = y satisfies( ) 22= =x y y.23. True: For every real number y, whether itis positive, zero, or negative, thecube root x = 3 y satisfies( ) 33= 3 =x y y24. True: For example x 2 ≤ 0 has solution[0].25. True: 2 2x + ax+ y + y = 02 22 a 2 1 a 1x + ax+ + y + y+ = +4 4 4 42 2 2⎛ a⎞ ⎛ 1⎞a + 1⎜x+ ⎟ + ⎜y+ ⎟ =⎝ 2⎠ ⎝ 2⎠4is a circle for all values of a.26. False: If a = b = 0 and c < 0 , the equationdoes not represent a circle.1 1 1≤ = .1+r 1– r 1– rInstructor’s Resource Manual Section 0.8 53

27. True;3y− b= ( x−a)43 3ay = x− + b;4 4If x = a + 4:3 3ay = ( a+ 4)– + b4 43a 3a = + 3– + b = b + 34 428. True: If the points are on the same line,they have equal slope. Then thereciprocals of the slopes are alsoequal.29. True: If ab > 0, a and b have the samesign, so (a, b) is in either the first orthird quadrant.30. True: Let x = ε /2. If ε > 0 , then x > 0and x < ε.31. True: If ab = 0, a or b is 0, so (a, b) lieson the x-axis or the y-axis. If a= b = 0,(a, b) is the origin.32. True: y1 y2,x1, y1 and x2,y 2are on the same horizontal line.= so ( ) ( )33. True: 2 2d = [( a+ b) – ( a– b)] + ( a– a)2= (2 b) = 2b34. False: The equation of a vertical linecannot be written in point-slopeform.35. True: This is the general linear equation.36. True: Two non-vertical lines are parallelif and only if they have the sameslope.37. False: The slopes of perpendicular linesare negative reciprocals.38. True: If a and b are rational and( a,0 ),( 0, b ) are the intercepts, thebslope is − which is rational.a39. False:ax + y = c ⇒ y =− ax + cax − y = c ⇒ y = ax −c( a)( −a) ≠ −1.(unless a =± 1)40. True: The equation is(3 + 2 mx ) + (6m− 2) y+ 4 − 2m=0which is the equation of a straightline unless 3+ 2 m and 6m− 2 areboth 0, and there is no real numberm such that3 + 2m= 0 and 6m− 2 = 0.41. True:f( x) = –( x + 4x+3)= –( x+ 3)( x+1)22− ( x + 4x+ 3) ≥ 0 on −3≤ x ≤ − 1.42. False: The domain does not includenπ + π where n is an integer.243. True: The domain is ( −∞, ∞)range is[ −6, ∞ ).44. False: The range is ( −∞, ∞ ) .45. False: The range ( −∞, ∞ ) .and the46. True: If f(x) and g(x) are even functions,f(x) + g(x) is even.f(–x) + g(–x) = f(x) + g(x)47. True: If f(x) and g(x) are odd functions,f(–x) + g(–x) = –f(x) – g(x)= –[f(x) + g(x)], so f(x) + g(x) is odd48. False: If f(x) and g(x) are odd functions,f(–x)g(–x) = –f(x)[–g(x)] = f(x)g(x),sof(x)g(x) is even.49. True: If f(x) is even and g(x) is odd,f(–x)g(–x) = f(x)[–g(x)]= –f(x)g(x), so f(x)g(x) is odd.50. False: If f(x) is even and g(x) is odd,f(g(–x)) = f(–g(x)) = f(g(x)); while iff(x) is odd and g(x) is even,f(g(–x)) = f(g(x)); so f(g(x)) is even.51. False: If f(x) and g(x) are odd functions,f ( g( − x))= f(–g(x)) = –f(g(x)), sof(g(x)) is odd.52. True:3 32(– x) + (– x) –2 x – xf(– x)= =2 2(– x) + 1 x + 132x+ x=−2x + 154 Section 0.8 Instructor’s Resource Manual

53. True:2(sin(– t)) + cos(– t)f(– t)=tan(– t) csc(– t)2 2( − sin t) + cos t (sin t) + cost= =– tan t(– csc t) tan tcsct54. False: f(x) = c has domain ( −∞, ∞)andthe only value of the range is c.Sample Test Problems1. a.n1 2⎛ 1 1 1 25⎜n+ ⎞ ⎟ ; ⎛ ⎜1+ ⎞ ⎟ = 2; ⎛ ⎜2 + ⎞⎟ = ;⎝ n ⎠ ⎝ 1⎠ ⎝ 2⎠4–2⎛ 1 ⎞ 4⎜–2+ ⎟ =⎝ –2⎠2555. False: f(x) = c has domain ( −∞, ∞ ), yetthe range has only one value, c.−1.8( − 1.8) = = − 0.9 =−12 56. True: g 57. True:3 ) 2( f g)(x)= ( x = x2 ) 3( g f )( x)= ( x = x66b.c.2 2 22n n ⎡ ⎤( – + 1) ; (1) – (1) + 1 = 1;⎣ ⎦⎡ 2(2) – (2) + 1⎤= 9;⎣⎦⎡ 2(–2) – (–2) + 1⎤= 49⎣⎦223/ n 3/1 3/2 –3/2 14 ; 4 = 64; 4 = 8; 4 =858. False:3 ) 2( f g)(x)= ( x = xf ( x)⋅ g(x)= x x = x23651 1 1 1 2d. n ; 1 = 1; = = ;n 1 2 2 259. False: The domain of f gexcludes any− =2 1 2−2values where g = 0.60. True: f(a) = 0Let F(x) = f(x + h), thenF(a – h) = f(a – h + h) = f(a) = 061. True:cos xcot x =sin xcos( −x)cot( − x)=sin( −x)cos x= = −cotx−sinx62. False: The domain of the tangent functionexcludes all nπ + π where n is an2integer.63. False: The cosine function is periodic, socos s = cos t does not necessarilyimply s = t; e.g.,cos 0 = cos 2π= 1, but 0≠2 π.2. a.b.c.1 1− 1 1+ +⎛ 1 1⎞⎛ 1 1⎞1 1m n⎜ + + ⎟⎜ − + ⎟ =⎝ m n⎠⎝ m n⎠ 1 11− +m nmn+ n+m=mn − n + m2 x 2 x−−x + 1 2x −x− 2 x+ 1 ( x− 2)( x+1)=3 2 3 2−−x+ 1 x− 2 x+ 1 x−23 22( x−2)− x=3 ( x−2) − 2( x+1)x − 4=x − 8( t −1) ( t− 1)( t + t+1) 2= = t + t+1t−1 t−13. Let a, b, c, and d be integers.ac+b d a c ad + bc= + = which is rational.2 2b 2d 2bdInstructor’s Resource Manual Section 0.8 55

4. x = 4.1282828…1000x= 4128.282828…10x= 41.282828…990x= 40874087x =99013.2 221 t –44t+ 12 ≤ –3; 21 t –44t+ 15≤0;244 ± 44 – 4(21)(15) 44 ± 26 3 5t = = = ,2(21) 42 7 3⎛ 3 5 3 5⎜t– ⎞⎛ ⎟⎜t– ⎞ ⎟≤0; ⎡ ,⎤7 3 ⎢7 3⎥⎝ ⎠⎝ ⎠ ⎣ ⎦5. Answers will vary. Possible answer:6.⎛⎜⎝130.50990...50 ≈8.15× 10 −1.32⎞⎟⎠3.243 42≈ 545.397. ( ) 2.5 3π– 2.0 – 2.0 ≈ 2.662 28. ( ) ( )sin 2.45 + cos 2.40 −1.00 ≈− 0.04959. 1–3x> 03x< 11x 2x−54x>−8x >−2; −2,∞( )2x−1 ⎛ 1⎞⎜ ⎟x − 2 ⎝ 2⎠14. > 0; −∞, ∪( 2, ∞)15.2( x+ 4)(2x−1) ( x−3) ≤0;[ − 4,3]16. 3x− 4 < 6; − 6< 3x− 4< 6; − 2< 3x

21. |t – 5| = |–(5 – t)| = |5 – t|If |5 – t| = 5 – t, then 5−t≥ 0.t ≤ 522. t− a = −( a− t)= a−tIf a− t = a− t,then a−t≥ 0.t≤ a23. If x ≤ 2, then2 20≤ 2x + 3x+ 2 ≤ 2x + 3x+ 2≤ 8+ 6+ 2=16alsox21 1+ 2 ≥ 2 so ≤ . Thus2x + 2 227.28.⎛2+ 10 0+4⎞center = ⎜ , ⎟=(6, 2)⎝ 2 2 ⎠1 2 2 1radius = (10 – 2) + (4 – 0) = 64 + 162 2= 2 52 2circle: ( x– 6) + ( y– 2) = 202 2x + y − 8x+ 6y= 02 2x − 8x+ 16+ y + 6y+ 9= 16+92 2( x− 4) + ( y+ 3) = 25;center = ( 4, − 3), radius = 52x 2x x2 22x+ 3 + 2 2 3 2 1 16 ⎛1⎞= + + ≤ ⎜ ⎟x + 2 x + 2 ⎝ 2 ⎠= 824. a. The distance between x and 5 is 3.25.b. The distance between x and –1 is less thanor equal to 2.c. The distance between x and a is greaterthan b.2 2d( A, B) = (1+ 2) + (2−6)= 9+ 16 = 52 2dBC ( , ) = (5− 1) + (5−2)= 16 + 9 = 52 2d( A, C) = (5+ 2) + (5−6)= 49 + 1 = 50 = 5 22 2 2( AB) ( BC) ( AC) ,+ = so Δ ABC is a righttriangle.29.2 2x − 2x+ y + 2y= 22 2x − 2x+ 1+ y + 2y+ 1= 2+ 1+12 2( x 1) ( y 1) 4− + + =center = (1, –1)2 2x + 6x+ y – 4 y = –72 2x + 6x+ 9+ y – 4y+ 4 = –7+ 9+42 2( x 3) ( y–2) 6+ + =center = (–3, 2)2 2d = (–3 –1) + (2 + 1) = 16 + 9 = 530. a. 3x+ 2y= 62y= − 3x+63y = − x+323m =−23y− 2 =− ( x−3)23 13y =− x+2 2⎛1+ 7 2+8⎞ ⎜ ⎟ =⎝ 2 2 ⎠26. midpoint: , ( 4,5)2 2d = (4 − 3) + (5 + 6) = 1+ 121 = 122Instructor’s Resource Manual Section 0.8 57

.2m = ;32y+ 1 = ( x−1)32 5y = x−3 3b. 3 x–2y= 5–2 y = –3x+53 5y = x– ;2 23m =23y–1 = ( x+2)23y = x+42c. y = 9c. 3x + 4y = 94y = –3x + 9;3 9 4y = – x+; m =4 4 34y–1 = ( x+2)34 11y = x+3 3d. x = –2e. contains (–2, 1) and (0, 3);y = x + 33–1m = ;0 + 2d. x = –33+ 1 4 11−3 8 432. m1 = = ; m2= = = ;5−2 3 11−5 6 311+1 12 4m3= = =11−2 9 3m1 = m2 = m3,so the points lie on the sameline.31. a.3–1 2m = = ;7+2 92y–1 = ( x+2)92 13y = x+9 933. The figure is a cubic with respect to y.3The equation is (b) x = y .34. The figure is a quadratic, opening downward,with a negative y-intercept. The equation is (c)2y = ax + bx+ c. with a < 0, b > 0, and c < 0.35.58 Section 0.8 Instructor’s Resource Manual

36.37.2 2x − 2x+ y = 32 2x − 2x+ 1+ y = 42 2( x− 1) + y = 440. 4x− y = 2y = 4x−2;1m = −4contains ( a ) ( b ),0 , 0, ;ab= 82ab = 1616b =ab− 0 b 1 =− =− ;0−a a 4a = 4b⎛16⎞a = 4 ⎜ ⎟⎝ a ⎠2a = 64a = 816 1b = = 2; y =− x+28 438.41. a.b.1 1 1f (1) = – = –1+1 1 2f ⎛ 1 1 1⎜– ⎞ ⎟ – 4⎝ 2 ⎠= – 1 + 1 – 1=2 2c. f(–1) does not exist.d.e.1 1 1 1f( t –1) = – = –t –1+1 t –1 t t –1⎛1⎞ 1 1 tf ⎜ ⎟ = – = – t⎝t⎠1+1 1 1+tt t39. y = x 2 – 2x + 4 and y – x = 4;x22x+ 4= x − 2x+4− 3x= 0xx ( –3) = 0points of intersection: (0, 4) and (3, 7)42. a.2+1 3g(2)= =2 212⎛ 1 ⎞ + 1b. g ⎜ ⎟ = = 31⎝ 2 ⎠c.g(2+ h)– g(2)=h=22h+6–3h–62( h+2)h=2 + h+1 2+1–2+h 2h–2( h+2)hh–1=2( h + 2)43. a. { x∈ : x ≠ –1, 1}b. { x∈ : x ≤ 2}Instructor’s Resource Manual Section 0.8 59

44. a.3(– x) 3xf(– x) = = – ;2 2(– x) + 1 x + 1odd46.b. g(– x) = sin(– x) + cos(– x)=− sin x + cos x = sin x + cos x;evenc.3 3h(– x) (– x) sin(– x) – x –sinx= + = ; oddd.2 2(– x) + 1 x + 1k(– x) = = ;4 4– x + (– x)x + x45. a. f (x) = x 2 –1even47. V(x) = x(32 – 2x)(24 – 2x)Domain [0, 12]48. a.⎛ 1⎞2 13( f + g)(2) = ⎜2– ⎟+ (2 + 1) =⎝ 2⎠2b.⎛3⎞15( f ⋅ g)(2) = ⎜ ⎟(5)=⎝2⎠2c.1( f g)(2)= f (5) = 5 – =5245b. g(x) =xx 2 + 1d.e.⎛ 3 ⎞ ⎛ 3 ⎞( g f )(2) = g⎜⎟ = ⎜ ⎟⎝ 2 ⎠ ⎝ 2 ⎠f3 ⎛ 1⎞(–1) = ⎜–1+ ⎟ = 0⎝ 1⎠3213+ 1 =4f.f22 2 ⎛3⎞ 2(2) + g (2) = ⎜ ⎟ + (5)⎝2⎠9 109= + 25 =4 449. a. y = 1 4 x2⎧c. h(x) = x2 if 0 ≤ x ≤ 2⎨⎩ 6–x if x > 260 Section 0.8 Instructor’s Resource Manual

. y = 1 4 ( x + 2)253. a. sin (–t) = –sin t = –0.8b.2 2sin t+ cos t = 12 2cos t = 1– (0.8) = 0.36cos t = –0. 6c. sin 2t = 2 sin t cos t = 2(0.8)(–0.6) = –0.96d.sin t 0.8 4tan t = = = – ≈ –1.333cos t –0.6 3c.1y = –1 + ( x+2)42⎛πe. cos – sin 0.82 t ⎞⎜ ⎟ = t =⎝ ⎠f. sin( π + t) =− sin t =− 0.854. sin 3t = sin(2 t+ t) = sin 2tcos t+cos 2tsint2 2= 2sintcos t+(1–2sin t)sint2 3= 2sin t(1– sin t) + sin t – 2sin t3 3= 2sin t – 2sin t+sin t – 2sin t3= 3sin t – 4sin t55. s = rt⎛ rev ⎞⎛ rad ⎞⎛ 1 min ⎞= 9⎜20 ⎟⎜2 π ⎟⎜ ⎟(1 sec) = 6π⎝ min ⎠⎝ rev ⎠⎝60 sec ⎠≈ 18.85 in.50. a. ( −∞ ,16]51.52. a.b. f g = 16 – x ; domain [–2, 2]44c. g f = ( 16 – x ) = (16 – x);domain ( −∞ ,16](note: the simplification4 2( 16– x) = (16– x)is only true giventhe restricted domain)f ( x) = x, g( x) = 1 + x, h( x) = x , k(x) = sin x,F(x)=b.c.1+sin2x = f g h k1sin(570 ° ) = sin(210 ° ) = – 2⎛9π⎞ ⎛π⎞cos⎜ ⎟= cos⎜ ⎟=0⎝ 2 ⎠ ⎝2⎠⎛ 13π⎞ ⎛ π⎞3cos ⎜–⎟= cos⎜− ⎟=⎝ 6 ⎠ ⎝ 6⎠222Review and Preview Problems1. a) 0< 2x< 4; 0< x

) x−7≤3 and x−7≥ −3x≤10 and x≥44≤x ≤10c) x−7≤1 and x−7≥−1x≤8 and x≥66≤x ≤8d) x− 7< 0.1 and x− 7 >−0.1x< 7.1 and x > 6.96.9 < x < 7.18. a) x− 2< 1 and x− 2>−1x< 3 and x > 11< x < 3b) x−2≥1 or x−2≤−1x≥3 or x≤1c) x− 2< 0.1 and x− 2>−0.1x< 2.1 and x > 1.91.9 < x < 2.1d) x− 2< 0.01 and x− 2> −0.01x< 2.01 and x > 1.991.99 < x < 2.019. a) x−1≠ 0; x ≠ 1b)22x −x−1≠ 0; x ≠1, − 0.510. a) x ≠ 0 b) x ≠ 011. a) f ( )0−10 = = 10−10.81−1f ( 0.9)= = 1.90.9 −10.9801−1f ( 0.99)= = 1.990.99 −10.998001−1f ( 0.999)= = 1.999.999 −11.002001−1f ( 1.001)= = 2.0011.001−11.0201−1f ( 1.01)= = 2.011.01−11.21−1f ( 1.1)= = 2.11.1−14−1f ( 2)= = 32−1b) g ( )g ( )g ( )g ( )g ( )g ( )g ( )12. a) ( )0 = − 10.9 =−0.03571430.99 =−0.00335570.999 =−0.0003335561.001 = 0.0003331111.01 = 0.003311261.1 = 0.031251g ( 2)=51F − 1 = =−1−10.1F ( − 0.1)= =−1−0.10.01F ( − 0.01)= =−1−0.010.001F ( − 0.001)= = −1−0.0010.001F ( 0.001)= = 10.0010.01F ( 0.01)= = 10.010.01F ( 0.1)= = 10.011F () 1 = = 11b) G ( − 1)= 0.841471G ( − 0.1)= 0.998334G ( − 0.01)= 0.999983G ( − ) =G ( ) =G ( 0.01)= 0.999983G ( 0.1)= 0.998334G () 1 = 0.8414710.001 0.999999830.001 0.9999998313. x− 5< 0.1 and x− 5>−0.1x< 5.1 and x > 4.94.9 < x < 5.114. x− 5< ε and x− 5> −εx< 5+ ε and x > 5−ε5− ε < x < 5+ε15. a. True. b. False: Choose a = 0.c. True. d. True16. ( )sin c+ h = sin ccos h+coscsinh62 Review and Preview Instructor’s Resource Manual

CHAPTER 1Limits1.1 Concepts Review1. L; c2. 63. L; right4. lim f ( x ) = Mx→c Problem Set 1.11.lim( x – 5) = –2x→39.10.x→–13 2x – 4x + x+6limx + 1( x+ 1)( x –5x+6)= limx→–1x + 1x→–122= lim ( x – 5x+6)= (–1) – 5(–1) + 6= 12x→0x→024 3 2x + 2 x – xlim2x2= lim( x + 2 x–1) = –12.3.lim (1 – 2 t) = 3t→–1x→−22 2lim ( x + 2x− 1) = ( − 2) + 2( −2) − 1 =− 111.x→–t2 2x – t ( x+t)( x– t)lim = limx→– t x+ t x→–t x+t= lim ( x– t)= –t – t = –2t4.x→−22 2lim ( x + 2t− 1) = ( − 2) + 2t− 1= 3+2t2( )25. ( t ) ( )lim − 1 = −1 − 1 = 0t→−16. ( ) ( )t→−12( )2 2 2 2lim t − x = −1 − x = 1−x12.x→32x –9limx→3x –3( x–3)( x+3)= limx→3x –3= lim( x + 3)= 3 + 3 = 67.2x – 4 ( x–2)( x+2)lim = limx–2 x–2= lim( x + 2)x→2 x→2x→2= 2 + 2 = 413.limt→2=limt→29( 2)24( t+ 4)( t−2)2(3t− 6)2( t− 2) t+4t −8.t→–72t + 4 t –21limt + 7( t+7)( t –3)= limt→–7t + 7= lim ( t – 3)t→–7= –7 – 3 = –1014.=t + 4limt→292+4 6= =9 9( t − 7)lim+t→7t − 7( t−7) t−7= lim+t→7t − 7= lim t −7+t→7= 7− 7 = 03Instructor’s Resource Manual Section 1.1 63

15.16.17.18.4 2 2 2x –18x + 81 ( x – 9)lim= lim( x– 3) ( x– 3)x→3 2 x→322 2( x–3) ( x+3)= lim = lim( x + 3) = (3 + 3)x→3 2( x –3) x→3= 362 23 3(3u+ 4)(2 u– 2) 8(3u+4)( u–1)lim= lim( u–1) ( u–1)u→1 2 u→12= lim 8(3u+ 4)( u– 1) = 8[3(1) + 4](1 – 1) = 0u→12 2(2 + h) − 4 4 + 4h+ h −4lim= limh→0 h h→0h2h + 4h= lim = lim( h + 4) = 4h→0 h h→02 2 2 2 2( x + h) − x x + 2xh+ h −xlim= limh→0 h h→0h2h + 2xh= lim = lim( h+ 2 x) = 2xh→0 h h→0x2x19. xsinx→01. 0.4207350.1 0.4991670.01 0.4999920.001 0.49999992–1. 0.420735–0.1 0.499167–0.01 0.499992–0.001 0.49999992sin xlim = 0.52x1−cos20. t1. 0.229849t2t0.1 0.02497920.01 0.002499980.001 0.00024999998–1. –0.229849–0.1 –0.0249792–0.01 –0.00249998–0.001 –0.000249999981−costlim = 0t→02t21. xx→02 2( x − sin x) / x1. 0.02513140.10.010.0012.775×10 −62.77775×10 −102.77778×10 −14–1. 0.0251314–0.1 2.775×10 −6–0.01–0.001( x–sin x)lim = 02x22. xx→02.77775×10 −102.77778×10 −1422 2(1 − cos x) / x1. 0.2113220.1 0.002495840.01 0.00002499960.001 2.5×10 −7–1. 0.211322–0.1 0.00249584–0.01 0.0000249996–0.001 2.5×10 −7(1 – cos x)lim = 02x23. t22( t −1) /(sin( t − 1))2. 3.565191.1 2.10351.01 2.010031.001 2.0010 1.18840.9 1.903170.99 1.990030.999 1.9992t −1lim = 2t→1sin( t −1)64 Section 1.1 Instructor’s Resource Manual

27. x2 2( x−π/4) /(tanx−1)x−sin( x−3) −324. xx−34. 0.1585293.1 0.001665833.01 0.00001666663.001 1.66667×10 −72. 0.1585292.9 0.001665832.99 0.00001666662.999 1.66667×10 −7x–sin( x–3)–3lim = 0x→3x –325. x (1+ sin( x−3 π / 2)) /( x−π )1. + π 0.45970.1 + π 0.05000.01 + π 0.00500.001 + π 0.0005–1. + π –0.4597–0.1 + π –0.0500–0.01 + π –0.0050–0.001 + π –0.00053π1+ sin( x −2 )lim = 0x→πx −π26. t (1 − cot t) /(1 / t)t→01. 0.3579070.1 –0.8966640.01 –0.9899670.001 –0.999–1. –1.64209–0.1 –1.09666–0.01 –1.00997–0.001 –1.0011–cottlim = –11tπ1. +40.03202440.1+ π4 0.2010020.01+ π4 0.2450090.001+ π 0.24954π− 1. +4 0.674117− 0.1+ π4 0.300668− 0.01+ π4 0.255008− 0.001+ π4 0.2505π2( x −4 )lim = 0.252x→π(tan x −1)428. u (2 − 2sin u) /3u29. a.π1. +20.119210.1+ π2 0.001993390.01+ π2 0.00002108620.001+π 2.12072×10 −72π− 1. +2 0.536908− 0.1+ π2 0.00226446− 0.01+ π2 0.0000213564− 0.001+π 2.12342×10 −722−2sinulim = 0u→ π 3u2lim f( x) = 2x→–3b. f(–3) = 1c. f(–1) does not exist.d.5lim f( x)=2x→–1e. f(1) = 2f.g.h.limx→ 1–x→1f(x) does not exist.lim f ( x ) = 2lim f( x) = 1+x→1i. lim f ( x)+x→−15=2Instructor’s Resource Manual Section 1.1 65

30. a.lim f ( x)x→–3does not exist.b.lim f ( x)x→1does not exist.b. f(–3) = 1c. f(–1) = 1d.lim f( x) = 2x→–1e. f(1) = 134.c. f(1) = 2d.lim f( x) = 2+x→1f.lim f ( x)x→1does not exist.g.lim f ( x ) = 1–x→1h.lim f ( x)does not exist.+x→1i. f ( x)lim = 2+x→−131. a. f(–3) = 2a.lim gx ( ) = 0x→1b. f(3) is undefined.c.d.e.f.lim f( x) = 2−x→–3lim f( x) = 4+x→–3lim f ( x)does not exist.x→–3lim f ( x)does not exist.+x→3b. g(1) does not exist.c.d.lim gx ( ) = 1x→2lim gx ( ) = 1+x→235. f ( x) x–⎡[ x]= ⎣ ⎤⎦32. a.b.c.lim f( x) =− 2−x→–1lim f( x) =− 2+x→–1lim f( x) =− 2x→–1d. f (–1) = –2e.lim f( x) = 0x→1a. f(0) = 033.f. f (1) = 0b.c.lim f ( x)does not exist.x→0lim f ( x ) = 1–x→0d.1lim f( x)=2x→12a.lim f( x) = 0x→066 Section 1.1 Instructor’s Resource Manual

36. f( x)=xx41. lim f ( x ) exists for a = –1, 0, 1.x→a 42. The changed values will not change lim f ( x )x→a any a. As x approaches a, the limit is still a2 .ata. f (0) does not exist.43. a.b.x −1lim does not exist.x→1x −1x − 1x − 1lim =− 1 and lim = 1−x→1x −1+x→1x −1x − 1lim =− 1x −1−x→137.b.c.d.lim f ( x)does not exist.x→0lim f ( x ) = –1–x→0lim f( x) = 1x→122x −1lim does not exist.x→1x −1−x→12x − 1lim =− 2 andx −1+x→12x − 1lim = 2x −1c.d.−x→12x − x−1 − 1lim =− 3x −1⎡lim⎢⎣1 1 ⎤−does not exist.−1 −1⎥⎦−x→1x x44. a. x xb.lim − = 0x→1+1limx does not exist.+x→038.limx→039. a.b.x + 2−2x( x+ 2− 2)( x+ 2+2)= limx→0x( x+ 2+2)x+ 2−2x= lim= limx→0 x( x+ 2 + 2) x→0x( x+ 2 + 2)1 1 1 2= lim= = =x→0x + 2+ 2 0+ 2+2 2 2 4lim f ( x)does not exist.x→1lim f( x) = 0x→0c.1/x lim x( − 1) = 0+x→01/x lim ( − 1) = 0+x→0d. x45. a) 1 b) 0c) − 1d) − 146. a) Does not exist b) 047.c) 1 d) 0.556limx→0xfor x < 0.does not exist sincex is not defined40.48.49.50.51.lim x = 1+x→0x→0xlim x = 0lim x = 1x→0xsin 2x1xlim =x →04 2Instructor’s Resource Manual Section 1.2 67

52.sin 5x5=limx →03 x 37. If x is within 0.001 of 2, then 2x is within 0.002of 4.53.⎛1⎞lim cos⎜⎟⎝ x ⎠x→0does not exist.54.55.56.57.58.59.⎛1⎞lim x cos⎜⎟ = 0⎝ x ⎠x→03x −1lim = 6x→12x+ 2−2xsin 2xlim = 22sin( x )x→0–x→22x – x–2lim = –3x –22lim = 01+2 x+ 1/( −1)x→1lim x;The computer gives a value of 0, butx→0lim−x→0x does not exist.8. If x is within 0.0005 of 2, then x 2 is within 0.002of 4.9. If x is within 0.0019 of 2, then 8xis within0.002 of 4.1.2 Concepts Review1. L – ε; L + ε2. 0 < x – a < δ; f( x)– L < ε3.ε310. If x is within 0.001 of 2, then x8 is within 0.002of 4.4. ma + bProblem Set 1.21. 0 < t – a < δ ⇒ f ( t)– M < ε2. 0 < u– b < δ ⇒ g( u)–L < ε3. 0 < z – d < δ ⇒ h(z)– P < ε4. 0 < y – e < δ ⇒ φ(y)– B < ε5. 0 < c – x < δ ⇒ f ( x)– L < ε6. 0 < t – a < δ ⇒ g(t)– D < ε11. 0 < x–0 < δ ⇒ (2 x–1)–(–1)< ε2 x–1+ 1 < ε ⇔ 2x< ε⇔ 2 x < εε⇔ x

12. 0 < x+ 21 < δ ⇒ (3 x–1) – (–64) < ε13.3 x–1+ 64 < ε ⇔ 3x+ 63 < ε⇔ 3( x + 21) < ε⇔ 3 x + 21 < εε⇔ x + 21

5⇔ x −4⋅

21.20< x+ 1 < ⇒ ( x –2 x–1)–2 0, ( x ) = x.x < ε ⇔ ( x ) = x < ε22222.4 40 < x < ⇒ x –0 = x 0. Then since lim f ( x ) = L , there isx→c some δ 1 > 0 such that0 < x– c < δ1⇒ f( x)–L < ε .Since lim f (x) = M, there is some δ 2 > 0 suchx→cthat 0 < x− c < δ2⇒ f( x)− M < ε .Let δ = min{δ 1 , δ 2 } and choose x 0 such that0 < x 0 – c < δ .Thus, f( x0)– L < ε ⇒ − ε < f( x0)− L 0 and the corresponding δ thatworks for G(x), then x–c < δ impliesF( x)–0 = F( x) ≤ G( x)< ε sincelim G(x) = 0.x→cThus, lim F( x) = 0.x→cInstructor’s Resource Manual Section 1.2 71

29. Choose ε > 0. Since lim f( x) = L, there is ax→ aδ > 0 such that for 0 < x – a < δ, f( x)– L < ε.That is, fora− δ < x< a or a < x< a+ δ ,L− ε < f( x)< L+ ε .Let f(a) = A,M = max { L− ε , L+ ε , A}, c = a – δ,d = a + δ. Then for x in (c, d), f ( x) ≤ M,sinceeither x = a, in which casef ( x) = f( a)= A ≤ M or 0 < x–a < δ soL− ε < f( x)< L+ ε and f ( x) < M.30. Suppose that L > M. Then L – M = α > 0. Nowtake ε < α 2 and δ = min{δ 1, δ 2 } where0 < x– a < δ1⇒ f( x)–L < ε and0 < x– a < δ2⇒ g( x)– M < ε.Thus, for 0 < x– a < δ ,L – ε < f(x) < L + ε and M – ε < g(x) < M + ε.Combine the inequalities and use the factthat f ( x) ≤ g( x)to getL – ε < f(x) ≤ g(x) < M + ε which leads toL – ε < M + ε or L – M < 2ε.However,L – M = α > 2εwhich is a contradiction.Thus L≤ M .31. (b) and (c) are equivalent to the definition oflimit.32. For every ε > 0 and δ > 0 there is some x with0 < x–c < δ such that f (x )– L > ε.x 3 – x 2 –2x –433. a. g(x) =x 4 – 4x 3 + x 2 + x + 6x + 6b. No, becausex 4 – 4x 3 + x 2 + x + 6 + 1 hasan asymptote at x ≈ 3.49.c. If δ ≤ 1 , then 2.75 < x < 34or 3 < x < 3.25 and by graphing3 2x −x −2x−4y = g( x)=4 3 2x − 4x + x + x+6on the interval [2.75, 3.25], we see that3 2x – x –2 x–40< < 34 3 2x –4x + x + x+6so m must be at least three.1.3 Concepts Review1. 482. 43. – 8; – 4 + 5c4. 0Problem Set 1.31. lim(2x + 1) 4x→1= limx→12x + lim 1 3x→11 2,1= 2limx →1= 2(1) + 1 = 3x + limx→12. lim (3x 2 –1) 5x→ –1= lim 3x 2 – lim 1 3x→ –1 x→–1= 3 lim x 2 – lim 1 8x→ –1 x→–12⎛= 3 lim x ⎞⎜ ⎟⎝ x→ –1 ⎠= 3(–1) 2 –1= 2– limx →–1 1 2, 13. lim [(2x +1)( x – 3)] 6x→0= lim (2x +1) ⋅ lim (x – 3) 4, 5x→ 0 x→ 0⎛⎞ ⎛⎞= ⎜ lim 2x + lim 1⎟ ⋅ ⎜ lim x – lim 3⎟ ⎝ x→ 0 x→ 0 ⎠ ⎝ x→0 x→ 0 ⎠3⎛⎞ ⎛⎞= ⎜ 2 lim x + lim 1⎟ ⋅ ⎜ lim x – lim 3⎟ ⎝ x →0 x→ 0 ⎠ ⎝ x→0 x→ 0 ⎠2, 1= [2(0) +1](0 – 3) = –34.2 2lim [(2x+ 1)(7 x + 13)]6x→2=2lim (2x+ 1) ⋅2lim (7x+ 13) 4, 3x→2 x→2⎛ 2 ⎞ ⎛ 2 ⎞= ⎜2 lim x + lim 1⎟⋅ ⎜7 lim x + lim 13⎟⎝ x→ 2 x→ 2 ⎠ ⎝ x→ 2 x→2 ⎠ 8,1⎡ ⎛= ⎢2⎜lim⎢⎣⎝ x→22⎞x⎟⎠2⎤⎡⎛+ 1⎥⎢7⎜lim⎥⎦⎢⎣⎝ x→2⎞x⎟⎠2⎤+ 13⎥⎥⎦= [2( 2) + 1][7( 2) + 13] = 1352272 Section 1.3 Instructor’s Resource Manual

5.6.7.8.2x+ 1limx →25–3 xlim(2x+ 1)x→2= 4, 5lim(5 – 3x)x→2lim 2x+ lim1x→2x→2= 3, 1lim 5 – lim 3xx→2x→22 lim x + 1x→ 2=5 – 3 lim x2x→2= 2(2) + 15–3(2) = –534x+ 1limx→–37–22x3lim (4x+ 1)=x→–32lim (7 – 2x)4, 5x→–33lim 4x+ lim 1x→–3x→–3=2lim 7 – lim 2x3, 1x→–3x→–334 lim x + 1x→–3=27 – 2 lim x8x→–334⎜⎛ lim x⎟⎞ + 1⎝ x→–3=⎠27 – 2⎜⎛ lim x⎟⎞⎝ x→–3⎠= 4(–3)3 + 17–2(–3) 2 = 10711lim 3 x – 5x→3= 5, 3lim (3 x – 5)x→3= 2, 13 lim x – lim 5x→3 x→3= 3(3) – 5 = 2lim25x+ 2x9x→–3=2lim (5x+ 2 x)4, 3x→–3= 5 lim2x + 2 lim x8x→–3 x→–32= 5⎛lim x⎞⎜ ⎟ + 2 lim x⎝x→–3 ⎠ x→–32= 5(–3) + 2(–3) = 39772929. lim (2t 3 +15) 13 8t→ –213⎡⎤= limt→–2 (2t3 + 15)⎣ ⎢⎦ ⎥4, 310.11.13⎡= 2 lim t 3 ⎤+ lim 15⎣ ⎢t→ –2 t→ –2 ⎦ ⎥313⎡⎤= 2⎛lim t⎞⎢ ⎜ ⎟ + lim 15⎥⎢ ⎝t→–2 ⎠ t→–2⎣⎥⎦= [2(–2) 3 + 15] 13 = –182, 1lim3 2–3w+ 7w9w→–2=3 2lim (–3w+ 7 w ) 4, 3w→–2= –3 lim3w + 7 lim2w 8w→–2 w→–2=3 2–3⎛lim w⎞7⎛lim w⎞⎜ ⎟ + ⎜ ⎟⎝w→–2 ⎠ ⎝w→–2⎠23 2= –3(–2) + 7(–2) = 2 1331/3⎛4y+ 8y⎞limy→2⎜y + 4 ⎟⎝ ⎠31/3⎛ 4y+ 8y⎞= lim⎜ y→2y + 4 ⎟⎝⎠313⎡ lim (4y+ 8 y)⎤⎢ y→2⎥= ⎢lim ( y 4) ⎥⎢+y→2⎥⎣⎦313⎛4 lim y + 8 lim y⎞⎜ y→2 y→2⎟= ⎜ ⎟⎜lim y + lim 4y→2 y→2⎟⎝⎠31/3⎡ ⎛ ⎞ ⎤⎢4⎜lim y⎟+ 8 lim y⎥⎢ y→2 y→2⎥=⎝ ⎠⎢lim y + 4 ⎥⎢ y→2⎥⎢⎥⎣⎦31/3⎡4(2) + 8(2) ⎤= ⎢ ⎥ = 2⎢⎣2+4 ⎥⎦974, 38, 12Instructor’s Resource Manual Section 1.3 73

12. lim (2w 4 –9w 3 +19) –1/2w→ 51= lim→54 32 − 9 + 19lim 1=w→5lim4 32 w – 9w+ 19w→5=14 3lim (2 w – 9w+ 19)w→5=14 3lim 2w− lim 9w+ lim 19w→5 w→5 w→5=14 32limw− 9limw+ 19w→5 w→51=4 32⎛lim w⎞9⎛lim w⎞⎜ ⎟ − ⎜ ⎟ + 19⎝w→5 ⎠ ⎝w→5⎠=14 32(5) − 9(5) + 19=1 1=144 1213.xlimxx→222− 4+ 4x→22( x − )2( x + )lim 4x→2=lim 471, 94,51,3824−4= = 04+418.19.2x + 7x+ 10 ( x+ 2)( x+5)lim= limx→2 x+ 2 x→2x+2= lim( x + 5) = 7x→22x + x− 2 ( x+ 2)( x−1)lim = limx→1 2 x→1( x+ 1)( x−1)x − 1x + 2 1+2 3= lim = =x→1x + 1 1+1 2x 2 –14x –5120. limx→ –3 x 2 –4x –21 = lim x→ –3= limx→ –3x –17x –7=–3 –17–3 – 7 = 2(x + 3)(x –17)(x + 3)(x –7)u 2 – ux + 2u –2x (u + 2)(u – x)21. limu→–2 u 2 = lim– u –6 u→ –2 (u + 2)(u –3)u – x= limu→ –2 u –3 = x + 25x 2 + ux – x – u22. limx→1 x 2 + 2x –3= limx→1(x –1)(x + u)= lim( x –1)(x + 3)x →1x + ux + 3 = 1+ u1+ 3 = u + 142 x 2 –6xπ+4π 223. lim x→π x 2 – π 2= limx→π2(x – 2π)x +π=2(x – π)( x –2π)= lim x→π (x – π)( x +π)2(π – 2π)π+π= –114.15.16.17.xlimx→22x→2− 5x+6= limx − 2 x→2( x )= lim − 3 =− 1limx→−1x2x→−1−2x−3= limx + 1( x )= lim − 3 =− 4limx→−12x + x=2x + 1limx→−1x→−1x→−1( x−3)( x−2)( x − 2)( x− 3)( x+1)( x + 1)2( x + x)2( x + )lim 1x→−1 x→−10= = 02( x−1)( x−2)( x−3) x−3lim= lim( x−1)( x− 2)( x+ 7) x+7−1−3 2= =−− 1+7 3(w + 2)(w 2 – w –6)24. limw→ –2 w 2 + 4w + 4(w + 2) 2 (w –3)25.26.= limw→ –2= –2 – 3 = –5(w + 2) 2 = limw→ –2lim2 2f ( x) + g ( x)x→a= lim2 2f ( x) + lim g ( x)x→a x→a=2 2⎛lim f( x) ⎞ ⎛lim g( x)⎞⎜ ⎟ + ⎜ ⎟⎝x→a ⎠ ⎝x→a⎠2 2= (3) + (–1) = 10(w –3)2 f ( x)– 3g(x)lim[2 f ( x)– 3g(x)]x→alim=x →af ( x)+ g(x)lim[ f ( x)+ g(x)]x→a2 lim f ( x)– 3 lim g(x)x →ax→a2(3) – 3(–1)===lim f ( x)+ lim g(x)3 + (–1)x→ax→a9274 Section 1.3 Instructor’s Resource Manual

27. lim 3 g(x)[f ( x)+ 3] = lim 3 g(x)⋅ lim[ f ( x)+ 3]28.x→ax→ax→a3= 3 lim g(x)⋅ ⎡ lim f ( x)+ lim 3⎤= – 1 ⋅ (3 + 3)x→a⎢⎣ x→ax→a⎥⎦= –64lim [ f( x) – 3] =⎡lim ( f( x) – 3)⎤x→a ⎢ ⎣x→a⎥ ⎦4=⎡lim f( x) – lim 3⎤⎢= (3 – 3) = 0⎣x→a x→a⎥⎦29. lim ⎡ ⎣ f( t) + 3 gt ( ) ⎤ ⎦ = lim f( t) + 3 lim gt ( )t→a t→a t→a= lim f( t) + 3 lim g( t)t→a t→a= 3 + 3 –1 = 630. lim u→a[ f (u) + 3g(u)] 3 =⎡= lim⎣ ⎢u→ af (u) + 3 limu→a43⎛⎞⎜ lim [ f(u) + 3g(u)] ⎟⎝ u →a ⎠⎤g(u)⎦ ⎥3x 2 –12 3( x –2)(x + 2)31. lim = limx→2 x –2 x→2 x –2= 3lim (x + 2) = 3(2 + 2) = 12x→234.32.33.34= [3+ 3( –1)] 3 = 02 2(3x + 2x+ 1) –17 3x + 2 x–16lim= limx→2 x–2 x→2x–2(3x+8)( x– 2)= lim = lim (3x+ 8)x –2x→2 x→2= 3limx+ 8= 3(2) + 8=14x→21122– xx– 2––x2x2xlim = lim = limx→2 x – 2 x→2x – 2 x→2x – 21 –1 –1 1= lim – = = = –x→22x2 lim x 2(2) 4x→233423(4– x )–3( x+2)( x –2)–222x4x4xlim = lim = limx→2 x – 2 x→2x – 2 x→2x – 2–3⎛lim x + 2⎞–3( x + 2)⎜ ⎟x→2–3(2 + 2)= lim =⎝ ⎠=x→242 2 4(2)2x4⎛lim x⎞⎜ ⎟⎝x→2⎠= – 3 435. Suppose lim f (x) = L and lim g(x) = M.x→cx→cf ( x)g(x)– LM ≤ g(x)f ( x)– L + L g(x)– Mas shown in the text. Choose ε 1 = 1. Sincelim ( ) = M , there is some δ 1 > 0 such that ifx→c 0 < x–c < δ1, gx ( )– M< ε1= 1 orM – 1 < g(x) < M + 1M –1 ≤ M + 1 and M + 1 ≤ M + 1 so for0 x– c δ1, g( x) M 1.< < < + Choose ε > 0.Since lim f (x) = L and lim g(x) = M, therex→cx→cexist δ 2 and δ 3 such that 0 < x–c < δ2⇒f( x)–L M and 0.Since lim g(x) = M there is δ 2 > 0 such thatx→c1 20 < x − c < δ2⇒ g( x)− M < M .2Let δ = min{δ 1 , δ 2 }, then1 1 M – gx ( )0 < x– c < δ ⇒ – =g( x) M g( x)M1 2 2 1 2= gx ( ) − M< g( x) − M= ⋅ MεM g( x) 2 2M M 2= εInstructor’s Resource Manual Section 1.3 75

1Thus, limx→c g(x) = 1 M = 1lim g(x) .x→cUsing statement 6 and the above result,f( x) 1lim = lim f( x) ⋅ limx→c g( x) x→c x→cg( x)37. limlim f ( x)1 x→c= lim f( x) ⋅ = .x→clim g( x) lim g( x)x→c⇔ limx→c x→cf (x) = L ⇔ limx→cx→ cf (x) – lim L = 0x →c⇔ lim [ f (x) – L] = 0x→c38. limx→cf (x) = 0 ⇔39.⇔ lim f 2 ( x) = 0x→c2⇔ lim f ( x ) = 0x→c ⇔ lim f ( x) = 0x→c2⇔ limx→cf( x) = 0f( x) = lim Lx→ c2⎡ ⎤lim f (x)⎣ ⎢ x→c ⎦ ⎥ = 02lim x =⎛lim x⎞⎜ ⎟ = lim x = lim xx→c ⎝x→c ⎠ x→c x→c22=⎛lim x ⎞⎜ ⎟ = c =⎝cx→c ⎠40. a. If41.42.limx→climx→c2 2x+1 x–5f( x) = , g( x)= and c = 2, thenx– 2 x–2[ f (x) + g(x)] exists, but neitherf (x) nor lim g(x) exists.x→c2b. If f ( x) = , g( x) = x,and c = 0, thenxlim [ f (x) ⋅ g( x)] exists, but lim f (x) doesx→cx→cnot exist.3+x 3–3lim = = 0x –3x→–3+x→–π+3 33 3π + x π + (– π)lim = = 0x – π43.44.45.x– 3 ( x–3) x –9lim = lim22x→3 + x –9 x→3+ x –92 2( x–3) x –9 x –9= lim= lim+ ( x–3)( x+ 3) + x+3x→3 x→323 –9= = 03+31+ x 1+1 2= =xlim→1– 4 + 4 x 4 + 4(1) 8x→2+x2 2 ( x + 1) (2 + 1) 2 52 ⋅ 2lim= = =2 2 2(3x−1) (3⋅2 −1) 5 546. x x x x47.48.49.lim ( − ) = lim − lim = 3 − 2 = 1x→3 −x→3 −x→3−xlim = –1xx→0– 2 2x x x→3+ lim + 2 = 3 + 2⋅ 3 = 15f( xgx ) ( ) = 1; gx ( ) =1f ( x)1lim gx ( ) = 0 ⇔ lim = 0x→a x→af( x)1⇔ = 0lim f( x)x→aNo value satisfies this equation, so lim f( x)x→ amust not exist.⎛50. R has the vertices ⎜ ± x 2 , ± 1 ⎟⎞⎝ 2⎠2Each side of Q has length x + 1 so the2perimeter of Q is 4 x + 1. R has two sides of2length 1 and two sides of length x so the2perimeter of R is 2+2 x .limperimeter of R 2 x + 2= limQx→0 + perimeter of x→0+42x + 122 0 + 2 2 1= = =24 0 + 1 4 22276 Section 1.3 Instructor’s Resource Manual

51. a.2 2NO = (0 – 0) + (1– 0) = 12 2 2 2OP = ( x –0) + ( y –0) = x + y2= x + x2 2 2 2NP = ( x –0) + ( y –1) = x + y –2y+ 12= x + x− 2 x + 14.5.3x tan x 3 x(sin x/ cos x) 3xlim = lim = limx→0 sin x x→0 sin x x→0cosx0= = 01sin xlim =x→02x12sin x 1lim = ⋅1=x→0x 2122 2MO = (1– 0) + (0 – 0) = 12 2 2 2MP = ( x–1) + ( y– 0) = y + x – 2x+16.θ θ θθ θ θsin 3 3 sin 3 3 sin 3lim = lim ⋅ = limθ→0 2 θ→02 3 2θ→032= x − x + 1perimeter of ΔNOPxlim→0+ perimeter of ΔMOP2 21 + x + x + x + x–2 x + 1= limx→0+2 21 + x + x + x – x+11+1= = 11+17.= 3 2 ⋅1 = 3 2sin 3θsin 3θcosθsin 3θlim = lim = lim→0 θtanθθ →0sin θ →0sinθθcosθ⎡ sin 3θ= lim ⎢cosθ⋅ 3⋅⋅θ →0⎢⎣3θ1sinθθ⎡ sin 3θ1 ⎤= 3lim⎢cosθ⋅ ⋅ = 3⋅1⋅1⋅ 1=3θ →03θsinθ⎥⎢⎣θ ⎥⎦⎤⎥⎥⎦b. Area of ΔNOP = 1 2 (1)( x) = x 21 xArea of Δ MOP = (1)( y ) =2 2xarea of ΔNOP2lim = lim = limx→0 + area of ΔMOPx→0 + xx→0+2= lim x = 0x→0+xx8.sin 5θtan 5θcos 5θsin 5θlim = lim = limθ →0 sin 2θθ →0sin 2θθ →0cos5θsin 2θ⎡ 1 sin5θ1 2θ⎤= lim ⎢ ⋅5⋅ ⋅ ⋅θ →0cos5θ 5θ 2 sin 2θ⎥⎣⎦5 ⎡ 1 sin5θ2θ⎤= lim2 ⎢ ⋅ ⋅θ →0cos5θ 5θ sin2θ⎥⎣⎦= 5 2 ⋅1⋅1⋅1 = 5 21.4 Concepts Review1. 02. 13. the denominator is 0 when t = 0 .4. 1Problem Set 1.49.cos πθcot πθsinθsinθsin πθlim= limθ →0θ θ 022sec →cosθcos πθ sinθ cosθ= limθ →02sinπθ⎡cos π cos sin 1 π= lim ⎢⋅ ⋅ ⋅θ →0⎣ 2 θ π sinπ1 ⎡sinθπ= lim cos πθcosθ⋅ ⋅2π⎢θ →0⎣θ sinπ= 12π ⋅1⋅1⋅1⋅1 = 12πθ θ θ θ ⎤θ ⎥⎦θ ⎤θ ⎥⎦1.cos x 1lim = = 1x→0x + 1 110.2sin 3t 9t sin 3t sin 3tlim = lim ⋅ ⋅ = 0 ⋅1⋅ 1 = 0t→0 2t t→02 3t 3t2.3.πlim θ cosθ= ⋅ 0 = 02θ→π/22 2cos t cos 0 1lim = = = 1t→01+ sint1+ sin0 1+011.2 2tan 3tsin 3tlim = limt t tt→0 0 22 t→(2 )(cos 3 )3(sin 3 t) sin 3t= lim ⋅ = 0⋅ 1 = 0t→022cos 3t3ttan 2t012. lim = = 0t→0 sin 2t−1−1Instructor’s Resource Manual Section 1.4 77

13.sin(3 t) + 4t ⎛sin 3t 4t⎞lim = lim⎜+ ⎟t→0 tsect t→0⎝tsect tsect⎠sin 3t4t= lim + limt→0tsect t→0tsectsin 3t= lim3cost⋅ + lim 4costt→0 3tt→0= 31 ⋅ + 4=719.sin xlim 1 2x+ =x→ 0sin2θ sinθ sinθ14. lim = limθ →0 θ2 θ →0θ θsinθsinθ= lim × lim = 1× 1 = 1θ →0 θ θ →0θ15. x ( x)lim sin 1/ = 0x→020. The result thatlim cost= 1 was established int→0the proof of the theorem. Thenlim cost = lim cos( c+h)t→c h→0= lim (cos ccos h−sin csin h)h→0= lim cos c lim cos h−sin c lim sin hh→0 h→0 h→0c= coslim sin tsin t t→csin c21. lim tan t = lim = = = tan ct→ct→ccos t lim cos t cos ct→clim cos tcos t t→ccos clim cot t = lim = = = cot ct→ct→csin t lim sin t sin ct→c216. x ( x )lim sin 1/ = 0x→0217. ( x)lim 1 cos / 0− x =x→01 122. lim sec t = lim = = sec ct → c t → c cos t cos c1 1lim csc t = lim = = csc ct → c t → c sin t sin c23. BP = sin t, OB = costarea( ΔOBP) ≤ area (sector OAP)≤ area ( Δ OBP) + area( ABPQ)1 1 2 1OB ⋅BP ≤ t(1) ≤ OB ⋅ BP + (1 – OB)BP2 2 21 1 1sin tcost ≤ t ≤ sin tcos t+(1– cos t)sint2 2 218.x→02lim cos x = 1tcost≤ ≤ 2 – costsin t1 sint1 π π≤ ≤ for − < t < .2–cost t cost2 21 sint1lim ≤ lim ≤ limt→02–cost t→0 t t→0costsin t1≤lim ≤ 1t→0tsin tThus, lim = 1.t→0t78 Section 1.4 Instructor’s Resource Manual

24. a. Written response1 1b. D = AB⋅ BP = (1 − cos t ) sin t2 2sin t(1 − cos t)=21 2 1 t sintcostE = t(1) – OB⋅ BP = –2 2 2 2D sin (1– cos )=t tE t – sin tcost6.7.8.x21lim = lim = 1x→∞x2–8x+15 x→∞ 1– 8 + 15xx2limx→∞2x3x3–100x521= limx→∞2 –100xπθπlim = lim =π1–θ→– ∞ 5 –54 θ – 5θ θ → ∞θ=12c.⎛D⎞lim ⎜ ⎟ = 0.75⎝ E ⎠t→0+9.3 2 3– 13 x – x 3lim= limx=x→∞πx3–5x2 x→∞π– 5 πx1.5 Concepts Review1. x increases without bound; f(x) gets close to L asx increases without bound2. f(x) increases without bound as x approaches cfrom the right; f(x) decreases without bound as xapproaches c from the left3. y = 6; horizontal4. x = 6; verticalProblem Set 1.51.2.3.4.5.x 1lim = lim = 1x→∞x –5 x→∞1–5x21xx3 x 5x →∞x3lim = lim = 0x→∞5––12tt→– ∞ 2 – 77 − t t→ ∞t21lim = lim = −1−11lim = lim = 1–5 1–tt→– ∞tt→–∞5t2 2xx= limx x x→∞2x xlimx→∞( – 5)(3 – ) − + 8 −151= lim = –1x →∞− 1+8 − 15xx210.11.12.13.14.15.2sin θ 2lim ; 0 ≤sin θ ≤ 1 for all θ and–5θ 2 →∞θ11θ2θ→∞21–5θ –5 θ→∞θ2lim = lim = 0 so3 3/2x + x x + x= lim3 x 3/2x →∞ 2x3 3 3 3limx→∞23 += limx →∞ 23x3=2θ 2 →∞θ3 3π x +3 3 x π x +lim3 lim3 x=x→∞3 x 3x x →∞ x x=lim3x→∞2 + 7 2 + 7π +2 +3x72x2 π2 23 1 + 8 x31 +limlim8 x=x→∞2 x 2x→∞ x=32+ 4 + 412+ 8x 33x→∞1+4x 2= lim = 8 = 2limx→∞2 2x x x x= limx x+x→∞2x+ + 3 + + 3( –1)( 1) –11x1x 23x21+ += lim = 1 = 1x→∞1–n 1 1lim = lim =n→∞2n+ 1 n→∞12 +2n2sin θlim = 0–5Instructor’s Resource Manual Section 1.5 79

16.2n 1 1lim = lim = = 1n→∞2n + 1 n→∞11+1+02n23.13 9y+9y+ 1y2y→– ∞ 2 – 1–2 2y –2y+2 y→ ∞ +yy2lim = lim = – ∞17.18.2n nlim nn→∞∞lim = lim = = =∞n→∞n + 1 n→∞1 11+ ⎛ ⎞ 1+0lim 1n⎜ + ⎟n→∞⎝n ⎠1n0lim = limn= = 0n→∞2n + 1 n→∞11+1+02n19. For x > 0,limx→∞2xx2= = 21x = x 2 .1+ 1 2 + 2x= lim = lim2 x→∞x+ 31x+1x2+ 3 x→∞3+22x24.25.26.n n–1ax 0 + ax 1 +…+ an–1x+anlimx→∞n n–1bx 0 + bx 1 +…+ bn–1x+bna1a –1 aa0 + +…+ n + nxxn–1xn a0= lim=x →∞ b1 bn–1bbn b0 + +…+ + 0xxn–1xnn1 1lim = lim = = 1n→∞2 n→∞n + 11 1+01+2nlimn→∞n2n3/2limn ∞= = =∞3n→∞n + 2n+ 1 2 1 11+ +2 3n n220.21.22.2x+1 2+12x1x2xx2x→∞ x + 4 x→∞ 1+ 4 x→∞1+4xx+lim = lim = lim = 02 2lim⎛2x3 – 2 x – 5⎞⎜ +⎟x→∞⎝⎠⎛ 2 2 2 22x 3– 2 x –5⎞⎛2x 3 2 x –5⎞⎜ + ⎟⎜ + + ⎟= lim⎝ ⎠⎝ ⎠x→∞2 22x+ 3+2 x –52 22x+ 3–(2 x –5)= limx→∞2 22x+ 3+2 x –588x= lim= limx →∞ 2 2 223 222x+ 3+ 2x−5x →∞ x + + x –5x2= limx→∞8x2+ + 2–3 52x2x2lim⎛x 2x x⎞⎜ + − ⎟x→∞⎝⎠= 0⎛ 2 ⎞⎛ 2 ⎞⎜ x + 2 x – x⎟⎜ x + 2x + x⎟= lim⎝ ⎠⎝ ⎠x→∞2x + 2x + x2 2x + 2 x– x 2x= lim= limx→∞2 2x + 2x + xx→∞x + 2x + x2 2= lim = = 1x→∞1+ 2+ 1 2x+27. As x→4 , x→ 4 while x – 4 → 0 + .28.xxxlim→4+ –4lim2= ∞t – 9 ( t+3)( t –3)= limt+ 3 t+3t→–3 +t→–3+= lim ( t – 3) = –6t→–3+– 229. As t →3 , t → 9 while 9– t → 0 .t →3–30. As2tlim9– t2= ∞+3 5 , 2 5 2/33 –x→ x → while 5– x → 0 .3+x→52lim x= – ∞35– x– 2 –31. As x→5 , x →25, x–5→ 0 , and3 – x → –2.x2xlim→5– ( x –5)(3– x )+2 2= ∞32. As θ →π , θ →π while sinθ → 0 − .2θ= −∞θlim→π+ sin θ2+80 Section 1.5 Instructor’s Resource Manual

33. Aslimx→3−−3x→3 , x → 27, while x − 3 → 0 − .3x=−∞x − 3+π π34. As θ → , πθ→ while cos θ → 0 – .2 2πθlim = – ∞+ cosθθ→ π2243.3 3lim = 0, lim = 0;x→∞x+ 1 x→–∞ x+1Horizontal asymptote y = 0.3 3lim , lim – ;x→–1 + x+ 1 = ∞ x→–1– x+1= ∞Vertical asymptote x = –135.lim2x – x–6 ( x+2)( x–3)= limx–3 x–3x→3 –x→3–= lim ( x + 2) = 5x→3–36.2x + 2 x–8 ( x+4)( x–2)lim = lim ( x+2)( x –2)2x→2 + x –4 x→2+x + 4 6 3= lim = =x + 2 4 2x→2+37. For 0≤ x < 1, x = 0 , so for 0 < x < 1, 0x =xthuslim = 0xx→0+38. For −1≤ x < 0, x =− 1, so for –1 < x < 0,x 1 x=− thus lim =∞ .x x x→0− x(Since x < 0, – 1 x > 0. )x44.3 3lim = 0, lim = 0;x→∞2 – 2( x+ 1) x→ ∞ ( x+1)Horizontal asymptote y = 0.3 3lim =∞ , lim =∞;2 – 2x→–1 + ( x+ 1) x→–1( x+1)Vertical asymptote x = –139. For x < 0, x = – x,thusx – xlim lim –1x= x=x→0 –x→0–40. For x > 0, x = x,thusx xlim = lim = 1x xx→0 +x→0+––41. As x→ 0 , 1+ cosx→ 2 while sin x → 0 .1+cosxlim = – ∞– sin xx→042. –1≤sinx≤ 1 for all x, and1 sinxlim = 0, so lim = 0.x→∞x x→∞x45.2x2lim = lim = 2,x→∞x –3 x→∞1–x3x2x2lim = lim = 2,→−∞ x –3 x→−∞1–3xHorizontal asymptote y = 22x2xlim , lim – ;x 3 x–3 = ∞ →+x→3– x–3= ∞Vertical asymptote x = 3Instructor’s Resource Manual Section 1.5 81

46.47.48.3 3lim = 0, lim = 0;x→∞2 – 29– x x→ ∞ 9– xHorizontal asymptote y = 03 3lim = – ∞ , lim =∞ ,2 – 2x→3 + 9– x x→39– x3 3lim =∞ , lim = – ∞ ;2 – 2x→–3 + 9– x x→–39– xVertical asymptotes x = –3, x = 314 14lim = 0, lim = 0;x→∞2 – 22x+ 7 x→ ∞ 2x+ 7Horizontal asymptote y = 0Since 2x 2 + 7 > 0 for all x, g(x) has no verticalasymptotes.2x2 2lim = lim = = 2,x→∞x12 5+ 5x→∞1+x22x2 2lim = lim = = –2– 1x→– ∞ 2 – – 15x + 5x→ ∞ +x22Since x + 5 > 0 for all x, g(x) has no verticalasymptotes.149. f (x) = 2x + 3–x 3 – 1 , thus ⎡ 1 ⎤lim [ f( x) – (2x+ 3)] = lim ⎢– 0x→∞x→∞3 ⎥ =⎣ x –1⎦The oblique asymptote is y = 2x + 3.50.4x+ 3f( x) = 3x+4– , thus2x + 1⎡ 4x+ 3⎤lim [ f( x) – (3x+ 4)] = lim ⎢–x→∞x→∞2 ⎥⎣ x + 1⎦⎡ 4+3 ⎤xx2= lim⎢–⎥= 0 .x→∞⎢ 1+1 ⎥⎢⎣x 2 ⎥⎦The oblique asymptote is y = 3x + 4.51. a. We say that lim f( x) = – ∞ if to eachx→c+negative number M there corresponds a δ > 0such that 0 < x – c < δ ⇒ f(x) < M.b. We say thatlim f( x)=∞ if to eachx→c–positive number M there corresponds a δ > 0such that 0 < c – x < δ ⇒ f(x) > M.52. a. We say that lim f( x)=∞ if to eachx→∞positive number M there corresponds anN > 0 such that N < x ⇒ f(x) > M.b. We say that lim f (x ) =∞ if to eachx→ –∞positive number M there corresponds anN < 0 such that x < N ⇒ f(x) > M.53. Let ε > 0 be given. Since lim f (x ) = A, there isx→∞a corresponding number M 1 such thatx > M1 ⇒ f( x)– A < ε . Similarly, there is a2εnumber M 2 such that x > M2 ⇒ g( x)– B < .2Let M = max{ M1, M2}, thenx > M ⇒ f( x) + g( x)–( A+B)= f ( x)– A+ g( x)– B ≤ f( x)– A + g( x)–Bε ε< + = ε2 2Thus, lim[ f ( x) + g( x)]= A+Bx→∞54. Written response82 Section 1.5 Instructor’s Resource Manual

55. a. lim sin x does not exist as sin x oscillatesx→∞between –1 and 1 as x increases.b. Let u = 1 , then as , 0 .x x →∞ u →+1lim sin = lim sin u = 0x→∞x u→0+1c. Let u = , then as x→∞, u → 0+.x1 1 sinulim xsin = lim sin u = lim = 1x + u + ux→∞ u→0 u→0m056. lim m(v)= lim= ∞−−v→cv→c2 21−v / c3x 2 + x +157. limx→∞ 2x 2 –1 = 3 258.59.limx→–∞22 x –3x2=25x+ 1 5⎛x x xx2 2lim ⎜ 2 + 3 – 2 – 5 ⎟=–→–∞⎝⎠ 2 2⎞3d. Let u = 1 x , then 3 / 23 / 2 1 ⎛ 1 ⎞lim x sin = lim ⎜ ⎟ sin ux→∞+x u→0⎝ u ⎠⎡⎛1 ⎞⎛sin u ⎞⎤= lim ⎢⎜⎟⎜⎟⎥= ∞+u→0⎣⎝u ⎠⎝u ⎠⎦e. As x →∞, sin x oscillates between –1 and 1,whilex–1/ 2 1–1/ 2= → 0.xlim x sin x = 0x→∞⎛ π 1 ⎞ ⎛ πlim sin⎜+ ⎟ = lim sin⎜+ ux→∞+⎝ 6 x ⎠ u→0⎝ 6⎞⎟⎠f. Let u = 1 x , thenπ 1= sin = 6 21⎛ 1 ⎞g. As x→∞ , x+ →∞ , so lim sin ⎜x+ ⎟xx→∞⎝ x ⎠does not exist. (See part a.)⎛ 1⎞1 1h. sin ⎜x+ ⎟= sin xcos + cos xsin⎝ x ⎠ x x⎡ ⎛ 1 ⎞ ⎤lim ⎢sin ⎜x+ ⎟– sin xx→∞x⎥⎣ ⎝ ⎠ ⎦⎡ ⎛ 1 ⎞ 1⎤= lim ⎢sin x⎜cos –1⎟+cos xsinx→∞xx⎥⎣ ⎝ ⎠⎦11As x →∞, cos → 1 so cos –1 → 0.xxFrom part b., lim sin 1x→∞ x = 0.As x →∞ both sin x and cos x oscillatebetween –1 and 1.⎡ ⎛ 1 ⎞ ⎤lim ⎢sin ⎜x+ ⎟– sin x = 0.x→∞x⎥⎣ ⎝ ⎠ ⎦60.61.62.63.64.65.66.67.68.69.70.71.limx→∞2x+ 1 2=23x+ 1 310⎛ 1 ⎞lim ⎜1+ ⎟ = 1x→∞⎝x ⎠x⎛ 1 ⎞lim ⎜1+ ⎟ = e ≈ 2.718x→∞⎝x ⎠x2⎛ 1 ⎞lim ⎜1+ ⎟ =∞x→∞⎝x ⎠sin x⎛ 1 ⎞lim ⎜1+ ⎟ = 1x→∞⎝x ⎠sin x – 3lim = –1x –3x→3–sin x – 3lim = –1tan( x – 3)x→3–cos( x – 3)lim = – ∞x –3x→3–cos xlim = –1x –+x→ π2x→0+π21xlim (1 + x) = e≈2.718x→0+1/ xlim (1 + x)=∞lim (1 + x) = 1x→0+xInstructor’s Resource Manual Section 1.5 83

1.6 Concepts Review1. lim f ( x )x→c 2. every integer3.lim f ( x) = f( a); lim f( x) = f( b)x→a +x→b–4. a; b; f(c) = W13.14.lim f( t) = lim (3 – t) = 0t→3 +t→3+lim f( t) = lim ( t – 3) = 0t→3 –t→3–lim f( t) = f(3);continuoust→3lim f( t) = lim (3 – t) = 0t→3 +t→3+lim f( t) = lim ( t – 9) = 0t→3 –t→3–lim f( t) = f(3);continuoust→322Problem Set 1.61. lim[( x–3)( x–4)] = 0 = f(3);continuousx→32.2lim ( x – 9) = 0 = g(3);continuousx→33. lim 3and h(3) do not exist, so h(x) is notx→3 x – 3continuous at 3.4. lim t – 4 and g(3) do not exist, so g(t) is nott→3continuous at 3.5.t – 3limt →3t – 3and h(3) do not exist, so h(t) is notcontinuous at 3.6. h(3) does not exist, so h(t) is not continuous at 3.7.8.lim t = 3 = f(3);continuoust→3lim t – 2 = 1 = g(3);continuoust→39. h(3) does not exist, so h(t) is not continuous at 3.10. f(3) does not exist, so f(x) is not continuous at 3.11.3 2t – 27 ( t – 3)( t + 3t+9)lim = limt –3 t –3t→3 t→32 2= lim( t + 3t+ 9) = (3) + 3(3) + 9 = 27 = r(3)t→3continuous12. From Problem 11, lim rt ( ) = 27, so r(t) is nott→3continuous at 3 becauselim rt ( ) ≠ r(3).t→315.lim f( x) = − 2 = f(3);continuoust→316. g is discontinuous at x = –3, 4, 6, 8; g is leftcontinuous at x = 4, 8; g is right continuous atx = –3, 617. h is continuous on the intervals( −∞, −5), −5, 4 , (4,6), 6,8 , (8, ∞ )18.19.20.21.22.2[ ] [ ]x –49 ( x–7)( x+7)lim = lim = lim ( x + 7)x→7 x–7 x→7 x–7x→7= 7 + 7 = 14Define f(7) = 14.22 x –18 2( x+3)( x– 3)lim = limx→3 3– x x→33– x= lim[–2( x + 3)] = –2(3 + 3) = –12x→3Define f(3) = –12.sin( θ )lim = 1θ →0θDefine g(0) = 1t –1 ( t –1)( t + 1)lim = limt –1 ( t –1)( t + 1)t→1 t→1t –1 1 1= lim= lim =t→1( t –1)( t + 1) t→1t + 1 2Define H(1) = 1 2 .4 2 2 2x + 2 x –3 ( x –1)( x + 3)lim= limx+ 1 x+1x→–1 x→–12( x+ 1)( x–1)( x + 3)= limx→–1x + 12= lim [( x–1)( x + 3)]x→–12= (–1–1)[(–1) + 3] = –8Define φ(–1) = –8.84 Section 1.6 Instructor’s Resource Manual

23.⎛ 2x –1 ⎞ ⎛( x–1)( x+1) ⎞limsin= limsinx→–1 ⎜ x 1 ⎟ ⎜ ⎟x→–1⎝+⎠ ⎝ x+1 ⎠= lim sin( x –1) = sin(–1–1) = sin( − 2) = – sin 2x→–1Define F(–1) = –sin 2.37.24. Discontinuous at x = π ,3025.233 – xf( x)=( π – x)( x–3)Discontinuous at x = 3, π38.26. Continuous at all points27. Discontinuous at all θ = nπ + π 2integer.where n is any28. Discontinuous at all u ≤− 529. Discontinuous at u = –139.30. Continuous at all points31.Gx ( ) =1(2 – x)(2 + x)Discontinuous on ( −∞, −2] ∪[2, ∞ )32. Continuous at all points sincelim f ( x) = 0 = f(0)and lim f ( x) = 1 = f(1).x→0x→140.33.lim gx ( ) = 0 = g(0)x→0lim gx ( ) = 1, lim gx ( ) = –1x→1 +x→1–limx→1at x = 1.g(x ) does not exist, so g(x) is discontinuous34. Discontinuous at every integer35. Discontinuous at t =36.1n + where n is any integer2Discontinuous at all points except x = 0, becauselim f ( x ) ≠ f ( c ) for c ≠ 0 . lim f ( x ) exists onlyx→c x→c at c = 0 and lim f ( x) = 0 = f(0).41. Continuous.x→042. Discontinuous: removable, define f (10) = 2043. Discontinuous: removable, define f (0) = 144. Discontinuous: nonremovable.45. Discontinuous, removable, redefine g (0) = 146. Discontinuous: removable, define F (0) = 0Instructor’s Resource Manual Section 1.6 85

47. Discontinuous: nonremovable.48. Discontinuous: removable, define f (4) = 449. The function is continuous on the intervals( 0,1 ],(1, 2], (2,3], …0.720.600.480.360.240.12Cost $1 2 3 4 5 6Length of call in minutes50. The function is continuous on the intervals[0,200], (200,300], (300,400], …352. Let f( x) = x + 3x− 2. f is continuous on [0, 1].f(0) = –2 < 0 and f(1) = 2 > 0. Thus, there is atleast one number c between 0 and 1 such thatx 3 + 3x − 2 = 0.53. Because the function is continuous on[ 0,2π ] and3 5(cos 0)0 + 6sin 0 – 3 = –3 < 0,3 5 3(cos2 π)(2 π ) + 6sin (2 π )–3= 8 π –3> 0, thereis at least one number c between 0 and 2π such3 5that (cos tt ) + 6sin t– 3 = 0.54. Let f ( x) = x3 − 7x2 + 14x− 8. f(x) iscontinuous at all values of x.f(0) = –8, f(5) = 12Because 0 is between –8 and 12, there is at leastone number c between 0 and 5 such thatf ( x) = x3 − 7x2 + 14x− 8 = 0 .This equation has three solutions (x = 1,2,4)Cost $80604020100 200 300 400 500Miles Driven51. The function is continuous on the intervals(0,0.25], (0.25,0.375], (0.375,0.5], …55. Let f ( x) = x − cos x.. f(x) is continuous at allvalues of x ≥ 0. f(0) = –1, f(π/2) = π /2Because 0 is between –1 and π /2, there is atleast one number c between 0 and π/2 such thatf ( x) = x − cos x = 0.The interval [0.6,0.7] contains the solution.Cost $43210.25 0.5 0.75 1Miles Driven5 356. Let f( x) = x + 4 x –7x+14f(x) is continuous at all values of x.f(–2) = –36, f(0) = 14Because 0 is between –36 and 14, there is at leastone number c between –2 and 0 such that5 3f( x) = x + 4 x –7x+ 14=0.86 Section 1.6 Instructor’s Resource Manual

57. Suppose that f is continuous at c, solim f ( x ) = f ( c ). Let x = c + t, so t = x – c, thenx→c as x → c , t → 0 and the statementlim f ( x ) = f ( c ) becomes lim f ( t+ c) = f( c).x→c t→0Suppose that lim f (t + c) = f (c) and let x = t +t→ 0c, so t = x – c. Since c is fixed, t → 0 means thatx → c and the statement lim f ( t+ c) = f( c)t→0becomes lim f ( x ) f ( c )x→c c.= , so f is continuous at58. Since f(x) is continuous at c,lim f ( x ) = f ( c ) > 0. Choose ε = f ( c), thenx→c there exists a δ > 0 such that0 < x− c < δ ⇒ f ( x) − f ( c)< ε .f x − f c > − ε = − f c , or f ( x ) > 0 .Thus, ( ) ( ) ( )Since also f ( c ) > 0 , f ( x ) > 0 for all x in( c− δ , c+δ ).59. Let g(x) = x – f(x). Then,g(0) = 0 – f(0) = –f(0) ≤ 0 and g(1) = 1 – f(1) ≥ 0since 0 ≤ f(x) ≤ 1 on [0, 1]. If g(0) = 0, thenf(0) = 0 and c = 0 is a fixed point of f. If g(1) = 0,then f(1) = 1 and c = 1 is a fixed point of f. Ifneither g(0) = 0 nor g(1) = 0, then g(0) < 0 andg(1) > 0 so there is some c in [0, 1] such thatg(c) = 0. If g(c) = 0 then c – f(c) = 0 orf(c) = c and c is a fixed point of f.60. For f(x) to be continuous everywhere,f(1) = a(1) + b = 2 and f(2) = 6 = a(2) + ba + b = 22a + b = 6– a = –4a = 4, b = –261. For x in [0, 1], let f(x) indicate where the stringoriginally at x ends up. Thus f(0) = a, f(1) = b.f(x) is continuous since the string is unbroken.Since 0 ≤ a, b≤ 1, f(x) satisfies the conditions ofProblem 59, so there is some c in [0, 1] withf(c) = c, i.e., the point of string originally at cends up at c.62. The Intermediate Value Theorem does not implythe existence of a number c between –2 and 2such that f() c = 0. The reason is that thefunction ( )− 2, 2 .f x is not continuous on [ ]63. Let f(x) be the difference in times on the hiker’swatch where x is a point on the path, and supposex = 0 at the bottom and x = 1 at the top of themountain.So f(x) = (time on watch on the way up) – (timeon watch on the way down).f(0) = 4 – 11 = –7, f(1) = 12 – 5 = 7. Since time iscontinuous, f(x) is continuous, hence there issome c between 0 and 1 where f(c) = 0. This c isthe point where the hiker’s watch showed thesame time on both days.⎡ ⎤64. Let f be the function on ⎢0,π ⎥⎣ 2 such that f(θ) is⎦the length of the side of the rectangle whichmakes angle θ with the x-axis minus the length ofthe sides perpendicular to it. f is continuous on⎡ ⎤⎢0,π ⎥⎣ 2 . If f(0) = 0 then the region is⎦circumscribed by a square. If f(0) ≠ 0, then⎛π⎞observe that f(0) =−f⎜ ⎟ . Thus, by the⎝ 2 ⎠Intermediate Value Theorem, there is an angleπθ 0 between 0 and such that f ( θ 0) = 0.2Hence, D can be circumscribed by a square.65. Yes, g is continuous at R .GMmlim g r = = lim g r2r→R − R r→R+( ) ( )66. No. By the Intermediate Value Theorem, if fwere to change signs on [a,b], then f must be0 at some c in [a,b]. Therefore, f cannotchange sign.67. a. f(x) = f(x + 0) = f(x) + f(0), so f(0) = 0. Wewant to prove that lim f (x) = f (c), or,x→cequivalently, lim [ f (x) – f (c)]= 0. Butx→cf(x) – f(c) = f(x – c), solim[ f ( x) – f( c)] = lim f( x– c).Letx→c x→ch = x – c then as x → c, h → 0 andlim f( x– c) lim f( h) f(0) 0.x→c h→0limx→c= = = Hencef (x) = f (c) and f is continuous at c.Thus, f is continuous everywhere, since cwas arbitrary.b. By Problem 43 of Section 0.5, f(t) = mt forall t in Q. Since g(t) = mt is a polynomialfunction, it is continuous for all realnumbers. f(t) = g(t) for all t in Q, thusf(t) = g(t) for all t in R, i.e. f () t = mt.Instructor’s Resource Manual Section 1.6 87

68. If f(x) is continuous on an interval thenlim f ( x ) = f ( c ) for all points in the interval:x→c lim f ( x ) = f ( c ) ⇒ lim f ( x)x→c x→c22= lim f ( x) =⎛lim f( x)⎞⎜ ⎟x→c ⎝x→c⎠2= ( f ( c)) = f( c)⎧ 1 if x ≥ 069. Suppose f( x) = ⎨ . f(x) is⎩ − 1 if x < 0discontinuous at x = 0, but g(x) = f ( x ) = 1 iscontinuous everywhere.70. a.⎡ 3 3⎤Domain: ⎢– , ;4 4⎥⎣ ⎦⎧ 3 3⎫Range: ⎨– , 0, ⎬⎩ 4 4⎭b. At x = 0c. If x = 0, f( x) = 0 , if3 3if x = , f( x) = , so4 4fixed points of f.1.7 Chapter Review3 3x = – , f( x) = – and4 43 3x =− ,0, are4 4Concepts Test1. False. Consider f ( x) = xat x = 2.2. False: c may not be in the domain of f(x), orit may be defined separately.3. False: c may not be in the domain of f(x), orit may be defined separately.b. If r is any rational number, then any deletedinterval about r contains an irrationalnumber. Thus, if f (r) = 1 , any deletedqinterval about r contains at least one point c1 1such that f()– r f() c = –0 .q= qHence,lim f (x) does not exist.x→rIf c is any irrational number in (0, 1), then aspx = → c (where p is the reduced formqqof the rational number) q →∞ , sof( x) → 0 as x → c.Thus,lim f ( x ) = 0 = f ( c ) for any irrationalx→c number c.71. a. Suppose the block rotates to the left. Using3geometry, f( x ) = – . Suppose the block4rotates to the right. Using geometry,3f( x ) = . If x = 0, the block does not rotate,4so f(x) = 0.4. True. By definition, where c = 0, L = 0.5. False: If f(c) is not defined, limexist; e.g.,x→c2x – 4f( x) = .x + 22x –4lim =−4.x→−2x + 2f(–2) does not exist, but6. True:f (x) might2x − 25 ( x− 5)( x+5)lim = limx→5 x−5 x→5x−5= lim ( x + 5) = 5 + 5 = 10x→57. True: Substitution Theoremsin x8. False: lim = 1x→0 x9. False: The tangent function is not defined forall values of c.sin x10. True: If x is in the domain of tan x = ,cos xthen cos x ≠ 0 , and Theorem A.7applies..88 Section 1.7 Instructor’s Resource Manual

11. True: Since both sin x and cos x arecontinuous for all real numbers, byTheorem C we can conclude that2f ( x)= 2 sin x − cos x is alsocontinuous for all real numbers.12. True. By definition, lim f ( x) f ( c)13. True. 2∈[ 1,3]14. False:lim−x→0x→cmay not exist15. False: Consider f ( x)= sin x.= .16. True. By the definition of continuity on aninterval.17. False: Since −1 ≤ sin x ≤ 1 for all x and1sin xlim = 0 , we get lim = 0 .x→∞xx→∞x18. False. It could be the case wherelim f x = 2x→−∞( )19. False: The graph has many verticalasymptotes; e.g., x = ± π/2, ± 3π/2,± 5π/2, …20. True: x = 2 ; x = –2+21. True: As x → 1 both the numerator anddenominator are positive. Since thenumerator approaches a constant andthe denominator approaches zero, thelimit goes to + ∞ .22. False: lim f ( x ) must equal f(c) for f to bex→c continuous at x = c.23. True: lim f ( x) = f⎛lim x⎞⎜ ⎟=f( c),so f isx→c ⎝x→c⎠continuous at x = c.xx→2.32 24. True: lim = 1 = f ( 2.3)25. True: Choose ε = 0. 001f (2) then sincef( x) = f (2), there is some δlimx→2such that 0 < x − 2 < δ ⇒f (x ) − f (2) < 0. 001f (2), or−0. 001f (2) < f (x) − f(2)< 0.001f(2)Thus, 0.999f(2) < f(x) < 1.001f(2) andf(x) < 1.001f(2) for 0 < x − 2 < δ.Since f(2) < 1.001f(2), as f(2) > 0,f(x) < 1.001f(2) on (2 − δ , 2+ δ ).26. False: That lim [ f (x) + g(x)] exists doesx→cnot imply that lim f (x) andx→clim g(x) exist; e.g., f( x)x→cx + 7gx ( ) = for c = − 2 .x + 227. True: Squeeze Theoremx – 3= andx + 228. True: A function has only one limit at apoint, so if lim f( x) = L andx→ alim f( x) = M, L = Mx→ a29. False: That f(x) ≠ g(x) for all x does notimply that lim f ( x) ≠ lim g( x).Forx→c x→c2x + x– 6example, if f( x)= andx – 25g( x) = x,then f(x) ≠ g(x) for all x,2but lim f( x) = lim g(x) = 5.x→2 x→ 230. False: If f(x) < 10, lim f( x) could equal 10x→2if there is a discontinuity point (2, 10).For example,3 2– x + 6x −2x−12f( x) = < 10 forx –2all x, but lim f( x) = 10.31. True:x→2lim f( x) = lim2f ( x)x→a x→a22=⎡lim f ( x) ⎤⎢= ( b)= b⎣x→a⎥⎦32. True: If f is continuous and positive on[a, b], the reciprocal is alsocontinuous, so it will assume all1values betweenf (a) and 1f (b) .Instructor’s Resource Manual Section 1.7 89

Sample Test Problems1.x −2 2−2 0lim = = = 0x + 2 2+2 4x→214.x −1 1−xlim = lim =−1x→1 − x−1 x→1− x−1−x → 1since x − 1< 0 as2.3.4.5.6.7.8.2 2u –1 1 −1lim = = 0u→1u + 1 1+12u –1 ( u–1)( u+1)lim = lim = lim( u + 1)u→1 u–1 u→1 u–1u→1= 1 + 1 = 2u+ 1 u+1 1lim = lim = lim ;u→1 2u –1 u→1( u+1)( u–1) u→1u–1does not exist1– 2x–2xx1lim = lim = limx→2 2x –4 x→2( x– 2)( x+ 2) x→2x( x+2)1=2(2+ 2) = 1 82z – 4 ( z+2)( z– 2)lim = limz→2 2z + z–6z→2( z+3)( z –2)= limz→ 2z + 2z + 3 = 2 + 22 + 3 = 4 5sin xtan xcos x1lim = lim= limx→0 sin 2xx→02sin x cos x x→02cos1=2 cos 2 0 = 1 23 2y –1 ( y–1)( y + y+1)lim = limy→1 2y –1 y→1( y –1)( y+1)2 2y + y+ 1 1 + 1+1 3= lim = =y→1y + 1 1+1 22x15.16.17.sin 5x5 sin 5xlim = limx→0 3xx→03 5x5 sin5x5 5= lim = × 1 =x3 x→05 3 31−cos 2x2 1−cos 2xlim = limx→0 3xx→03 2x2 1−cos2x2= lim = × 0 = 03 x→02x311−x − 1 1 0lim limx += = = 1x→∞x + 2 x→∞21+1+0x18. Since −1≤sint≤ 1 for all t and19.20.21.sin tget lim = 0 .t→∞t1lim 0t→∞ t= , wet + 2limt→2( t 2) 2 = ∞ because as t → 0, t + 2→4−while the denominator goes to 0 from the right.cos x+lim = ∞ , because as x → 0 , cos x → 1x→0+ xwhile the denominator goes to 0 from the right.lim tan 2xx→π/4−( π )= ∞ because as x → ( π )−2 x → /2 , so tan 2 x →∞ .−/4 ,9.x– 4 ( x –2)( x + 2)lim = limx→4 x –2 x→4x –2= lim ( x + 2) = 4 + 2 = 4x→422.1+sinx+lim = ∞ , because as x → 0 ,xx→0+1+ sinx→ 1 while the denominator goes to0 from the right.10.11.cos xlim does not exist.x→0xx – xlim lim lim (–1) –1x 0– x= x 0– x= =→ → x→0–12. xlim 4 = 2x→(1/ 2)+13. ( ) lim t − t = lim t − lim t = 1− 2 = − 1t→2 –t→2 –t→2–23. Preliminary analysis: Let ε > 0. We need to finda δ > 0 such that0 < | x− 3| < δ ⇒ | 2x+ 1 − 7| < ε.( )|2x− 6| < ε ⇔ 2| x− 3| < εεε⇔| x − 3 | < . Choose δ = .2 2Let ε > 0. Choose δ = ε /2. Thus,( ) ( )2x+ 1 − 7 = 2x− 6 = 2 x− 3 < 2 ε /2 = ε.90 Section 1.7 Instructor’s Resource Manual

24. a. f(1) = 028.b.c.lim f( x) = lim (1 – x) = 0x→1 +x→1+lim f( x) = lim x = 1x→1 –x→1–d.lim f( x) –1x→–13lim f( x) = lim x = –1 andx→–1 –x→–1–lim f( x) = lim x = –1x→–1 +x→–1+= because25. a. f is discontinuous at x = 1 because f(1) = 0,but lim f (x ) does not exist. f isx→1discontinuous at x = –1 because f(–1) doesnot exist.b. Define f(–1) = –126. a. 0 < u– a < δ ⇒ g( u)–M < ε27. a.b. 0 < a– x < δ ⇒ f( x)–L < εb.lim[2 f ( x) – 4 g( x)]x→3= 2lim f ( x)–4lim g( x)x→3 x→3= 2(3) – 4(–2) = 142x –9lim gx ( ) = lim gx ( )( x+3)x→3 x –3 x→3= lim gx ( ) ⋅ lim ( x+ 3) = –2 ⋅ (3 + 3) = –12x→3 x→3c. g(3) = –2d.e.lim g( f( x)) = g⎛lim f( x) ⎞⎜ ⎟= g(3) = –2x→3 ⎝x→3⎠x→32lim f ( x) – 8 g( x)=⎡lim f ( x) ⎤⎢– 8 lim g( x)⎣x→3 ⎥ ⎦ x→32= (3) – 8(–2) = 5229. a(0) + b = –1 and a(1) + b = 1b = –1; a + b = 1a – 1 = 1a = 25 330. Let f ( x) = x –4 x –3x+1f(2) = –5, f(3) = 127Because f(x) is continuous on [2, 3] andf(2) < 0 < f(3), there exists some number cbetween 2 and 3 such that f(c) = 0.31. Vertical: None, denominator is never 0.xxHorizontal: lim = lim = 0 , sox→∞2 2x + 1 x→−∞x + 1y = 0 is a horizontal asymptote.32. Vertical: None, denominator is never 0.2 2xxHorizontal: lim = lim = 1 , sox→∞2 2x + 1 x→−∞x + 1y = 1 is a horizontal asymptote.33. Vertical: x = 1, x =− 1 becauseand2xlim =∞2x→−1− x −12 22xlim = ∞2x→1+ x −1xxHorizontal: lim = lim = 1, sox→∞2 2x −1 x→−∞x −1y = 1 is a horizontal asymptote.f.limx→3= 0gx ( )– g(3) –2– g(3) −2 −( −2)= =f( x) 3 3Instructor’s Resource Manual Section 1.7 91

34. Vertical: x = 2, x = − 2 because3xlim =∞2x→2+ x − 4Horizontal:and3xlim =∞x→∞2x − 43xlim =−∞x→−∞2x − 4asymptotes.3xlim =∞2x→−2− x − 4and, so there are no horizontal35. Vertical: x =± π /4, ± 3 π /4, ± 5 π /4,… becauselim tan 2xx→π/4−multiples of π / 4.=∞ and similarly for other oddHorizontal: None, because lim tan 2xandx→∞lim tan 2xdo not exist.x→−∞36. Vertical: x = 0, becausesin x 1 sin xlim = lim =∞.2x→0 + x x→0+ x xHorizontal: y = 0, becausesin x sin xlim = lim = 0.x→∞2 x 2x →−∞ xReview and Preview Problems21. a. f ( 2) = 2 = 42b. f ( 2.1) = 2.1 = 4.41c. f ( ) f ( )d.f2.1 − 2 = 4.41− 4 = 0.41( ) − f ( )2.1 2 0.41= = 4.12.1−2 0.1f a+ h = a+ h = a + 2ah+he. ( ) ( ) 2 2 2f.g.h.2 2 2( + ) − ( ) = + 2 + −f a h f a a ah h a( ) ( )( + ) −= 2ah+ h2f a+ h − f a 2ah + h= = 2a+ha h a h( + ) − ( )( a+ h)−af a h f alim = lim 2a+ h = 2ah→0 h→02( )2. a. g ( 2)= 1/2b. g ( 2.1)= 1/ 2.1 ≈ 0.476c. g( ) g( )d.2.1 − 2 = 0.476 − 0.5 =− 0.024( ) − g( )g 2.1 2 −0.024= = −0.242.1−2 0.1e. g ( a+ h) = 1/ ( a+h)−hg a+ h − g a = 1/ a+ h − 1/ a=a af. ( ) ( ) ( )g.h.−hg a+ h − g a a( a+ h)−1= =a+ h − a h a a+h( ) ( )( )h→0( + ) − ( ) −1=2( a+ h) −a ag a h g alim3. a. F ( 2)= 2 ≈ 1.414b. F ( 2.1)= 2.1 ≈ 1.449c. F( ) F( )d.( )2.1 − 2 = 1.449 − 1.414 = 0.035( ) − F( )F 2.1 2 0.035= = 0.352.1−2 0.1e. F ( a+ h)= a+hf. F ( a+ h) − F( a)= a+ h − ag.h.( ) ( )( + ) −F a+ h − F a a+ h − a=a h a h( ) ( )( + ) −F a+ h − F a a+ h − alim= lima h a hh→0 h→0= limh→0( a+ h − a)( a+ h + a)h( a+ h + a)a+ h−a= limh→0h a h a( + + )h= limh→0h a h a( + + )1 1 a= lim= =h→0a+ h + a 2 a 2a( + h)92 Review and Preview Instructor’s Resource Manual

4. a. G ( 2) = ( 2) 3+ 1= 8+ 1=9b. G ( 2.1) = ( 2.1) 3+ 1 = 9.261+ 1 = 10.261c. G( ) G( )d.2.1 − 2 = 10.261− 9 = 1.261( ) − G( )G 2.1 2 1.261= = 12.612.1−2 0.1G a+ h = a+ h + 1e. ( ) ( ) 33 2 2 3= a + 3a h+ 3ah + h + 1f. ( ) ( ) ( )g.⎡3⎤ ⎡3⎤G a+ h − G a = a+ h + 1 −⎣a + 1⎣ ⎦ ⎦= + + + + − +3 2 2 3 3( a 3a h 3ah h 1) ( a 1)2 2 3= 3ah+ 3ah + h( ) ( )( + ) −G a+ h − G a 3ah+ 3ah + h=a h a h2 2 32 2= 3a + 3ah+h4 3 32π30 = 2 = cm3 34 62.5π125πV1= π ( 2.5 ) = = cm3 3 6125π3 32π3Δ V = V1− V0= cm − cm6 361 cm3 31.940cm3= π ≈610. V π ( )3 311. a. North plane has traveled 600miles. Eastplane has traveled 400 miles.b.c.2 2d = 600 + 400= 721 miles2 2d = 675 + 500= 840 milesh.( + ) − ( )( a+ h)−aG a h G alim = lim 3a + 3ah+hh→0 h→0= 3a22 2a+ b = a + 3a b+5. a. ( ) 3 3 2b. ( ) 4 4 3a+ b = a + 4a b+c. ( ) 5 5 456. ( )a+ b = a + a b+n n n−1a+ b = a + na b+7. ( )sin x + h = sin xcos h+cos xsinh8. ( )cos x + h = cos xcos h−sin xsinh9. a. The point will be at position ( 10,0 ) in allthree cases ( t = 1, 2,3 ) because it will havemade 4, 8, and 12 revolutions respectively.b. Since the point is rotating at a rate of 4revolutions per second, it will complete 1revolution after 1 second. Therefore, the4point will first return to its starting position1at time t = .4Instructor’s Resource Manual Review and Preview 93

CHAPTER 2 The Derivative2.1 Concepts Review4.1. tangent line2. secant line3.f ( c+ h) − f( c)h4. average velocityProblem Set 2.11. Slope5–3= =2– 324Slope ≈ 1.52.3.6–4Slope = = –24–65.5Slope ≈2Slope ≈− 26.3Slope ≈ – 294 Section 2.1 Instructor’s Resource Manual

7. y = x 2 + 1a., b.d.3[(2.01) −1.0] −7msec=2.01−20.120601=0.01= 12.0601c. m tan = 2d.2(1.01) + 1.0 −2msec=1.01−10.0201=.01= 2.01f (1 + h) – f(1)e. mtan= limh→0h2 2[(1 + h) + 1] – (1 + 1)= limh→0h22+ 2h+ h −2= limh→0hh(2 + h)= limh→0h= lim (2 + h) = 2h→08. y = x 3 –1a., b.c. m tan = 12f (2 + h) – f(2)e. mtan= limh→0h3 3[(2 + h) –1] – (2 −1)= limh→0h2 312h+ 6h + h= limh→0h2h(12 + 6 h+h )= limh→0h= 129. f (x) = x 2 –1f ( c+h)– f( c)mtan= limh→0h2 2[( c+h) –1] – ( c –1)= limh→0h2 2 2c + 2 ch+ h –1– c + 1= limh→0hh(2 c+h)= lim = 2ch→0hAt x = –2, m tan = –4x = –1, m tan = –2x = 1, m tan = 2x = 2, m tan = 410. f (x) = x 3 –3xf ( c+h)– f( c)mtan= limh→0h3 3[( c+ h) – 3( c+h)] – ( c – 3 c)= limh→0h3 2 2 3 3c + 3c h+ 3 ch + h –3 c–3 h– c + 3c= limh→0h2 2h(3c + 3ch+ h −3)2= lim = 3 c – 3h→0hAt x = –2, m tan = 9x = –1, m tan = 0x = 0, m tan = –3x = 1, m tan = 0x = 2, m tan = 9Instructor’s Resource Manual Section 2.1 95

11.13. a.b.2 216(1 ) –16(0 ) = 16 ft2 216(2 ) –16(1 ) = 48 ft144 – 64V = = 80 ft/sec3–2c. ave1f( x)=x + 1f (1 + h) – f(1)mtan= limh→0h1−12+h 2= limh→0h−h2(2 + h)= limh→0h1= lim −h→02(2 + h )1= – 41 1y– = – ( x–1)2 412. f (x) = 1x –1f(0 + h) − f(0)mtan= limh→0h1+ 1h−1= limh→0hhh−1limh→0=h1= limh→0h −1=−1y + 1 = –1(x – 0); y = –x – 1d.e.14. a.b.c.2 216(3.01) −16(3)Vave=3.01−30.9616=0.01= 96.16 ft/s2f ( t) 16 t ; v 32c= =v = 32(3) = 96 ft/s2 2(3 + 1) – (2 + 1)Vave53–2= = m/sec2 2[(2.003) + 1] − (2 + 1)Vave=2.003 − 20.012009=0.003= 4.003 m/sec2 2[(2 + h) + 1] – (2 + 1)Vave=2 + h –224h+h=h= 4 + hd. f (t) = t 2 + 1f (2 + h) – f(2)v = limh→0h2 2[(2 + h) + 1] – (2 + 1)= limh→0h24h+h= limh→0h= lim (4 + h)h→0= 496 Section 2.1 Instructor’s Resource Manual

15. a.b.ααf( + h)– f( )v = limh→0h2( α + h) + 1 – 2α+ 1= limh→0h2α+ 2h+ 1– 2α+ 1= limh→0h( 2α + 2h+ 1– 2α + 1)( 2α + 2h+ 1+ 2α+ 1)= limh→0h( 2 + 2h+ 1+ 2 + 1)=2hαlimh →0h ( 2 + 2 h+ 1 + 2 + 1)αα2 1= = ft/s2α + 1+ 2α + 1 2α+ 11 1=2α+ 1 22α + 1=2α2α + 1= 4; α = 3 2The object reaches a velocity of 1 2 ft/s when t = 3 2 .16. f (t) = –t 2 + 4t2 2[–( c+ h) + 4( c+ h)] – (– c + 4 c)v = limh→0h2 2 2– c –2 ch– h + 4c+ 4 h+c –4c= limh→0hh(–2 c– h+4)= lim = –2c+ 4h→0h–2c + 4 = 0 when c = 2The particle comes to a momentary stop att = 2.17. a.⎡1 2 ⎤ ⎡12 ⎤⎢ (2.01) + 1 – (2) + 1 = 0.020052 ⎥ ⎢2⎥⎣ ⎦ ⎣ ⎦0.02005b. r ave = = 2.005 g/hr2.01– 2g18. a.b.19. a.2 21000(3) –1000(2) = 50002 21000(2.5) –1000(2) 2250= = 45002.5 – 2 0.5c. f (t) = 1000t 2 2 21000(2 + h) −1000(2)r = limh→0h24000 + 4000h+1000 h – 4000= limh→0hh(4000 + 1000 h)= lim = 4000h→0h3 35 –3 98d ave = = = 49 g/cm5–3 2c.1 2f() t = t + 12⎡1 2 1 2(2 + h) + 1 ⎤– ⎡ 2 + 1⎤2 2r = lim⎣ ⎦ ⎣ ⎦h→0h1 22+ 2h+ 2h + 1−2−1= limh→0h( 2 + h)h 1= lim 2= 2h→0hAt t = 2, r = 2b. f (x) = x 3 3 3(3 + h) – 3d = limh→0h2 327 + 27h+ 9 h + h – 27= limh→0h2h(27 + 9 h+h )= lim = 27 g/cmh→0hInstructor’s Resource Manual Section 2.1 97

20.Rc ( + h)– Rc ( )MR = limh→0h2 2[0.4( c+ h) – 0.001( c+h) ] – (0.4 c– 0.001 c )= limh→0h2 2 20.4c+ 0.4 h– 0.001 c – 0.002 ch– 0.001 h – 0.4c+0.001c= limh→0hh(0.4–0.002 c–0.001 h)= lim = 0.4 – 0.002ch→0hWhen n = 10, MR = 0.38; when n = 100, MR = 0.221.22.23. ave2 22(1 + h) – 2(1)a = limh→0h22+ 4h+2 h –2= limh→0hh(4 + 2 h)= lim = 4h→0hpc ( + h)– pc ( )r = limhh→02 3 2 3[120( c+ h) – 2( c+h) ] – (120 c – 2 c )= limh→0h2 2h(240 c– 6c + 120 h– 6 ch– 2 h )= limh→0h2= 240 c– 6cWhen t = 10,t = 20,t = 40,2r = 240(10) – 6(10) = 18002r = 240(20) – 6(20) = 24002r = 240(40) – 6(40) = 0100 – 800 175r = = – ≈ –29.16724 – 0 629,167 gal/hr700 – 400At 8 o’clock, r ≈ ≈−756−1075,000 gal/hr24. a. The elevator reached the seventh floor at timet = 80 . The average velocity isv = ( 84 − 0) / 80 = 1.05 feet per secondavgb. The slope of the line is approximately60 −12= 1.2 . The velocity is55 −15approximately 1.2 feet per second.c. The building averages 84/7=12 feet fromfloor to floor. Since the velocity is zero fortwo intervals between time 0 and time 85, theelevator stopped twice. The heights areapproximately 12 and 60. Thus, the elevatorstopped at floors 1 and 5.25. a. A tangent line at t = 91has slopeapproximately ( 63 − 48) /(91−61) = 0. 5 . Thenormal high temperature increases at the rateof 0.5 degree F per day.b. A tangent line at t = 191has approximateslope ( 90 − 88) / 30 ≈ 0. 067 . The normalhigh temperature increases at the rate of0.067 degree per day.c. There is a time in January, about January 15,when the rate of change is zero. There is alsoa time in July, about July 15, when the rate ofchange is zero.d. The greatest rate of increase occurs aroundday 61, that is, some time in March. Thegreatest rate of decrease occurs between day301 and 331, that is, sometime in November.26. The slope of the tangent line at t = 1930 isapproximately (8 − 6) /(1945 −1930) ≈ 0.13 . Therate of growth in 1930 is approximately 0.13million, or 130,000, persons per year. In 1990,the tangent line has approximate slope(24 −16) /(20000 −1980) ≈ 0.4 . Thus, the rate ofgrowth in 1990 is 0.4 million, or 400,000,persons per year. The approximate percentagegrowth in 1930 is 0.107 / 6 ≈ 0.018 and in 1990 itis approximately 0.4 / 20 ≈ 0. 02 .27. In both (a) and (b), the tangent line is alwayspositive. In (a) the tangent line becomes steeperand steeper as t increases; thus, the velocity isincreasing. In (b) the tangent line becomes flatterand flatter as t increases; thus, the velocity isdecreasing.98 Section 2.1 Instructor’s Resource Manual

28.29.30.31.32.1 3f () t = t + t3f ( c+h)– f( c)current = limh→0h⎡1 3 1 3( c+ h) + ( c+ h) ⎤–c + c3 3= lim⎣⎦h→0h2 2h( c + ch+ 1 h + 1)= lim 3 = c2 + 1h→0hWhen t = 3, the current =102c + 1=20c 2 = 19c = 19 ≈ 4.4A 20-amp fuse will blow at t = 4.4 s.A =π r2 , r = 2t( )A = 4πt 2 2 24 π (3 + h) –4 π(3)rate = lim h →0hh(24π+ 4 πh)= lim = 24π km 2 /dayh→0h4 3 1V = π r , r = t3 41 3V = πt483 31 (3 + h) −3 27π lim48 h→0h 483rate = =9= π inch / sec163 2y = f( x) = x –2x+ 1a. m tan = 7 b. m tan = 0c. m tan = –1 d. m tan = 17. 92y = f( x) = sinxsin 2x2a. m tan = –1.125 b. m tan ≈ –1.0315π2.2 Concepts Review1.f ( c + h )– f ( c ) ( )– ( );f t f ch t – c2. f ′( c)3. continuous; f ( x)4. f '( x); dydxProblem Set 2.21.2.3.4.f ′(1) = lim=f (1 + h) – f(1)h →0h2 2 2(1 + h) – 1 2h+h= lim= limh→0 h h→0h= lim (2 + h) = 2h→0f (2 + h) – f(2)f ′(2) = limh→0h2 2[2(2 + h)] –[2(2)]= limh→0h216h+4h= lim = lim (16 + 4 h) = 16h→0 h h→0f ′(3) = limf (3 + h) – f(3)h →0h2 2[(3 + h) – (3 + h)] – (3 – 3)= limh→0h25h+h= lim = lim (5 + h) = 5h→0 h h→0ff (4 + h) – f(4)′(4) = lim h →0h1 13–(3 + h)3+h 4–1 3(3 + h)h→0 h→0 h→0x– –1= lim = lim = limh h 3(3 + h )33.34.c.m tan = 0d.m tan ≈ 1.1891s = f() t = t+tcost2At t = 3, v ≈ 2.818( t + 1)s = f()t =t + 2At t = 1.6, v ≈ 4.27735.= – 1 9s( x+h)– s( x)s′ ( x) = limh →0h[2( x+ h) + 1] – (2x+1)= limh→0h2h= lim = 2h→0hInstructor’s Resource Manual Section 2.2 99

6.7.8.9.10.11.f ( x+h)– f( x)f′ ( x) = limh →0h[ α( x+ h) + β]–( α x+β )= limh→0hαh= lim = αh→0hrx ( + h)– rx ( )r′ ( x) = limh→0h2 2[3( x+ h) + 4] – (3x+ 4)= limh→0h26xh+ 3h= lim = lim (6x + 3 h) = 6xh→0 h h→0f′ ( x) = limf ( x+h)– f( x)h →0h2 2[( x+ h) + ( x+ h) + 1] – ( x + x+1)= limh→0h22xh + h + h= lim = lim (2x + h+ 1) = 2x+1h→0 h h→0f′ ( x) = limf ( x+h)– f( x)h →0h2 2[ ax ( + h) + bx ( + h) + c]–( ax + bx+c)= limh→0h22axh + ah + bh= lim = lim (2 ax + ah + b)h→0 hh→0= 2ax + bf′ ( x) = limf x h f x +h →0h4 4( )– ( )( x + h) – x= limh→0h3 2 2 3 44hx + 6h x + 4h x + h= limh→0h= lim (4x + 6hx + 4 h x + h ) = 4xh→03 2 2 3 3f ( x+h)– f( x)f′ ( x) = limh→0h3 2 3 2[( x+ h) + 2( x+ h) + 1] – ( x + 2x+ 1)= limh→0h2 2 3 23hx + 3h x + h + 4hx + 2h= limh→0h2 2 2= lim (3x + 3hx + h + 4x + 2 h) = 3x + 4xh→012.13.14.15.16.g′ ( x) = limg( x+h)– g( x)h →0h4 2 4 2[( x + h) + ( x+ h) ] – ( x + x )= limh→0h4hx + 6h x + 4h x + h + 2hx + h= limh→0h3 2 2 3= lim (4x + 6hx + 4h x + h + 2 x + h)h→0= 4x 3 + 2xh′ ( x) = lim3 2 2 3 4 2hx+h hxh →0( )– ( )h⎡⎛2 2⎞1⎤= lim ⎢⎜– ⎟⋅h→0x + h x h⎥⎣⎝⎠ ⎦⎡ –2h1 ⎤ –2= lim ⎢ ⋅ limh 0 x( x h) h⎥ =→ ⎣ + ⎦ h→0x( x+h)Sx ( + h)– Sx ( )S′ ( x) = limh→0h⎡⎛1 1 ⎞ 1⎤= lim ⎢⎜– ⎟⋅h→0x + h+ 1 x+1 h⎥⎣⎝⎠ ⎦⎡ – h 1 ⎤= lim ⎢⋅ ⎥h →0⎣ ( x + 1)( x+ h+1) h⎦–1 1= lim = −h→0( x+ 1)( x+ h+ 1) ( x + 1)F′ ( x) = limF x+h F xh →0( )– ( )h⎡⎛6 6 ⎞ 1⎤= lim ⎢–⋅ ⎥h→02 2⎢⎜( x+ h) + 1 x + 1⎟⎣⎝⎠ h ⎥⎦22= –x2⎡ 2 2 26( x + 1) – 6( x + 2hx+ h + 1) 1 ⎤= lim ⎢⋅ ⎥h→02 2 2⎢⎣( x + 1)( x + 2hx+ h + 1) h ⎥⎦⎡2–12 hx – 6h1 ⎤= lim ⎢⋅ ⎥h→0(2 1)(2 22⎢⎣x + x + hx+ h + 1) h ⎥⎦–12 x – 6h 12x= lim = −h→0(2 1)(2 22 1) (2 1)2x + x + hx+ h + x +F( x+h)– F( x)F′ ( x) = lim h →0h⎡ ⎛ x+ h–1 x–1⎞1⎤= lim ⎢⎜– ⎟⋅h→0x + h+ 1 x+1 h⎥⎣⎝⎠ ⎦⎡ 2 2x + hx+ h–1–( x + hx– h–1) 1⎤= lim ⎢⋅ ⎥h→0⎢⎣( x + h+ 1)( x+1) h⎥⎦⎡ 2h1⎤2= lim ⎢⋅h→0( x h 1)( x 1) h⎥ =2⎣ + + + ⎦ ( x + 1)100 Section 2.2 Instructor’s Resource Manual

17.18.Gx ( + h)– Gx ( )G′ ( x) = lim h →0h⎡ ⎛2( x+ h) –1 2 x–1⎞1 ⎤= lim ⎢⎜– ⎟⋅h→0x + h–4 x–4h⎥⎣⎝⎠ ⎦⎡ 2 22x + 2hx−9x− 8h+ 4 − (2x + 2hx−9x− h+4) 1⎤= lim ⎢⋅ ⎥h→0⎢( x+ h−4)( x−4)h⎣ ⎥⎦ h →0–7 7= lim = –h→0( x+h–4)( x–4) ( x –4)Gx ( + h)– Gx ( )G′ ( x) = lim h →0h2⎡ –7h1⎤= lim ⎢⋅( x + h –4)( x –4) h⎥⎣⎦⎡⎛2( x+h) 2x⎞ 1 ⎤ ⎡2 2 2(2x+ 2 h)( x – x) – 2 x( x + 2 xh+h – x– h) 1⎤= lim ⎢–⋅ ⎥ = limh→0⎜ 2 2⎢ ( x+ h) –( x+h) x – x⎟⎢⋅ ⎥⎣⎝⎠ h2 2 2⎥ h→0⎦ ⎢⎣( x + 2 hx+h – x– h)( x – x)h ⎥⎦⎡2 2–2 h x–2hx1⎤= lim ⎢⋅ ⎥h→0(2 22 – – )(2⎢⎣x + hx+h x h x – x ) h ⎥⎦2–2 hx – 2x= limh→0(2 22 – – )(2x + hx + h x h x – x )2–2x2= = –2 2 2( x – x) ( x–1)19.20.g( x+h)– g( x)g′ ( x) = limh→0h3( x + h) – 3x= limh→0h( 3x + 3 h – 3 x)( 3x+ 3h + 3 x)= limh→0h( 3x+ 3h + 3 x)3h3= lim= limh→0h( 3x+ 3h + 3 x) h→03x+ 3h + 3xg( x+h)– g( x)g′ ( x) = limh→0h⎡⎛1 1 ⎞ 1⎤= lim ⎢–⋅ ⎥h→0⎢⎜ 3( x+h) 3x⎟⎣⎝⎠ h ⎥⎦⎡ 3 x – 3x+3h1⎤= lim ⎢⋅ ⎥h→0⎢⎣9 xx ( + h)h ⎥⎦⎡( 3 x – 3x+ 3 h)( 3x + 3x+3 h) 1⎤= lim ⎢⋅ ⎥h→0⎢⎣9 xx ( + h)( 3x+ 3x+3 h)h ⎥⎦–3 h–3= lim =h →0h 9 x ( x+ h )( 3 x + 3 x+ 3 h ) 3 x⋅2 3 x3=2 3x1= –2x3xInstructor’s Resource Manual Section 2.2 101

21.22.H ( x+h)– H( x)H′ ( x) = lim h →0h⎡⎛3 3 ⎞ 1⎤= lim ⎢⎜– ⎟⋅⎥h→0⎣⎝x+h–2 x–2⎠h ⎦⎡3 x–2 –3 x+h–2 1⎤= lim ⎢⋅ ⎥h→0⎢⎣( x+h–2)( x–2)h ⎥⎦3( x– 2 – x+ h– 2)( x– 2 + x+h– 2)= limh→0h ( x+ h–2)( x–2)( x–2 + x+h–2)−3h= lim h →0h [( x –2) x+ h –2 + ( x+h –2) x –2]–3=h x x + h + x + h xlim→ 0 ( – 2) – 2 ( – 2) – 23 3= – = −2( x– 2) x– 2 2( x − 2)H ( x+h)– H( x)H′ ( x) = limh→0h322 2( x+ h) + 4 – x + 4= limh→0h⎛ 2 2 2 2 2 2x 2hx h 4 – x 4⎞⎛x 2hx h 4 x 4⎞⎜ + + + + ⎟⎜ + + + + + ⎟= lim⎝ ⎠⎝ ⎠h→0 2 2 2h⎛x + 2hx+ h + 4+ x + 4⎞⎜⎟⎝⎠22hx+ h= limh→0 2 2 2h⎛x + 2hx+ h + 4+ x + 4⎞⎜⎟⎝⎠2x+h= limh→02 2 2x + 2hx+ h + 4+ x + 42xx= =2 22 x + 4 x + 4f ()– t f( x)23. f′ ( x) = limt→xt – x2 2( t − 3 t)–( x –3 x)= limt→xt – x2 2t – x –(3 t –3 x)= limt→xt – x( t – x)( t+x)–3( t – x)= limt→xt – x( t – x)( t+x–3)= lim = lim( t+x– 3)t→x t – x t→x= 2 x – 324.f ()– t f( x)f′ ( x) = limt→xt – x3 3( t + 5 t)–( x + 5 x)= limt→xt – x3 3t – x + 5 t –5x= limt→xt – x2 2( t – x)( t + tx+ x ) + 5( t – x)= limt→xt – x2 2( t – x)( t + tx+ x + 5)= limt→xt – x2 2 2= lim( t + tx + x + 5) = 3 x + 5t→x 102 Section 2.2 Instructor’s Resource Manual

25.f ()– t f( x)f′ ( x) = limt→xt – x⎡⎛ t x ⎞⎛ 1 ⎞⎤= lim ⎢⎜ – ⎟⎜ ⎟t→xt –5 x–5 t – x⎥⎣⎝ ⎠⎝ ⎠⎦lim tx –5 t – tx + 5x=t → x ( t – 5)( x – 5)( t – x )–5( t – x) –5= lim= limt→x( t –5)( x–5)( t – x) t→x( t –5)( x–5)=−5( x − 5) 238. The slope of the tangent line is always − 1 .39. The derivative is positive until x = 0 , thenbecomes negative.f ()– t f( x)26. f′ ( x) = limt→xt – x⎡ ⎛t+ 3 x+ 3⎞⎛ 1 ⎞⎤= lim ⎢⎜ – ⎟⎜ ⎟t→xt x t – x⎥⎣⎝ ⎠⎝ ⎠⎦3 x–3 t –3 3= lim = lim = –t→x xt( t – x)t→xxt 2x27. f (x) = 2x 3 at x = 540. The derivative is negative until x = 1 , thenbecomes positive.28. f (x) = x 2 + 2x at x = 329. f (x) = x 2 at x = 230. f (x) = x 3 + x at x = 331. f (x) = x 2 at x32. f (x) = x 3 at x33. f (t) = 2 tat t41. The derivative is − 1 until x = 1 . To the right ofx = 1 , the derivative is 1. The derivative isundefined at x = 1 .34. f(y) = sin y at y35. f(x) = cos x at x36. f(t) = tan t at t37. The slope of the tangent line is always 2.42. The derivative is − 2 to the left of x = − 1; from− 1 to 1, the derivative is 2, etc. The derivative isnot defined at x =− 1, 1, 3 .Instructor’s Resource Manual Section 2.2 103

43. The derivative is 0 on ( −3, − 2), 2 on ( −2, − 1), 0on ( − 1, 0), − 2 on ( 0,1 ), 0 on ( 1, 2 ), 2 on ( 2,3 )and 0 on ( 3, 4 ). The derivative is undefined atx =−2, − 1, 0, 1, 2, 3 .44. The derivative is 1 except at x =− 2, 0, 2 whereit is undefined.45. Δy = [3(1.5) + 2] – [3(1) + 2] = 1.546.2 2Δ y = [3(0.1) + 2(0.1) + 1] – [3(0.0) + 2(0.0) + 1]= 0.2347. Δy = 1/1.2 – 1/1 = – 0.166748. Δy = 2/(0.1+1) – 2/(0+1) = – 0.181849.3 3Δ y = – ≈0.00812.31+ 1 2.34 + 150. Δ y = cos[2(0.573)] – cos[2(0.571)] ≈ –0.003653.54.55.1 1Δ y –=x+Δ x+ 1 x+1ΔxΔx⎛ x+ 1–( x+Δ x+ 1) ⎞⎛ 1 ⎞= ⎜ ⎟⎜ ⎟⎝ ( x + Δ x+ 1)( x+ 1) ⎠⎝Δx⎠– Δx=( x + Δ x+ 1)( x+ 1) Δx1= – ( x + Δ x + 1)( x + 1)dy ⎡ 1 ⎤ 1= limdx⎢− Δ→ x 0 ( x x 1)( x 1)⎥ = −2⎣ +Δ + + ⎦ ( x + 1)1 ⎛ 1⎞1+ − 1+Δ y⎜ ⎟x+Δx x=⎝ ⎠ΔxΔx1 1 −Δx−x x x x( x+Δx)1=+Δ= = −Δx Δ x x x+Δxdy1 1= lim − = −2( +Δ )dx Δ→ x 0 x x x x( )x+ Δx−1 x−1−Δ y=x+Δ x+ 1 x+1ΔxΔx1 1 1 1 1= ×x+Δ x+ x+ Δx( x+ )( x+Δx− ) −( x− )( x+Δ x+)( 1)( 1)2 2x + xΔx− x+ x+Δx−1− ⎡x + xΔx− x+ x−Δx−1⎤⎣⎦ 1x + xΔ x+ x+ x+Δ x+1x2Δx1 2= × =2 2x + xΔ x+ x+ x+Δ x+ 1 Δxx + xΔ x+ x+ x+Δ x+1dy2 2 2= lim= == ×2Δ( )dx Δ→ x 0 2 2 2x + xΔ x+ x+ x+Δ x+ 1 x + 2x+ 1 x + 151.52.2 2 2Δ y ( x+Δx) – x 2 xΔ x+ ( Δx)= = = 2x+ΔxΔx Δx Δxdy= lim (2 x +Δ x) = 2xdx Δ→ x 03 2 3 2Δ y [( x+Δ x) – 3( x+Δx) ] – ( x – 3 x )=ΔxΔx2 2 2 33x Δ x+ 3 x( Δx) –6 xΔx–3( Δ x)+Δx=Δx2 2= 3x + 3 xΔx–6 x–3 Δ x+ ( Δ x)dy2 2= lim (3x + 3 xΔx– 6 x– 3 Δ x+ ( Δ x) )dxΔ→ x 02= 3 x –6x56.( x x) 2 2x+Δ −1 −1−Δ y=x +Δx xΔxΔx2 2( +Δ ) − − ( +Δ )( − )⎡x x x x x x x 1 ⎤1= ⎢⎥×⎢ x( x+Δx)⎥ Δx⎣⎦2 2 3 2( + 2 Δ + ( Δ )) − − ( + Δ − −Δ )⎡x x x x x x x x x x x ⎤1=⎢⎥×⎢ 2x + xΔx⎥ Δx⎢⎣⎥⎦( )2 22x Δ x+ x Δ x +Δ x 1 x + xΔ x+1= × =2x2x + xΔ x Δ x + xΔx2 2dy x + xΔ x + 1 x + 1= lim=x 0 2 2dx Δ→ x + xΔxx104 Section 2.2 Instructor’s Resource Manual

157. f′ (0) ≈ – ; f′(2) ≈ 122f′ (5) ≈ ; f′(7) ≈ –3358. g′ (–1) ≈ 2; g′(1) ≈ 01g′ (4) ≈ –2; g′(6) ≈ – 363. The derivative is 0 at approximately t = 15 andt = 201 . The greatest rate of increase occurs atabout t = 61 and it is about 0.5 degree F per day.The greatest rate of decrease occurs at aboutt = 320 and it is about 0.5 degree F per day. Thederivative is positive on (15,201) and negative on(0,15) and (201,365).59.60.61. a.5 3f(2) ≈ ; f′ (2) ≈2 2f(0.5) ≈1.8; f′ (0.5) ≈ –0.6b.2.9 −1.9= 0.52.5 − 0.5c. x = 5d. x = 3, 5e. x = 1, 3, 5f. x = 0g.3x ≈− 0.7, and 5 < x < 7262. The derivative fails to exist at the corners of thegraph; that is, at t = 10,15, 55, 60, 80 . Thederivative exists at all other points on the interval( 0,85) .64. The slope of a tangent line for the dashedfunction is zero when x is approximately 0.3 or1.9. The solid function is zero at both of thesepoints. The graph indicates that the solidfunction is negative when the dashed functionhas a tangent line with a negative slope andpositive when the dashed function has a tangentline with a positive slope. Thus, the solidfunction is the derivative of the dashed function.65. The short-dash function has a tangent line withzero slope at about x = 2. 1, where the solidfunction is zero. The solid function has a tangentline with zero slope at about x = 0.4, 1. 2 and 3.5.The long-dash function is zero at these points.The graph shows that the solid function ispositive (negative) when the slope of the tangentline of the short-dash function is positive(negative). Also, the long-dash function ispositive (negative) when the slope of the tangentline of the solid function is positive (negative).Thus, the short-dash function is f, the solidfunction is f ' = g , and the dash function is g ' .66. Note that since x = 0 + x, f(x) = f(0 + x) = f(0)f(x),hence f(0) = 1.f ( a+h)– f( a)f′ ( a) = limh→0hf ( a) f( h)– f( a)= limh→0hf( h)–1 f( h)– f(0)= f( a) lim = f( a) limh→0 hh→0h= f( a) f′ (0)f ′(a) exists since f ′(0) exists.Instructor’s Resource Manual Section 2.2 105

67. If f is differentiable everywhere, then it iscontinuous everywhere, solim f( x) = lim− −(mx+ b) = 2m + b = f (2) = 4x→2 x→ 2and b = 4 – 2m.For f to be differentiable everywhere,f ( x) − f(2)f ′(2) = limmust exist.x→2x − 22f( x) − f(2) x −4lim = lim = lim ( x + 2) = 4+ x−2 + x−2+68.x→2 x→2 x→2f( x) − f(2) mx+ b−4lim= limx→2 − x−2 x→2− x−2mx + − m − m x −limlimx→2 − x−2 x→2− x−24 2 4 ( 2)= = = mThus m = 4 and b = 4 – 2(4) = –4fs( x) = lim( + )– ( ) + ( )– ( – )f x h f x f x f x hh→02hf( x+h)– f( x) f( x– h)– f( x)⎤hh ⎥⎡= lim ⎢+h→0⎣ 2 –2 ⎦1 f( x+ h)– f( x) 1 f[ x+(– h)]– f( x)= lim+ lim2 h→0 h 2 – h→0– h1 1= f′ ( x) + f′ ( x) = f′( x).2 2For the converse, let f (x) = x . Thenh –– h h – hfs(0) = lim = lim = 0h→0 2hh→02hbut f ′(0) does not exist.b. If f is an even function,f() t − f( x0)f′ (– x0) = lim. Let u = –t, ast→−xt+x0 0f( −u) − f( x0)f′ ( − x0) = limu→x0 − u+x0f u − f x0 f u − f x0limlimu→x0 −( u−x0)u→x0u−x0above, then( ) ( ) ( ) ( )= =−= − f ′(x 0 ) = −m.70. Say f(–x) = –f(x). Thenf (– x+h) – f(– x)f′ (– x) = limh→0h– f( x– h) + f( x) f( x– h)– f( x)= lim= – limh→0 hh→0hf[ x+ (– h)] − f( x)= lim = f′( x)so f ′(x) is– h→0– han even function if f(x) is an odd function.Say f(–x) = f(x). Thenf(– x+h) – f(– x)f′ (– x) = limh →0hf ( x– h)– f( x)= limh→0hf[ x+(– h)]– f( x)= – lim = – f′( x)so f ′(x)– h→0– his an odd function if f(x) is an even function.71.69.f() t − ( )f′ ( x0) = lim , sot−f x0t→xx0 0f t − f −x0f′ ( − x0) = limt→−xt−( −x00)f () t − f( −x0)= limt→−xt+x0 0() ( )a. If f is an odd function,f() t −− [ f( −x0)]f′ ( − x0) = limt→−xt+x=0 0f() t + f( −x0)lim.t→−x0 t+x0Let u = –t. As t →−x 0 , u → x 0 and sof( − u) + f( x0)f′ ( − x0) = limu→x− u+x0 0− f u + f x0 − f u − f x0limlimu→x0 −( u−x0) u→x0−( u−x0)f( u) − f( x0)lim f′( x0) m.u→x0 u−x0( ) ( ) [ ( ) ( )]= == = =72.a.b.80 < x < ;380 ≤ x ≤ ;3⎛ 8 ⎞⎜0, ⎟⎝ 3 ⎠⎡ 8⎤⎢0, 3 ⎥⎣ ⎦c. A function f(x) decreases as x increases whenf ′(x) < 0.a. π < x < 6.8 b. π < x < 6.8c. A function f(x) increases as x increases whenf ′(x) > 0.106 Section 2.2 Instructor’s Resource Manual

2.3 Concepts Review1. the derivative of the second; second;f (x) g ′(x ) + g(x) f ′( x)2. denominator; denominator; square of theg( x) f′ ( x)– f( x) g′( x)denominator;2g ( x)3. nx n–1 h; nx n–14. kL(f); L(f) + L(g); D xProblem Set 2.31.2.2 2Dx(2 x ) = 2 Dx( x ) = 2⋅ 2x = 4x3 3 2 2Dx(3 x ) = 3 Dx( x ) = 3⋅ 3x = 9x3. Dx( π x) = π Dx( x) =π⋅ 1=π4.5.6.7.3 3 2 2Dx( π x ) = π Dx( x ) =π⋅ 3x = 3πx–2 –2 –3 –3Dx(2 x ) = 2 Dx( x ) = 2(–2 x ) = –4x–4 –4 –5 –5Dx(–3 x ) = –3 Dx( x ) = –3(–4 x ) = 12x⎛π⎞ –1 –2 –2Dx⎜⎟ =π Dx ( x ) =π (–1 x ) = – π x⎝ x ⎠13.14.15.16.17.4 3 2Dx ( x + x + x + x+1)4 3 2= Dx( x ) + Dx( x ) + Dx( x ) + Dx( x) + Dx(1)3 2= 4x + 3x + 2x+14 3 2 2Dx (3 x – 2 x – 5 x +π x+π)x4x3x2Dx( x) Dx( 2)3 2= 3 D ( x )–2 D ( x )–5 D ( x )+π + π= 3(4 x ) – 2(3 x ) – 5(2 x) +π (1) + 03 2= 12 x – 6 x –10x+π7 5 –2Dx ( π x –2 x –5 x )7 5 –2x x x6 4 –3=π D ( x )–2 D ( x )–5 D ( x )=π (7 x ) – 2(5 x ) – 5(–2 x )6 4 –3= 7 π x –10x + 10x12 − 2 −10Dx ( x + 5 x −π x )x12x−2 x−1011 −3 −11= D ( x ) + 5 D ( x ) −π D ( x )= 12x + 5( −2 x ) −π( − 10 x )11 −3 −11= 12x − 10x + 10πx⎛ 3 –4 ⎞ –3 –4Dx⎜+ x 3 D ( ) ( )3 ⎟= x x + Dxx⎝ x ⎠–4 –5 9 –5= 3(–3 x ) + (–4 x ) = – – 4x4x8.9.10.11.= – π x 2⎛ α ⎞Dx⎜3 ⎟ = αDx ( x⎝ x ⎠) = α(–3 x ) = –3αx3α= –x4–3 –4 –4⎛100⎞ –5 –6Dx⎜100 D ( ) 100(–5 )5 ⎟ = x x = x⎝ x ⎠–6 500= –500 x = –6xα α α⎛ 3 ⎞ 3 –5 3 –6Dx⎜( ) (–5 )5 ⎟ = Dxx = x⎝4x⎠ 4 415 α –6– –15 α= x =4 64x2 2Dx( x + 2 x) = Dx( x ) + 2 Dx( x) = 2x+218.19.20.21.–6 –1 –6 –1x + = x + x–7 –2 –7 –2D (2 x x ) 2 D ( x ) D ( x )= 2(–6 x ) + (–1 x ) = –12 x – x⎛2 1 ⎞ –1 –2Dx⎜– 2 D ( )– ( )2 ⎟ = x x Dxx⎝ x x ⎠–2 –3 2 2= 2(–1 x ) – (–2 x ) = – +2 3x x⎛ 3 1 ⎞ –3 –4Dx⎜– 3 D ( )– ( )3 4 ⎟ = x x Dxx⎝ x x ⎠–4 –5 9 4= 3(–3 x ) – (–4 x ) = – +4 5x x⎛ 1 ⎞ 1 –1D x ⎜ + 2 x⎟= D ( ) 2 ( )2 2x x + D x x⎝ x ⎠–2= 1 (–1 x ) + 2(1) = – 1 + 22 22x12.4 3 4 3Dx(3 x + x ) = 3 Dx( x ) + Dx( x )3 2 3 2= 3(4 x ) + 3x = 12x + 3xInstructor’s Resource Manual Section 2.3 107

22.23.24.⎛ 2 2⎞ 2 –1 ⎛2⎞D x ⎜ – ⎟=D ( )–3 3 3 x x Dxx ⎜ ⎟⎝ ⎠ ⎝3⎠–2= 2 (–1 x ) – 0 = –23 23x2 2 2x + = x + + + x2 2D [ x( x 1)] xD ( x 1) ( x 1) D ( x)= x(2 x) + ( x + 1)(1) = 3x+ 13 3 3x = x + x2 3 3D [3 x( x –1)] 3 xD ( x –1) ( x –1) D (3 x)= 3 x(3 x ) + ( x –1)(3) = 12 x –326.27.2Dx[(–3x+2) ]= (–3x+ 2) Dx(–3x+ 2) + (–3x+ 2) Dx(–3x+2)= (–3x + 2)(–3) + (–3x + 2)(–3) = 18x – 122 3Dx[( x + 2)( x + 1)]2 3 3 2xx2 2 3= ( x + 2) D ( x + 1) + ( x + 1) D ( x + 2)= ( x + 2)(3 x ) + ( x + 1)(2 x)4 2 4= 3x + 6x + 2x + 2x4 2= 5x + 6x + 2x25.2Dx[(2x+1) ]= (2x+ 1) D (2x+ 1) + (2x+ 1) D (2x+1)x= (2x+ 1)(2) + (2x+ 1)(2) = 8x+4x28.4 2Dx[( x –1)( x + 1)]4 2 2 4xx4 2 3= ( x – 1) D ( x + 1) + ( x + 1) D ( x – 1)= ( x – 1)(2 x) + ( x + 1)(4 x )5 5 3 5 3= 2 x –2x+ 4x + 4x = 6x + 4 x –2x29.2 3Dx[( x + 17)( x – 3x+1)]2 3 3 2xx2 2 3= ( x + 17) D ( x – 3x+ 1) + ( x – 3x+ 1) D ( x + 17)= ( x + 17)(3 x –3) + ( x –3x+1)(2 x)4 2 4 2= 3x + 48 x –51+ 2 x –6x + 2x4 2= 5x + 42x + 2 x–5130.31.32.33.34.4 3 24 3 2 3 2 4Dx[( x + 2 x)( x + 2x+ 1)] = ( x + 2 x) Dx( x + 2x + 1) + ( x + 2x + 1) Dx( x + 2 x)4 2 3 2 3= ( x + 2 x)(3x + 4 x) + ( x + 2x + 1)(4x+ 2)6 5 3 2= 7x + 12x + 12x + 12x+ 22 22 2 2 2Dx[(5 x – 7)(3 x – 2x+1)] = (5 x –7) Dx(3 x –2x+ 1) + (3 x –2x+1) Dx(5 x –7)2 2= (5 x – 7)(6 x– 2) + (3 x – 2x+1)(10 x)3 2= 60 x – 30 x – 32x+142 42 4 4 2Dx[(3x + 2 x)( x – 3x+1)] = (3x + 2 x) Dx( x –3x+ 1) + ( x –3x+ 1) Dx(3x + 2 x)2 3 4= (3x + 2 x)(4 x –3) + ( x –3x+ 1)(6x+2)5 4 2= 18x + 10 x – 27 x – 6x+22 21 x DxDxx2 ⎟2 2⎛ ⎞ (3 + 1) (1) – (1) (3 + 1)Dx⎜ =⎝3x+ 1 ⎠(3x+ 1)2(3x + 1)(0) – (6 x) 6x= = –(3x+ 1) (3x+ 1)2 2 2 22 22 x DxDxx2 ⎟2 2⎛ ⎞ (5 –1) (2)–(2) (5 –1)Dx⎜ =⎝5 x –1 ⎠(5 x –1)2(5 x –1)(0) – 2(10 x) 20x= = –(5 x –1) (5 x –1)2 2 2 2108 Section 2.3 Instructor’s Resource Manual

35.36.37.38.39.40.41.2 2⎛ 1 ⎞ (4 x – 3x+ 9) Dx(1) – (1) Dx(4 x – 3x+9)D x ⎜ 2 ⎟ =2 2⎝4 x –3x+9⎠(4 x – 3x+9)2(4 x –3x+ 9)(0)–(8 x–3) 8x−3= = –2 2 2 2(4 x – 3x+ 9) (4 x – 3x+9)− 8x+ 3=(4 x – 3x+9)2 23 3⎛ 4 ⎞ (2 x –3 xD ) x(4)–(4) Dx(2 x –3 x)D x ⎜ 3 ⎟ =3 2⎝2 x –3x⎠(2 x – 3 x)3 2 2(2 x –3 x)(0)–4(6 x –3) –24x+ 12= =3 2 3 2(2 x – 3 x) (2 x – 3 x)⎛ x –1⎞ ( x+ 1) Dx( x–1)–( x–1) Dx( x+1)Dx⎜ ⎟ =⎝ x + 1 ⎠2( x + 1)( x+1)(1) – ( x–1)(1) 2= =( x+ 1) ( x+1)2 2⎛2 x –1⎞( x–1) Dx(2 x–1)–(2 x–1) Dx( x–1)Dx⎜ ⎟ =⎝ x –1 ⎠2( x –1)( x– 1)(2) – (2 x– 1)(1) 1= = –( x–1) ( x–1)2 2⎛ 22 x –1⎞xDx=⎜ 3x+ 5 ⎟⎝ ⎠2(3x+5)(4 x) – (2 x –1)(3)=2(3x+ 5)26x+ 20x+3=2(3x+ 5)2 2(3x+ 5) D (2 x –1) – (2 x –1) Dx(3x+5)2(3x+ 5)2 2⎛ 5 x –4 ⎞ (3x + 1) Dx(5 x– 4) – (5 x– 4) Dx(3x+ 1)Dx⎜ 2 ⎟ =2 2⎝3x+ 1⎠(3x+ 1)2(3x + 1)(5) – (5 x– 4)(6 x)=2 2(3x+ 1)2− 15x+ 24x+5=2 2(3x+ 1)⎛ 22 x –3x1⎞2 2+xDx=⎜ 2x+ 1 ⎟2⎝⎠(2x+ 1)2(2x+ 1)(4 x– 3) – (2 x – 3x+1)(2)=2(2x+ 1)24x+ 4 x–5=2(2x+ 1)(2x+ 1) D (2 x – 3x+ 1) – (2 x – 3x+ 1) D (2x+1)xInstructor’s Resource Manual Section 2.3 109

42.43.44.⎛ 25x2 x–6⎞2 2+ (3 x–1) Dx(5x + 2 x– 6) – (5x + 2 x– 6) Dx(3 x–1)Dx=⎜ 3 x –1 ⎟2⎝⎠(3 x –1)2(3 x–1)(10x+ 2) – (5x + 2 x– 6)(3)=2(3 x –1)215 x –10x+16=2(3 x –1)⎛ 2x – x 1⎞2 2 2 2+ ( x + 1) Dx( x – x+ 1)–( x – x+ 1) Dx( x + 1)Dx=⎜ 2x + 1 ⎟2 2⎝ ⎠( x + 1)2 2( x + 1)(2 x–1)–( x – x+1)(2 x)=2 2( x + 1)2x –1=2 2( x + 1)⎛ 2x – 2x5⎞2 2 2 2+ ( x + 2 x–3) Dx( x –2x+ 5)–( x –2x+ 5) Dx( x + 2 x–3)Dx=⎜ x2+ 2 x –3⎟2 2⎝ ⎠( x + 2 x–3)2 2( x + 2 x– 3)(2 x– 2) – ( x – 2x+ 5)(2x+2)=2 2( x + 2 x–3)24 x –16 x–4=2 2( x + 2 x–3)45. a. ( f ⋅ g) ′(0) = f(0) g′ (0) + g(0) f′(0)= 4(5) + (–3)(–1) = 23b. ( f + g)(0) ′ = f′ (0) + g′(0) = –1+ 5=4c.g(0) f′ (0) – f(0) g′(0)( f g)(0)′ =2g (0)–3(–1) – 4(5) 17= = –2(–3) 946. a. ( f – g) ′(3) = f′ (3) – g′(3) = 2 – (–10) = 12b. ( f ⋅ g) ′(3) = f(3) g′ (3) + g(3) f′(3) = 7(–10) + 6(2) = –58c.( g f)(3)′ =f(3) g′ (3) – g(3) f′(3)f2(3)7(–10) – 6(2) 82= = –2(7) 4947.2Dx[ f( x)] = Dx[ f( x) f( x)]= f ( xD ) x[ f( x)] + f( xD ) x[ f( x)]= 2 ⋅ f ( x) ⋅ Dxf( x)48. Dx[ f( xgxhx ) ( ) ( )] = f( xD ) x[ gxhx ( ) ( )] + gxhxD ( ) ( ) x f( x)= f ( x)[ gxDhx ( ) x ( ) + hxDgx ( ) x ( )] + gxhxD ( ) ( ) x f( x)= f ( xgxDhx ) ( ) x ( ) + f( xhxDgx ) ( ) x ( ) + gxhxD ( ) ( ) x f( x)110 Section 2.3 Instructor’s Resource Manual

49.50.2Dx ( x –2x+ 2) = 2 x–2At x = 1: m tan = 2(1) – 2 = 0Tangent line: y = 12 2⎛ 1 ⎞ ( x + 4) Dx(1)–(1) Dx( x + 4)Dx⎜ x2 ⎟ =2 2⎝ + 4 ⎠( x + 4)2( x + 4)(0)–(2 x) 2x= = –2 2 2 2( x + 4) ( x + 4)At x = 1: m tan =− 2(1) = – 22 2(1 + 4) 251 2Tangent line: y– = – ( x–1)5 252 7y = – x+25 2554. Proof #1:x[ ( ) − ( )] = x[ ( ) + ( −1) ( )]= D [ f( x) ] + D [( −1) g( x)]D f x g x D f x g xProof #2:x= D f( x) −D g( x)Let F( x) = f( x) − g( x). Thenx[ f ( x+ h) − g( x+ h) ] −[ f( x) −g( x)]F'( x)= limh→0h⎡ f ( x+ h) − f( x) g( x+ h) −g( x)⎤= limh→0⎢−hh⎥⎣⎦= f '( x) −g'( x)xx51.3 2 2Dx ( x – x ) = 3 x –2xThe tangent line is horizontal when m tan = 0:2mtan = 3 x –2x= 0x(3x− 2) = 055. a.2Dt(–16t + 40t+ 100) = –32t+40v = –32(2) + 40 = –24 ft/sb. v = –32t + 40 = 0t = 5 4 s52.x = 0 and x = 2 3⎛2 4 ⎞(0, 0) and ⎜ , – ⎟⎝3 27⎠⎛13 2 ⎞ 2Dx⎜ x + x – x⎟= x + 2 x–1⎝3⎠2mtan = x + 2 x–1=12x + 2 x–2=0–2 ± 4 – 4(1)(–2) –2±12x = =2 2= –1– 3, –1+3x = –1±3⎛ 5 5⎜− 1+ 3, − 3 ⎞ ⎟, ⎛ ⎜−1− 3, + 3⎞⎟⎝ 3 ⎠ ⎝ 3 ⎠53. 5 −5y = 100 / x = 100x−6y' =−500xSet y ' equal to − 1 , the negative reciprocal ofthe slope of the line y = x . Solving for x gives1/6x =± 500 ≈± 2.817−5/6y =± 100(500) ≈± 0.563The points are (2.817,0.563) and( − 2.817, −0.563).56.57.2Dt(4.5t + 2 t) = 9t+29t + 2 = 30t = 289 s2mtan = Dx(4 x– x ) = 4 – 2xThe line through (2,5) and (x 0 , y 0 ) has slopey0− 5 .x − 2024 x0 – x0–5x0=x0–22 2x0 + x0 = x0 + x020 – 4 0 3 04–2–2 8 –8 – 4 –5x x + =( x0 –3)( x 0 –1) = 0x 0 = 1, x 0 = 32At x 0 = 1: y 0 = 4(1) – (1) = 3m tan = 4–2(1) = 2Tangent line: y – 3 = 2(x – 1); y = 2x + 12At x = 3 : y = 4(3) – (3) = 30 0m tan = 4–2(3) = –2Tangent line: y – 3 = –2(x – 3); y = –2x + 9Instructor’s Resource Manual Section 2.4 111

58.59.2Dx ( x ) = 2xThe line through (4, 15) and ( x0, y 0)has slopey0−15 . If (x 0 , y 0 ) is on the curve y = x 2 , thenx − 40tan 2x02x0x02x0 2x0 = x020 – 8 0 15 0m–15= = .– 42 –8 –15x x + =( x0 –3)( x 0 –5) = 02At x0 = 3: y0= (3) = 9She should shut off the engines at (3, 9). (Atx 0 = 5 she would not go to (4, 15) since she ismoving left to right.)2Dx (7 – x ) = –2xThe line through (4, 0) and ( x0, y 0)hasyslope 0 − 0 . If the fly is at ( x0, y 0)when thex0− 4spider sees it, then m2 20 + 0 = 0–2x 8x 7 – xx 2 0 –8x 0 + 7 = 0( x0 –7)( x 0 –1) = 0x = 1: y = 6At 0 0tan 00207– x –0= –2x= .x – 42 2d = (4 –1) + (0 – 6) = 9 + 36 = 45 = 3 5≈ 6. 7They are 6.7 units apart when they see eachother.⎛ 1 ⎞ 160. P(a, b) is ⎜a, ⎟.Dxy = – so the slope of⎝ a ⎠ 2x1the tangent line at P is – . The tangent line is2a1 11y – = – ( x– a)or y = – ( x–2 a)whicha 22aahas x-intercept (2a, 0).2 1 2 1dOP ( , ) = a + , dPA ( , ) = ( a–2 a)+2 2aa2 1= a + = d( O, P)so AOP is an isosceles2atriangle. The height of AOP is a while the base,OA has length 2a, so the area is 1 2 (2a)(a) = a2 .61. The watermelon has volume 4 3 πr3 ; the volumeof the rind is34 3 4 r 271 3V r – ⎛ ⎞= π π ⎜r – ⎟ = πr.3 3 ⎝ 10 ⎠ 750At the end of the fifth week r = 10, so271 2 271 2 542πDV r = π r = π (10) = ≈ 340 cm 3250 250 5per cm of radius growth. Since the radius isgrowing 2 cm per week, the volume of the rind is542 πgrowing at the rate of (2) 681 cm3≈ per5week.2.4 Concepts Review1.sin( x + h) – sin( x)h2. 0; 13. cos x; –sin x4.π 1 3 1⎛π ⎞cos = ; y– = ⎜ x–⎟3 2 2 2⎝3⎠Problem Set 2.41. D x (2 sin x + 3 cos x) = 2 D x (sin x) + 3 D x (cos x)= 2 cos x – 3 sin x2.3.4.5.2Dx(sin x) = sin xDx(sin x) + sin xDx(sin x)= sin x cos x + sin x cos x = 2 sin x cos x = sin 2x2 2Dx(sin x+ cos x) = Dx(1) = 02 2Dx(1 – cos x) = Dx(sin x)= sin x Dx(sin x) + sin xDx(sin x)= sin x cos x + sin x cos x= 2 sin x cos x = sin 2x⎛ 1 ⎞Dx(sec x)= Dx⎜ ⎟⎝cosx ⎠cos x Dx(1) – (1) Dx(cos x)=2cos x= sin x 1 sin xsec x tan x2cos x= cos x⋅ cos x=112 Section 2.4 Instructor’s Resource Manual

6.7.8.9.⎛ 1 ⎞Dx(csc x)= Dx⎜ ⎟⎝sinx ⎠sin x Dx(1) − (1) Dx(sin x)=2sin x= – cos x –1 cos x– csc x cot x2sin x= sin x⋅ sin x=⎛ sin x ⎞Dx(tan x)= Dx⎜ ⎟⎝cosx ⎠cos x Dx(sin x) − sin xDx(cos x)=2cos x2 2cos x+sin x 1= = = sec2 2cos x cos x⎛cosx ⎞Dx(cot x)= Dx⎜ ⎟⎝ sin x ⎠sin x Dx(cos x) − cos xDx(sin x)=2sin x2 2 2 2− sin x – cos x –(sin x+cos x)= =2 2sin xsin x12= – = –csc x2sin x⎛sinx+cos x⎞Dx⎜⎟⎝ cos x ⎠cos x Dx(sin x+ cos x) − (sin x+cos x) Dx(cos x)=2cos xcos x(cos x– sin x) – (– sin x– sin xcos x)=2cos x2 2cos x+sin x 1= = = sec2 2cos x cos x222xx⎛sinx+cos x⎞10. Dx⎜⎟⎝ tan x ⎠tan x Dx(sin x+ cos x) − (sin x+cos x) Dx(tan x)=2tan xtan x(cos x– sin x) – sec x(sin x+cos x)=2tan x2 2⎛ ⎞ ⎛ ⎞sin x sin x 1 sin x= ⎜sin x – – –2 ⎟÷⎜ 2 ⎟⎝ cos x cos x cos x⎠ ⎝cosx ⎠2 2⎛ ⎞⎛ ⎞sin x sin x 1 cos x= ⎜sinx − − −2 ⎟⎜ 2 ⎟⎝ cos x cos x cos x⎠⎝ sin x ⎠2cos x 1 cos x= −cosx − −sin x sin x 2sin x11. D ( sin xcos x) = sin xD [ cos x] + cos xD [ sin x]x x x( ) ( )2 2= sin x − sin x + cos x cos x = cos x−sinx12. D ( sin xtan x) = sin xD [ tan x] + tan xD [ sin x]x x x2( ) ( )= sin x sec x + tan x cos x⎛ 1 ⎞ sinx= sin x⎜2 ⎟+⎝cosx ⎠ cos x= tan xsec x+sin x2( cos x)( ) ( )⎛sinx ⎞ xDxsin x − sin xDxx13. Dx⎜ ⎟ =x2⎝ ⎠xxcosx−sin x=2x⎛1−cosx⎞ xDx1−cosx − 1−cosx Dxx14. Dx⎜ ⎟ =2⎝ x ⎠xxsin x+ cos x−1=2x( ) ( ) ( )15.2 2 2x = x + x2D ( x cos x) x D (cos x) cos xD ( x )=− x sin x+2xcosx16.⎛ xcosx+sin x⎞Dx⎜ 2 ⎟⎝ x + 1 ⎠2 2xx2 2( x + 1) D ( xcos x+ sin x) − ( xcos x+ sin x) D ( x + 1)=( x + 1)2( x + 1)(– xsin x+ cos x+ cos x) – 2 x( xcos x+sin x)=2 2( x + 1)3– x sin x– 3xsin x+2cos x=2 2( x + 1)Instructor’s Resource Manual Section 2.4 113

17.18.2y = tan x = (tan x)(tan x)2 2Dxy = (tan x)(sec x) + (tan x)(sec x)2= 2tanxsecx3 2y = sec x = (sec x)(sec x)2 2Dxy = (sec x)sec xtan x+(sec x) Dx(sec x)3= sec x tan x+ sec x(sec x⋅sec xtanx+ sec x⋅sec xtan x)3 3= sec xtan x+2sec xtanx2= 3sec xtanx19. D x (cos x) = –sin xAt x = 1: m tan = – sin1 ≈ –0.8415y = cos 1 ≈ 0.5403Tangent line: y – 0.5403 = –0.8415(x – 1)20. Dx (cot x) = – csc xπAt x = : mtan= –2;4y = 1⎛ π ⎞Tangent line: y–1 = –2 ⎜x– ⎟⎝ 4 ⎠21.22.23.2Dxsin 2 x = Dx(2sin xcos x)= 2sin ⎡⎣x Dxcosx+cosxDxsinx⎤⎦2 2=− 2sin x+2cos x2 2x = x − = x − xD cos 2 x D (2cos x 1) 2D cos x D 1=−2sinxcosxD (30sin 2 t) = 30 D (2sin tcos t)tt2 2( t t)= 30 − 2sin + 2cos= 60cos 2t30sin 2t= 151sin 2t=2π π2 t = → t =6 12π ⎛ π ⎞At t = ; 60cos⎜2⋅ ⎟=30 3 ft/sec12 ⎝ 12 ⎠The seat is moving to the left at the rate of 30 3ft/s.24. The coordinates of the seat at time t are(20 cos t, 20 sin t).⎛ π π⎞ a. ⎜20cos , 20sin ⎟ = (10 3, 10)⎝ 6 6⎠≈ (17.32, 10)25.26.27.b. D t (20 sin t) = 20 cos tππAt t = : rate = 20cos = 10 3 ≈ 17. 32 ft/s6 6c. The fastest rate 20 cos t can obtain is20 ft/s.y = tan x2y'= sec xWhen y = 0 , y = tan 0 = 0 andThe tangent line at x = 0 is y = x.2y = tan x = (tan x)(tan x)2 2y' = (tan x)(sec x) + (tan x)(sec x)2= 2tanxsecx2Now, sec x is never 0, but tan x = 0 atx = kπ where k is an integer.y = 9sin xcosxy' = 9 sin x( − sin x) + cos x(cos x)2 2= 9⎡sin x−cosx⎤⎣⎦= 9 −cos2[ ][ x]2y ' = sec 0= 1.The tangent line is horizontal when y ' = 0or, inthis case, where cos 2x = 0 . This occurs whenπ πx = + k where k is an integer.4 228. f ( x) = x−sinxf '( x) = 1−cosxf '( x ) = 0 when cos x = 1; i.e. when x = 2kπwhere k is an integer.f '( x ) = 2 when x = (2k + 1) π where k is aninteger.29. The curves intersect when 2sinx= 2cos x,sin x = cos x at x = π 4 for 0 < x < π 2 .πDx ( 2sin x) = 2cosx; 2cos = 14πDx ( 2cos x) = – 2sinx; − 2sin = − 141(–1) = –1 so the curves intersect at right angles.30. v = Dt(3sin 2 t) = 6cos 2tAt t = 0: v = 6 cm/st = π 2 : v = − 6 cm/st= π : v = 6 cm/s114 Section 2.4 Instructor’s Resource Manual

31.2 22 sin( x + h) – sin xDx(sin x ) = limh→0h2 2 2sin( x + 2 xh+h ) – sin x= limh→0hsin x cos(2 xh + h ) + cos x sin(2 xh + h ) – sin x= limh →0h2 2 2 2 22 2 2 2sin x [cos(2 xh + h ) –1] + cos x sin(2 xh + h )= limh→0h⎡2 22 cos(2 xh + h ) –1 2 sin(2 xh + h ) ⎤= lim(2 x+ h) sin x + cos x2 2 2⎢h→02 2⎥ = 2 x(sinx ⋅ 0+ cosx ⋅ 1) = 2xcosx⎢⎣2xh + h 2xh + h ⎥⎦32.sin(5( x + h)) – sin 5xDx (sin 5 x) = limh→0hsin(5x + 5 h) – sin 5x= limhh→0sin 5x cos5h+cos5xsin 5 h– sin 5x= limh→0h⎡ cos5 h–1 sin 5h⎤= lim ⎢sin 5x+ cos5xh→0hh ⎥⎣⎦⎡ cos5 h–1 sin 5h⎤= lim ⎢5sin 5x+ 5cos5xh→05h5h⎥⎣⎦= 0 + 5cos5x⋅ 1 = 5cos5x33. f(x) = x sin xa.34. ( )3 2f x = cos x− 1.25cos x+0.225x0 ≈ 1.95f ′(x 0 ) ≈ –1. 242.5 Concepts Review1. Du t ; f′ ( g( t)) g′( t)2. DwG v ; ′( Hs ( )) H′( s)b. f(x) = 0 has 6 solutions on [ π ,6 π ]f ′(x) = 0 has 5 solutions on [ π ,6 π ]c. f(x) = x sin x is a counterexample.0,π .Consider the interval [ ]f ( − π) = f ( π) = 0 and ( ) 0f x = hasexactly two solutions in the interval (at 0 andf ' x = 0 has two solutionsπ ). However, ( )in the interval, not 1 as the conjectureindicates it should have.d. The maximum value of f ( x)– f′ ( x)on[ π ,6 π ] is about 24.93.3.4.2 2( f ( x)) ;( f( x ))2 22xcos( x );6(2x+1)Problem Set 2.51. y = u 15 and u = 1 + xD y = D y⋅D ux u x14= (15 u )(1)14= 15(1 + x)2. y = u 5 and u = 7 + xD y = D y⋅D ux u x= (5u 4 )(1)4= 5(7 + x)3. y = u 5 and u = 3 – 2xD y = D y⋅D ux u x4 4= (5 u )(–2) = –10(3 – 2 x)Instructor’s Resource Manual Section 2.4 115

4. y = u 7 and u = 4+2xD y = D y⋅D ux u x6 2 6= (7 u )(4 x) = 28 x(4 + 2 x )5. y = u 11 and u = x – 2x + 3x+1D y = D y⋅D u6.7.8.23 2x u x10 2= (11 u )(3 x – 4x+3)2 3 2 10= 11(3 x – 4x+ 3)( x – 2x + 3x+1)–7 2y = u and u = x – x+1Dx y = Duy⋅Dxu–8= (–7 u )(2 x–1)2 –8= –7(2 x– 1)( x – x+1)y = u–5 and u = x+3Dx y = Duy⋅Dxu–6 –6 5= (–5 u )(1) = –5( x+ 3) = –( x + 3)–9 2y = u and u = 3 x + x– 3Dx y = Duy⋅Dxu–10= (–9 u )(6x+1)2 –10= –9(6x+ 1)(3 x + x– 3)9(6x+ 1)= –(32 – 3)10x + x9. y = sin u and u = x + xDx y = Duy⋅Dxu= (cos u)(2x + 1)2= (2x + 1)cos( x + x)2610. y = cos u and u = 3 x –2xDx y = Duy⋅Dxu= (–sin u)(6x – 2)2= –(6 x –2)sin(3 x –2 x)11.12.13.14.3y = u and u = cos xDx y = Duy⋅Dxu2= (3 u )(– sin x)2= –3sin x cos x42y = u , u = sin v, and v = 3xDx y = Duy⋅Dvu⋅Dxv3= (4 u )(cos v)(6 x)3 2 2= 24xsin (3 x )cos(3 x )3 x + 1y = u and u =x –1Dx y = Duy⋅Dxu22 ( x–1) Dx( x+ 1)–( x+1) Dx( x–1)= (3 u )2( x –1)2 2⎛ x+ 1 ⎞ ⎛ –2 ⎞ 6( x+1)= 3⎜ ⎟ = −⎝ x –1 ⎠ ⎜ 2 4( x–1) ⎟ ⎝ ⎠ ( x–1)y = u −3 x 2and u = −x − πDx y = Duy⋅Dxu−4( x− π) Dx( x−2) −( x−2) Dx( x−π)=(−3 u ) ⋅2( x −π)−4 2⎛ x−2 ⎞ (2 −π) ( x−π)= − 3⎜⎟ =−3 (2 −π)⎝ x −π⎠ 2 4( x−π) ( x−2)3x15. y = cos u and u =x + 222 2( x+ 2) Dx(3 x )–(3 x ) Dx( x+2)Dx y = Duy⋅Dxu= (– sin u)2( x + 2)⎛ 2 23 x ⎞ 2 2( x+2)(6 x)–(3 x )(1) 3x + 12x ⎛ 3x⎞= –sin= – sin⎜ x + 2 ⎟22⎝ ⎠ ( x + 2)( x + 2) ⎜ x + 2 ⎟⎝ ⎠16.23 xy = u , u = cos v , and v =1– xDx y = Duy⋅Dvu⋅Dxv2 22 (1 – x) Dx( x ) – ( x ) Dx(1 − x)= (3 u )( − sin v)2(1 – x)⎛ 2 2 22(1 – )(2 ) – ( )(–1)–3cosx ⎞ ⎛sinx ⎞ x x x2 2 2–3(2 x– x ) ⎛2 x ⎞ ⎛ x ⎞= = cos sin⎜1– x⎟ ⎜1– x⎟22⎝ ⎠ ⎝ ⎠ (1 – x)(1 – x)⎜1– x ⎟ ⎜1–x ⎟⎝ ⎠ ⎝ ⎠116 Section 2.5 Instructor’s Resource Manual

17.18.2 2 2 2 2 2 2 2 2Dx[(3 x– 2) (3 – x ) ] = (3 x– 2) Dx(3 – x ) + (3 – x ) Dx(3 x– 2)2 2 2 2= (3 x– 2) (2)(3 – x )(–2 x) + (3 – x ) (2)(3 x– 2)(3)2 22 2= 2(3x−2)(3 −x )[(3x−2)( − 2 x) + (3 −x)(3)] = 2(3x −2)(3 − x )(9 + 4x−9 x )2 4 7 3 2 4 7 3 7 3 2 4Dx[(2–3 x ) ( x + 3) ] = (2–3 x ) Dx( x + 3) + ( x + 3) Dx(2–3 x )2 4 7 2 6 7 3 2 32 3 7 2 7 5= (2–3 x ) (3)( x + 3) (7 x ) + ( x + 3) (4)(2–3 x ) (–6 x)= 3 x(3 x – 2) ( x + 3) (29 x –14x+ 24)19.⎡ 22 2( x + 1) ⎤ (3 x– 4) Dx( x+ 1) – ( x+1) Dx(3 x– 4)Dx⎢ ⎥ =3 x –4 2⎢⎣⎥⎦(3 x – 4)( x+ 1)(3x−11)=2(3x− 4)2 2(3 x– 4)(2)( x+ 1)(1) – ( x+1) (3) 3 x – 8 x–11= =2 2(3 x– 4) (3 x– 4)20.⎡ 2 2 2 22 x –3 ⎤ ( x + 4) Dx(2 x–3)–(2 x–3) Dx( x + 4)Dx⎢ ⎥ =2 2 2 4⎢⎣( x + 4) ⎥⎦( x + 4)2 2 22( x + 4) (2) – (2 x– 3)(2)( x + 4)(2 x)− 6x+ 12x+8==2 42 3( x + 4)( x + 4)2 2 ′ 2221. y ′ = 2( x + 4)( x + 4) = 2( x + 4)( 2x) = 4x( x + 4)22. y′= 2 ( x + sin x)( x + sin x) = 2( x + sin x)( 1+cos x)′23.3 2⎛3–2 t ⎞ ⎛3–2t ⎞ ( t+ 5) Dt(3 t –2)–(3 t –2) Dt( t+5)Dt⎜ `⎟ = 3⎜ ⎟⎝ t + 5 ⎠ ⎝ t + 5 ⎠ 2( t + 5)2⎛3 t – 2 ⎞ ( t+5)(3) – (3 t – 2)(1)= 3 ⎜ ⎟⎝ t + 5 ⎠2( t + 5)251(3 t – 2)=4( t + 5)24.⎛ 22 2s –9⎞ ( s+ 4) Ds( s –9)–( s –9) Ds( s+4)Ds=⎜ s 4 ⎟2⎝+⎠( s + 4)2 2( s+ 4)(2 s) – ( s – 9)(1) s + 8s+9= =2 2( s+ 4) ( s+4)25.26.d 3 3 d3 ( t 5) (3t 2) (3t 2) ( t 5)d ⎛+ − − − +(3t−2) ⎞=dt dtdt ⎜ t 5 ⎟2⎝+⎠( t + 5)2(6t+47)(3 t – 2)=2( t + 5)d 3 2(sin θ ) = 3sin θcosθdθ2 3( t+5)(3)(3 t – 2) (3) – (3 t – 2) (1)=2( t + 5)27.dd3 2 2 (cos 2 x) (sin x) − (sin x) (cos 2 x)dy d ⎛ sin x ⎞ ⎛ sin x ⎞ d sin x ⎛ sin x ⎞= 3 3dx dx⎜ ⎟ = ⎜ ⎟ ⋅ = ⎜ ⎟ ⋅dx dx ⎝cos 2x ⎠ ⎝cos 2x ⎠ dx cos 2x ⎝cos 2x ⎠2cos 2x2 2 33 ⎛ sin x ⎞ cos xcos 2x+ 2sin xsin 2x 3sin xcos xcos 2x+6sin xsin 2x= ⎜ ⎟ cos 2 x cos 2 2 =⎝ ⎠cos 4x 2 x23(sin x)(cos xcos 2x+2sin xsin 2 x)=4cos 2xInstructor’s Resource Manual Section 2.5 117

28.dy d 2 d 2 2 d= [sin ttan( t + 1)] = sin t⋅ [tan( t + 1)] + tan( t + 1) ⋅ (sin t)dt dt dt dt2 2 2 2 2 2= (sin t)[sec ( t + 1)](2 t) + tan( t + 1)cost = 2tsin tsec ( t + 1) + costtan( t + 1)29.2⎛ 22 2x + 1⎞ ( x+ 2) Dx( x + 1) – ( x + 1) Dx( x+2)f′ ( x) = 3⎜ x 2 ⎟2⎝+⎠( x + 2)f ′(3) = 9.622⎛2 2x + 1⎞2x + 4 x– x –1= 3⎜ x 2 ⎟2⎝+⎠ ( x + 2)2 2 23( x + 1) ( x + 4 x–1)=4( x + 2)30.2 3 2 4 2 4 2 3G′ ( t) = ( t + 9) Dt( t –2) + ( t –2) Dt( t + 9)2 2 2 2 3= 2 t(7t + 30)( t + 9) ( t – 2)G′ (1) = –74002 3 2 3 2 4 2 2= ( t + 9) (4)( t – 2) (2 t) + ( t – 2) (3)( t + 9) (2 t)31.2F′ ( t) = [cos( t + 3t+ 1)](2t+3)2(2t 3)cos( t 3t1)= + + + ; F′ (1) = 5cos5 ≈ 1.418332.2 2g′ ( s) = (cos πs) Ds(sin π s) + (sin πs) Ds(cos π s)2 2=πsin πs[2cos πs– sin π s]1g′ ⎛⎜⎞⎟= – π⎝2⎠2= (cos πs)(2sin πs)(cos πs)( π ) + (sin πs)(– sin πs)( π )33.34.35.36 .37.38.4 2 3 2 23 2 2 2Dx[sin ( x + 3 x)] = 4sin ( x + 3 x) Dxsin( x + 3 x)= 4sin ( x + 3 x)cos( x + 3 x) Dx( x + 3 x)3 2 23 2 2= 4sin ( x + 3 x)cos( x + 3 x)(2x+3) = 4(2x + 3)sin ( x + 3 x)cos( x + 3 x)5 4Dt[cos (4 t –19)] = 5cos (4 t –19) Dtcos(4 t –19)4= –5cos (4 t –19)sin(4 t –19)(4)3 2Dt[sin (cos t)] = 3sin (cos t) Dtsin(cos t)2= 3sin (cos t)cos(cos t)(– sin t)4= –20cos (4 t –19)sin(4 t –19)4= 5cos (4 t –19)[– sin(4 t –19)] Dt(4 t –19)2= 3sin (cos t)cos(cos t) Dt(cos t)2= –3sin tsin (cos t) cos(cos t)4 1 3 1 1D⎡ ⎛cos u+ ⎞ ⎤ ⎛u4cos u+ ⎞ ⎛Ducosu+⎞ 3 ⎛ 1 1 1⎢ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟u–1 ⎥4cosu+ ⎞ ⎡ ⎛–sinu+ ⎞ ⎤ ⎛Du+⎞= ⎜ ⎟ u⎣ ⎝ ⎠⎦⎝u–1 ⎠ ⎝u–1⎠u–1 ⎢ ⎜ ⎟u–1 ⎥ ⎜ ⎟⎝ ⎠⎣⎝ ⎠⎦⎝u–1⎠3 ⎛u+ 1⎞ ⎛u+1⎞( u–1) Du( u+ 1)–( u+1) Du( u–1)= –4cos ⎜ ⎟sin⎜ ⎟8 3 ⎛u+ 1 ⎞ ⎛u+1 ⎞= cos sin⎝u–1 ⎠ ⎝u–1 ⎠22 ⎜ ⎟ ⎜ ⎟( u –1)( u –1) ⎝u–1 ⎠ ⎝u–1⎠4 2 3 2 2Dθ[cos (sin θ )] = 4cos (sin θ ) Dθcos(sin θ )3 2 2 2 2= –4cos (sin θ )sin(sin θ )(cos θ ) D θ ( θ )3 2 2 2= 4cos (sin θ )[– sin(sin θ )] D θ (sin θ )3 2 2 2= –8θ cos (sin θ )sin(sin θ )(cos θ )2 2 22Dx[ xsin (2 x)] = xDxsin (2 x) + sin (2 x)Dxx= x[2sin(2 x) Dxsin(2 x)] + sin (2 x)(1)222= x[2sin(2 x)cos(2 x) Dx(2 x)] + sin (2 x)= x[4sin(2 x)cos(2 x)] + sin (2 x)= 2xsin(4 x) + sin (2 x)39. Dx{sin[cos(sin 2 x)]} = cos[cos(sin 2 x)] Dxcos(sin 2 x)= cos[cos(sin 2 x)][– sin(sin 2 x)] Dx(sin 2 x)= – cos[cos(sin 2 x)]sin(sin 2 x)(cos 2 x) Dx(2 x)= –2cos[cos(sin 2 x)]sin(sin 2 x)(cos 2 x)40.2Dt{cos [cos(cos t)]} = 2cos[cos(cos t)] Dtcos[cos(cos t)]= 2cos[cos(cos t)]{– sin[cos(cos t)]} Dtcos(cos t)= –2cos[cos(cos t)]sin[cos(cos t)][– sin(cos t)] Dt(cos t)= 2cos[cos(cos t)]sin[cos(cos t)]sin(cos t)(– sin t)= –2sin tcos[cos(cos t)]sin[cos(cos t)]sin(cos t)118 Section 2.5 Instructor’s Resource Manual

41. ( f + g) ′(4) = f′ (4) + g′(4)1 3≈ + ≈ 22 2′ ′42. ( f − 2g) ( 2) = f′( 2) −( 2g) ( 2)= f′( 2) −2g′( 2)= 1− 2( 0)= 1′′43. ( fg ) ( 2 ) = ( fg + gf )( 2) = 2( 0) + 11 ( ) = 144.g(2) f′ (2) – f(2) g′(2)( f g)(2)′ =2g (2)(1)(1) – (3)(0)≈ = 12(1)45. ( f g) ′(6) = f′ ( g(6)) g′(6)= f′ (2) g′(6) ≈(1)( − 1) = –146. ( gf) ′(3) = g′ ( f(3)) f′(3)⎛3⎞3= g′ (4) f′(3) ≈ ⎜ ⎟(1)=⎝2⎠247. D F( 2 x) = F ′( 2x) D ( 2x) = 2F′( x)48.Dx x 2xF= 2xF′222( x + 1) = F ′( x + 1) Dx( x + 1)2′( x + 1)[ ] ()−2−349. D t ( F()t ) = −2F t F ′ t50.ddz⎡⎢⎢⎣1( F( z))2⎤⎥ = −2⎥⎦( ) ( )−3( F( z)) F ′( z)dddz ⎢⎣⎥⎦dz= 21+ 2 2 2 = 41+2 251. ⎡2( 1+ F ( 2z))⎤ = 2( 1+ F( 2z)) ( 1+F( 2z))52.( F ( z))( F′ ( z)) ( F( z)) F′( z)⎡ ⎤d12 1 d−⎢⎡ 2 ⎤y + ⎥ = 2y+ 2( F( y ))dy ⎢ F( y )⎥ dy⎢ ⎥⎣ ⎦⎣ ⎦( )2 d 2= 2y− F′y y = 2y−dy⎛⎜= 2y⎜1−⎜⎝2F′( y )2( F( y ))2⎞⎟⎟⎟⎠22yF′( y )22( F( y ))ddx=−sinxF′53. F ( cos x) = F′( cos x) ( cos x)( cos x)ddx=−F′x F xddx54. cos( F ( x)) =−sin( F( x)) F( x)( ) sin ( ( ))ddx255. Dx⎣tan ( F( 2x)) ⎦sec ( F( 2x)) Dx⎣F( 2x)2= sec ( F( 2x)) × F′( 2x) × Dx[ 2x]2= 2F′( 2x) sec ( F( 2x))⎡ ⎤ = ⎡ ⎤⎦ddg tan 2 x g' tan 2x tan 2xdx⎣ ⎦ = ⋅dx56. ⎡ ( ) ⎤ ( )257. D ⎡F( x) sin F( x)x⎣2( )( )= g' tan 2x sec 2x⋅2( )= 2 g ' tan2x sec 2x⎤⎦( ) ⎡2 2sin ( ) ⎤ sin ( ) ( )= F x × Dx⎣F x⎦+ F x × DxF x( ) 2sin ( ) sin ( )= F x × F x × Dx⎡⎣F x ⎤⎦+2F′( x) sin F( x)( ) 2sin ( ) cos( ( )) x ⎣ ( )2F′( x) sin F( x)( ) ′( ) ( ) ( )2F′( x) sin F( x)= F x × F x × F x × D ⎡F x ⎤⎦+= 2F x F x sinF x cosF x+3 258. D sec F( x) 3sec F( x) D sec F( x)x⎡ ⎤⎣ ⎦= ⎡ ⎤ ⎡ ⎤2( ) ( ) ( ) [ ]= 3sec ⎡⎣F x ⎤⎦sec F x tan F x Dxx3( ) ( ) ( )= 3F′x sec F x tanF x2⎣ ⎦ x ⎣ ⎦59. g '( x) =− sin f ( x) Dxf ( x) =−f′( x) sin f ( x)g′ ( 0) =− f′( 0) sin f ( 0)=−2sin1 ≈−1.683( 1+ secF ( 2x)) x− x ( 1+secF( 2x))60. G′ ( x)=dx dx2( 1+secF( 2x))( 1+ sec F ( 2x)) −2xF′( 2x) sec F( 2x) tan F( 2x)=2( 1+secF( 2x))1+ secF( 0)− 0 1+secF( 0)G′ ( 0)= =2 2( 1+ secF( 0)) ( 1+secF( 0))1 1= = ≈−0.7131+ secF0 1+sec2( )ddInstructor’s Resource Manual Section 2.5 119

61. F′ ( x) =− f ( x) g′ ( x) sin g( x) + f′( x) cos g( x)F′ () 1 =− f () 1 g′ () 1 sing() 1 + f′() 1 cosg()1=− 21sin0 () +− 1cos0=−162.63.y = 1+ xsin 3 x; y′= 3xcos3x+sin 3xπ π πy′ ( π /3)= 3 cos3 + sin = − π + 0= −π3 3 3y− 1 = −πx−π/3y =−πx− π /3+1The line crosses the x-axis at3 −πx = .32y = sin x; y′= 2sin xcos x = sin 2x= 1x = /4 + k , k = 0, ± 1, ± 2,...ππ3 2 22 4 3 2 464. y′ = ( x + 1) 2( x + 1) x + 3( x + 1) x( x + 1)3 4 234222= 2x ( x + 1)( x + 1) + 3x( x + 1) ( x + 1)( ) ( )( ) ( )( ) ( )3 2 2y′ 1 = 2 2 2 + 3 1 2 2 = 32 + 48 = 80y− 32 = 80x− 1, y = 80x+312−265. y′ =− 2( x + 1) ( 2x) =− 4x( x + 1)() ()( )3 −3−3y′ 1 =− 4 1 1+ 1 =−1/21 1 1 1 3y− =− x+ , y = − x+4 2 2 2 42 266. y′ = 32 ( x+ 1) ( 2) = 62 ( x+1)2y′ ( 0) = 6( 1)= 6y− 1= 6x− 0, y = 6x+1The line crosses the x-axis at x =− 1/6 .2−267. y′ =− 2( x + 1) ( 2x) =− 4x( x + 1)68. a.() ( )−33 −3y′ 1 =− 4 2 =−1/21 1 1 1 3y− =− x+ , y = − x+4 2 2 2 4Set y = 0 and solve for x. The line crosses thex-axis at x = 3/2.b.2 2 2 2⎛ x ⎞ ⎛ y⎞ ⎛4cos2t⎞ ⎛7sin2t⎞⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟⎝4⎠ ⎝7⎠ ⎝ 4 ⎠ ⎝ 7 ⎠2 2= cos 2t+ sin 2t= 12 2 2 2L = ( x–0) + ( y–0)= x + y2 2= (4cos2 t) + (7sin 2 t)2 2= 16cos 2t+49sin 2tc. 12 2DL= t Dt(16 cos 2t+49sin 2 t)2 22 16cos 2t+49sin 2t32cos 2 tDt(cos 2 t) + 98sin 2 tDt(sin 2 t)=2 22 16cos 2t+49sin 2t–64cos 2tsin 2t+196sin 2tcos 2t=2 22 16cos 2t+49sin 2t− 16sin4t+49sin4t=2 216cos 2t+49sin 2t33sin 4t=2 216cos 2t+49sin 2tAt t = π 8 : rate = 3316⋅ 1 + 49⋅169. a. (10cos8 πt,10sin 8 π t)2 2b. Dt(10sin8 π t) = 10cos(8 πt) Dt(8 π t)= 80πcos(8 π t)At t = 1: rate = 80π ≈ 251 cm/sP is rising at the rate of 251 cm/s.70. a. (cos 2t, sin 2t)b.c.2 2 2(0–cos2 t) + ( y–sin2 t) = 5 , so2y = sin 2t+25 – cos 2t≈ 5.8 ft/sec.2D⎛t sin 2t 25 cos 2t⎞⎜ + − ⎟⎝⎠1= 2cos2t+ ⋅4cos2tsin2t22 25−cos 2t⎛ sin 2t⎞= 2cos2t1+⎜2 ⎟⎝ 25 – cos 2t⎠71. 60 revolutions per minute is 120π radians perminute or 2π radians per second.a. (cos 2 πt,sin 2 π t)b.c.2 2 2(0–cos2 π t) + ( y–sin2 π t) = 5 , soy = sin 2π+ t 25 – cos 2πt2D⎛t sin 2 t 25 cos 2 t⎞⎜ π+ − π⎟⎝⎠= 2πcos2πt1+ ⋅4πcos2πtsin2πt22 25−cos 2πt⎛= 2πcos2π t1+⎜⎝2sin 2πt⎞2 ⎟25 – cos 2πt⎠120 Section 2.5 Instructor’s Resource Manual

72. The minute hand makes 1 revolution every hour,so at t minutes after the hour, it makes an angleπtof radians with the vertical. By the Law of30Cosines, the length of the elastic string is2 2πts = 10 + 10 – 2(10)(10)cos 30πt= 10 2 – 2cos 30ds1 π πt= 10⋅ ⋅ sindtπt15 302 2–2cos 30πsinπt30=3 2–2cosπt30At 12:15, the string is stretching at the rate ofπsinπ2 π= ≈ 0.74 cm/min3 2–2cosπ 3 2273. The minute hand makes 1 revolution every hour,so at t minutes after noon it makes an angle ofπtradians with the vertical. Similarly, at t30minutes after noon the hour hand makes an angleπtof with the vertical. Thus, by the Law of360Cosines, the distance between the tips of thehands is2 2 ⎛ πtπt⎞s = 6 + 8 –2⋅6⋅8cos ⎜ – ⎟⎝30 360 ⎠=11πt100 – 96cos 360ds 1 44π11πt= ⋅ sindt 2 100–96cost 15 36011π36011πt36011πt36022πsin=15 100 – 96cosAt 12:20,11ds 22πsinπ18= ≈ 0.38 in./mindt 15 100 – 96cos 11π1874. From Problem 73,dsdt22πsin11πt360=.15 100 – 96cos 11πt360Using a computer algebra system or graphingutility to view ds for 0≤t≤ 60, ds is largestdtdtwhen t ≈ 7.5. Thus, the distance between the tipsof the hands is increasing most rapidly at about12:08.75. sin x0 = sin 2x0sin x = 2sin x cos x76.0 0 01cos x0 = [if sin x0≠ 0]2x 0 = π 3D x (sin x) = cos x, D x (sin 2x) = 2cos 2x, so at x 0 ,the tangent lines to y = sin x and y = sin 2x haveslopes of m 1 = 1 2 and ⎛ 1 ⎞m2= 2 ⎜– ⎟=–1,⎝ 2 ⎠respectively. From Problem 40 of Section 0.7,m2 – mtanθ = 1where θ is the angle between1 + mm 1 21 32 2–1– –the tangent lines. tan θ = = = –3,1 11 + ( 2 )(–1)2so θ ≈ –1.25. The curves intersect at an angle of1.25 radians.1tAB = OAsin2 212D = OAcos t ⋅ AB = OA cos t sint2 2 2 2E = D + area (semi-circle)22 t t 1 ⎛1⎞= OA cos sin + π ⎜ AB ⎟2 2 2 ⎝2⎠2 1 2 2= OA cos t sin t + π OA sint2 2 2 22 t ⎛ t 1 t ⎞= OA sin ⎜cos + πsin⎟2⎝2 2 2⎠cos tD2E = cos t 1 t+ πsin 2 2 2D 1lim = = 1t→0+ E 1+0Dcos( t/ 2)lim = limt→π − E t→π − πcos( t/ 2) + sin( t/ 2)20= = 0π0 +2Instructor’s Resource Manual Section 2.5 121

77.78.79.2y = u and u = xD x y = D u y ⋅ D x u1 2xx x= ⋅ 2x= = =2 u22 x x x2x –12 2Dxx –1 = D ( –1)2 x xx –12x –122 x x –1= (2 x)=2x –12x –1sin xDxsin x = Dx(sin x)sin xsin x= cos x = cot x sin xsin x1 2DLx = L' x D x = ⋅ 2x=2x x2 2 280. a. x ( ) ( ) x( )81. [ f( f( f( f(0))))]′= f′ ( f( f( f(0)))) ⋅ f′ ( f( f(0))) ⋅ f′ ( f(0)) ⋅ f′(0)= 2 ⋅ 2 ⋅2 ⋅ 2 = 1682. a.b.d f[2] f f x f xdx= '( ( )) ⋅ '( )[1] d [1]= f '( f ) ⋅ f ( x)dxd f[3]= f f f x ⋅ f f x ⋅ f xdx'( ( ( ))) '( ( )) '( )[2] [1] d [1]= f '( f ( x)) ⋅ f '( f ( x)) ⋅ f ( x)dx[2] d [2]= f '( f ( x)) ⋅ f ( x)dxc. Conjecture:d [ n] [ n−1] d [ n−1]f ( x) = f '( f ( x)) ⋅ f ( x)dxdxb.4 4 4DL x (cos x) = sec xDx(cos x)4 3= sec x(4cos xD ) x (cos x)4 3= 4sec xcos x( −sin x)1 3= 4⋅ ⋅cos x ⋅ −sin4cos x= –4secx sinx = − 4tan x( x)⎛ f( x) ⎞ ⎛ 1 ⎞83. ⎜ ⎟ ⎜ ⎟ ( ) ( )−1 −1 −1x = x ⋅ = x ⋅ = x +xD D f( x) D f( x) ( gx ( )) f( xD ) ( gx ( )) ( gx ( )) Df( x)⎝ gx ( ) ⎠ ⎝ gx ( ) ⎠84. ( ) = ( ( ))( ) = ( ( ))( ( ))−2 −1 −2 −1x x x x= f( x) ⋅( − 1)( g( x)) D g( x) + ( g( x)) D f( x) = − f( x)( g( x)) D g( x) + ( g( x)) D f( x)−f( x) Dxg( x) Dx f( x) −f( x) Dxg( x)gx ( ) Df x ( x) − f( xDgx ) x ( ) gxDf ( ) x ( x)= + =+ ⋅ = +2 22 2g ( x) gx ( ) g ( x)gx ( ) gx ( ) g ( x) g ( x)gxDf ( ) x ( x) − f( xDgx ) x ( )=2g ( x)( ( )) ( ( ( ))) ( ( )) ( )( ( )) ( ( ( ))) ( ( )) ( )′( 2 ) ′( ( 2)) ′( 2) ′( 1) ′( ( 1)) ′( 1) ′( 2) ′( 1)g′ x f′ f f f x f′ f f x f′ f x f′xg′ x f′ f f f x f′ f f x f′ f x f′x1 1 1 1 1= f f f x f f x f x f x = f f x f x f x f x2 2= ⎡⎣f′ ( x1) ⎤⎦ ⎡⎣f′( x2)⎤⎦( 2) = ( ( ( ( 2)))) ( ( ( 2))) ( ( 2)) ( 2)= f ′ f ( f ( x ))f′ ( f x ) f′ ( x ) f′ ( x ) = f′ ( f ( x )) f′ ( x ) f′ ( x ) f′( x )g′ x f′ f f f x f′ f f x f′ f x f′x( ) ( )1 1 1 2 2 2 1 22 2( ) ( ) ( )= ⎡⎣f′ x ⎤⎦ ⎡⎣f′ x ⎤⎦= g′x1 2 1122 Section 2.5 Instructor’s Resource Manual

2.6 Concepts Review1.2.3 d yf′′′ ( x), Dxy,, y '''dx32ds ds d s; ;dt dt 2dt33.dy3(3x5) (3) 9(3x5)dx = + = +22 2d y= 18(3x+ 5)(3) = 162x+2702dx3d y= 1623dx′ >3. f () t 04. 0; < 0Problem Set 2.61.dy 23x6x6dx = + +2d y= 6x+ 62dx3d y= 63dxdy 4 32. 5x4xdx = +d 2 ydx 2 = 20x3 +12x 23d y 2= 60x+ 24x3dx4.dy4 45(3 – 5 x) (–5) –25(3 – 5 x)dx = =2d y3 3= –100(3 – 5 x) (–5) = 500(3 – 5 x)2dx3d y= 1500(3 – 5 x) (–5) = –7500(3 – 5 x)3dxdy5. 7cos(7 x)dx =22 2d y 2= –7 sin(7 x)2dx3d y 3= –7 cos(7 x) = –343cos(7 x)3dx6.dy 2 33xcos( x )dx =2d y 2 2 3 3= 3 x [–3x sin( x )] + 6xcos( x )2dx34 3 3= –9x sin( x ) + 6xcos( x )d y 4 3 2 3 3 3 2 3= –9x cos( x )(3 x ) + sin( x )(–36 x ) + 6 x[– sin( x )(3 x )] + 6cos( x )3dx6 3 3 3 3 3 3= –27x cos( x ) – 36x sin( x ) –18x sin( x ) + 6cos( x )6 3 3 3= (6 – 27 x )cos( x ) – 54x sin( x )7.dy ( x –1)(0)–(1)(1) 1= = –dx 2 2( x–1) ( x–1)2 2d y ( x–1) (0)–2( x–1) 2=− =2 4 3dx ( x –1) ( x –1)3 3 2d y ( x−1) (0) −2[3( x−1) ]=3 6dx( x −1)6=−4( x −1)8.dy (1– x)(3) – (3 x)(–1) 3= =dx 2 2(1 – x) ( x– 1)2 2d y ( x–1) (0)–3[2( x–1)] 6= = –2 4 3dx ( x –1) ( x –1)3 3 2d y ( x−1) (0) −6(3)( x−1)=−3 6dx( x −1)18=4( x −1)Instructor’s Resource Manual Section 2.6 123

9. f′ ( x) = 2 x; f′′ ( x) = 2; f′′(2) = 210.11.2f ′( x) = 15x + 4x+1f′′ ( x) = 30x+4f ′′(2) = 64f′ () t = –22t4f′′ () t =3t4 1f ′′(2)= =8 212.2 2(5 – u)(4 u) – (2 u )(–1) 20 u– 2uf′ ( u)= =2 2(5 – u) (5 – u)2 2(5– u) (20–4 u)–(20 u–2 u )2(5– u)(–1)f′′ ( u)=4(5 – u)100=3(5 – u)100 100f ′′(2)= =33 2713.′ –3( ) = –2(cos π) (– sin π)π = 2π(cosθπ) –3 (sin θπ)f θ θ θ′′ –3 –4( ) = 2 π[(cos π) ( π)(cos π ) + (sin π)(–3)(cos π) (– sin π)( π )]f θ θ θ θ θ θ2 2f ′′(2) = 2 π [1+ 3(0)(1)] = 2π2 −2 2 −4= 2 π [(cos π ) + 3sin π(cos π ) ]θ θ θ14.π ⎛ π ⎞ π′⎛ ⎞ ⎛ ⎞ ⎛ π⎞ ⎛π⎞ ⎛π⎞() = cos ⎜ ⎟⎜– + sint 2 ⎟ ⎜ ⎟ = ⎜– ⎟cos⎜ ⎟+sin⎜ ⎟⎝ ⎠⎝t ⎠ ⎝ t ⎠ ⎝ t ⎠ ⎝ t ⎠ ⎝ t ⎠f t tff⎛ π⎞⎡⎛π⎞⎛ π ⎞⎤⎛ π ⎞ ⎛π⎞ ⎛ π ⎞ ⎛π⎞′′( t) = ⎜– ⎟⎢– sin ⎜ ⎟⎜– cos – cost t 2 ⎟⎥+ ⎜ +2 ⎟ ⎜ ⎟ ⎜t 2 ⎟ ⎜ ⎟⎝ ⎠⎣⎝ ⎠⎝ t ⎠⎦⎝t ⎠ ⎝ ⎠ ⎝ t ⎠ ⎝ t ⎠2 2π π π′′⎛ ⎞(2) = – sin ⎜ ⎟= – ≈ –1.238 ⎝2⎠82π ⎛π⎞= – sin3 ⎜ ⎟⎝ t ⎠t15.′ 2 2 2 3( ) = (3)(1– ) (–2 ) + (1– )f s s s s s5 3f ′′( s) = –42s + 60 s –18sf ′′(2) = –9002 2 2 2 3= –6 (1– ) + (1– )s s s6 4 2= –7s + 15 s –9s+ 116.( x– 1)2( x+ 1) – ( x+1)f′ ( x)=2( x –1)2 222x – 2 x–3=2( x –1)( x–1) (2 x–2)–( x –2 x–3)2( x–1)f′′ ( x)=4( x –1)8f ′′(2) = = 8312( x–1)(2 x– 2) – ( x – 2 x– 3)(2) 8= =( x–1) ( x–1)3 317.xn n–12x (n) = ( –1)n–23x (n) = ( –1)( –2)n–34x (n) = ( –1)( –2)( –3)n–4n−xn=nxnD ( x ) = nxD x n n xD x n n n xD x n n n n xD 1 ( x ) n( n–1)( n–2)( n–3)...(2)xD ( x ) = n( n–1)( n–2)( n–3)...2(1)x = n!018. Let k < n.n k n−k k kx x x xn n–1Dxanx +…+ a1x+ a0=D ( x ) = D [ D ( x )] = D ( k!) = 0so19. a.b.c.[ ] 04 3xD (3x + 2 x–19) = 012 11 10xD (100x − 79 x ) = 011 2 5xD ( x –3) = 0124 Section 2.6 Instructor’s Resource Manual

20.D ⎛1⎞ 1x ⎜ ⎟ = –⎝ x ⎠ x 22 ⎛1⎞ –2 –3 2Dx ⎜ ⎟ = Dx(– x ) = 2x=⎝ x ⎠3x3 ⎛1⎞ –3 3(2)Dx ⎜ ⎟ = Dx (2 xx) =⎝ ⎠– 4xD 4 ⎛1 ⎞ 4(3)(2)x ⎜ ⎟ =⎝ x ⎠ x 5nD n ⎛1 ⎞ ( −1) n!x ⎜ ⎟ =⎝ x ⎠ x n + 1b.c.23 t –12t > 03t(t – 4) > 0; ( −∞,0) ∪ (4, ∞ )23 t –12t < 0(0, 4)d. 6t – 12 < 06t < 12t < 2; ( −∞ ,2)e.21.2f′ ( x) = 3x + 6 x–45= 3( x+ 5)( x−3)3(x + 5)(x – 3) = 0x = –5, x = 3f′′ ( x) = 6x+6f ′′(–5) = –24f ′′(3) = 2425. a.2vt () = ds = 3 t –18t+24dt2d sat () = = 6 t–182dt22. g′ () t = 2at+bg′′ () t = 2ag′′ (1) = 2a=−4a =−2g′ (1) = 2a+ b=32(–2) + b = 3b = 7g()1 = a+ b+ c = 5( − 2) + ( 7)+ c = 5c = 0ds23. a. vt () = = 12–4tdt2at () = d s = –42dtb. 12 – 4t > 04t < 12−∞ ,3t < 3; ( )c. 12 – 4t < 0t > 3; (3, ∞ )d. a(t) = –4 < 0 for all te.b.23 t –18t+ 24 > 03(t – 2)(t – 4) > 0( −∞,2) ∪(4, ∞ )c. 3t 2 –18t + 24 < 0(2, 4)d. 6t – 18 < 06t < 18t < 3; ( −∞ ,3)e.26. a.b.c.ds 2vt () = = 6 t –6dt2d sat () = = 12t2dt26 t –6>06(t + 1)(t – 1) > 0( −∞, −1) ∪(1, ∞ )26 t –6

e.27. a.b.ds 16vt () = = 2 t–dt 2t2d s 32at () = = 2+2 3dt t162 t – > 02t32 t –16> 0; (2, ∞ )2tc. 2t – 16 < 0; (0, 2)2td. 2 + 32t 3 < 02t 3 + 32< 0; The acceleration is note.t 3negative for any positive t.29.30.31.3 2vt () = ds = 2 t –15t + 24tdt22at ( ) = d s = 6 t – 30t+242dt26 t – 30t+ 24 = 06(t – 4)(t – 1) = 0t = 4, 1v(4) = –16, v(1) = 11ds 1 3 2vt () = = (4 t –42t + 120) tdt 102d s 1 2at ( ) = = (12 t – 84t+120)2dt 101 (12 2t – 84 t+ 120) = 01012 ( 2)( 5) 010 t− t−=t = 2, t = 5v(2) = 10.4, v(5) = 5ds1v1( t) = = 4–6tdt2v2( t) = ds = 2 t –2dt28. a.b.c.dsvt () = = 1–dt t2d s 8at () = =2 3dt t421– > 0t242t –4> 0; (2, ∞ )2t421– < 0; (0, 2)ta. 4 – 6t = 2t – 28t = 63t = sec4b. 4–6t= 2 t –2; 4 – 6t = –2t + 2c.1 3t = sec and t = sec2 42 24 t –3 t = t –2t24 t –6t = 02t(2t – 3) = 03t = 0 sec and t = sec2d.e.83< 0; The acceleration is not negative fortany positive t.32.1 2v1( t) = ds = 9 t –24t+18dt2 2v2 () t = ds = –3t + 18 t –12dt2 29 t – 24t+ 18 = –3t + 18 t –12212 t – 42t+ 30 = 022 t –7t+ 5=0(2t – 5)(t – 1) = 0t = 1, 5 2126 Section 2.6 Instructor’s Resource Manual

33. a. v(t) = –32t + 48initial velocity = v 0 = 48 ft/secb. –32t + 48 = 0c.d.t = 3 2 sec2s = –16(1.5) + 48(1.5) + 256 = 292 ft2–16t+ 48t+ 256 = 02–48 ± 48 – 4(–16)(256)t = ≈ –2.77, 5.77–32The object hits the ground at t = 5.77 sec.e. v(5.77) ≈ –137 ft/sec;speed = − 137 = 137 ft/sec.34. v(t) = 48 –32ta. 48 – 32t = 0t = 1.52s = 48(1.5) –16(1.5) = 36 ftb. v(1) = 16 ft/sec upwardc.248 t –16t = 0–16t(–3 + t) = 0t = 3 sec35. vt () = v0–32tv0 – 32t = 0v0t =32v0 v0v ⎛ 0 ⎜ ⎞ –16 ⎛ ⎞ = 528032⎟ ⎜32⎟⎝ ⎠ ⎝ ⎠2 20 v0v– = 528032 6420v= 528064v 0 = 337,920 ≈ 581 ft/sec36. vt () = v0+ 32tv0 + 32t= 140v 0 + 32(3) = 140v 0 = 442s = 44(3) + 16(3) = 276 ft237.2vt t t() = 3 –6 –242d3 t –6 t –242t t =tdt 23 t –6 t –24( t –4)( t+2)=(6 t – 6)t t3 –6 –24 (6 –6)( –4)( + 2)( t – 4)( t+2) (6 t – 6)< 0( t –4)( t+2)t < –2, 1 < t < 4; ( −∞, −2)∪ (1, 4)38. Point slowing down whend vt () < 0dtd vt () at ()vt () =dt v()tvt () at ()< 0 when a(t) and v(t) have oppositevt ()signs.39. Dx( uv)= uv′ + uv ′Dx2 ( uv)= uv′′ + uv ′ ′ + uv ′ ′ + u′′v= uv′′ + 2u′ v′ + u′′vDx3 ( uv) = uv′′′ + uv ′ ′′ + 2( uv ′ ′′ + u′′ v′ ) + u′′ v′ + u′′′v= uv′′′ + 3u′′′ v + 3u′′′ v + u′′′vnn ⎛n⎞n−k kDx( uv) = ∑ ⎜ ⎟Dx ( u) Dx( v)kk=0⎝⎠⎛ nwhere ⎜ ⎟ ⎞ is the binomial coefficient⎝ k⎠40.n!.( n– k)! k!⎛4⎞( sin ) = ⎜ ⎟ ( ) (sin )⎝0⎠⎛4⎞3 4 1 ⎛4⎞2 4 2+ ⎜ ⎟Dx( x ) Dx(sin x) + ⎜ ⎟Dx( x ) Dx(sin x)⎝1⎠⎝2⎠⎛4⎞1 4 3 ⎛4⎞0 4 4+ ⎜ ⎟Dx( x ) Dx(sin x) + ⎜ ⎟Dx( x ) Dx(sin x)⎝3⎠⎝4⎠2= 24sin x + 96xcos x−72x sin x3 4− 16x cos x+x sin x4 4 4 4 0Dx x x Dx x DxxInstructor’s Resource Manual Section 2.6 127

41. a.42. a.b. f ′′′(2.13) ≈ –1. 2826b. f ′′′(2.13) ≈ 0.02712.7 Concepts Review1.2.3.4.93x – 33y2dydx2 2 2x(2 y) dy + y + 3 y dy – dy = 3xdx dx dxp –1 5 xpq 2 2 / 3; ( x – 5 x ) (2 x – 5)q 3Problem Set 2.71. 2 yDxy–2x=02xxDxy = =2yy2. 18x+ 8yD y = 0–18x9xDxy = = –8y4y3. xD y+ y=0xyDxy = –xx4.5.6.7.8.9.10.22x+ 2αyDxy = 02xxDxy = – = –2 22αy α y2x(2 y) Dxy+ y = 121– yDxy =2xy22x+ 2x Dxy+ 4xy+ 3xDxy+ 3y= 02Dxy(2x + 3 x) = –2 x– 4 xy– 3y–2 x –4 xy –3yDxy =22x+ 3x2 2 212x + 7 x(2 y) Dxy+ 7 y = 6y Dxy2 2 212x + 7 y = 6 y Dxy –14xyDxy2 212x+ 7 yDxy =26 y –14xy2 2x Dxy+ 2 xy = y + x(2 y)Dxy2 2x Dxy – 2 xyDxy = y –2xy2y – 2xyDxy =2x – 2xy1⋅ (5x Dxy+ 5 y) + 2Dxy2 5xy2 3= 2 yDxy+ x(3 y ) Dxy+y5x2Dx y+2 Dxy–2 yDxy–3xy Dxy2 5xy3 5y= y –2 5 xy3 5yy –2 5xyDxy =5x2+ 2–2 y –3xy2 5xy1x D 12 1 x y+ y+ = xD x y+yy +xD – – 12 1 x y xD x y = y y+y +y– y+1Dxy =x – x2 y+1128 Section 2.7 Instructor’s Resource Manual

11. xDxy+ y+ cos( xy)( xDxy+ y) = 0x Dxy+ xcos( xy) Dxy = – y– ycos( xy)– y– ycos( xy)yDxy = = –x + xcos( xy)x12.13.14.2 2– sin( xy )(2 xyDxy+ y ) = 2yDxy+12 2 2–2xysin( xy ) Dxy–2yDxy = 1+y sin( xy )2 21+y sin( xy )Dxy =2–2xysin( xy )–2y3 2 3 2x y′ + 3x y+ y + 3xy y′= 03 2 2 3y′ ( x + 3 xy ) = –3 x y–y2 3–3 x y–yy′ =3 2x + 3xy36 9At (1, 3), y′ = – = –28 79Tangent line: y– 3 = – ( x–1)72 2x (2 y) y′ + 2xy + 4xy′ + 4y = 12y′2 2y′ (2x y+ 4 x–12) = –2 xy – 4y2 2–2 xy –4 y – xy –2yy′ = =2 22x y 4 x–12 x y 2 x–6+ +At (2, 1), y′ = –2Tangent line: y–1 = –2( x–2)15. cos( xy)( xy′ + y)= y′y′ [ xcos( xy)–1] = – ycos( xy)– ycos( xy) ycos( xy)y′ = =x cos( xy) –1 1– x cos( xy)⎛π⎞At ⎜ , 1 ⎟, y′ = 0⎝2⎠⎛ π ⎞Tangent line: y–1= 0 ⎜ x– ⎟⎝ 2 ⎠y = 116.2 2y′ + [– sin( xy )][2 xyy′+ y ] + 6x= 02 2 2y′ [1– 2xysin( xy )] = y sin( xy ) – 6x2 2y sin( xy ) – 6xy′ =21–2xysin( xy )6At (1, 0), y′ = – = –61Tangent line: y – 0 = –6(x – 1)17.18.19.20.21.22.23.24.25.2 –1/ 3 2 –1/ 3x – y y′ –2y′=03 32 –1/ 3 ⎛2–1/ 3 ⎞x = y′⎜ y + 2⎟3 ⎝3⎠2 –1/3x3y′ =2 –1/ 3y + 2323 1At (1, –1), y′ = =4 23Tangent line:1y+ 1 = ( x–1)212y′ + 2xyy′+ y = 02 y⎛ 1 ⎞ 2y′ + 2 xy = – y⎜2 y ⎟⎝ ⎠2– yy′ =1 + 2xy2 y–1 2y′ = = –17 172At (4, 1),Tangent line:dy 2/3 1= 5x+dx 2 x2y–1 = – ( x–4)17dy 1 –2/3 5/2 1 5/2= x –7 x = –7xdx 3 3 23 xdy 1 –2/3 1 –4/3 1 1= x – x = –dx 3 3 3 2 3 43 x 3 xdy 1 –3/ 4 1= (2x+ 1) (2) =dx 4 4 32 (2x+ 1)dy 1 (32 – 4 )–3/4x x (6 x – 4)dx = 46 x–4 3 x–2= =4 2 3 4 2 34 (3 x –4 x) 2 (3 x –4 x)dydx = 31 ( 3 –2 ) –2/3 (3 2x x x –2)dy d 3 −2/3= [( x + 2 x) ]dx dx22 3 –5/3 2 6x+ 4= – ( x + 2 x) (3x+ 2) =−3 3 3 53 ( x + 2 x)Instructor’s Resource Manual Section 2.7 129

26.27.dy 5 –8/3 –8/3– (3 x–9) (3) –5(3 x–9)dx = 3=dy 1= (2x+ cos x)dx 22 x + sinx2x+ cosx=22 x + sinx34.35.2 dx 2 dx1= cos( x )(2 x) + 6xdy dydx 1=dy 2 22x cos( x ) + 6x(x + 2) 2 + y 2 = 15y28.dy 1 2= [ x (–sin x) + 2xcos x]dx 22 x cosx22x cos x– x sinx=22 x cosx−5−55 x29.30.31.32.dy d 2 –1/3= [( x sin x) ]dx dx= 1 2 –4/3 2– ( sin ) ( cos 2 sin )3 x x x x+x x2x cos x+2xsinx= –3 2 43 ( x sin x)dy 1 (1 sin 5 ) –3/ 4x (cos5 x )(5)dx = 4+5cos5x=44(1+sin5 x)32 –3/4 2dy [1 + cos( x + 2 x)] [– sin( x + 2 x)(2x+2)]=dx42( x + 1)sin( x + 2 x)=−4 2 32 [1+ cos( x + 2 x)]2 2 –1/2 2dy (tan x + sin x) (2 tan xsec x+2sin xcos x)=dx22tan xsec x+sin xcosx=2 2tan x+sin x36.dy2x+ 4+ 2y = 0 dxdy 2x + 4 x + 2=− =−dx 2y yThe tangent line at ( x0, y 0)has equationx0+ 2y– y0 =− ( x– x0)which simplifies toy02 22 x0 – yy0 –2 x– xx0 + y0 + x0= 0. Since2 2x0 + y0 = –3–4 x0,( x0, y 0)is on the circle,so the equation of the tangent line is– yy0 –2 x0 –2 x – xx 0 = 3.3If (0, 0) is on the tangent line, then x 0 = – .2Solve for y 0 in the equation of the circle to get3y 0 =± . Put these values into the equation of2the tangent line to get that the tangent lines are3y+ x = 0 and 3 y– x = 0.2 216( x + y )(2x+ 2 yy′ ) = 100(2 x– 2 yy′)3 2 2 332x + 32x yy′ + 32xy + 32y y′ = 200 x– 200yy′33.2 ds 2s + 2st + 3t= 0dt2 2 2 2ds – s –3t s + 3t= = −dt 2st 2st2 dt2 dts + 2st+ 3t= 0ds dsdt 2 2( s + 3 t ) = –2stdsdt 2st=−ds 2 2s + 3t2 3 3 2y′ (4x y+ 4y + 25 y) = 25 x–4 x –4xy3 225 x –4 x –4xyy′ =2 34x y+ 4y + 25y1The slope of the normal line = – y ′2 34x y+ 4y + 25y=3 24x + 4 xy –25x65 13At (3, 1), slope = =45 913Normal line: y–1 = ( x–3)9130 Section 2.7 Instructor’s Resource Manual

37. a.38.39.b.2xy′ + y + 3y y′= 02y′ ( x+ 3 y ) = – yyy′ = –2x + 3y⎛ – y ⎞ ⎛ – y ⎞2xy′′ + + + 3y y′′⎜ 2 2x 3y ⎟ ⎜ x 3y⎟⎝ + ⎠ ⎝ + ⎠2⎛ – y ⎞+ 6y = 0⎜x2+ 3 y⎟⎝ ⎠32 2y6yxy′′ + 3 y y′′– + = 02 2 2x+ 3 y ( x+3 y )2 2y6yy′′ ( x+ 3 y ) = –x + 3 y ( x + 3 y )2 2xyy′′ ( x+ 3 y ) =2 2( x+3 y )2xyy′′ =( x+3 y )23 x –8yy′ = 023xy′ =8y2 36 x–8( yy′′ + ( y′) ) = 022⎛3x⎞6 x–8 yy′′ –8 = 0⎜ 8y⎟⎝ ⎠49x6 x–8 yy′′ – = 028y2 448xy− 9x= 8yy′′28y2 448 xy – 9xy′′ =364y2 22( x y′ + 2 xy) –12y y′= 02 22 x y′ –12 y y′=–4xy2xyy′ =6 y – x2 2232 2 22 2 22( x y′′ + 2xy′ + 2xy′ + 2 y) –12[ y y′′ + 2 y( y′) ] = 02 2 22x y′′ − 12y y′′ =−8xy′ − 4y+24 y( y′)2 2 32 2 16x y 96x yy′′ (2 x –12 y ) =− – 4y+2 2 2 2 26 y – x (6 y − x )4 2 3 52 2 12x y+ 48x y −144yy′′ (2 x –12 y ) =2 2 2(6 y – x )5 4 2 32 2 72y −6x y−24x yy′′ (6 y – x ) =2 2 2(6 y – x )5 4 2 372y −6x y−24x yy′′ =2 2 3(6 y – x )At (2, 1),40. 2x+ 2yy′= 041.2xxy′ = – = –2yy−120y′′ = = − 15822+ 2[ yy′′ + ( y′) ] = 02⎛ x ⎞2+ 2yy′′+ 2 ⎜– ⎟ = 0⎝ y ⎠22x2yy′′ =−2−2y2 2 21 x y + xy′′ =− − =−y 3 3y yAt (3, 4),25y′′ =−642 23x + 3y y′ = 3( xy′+ y)2 2y′ (3 y –3 x) = 3 y–3x2y–xy′ =2y – x⎛3 3⎞At ⎜ , ⎟,y′ = –1⎝2 2⎠Slope of the normal line is 1.3 3Normal line: y – 1 ⎛ ⎞= ⎜x– ⎟;y = x2 ⎝ 2⎠This line includes the point (0, 0).42. xy′ + y = 0y′ = – yx2x− 2yy′= 0xy′ =yThe slopes of the tangents are negativereciprocals, so the hyperbolas intersect at rightangles.Instructor’s Resource Manual Section 2.7 131

43. Implicitly differentiate the first equation.4x+ 2yy′= 02y′ = – xyImplicitly differentiate the second equation.2yy′ = 42y′ =ySolve for the points of intersection.22x+ 4x= 622( x + 2 x– 3) = 0(x + 3)(x – 1) = 0x = –3, x = 1x = –3 is extraneous, and y = –2, 2 when x = 1.The graphs intersect at (1, –2) and (1, 2).At (1, –2): m1 = 1, m2= –1At (1, 2): m1 = –1, m2= 144. Find the intersection points:x + y = 1 → y = 1−x2 2 2 2( x )2 2− 1 + y = 12 2( x ) ( x )− 1 + 1− = 12 21x − 2x+ 1+ 1− x = 1 ⇒ x =2⎛1 3⎞ ⎛1 3⎞Points of intersection:⎜, and , –2 2 ⎟ ⎜ 2 2 ⎟⎝ ⎠ ⎝ ⎠Implicitly differentiate the first equation.2x+ 2yy′= 0y′ = – xyImplicitly differentiate the second equation.2( x–1) + 2yy′= 01– xy′ =y⎛ 1 3 ⎞ 1 1⎜2 2 ⎟= =⎝ ⎠ 3 3At ⎜ , ⎟: m1 – , m21+1 23 3 3233 3πtanθ= = = 3 → θ =1+ −31 1( )( )⎛ 1 3 ⎞, – : 1 , –1⎜2 2 ⎟= =⎝ ⎠ 3 3−1−1−23 3 3tanθ= = = − 31 1 21 + –At ⎜ ⎟ m1 m2θ =2π3( )( )3 3345.2 2x – x(2 x) + 2(2 x) = 2827x = 282x = 4x = –2, 2Intersection point in first quadrant: (2, 4)y′ 1 = 22 x– xy2′ – y+ 4yy2′= 0y2(4 ′ y– x) = y– 2xy– 2xy2′ =4 y–xAt (2, 4): m1 = 2, m2= 00–2–1tan θ = = –2; θ =π+ tan (–2) ≈ 2.0341 + (0)(2)2 2 2 246. The equation is mv – mv0 = kx0– kx .Differentiate implicitly with respect to t to getdv dx dx2 mv = –2 kx . Since v = this simplifiesdt dtdtdvto 2 mv –2kxvdt = or dvm – kxdt = .47.2 2x – xy+ y = 16 , when y = 0,2x = 16x = –4, 4The ellipse intersects the x-axis at (–4, 0) and(4, 0).2 x– xy′ – y+ 2yy′= 0y′ (2 y– x) = y– 2xy– 2xy′ =2 y–xAt (–4, 0), y′ = 2At (4, 0), y′ = 2Tangent lines: y = 2(x + 4) and y = 2(x – 4)132 Section 2.7 Instructor’s Resource Manual

48.2 dx2 dxx + 2 xy –2 xy– y 0dy dy=dx 2 2(2 xy– y ) = 2 xy– x ;dy2dx 2 xy – x=dy 22 xy–y22 xy – x= 0 if x(2y – x) = 0, which occurs22 xy – yxwhen x = 0 or y = . There are no points on22 2xx y– xy = 2 where x = 0. If y = , then22 3 3 32 ⎛2 x ⎞ ⎛– x ⎞= x ⎜ ⎟ x⎜ ⎟ = x –x =x⎝2⎠ ⎝2⎠2 4 42y = = 1.2The tangent line is vertical at (2, 1).so x = 2,49. 2x+ 2y dy = 0; dy = –xdx dx y0The tangent line at ( x0, y 0)has slope – xy,0hence the equation of the tangent line isx0y– y0 = – ( x– x0)which simplifies toy00 0 – (2020 ) 02 2yy + xx x + y = or yy0 + xx0 = 1since ( x0, y 0)is on x + y = 1. If (1.25, 0) ison the tangent line through ( x0, y 0), x 0 = 0.8.Put this intox2 2+ y = 1 to get y 0 = 0.6, sincey 0 > 0. The line is 6y + 8x = 10. When x = –2,13y = , so the light bulb must be 13 units high.332.8 Concepts Reviewdu1. ; t = 2dt2. 400 mi/hr3. negative4. negative; positiveProblem Set 2.81.2.3.4.5.3 dxV = x ; = 3dtdV 2 dx= 3xdt dtdV 2When x = 12, = 3(12) (3) = 1296 in. 3 /s.dt4 3 dVV = π r ; = 33 dtdV 2 dr= 4πrdt dt2 drWhen r = 3, 3= 4 π (3) .dtdr 10.027dt = 12π≈ in./s2 2 2 dxy = x + 1; = 400dtdy dx2y= 2xdt dtdy x dx= mi/hrdt y dtdy 5When x = 5, y = 26, = (400)dt 26≈ 392 mi/h.1 2 r 3 3hV = π r h; = ; r =3 h 10 102 31 ⎛3h⎞3πh dVV = π ⎜ ⎟ h = ; = 3, h = 53 ⎝10⎠100 dt2dV 9πh dh=dt 100 dt29 π(5)dhWhen h = 5, 3 =100 dtdh 40.42dt = 3π≈ cm/s2 2 2 dx dys = ( x + 300) + y ; 300, 400,dt= dt=2s ds = 2( x+ 300) dx + 2ydydt dt dts ds = ( x+ 300)dx + ydydt dt dtWhen x = 300, y = 400, s = 200 13 , sods200 13 (300 300)(300) 400(400)dt = + +ds471dt ≈ mi/hInstructor’s Resource Manual Section 2.8 133

6.7.2 2 2 dyy = x + (10) ; = 2dtdy dx2y= 2xdt dtWhen y = 25, x ≈ 22.9, sodx y dy 25= ≈ (2) ≈ 2.18 ft/sdt x dt 22.92 2 2 dx20 = x + y ; = 1dtdx dy0= 2x+ 2ydt dtWhen x = 5, y = 375 = 5 15 , sody x dx 5= – = – (1) ≈ –0.258 ft/sdt y dt 5 15The top of the ladder is moving down at0.258 ft/s.11.12.hx 40 xV = (20); = , x = 8h2 5 h2 dVV = 10 h(8 h) = 80 h ; = 40dtdV dh= 160hdt dtWhen h = 3, 40 = 160(3) dhdtdh 1dt = 12ft/min2 dxy = x – 4; = 5dtdy 1 dx x dx= (2 x)=dt 2 22 x –4 dt x –4 dtdy 3 15When x = 3,(5) 6.7dt = = 23 –4 5≈ units/sdV8. = –4 ft 3 /h;2 dhV =π hr ; = –0.0005 ft/hdtdt2 V –1 dA –1 dV V dhA =π r = = Vh , so = h – .hdt dt 2h dt2When h = 0.001 ft, V =π (0.001)(250) = 62.5πdAand 1000(–4) –1,000,000(62.5 )(–0.0005)dt = π= –4000 + 31,250π ≈ 94,175 ft 2 /h.(The height is decreasing due to the spreading ofthe oil rather than the bacteria.)9.1 2 d rV = π r h; h = = , r = 2h3 4 21 2 4 3 dVV = π (2 h) h = π h ; = 163 3 dtdV 2 dh= 4πhdt dt2 dhWhen h = 4, 16 = 4 π (4)dtdh 10.0796dt = 4π≈ ft/s13.14.2 drA=π r ; = 0.02dtdA dr= 2πrdt dtdAWhen r = 8.1, 2 (0.02)(8.1) 0.324dt = π = π≈ 1.018 in. 2 /s2 2 2 dx dys = x + ( y + 48) ; 30, 24dt= dt=2s ds = 2x dx + 2( y+48)dydt dt dts ds = x dx + ( y+48)dydt dt dtAt 2:00 p.m., x = 3(30) = 90, y = 3(24) = 72,so s = 150.ds(150) 90(30) (72 48)(24)dt = + +ds 558037.2dt = 150= knots/h10.2 2 2 dxy = x + (90) ; = 5dtdy dx2y= 2xdt dtWhen y = 150, x = 120, sody x dx 120= = (5) = 4 ft/sdt y dt 150134 Section 2.8 Instructor’s Resource Manual

15. Let x be the distance from the beam to the pointopposite the lighthouse and θ be the anglebetween the beam and the line from thelighthouse to the point opposite.x dθtan θ = ; = 2(2 π ) = 4π rad/min,1 dt2 dθdxsec θ =dt dt1 –1 1 2 5At x = , θ = tan and sec θ = .2 2 4dx 5 (4 ) 15.71dt = 4π ≈ km/min16.4000tanθ =x2 dθ4000 dxsec θ =−dt 2x dt1 dθ1 4000When θ = , = and x = ≈ 7322.2 dt 10 tan2⎛ ⎞ ⎡ ⎤dx 2 1 1 (7322)≈sec ⎜ ⎟⎢−⎥dt 2 ⎝ 10 ⎠⎢⎣4000 ⎥⎦≈ –1740 ft/s or –1186 mi/hThe plane’s ground speed is 1186 mi/h.17. a. Let x be the distance along the ground fromthe light pole to Chris, and let s be thedistance from Chris to the tip of his shadow.By similar triangles, 6 30 x= , so s =s x + s 4ds 1 dx dxand = . = 2 ft/s, hencedt 4 dt dtds 1= ft/s no matter how far from the lightdt 2pole Chris is.b. Let l = x + s, thendl dx ds 1 5= + = 2 + = ft/s.dt dt dt 2 2c. The angular rate at which Chris must lift hishead to follow his shadow is the same as therate at which the angle that the light makeswith the ground is decreasing. Let θ be theangle that the light makes with the ground atthe tip of Chris' shadow.6 2 dθ6 dstanθ = so sec θ = – ands dt 2s dt2dθ6cos θ ds ds 1= – .dt 2s dt dt = 2ft/sπWhen s = 6, θ = , so4121( ) 26dθ 2 ⎛1⎞1= – – .dt 2 ⎜ ⎟=6 ⎝2⎠24Chris must lift his head at the rate of124 rad/s.18. Let θ be the measure of the vertex angle, a be themeasure of the equal sides, and b be the measureθof the base. Observe that b= 2asin and the 2θheight of the triangle is a cos .21⎛ θ ⎞⎛ θ ⎞ 1 2A= ⎜2asin ⎟⎜acos ⎟=a sinθ2⎝ 2⎠⎝ 2⎠21 2dθ1A = (100) sinθ= 5000sin θ;=2 dt 10dAdθ= 5000cosθdtdtπ dA ⎛ π⎞⎛ 1 ⎞When θ = , = 5000⎜cos ⎟⎜ ⎟=250 36 dt ⎝ 6⎠⎝10⎠2≈ 433 cm min .19. Let p be the point on the bridge directly abovethe railroad tracks. If a is the distance between pdaand the automobile, then = 66 ft/s. If l is thedtdistance between the train and the point directlydlbelow p, then = 88 ft/s. The distance from thedt2 2train to p is 100 + l , while the distance fromp to the automobile is a. The distance betweenthe train and automobile is22 2 2 2 2 2D = a +⎛100 l⎞⎜ + ⎟ = a + l + 100 .⎝ ⎠dD 1 ⎛ da dl ⎞= ⋅ ⎜2a+ 2l⎟dt 2 2 22 a + l + 100 ⎝ dt dt ⎠ada+ ldldt dt=. After 10 seconds, a = 6602 2 2a + l + 100and l = 880, sodD 660(66) + 880(88)= ≈110ft/s.dt 2 2 2660 + 880 + 100Instructor’s Resource Manual Section 2.8 135

20.21.22.1 2 2hV = πh⋅ ( a + ab+ b ); a = 20, b = + 20,3 421 ⎛h⎞V = π h 400 + 5h+ 400 + + 10h+4003 ⎜16 ⎟⎝⎠31 ⎛2 h ⎞= π 1200h+ 15h+3 ⎜16⎟⎝⎠dV 21 ⎛31200 30h ⎞= π + h +dhdt 3 ⎜16 ⎟⎝⎠dtdVWhen h = 30 and = 2000,dt1 ⎛675 ⎞dh3025πdh2000 = π ⎜1200 + 900 + ⎟ =3 ⎝4 ⎠ dt 4 dtdh 3200.84dt = 121π≈ cm/min.2 ⎡ h⎤dVV =π h ⎢r – ; = –2, r = 83⎥⎣ ⎦ dt3 32 h 2 hV rh – ππ=π = 8 π h –3 3dV 2= 16 πhdh – π hdhdt dt dt2When h = 3, –2 = dh [16 π(3)– π (3) ]dtdh –2–0.016dt = 39π≈ft/hr2 2 2s = a + b − 2abcos θ;d11a = 5, b = 4,θ 2 – π πdt 6 62s = 41– 40cos= π = rad/hθds dθ2s = 40sinθdt dtπAt 3:00, θ = and s = 41 , so22 41 ds40sin ⎛π⎞⎛11π⎞220π= ⎜ ⎟⎜ ⎟=dt ⎝ 2 ⎠⎝ 6 ⎠ 3ds18dt ≈ in./hr24.48 10 + s=264 –16ts210 t – 40s = .21– t2 2ds 20 t(1– t ) – (10 t – 40)(–2 t)60t== –dt2 22 2(1 – t )(1 – t )By similar triangles,(for t > 1), sods 120The ball hits the ground when t = 2, – .dt = 9The shadow is moving 120 13.339 ≈ ft/s.2 ⎛ h ⎞V = π h ⎜r – ⎟; r = 20⎝ 3 ⎠2⎛h ⎞ 2 π 3V =π h ⎜20 – ⎟= 20πh − h⎝ 3⎠3dV2 dh= (40 πh−πh )dtdtdhAt 7:00 a.m., h = 15, 3,dt ≈ − sodV2= (40 π(15) −π(15) )( −3) ≈ −1125π≈ −3534.dtWebster City residents used water at the rate of2400 + 3534 = 5934 ft 3 /h.25. Assuming that the tank is now in the shape of anupper hemisphere with radius r, we again let t bethe number of hours past midnight and h be theheight of the water at time t. The volume, V, ofwater in the tank at that time is given by2V = π r 3 − π ( r− h) 2 ( 2r+h)3 316000and so V = π − π (20 − h) 2 ( 40 + h)3 3from whichdV π 2 2(20 ) dh π=− − h + (20 − h) ( 40+h)dhdt 3 dt 3dtdVAt t = 7 , ≈−525π1649dtThus Webster City residents were using water atthe rate of 2400 + 1649 = 4049 cubic feet perhour at 7:00 A.M.23. Let P be the point on the ground where the ballhits. Then the distance from P to the bottom ofthe light pole is 10 ft. Let s be the distancebetween P and the shadow of the ball. The heightof the ball t seconds after it is dropped is264 –16 t .26. The amount of water used by Webster City canbe found by:usage = beginning amount + added amount− remaining amountThus the usage is≈ π(20) (9) + 2400(12) −π(20) (10.5) ≈ 26,915 ftover the 12 hour period.2 2 3136 Section 2.8 Instructor’s Resource Manual

dx27. a. Let x be the distance from the bottom of the wall to the end of the ladder on the ground, so = 2 ft/s. Let ydty 18216be the height of the opposite end of the ladder. By similar triangles, = , so y = .12 22144 + x144 + xdy 216 dx 216xdx= – 2x= –dt2 3/22(144 + x ) dt (1442 )3/2+ x dtWhen the ladder makes an angle of 60° with the ground, x = 4 3 anddy 216(4 3)– 2 –1.125dt = 3/2(144 + 48)⋅ =ft/s.b.2d y d ⎛ 216x dx⎞2d ⎛ 216x ⎞dx 216x d x= – = – –2 dt (1442 )3/2dt⎜+ x dt ⎟⋅2 3/2 2 3/2 2⎝⎠ dt ⎜(144 x )⎟⎝ + ⎠ dt (144 + x ) dt2dx d x= 2, = 0, thusdt 2dt2 ⎡2 3/2 32–216(144 ) dx 216 ( 2 ) 144 (2 ) dxd y+ x + x + x x⎤⎢dtdt ⎥2 2 3Since=dt ⎢(144 + x )⎥⎢⎣⎥⎦2 2 2 22–216(144 + x ) + 648x ⎛dx⎞432 x – 31,104 ⎛dx⎞=2 5/2 ⎜ ⎟ =(144 + x ) ⎝ dt ⎠2 5/2 ⎜ ⎟(144 + x ) ⎝ dt ⎠When the ladder makes an angle of 60° with the ground,2d y 432⋅48 – 31,104 (2)2= ≈ –0.08 ft/s 22 5/2dt (144 + 48)dxdt28. a. If the ball has radius 6 in., the volume of thewater in the tank is332 πh4 ⎛1⎞V = 8 πh– – π ⎜ ⎟3 3 ⎝2⎠32 πhπ= 8 π h – –3 6dV 2= 16 πhdh – π hdhdt dt dtThis is the same as in Problem 21, so dh isdtagain –0.016 ft/hr.b. If the ball has radius 2 ft, and the height ofthe water in the tank is h feet with 2≤ h ≤ 3,the part of the ball in the water has volume24 3 2 ⎡ 4– h⎤(6– h)h ππ(2) – π (4 – h) 2 – .3 ⎢ =3 ⎥⎣ ⎦ 3The volume of water in the tank is3 22 πh(6 – h)h π 2V = 8 π h – – = 6hπ3 3dV dh= 12hπdt dtdh 1 dV=dt 12hπdtWhen h = 3,dh 1 (–2) –0.018dt = 36π≈ft/hr.29.dV 2= k(4 π r )dta.4 3V = π r3dV 2 dr= 4πrdt dt2 2k(4 π r ) = 4πrdrkdt =drdtb. If the original volume was V 0 , the volume8after 1 hour is V 0.The original radius273was r 30 = V0while the radius after 14πhour is r 8 3 231 = 0 0 .27 V ⋅ =4π3r Since dr isdtdr 1constant, = – r0unit/hr. The snowballdt 3will take 3 hours to melt completely.Instructor's Resource Manual Section 2.8 137

30. PV = kdV dPP + V = 0dt dtdPAt t = 6.5, P ≈ 67, –30,dt ≈ V = 300dV V dP 300= – = – (–30) ≈ 134 in. 3 /mindt P dt 6731. Let l be the distance along the ground from thebrother to the tip of the shadow. The shadow iscontrolled by both siblings when 3 = 5 orl l + 4l = 6. Again using similar triangles, this occursy 6when = , so y = 40. Thus, the girl controls20 3the tip of the shadow when y ≥ 40 and the boycontrols it when y < 40.Let x be the distance along the ground from thedxlight pole to the girl. –4dt =When y ≥ 40, 20 5 4= or y = x .y y–x 3When y < 40, 20 320= or y = ( x+4).y y– ( x+4) 17x = 30 when y = 40. Thus,⎧ 4x if x≥30⎪ 3y = ⎨⎪ 20 ( x+ 4) if x

13.20.3 1 –2 / 3 1y = x; dy = x dx = dx;3 3 23 xx = 27, dx = –0.091dy = (–0.09) ≈ –0.00333 23 (27)3 326.91 ≈ 27 + dy = 3 – 0.0033 = 2.996714. a.b.3 3Δ y = (1.5) – (0.5) = 3.253 3Δ y = (–0.25) – (–1) = 0.98437521.22.4 3V = π r ; r = 5, dr = 0.12532 2dV = 4π r dr = 4 π(5) (0.125) ≈ 39.27 cm 33 3V = x ; x = 40, dx = 0.52 3 2dV = 3x dx = 3( 40) (0.5) ≈ 17.54 in. 315. a.b.16. a.b.17. a.18.b.1 1 1Δ y = – = –1.5 1 31 1Δ y = + = –0.3–1.25 22 2Δ y = [(2.5) – 3] – [(2) – 3] = 2.25dy = 2xdx = 2(2)(0.5) = 22 2Δ y = [(2.88) – 3] – [(3) – 3] = –0.7056dy = 2xdx = 2(3)(–0.12) = –0.724 4Δ y = [(3) + 2(3)] – [(2) + 2(2)] = 673 3dy = (4x + 2) dx = [4(2) + 2](1) = 344 4Δ y = [(2.005) + 2(2.005)] – [(2) + 2(2)]≈ 0.17063 3dy = (4x + 2) dx = [4(2) + 2](0.005) = 0.171y = x; dy = dx; x = 400, dx = 22 x1dy = (2) = 0.052 400402 ≈ 400 + dy = 20 + 0.05 = 20.0523.24.4 3V = π r ; r = 6 ft = 72in., dr = –0.332 2dV = 4π r dr = 4 π(72) (–0.3) ≈ –19,5434 3V ≈ π (72) –19,54333 3≈1,543,915 in ≈893 ftV =π r 2 h; r = 6ft = 72in., dr =− 0.05,h = 8ft = 96in.dV = 2π rhdr = 2 π(72)(96)( −0.05) ≈ − 2171in.About 9.4 gal of paint are needed.25. C = 2πr ; r = 4000 mi = 21,120,000 ft, dr = 2dC = 2π dr = 2 π(2) = 4π≈ 12.6 ft26. T = 2 πL; L = 4, dL = –0.0332dT =2π132 π⋅ ⋅ =32L2 L32πdT = (–0.03) ≈ –0.008332(4)The time change in 24 hours is(0.0083)(60)(60)(24) ≈ 717 sec319.1y = x; dy = dx; x = 36, dx = –0.12 x1dy = (–0.1) ≈ –0.00832 3635.9 ≈ 36 + dy = 6 – 0.0083 = 5.991727.4 3 4 (10) 3r4189V = π = π ≈3 32 2dV = 4π r dr = 4 π(10) (0.05) ≈ 62.8 Thevolume is 4189 ± 62.8 cm 3 .The absolute error is ≈ 62.8 while the relativeerror is 62.8 / 4189 ≈ 0.015 or 1.5% .Instructor’s Resource Manual Section 2.9 139

28.29.2 2V =π r h =π(3) (12) ≈ 339dV = 24π rdr = 24 π(3)(0.0025) ≈ 0.565The volume is 339 ± 0.565 in. 3The absolute error is ≈ 0.565 while the relativeerror is 0.565/ 339 ≈ 0.0017 or 0.17% .2s = a + b 2 – 2abcosθ2 2= 151 + 151 – 2(151)(151) cos 0.53 ≈ 79.097s =45,602 – 45,602cosθ1ds = ⋅ 45,602sinθdθ2 45, 602 – 45, 602 cosθ=22,801sinθdθ45,602 – 45,602cosθ22,801sin 0.53= (0.005) ≈ 0.72945,602 – 45,602cos 0.53s ≈ 79.097 ± 0.729 cmThe absolute error is ≈ 0.729 while the relativeerror is 0.729 / 79.097 ≈ 0.0092 or 0.92% .33. Using the approximationf ( x+ Δx) ≈ f( x) + f '( x)Δ xwe let x = 3.05 and Δ x =− 0.05 . We can rewritethe above form asf ( x) ≈ f( x+Δx) − f '( x)Δ xwhich givesf(3.05) ≈ f(3) − f '(3.05)( −0.05)1= 8 + (0.05) = 8.0125434. From similar triangles, the radius at height h is2 .5 h Thus, 1 2 4 3V = π r h= π h , so3 754 2dV = π h dh.h = 10, dh = –1:254 (100)( 1) 50cm3dV = π − ≈ −253 3The ice cube has volume 3 = 27cm , so there isroom for the ice cube without the cupoverflowing.30.1 1A= absin θ = (151)(151)sin 0.53 ≈ 5763.332 222,801A= sin θθ ; = 0.53, dθ=0.005222,801dA = (cos θ ) dθ222,801= (cos 0.53)(0.005) ≈ 49.182A ≈ 5763.33 ± 49.18 cm 2The absolute error is ≈ 49.18 while the relativeerror is 49.18 / 5763.33 ≈ 0.0085 or 0.85% .2= 3 – 2 + 11; = 2, = 0.00131. y x x x dxdy = (6x – 2)dx = [6(2) – 2](0.001) = 0.012d y= 6, so with Δx = 0.001,2dx12Δy– dy ≤ (6)(0.001) = 0.000003232. Using the approximationf ( x+Δx) ≈ f( x) + f '( x)Δ xwe let x = 1.02 and Δ x = − 0.02 . We can rewritethe above form asf ( x) ≈ f( x+Δx) − f '( x)Δ xwhich givesf(1.02) ≈ f(1) − f '(1.02)( −0.02)= 10 + 12(0.02) = 10.2435.2 4 3= π + π32 4 3= 100 π + π ;3= 10, = 0.12= (200π + 4 π )V r h rV r r r drdV r r dr= (2000π+ 400 π )(0.1) = 240π≈ 754 cm 3dm36. The percent increase in mass is .m2–3/ 2m ⎛0 v ⎞ ⎛ 2v⎞dm = – 1– – dv2 ⎜ 2 ⎜ 2 ⎟c ⎟⎝ ⎠ ⎝ c ⎠2–3/ 2mv⎛0 v ⎞= 1–dv2 2c ⎜ c ⎟⎝ ⎠2–12dm v ⎛ v ⎞ v ⎛ c ⎞= 1–dv =dvm 2 2 2 2 2c ⎜ c ⎟ c ⎜c − v ⎟⎝ ⎠ ⎝ ⎠v= dv2 2c − vv = 0.9c, dv = 0.02cdm 0.9c0.018= (0.02 c) = ≈0.095m 2 2c − 0.81c0.19The percent increase in mass is about 9.5.140 Section 2.9 Instructor’s Resource Manual

37.2f( x) = x ; f '( x) = 2 x; a = 2The linear approximation is thenLx ( ) = f(2) + f'(2)( x−2)= 4+ 4( x− 2) = 4x−439. hx ( ) = sin x; h'( x) = cos x; a=0The linear approximation is thenLx ( ) = 0+ 1( x− 0) = x38.2 2g( x) = x cos x; g'( x) = − x sinx+2xcosxa = π /2The linear approximation is then2⎛π⎞ ⎛ π ⎞Lx ( ) = 0+−⎜ ⎟ ⎜x−⎟⎝ 2⎠ ⎝ 2⎠2 3π π=− x +4 82π ⎛ π ⎞Lx ( ) = 0+− ⎜x−⎟4 ⎝ 2⎠2 3π π=− x +4 840. F( x) = 3x+ 4; F'( x) = 3; a = 3The linear approximation is thenLx ( ) = 13 + 3( x− 3) = 13 + 3x−9= 3x+ 4Instructor’s Resource Manual Section 2.9 141

2f x = 1 −x;41. ( )1 −1/2f′ x = 1 −x ( − 2 x )2−x= , a = 021−x2( ) ( )The linear approximation is thenL x = 1+ 0 x− 0 = 1( ) ( )x= ;1 − x42. g( x) 2(2−x ) −x( − x)2+ x2( 1−x)2( 1−x)1 2 1 1g' ( x)= = , a =2 22The linear approximation is then2 20 ⎛ 1 ⎞ 20 4L ( x)= + ⎜ x − ⎟ = x −3 9 ⎝ 2 ⎠ 9 945. f ( x) = mx + b;f ( x) = m′The linear approximation is thenL( x) = ma + b + m( x − a)= am + b + mx − ma= mx + bf x = L x( ) ( )46. ( ) ( ) ( )L x − f x = a + x−a − xax a x− 2 a x + a= − + =2 2 a12xa 2( x − a) 2= ≥02 a47. The linear approximation to f ( x ) at a isLx ( ) = f( a) + f'( a)( x−a)2= a + 2 a( x−a)2= 2ax−aThus,2 2f ( x) − L( x) = x − 2ax−a( )2 2= x − 2ax+a2= ( x−a)≥ 0α′α −148. f ( x) = ( 1+x) , f ( x) = α ( 1+x) , a = 0The linear approximation is thenL x = 1 + α x = αx+( ) ( ) 1y′43. h ( x) = x sec x;h ( x) = sec x + x sec x tan x,a = 0The linear approximation is thenL x = 0 + 1 x − 0 =( ) ( ) x5−5 5 x−5α =− 2y′44. G ( x) = x + sin 2x;G ( x) = 1+2 cos 2x, a = π / 2The linear approximation is thenπ ⎛ π ⎞L x = + −1⎜ x − ⎟ = −x+2 ⎝ 2 ⎠( ) ( ) π5−5 5 x−5α =− 1142 Section 2.9 Instructor’s Resource Manual

5y49. a. lim ε ( h) = lim ( f ( x+ h) − f ( x) − f′( x)h)h→0 h→0( ) ( ) ′( )= f x − f x − f x 0=0−5 5 xb.( h) ⎡ f ( x+ h) − f ( x)εlim = lim ⎢− f ′h→0h ⎣ h= f′ x − f′x = 0( ) ( )( x)⎤⎥⎦−5α = − 0.5y2.10 Chapter Review5−5 5 xConcepts Test3 21. False: If f ( x) = x , f '( x) = 3xand thetangent line y = 0 at x = 0 crosses thecurve at the point of tangency.−5yα = 02. False: The tangent line can touch the curveat infinitely many points.3. True:3mtan 4 x ,= which is unique for eachvalue of x.54. False: mtan = – sin x,which is periodic.−5 5 x−5yα = 0.55. True: If the velocity is negative andincreasing, the speed is decreasing.6. True: If the velocity is negative anddecreasing, the speed is increasing.7. True: If the tangent line is horizontal, theslope must be 0.58. False:2 2f ( x) = ax + bgx , ( ) = ax + c,≠ c. Then f ′( x) 2 ax g′( x),bf(x) ≠ g(x).= = but−5 5 x−55y−5 5 xα = 19. True: Dxf( g( x)) = f′ ( g( x)) g′( x);sinceg(x) = x, g′ ( x) = 1, soDxf( g( x)) = f′( g( x)).10. False: Dxy = 0 because π is a constant, nota variable.11. True: Theorem 3.2.A12. True: The derivative does not exist when thetangent line is vertical.13. False: ( f ⋅ g)( ′ x) = f( x) g′ ( x) + g( x) f′( x)−5α = 214. True: Negative acceleration indicatesdecreasing velocity.Instructor’s Resource Manual Section 2.10 143

15. True: If16. False:3f ( x) x g( x),= then3 2Dxf( x) = x g′( x) + 3 x g( x)= x2 [ xg′( x) + 3 g( x)].Dxy23 x ;= At (1, 1):2m tan = 3(1) = 3Tangent line: y – 1 = 3(x – 1)17. False: Dxy = f( x) g′ ( x) + g( x) f′( x)2Dxy = f( x) g′′ ( x) + g′ ( x) f′( x)+ g( x) f′′ ( x) + f′ ( x) g′( x)= f ( xg ) ′′( x) + 2 f′ ( xg ) ′( x) + f′′( xgx ) ( )18. True: The degree of19. True:20. True:25Dxy = 0.3 8y ( x x)= + is 24, sonn–1f ( x) = ax ; f′( x)= anxf ( x) g( x) f′ ( x)– f( x) g′( x)( ) ( )Dx= gx 2g x21. True: h′ ( x) = f( x) g′ ( x) + g( x) f′( x)h′ () c = f() c g′ () c + g() c f′() c= f(c)(0) + g(c)(0) = 022. True:23. True:⎛π⎞ sin x – sinf ′ ⎜ ⎟=lim⎝2 ⎠ x –x→ π2x→ π2sin x –1= limx –π22 2D ( kf)kD f= andπ22 2 2D ( f + g)= D f + D gπ( )24. True: h′ ( x) = f′ ( g( x)) ⋅ g′( x)h′ () c = f′ ( g()) c ⋅ g′() c = 025. True: ( f g) ′(2) = f′ ( g(2)) ⋅g′(2)= f′ (2) ⋅ g′(2) = 2⋅ 2 = 426. False: Consider f ( x) = x.The curvealways lies below the tangent.27. False: The rate of volume change dependson the radius of the sphere.dr28. True: c = 2πr ; 4dt =dc dr= 2π = 2 π (4) = 8πdt dt229. True: Dx(sin x) = cos x;2Dx(sin x ) = – sin x ;3Dx(sin x ) = – cos x ;4Dx(sin x ) = sin x ;5Dx(sin x ) = cos x30. False: Dx(cos x) = – sin x;2Dx(cos x ) = – cos x ;3Dx(cos x ) = sin x ;4 3Dx(cos x) = Dx[ Dx(cos x)] = Dx(sin x)1+3 1Since Dx(cos x) = Dx(sin x),n+3 (cos )nDxx = Dx(sin x ).31. True:32. True:33. True:tan x 1 sin xlim = limx→0 3x 3 x→0xcosx1 1= ⋅ 1 =3 3ds 2v = = 15t+ 6 which is greaterdtthan 0 for all t.4 3V = π r3dV 2 dr= 4πrdt dtdVIf = 3, thendtdr0.dt >2d r 3= –dt 2 πr2 3drdt34. True: When h > r, then35. True:36. False:drdtso4 32V = π r , S = 4πr32dV = 4π r dr = S ⋅ dr3= so24 π r2d r< 02dt2d h> 02dtIf Δr = dr, then dV = S ⋅Δ r4dy = 5 x dx,so dy > 0 when dx > 0,but dy < 0 when dx < 0.37. False: The slope of the linear approximationis equal tof '( a) = f '(0) =− sin(0) = 0 .144 Section 2.10 Instructor’s Resource Manual

Sample Test Problems1. a.3 33( x + h) – 3xf′ ( x) = limh →0h2 2 39x h+ 9xh + 3h= limh→0hh→02 2 2= lim (9x + 9xh + 3 h ) = 9xb.5 5[2( x + h) + 3( x+ h)] – (2x + 3 x)f′ ( x) = limh hh→0→04 3 2 2 3 4= lim (10 x + 20x h+ 20x h + 10xh + 2h+ 3) = 10x+ 34 3 2 2 3 4 510x h+ 20x h + 20x h + 10xh + 2h + 3h= limh→0h4c.1 – 13( x+h) 3x⎡ h ⎤ 1f′ ( x) = lim = lim ⎢– ⎥h→0 h h→0⎣3( x+h ) x⎦h⎛ 1 ⎞ 1= lim – ⎜ ⎟=–⎝3( ) ⎠ 3h→0xx+h 2xd.e.f.g.h.2. a.⎡⎛1 1 ⎞1⎤⎡ 2 23x + 2–3( x+h) –2 1⎤f′ ( x) = lim ⎢ –⎥ = limh→0⎜ 2 2⎢ 3( x+ h) + 2 3x+ 2⎟⎢⋅ ⎥⎣⎝⎠h2 2⎥ h→0⎦ ⎢⎣(3( x+ h ) + 2)(3 x + 2) h ⎥⎦⎡2–6 xh –3h1⎤–6 x– 3h 6x= lim ⎢⋅ ⎥ = lim = –h→0(3( )2 2)(32⎢⎣x+ h + x + 2) h ⎥⎦h→0(3( x+ h) + 2)(3x + 2) (3x+ 2)f′ ( x) = lim3( x + h) – 3xh →0h3h3= lim= limh→0h( 3x+ 3h + 3 x) h→03x+ 3h + 3x2 2 2 2( 3x + 3 h – 3 x)( 3x+ 3h + 3 x)= limh→0h( 3x+ 3h + 3 x)3=2 3xsin[3( x + h)] – sin 3xsin(3x + 3 h) – sin 3xf′ ( x) = lim= limh →0hh→0hsin 3x cos3h+sin 3hcos3 x– sin 3xsin 3 x(cos3 h–1)sin 3hcos3x= lim= lim+ limh→0hh→0 hh→0hcos3 h–1 sin 3hsin 3h= 3sin3xlim + cos3xlim = (3sin 3 x)(0) + (cos3 x)3 lim = (cos3 x)(3)(1) = 3cos3xh→0 3hh→0hh →03 h2 2( x+ h) + 5 – x + 5f′ ( x) = lim h →0h22xh+ h= limh→0 2 2h⎛( x+ h) + 5+ x + 5⎞⎜⎟⎝⎠⎛ 2 2 2 2( x h) 5 – x 5⎞⎛( x h) 5 x 5⎞⎜ + + + ⎟⎜ + + + + ⎟= lim⎝ ⎠⎝ ⎠h→0 2 2h⎛( x+ h) + 5+ x + 5⎞⎜⎟⎝⎠2x+h 2x x= lim= =h→02 2 2 2( x+ h) + 5+ x + 5 2 x + 5 x + 5cos[ π ( x + h)] – cos πxcos( π x +πh) – cos πxcos πx cos πh– sin πxsin πh– cos πxf′ ( x) = lim= lim= limh →0hh→0hh→0h⎛ 1–cosπh⎞ ⎛ sinπh⎞= lim ⎜– πcos πx⎟− lim ⎜πsinπx⎟ = (– π cos πx)(0) – ( πsin π x) = – πsinπ xh→0⎝ πh⎠ h→0⎝ πh⎠2 22 t –2x 2( t – x)( t+x)g′ ( x) = lim = limt→x t – x t→xt – x= 2lim( t + x ) = 2(2 x ) = 4 xt→x Instructor’s Resource Manual Section 2.10 145

.c.d.e.f.3 3( t + t)–( x + x)g′ ( x) = limt→xt – x2 2( t – x)( t + tx+ x ) + ( t – x)= limt→xt – x2 2lim( t tx x2= + + + 1) = 3x+ 1t→x 1 1t x– x – tg′ ( x) = lim = limt→x t – x t→xtx( t – x)–1 1= lim = –t→x tx 2x⎡⎛ 1 1 ⎞⎛ 1 ⎞⎤g′ ( x) = lim ⎢⎜ –t→x2 2 ⎟⎜ ⎟⎥⎣⎝t+ 1 x + 1⎠⎝t– x⎠⎦2 2lim x – t=t→x( 2 1)( 2t + x + 1)( t – x )lim –( x+t)( t – x)=t→x( 2 1)( 2t + x + 1)( t – x )–( x+t) 2x= lim = –t→x2 2 2 2( t + 1)( x + 1) ( x + 1)t – xg′ ( x) = limt→xt – x( t – x)( t + x)= limt → x ( t – x )( t + x )t – x1= lim= limt→x( t – x)( t + x)t→xt + x1=2 xsin πt– sin πxg′ ( x) = limt→xt – xLet v = t – x, then t = v + x and ast → x, v→0.sin πt – sin πx sin π ( v+ x) – sin πxlim= limt→x t – x v→0vsin πvcos π x+ sin πxcos πv– sin πx= limv→0v⎡ sin πvcos πv–1⎤= lim ⎢πcos π x +πsinπxv→0πvπv⎥⎣⎦=πcos πx ⋅ 1+πsin πx⋅ 0 =πcosπ xOther method:Use the subtraction formulaπ ( t+ x) π( t−x)sin πt– sin π x = 2cos sin2 2g.h.3 3t + C – x + Cg′ ( x) = limt→xt – x⎛ 3 3 3 3t C – x C⎞⎛t C x C⎞⎜ + + ⎟⎜ + + + ⎟= lim⎝ ⎠⎝ ⎠t→x3 3( t – x)⎛t + C + x + C⎞⎜⎟⎝⎠3 3t – x= limt→x3 3( t – x)⎛t + C + x + C⎞⎜⎟⎝⎠2 2 2t + tx+x 3x= lim=t→x3 3 3t + C + x + C 2 x + Ccos 2 t – cos 2xg′ ( x) = limt→xt – xLet v = t – x, then t = v + x and ast → x, v→0.cos 2 t – cos 2x cos 2( v+x) – cos 2xlim= limt→x t – x v→0vcos 2vcos 2 x– sin 2vsin 2 x– cos 2x= limv→0v⎡ cos 2 v–1 sin 2v⎤= lim ⎢2cos 2 x – 2sin 2xv→02v2v⎥⎣⎦= 2cos2x ⋅0–2sin2x⋅ 1 = –2sin2xOther method:Use the subtraction formulacos 2t− cos 2x =− 2sin( t+ x)sin( t−x).3. a. f(x) = 3x at x = 1b.c.3f ( x) = 4xat x = 2f ( x)3x= at x = 1d. f(x) = sin x at x = πe.f( x)4= at xxf. f(x) = –sin 3x at xg. f(x) = tan x ath.4. a.f( x)πx =41= at x = 5x3f ′(2) ≈ – 4b.3f ′(6)≈2146 Section 2.10 Instructor’s Resource Manual

5.c.d.e.6– 32 9V avg = =7–3 8d f 2 2( t ) = f ′( t )(2 t )dt2 84 f ′⎛ ⎞(4) ≈ 4⎜⎟=⎝ 3 ⎠ 3At t = 2,d 2[ f ( t )] = 2 f ( t ) f ′( t )dtAt t = 2,2 f(2) f′ (2)⎛ 3 ⎞≈ 2(2) ⎜– ⎟=–3⎝ 4 ⎠f. d ( f ( f ( t ))) = f ′( f ( t )) f ′( t )dtAt t = 2, f′ ( f(2)) f′ (2) = f′ (2) f′(2)3 3 9≈ ⎛ ⎜– ⎞⎛ ⎟⎜–⎞⎟ =⎝ 4⎠⎝ 4⎠165 4Dx (3 x ) = 15x11.12.13.14.2 3 2 2d ⎛4 x –2 ⎞ ( x + x)(8 x)–(4 x –2)(3x+ 1)=dx ⎜ 3 3 2x x ⎟⎝ + ⎠( x + x)4 2− 4x+ 10x+ 2=3 2( x + x)1Dt( t 2t+ 6) = t (2) + 2t+62 2t+ 6t= + 2t+ 62t+ 6d ⎛ 1 ⎞ d 2 –1/2⎟= ( x + 4)dx ⎜ 2 ⎟⎝ x + 4 ⎠dx1 2 –3/2= – ( x + 4) (2 x)2= –x2 3( x + 4)2d x –1 d 1 d 12 1x−= = = −dx 3 3 2x – x dx x dx 2x6.7.3 2 –2 2 –3Dx ( x –3 x + x ) = 3 x –6 x+(–2) x2 –3= 3 x –6 x–2x3 2 2Dz( z + 4z + 2 z) = 3z + 8z+215.3 2(sin + cos ) = cos + 3cos (– sin )D θ θ θ θ θ θ= cos θ – 3sinθ cos θ2 3(sin + cos )D θ θ θ23= – sin – 3[sin (2)(cos )(– sin ) + cos ]θ θ θ θ θ8.2⎛3 x – 5 ⎞ ( x + 1)(3) – (3 x– 5)(2 x)Dx⎜ 2 ⎟ =x2 2⎝ + 1 ⎠ ( x + 1)2− 3x+ 10x+3=2 2( x + 1)16.2 3= – sin + 6sin cos – 3cosθ θ θ θd 2 2 2[sin( t ) – sin ( t)] = cos( t )(2 t) – (2sin t)(cos t)dt2= 2cos( t t )–sin(2) t9.10.2⎛ 4t− 5 ⎞ (6t + 2 t)(4) – (4 t – 5)(12t+2)Dt⎜ 2 ⎟ =2 2⎝6t + 2 t⎠(6t + 2 t)2− 24t+ 60t+10=2 2(6t+ 2 t)2/3 2 –1/3Dx(3x+ 2) = (3x+2) (3)3–1/3= 2(3x+ 2)2 2/3 2 –4/3Dx(3x+ 2) = – (3x+2) (3)3= –2(3x+ 2)–4/317.18.2 2 2[sin( )] = cos( )(2 ) = 2 cos( )D θ θ θ θ θ θd 3 2(cos 5 x) = (3cos 5 x)(– sin 5 x)(5)dx2= –15cos 5xsin 5xInstructor’s Resource Manual Section 2.10 147

19.d 2[sin (sin( π θ))] = 2sin(sin( πθ))cos(sin( πθ))(cos( πθ))( π)= 2πsin(sin( πθ ))cos(sin( πθ))cos( π θ)dθd[sin (cos 4 t)] = 2sin(cos 4 t) cos(cos 4 t) (– sin 4 t)(4)= –8sin(cos 4 t)cos(cos 4 t)sin 4tdt220. ( )21.22.2 2D θ tan 3 θ = (sec 3 θ)(3) = 3sec 3θ2 2d ⎛ sin 3 x ⎞ (cos5 x )(cos3 x)(3) – (sin 3 x)(– sin 5 x )(10 x)⎜ 2 ⎟ =dx 2 2⎝cos5x ⎠cos 5x2 23cos5x cos3x+10xsin 3xsin 5x=2 2cos 5x23.2 2 2 3 2f ′( x) = ( x –1) (9 x – 4) + (3 x – 4 x)(2)( x –1)(2 x)f ′(2) = 6722 2 2 2 3= ( x –1) (9 x –4) + 4 x( x –1)(3 x –4 x)24. g′ ( x) = 3cos3x+ 2(sin 3 x)(cos3 x)(3)= 3cos3x+ 3sin6xg′′ ( x) = –9sin 3x+18cos 6xg′′ (0) = 1825.2 2 2 2d ⎛ cot x ⎞ (sec x )(– csc x) – (cot x)(sec x )(tan x )(2 x)⎜ 2 ⎟ =dx 2 2⎝secx ⎠sec x2 2– csc x – 2xcot xtanx=2sec x26.27.28.29.⎛ 4tsin t ⎞ (cos t – sin t)(4tcost+4sin t) – (4tsin t)(– sin t – cos t)Dt⎜ ⎟ =⎝cos t – sin t⎠2(cos t – sin t)2 2 224tcos t+ 2sin 2 t – 4sin t+4tsint 4t+2sin 2 t – 4sin t==22(cos t – sin t)(cos t – sin t)3 2 2f′ ( x) = ( x– 1) 2(sin πx– x)( πcos π x– 1) + (sin π x– x) 3( x– 1)3 2 2= 2( x–1) (sin πx– x)( πcos π x–1) + 3(sin π x– x) ( x–1)f ′(2) = 16 −4π≈3.434h′ ( t) = 5(sin(2 t) + cos(3 t)) (2cos(2 t) – 3sin(3 t))4 3 2h′′ ( t) = 5(sin(2 t) + cos(3 t)) ( − 4sin(2 t) – 9cos(3 t)) + 20(sin(2 t) + cos(3 t)) (2cos(2 t) – 3sin(3 t))4 3 2h′′ (0) = 5⋅1 ⋅( − 9) + 20⋅1 ⋅ 2 = 352g′ 2( r) = 3(cos 5 r)(– sin 5 r)(5)= –15cos 5rsin 5r2g′′ 3 2( r) = –15[(cos 5 r)(cos5 r)(5) + (sin 5 r)2(cos5 r)(– sin 5 r)(5)]= –15[5cos 5 r –10(sin 5 r)(cos5 r)]2 2g′′′ ( r) = –15[5(3)(cos 5 r)(– sin 5 r)(5) −(10sin 5 r)( −sin 5 r)(5) − (cos5 r)(20sin 5 r)(cos5 r)(5)]2 3= –15[ − 175(cos 5 r)(sin 5 r) + 50sin 5 r]g′′′ (1) ≈ 458.830. f ′() t = h′ ( gt ()) g′ () t + 2 gtg () ′()t31. G′ ( x) = F′ ( r( x) + s( x))( r′ ( x) + s′ ( x)) + s′( x)G′′ ( x) = F′ ( r( x) + s( x))( r′′ ( x) + s′′ ( x)) + ( r′ ( x) + s′ ( x)) F′′ ( r( x) + s( x))( r′ ( x) + s′ ( x)) + s′′( x)2= F′ ( rx ( ) + sx ( ))( r′′ ( x) + s′′ ( x)) + ( r′ ( x) + s′ ( x)) F′′ ( rx ( ) + sx ( )) + s′′( x)148 Section 2.10 Instructor’s Resource Manual

32.33.2F′ ( x) = Q′ ( R( x)) R′( x) = 3[ R( x)] (–sin x)=2–3cosxsinx2F′ ( z) = r′ ( sz ( )) s′( z) = [3cos(3 sz ( ))](9 z )2 3= 27zcos(9 z )dy34. 2( x – 2)dx =2x – y + 2 = 0; y = 2x + 2; m = 212( x – 2) = – 27x =42⎛7 ⎞ 1 ⎛7 1 ⎞y = ⎜ – 2 ⎟ = ; ⎜ , ⎟⎝4 ⎠ 16 ⎝4 16⎠4 335. V = π r3dV 24 rdr = πdV 2When r = 5, 4 (5) 100 314dr = π = π≈ m3 permeter of increase in the radius.36.37.4 3 dVV = π r ; = 103 dtdV 2 dr= 4πrdt dtWhen r = 5, 10 = 4 π (5)dr 10.0318dt = 10π≈ m/h2drdt1 6 b 3hV = bh(12); = ; b =2 4 h 2⎛3h⎞2 dVV = 6⎜⎟h = 9 h ; = 9⎝ 2 ⎠ dtdV dh= 18hdt dtWhen h = 3, 9=18(3) dhdtdh 10.167dt = 6≈ ft/min38. a. v = 128 – 32tv = 0, when t = 4s2s = 128(4) –16(4) = 256 ft39.b.2128 t –16t = 0–16t(t – 8) = 0The object hits the ground when t = 8sv = 128 – 32(8) = –128 ft/s3 2s = t – 6t + 9t2vt () = ds = 3 t –12t+9dt2d sat () = = 6 t–122dta.b.40. a.23 t –12t+ 9 0t > 2; (2, ∞ )b.c.d.e.f.20 19 12 5Dx( x + x + x + 100) = 020 20 19 18Dx( x + x + x ) = 20!20 21 20Dx(7x + 3 x ) = (7 ⋅ 21!) x+ (3⋅20!)20 4Dx(sin x+ cos x) = Dx(sin x+cos x)= sin x + cos x20 20Dx(sin 2 x) = 2 sin 2x= 1,048,576 sin 2x20D x20⎛1 ⎞ (–1) (20!) 20!⎜ ⎟ = =⎝ x ⎠21 21x xdy41. a. 2( x–1) + 2y = 0 dxdy –( x –1) 1– x= =dx y yb.dy 2 2 dyx(2 y) + y + y(2 x) + x = 0dxdxdy 2 2(2 xy+ x ) = –( y + 2 xy)dx2dy y + 2xy=−dx 2x + 2xyInstructor’s Resource Manual Section 2.10 149

2 2 dy 3 2 dy 2 3c. 3x + 3 y = x (3 y ) + 3x ydx dxdy 2 3 2 2 3 2(3 y –3 x y ) = 3 x y –3xdx2 3 2 2 3 2dy 3 x y –3 x x y – x= =dx 2 3 2 2 3 23 y –3 x y y – x y⎡ dy ⎤d. x cos( xy) ⎢x + y + sin( xy) = 2xdx ⎥⎣ ⎦2 dyx cos( xy ) 2 x – sin( xy ) – xy cos( xy )dx =dy 2 x – sin( xy) – xy cos( xy)=dx 2x cos( xy)a.b.45. a.2(1) + 4(–2)(1) + 8(1)dy = – (–0.01)22(–2)(1) + 2(–2 + 2)= –0.00252(–1) + 4(–2)(–1) + 8(–1)dy = – (–0.01)22(–2)(–1) + 2(–2 + 2)= 0.0025d 2 3[ f ( x ) + g ( x )]dx2= 2 f ( x) f′ ( x) + 3 g ( x) g′( x)22 f(2) f′ (2) + 3 g (2) g′(2)2= 2(3)(4) + 3(2) (5) = 84e.2 ⎛ dy ⎞xsec ( xy) ⎜x + y⎟+ tan( xy) = 0⎝ dx ⎠2 2 dy2x sec ( xy) –[tan( xy) xy sec ( xy)]dx = +2dy tan( xy) + xy sec ( xy)= –dx 2 2x sec ( xy)b. d [ f ( xgx ) ( )] = f ( xg ) ′( x ) + gx ( ) f ′( x )dxf(2) g′ (2) + g(2) f′(2) = (3)(5) + (2)(4) = 23c. d [ f ( gx ( ))] = f ′( gx ( )) g ′( x )dxf′ ( g(2)) g′ (2) = f′ (2) g′(2) = (4)(5) = 2042.22yy1′ = 12x26xy1′ =yAt (1, 2): y′ 1 = 34x+ 6yy′2 = 02xy2′ = – 3 y1At (1, 2): y′ 2 = – 3Since ( y1′ )( y2′ ) = –1 at (1, 2), the tangents areperpendicular.43. dy = [ π cos( π x) + 2 x]dx ; x = 2, dx = 0.0144.dy = [ π cos(2 π) + 2(2)](0.01) = (4 + π)(0.01)≈ 0.0714dy 2 2 dyx(2 y) + y + 2 y[2( x+ 2)] + ( x+ 2) (2) = 0dxdxdy2 2[2xy + 2( x + 2) ] = –[ y + 2 y(2x+ 4)]dx2dy –( y + 4xy + 8 y)=dx 22xy + 2( x + 2)2y + 4xy+8ydy = – dx2 2( 2)2xy + x +When x = –2, y = ±146.d.2Dx[ f ( x)] = 2 f( x) f′( x)2 2Dx [ f ( x)] = 2[ f( x) f′′ ( x) + f′ ( x) f′( x)]2= 2 f(2) f′′ (2) + 2[ f′(2)]2= 2(3)(–1) + 2(4) = 262 2 2 dx(13) = x + y ; = 2dtdx dy0= 2x+ 2ydt dtdy x dx= –dt y dtWhen y= 5, x = 12, sody 12 24– (2) – –4.8dt = 5 = 5= ft/s47. sin15 °= y , dx 400x dt=y = xsin15°48. a.dy dx= sin15°dt dtdy400sin15dt = ° ≈ 104 mi/hrb.2 22 2( ) 2Dx ( x ) = 2 x ⋅ x = x = x = 2xx x xxx⎛ ⎞2x − x⎛ x ⎞ ⎜ ⎟ x − xDxx = D⎝ ⎠x⎜⎟= = = 0x 2 2⎝ ⎠ x x150 Section 2.10 Instructor’s Resource Manual

c.d.49. a.b.3 2x x x xD x = D ( D x) = D (0) = 02 2Dx( x ) = Dx(2 x) = 2sinθD θ sinθ = cosθ = cotθ sinθsinθcosθD θ cos θ = ( − sin θ) = − tanθ cosθcosθ3. x( x )( x )x( x )( x )−1 −2 ≤ 0−1 − 2 = 0x = 0, x = 1 or x = 2The split points are 0, 1, and 2. The expressionon the left can only change signs at the split−∞ ,0 ,points. Check a point in the intervals ( )( 0,1 ) , ( 1, 2 ) , and ( 2,∞ ). The solution set is{ x| x≤ 0 or 1≤ x≤ 2}, or ( −∞,0] ∪ [ 1,2].1f( x) = x+ 1; f '( x) = − x+ 1 ; a = 32Lx ( ) = f(3) + f'(3)( x−3)−50. a. ( ) 1/21 −1/2= 4 +− (4) ( x −3)21 3 1 11= 2 − x+ = − x+4 4 4 4b. f( x) = xcos x; f '( x) = − xsin x+ cos x; a = 1Lx ( ) = f(1) + f'(1)( x−1)= cos1 + ( − sin1+ cos1)( x −1)= cos1 − (sin1) x+ sin1 + (cos1) x−cos1= (cos1− sin1) x + sin1≈− 0.3012x+ 0.8415Review and Preview Problems1. ( x )( x )( x )( x )2.−2 − 3 < 0−2 − 3 = 0x = 2 or x = 3The split points are 2 and 3. The expression onthe left can only change signs at the split points.,2 2,3 ,Check a point in the intervals ( −∞ ), ( )and ( 3, ∞ ). The solution set is { x|2 x 3}( 2,3 ).−2 −1 0 1 2 3 4 5 6 7 82< < orx −x− 6>0( x− 3)( x+ 2)> 0( x− 3)( x+ 2)= 0x = 3 or x =−2The split points are 3 and − 2 . The expression onthe left can only change signs at the split points., 2 − 2,3 ,Check a point in the intervals ( −∞ − ), ( )and ( 3, ∞ ). The solution set is{ x| x 3}, or ( −∞, −2) ∪ ( 3, ∞ ) .4.5.−5−4 −3 −2 −1 0 1 2 3 4 53 2x + 3x + 2x≥0x x + 3 + 2 ≥02( x )( )( x )( )( x )x x+ 1 + 2 ≥0x x+ 1 + 2 = 0x = 0, x =− 1, x =−2The split points are 0, − 1 , and − 2 . Theexpression on the left can only change signs atthe split points. Check a point in the intervals( −∞, − 2), ( −2, − 1), ( − 1, 0), and ( 0,∞ ) . Thex| −2≤ x≤ −1 or x≥ 0 , orsolution set is { }[ −2, −1] ∪[ 0, ∞ ) .−5−4 −3 −2 −1 0 1 2 3 4 5( − 2)x x2x − 4x( x−2)( x− 2)( x+2)≥ 0≥ 0The expression on the left is equal to 0 orundefined at x = 0 , x = 2 , and x = − 2 . Theseare the split points. The expression on the left canonly change signs at the split points. Check a, 2 2,0 0, 2 ,point in the intervals: ( −∞ − ), ( − ) , ( )and ( 2,∞ ) . The solution set is{ x| x< −2 or 0≤ x< 2 or x > 2}, or( −∞, −2) ∪[ 0,2) ∪( 2, ∞ ).−5−4 −3 −2 −1 0 1 2 3 4 5−5−4 −3 −2 −1 0 1 2 3 4 5Instructor’s Resource Manual Review and Preview 151

6.2x − 902 >x + 2( x− 3)( x+3)> 02x + 2The expression on the left is equal to 0 at x = 3 ,and x =− 3 . These are the split points. Theexpression on the left can only change signs atthe split points. Check a point in the intervals:( −∞, − 3), ( − 3,3), and ( 3, ∞ ). The solution setx| x 3 or x 3 −∞, −3 ∪ 3, ∞ .is { }, or ( ) ( )−5−4 −3 −2 −1 0 1 2 3 4 53 37. f '( x) = 4( 2x+ 1) ( 2) = 8( 2x+1)8. f '( x) = cos( π x) ⋅ π = π cos( πx)29. f '( x) = ( x −1) ⋅−sin( 2x) ⋅ 2+ cos( 2x) ⋅( 2x)2=−2( x − 1) sin( 2x) + 2xcos( 2x)10. f '( x)x ⋅sec xtan x−sec x⋅1=2xsec x( xtan x−1)=2x211. f ( x) = ( x)⋅ x⋅2= 6( sec 3x)( tan 3x)' 2 tan3 sec 3 312sin xcosx=21+sin x−1/2212. f '( x) = ( 1+sin x) ( 2sin x)( cos x)1 cos x= ⋅ =2 2 x−1/213. f '( x) cos( x) x(note: you cannot cancel the x here because itis not a factor of both the numerator anddenominator. It is the argument for the cosine inthe numerator.)' 1 sin2 cos2 2cos2 xf x = x ⋅ x⋅ =2 sin 2x−14. ( ) ( ) 1/215. The tangent line is horizontal when the derivativeis 0.2y ' = 2tanx⋅sec x2tanxsecx= 02sin x= 02cos xThe tangent line is horizontal wheneversin x = 0 . That is, for x = kπ where k is aninteger.16. The tangent line is horizontal when the derivativeis 0.y ' = 1+cosxThe tangent line is horizontal whenevercos x = − 1 . That is, for x = ( 2k+ 1)π where k isan integer.17. The line y = 2 + x has slope 1, so any line parallelto this line will also have a slope of 1.For the tangent line to y = x+ sin x to be parallelto the given line, we need its derivative to equal 1.y' = 1+ cosx= 1cos x = 0The tangent line will be parallel to y = 2 + xπwhenever x ( k )= 2 + 1 . 218. Length: 24 − 2xWidth: 9−2xHeight: xVolume: l⋅w⋅ h = ( 24 −2x)( 9 −2x)x= x 9−2x 24−2x19. Consider the diagram:1x4 − x( )( )His distance swimming will be2 2 21 + x = x + 1 kilometers. His distancerunning will be 4 − x kilometers.Using the distance traveled formula, d = r⋅ t , wedsolve for t to get t = . Andy can swim at 4rkilometers per hour and run 10 kilometers perhour. Therefore, the time to get from A to D will2x + 1 4−xbe + hours.4 10152 Review and Preview Instructor’s Resource Manual

20. a. f ( 0) = 0− cos( 0)= 0− 1= −1f ( π) = π − cos( π) = π −( − 1)= π + 1Since x − cos x is continuous, f ( 0)< 0,and f ( π ) > 0 , there is at least one point c.in the interval ( 0,π ) where f ( c ) = 0 .(Intermediate Value Theorem)⎛π ⎞ π ⎛π ⎞ πb. f ⎜ ⎟= − cos⎜ ⎟=⎝ 2 ⎠ 2 ⎝ 2⎠2f ' x = 1+sinx( )⎛π⎞ ⎛π⎞f ' ⎜ ⎟= 1+ sin⎜ ⎟= 1+ 1=2⎝ 2 ⎠ ⎝ 2⎠The slope of the tangent line is m = 2 at the⎛π π ⎞point ⎜ , ⎟⎝ 2 2 ⎠ . Therefore,π ⎛ π ⎞y− = 2⎜x−⎟2 ⎝ 2 ⎠ or 2 πy = x− .2πc. 2x− = 0.2π2x=2πx =4The tangent line will intersect the x-axis atπx = .4221. a. The derivative of x is 2x and thederivative of a constant is 0. Therefore, one2possible function is f ( x) = x + 3 .b. The derivative of − cos x is sin x and thederivative of a constant is 0. Therefore, onef x =− cos x + 8 .possible function is ( ) ( )3 2c. The derivative of x is 3x , so the1 3derivative of3 x is 2x . The derivative of21 2x is 2x , so the derivative of2 x is x .The derivative of x is 1, and the derivative ofa constant is 0. Therefore, one possible1 3 1 2function is x + x + x+ 2 .3 222. Yes. Adding 1 only changes the constant term inthe function and the derivative of a constant is 0.Therefore, we would get the same derivativeregardless of the value of the constant.Instructor’s Resource Manual Review and Preview 153

CHAPTER 3Applications of theDerivative7. ) 2 3;3.1 Concepts ReviewΨ ′( x = x+ 32x + 3 = 0 when x = – .21Maximum value = 6, minimum value = – 41. continuous; closed and bounded3Critical points: –2, – ,2 12. extreme⎛ 3⎞9Ψ (–2) = –2, Ψ ⎜– ⎟= – , Ψ (1) = 43. endpoints; stationary points; singular points⎝ 2⎠49Maximum value = 4, minimum value = –4. f ′() c = 0; f′() c does not exist4Problem Set 3.18.1 2 6 2G′ ( x) = (6x + 6 x–12) = ( x + x–2);5 521. Endpoints: − 2 , 4x + x– 2= 0 when x = –2, 1Singular points: noneCritical points: –3, –2, 1, 3Stationary points: 0, 29 7G(–3) = , G(–2) = 4, G(1) = – , G(3) = 9Critical points: − 2,0, 2, 45 5Maximum value = 9,2. Endpoints: − 2 , 47Singular points: 2minimum value = –Stationary points: 05Critical points: − 2,0, 2, 43. Endpoints: − 2 , 4Singular points: noneStationary points: − 1,0,1, 2,3Critical points: −2, − 1, 0,1, 2, 3, 49.2 2f′ ( x) = 3 x –3; 3 x –3= 0 when x = –1, 1.Critical points: –1, 1f(–1) = 3, f(1) = –1No maximum value, minimum value = –1(See graph.)4. Endpoints: − 2 , 4Singular points: noneStationary points: noneCritical points: − 2, 45. f′ ( x) = 2x+ 4; 2x+ 4= 0 when x = –2.Critical points: –4, –2, 0f(–4) = 4, f(–2) = 0, f(0) = 4Maximum value = 4, minimum value = 010.2 2f′ ( x) = 3 x –3; 3 x –3= 0 when x = –1, 1.16. h′ ( x) = 2x+ 1; 2x+ 1= 0 when x = – .32Critical points: – ,–1,1,321Critical points: –2, – ,2 2⎛ 3⎞ 17⎜– ⎟ = , (–1) 3, (1) –1, (3) 192 8= f = f =1 1h(–2) = 2, ⎜– ⎞ ⎝ ⎠⎟ – ,⎝ 2⎠h(2) = 6Maximum value = 19, minimum value = –14154 Section 3.1 Instructor’s Resource Manual

11.1h′ () r =− ; h′() r is never 0; h′ () r is not defined2rwhen r = 0, but r = 0 is not in the domain on[–1, 3] since h(0) is not defined.Critical points: –1, 3Note that lim hr ( ) =−∞ and lim hx ( ) =∞ .x→0 −x→0+No maximum value, no minimum value.15.2x′( ) =− ;+2 22 2g x(1 x )2x− = 0 when x = 0.(1 + x )Critical point: 0g(0) = 1++As x→∞, g( x) → 0 ; as x→−∞, g( x) → 0 .Maximum value = 1, no minimum value(See graph.)2x12. g′ ( x) =− ;2 2(1 +2 2x ) (1 + x )Critical points: –3, 0, 1g(–3) = 110 , g(0) = 1, g(1) = 1 22x− = 0 when x = 0Maximum value = 1, minimum value = 11013. ( )f ' x = 4x −4x32( )= 4x x −1= 4x( x− 1)( x+1)( )( x )4x x− 1 + 1 = 0 when x = 0,1, − 1.Critical points: −2, − 1,0,1, 2f ( − 2)= 10; f ( − 1)= 1; f ( 0)= 2; ( )f ( 2)= 10Maximum value: 10Minimum value: 1f 1 = 1;16.21−xf′ ( x) = ;2 2(1 + x )21−x= 02 2(1 + x )when x = –1, 1Critical points: –1, 1, 41 1 4f( − 1) = − , f(1) = , f(4)=2 2 17Maximum value = 1 2 ,minimum value1= – 214. ( )4 2f ' x = 5x − 25x+ 204 2( x x )2 2( x )( x )= 5 − 5 + 4= 5 −4 −1= 5( x− 2)( x+ 2)( x− 1)( x+1)( x )( x )( x )( x )5 − 2 + 2 − 1 + 1 = 0 whenx =−2, − 1,1, 2Critical points: −3, −2, − 1,1, 2193f ( − 3)= − 79; f ( − 2)= − ; ( 1)35313f =3f () 1 = ; ( 2)Maximum value: 35 3Minimum value: − 7941f − = − ;3π17. r′ ( θ ) = cos θ;cosθ = 0 when θ = + kπ2π πCritical points: – ,4 6⎛ π⎞ 1 ⎛π⎞1r⎜− ⎟=− , r⎜ ⎟=⎝ 4⎠ 2 ⎝6⎠21 1Maximum value = , minimum value = –2218. s′ () t = cost+ sin; t cos t + sin t = 0 whenπtan t = –1 or t = – + kπ.43π Critical points: 0, , π4⎛3π⎞s(0) = –1, s ⎜ ⎟ =⎝ 4 ⎠2, s( π ) = 1.Maximum value = 2,minimum value = –1Instructor’s Resource Manual Section 3.1 155

x –119. a′ ( x) = ; a′( x)does not exist when x = 1.x –1Critical points: 0, 1, 3a(0) = 1, a(1) = 0, a(3) = 2Maximum value = 2, minimum value = 03(3 s – 2)20. f ′() s = ; f′() s does not exist when s = 2 .3 s –23Critical points: − 1, 2 , 43f(–1) = 5, f 2( ) = 0, f(4) = 103Maximum value = 10, minimum value = 0121. g′ ( x) = ; f′( x)does not exist when x = 0.2/33xCritical points: –1, 0, 27g(–1) = –1, g(0) = 0, g(27) = 3Maximum value = 3, minimum value = –1222. s′ () t = ; s′() t does not exist when t = 0.3/55tCritical points: –1, 0, 32s(–1) = 1, s(0) = 0, s(32) = 4Maximum value = 4, minimum value = 023. H '()t =− sint− sin t = 0 whent = 0, π ,2 π,3 π,4 π,5 π,6 π,7 π,8πCritical points: 0, π ,2 π,3 π,4 π,5 π,6 π,7 π,8πH 2π = 1;H ( 0)= 1; H ( π ) =− 1; ( )H ( 3π ) =− 1; H ( 4π ) = 1; H ( 5π ) =− 1;H ( 6π ) = 1; H ( 7π ) =− 1; H ( 8π ) = 1Maximum value: 1Minimum value: − 124. ( )g ' x = 1−2cosx11− 2cos x = 0 → cos x = when25π π π 5πx =− , − , ,3 3 3 35π π π 5πCritical points: −2 π , − , − , , ,2π3 3 3 3⎛ 5π⎞ −5πg ( − 2π ) =− 2π; g ⎜− ⎟= −⎝ 3 ⎠ 33 ;⎛ π ⎞ π ⎛g ⎜− ⎟=− + 3 ; g π ⎞⎜ ⎟π = −⎝ 3⎠3 ⎝ 3⎠33 ;⎛5π⎞ 5πg ⎜ ⎟ = +⎝ 3 ⎠ 33 ; g ( 2π ) = 2πMaximum value: 5 π +335πMinimum value: − −33225. g '( θ ) = θ ( secθ tanθ)+ 2θsecθ= θsecθ( θ tanθ+ 2)θsecθ( θ tanθ + 2)= 0 when θ = 0 .Consider the graph:− π41−1yπ4π πCritical points: − ,0,4 4⎛ π ⎞ π 2g ⎜− ⎟=⎝ 4⎠16Maximum value:26. h'() t =27. a.2⎛5⎜⎝3; g ( 0)= 0;2π 216⎞⎟⎠2/3 5/3( 2+ t) t −t() 1( 2 + t)x2⎛ 2g π ⎞⎜ ⎟π =⎝ 4⎠16; Minimum value: 0( 2+ t) ( 2+t)2/32t( t+5)=232 ( + t)h'( t ) is undefined when 222/3⎛5 ⎞ 2/3⎛10 2 ⎞t ⎜ ( 2 + t)− t⎟ t ⎜ + t⎟⎝3 ⎠ ⎝ 3 3= =⎠2 2t =− and h' () t = 0when t = 0 or t = − 5 . Since − 5 is not in theinterval of interest, it is not a critical point.Critical points: − 1, 0, 8h( − 1)=− 1; h ( 0)= 0; h ( 8) = 165Maximum value: 16 5; Minimum value: − 12 2f′ ( x) = 3 x –12x+ 1;3 x –12x+ 1=03333when x = 2– and x = 2 + .3333 33Critical points: –1, 2 – , 2 + , 53 333f(–1) = –6, f ⎛2– ⎞ ⎜≈ 2.04,3 ⎟⎝ ⎠33f ⎛2 ⎞⎜+ ≈ –26.04,3 ⎟f(5) = –18⎝ ⎠Maximum value ≈ 2.04;minimum value ≈ − 26.04156 Section 3.1 Instructor’s Resource Manual

.3 2 2( x –6x + x+ 2)(3 x –12x+1)g′ ( x) =;3 2x –6x + x+2g'( x ) = 0 when33x = 2– and 333x = 2 + . g′ ( x)does not exist when3f(x) = 0; on [–1, 5], f(x) = 0 whenx ≈ –0.4836 and x ≈ 0.717233Critical points: –1, –0.4836, 2– ,3330.7172, 2 + , 53g(–1) = 6, g(–0.4836) = 0,33g ⎛2 – ⎞ ⎜≈ 2.04,3 ⎟g(0.7172) = 0,⎝ ⎠33g ⎛2 ⎞⎜+ ≈ 26.04,3 ⎟g(5) = 18⎝ ⎠Maximum value ≈ 26.04,minimum value = 029. Answers will vary. One possibility:y5−55 x−530. Answers will vary. One possibility:y55 x−528. a. f ′( x) = xcos x;on [–1, 5], x cos x = 0 whenb.π 3πx = 0, x = , x =2 2π 3πCritical points: –1, 0, , , 52 2⎛π⎞f(–1) ≈3.38, f(0) = 3, f ⎜ ⎟ ≈ 3.57,⎝2⎠⎛3π⎞f ⎜ ⎟ ≈ –2.71, f(5) ≈− 2.51⎝ 2 ⎠Maximum value ≈ 3.57,minimum value ≈–2.71(cos x + xsin x+2)( xcos x)g′ ( x) =;cos x+ xsin x+23g ( x) 0 when x 0, x π π′ = = = , x =2 2g′ ( x)does not exist when f(x) = 0;on [–1, 5], f(x) = 0 when x ≈ 3.45π 3πCritical points: –1, 0, , 3.45, , 52 2g(–1) ≈ 3.38, g(0) = 3, g⎛⎜π ⎞ ⎟≈3.57,⎝2⎠⎛3π⎞g(3.45) = 0, g ⎜ ⎟ ≈ 2.71, g(5) ≈ 2.51⎝ 2 ⎠Maximum value ≈ 3.57;minimum value = 031. Answers will vary. One possibility:y55 x−532. Answers will vary. One possibility:y55 x−5Instructor’s Resource Manual Section 3.1 157

33. Answers will vary. One possibility:y55 x3.2 Concepts Review1. Increasing; concave up2. f′ ( x) > 0; f′′( x) < 03. An inflection point4. f ′′() c = 0; f′′() c does not exist.−534. Answers will vary. One possibility:y55 x−535. Answers will vary. One possibility:yProblem Set 3.21. f′ ( x) = 3; 3 > 0 for all x. f(x) is increasingfor all x.2. g′ ( x) = 2 x–1;2x – 1 > 0 when1x > . g(x) is2⎡1 ⎞increasing on ⎢ , ∞ ⎟ and decreasing on⎣ 2 ⎠⎛ 1 ⎤⎜ – ∞, .2⎥⎝ ⎦3. h′ () t = 2t+ 2; 2t + 2 > 0 when t > –1. h(t) isincreasing on [–1, ∞ ) and decreasing on( −∞ , –1].55 x−536. Answers will vary. One possibility:y4.5.2 2f ( x) 3 x ; 3x0′ = > for x ≠ 0 .f(x) is increasing for all x.2G′ ( x) = 6 x –18x+ 12 = 6( x– 2)( x–1)Split the x-axis into the intervals (– ∞ , 1), (1, 2),(2, ∞ ).33 3Test points: x = 0, , 3; G′ (0) = 12, G′ ⎛⎜⎞⎟=– ,2⎝2⎠2G′ (3) = 12G(x) is increasing on (– ∞ , 1] ∪ [2, ∞ ) anddecreasing on [1, 2].5−55 x6.2f′ () t = 3t + 6t = 3( t t+2)Split the x-axis into the intervals (– ∞ , –2),(–2, 0), (0, ∞ ).Test points: t = –3, –1, 1; f ′(–3) = 9,f ′(–1) = –3, f ′(1) = 9f(t) is increasing on (– ∞ , –2] ∪ [0, ∞ ) anddecreasing on [–2, 0].7.3 2 2h′ ( z) = z –2 z = z ( z–2)Split the x-axis into the intervals (– ∞ , 0), (0, 2),(2, ∞ ).Test points: z = –1, 1, 3; h′ (–1) = –3, h′ (1) = –1,h′ (3) = 9h(z) is increasing on [2, ∞ ) and decreasing on(– ∞ , 2].158 Section 3.2 Instructor’s Resource Manual

8.2– xf′ ( x)=3xSplit the x-axis into the intervals (– ∞ , 0), (0, 2),(2, ∞ ).Test points: –1, 1, 3; f ′(–1) = –3, f ′(1) = 1,1f ′(3) = – 27f(x) is increasing on (0, 2] and decreasing on(– ∞ , 0) ∪ [2, ∞ ).16.17.2f′′ ( x) = 12x + 48x = 12 x( x+ 4); f′′( x) > 0 whenx < –4 and x > 0.f(x) is concave up on (– ∞ , –4) ∪ (0, ∞ ) andconcave down on (–4, 0); inflection points are(–4, –258) and (0, –2).2 2 2F′′ ( x) = 2sin x–2cos x+ 4=6–4cos x;26–4cos x > 0 for all x since 0 ≤ cos x ≤ 1.F(x) is concave up for all x; no inflection points.2π9. H′ () t = cos; t H′() t > 0when 0 ≤ t < and23π < t ≤ 2 π .2⎡ π⎤ ⎡3π⎤H(t) is increasing on ⎢0, ∪ , 2π2⎥ ⎢ 2 ⎥⎣ ⎦ ⎣ ⎦ and⎡π3π⎤decreasing on ⎢ , .2 2 ⎥⎣ ⎦π10. R′ ( θ ) = –2cosθsin θ; R′( θ) > 0 when < θ 0 for all x.G(x) is concave up for all x; no inflection points.2 2f′ ( x) = 3 x –12; 3 x –12> 0 whenx < –2 or x > 2.f(x) is increasing on (– ∞ , –2] ∪ [2, ∞ ) anddecreasing on [–2, 2].f ′′( x) = 6 x;6x > 0 when x > 0. f(x) is concave upon (0, ∞ ) and concave down on (– ∞ , 0).and 3 π < θ < 2 π .2⎡π⎤ ⎡3π⎤R(θ ) is increasing on ⎢ , π ∪ , 2π2 ⎥ ⎢ 2 ⎥⎣ ⎦ ⎣ ⎦ and⎡ π⎤ ⎡ 3π⎤decreasing on ⎢0, ∪ π, .2⎥ ⎢ 2 ⎥⎣ ⎦ ⎣ ⎦11. f′′ ( x) = 2; 2 > 0 for all x. f(x) is concave up for allx; no inflection points.12. G′′ ( w) = 2; 2 > 0 for all w. G(w) is concave up forall w; no inflection points.13. T′′ () t = 18; t 18t > 0 when t > 0. T(t) is concave upon (0, ∞ ) and concave down on (– ∞ , 0);(0, 0) is the only inflection point.14.6 2 4f′′ ( z) = 2– = ( z –3); z4 – 3> 0 for4 4z z4z < – 3 and z >4 3.4 4f(z) is concave up on (– ∞, – 3) ∪( 3, ∞ ) and4 4concave down on (– 3, 0) ∪ (0, 3); inflection⎛ 4 1 ⎞points are ⎜– 3, 3– ⎟⎝ 3 ⎠ and ⎛4 1 ⎞⎜ 3, 3 – ⎟.⎝ 3 ⎠20.2g′ ( x) = 12 x –6 x–6= 6(2x+ 1)( x–1);g′ ( x) > 01when x < – or x > 1. g(x) is increasing on2⎛ 1 ⎤⎡ 1 ⎤⎜ – ∞, – ∪[1, ∞)2⎥and decreasing on ⎢– , 1 .⎝ ⎦2 ⎥⎣ ⎦g′′ ( x) = 24 x–6= 6(4 x–1);g′′ ( x) > 0 when1x > .4⎛1 ⎞g(x) is concave up on ⎜ , ∞⎟and concave down⎝4⎠⎛ 1 ⎞on ⎜– ∞, ⎟.⎝ 4 ⎠15.2q′′ ( x) 12 x – 36 x– 48; q′′( x) 0and x > 4.q(x) is concave up on (– ∞ , –1) ∪ (4, ∞ ) andconcave down on (–1, 4); inflection points are(–1, –19) and (4, –499).= > when x < –1Instructor’s Resource Manual Section 3.2 159

21.3 2 2g′ ( x) = 12 x –12x = 12 x ( x–1);g′ ( x) > 0when x > 1. g(x) is increasing on [1, ∞ ) anddecreasing on ( −∞ ,1].2g′′ ( x) = 36 x – 24x = 12 x(3 x– 2); g′′ ( x) > 02when x < 0 or x > . g(x) is concave up on3⎛2⎞(– ∞, 0) ∪⎜, ∞⎟⎝3⎠ and concave down on ⎛ 2 ⎞⎜0, ⎟.⎝ 3 ⎠23.4 2 2 2G′ ( x) = 15 x –15x = 15 x ( x –1); G′ ( x) > 0when x < –1 or x > 1. G(x) is increasing on(– ∞ , –1] ∪ [1, ∞ ) and decreasing on [–1, 1].3 2G′′ ( x) = 60 x –30x = 30 x(2 x –1);⎛ 1 ⎞Split the x-axis into the intervals ⎜−∞, − ⎟,⎝ 2 ⎠⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞⎜−, 0 ⎟, ⎜0, ⎟, ⎜ , ∞⎟.⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠1 1Test points: x = –1, – , , 1; G′′ (–1) = –30,2 21 15 1 15G′′ ⎛ ⎞ ⎛ ⎞⎜– ⎟= , G′′ ⎜ ⎟= – , G′′(1) = 30.⎝ 2⎠ 2 ⎝2⎠2⎛ 1 ⎞ ⎛ 1 ⎞G(x) is concave up on ⎜– , 0 ⎟∪⎜ , ∞⎟⎝ 2 ⎠ ⎝ 2 ⎠ and⎛ 1 ⎞ ⎛ 1 ⎞concave down on ⎜– ∞, – ⎟∪⎜0, ⎟.⎝ 2 ⎠ ⎝ 2 ⎠22.5 3 3 2F′ ( x) = 6 x –12x = 6 x ( x –2)Split the x-axis into the intervals (– ∞ , − 2) ,( − 2, 0), (0, 2), ( 2, ∞ ) .Test points: x = –2, –1, 1, 2; F′ (–2) = –96,F′ (–1) = 6, F′ (1) = –6, F′(2) = 96F(x) is increasing on [– 2, 0] ∪[ 2, ∞ ) anddecreasing on (– ∞, – 2] ∪ [0, 2]4 2 2 2F′′ ( x) = 30 x – 36x = 6 x (5 x – 6);when6x < – or 5F(x) is concave up on⎛concave down on⎜⎝25 x –6>06x > .5⎛ 6 ⎞ ⎛ 6 ⎞⎜– ∞, – ∪ , ∞5 ⎟ ⎜ 5 ⎟⎝ ⎠ ⎝ ⎠6 6 ⎞– , .5 5 ⎟⎠and24.2xH′ ( x) = ; H′ ( x) > 0 when x > 0.2 2( x + 1)H(x) is increasing on [0, ∞ ) and decreasing on(– ∞ , 0].22(1– 3 x )H′′ ( x) = ; H′′( x) > 0 when2 3( x + 1)1 1– < x < .3 3⎛ 1 1 ⎞H(x) is concave up on ⎜– , ⎟ and concave⎝ 3 3⎠ ⎛ 1 ⎞ ⎛ 1 ⎞down on ⎜– ∞, – ⎟∪⎜ , ∞⎟.⎝ 3⎠ ⎝ 3 ⎠160 Section 3.2 Instructor’s Resource Manual

25.cos xπf′ ( x) = ; f′( x) > 0 when 0 < x < . f(x)2 sinx2⎡ π⎤is increasing on ⎢0, 2 ⎥ and decreasing on⎣ ⎦⎡π⎤⎢ , π .2 ⎥⎣ ⎦2 2–cos x–2sinxf′′ ( x) = ; f′′( x) < 0 for all x in3/24sin x(0, ∞ ). f(x) is concave down on (0, π ).–2(5x+ 1)f ( x) ;4/39x1x – , f′′( x)5′′ = –2(5x + 1) > 0 when< does not exist at x = 0.1 8Test points: –1, – , 1; f ′′(–1) = ,10 94/31 10 4f′′ ⎛ ⎞⎜– ⎟= – , f(1) = – .⎝ 10 ⎠ 9 3⎛ 1 ⎞⎜– ∞, – ⎟⎝ 5 ⎠⎛ 1 ⎞⎜– , 0 ⎟ ∪ (0, ∞ ).⎝ 5 ⎠f(x) is concave up ondown onand concave26.27.4x > .33 x –4g′ ( x) = ; 3x – 4 > 0 when2 x –2g(x) is increasing on [2, ∞ ).3 x –8g′′ ( x) = ; 3x – 8 > 0 when3/24( x – 2)8x > .3⎛8 ⎞g(x) is concave up on ⎜ , ∞⎟and concave down⎝3⎠⎛ 8 ⎞on ⎜2, ⎟.⎝ 3 ⎠2–5x2f′ ( x) = ; 2 – 5x > 0 when x < , f ′( x)1/33x5does not exist at x = 0.Split the x-axis into the intervals ( − ∞ , 0),⎛ 2⎞ ⎛2⎞⎜0, ⎟, ⎜ , ∞⎟.⎝ 5⎠ ⎝5⎠1 7Test points: –1, , 1; f ′( − 1) = – ,5 31 3f′ ⎛ ⎞⎜ ⎟= 5 , f′(1) = –1.⎝5⎠3⎡ 2⎤f(x) is increasing on ⎢0, 5 ⎥ and decreasing on⎣ ⎦⎡2⎞(– ∞, 0] ∪⎢, ∞ ⎟ .⎣5⎠28.4( x + 2)g ( x) ;2/33x′ = x + 2 > 0 when x > –2, g′( x)does not exist at x = 0.Split the x-axis into the intervals ( , 2)−∞ − ,(–2, 0), (0, ∞ ).4Test points: –3, –1, 1; g′ (–3) = – ,5/334g′ (–1) = , g′(1) = 4.3g(x) is increasing on [–2, ∞ ) and decreasing on(– ∞ , –2].4( x – 4)g′′ ( x) = ; x – 4 > 0 when x > 4, g′′( x)5/39xdoes not exist at x = 0.20Test points: –1, 1, 5; g′′ (–1) = ,9g′′ (1) = – 4 , g′′(5) =4 .3 5/39(5)g(x) is concave up on (– ∞ , 0) ∪ (4, ∞ ) andconcave down on (0, 4).Instructor’s Resource Manual Section 3.2 161

29.35.2f ( x) = ax + bx+ c; f′( x) = 2 ax+b;f ′′( x) = 2aAn inflection point would occur where f′′ ( x) = 0,or 2a = 0. This would only occur when a = 0, butif a = 0, the equation is not quadratic. Thus,quadratic functions have no points of inflection.30.31.32.33.36.3 2f ( x) = ax + bx + cx+d;2f ′( x) = 3ax + 2 bx+ c;f ′′( x) = 6ax+2bAn inflection point occurs where f ( x) 0′′ = , or6ax + 2b = 0.The function will have an inflection point atbx = – , a ≠ 0.3a37. Suppose that there are points x 1 and x 2 in Iwhere f′ ( x1) > 0 and f′ ( x2) < 0. Since f ′ iscontinuous on I, the Intermediate Value Theoremsays that there is some number c between x 1 andx 2 such that f′ () c = 0, which is a contradiction.Thus, either f′ ( x) > 0 for all x in I and f isincreasing throughout I or f′ ( x) < 0 for all x in Iand f is decreasing throughout I.38. Since x 2 + 1= 0 has no real solutions, f ′( x)exists and is continuous everywhere.x2 – x+ 1= 0 has no real solutions. x2 – x+ 1>02and x + 1> 0 for all x, so f′ ( x) > 0 for all x.Thus f is increasing everywhere.39. a. Letf ( x)2= x and let I = [ 0, a], a > y.f′ ( x) = 2x> 0 on I. Therefore, f(x) isincreasing on I, so f(x) < f(y) for x < y.b. Let f ( x)= x and let I = [ 0, a], a > y.1f′ ( x)= > 0 on I. Therefore, f(x) is2 xincreasing on I, so f(x) < f(y) for x < y.34.1c. Let f( x)= and let I = [0, a], a > y.x1f′ ( x) = − < 0 on I. Therefore f(x) is2xdecreasing on I, so f(x) > f(y) for x < y.40.2f ′( x) = 3ax + 2bx+cIn order for f(x) to always be increasing, a, b, andc must meet the condition 3ax + 2bx + c > 0 forall x. More specifically, a > 0 and b22− 3ac

41.3 b–axf ( x) .5/24xbthen 13 2a2′′ = If (4, 13) is an inflection point= + and 3 b –4 a = 0. Solving these432 ⋅39 13equations simultaneously, a = and b = .8 242. f ( x) = a( x−r1)( x−r2)( x−r3)f ′( x) = a[( x−r1)(2 x−r2 − r3) + ( x−r2)( x−r3)]2f ′( x) = a[3x − 2 x( r1+ r2 + r3) + rr 1 2 + r2r3 + rr 1 3]f ′′( x) = a[6x− 2( r1+ r2 + r3)]a[6x− 2( r1+ r2 + r3)] = 0r1+ r2 + r36x = 2( r1+ r2 + r3);x =345. a.b. f′ ( x) < 0 :(1.3, 5.0)c. f′′ ( x) < 0 :( −0.25, 3.1) ∪ (6.5, 7]d.1 xf′ ( x) = cos x– sin2 243. a. [ f ( x) + g( x)] ′ = f′ ( x) + g′( x).Since f′ ( x) > 0 and g′ ( x) > 0 for all x,f′ ( x) + g′( x) > 0 for all x. No additionalconditions are needed.b. [ f ( x) ⋅ gx ( )]′ = f( xg ) ′( x) + f′( xgx ) ( ).f( x) g′ ( x) + f′( x) g( x) > 0 iff′( x)f ( x) >− g( x)for all x.g ′ ( x)c. [ f ( gx ( ))]′ = f′ ( gx ( )) g′( x).Since f′ ( x) > 0 and g′ ( x) > 0 for all x,f′ ( g( x)) g′ ( x) > 0 for all x. No additionalconditions are needed.e.46. a.1 xf′′ ( x) =−sinx−cos4 244. a. [ f ( x) + g( x)] ′′ = f′′ ( x) + g′′( x).Since f′′ ( x) > 0 and g′′ > 0 for all x,f′′ ( x) + g′′( x) > 0 for all x. No additionalconditions are needed.b. [ f ( x) ⋅ g( x)] ′′ = [ f( x) g′ ( x) + f′ ( x) g( x)]′= f ( xg ) ′′( x) + f′′ ( xgx ) ( ) + 2 f′ ( xg ) ′( x).The additional condition is thatf( x) g′′ ( x) + f′′ ( x) g( x) + 2 f′ ( x) g′( x) > 0for all x is needed.c. [ f ( gx ( ))]′′ = [ f′ ( gx ( )) g′ ( x)]′2f′ ( g( x)) g′′ ( x) f′′ ( g( x))[ g′( x)] .= +The additional condition is that2f ′′( gx ( ))[ g′( x)]f′ ( g( x))>−g′′( x)for all x.b. f′ ( x) < 0 :(2.0, 4.7) ∪ (9.9, 10]c. f′′ ( x) < 0 :[0, 3.4) ∪ (7.6, 10]2 x x 2 xd. f′ ( x) = x ⎡ ⎢− cos ⎛ ⎜ ⎞ ⎟sin ⎛ ⎜ ⎞ ⎟ ⎤ + cos⎛ ⎜ ⎞⎟3 3 3⎥⎣ ⎝ ⎠ ⎝ ⎠⎦⎝3⎠2 ⎛ 2cosx ⎞ x ⎛sinx ⎞= ⎜ ⎟−⎜ ⎟⎝3⎠ 3 ⎝ 3 ⎠Instructor’s Resource Manual Section 3.2 163

e.2x 2x 2 2xf′′ ⎛ ⎞ ⎛ ⎞( x) =− cos⎜ ⎟−sin⎜ ⎟9 ⎝ 3 ⎠ 3 ⎝ 3 ⎠c.3 2d s d s< 0, > 03 2dt dts47. f′ ( x) > 0 on (–0.598, 0.680)f is increasing on [–0.598, 0.680].48. f′′ ( x) < 0 when x > 1.63 in [–2, 3]f is concave down on (1.63, 3).49. Let s be the distance traveled. Then dsdtspeed of the car.is thed.Concave up.2d s= 10 mph/min2dtts= k a constant a.dsdtks,stConcave up.tConcave up.e.dsdtand2d sdt2are approaching zero.b.2d s> 02dtssConcave down.tConcave up.t164 Section 3.2 Instructor’s Resource Manual

f.dsdtsis constant.c.dV = k, dh > 0, d h < 0dt dt 2dtConcave down.ht ()250. a.Neither concave up nor down.dV= k 0 in the futuredt 2dtwhere I is inflation.I()t2ttb.dV 1 1= 3– = 2 gal/mindt 2 2Neither concave up nor down.vt ()te.2dp d pdp< 0, but > 0 and at t = 2: 0dt2dtdt > .where p is the price of oil.Concave up.P()tt2tInstructor’s Resource Manual Section 3.2 165

f.2dT d T> 0, < 0 , where T is David’sdt 2dttemperature.Concave down.Tt ()c.2dP d P> 0, < 0 , where P is worlddt 2dtpopulation.Concave down.P()ttt51. a.2dC d C> 0, > 0 , where C is the car’s cost.dt 2dtConcave up.Ct ()d.2dθd θ> 0, > 0 , where θ is the angle thatdt 2dtthe tower makes with the vertical.Concave up.θ( t)b. f(t) is oil consumption at time t.2df < 0, d f > 0dt 2dtConcave up.f()tte. P = f(t) is profit at time t.2dP d P> 0, < 0dt 2dtConcave down.P()tttt166 Section 3.2 Instructor’s Resource Manual

f. R is revenue at time t.dPP < 0, 0dt >Could be either concave up or down.Pt54. The height is always increasing so h'( t ) > 0 . Therate of change of the height decreases for the first50 minutes and then increases over the next 50minutes. Thus h''( t ) < 0 for 0≤ t ≤ 50 andh''( t ) > 0 for 50 < t ≤ 100 .P52. a. R(t) ≈ 0.28, t < 1981b. On [1981, 1983],R(1983) ≈ 0.362dR d R> 0, > 0 ,dt 2dtdV 353. = 2 in /secdtThe cup is a portion of a cone with the bottom cutoff. If we let x represent the height of the missingcone, we can use similar triangles to show thatx x+5=3 3.53.5x= 3x+150.5x= 15x = 30Similar triangles can be used again to show that, atany given time, the radius of the cone at waterlevel ish + 30r =20Therefore, the volume of water can be expressedas3( h 30) 45V = π + − π .1200 2We also know that V = 2tfrom above. Setting thetwo volume equations equal to each other andt55. V = 3, t 0≤t≤ 8. The height is always increasing,so h'( t ) > 0. The rate of change of the heightdecreases from time t = 0 until time t 1 when thewater reaches the middle of the rounded bottompart. The rate of change then increases until timet 2 when the water reaches the middle of the neck.Then the rate of change decreases until t = 8 andthe vase is full. Thus, h''( t)> 0 for t1 < t < t2andh''( t) < 0 for t < t < 8 .24ht ()2t1t 2 8tsolving for h gives h= 3 2400 t+ 27000 − 30 .πInstructor’s Resource Manual Section 3.2 167

56. V = 20 − .1t, 0≤t≤ 200. The height of the wateris always decreasing so h'( t ) < 0 . The rate ofchange in the height increases (the rate is negative,and its absolute value decreases) for the first 100days and then decreases for the remaining time.Therefore we have h''( t ) > 0 for 0< t < 100, andh''( t ) < 0 for 100 < t < 200 .3.3 Concepts Review1. maximum2. maximum; minimum3. maximum4. local maximum, local minimum, 0Problem Set 3.357. a. The cross-sectional area of the vase isapproximately equal to ΔV and thecorresponding radius is r = Δ V / π . Thetable below gives the approximate values for r.The vase becomes slightly narrower as youmove above the base, and then gets wider asyou near the top.Depth V A ≈ ΔV r = Δ V / π1 4 4 1.132 8 4 1.133 11 3 0.984 14 3 0.985 20 6 1.386 28 8 1.60b. Near the base, this vase is like the one in part(a), but just above the base it becomes larger.Near the middle of the vase it becomes verynarrow. The top of the vase is similar to theone in part (a).Depth V A ≈ ΔV r = Δ V / π1 4 4 1.132 9 5 1.263 12 3 0.984 14 2 0.805 20 6 1.386 28 8 1.602f′ ( x) = 3 x –12x = 3 x( x–4)1.Critical points: 0, 4f ′( x)> 0 on (– ∞ , 0), f′ ( x) < 0 on (0, 4),f′ ( x) > 0 on (4, ∞ )f′′ ( x) = 6 x–12; f′′ (0) = –12, f′′(4) = 12.Local minimum at x = 4;local maximum at x = 02 2f′ ( x) = 3 x –12=3( x –4)2.Critical points: –2, 2f′ ( x) > 0 on (– ∞ , –2), f′ ( x) < 0 on (–2, 2),f′ ( x) > 0 on (2, ∞ )f′′ ( x) = 6 x; f′′ (–2) = –12, f′′(2) = 12Local minimum at x = 2;local maximum at x = –23. f ′⎛ π ⎞( θ ) = 2cos 2 θ; 2cos 2θ≠ 0 on ⎜0, ⎟⎝ 4 ⎠No critical points; no local maxima or minima on⎛ π ⎞⎜0, ⎟.⎝ 4 ⎠4.1 11f′ ( x) = + cos x; + cos x = 0 when cos x = – .2 22Critical points: 2 π,4 π3 3⎛ 2π⎞2π4πf′ ( x) > 0 on ⎜0, ⎟,f′ ⎛ ⎞( x) < 0 on ⎜ , ⎟,⎝ 3 ⎠⎝ 3 3 ⎠4πf′ ⎛ ⎞( x) > 0 on ⎜ , 2π⎟⎝ 3 ⎠2 3 4 3f ( x) –sin x; f ππ′′⎛ ⎞ ⎛ ⎞= ′′ ⎜ ⎟= – , f′′⎜ ⎟=⎝ 3 ⎠ 2 ⎝ 3 ⎠ 24 πLocal minimum at x = ; local maximum at32 πx = .3168 Section 3.3 Instructor’s Resource Manual

5. Ψ ′( θ ) = 2sinθ cosθ6.π π− < θ 0 on ⎜0, ⎟,⎝ 2 ⎠⎝ 2 ⎠2 2Ψ ′′( θ ) = 2cos θ –2sin θ;Ψ ′′(0) = 2Local minimum at x = 03r′ ( z) = 4zCritical point: 0r′ ( z) < 0 on ( −∞ , 0);r′ ( z) > 0 on (0, ∞ )2r′′ ( x) 12 x ; r′′(0) 0;= = the Second DerivativeTest fails.Local minimum at z = 0; no local maxima7. f '( x)2( x + 4) ⋅1−x( 2x)24 − x2( x + 4) 2( x + 4)= =Critical points: − 2,22 2f '( x ) < 0 on ( −∞, − 2)and ( )f '( x ) > 0 on ( − 2, 2)f ''( x)2x( x − )23( x + 4)2 12=2,∞ ;11f ''( − 2)= ; f ''( 2)=−1616Local minima at x =− 2 ; Local maxima at x = 28. g'( z)2 2( + )( ) − ( )1 z 2z z 2z 2z= =Critical point: z = 0−∞ ,0g' ( z ) < 0 on ( )g' ( z ) > 0 on ( 0,∞ )g''( z)−=2 2( 1+ z ) ( 1+z )223 ( z −1)23( z + 1)g ''( 0)= 2Local minima at z = 0 .2 219. h' ( y) = 2y+2y3 4Critical point: −2⎛h' ( y )3 4 ⎞< 0 on ⎜−∞,− ⎟⎝ 2 ⎠⎛h' ( y )3 4 ⎞> 0 on ⎜−,0 ⎟⎝ 2 ⎠2h'' ( y) = 2 −3y34 2 16h ⎛⎜− ⎞⎟= 2− = 2+ = 63⎝ 2 ⎠ −434( 2 )3 4Local minima at −210. f '( x)and ( 0,∞ )2( )( ) ( )( )x + 1 3 − 3x+ 1 2x 3−2x −3x= =2 2( x + 1) ( x + 1)2 2The only critical points are stationary points. Findthese by setting the numerator equal to 0 andsolving.23− 2x− 3x= 0a = − 3, b = − 2, c = 32( ) ( )( )( − )2± −2 −4 − 3 3 2± 40 − 1±10x = = =2 3 −6 3Critical points:f '( x ) < 0 on⎛− +⎜⎝1 10 ,3f '0 ( ) > 0onf ''( x)−1−103− 1+10and3⎞⎟and⎠⎛ −1−10⎜−∞,⎝ 3⎞∞⎟.⎠⎛−1− 10 − 1+10 ⎞⎜,3 3 ⎟⎝⎠3 2( x + x − x−)23( x + 1)23 3 9 1=⎛−1−10 ⎞f '' ⎜≈ 0.7393 ⎟⎝ ⎠⎛− 1+10 ⎞f '' ⎜≈−2.7393 ⎟⎝ ⎠−1−10Local minima at x = ;3− 1+10Local maxima at x =32Instructor’s Resource Manual Section 3.3 169

11.2 2f′ ( x) = 3 x –3=3( x –1)Critical points: –1, 1f′′ ( x) = 6 x; f′′ (–1) = –6, f′′(1) = 6Local minimum value f(1) = –2;local maximum value f(–1) = 26 ⎛ ⎛ 2 ⎞ ⎞ 6 ⎛15⎞r′′ () s = – ; r′′ – = –8/525s⎜ ⎜ ⎟ ⎜ ⎟⎝15 ⎠ ⎟ 25 ⎝ 2⎝ ⎠ ⎠⎛ 53⎛ 2 ⎞ ⎞r′ ( s) < 0 on − , 0 , r′( s) > 0 on (0, ∞)⎜ ⎜ ⎟⎝15⎠ ⎟⎝⎠5/3 8/312.3 2g′ ( x) = 4x + 2x = 2 x(2x+ 1)Critical point: 02g′′ ( x) = 12x + 2; g′′(0) = 2Local minimum value g(0) = 3; no local maximumvaluesLocal minimum value r(0) = 0; local maximumvalue2 2 2 3 2r ⎛ ⎛ ⎞– ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎟= –3 + =⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝15 ⎠ ⎟ ⎝15 ⎠ ⎝15 ⎠ 5 ⎝15⎝ ⎠⎠5/3 5/3 2/3 2/313.14.15.16.3 2 2H′ ( x) = 4 x –6x = 2 x (2 x–3)3Critical points: 0, 22H′′ ( x) = 12 x –12x = 12 x( x–1);H ′′(0) = 0,3H ′′ ⎜⎛⎟⎞ = 9⎝2⎠3H′ ⎛ ⎞( x) < 0 on ( −∞ , 0), H′( x) < 0 on ⎜0,⎟⎝ 2 ⎠3 27Local minimum value H ⎛ ⎜⎞ ⎟ – ;2 = no local⎝ ⎠ 16maximum values (x = 0 is neither a localminimum nor maximum)4f′ ( x) = 5( x–2)Critical point: 23f′′ ( x) = 20( x– 2) ; f′′(2) = 0f′ ( x) > 0 on ( −∞ , 2), f′( x) > 0 on (2, ∞ )No local minimum or maximum values2g′ () t – ; g′() t1/33( t – 2)= does not exist at t = 2.Critical point: 22 2g′ (1) = , g′(3) = –3 3No local minimum values; local maximum valueg(2) = π .3/52 15s+ 2r′ () s = 3 + = ; r′() s = 0when3/5 3/55s5s5/3⎛ 2 ⎞s = – ⎜ ⎟ , r′( s)does not exist at s = 0.⎝15⎠5/3⎛ 2 ⎞Critical points: – ⎜ ⎟ , 0⎝15⎠17.18.19.20.12f′ () t = 1+tNo critical pointsNo local minimum or maximum values2xx ( + 8)f′ ( x)=2 3/2( x + 4)Critical point: 0f′ ( x) < 0 on ( −∞ , 0), f′( x) > 0 on (0, ∞ )Local minimum value f(0) = 0, no local maximumvalues1Λ ′( θ ) = – ; Λ′( θ ) does not exist at1+sinθ3 πθ = , but Λ ( θ ) does not exist at that point2either.No critical pointsNo local minimum or maximum valuessinθcosθπ 3πg′ ( θ) = ; g′( θ) = 0 when θ = , ;sinθ2 2g′ ( θ ) does not exist at x = π .⎛ π ⎞Split the x -axis into the intervals ⎜0, ⎟,⎝ 2 ⎠⎛π ⎞ ⎛ 3π⎞ ⎛3π⎞⎜ , π⎟, ⎜π, ⎟, ⎜ , 2 π⎟.⎝2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠π 3π 5π 7π π 1Test points: , , , ; g′ ⎛ ⎞⎜ ⎟=,4 4 4 4 ⎝4⎠2g 3 1 5 1 7 1′⎛ π ⎞– , , –4 2 g ⎛ π ⎞4 2 g ⎛ π ⎞⎜ ⎟= ′ ⎜ ⎟= ′ ⎜ ⎟=⎝ ⎠ ⎝ ⎠ ⎝ 4 ⎠ 2Local minimum value g(π ) = 0; local maximumvalues g⎛π⎞ ⎛3π⎞⎜ ⎟=1 and g ⎜ ⎟ = 1⎝2⎠ ⎝ 2 ⎠170 Section 3.3 Instructor’s Resource Manual

21. f '( x) = 4( sin2x)( cos2x)4sin2 ( x)( cos2x ) = 0when( 2k−1)πx =4orkπx = where k is an integer.2Critical points: 0, π4, π2, 2⎛π⎞ ⎛π⎞f ( 0)= 0; f ⎜ ⎟ = 1; f ⎜ ⎟ = 0 ;⎝ 4 ⎠ ⎝ 2 ⎠f ( 2)≈ 0.5728⎛π⎞Minimum value: f ( 0)= f ⎜ ⎟=0⎝ 2 ⎠⎛π⎞Maximum value: f ⎜ ⎟ = 1⎝ 4 ⎠22. f '( x)−=22( x −4)22( x + 4)f '( x ) = 0 when x = 2 or x =− 2 . (there are nosingular points)Critical points: 0, 2 (note: − 2 is not in the givendomain)1f ( 0)= 0; f ( 2)= ; f ( x) → 0 as x →∞.2Minimum value: f ( 0)= 01Maximum value: f ( 2)=223. g'( x)−=3x( x −64)32( x + 32)g' ( x ) = 0 when x = 0 or x = 4 .Critical points: 0, 41g ( 0)= 0; g ( 4)=6As x approaches ∞ , the value of g approaches 0but never actually gets there.1Maximum value: g ( 4)=6Minimum value: g ( 0)= 025.3 39F′ ( x) = –4; –4= 0 when x =x x169Critical points: 0, , 4169 9F(0) = 0, F ⎛ ⎜⎞ ⎟ ,16 = F(4) = –4⎝ ⎠ 4Minimum value F(4) = –4; maximum valueF 9 9⎜⎛ ⎟⎞ ⎝16 ⎠= 426. From Problem 25, the critical points are 0 and 9 .169 9F′ ( x) > 0 on ⎜ ⎛ 0, ⎟ ⎞ , F′( x) < 0 on ⎜ ⎛ , ∞⎟⎞⎝ 16 ⎠ ⎝16⎠⎛ 9 ⎞F decreases without bound on ⎜ , ∞⎟.⎝16⎠ No9 9minimum values; maximum value F ⎜⎛ ⎟⎞ ⎝16 ⎠= 427.2f ′−( x) = 64( −1)(sin x) cos x−2+ 27( −1)(cos x) ( −sin x)64cos x 27sin x=− +2 2sin x cos x2 2(3sin x − 4cos x)(9sin x+ 12 cos xsin x+16cos x)=2 2sin xcosx⎛ π ⎞On ⎜0, ⎟, f′ ( x) = 0 only where 3sin x = 4cos x;⎝ 2 ⎠4tan x = ;3−1 4x = tan ≈ 0.92733Critical point: 0.9273For 0 < x < 0.9273, f′ ( x) < 0, while forπ0.9273 < x < , f '( x ) > 0 2Minimum valueno maximum value⎛ −14⎞ 64 27f ⎜tan ⎟ = + = 125;⎝ 3 ⎠ 4 35 524. h'( x)='( ) 0−2x2( x + 4) 2h x = when x = 0 . (there are no singularpoints)Critical points: 0Since h' ( x ) < 0 for x > 0 , the function is alwaysdecreasing. Thus, there is no minimum value.1Maximum value: h ( 0)=4Instructor’s Resource Manual Section 3.3 171

28.2 2(8 −x) (32 x) −(16 x )2(8 −x)( −1)g′ ( x) = 2x+4(8 − x)3256x 2 x[(8 − x) + 128]= 2x+ =3 3(8 −x) (8 −x)For x > 8, g′ 3( x) = 0 when (8 − x) + 128 = 0;3 3(8 − x) = −128; 8 − x =− 128 ;3x = 8 + 4 2 ≈ 13.043g′ ( x) < 0 on (8, 8 + 4 2),3g′ ( x) > 0 on (8+ 4 2, ∞ )g(13.04) ≈ 277 is the minimum value29. H'( x)2( − )2x x 1=2x −1H' ( x ) = 0 when x = 0 .H '( x ) is undefined when x =− 1 or x = 1Critical points: − 2 , − 1 , 0, 1, 22 3 1 0 0 1H ( − ) = ; H ( − ) = ; H ( ) = ; ()H ( 2)= 3Minimum value: H( − 1) = H( 1)= 0Maximum value: H( ) H( )− 2 = 2 = 3H 1 = 0;230. h'() t = 2tcost2π6πh' () t = 0 when t = 0 , t = , t = , and2 210πt =22 π 2 3π2 5π(Consider t = , t = , and t = )2 222π 6π 10π Critical points: 0, , , , π2 2 2⎛ 2π⎞ ⎛ 6π⎞h ( 0)= 0;h⎜⎟ = 1;h⎜⎟ =− 1 ;⎝ 2 ⎠ ⎝ 2 ⎠⎛ 10π⎞h⎜⎟ = 1⎝ 2 ⎠; h( π ) ≈− 0.4303⎛ 6π⎞Minimum value: h⎜⎟ =− 1⎝ 2 ⎠⎛ 2π⎞ ⎛ 10π⎞Maximum value: h⎜ ⎟= h⎜ ⎟=1⎝ 2 ⎠ ⎝ 2 ⎠31. f '( x ) = 0 when x = 0 and x = 1 . On the interval( −∞,0)we get f '( x ) < 0 . On (0, ∞ ) , we getf '( x ) > 0 . Thus there is a local min at x = 0 butno local max.32. f '( x ) = 0 at x = 1, 2, 3, 4 ; f '( x)is negative on( −∞,1) ∪(2,3) ∪(4, ∞)and positive on(1, 2) ∪ (3, 4). Thus, the function has a localminimum at x = 1,3 and a local maximum atx = 2, 4 .33. f '( x ) = 0 at x = 1, 2, 3, 4 ; f '( x)is negative on(3, 4) and positive on( −∞,1) ∪(1,2) ∪(2,3) ∪(4, ∞ ) Thus, the functionhas a local minimum at x = 4 and a localmaximum at x = 3 .34. Since f '( x)≥ 0 for all x, the function is alwaysincreasing. Therefore, there are no local extrema.35. Since f '( x)≥ 0 for all x, the function is alwaysincreasing. Therefore, there are no local extrema.36. f '( x ) = 0 at x = 0, A, and B.f '( x ) is negative on ( −∞ ,0)and ( A,B )f '( x ) is positive on ( 0, A ) and ( B,∞ )Therefore, the function has a local minimum atx = 0 and x = B , and a local maximum at x = A .37. Answers will vary. One possibility:5−5y3 638. Answers will vary. One possibility:5−5y3 6xx172 Section 3.3 Instructor’s Resource Manual

39. Answers will vary. One possibility:y53 6 x−540. Answers will vary. One possibility:y53 6 x−541. Answers will vary. One possibility:y53 6 x−542. Answers will vary. One possibility:y53 6 x−543. The graph of f is a parabola which opens up.Bf '( x)= 2Ax+ B = 0 → x =−2Af ''( x)= 2ASince A > 0 , the graph of f is always concave up.There is exactly one critical point which yields theminimum of the graph.⎛ B ⎞ ⎛ B ⎞ ⎛ B ⎞f ⎜− ⎟= A⎜− ⎟ + B⎜− ⎟+C⎝ 2A⎠ ⎝ 2A⎠ ⎝ 2A⎠2 2B B= − + C4A2A2 2B − 2B + 4AC=4A2 24AC−B B −4AC= = −4A4AIf f ( x) ≥ 0 with 0222A > , then ( B AC)− −4 ≥ 0,or B − 4AC≤ 0.2⎛ B ⎞If B − 4AC≤ 0, then we get f ⎜−⎟≥0⎝ 2A⎠⎛ B ⎞Since 0 ≤ f ⎜− ⎟≤f ( x)for all x, we get⎝ 2A⎠f ( x) ≥ 0 for all x.44. A third degree polynomial will have at most twoextrema.2f '( x)= 3Ax + 2Bx+Cf '' x = 6Ax+2B( )Critical points are obtained by solving f '( x ) = 0.23Ax + 2Bx + C = 02− 2B± 4B −12ACx =6A2− 2B± 2 B −3AC=6A2− B± B −3AC=3ATo have a relative maximum and a relativeminimum, we must have two solutions to theabove quadratic equation. That is, we must have2B − 3AC> 0.2−B− B −3ACThe two solutions would be3A2− B+ B −3ACand. Evaluating the second3Aderivative at each of these values gives:Instructor’s Resource Manual Section 3.3 173

⎛2−B− B −3AC⎞f ''⎜ 3A⎟⎝⎠⎛2−B− B −3AC⎞= 6A+ 2B⎜ 3A⎟⎝⎠2=−2B−2 B − 3AC + 2B2=−2 B −3ACand⎛2− B+ B −3AC⎞f ''⎜ 3A⎟⎝⎠⎛2− B+ B −3AC⎞= 6A+ 2B⎜ 3A⎟⎝⎠=− 2B+ 22B − 3AC + 2B2= 2 B −3AC2If B − 3AC> 0, then −2 B − 3ACexists and2is negative, and 2 B − 3ACexists and ispositive.Thus, from the Second Derivative Test,−B− B −3AC3A2− B+ B −3ACand3Aminimum.22would yield a local maximumwould yield a local45. f′′′ () c > 0implies that f ′′ is increasing at c, so fis concave up to the right of c (since f′′ ( x) > 0 tothe right of c) and concave down to the left of c(since f′′ ( x) < 0 to the left of c). Therefore f has apoint of inflection at c.3.4 Concepts Review1. 0 < x < ∞2.2002x +xn3. S = ∑ ( y ) 2i −bxii=14. marginal revenue; marginal costProblem Set 3.41. Let x be one number, y be the other, and Q be thesum of the squares.xy = –1616y = –xThe possible values for x are in (– ∞ , 0) or (0, ∞ ) .2 2 2 256Q = x + y = x +2xdQ 512= 2 x –dx3x5122 x – = 03x4x = 256x = ±4The critical points are –4, 4.dQdQ< 0 on (– ∞ , –4) and (0, 4). 0dx dx > on(–4, 0) and (4, ∞ ).When x = –4, y = 4 and when x = 4, y = –4. Thetwo numbers are –4 and 4.2. Let x be the number.Q = x – 8xx will be in the interval (0, ∞ ).dQ 1 –1/ 2x – 8dx = 21 –1/ 2– 8 02 x =–1/ 2x = 161x =256dQ0dx > on ⎛ 1 ⎞⎜0, ⎟⎝ 256 ⎠ and dQdx < 0 on ⎛ 1 ⎞⎜ , ∞⎟.⎝256⎠1Q attains its maximum value at x = .2563. Let x be the number.4Q = x – 2xx will be in the interval (0, ∞ ).dQ 1 –3/ 4x – 2dx = 41 –3/ 4– 2 04 x =–3/ 4x = 81x =16dQ0dx > on ⎛ 1 ⎞⎜0, ⎟⎝ 16 ⎠ and dQdx < 0 on ⎛ 1 ⎞⎜ , ∞⎟⎝16⎠Q attains its maximum value at1x = .16174 Section 3.4 Instructor’s Resource Manual

4. Let x be one number, y be the other, and Q be thesum of the squares.xy = –1212y = –xThe possible values for x are in (– ∞ , 0) or (0, ∞ ) .2 2 2 144Q = x + y = x +2xdQ 288= 2 x –dx3x2882 x – = 03x4x = 144x =± 2 3The critical points are –2 3, 2 3dQ< 0 on (– ∞ , – 2 3) and (0, 2 3).dxdQ> 0 on (–2 3, 0) and (2 3, ).dx ∞When x = –2 3, y = 2 3 and whenx = 2 3, y = –2 3.The two numbers are –2 3 and 2 3.5. Let Q be the square of the distance between (x, y)and (0, 5).2 2 2 2 2Q = ( x– 0) + ( y– 5) = x + ( x – 5)4 2= x – 9x+ 25dQ 34 x –18xdx =34 x –18x = 022 x(2 x –9) = 0x = 0, ±32dQ0dx < on ⎛ 3 ⎞⎜– ∞, – ⎟⎝ 2 ⎠ and ⎛ 3 ⎞⎜0, ⎟.⎝ 2 ⎠dQ0dx > on ⎛ 3 ⎞ ⎛ 3 ⎞⎜– , 0 ⎟ and ⎜ , ∞⎟.⎝ 2 ⎠ ⎝ 2 ⎠3 9 3When x = – , and when ,2 y =2x =29y = .2⎛ 3 9⎞ ⎛ 3 9⎞The points are ⎜– , ⎟ and ⎜ , ⎟.⎝ 2 2⎠ ⎝ 2 2⎠6. Let Q be the square of the distance between (x, y)and (10, 0).2 2 2 2 2Q = ( x–10) + ( y– 0) = (2 y –10) + y7.4 2= 4 y – 39y+ 100dQ 316 y – 78ydy =316 y – 78y = 022 y(8 y –39) = 039y = 0, ±2 2dQ ⎛ 39 ⎞ ⎛ 39 ⎞< 0 on – , – and 0, .dy ⎜∞2 2 ⎟ ⎜ 2 2 ⎟⎝ ⎠ ⎝ ⎠dQ ⎛ 39 ⎞ ⎛ 39 ⎞> 0 on – , 0 and , .dy ⎜ ∞2 2 ⎟ ⎜ 2 2 ⎟⎝ ⎠ ⎝ ⎠When39 39y = – , x = and when2 2 439 39y = , x = .2 2 4⎛39 39 ⎞ ⎛39 39 ⎞The points are⎜, – and , .4 2 2 ⎟ ⎜ 4 2 2 ⎟⎝ ⎠ ⎝ ⎠2x ≥ x if 0≤x ≤ 12f( x) = x− x ; f′( x) = 1−2 x;1f′ ( x) = 0 when x =21Critical points: 0, , 121 1f(0) = 0, f(1) = 0, f ⎛ ⎜⎞ ⎟ ;2 = therefore, 1⎝ ⎠ 42exceeds its square by the maximum amount.8. For a rectangle with perimeter K and width x, theKlength is − x . Then the area is2K Kx 2A x ⎛ ⎞= ⎜ − x⎟= −x.⎝ 2 ⎠ 2dA = K − 2 x; dA = 0 when x =Kdx 2 dx4K KCritical points: 0, ,4 22K K KAt x = 0 or , A = 0; at x = , A = .2 4 16The area is maximized when the width is onefourth of the perimeter, so the rectangle is asquare.Instructor’s Resource Manual Section 3.4 175

9. Let x be the width of the square to be cut out and Vthe volume of the resulting open box.2 3 2V = x(24 − 2 x) = 4x − 96x + 576xdV 212x 192x 576 12( x 12)( x 4);dx = − + = − −12(x – 12)(x – 4) = 0; x = 12 or x = 4.Critical points: 0, 4, 12At x = 0 or 12, V = 0; at x = 4, V = 1024.The volume of the largest box is31024 in.10. Let A be the area of the pen.2 dAA = x(80 − 2 x) = 80x− 2 x ; = 80 − 4 x;dx80 − 4x= 0; x = 20Critical points: 0, 20, 40.At x = 0 or 40, A = 0; at x = 20, A = 800.The dimensions are 20 ft by 80 – 2(20) = 40 ft,with the length along the barn being 40 ft.11. Let x be the width of each pen, then the lengthalong the barn is 80 – 4x.2 dAA = x(80 − 4 x) = 80x− 4 x ; = 80 − 8 x;dxdA0 when x 10.dx = =Critical points: 0, 10, 20At x = 0 or 20, A = 0; at x = 10, A = 400.The area is largest with width 10 ftand length 40 ft.12. Let A be the area of the pen. The perimeter is100 + 180 = 280 ft.y + y – 100 + 2x = 180; y = 140 – x2 dAA = x(140 − x) = 140 x− x ; = 140 − 2 x;dx140 − 2x= 0; x = 70Since 0≤ x ≤ 40, the critical points are 0 and 40.When x = 0, A = 0. When x = 40, A = 4000. Thedimensions are 40 ft by 100 ft.13.900xy = 900; y =xThe possible values for x are in (0, ∞ ).⎛900 ⎞ 2700Q = 4x+ 3y = 4x+ 3⎜⎟= 4x+⎝ x ⎠ xdQ 2700= 4 −dx 2x27004– = 02x2x = 675x =± 15 3x = 15 3 is the only critical point in (0, ∞ ).dQ< 0 on (0, 15 3) anddxdQ0 on (15 3, ).dx >∞900When x = 15 3, y = = 20 3.15 3Q has a minimum when x = 15 3 ≈ 25.98 ft andy = 20 3 ≈ 34.64 ft.30014. xy = 300; y =xThe possible values for x are in (0, ∞ ).1200Q = 6x+ 4y = 6x+xdQ 1200= 6–dx 2x12006– = 02x2x = 200x =± 10 2x = 10 2 is the only critical point in (0, ∞ ).dQdQ< 0 on (0, 10 2) and 0 on (10 2, )dx dx >∞300When x = 10 2, y = = 15 2.10 2Q has a minimum when x = 10 2 ≈ 14.14 ft andy = 15 2 ≈ 21.21 ft.30015. xy = 300; y =xThe possible values for x are in (0, ∞).3000Q = 3(6x + 2y) + 2(2y) = 18x + 10y = 18x +xdQ 3000= 18 –dx2x300018 – = 02x2 500x =310 5x =±310 5x = is the only critical point in (0, ∞).3dQ ⎛ 10 5 ⎞< 0 on 0,dx ⎜3 ⎟and⎝ ⎠dQ ⎛10 5 ⎞> 0 on , ∞ .dx ⎜3 ⎟⎝ ⎠176 Section 3.4 Instructor’s Resource Manual

When10 5 300x = , y = = 6 153Q has a minimum wheny = 6 15 ≈ 23.24 ft.10 5310 5x = ≈ 12.91 ft and390016. xy = 900; y =xThe possible values for x are in (0, ∞ ).3600Q = 6x+ 4y = 6x+xdQ 3600= 6–dx 2x36006– = 02x2x = 600x =± 10 6x = 10 6 is the only critical point in (0, ∞ ).dQdQ< 0 on (0, 10 6) and 0 on (10 6, ).dx dx > ∞900When x = 10 6, y = = 15 610 6Q has a minimum when x = 10 6 ≈ 24.49 ft andy = 15 6 ≈ 36.74.x 2It appears that .y = 3Suppose that each pen has area A.Axy = A; y =xThe possible values for x are in (0, ∞ ).4AQ = 6x+ 4y = 6x+xdQ 4A= 6–dx 2x4A6– = 02x2 2Ax =32Ax =±32Ax = is the only critical point on (0, ∞ ).3dQ ⎛ 2A ⎞< 0 on 0,dx ⎜3 ⎟and⎝ ⎠dQ ⎛ 2A⎞> 0 on , ∞ .dx ⎜3 ⎟⎝ ⎠When2A A 3Ax = , y = =3 2A232Ax 3 2y = = 3A3217. Let D be the square of the distance.22⎛2 2 2 x ⎞D = ( x− 0) + ( y− 4) = x + −4⎜ 4 ⎟⎝ ⎠4x 2= − x + 16163 3dD x x22; x 2x 0; x( x 8) 0dx = 4 − 4− = − =x = 0, x = ± 2 2Critical points: 0, 2 2, 2 3Since D is continuous and we are considering aclosed interval for x, there is a maximum andminimum value of D on the interval. Theseextrema must occur at one of the critical points.At x = 0, y = 0, and D = 16. At x = 2 2, y = 2,and D = 12. At x = 2 3, y = 3, and D = 13.2xTherefore, the point on y = closest to ( 0, 4 ) is4( 2 2,2)P and the point farthest from ( )Q ( 0,0).0, 4 is18. Let r 1 and h 1 be the radius and altitude of theouter cone; r 2 and h 2 the radius and altitude ofthe inner cone.1 2V1 r1 h13= π is fixed. r1 =3V 1πh 1h – h rh1 r1By similar triangles1 2 = 2(see figure).⎛ h2 ⎞ 3V1 ⎛ h2⎞r2 = r1⎜1– ⎟=⎜1–⎟⎝ h1 ⎠ πh1 ⎝ h1⎠Instructor’s Resource Manual Section 3.4 177

21 2 1 ⎡ 3V1 ⎛ h2⎞⎤V2 = π r2 h2 = π⎢ ⎜1–⎟⎥h23 3 ⎢⎣πh1 ⎝ h1⎠⎥⎦2 2π 3Vh 1 2⎛ h2⎞ h2⎛ h2⎞= ⋅ ⎜1– ⎟ = V1⎜1–⎟3 πh1 ⎝ h1 ⎠ h1 ⎝ h1⎠hLet k 2 ,h12then V2 = V1 k(1 – k) .dV22V1(1– k) –2 kV1(1– k) V1(1– k)(1–3 k)dk = =0 < k < 1 so= the ratio of the altitudes of the cones,dV20dk = when 1k = .32 2dV2 dV21= V2 1(6k− 4); < 0 when k =2dkdk3The altitude of the inner cone must be 1 3 thealtitude of the outer cone.19. Let x be the distance from P to where the womanlands the boat. She must row a distance of2x + 4 miles and walk 10 – x miles. This will2x + 4 10– xtake her T( x)= + hours;3 4x 10 ≤ x ≤ 10. T′ ( x) = – ; T′( x) = 023 x + 4 46when x = .719T (0) = hr = 3 hr 10 min ≈ 3.17 hr ,6⎛ 6 ⎞ 15+7T ⎜ ⎟ = ≈ 2.94 hr,⎝ 7 ⎠ 6104T (10) = ≈ 3.40 hr3She should land the boatshore from P.6≈ 2.27 mi down the72x + 4 10– x20. T( x) = + , 0≤ x≤10.3 50x 1T′ ( x) = – ; T′( x) = 0 when23 x –4 506x =2491136T (0) = ≈ 0.867 hr; T ⎛⎜⎞ ⎟≈0.865 hr;15⎝ 2491 ⎠T (10) ≈ 3.399 hr6She should land the boat ≈ 0.12 mi down2491the shore from P.21.2x + 4 10– xT( x) = + , 0≤ x≤10.20 4x 1T′ ( x) = – ; T′( x) = 0220 x + 4 42 10 13(0)20 4 5T = + = hr = 2 hr, 36 minhas no solution.104T (10) = ≈ 0.5 hr20She should take the boat all the way to town.22. Let x be the length of cable on land, 0 ≤ x ≤ L.Let C be the cost.2 2C = a ( L− x)+ w + bxdC a( L − x)= − + bdx 2 2( L− x)+ waL ( − x)− + b = 0 when2 2( L− x)+ w2 2 2 2 2b [( L− x) + w ] = a ( L−x)2 2 2 2 2( a −b )( L− x)= b wbwx = L− ft on land;2 2a – bawft under water2 2a – b2 2d C aw= > 0 for all x, so this2 2 2 3 2dx [( L − x) + w ]minimizes the cost.178 Section 3.4 Instructor’s Resource Manual

23. Let the coordinates of the first ship at 7:00 a.m. be(0, 0). Thus, the coordinates of the second ship at7:00 a.m. are (–60, 0). Let t be the time in hourssince 7:00 a.m. The coordinates of the first andsecond ships at t are (–20t, 0) and( 60 15 2 t , 15 2t)− + − respectively. Let D be thesquare of the distances at t.2 2= ( − 20 + 60 − 15 2 ) + ( 0 + 15 2 )2( ) t ( ) t= 2( 1300 + 600 2 ) t − ( 2400 + 1800 2 )( ) t ( )D t t t= 1300 + 600 2 − 2400 + 1800 2 + 3600dDdt2 1300 + 600 2 − 2400 + 1800 2 = 0 when12 + 9 2t = ≈1.15hrs or 1 hr, 9 min13 + 6 212 + 9 2D is the minimum at t = since13 + 6 2for all t.The ships are closest at 8:09 A.M.2d D> 02dtb 2 224. Write y in terms of x: y = a − x (positiveasquare root since the point is in the first quadrant).Compute the slope of the tangent line:bxy′ =− .2 2a a − xFind the y-intercept, y 0 , of the tangent linethrough the point (x, y):y0− y bx =−0 − x 2 2a a − x2 2bx bx b 2 2y0= + y = + a −x2 2 2 2a a −x a a −xaab=2 2a − xFind the x-intercept, x 0, of the tangent linethrough the point (x, y):y–0bx= –x– x02 2a a – x2 2 2 2 2ay a – x a – x ax0= + x = + x =bx x xCompute the Area A of the resulting triangle andmaximize:3 3 −11 ab ab 2 2A= x0y0= =⎛x a x⎞⎜ − ⎟2 2 22x a − x 2 ⎝ ⎠3 −22dA a b ⎛2 2 2 2 x ⎞=−⎛x a x⎞⎜ − ⎟a −x−dx 2 ⎝ ⎠ ⎜2 2 ⎟⎝a − x ⎠3ab2 2=2 2 2 3/22 ( )(2 x – ax a − x)3ab2 22 2 2 3/2 (2 x − a2 x ( a x )) =−0a b 2 ⎛ a ⎞ bx = ; y = a − ⎜ ⎟ =2 a ⎝ 2 ⎠ 2⎛ a ⎞b⎜⎟2 by′ = −⎝ ⎠= −2 a2 ⎛ a ⎞a a − ⎜ ⎟⎝ 2 ⎠dA ⎛ a ⎞Note that < 0 on ⎜0,⎟dx ⎝ 2 ⎠ and2whendA ⎛ a ⎞> 0 on ⎜ , a ⎟,so A is a minimum atdx ⎝ 2 ⎠ax = . Then the equation of the tangent line is2b⎛a ⎞ by = − ⎜x − ⎟+ or bx + ay − ab 2 = 0 .a ⎝ 2 ⎠ 225. Let x be the radius of the base of the cylinder andh the height.2 22 2 2 ⎛h⎞2 2 hV =π x h; r = x + ⎜ ⎟ ; x = r −⎝2⎠4⎛ 2 32 h ⎞2 πhV =π r − h =πhr−⎜ 4 ⎟⎝ ⎠4dV 22 3π2 3=πr − h ; V′= 0 when h=±rdh 4 32dV 3πhSince,2dh 22 3rwhen h = .3=− the volume is maximized2 3( r) 33⎛2 3 ⎞ π2V =π ⎜r r −3 ⎟⎝ ⎠ 42π 3 3 2π 3 3 4π3 3= r − r = r3 9 926. Let r be the radius of the circle, x the length of therectangle, and y the width of the rectangle.P = 2x + 2y; r2 2y = 4 r − x ;dP 2x= 2 − ;dx 2 24r − x2 2 2 22 ⎛ x ⎞ ⎛ y ⎞ 2= ⎜ ⎟ + ⎜ ⎟ ; r = x +y ;⎝2⎠ ⎝ 2⎠4 42 2P = 2x+ 2 4r − xInstructor’s Resource Manual Section 3.4 179

2x2 22− = 0; 2 4r − x = 2 x;2 24r− x2 2 216r − 4x = 4 x ; x = ± 2r2 2d P 8r=− < 0 x =2 2 2 3 2dx (4 r − x )300 3 400A' ( x ) = 0 when x = − ≈ 10.874 .11 11Critical points: x = 0, 10.874, 25At x = 0, A ≈ 481; at x = 10.874, A ≈ 272; atx = 25, A = 625.when 2; ra. For minimum area, the cut should be2 2y = 4r − 2r = 2rapproximately 4(10.874) ≈ 43.50 cm fromone end and the shorter length should be bentThe rectangle with maximum perimeter is a squareto form the square.with side length 2r .b. For maximum area, the wire should not be27. Let x be the radius of the cylinder, r the radius ofcut; it should be bent to form a square.the sphere, and h the height of the cylinder.2 230. Let x be the length of the sides of the base, y be2 2 h 2 hA =2π xh ; r = x + ; x = r −the height of the box, and k be the cost per square4 4inch of the material in the sides of the box.2 42 h2 2 hV = x 2 y ;A= 2π r − h = 2π h r −4 42 2The cost is C = 1.2kx + 1.5kx + 4kxy2 3dAπ( 2r h−h)2 ⎛ V ⎞ 2 4kV= ; A′= 0 when h= 0, ± 2r= 2.7kx + 4kx⎜dh 2 2 h42 ⎟= 2.7kx+⎝ x ⎠xhr −4dC 4kV dC3dAdA5.4 kx ; 0 when x 0.905 V> 0 on (0, 2 r) and < 0 on ( 2 r, 2 r),dx= − 2x dx= ≈dhdhV3so A is a maximum when h = 2. ry ≈≈1.22V3 2(0.905 V )rThe dimensions are h= 2, r x = .231. Let r be the radius of the cylinder and h the heightof the cylinder.28. Let x be the distance from I 1 .2 32 2 V − πr 3 3 V 2kI1 kI2V =π r h+ π r ; h = = − rQ = +x2 23 2 2πrπr3( s − x )Let k be the cost per square foot of the cylindricaldQ −2kI1 2kI2= +wall. The cost isdx 3 3x ( s−x)2C = k(2 π rh) + 2 k(2 π r )32kI1 2kI2 x I1− + = 0; = ;⎛ 2⎛ V 2 ⎞ 2 ⎞ ⎛2V 8πr⎞3 3 3= kx ( s−x) ( s−x)I⎜2πr⎜ − r22 ⎟+ 4π r ⎟= k+⎝ ⎝πr3 ⎠ ⎠⎜ r 3 ⎟⎝ ⎠s3 Ix = 1dC ⎛ 2 V 16πr ⎞ ⎛ 2 16; V π r ⎞= k3 I 31 + I⎜− +2 ⎟ k⎜− +2 ⎟=02dr ⎝ r 3 ⎠ ⎝ r 3 ⎠21/3d Q 6kI1 6kI23 3V1⎛3V⎞= + > 0, so this pointwhen r = , r =2 4 4⎜ ⎟dx x ( s − x)8π2⎝π ⎠minimizes the sum.1/3 1 34V 1⎛3V ⎞ ⎛3V⎞h = −2/3 ⎜ ⎟ = ⎜ ⎟29. Let x be the length of a side of the square, soπ 3V3 ⎝ π ⎠ ⎝ π ⎠( π )100 − 4xis the side of the triangle, 0 ≤ x ≤ 25For a given volume V, the height of the cylinder is31/31/3⎛3V⎞1⎛3V⎞2 1 ⎛100 −4x⎞ 3 ⎛100 −4x⎞⎜ ⎟ and the radius is ⎜ ⎟ .A= x + ⎜ ⎟ ⎜ ⎟⎝ π ⎠2 ⎝ π ⎠2⎝ 3 ⎠ 2 ⎝ 3 ⎠22 3 ⎛10,000 − 800x+ 16x⎞= x + 4 ⎜ 9 ⎟⎝⎠dA 200 3 8 32xxdx = − 9 + 9180 Section 3.4 Instructor’s Resource Manual

dx32. 2cos2t2 3sin2 t;dt = −33.dx10 when tan2t= ;dt =3π2t= +π n for any integer n6π πt = + n12 2π πWhen t = + n,12 2⎛π⎞ ⎛π⎞x = sin ⎜ +π n⎟+ 3 cos⎜ +πn⎟⎝6 ⎠ ⎝6⎠ππ= sin cos π n+ cos sin π n6 6π π+ 3(cos cosπn−sin sin π n)6 6n 1 n 3= ( − 1) + ( − 1) = 2.2 2The farthest the weight gets from the origin is 2units.2r θ 2AA = ; θ =2 2rThe perimeter is2Ar2AQ = 2r+ rθ= 2r+ = 2r+2r rdQ 2= 2 − A ; Q ′= 0 when r = Adr 2r2Aθ = = 22( A)2d Q 4A= > 0, so this minimizes the perimeter.2 3dr r34. The distance from the fence to the base of theladder is tanhθ .The length of the ladder is x.h + wtanθhcos θ = ; x cos θ = + w;xtanθh wx = +sinθcosθ3 3dx h cosθ wsinθ wsin θ − h cos θ=− + ; = 0dθ 2 2 2 2sin θ cos θ sin θ cos θ3 hwhen tan θ =wtan−1 3hθ =w3 3hhtan θ = ; sin θ =,3w 2/3 2/3h + w3wcosθ =2/3 2/3h + w⎛ 2/3 2/3 2/3 2/3h + w⎞ ⎛h + w⎞x = h⎜ ⎟+w⎜ ⎟⎜ 3 3h ⎟ ⎜ w ⎟⎝ ⎠ ⎝ ⎠2/3 2/3 3/2= ( h + w )35. x is limited by 0≤ x ≤ 12 .2 3 dA2A = 2 x(12 − x ) = 24x− 2 x ; = 24 − 6 x ;dx224 − 6x= 0; x = − 2, 2Critical points: 0, 2, 12.When x = 0 or 12, A = 0.2When x = 2, y = 12 − (2) = 8.The dimensions are 2x = 2(2) = 4 by 8.36. Let the x-axis lie on the diameter of the semicircleand the y-axis pass through the middle.2 2Then the equation y = r − x describes thesemicircle. Let (x, y) be the upper-right corner ofthe rectangle. x is limited by 0 ≤ x ≤ r .2 2A = 2xy = 2x r − x2dA 2 2 2x2 2 2dx = 2 r −x − = ( r −2 x )r 2 x 2 r 2 x 2− −2 2 2 r( r − 2 x ) = 0; x =2 2r − x2rCritical points: 0, , r2r 2When x = 0 or r, A = 0. When x = , A = r .222 ⎛ r ⎞ ry = r − ⎜ ⎟ =⎝ 2 ⎠ 2The dimensions arer 2r by .2 2Instructor’s Resource Manual Section 3.4 181

37. If the end of the cylinder has radius r and h is the⎛ d ⎞height of the cylinder, the surface area isA′ ( w) < 0 on ⎜ , d⎟. Maximum area is for a⎝ 2 ⎠2AA = 2π r + 2π rh so h = – r.square.2 πrThe volume ish2 2⎛A ⎞ Ar40. Note that cos t = , so h = r cos t,3V =π r h =π r ⎜ – r⎟= – πr.r⎝2πr⎠ 21 2 2 2 2sin t = r – h , and r – h = rsintA 2ArV′ () r = –3 π r ; V′() r = 0when r = ,26π2 2 2Area of submerged region = tr – h r – hV′′ () r =−6 πr,so the volume is maximum when2 2= tr – ( r cos t)( r sin t) = r ( t –cost sin t)Ar = .A = area of exposed wetted region6π2 2 2=πr – π h – r ( t –costsin t)A Ah = – r 2 2r2πr= 6π=2 2= r ( π– π cos t – t+costsin t)dA r2 2 2(2 cos t sin t –1 cos t – sin t )38. The ellipse has equationdt = π +2 22 22 b x b= r (2πcostsin t – 2sin t)2 2y =± b – =± a – x2a a2= 2r sin t( π cos t –sin t)⎛ b 2 2 ⎞Let ( x, y) = ⎜ x, a – x ⎟ be the upper righthandcorner of the rectangle (use a and b positive).dt⎝ a ⎠dASince 0 < t < π , = 0 only whenThen the dimensions of the rectangle are 2x byπ cost = sin t or tan t = π . In terms of r and h,2b a2 21 2 2– x and the area isr – hathis isrr= π or h = .h24bx2 2r1+πA( x) = a – x .a41. The carrying capacity of the gutter is maximized2 2 24b 2 2 4bx 4 b( a –2 x )A′ ( x) = a – x – =;when the area of the vertical end of the gutter isa 2 2 2 2a a – x a a – xmaximized. The height of the gutter is 3sinθ . Thearea isaA′ ( x) = 0 when x = , so the corner is at⎛1⎞2A = 3(3sin θ ) + 2 ⎜ ⎟ (3cos θ)(3sin θ)⎝2⎠⎛ a b ⎞⎜ , ⎟.The corners of the rectangle are at= 9sinθ + 9cosθsinθ.⎝ 2 2 ⎠ dA9cosθ 9( sin θ)sinθ 9cosθ cosθ⎛ a b ⎞ ⎛ a b ⎞ ⎛ a b ⎞⎜ , ⎟, ⎜– , ⎟, ⎜– , – ⎟⎝ 2 2 ⎠ ⎝ 2 2 ⎠ ⎝ 2 2 ⎠ ,dθ = + − +2 2= 9(cosθ − sin θ + cos θ)⎛ a b ⎞⎜ , − ⎟⎝ 2 2⎠ .2= 9(2cos θ + cosθ− 1)21 πThe dimensions are a 2 and b 2 .2cos θ + cosθ− 1=0; cosθ=− 1, ; θ =π ,2 3π39. If the rectangle has length l and width w, theSince 0 ≤θ≤ , the critical points are2diagonal is d = l + w2 2 22, so l = d – w . Theπ π2 20, , and .area is A = lw = w d – w .3 22 2 22 2 w d –2wWhen θ = 0 , A = 0.A′ ( w) = d – w – = ;2 2 2 2π 27 3d – w d – wWhen θ = , A = ≈ 11.7.3 4dA′ ( w) = 0 when w= and soπ2When θ = , A = 9.22 2d d⎛ d ⎞l = d – = . A′ ( w) > 0 on 0,2⎜ ⎟2⎝ 2 ⎠ and πThe carrying capacity is maximized when θ = .3182 Section 3.4 Instructor’s Resource Manual

42. The circumference of the top of the tank is thecircumference of the circular sheet minus the arclength of the sector,20π− 10θ meters. The radius of the top of the20π−10θ5tank is r = = (2 π−θ) meters. The2ππslant height of the tank is 10 meters, so the heightof the tank is22 ⎛ 5θ⎞ 52h = 10 −⎜10 − ⎟ = 4πθ−θ⎝ π ⎠ πmeters.22 21 1 ⎡ 5 ⎤ ⎡5⎤V = π r h = π (2 π−θ ) 4πθ −θ3 3 ⎢π⎥ ⎢π⎥⎣ ⎦ ⎣ ⎦125 2 2= (2 π−θ ) 4 πθ −θ23πdV 125 2=⎛2 ⎜ 2(2 π−θ )( − 1) 4 πθ −θdθ3π⎝2(2 π−θ) 1( 2 )(4π−2 θ ) ⎞+⎟24πθ−θ⎟⎠125(2 π−θ) 2 2= (3θ−12π θ + 4 π ) ;2 23π 4πθ−θ125(2 π−θ) 2 2(3θ−12π θ + 4 π ) = 02 23π 4πθ−θ2 22π− θ = 0 or 3θ−12π θ + 4π = 02 6 2 6θ = 2 π , θ = 2 π− π , θ = 2π+ π3 3Since 0 θ 2 ,2 62π−3the volume.< < π the only critical point isπ. A graph shows that this maximizes44. Let x be the length of the edges of the cube. The21surface area of the cube is 6x so 0 ≤ x ≤ .62The surface area of the sphere is 4 π r , so22 2 1–6x6x + 4π r = 1, r =4π3 4 3 3 1 (1 – 62 )3/2V = x + π r = x + x3 6 π2dV 2 3⎛2 1–6x⎞= 3 x – x 1–6x = 3 x⎜x–⎟dx π⎜ π ⎟⎝⎠dV10 when x 0,dx = = 6 +π13V (0) = ≈0.094 m .6 π3/2V ⎛ 1 –3/ 2 1 6⎜ ⎞ ⎟= (6 +π ) +⎛ ⎜1– ⎞⎟⎝ 6 +π⎠6 π⎝ 6 +π⎠⎛ π ⎞ –3/2 13= ⎜1 + ⎟(6 +π ) = ≈0.055 m⎝ 6 ⎠6 6+πFor maximum volume: no cube, a sphere of radius1≈ 0.282 meters.2 πFor minimum volume: cube with sides of length1≈ 0.331 meters,6 +π1sphere of radius ≈ 0.165 meters2 6+π45. Consider the figure below.43. Let V be the volume. y = 4 – x and z = 5 – 2x.x is limited by 0≤ x ≤ 2.5.2 3V = x(4 −x)(5 − 2 x) = 20x− 13x + 2xdV2 220 26x 6 x ; 2(3x 13x10) 0;dx = − + − + =2(3x−10)( x− 1) = 0;10x = 1, 3Critical points: 0, 1, 2.5At x = 0 or 2.5, V = 0. At x = 1, V = 9.Maximum volume when x = 1, y = 4 – 1 = 3, andz = 5 – 2(1) = 3.a.2 2 2y = x −( a− x) = 2ax−a1Area of A = A = ( a−x ) y21 ( ) 222 a x ax a= − − 1 1dA− ( )1 ( a x) (2 a)22 2 2=− ax − a +dx 2 22ax − a2a − 3 ax2=22ax− aInstructor’s Resource Manual Section 3.4 183

2−32a ax 2a= 0 when x = .22ax− a3dA ⎛a 2a ⎞ dA ⎛2a ⎞> 0 on ⎜ , ⎟ and < 0 on ⎜ , a ⎟,dx ⎝2 3 ⎠ dx ⎝ 3 ⎠2aso x = maximizes the area of triangle A.3b. Triangle A is similar to triangle C, soax axw = =y 22ax − a21 axArea of B = B = xw=2 22 2ax− a⎛2 22x 2ax a x a ⎞dB a ⎜− −2ax a2 ⎟−= ⎜⎟dx 2 22ax − a⎜⎟⎝⎠2 2 2 2a ⎛2 x(2 ax −a ) −ax ⎞ a ⎛ 3ax −2xa⎞= =2⎜ 2 3/2 22 3/2(2 ax −a ) ⎟ ⎜(2 ax −a) ⎟⎝ ⎠ ⎝ ⎠2 2a ⎛ 3x − 2xa ⎞2a= 0 when x = 0,2 ⎜ 2 3/2(2 ax a ) ⎟⎝ − ⎠32aSince x = 0 is not possible, x = .3dB ⎛a 2a ⎞ dB ⎛2a ⎞< 0 on ⎜ , ⎟ and > 0 on ⎜ , a ⎟,dx ⎝2 3 ⎠ dx ⎝ 3 ⎠2aso x = minimizes the area of triangle B.3c.z =2 2x + w =2 22 a xx +22 ax – a=32ax22 ax – a2 2 2 3dz 1 2ax −a ⎛6 ax (2 ax −a ) −2 ax (2 a)⎞= dx 2 3 2 22 ax ⎜ (2 ax − a ) ⎟⎝⎠=2 3 3 24a x − 3a x3 2 32 ax (2 ax − a )dz3a0 when x 0,dx = = → ax =434dz ⎛a 3a ⎞ dz ⎛3a ⎞< 0 on ⎜ , ⎟ and > 0 on ⎜ , a ⎟,dx ⎝2 4 ⎠ dx ⎝ 4 ⎠3aso x = minimizes length z.446. Let 2x be the length of a bar and 2y be the width of a bar.⎛π θ ⎞ ⎛ 1 θ 1 θ ⎞ a ⎛ θ θ ⎞x = acos⎜ − ⎟= a⎜cos + sin ⎟= ⎜cos + sin ⎟⎝4 2⎠ ⎝ 2 2 2 2⎠2⎝ 2 2⎠⎛π θ ⎞ ⎛ 1 θ 1 θ ⎞ a ⎛ θ θ ⎞y = asin ⎜ − ⎟= a⎜cos − sin ⎟= ⎜cos −sin⎟⎝4 2⎠ ⎝ 2 2 2 2⎠2⎝ 2 2⎠Compute the area A of the cross and maximize.2 ⎡ a ⎛ θ θ ⎞⎤⎡ a ⎛ θ θ ⎞⎤ ⎡ a ⎛ θ θ ⎞⎤A = 2(2 x)(2 y) −(2 y)= 8⎢ ⎜cos + sin ⎟ cos sin 4 cos sin2 2 2⎥⎢ ⎜ − ⎟ − −2 2 2⎥ ⎢ ⎜ ⎟2 2 2⎥⎣ ⎝ ⎠⎦⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦2⎛ 2θ 2θ ⎞ 2⎛ θ θ ⎞ 2 2= 4a ⎜cos −sin ⎟−2a ⎜1− 2cos sin ⎟= 4a cosθ−2 a (1 −sin θ )⎝ 2 2⎠ ⎝ 2 2⎠dA 2 24asinθ2acosθdθ =− + ; 2 2 1− 4asinθ + 2acosθ = 0 when tanθ= ;21 2sin θ = , cosθ=5 52d A02dθ < when 1tanθ = , so this maximizes the area.222⎛ 2 ⎞ 2⎛ 1 ⎞ 10a2 2A= 4 a ⎜ ⎟–2a ⎜1– ⎟= –2a = 2 a ( 5 –1)⎝ 5⎠ ⎝ 5⎠52184 Section 3.4 Instructor’s Resource Manual

47. a.b.′ 1/2 1/2( ) 15(9 25 30cos ) −−= + − sin = 15(34 − 30cos ) sinL θ θ θ θ θ15′′ 3/2 1/2( ) (34 30cos ) −−=− − (30sin )sin + 15(34 − 30cos ) cos2−3/2 2 −1/2=−225(34 − 30cos ) sin + 15(34 − 30cos ) cosL θ θ θ θ θ θθ θ θ θ−3/2 2= 15(34 −30cos ) [ − 15sin + (34 − 30cos )cos ]θ θ θ θ−3/2 2 2= 15(34 −30cos ) [ − 15sin + 34cos − 30cos ]θ θ θ θ−3/2 2= 15(34 −30cos ) [ − 15 + 34cos − 15cos ]θ θ θ−3/2 2=−15(34 −30cos ) [15cos − 34cos + 15]θ θ θ234 ± (34) −4(15)(15) 5 3L′′ = 0 when cos θ = = ,2(15) 3 5−1 ⎛3⎞θ = cos ⎜ ⎟⎝5⎠−1/2⎛ −1 3 ⎞ ⎛ 3 ⎞ 4L′ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜cos ⎜ ⎟⎟= 15⎜9 + 25 − 30⎜ ⎟⎟⎜ ⎟=3⎝ ⎝5⎠⎠ ⎝ ⎝5⎠⎠⎝5⎠1/2⎛ −1 ⎛3⎞⎞ ⎛ ⎛3⎞⎞L⎜cos ⎜ ⎟⎟= ⎜9 + 25 − 30⎜ ⎟⎟= 4⎝ ⎝5⎠⎠ ⎝ ⎝5⎠⎠φ = 90° since the resulting triangle is a 3-4-5 right triangle.′ 1/2 1/2( ) 65(25 169 130cos ) −−= + − sin = 65(194 − 130cos ) sinL θ θ θ θ θ65′′ 3/2 1/2( ) (194 130cos ) −−=− − (130sin )sin + 65(194 − 130cos ) cos2−3/2 2 −1/2=−4225(194 − 130cos ) sin + 65(194 − 130cos ) cosL θ θ θ θ θ θθ θ θ θ−3/2 2= 65(194 −130cos ) [ − 65sin + (194 − 130cos )cos ]θ θ θ θ−3/2 2 2= 65(194 −130cos ) [ − 65sin + 194cos − 130cos ]θ θ θ θ−3/2 2= 65(194 −130cos ) [ − 65cos + 194cos − 65]θ θ θ−3/2 2=−65(194 −130cos ) [65cos − 194cos + 65]L′′ = 0θ θ θ194 ± (194) −4(65)(65) 13 5when cos θ = = ,2(65) 5 13−1 ⎛ 5 ⎞θ = cos ⎜ ⎟⎝13⎠1/2⎛ −1 5 ⎞ ⎛ 5 ⎞ 12L′ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜cos ⎜ ⎟⎟= 65⎜25 + 169 − 130⎜ ⎟⎟⎜ ⎟=5⎝ ⎝13 ⎠⎠ ⎝ ⎝13 ⎠⎠⎝13⎠1/2⎛ −1 ⎛ 5 ⎞⎞ ⎛ ⎛ 5 ⎞⎞L⎜cos ⎜ ⎟⎟= ⎜25 + 169 − 130⎜ ⎟⎟= 12⎝ ⎝13 ⎠⎠ ⎝ ⎝13⎠⎠φ = 90° since the resulting triangle is a 5-12-13 right triangle.22 2c. When the tips are separating most rapidly, φ = 90 ° , L = m − h , L′= hInstructor’s Resource Manual Section 3.4 185

d.2 2 1/2L′ −( ) = hm( h + m − 2hmcos ) sinθ θ θ2 2 2 2 3/2 2 2 2 1/2L′′ −−( ) =− h m ( h + m − 2hmcos ) sin + hm( h + m − 2hmcos ) cosθ θ θ θ θ2 2 −3/2 2 2 2 2= hm( h + m −2hmcos ) [ − hmsin + ( h + m )cos − 2hmcos ]θ θ θ θ2 2 −3/2 2 2 2= hm( h + m −2hm cos ) [ − hm cos + ( h + m )cos − hm]θ θ θ2 2 −3/2 2 2 2=− hm( h + m −2hm cos ) [ hm cos − ( h + m )cos + hm]θ θ θ2 2 2L′′ = 0 when hmcos − ( h + m )cos + hm=0θ( hcos θ −m)( mcos θ − h) = 0m hcos θ = ,h mh−1⎛ h ⎞Since h < m, cos θ = so θ = cos ⎜ ⎟m⎝m⎠ . −1/2 2 2⎛ −1⎛ h ⎞⎞ ⎛ 2 2 ⎛ h ⎞⎞ m −hL′ ⎜cos ⎜ ⎟⎟= hm⎜h + m −2hm⎜ ⎟⎟⎝ ⎝m⎠⎠ ⎝ ⎝m⎠⎠m1/2⎛ −1⎛ h ⎞⎞ ⎛ 2 2 ⎛ h ⎞⎞2 2L⎜cos ⎜ ⎟⎟= ⎜h + m − 2hm⎜ ⎟⎟= m −h⎝ ⎝m⎠⎠ ⎝ ⎝m⎠⎠h2 + L2 = m2 , φ = 90 ° .Sinceθ2 22 2 −1/2m − h= hm( m − h )= hm48. We are interested in finding the global extrema forthe distance of the object from the observer. Wewill obtain this result by considering the squareddistance instead. The squared distance can beexpressed as22 1 2Dx ( ) ( x 2) ⎛⎞= − + ⎜100+ x−x ⎟⎝ 10 ⎠The first and second derivatives are given by1 3 3 2D'( x) = x − x − 36x+196 and25 53 2D''( x) = ( x −10x−300)25Using a computer package, we can solve theequation D'( x ) = 0 to find the critical points. Thecritical points are x ≈ 5.1538,36.148 . Using thesecond derivative we see thatD ''(5.1538) ≈− 38.9972 (max) andD ''(36.148) ≈ 77.4237 (min)Therefore, the position of the object closest to theobserver is ≈ ( 36.148,5.48)while the position ofthe object farthest from the person is≈ (5.1538,102.5) .(Remember to go back to the original equation forthe path of the object once you find the criticalpoints.)49. Here we are interested in minimizing the distancebetween the earth and the asteroid. Using thecoordinates P and Q for the two bodies, we canuse the distance formula to obtain a suitableequation. However, for simplicity, we willminimize the squared distance to find the criticalpoints. The squared distance between the objectsis given by2Dt ( ) = (93cos(2 πt) −60cos[2 π(1.51t−1)])2+ (93sin(2 πt) −120sin[2 π(1.51t−1)])The first derivative isD'( t) ≈−34359[ cos(2 πt) ][ sin(9.48761 t)]+ [ cos(9.48761 t) ][(204932sin(9.48761 t)−141643sin(2 πt))]Plotting the function and its derivative reveal aperiodic relationship due to the orbiting of theobjects. Careful examination of the graphs revealsthat there is indeed a minimum squared distance(and hence a minimum distance) that occurs onlyonce. The critical value for this occurrence ist ≈ 13.82790355 . This value gives a squareddistance between the objects of ≈ 0.0022743million miles. The actual distance is ≈ 0.047851million miles ≈ 47,851 miles.186 Section 3.4 Instructor’s Resource Manual

50. Let x be the width and y the height of the flyer.51. Consider the following sketch.2 inches1 inch1 inch2 inchesWe wish to minimize the area of the flyer,A = xy .As it stands, A is expressed in terms of twovariables so we need to write one in terms of theother.The printed area of the flyer has an area of 50square inches. The equation for this area is( x−2)( y− 4)= 50We can solve this equation for y to obtain50y = 4x − 2+Substituting this expression for y in our equationfor A, we get A in terms of a single variable, x.A = xy⎛ 50 ⎞ 50x= x⎜+ 4⎟= + 4x⎝ x−2 ⎠ x−2The allowable values for x are 2 < x for x in ( 7,∞ ), we conclude that A attains itsminimum value at x = 7 . This value of x makesy = 14 . So, the dimensions for the flyer that willuse the least amount of paper are 7 inches by 14inches.52. a.By similar triangles,27tx = − t2t + 64xt= .2 227 − t + 64 t + 642 272+ −tt2+ 64 17282 2 3/227 t 64dx= − 1= −1dt t + 64 ( t + 64)17282 3/2( t + 64)− 1= 0 when t = 4 52 2d x −5184t d x= ; < 02 2 5/2 2dt ( t + 64) dtTherefore27 ( 4 5 )( ) 2 +t=4 5x = − 4 5 = 5 5 ≈11.18ft is the4 5 64maximum horizontal overhang.b. There are only a few data points, but they doseem fairly linear.c. The data values can be entered into mostscientific calculators to utilize the LeastSquares Regression feature. Alternately onecould use the formulas for the slope andintercept provided in the text. The resultingline should be y = 0.56852 + 2.6074xInstructor’s Resource Manual Section 3.4 187

d. Using the result from c., the predicted numberof surface imperfections on a sheet with area2.0 square feet isy = 0.56852 + 2.6074(2.0) = 5.7833 ≈6since we can't have partial imperfectionsdSdbddb53. a. = ∑[ yi− (5 + bxi)]i=1nni=1nd= ∑ y i − + bx idb∑i=1n[ (5 )]= 2( y −5 −bx )( −x)i i i2( xy i i xi bxi)⎡= 2⎢∑− + 5 +⎢⎣i=1n n n2∑xiyi ∑xi b∑xii= 1 i= 1 i=1=− 2 + 10 + 222⎤⎥⎥⎦dSSetting 0db = givesn n n20=− 2 x y + 10 x + 2b x∑ ∑ ∑i i i ii= 1 i= 1 i=1n n n2=− ∑xy i i + ∑xi + b∑xii= 1 i= 1 i=1n n n2∑ i = ∑ i i −5∑ii= 1 i= 1 i=1n n∑xyi i − 5∑xii= 1 i=1=n2∑ xii=10 5b x x y xbYou should check that this is indeed the valueof b that minimizes the sum. Taking thesecond derivative yields2n2xii 1d S= 22 ∑db =which is always positive (unless all the xvalues are zero). Therefore, the value for babove does minimize the sum as required.b. Using the formula from a., we get that(2037) − 5(52)b = ≈ 3.0119590c. The Least Squares Regression line isy = 5+3.0119xUsing this line, the predicted total number oflabor hours to produce a lot of 15 brassbookcases isy = 5 + 3.0119(15) ≈ 50.179 hours54. C(x) = 7000 + 100x55.56.57.58.59.60. a.250 – pn ( )nn = 100 + 10so pn ( ) = 300–52nRn ( ) = npn ( ) = 300 n– 2Pn ( ) = Rn ( )– Cn ( )2n= 300 n– – (7000 + 100 n)2n=− 7000 + 200 n – 2Estimate n ≈ 200P′ ( n) = 200– n; 200– n = 0 when n = 200.P′′ ( n) = –1, so profit is maximum at n = 200.2Cx ( ) 100= + 3.002 – 0.0001 xx xCx ( )When x = 1600, = 2.9045 or $2.90 per unit.xdC3.002 0.0002xdx = −C′ (1600) = 2.682 or $2.68Cn ( ) 1000 n= +n n 1200Cn ( )When n = 800, ≈ 1.9167 or $1.92 per unit.ndC ndn = 600C′ (800) ≈ 1.333 or $1.33b.2dC233 18x3xdx = − +2 2d Cd C= − 18 + 6 x; = 0 when x = 32 2dxdx22d Cd C< 0 on (0, 3), > 0 on (3, ∞ )22dxdxThus, the marginal cost is a minimum whenx = 3 or 300 units.233 − 18(3) + 3(3) = 6188 Section 3.4 Instructor’s Resource Manual

61. a.62.63.64.2 xRx ( ) = xpx ( ) = 20x+ 4x−3dR220 8x x 10 x x 2dx = + − = − +dRb. Increasing when 0dx >c.23( )( )20 + 8x− x > 0 on [0, 10)Total revenue is increasing if 0 ≤ x ≤ 10.2 2d R d R= 8–2 x; = 0 when x = 42 2dx dx3d R dR=− 2; is maximum at x = 4.3dx dxxRx ( ) = x ⎛ ⎜182−⎞⎟⎝ 36 ⎠1/2−1/2 1/2dR 1⎛ 1182 x ⎞ ⎛ ⎞ ⎛x182x ⎞= ⎜ − ⎟ ⎜− ⎟+ ⎜ − ⎟dx 2⎝ 36⎠ ⎝ 36⎠ ⎝ 36⎠−12⎛ x ⎞ ⎛ x ⎞= ⎜182 − ⎟ ⎜182− ⎟⎝ 36 ⎠ ⎝ 24 ⎠dR0 when x 4368dx = =x1 = 4368; R(4368) ≈ 34,021.83dRAt x1, 0dx = .800xRx ( ) = −3xx + 3dR ( x + 3)(800) −800x2400= − 3= −3;dx 2 2( x+ 3) ( x+3)dR0 when x 20 2 3 25dx = = − ≈x1 = 25; R(25) ≈ 639.29At 1 , dRxdx = 0 .( x − 400)px ( ) = 12 − (0.20) = 20 − 0.02x10Rx ( ) = 20x−0.02x2dRdR= 20 − 0.04 x; = 0 when x = 500dxdxTotal revenue is maximized at x 1 = 500 .65. The revenue function would be2R( x) = x⋅ p( x) = 200x− 0.15x. This, togetherwith the cost function yields the following profitfunction:2⎧− ⎪ 5000 + 194x−0.148 x if 0 ≤ x≤500P( x)= ⎨2⎪ ⎩ − 9000 + 194x − 0.148 x if 500 < x ≤ 750a. The only difference in the two pieces of theprofit function is the constant. Since thederivative of a constant is 0, we can say thaton the interval 0 < x < 750 ,dP194 0.296xdx = −There are no singular points in the giveninterval. To find stationary points, we solvedP= 0dx194 − 0.296x= 0− 0.296x=−194x ≈ 655Thus, the critical points are 0, 500, 655, and750.0 5000 P 500 = 55,000 ;P ( ) =− ; ( )P ( 655)= 54,574.30 ; ( )P 750 = 53,250The profit is maximized if the companyproduces 500 chairs. The current machine canhandle this work, so they should not buy thenew machine.b. Without the new machine, a production levelof 500 chairs would yield a maximum profitof $55,000.66. The revenue function would be2R( x) = x⋅ p( x) = 200x− 0.15x. This, togetherwith the cost function yields the following profitfunction:2⎧− ⎪ 5000 + 194x−0.148 x if 0 ≤ x≤500P( x)= ⎨2⎪ ⎩ − 8000 + 194x − 0.148 x if 500 < x ≤ 750a. The only difference in the two pieces of theprofit function is the constant. Since thederivative of a constant is 0, we can say thaton the interval 0 < x < 750 ,dP194 0.296xdx = −There are no singular points in the giveninterval. To find stationary points, we solvedP= 0dx194 − 0.296x= 0− 0.296x= −194x ≈ 655Instructor’s Resource Manual Section 3.4 189

67.68.69. a.Thus, the critical points are 0, 500, 655, and 750.0 5000 P 500 = 55,000 ;P ( ) =− ; ( )P ( 655)= 55,574.30 ; ( )P 750 = 54, 250The profit is maximized if the company produces655 chairs. The current machine cannot handlethis work, so they should buy the new machine.b. With the new machine, a production level of655 chairs would yield a maximum profit of$55,574.30.Rx ( ) = 10x−0.001 x; 0 ≤ x≤30022 2Px ( ) = (10 x– 0.001 x ) – (200 + 4 x– 0.01 x )= –200 + 6x+ 0.009xdPdP= 6 + 0.018 x; = 0 when x ≈ − 333dxdxCritical numbers: x = 0, 300; P(0) = –200;P(300) = 2410; Maximum profit is $2410at x = 300.2⎧ ⎪ + x− 2x ≤ x≤2200 4 0.01 if 0 300Cx ( ) = ⎨⎪⎩ 800 + 3x− 0.01 x if 300 < x≤4502⎧− ⎪ 200 + 6x+ 0.009 x if 0 ≤ x≤300Px ( ) = ⎨2⎪ ⎩ − 800 + 7x + 0.009 x if 300 < x ≤ 450There are no stationary points on the interval[0, 300]. On [300, 450]:dPdP= 7+ 0.018 x; = 0 when x ≈ − 389dxdxThe critical numbers are 0, 300, 450.P(0) = –200, P(300) = 2410, P(450) = 4172.5Monthly profit is maximized at x = 450,P(450) = 4172.502 2 2⎛a+ b⎞a + 2ab+bab ≤ ⎜ ⎟ =⎝ 2 ⎠ 42 2a 1 b= + ab +4 2 4This is true if2 22 2a 1 b ⎛a b⎞ ⎛a–b⎞0 ≤ – ab + = ⎜ – ⎟ = ⎜ ⎟4 2 4 ⎝2 2⎠ ⎝ 2 ⎠Since a square can never be negative, this isalways true.b.2 2a + 2ab+bFb ( ) =4b+ 2 2 2As b→ 0 , a + 2ab+ b → a while+4b→ 0 , thusclose to a.lim Fb ( ) =∞ which is notb→0+2+ 2 +2 a2blim2b→∞a ab b + a+blim= =∞ ,b→∞4b4so when b is very large, F(b) is not close to a.22( a+ b)(4 b) – 4( a+b)F′ ( b)=216b2 2 2 24 b –4 a b – a= = ;2 216b4b2 2F′ ( b) = 0 when b = a or b = a since a andb are both positive.2 2( a+a) 4aF( a)= = = a4a4a2( a+b)Thus a ≤ for all b > 0 or4b2( a+b)a+bab ≤ which leads to ab ≤ .423 31 ⎛a+ b+ c⎞( a+ b+c)c. Let Fb ( ) = ⎜ ⎟ =b⎝3 ⎠ 27b3( a+ b+ c) (27 b) – 27( a+ b+c)F′ ( b)=2 227 b2( a+ b+ c) [3 b–( a+ b+c)]=227b2( a+ b+c) (2 b– a– c) =;227ba+cF′ ( b) = 0 when b = .22 3⎛a c 2 a c a cF+ ⎞ ⎛ + + ⎞⎜ ⎟= ⋅ ⎜ + ⎟⎝ 2 ⎠ a+c ⎝ 3 6 ⎠3 3 22 ⎛3( a+ c) ⎞ 2 ⎛a+ c⎞ ⎛a+c⎞= ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟a+ c⎝ 6 ⎠ a+c⎝ 2 ⎠ ⎝ 2 ⎠2 3⎛a 1Thus+ c⎞ ⎛a + b + c⎞⎜ ⎟ ≤ ⎜ ⎟⎝ 2 ⎠ b ⎝ 3 ⎠⎛a+c⎞From (b), ac ≤ ⎜ ⎟⎝ 2 ⎠ , thus323for all b > 0.1 ⎛a+ b+c⎞⎛a+ b+c⎞ac ≤ ⎜ ⎟ or abc ≤ ⎜ ⎟b ⎝ 3 ⎠ ⎝ 3 ⎠which gives the desired result1/3 a+ b+c( abc)≤ .33190 Section 3.4 Instructor’s Resource Manual

70. Let a = lw, b = lh, and c = hw, then2S = 2(a + b + c) while V = abc. By problem1/3 a+ b+c69(c), ( abc)≤ so321/3 2( a+ b+c)S( V ) ≤ = .23 ⋅ 6In problem 1c, the minimum occurs, hencea+cequality holds, when b = . In the result used2from Problem 69(b), equality holds when c = a,a+athus b= = a,so a = b = c. For the boxes,2this means l = w = h, so the box is a cube.3.5 Concepts Review1. f(x); –f(x)1 1Critical points: – ,2 21 1f′ ( x) > 0 when x < – or x >2 2⎛ 1 ⎤ ⎡ 1 ⎞f(x) is increasing on ⎜– ∞, – ⎥∪⎢, ∞⎟⎝ 2⎦ ⎣ 2 ⎠ and⎡ 1 1 ⎤decreasing on ⎢– , ⎥⎣ 2 2⎦ .Local minimumf ⎛ 1⎜⎞ ⎟ – 2 –10 –11.4⎝ 2 ⎠= ≈1Local maximum f ⎛ ⎜– ⎞ ⎟ 2 –10 –8.6⎝ 2 ⎠= ≈f′′ ( x) = 12 x; f′′( x) > 0 when x > 0. f(x) is concaveup on (0, ∞ ) and concave down on(– ∞ , 0); inflection point (0, –10).2. decreasing; concave up3. x = –1, x = 2, x = 3; y = 14. polynomial; rational.Problem Set 3.51. Domain: ( −∞, ∞ ); range: ( −∞, ∞ )Neither an even nor an odd function.y-intercept: 5; x-intercept: ≈ –2.32 2f′ ( x) = 3 x –3; 3 x –3= 0 when x = –1, 1Critical points: –1, 1f′ ( x) > 0 when x < –1 or x > 1f(x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) anddecreasing on [–1, 1].Local minimum f(1) = 3;local maximum f(–1) = 7f′′ ( x) = 6 x; f′′( x) > 0 when x > 0.f(x) is concave up on (0, ∞ ) and concave downon (– ∞ , 0); inflection point (0, 5).3. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )Neither an even nor an odd function.y-intercept: 3; x-intercepts: ≈ –2.0, 0.2, 3.22f′ ( x) = 6 x – 6 x–12 = 6( x– 2)( x+1);f′ ( x) = 0 when x = –1, 2Critical points: –1, 2f′ ( x) > 0 when x < –1 or x > 2f(x) is increasing on (– ∞ , –1] ∪ [2, ∞ ) anddecreasing on [–1, 2].Local minimum f(2) = –17;local maximum f(–1) = 10f′′ ( x) = 12x− 6= 6(2x− 1); f′′ ( x) > 0 when1x > .2⎛1 ⎞f(x) is concave up on ⎜ , ∞⎟and concave down⎝2⎠⎛ 1 ⎞⎛1 7⎞on ⎜– ∞, ⎟;inflection point: ⎜ , – ⎟⎝ 2 ⎠⎝2 2⎠2. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )Neither an even nor an odd function.y-intercept: –10; x-intercept: 22 2 2f′ ( x) = 6 x –3= 3(2 x –1); 2 x –1= 0 when1 1x = – ,2 2Instructor’s Resource Manual Section 3.5 191

4. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )Neither an even nor an odd functiony-intercept: –1; x-intercept: 12f′ ( x) = 3( x–1) ; f′( x) = 0 when x = 1Critical point: 1f′ ( x) > 0 for all x ≠ 1f(x) is increasing on (– ∞ , ∞ )No local minima or maximaf′′ ( x) = 6( x–1); f′′( x) > 0 when x > 1.f(x) is concave up on (1, ∞ ) and concave downon (– ∞ , 1); inflection point (1, 0)5. Domain: (– ∞ , ∞ ); range: [0, ∞ )Neither an even nor an odd function.y-intercept: 1; x-intercept: 13G′ ( x) = 4( x–1) ; G′( x) = 0 when x = 1Critical point: 1G′ ( x) > 0 for x > 1G(x) is increasing on [1, ∞ ) and decreasing on(– ∞ , 1].Global minimum f(1) = 0; no local maxima2G′′ ( x) = 12( x–1) ; G′′( x) > 0 for all x ≠ 1G(x) is concave up on (– ∞ , 1) ∪ (1, ∞ ); noinflection points⎡ 1 ⎞6. Domain: (– ∞ , ∞ ); range: ⎢– , ∞ ⎟⎣ 4 ⎠2 2 2 2H (– t) = (– t) [(– t) –1] = t ( t –1) = H( t);evenfunction; symmetric with respect to the y-axis.y-intercept: 0; t-intercepts: –1, 0, 13 2H′ () t = 4 t –2t = 2(2 t t –1); H′() t = 0 when1 1t = – , 0,2 21 1Critical points: – , 0,2 211H′ () t > 0 for – < t < 0 or < t.22⎡ 1 ⎤ ⎡ 1 ⎞H(t) is increasing on ⎢ – , 0 ⎥ ∪ ⎢ , ∞ ⎟⎣ 2 ⎦ ⎣ 2 ⎠ and⎛ 1 ⎤ ⎡ 1 ⎤decreasing on ⎜ – ∞, – ⎥∪⎢0,⎥⎝ 2⎦ ⎣ 2⎦⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1Global minima f ⎜– ⎟= – , f ⎜ ⎟=– ;⎝ 2 ⎠ 4 ⎝ 2 ⎠ 4Local maximum f(0) = 02 2H′′ ( t) = 12 t – 2 = 2(6 t –1); H′′> 0 when1 1t < – or t >6 6⎛ 1 ⎞ ⎛ 1 ⎞H(t) is concave up on ⎜– ∞, – ⎟∪⎜ , ∞⎟⎝ 6 ⎠ ⎝ 6 ⎠⎛ 1 1 ⎞and concave down on ⎜– , ⎟;⎝ 6 6 ⎠ inflectionpoints⎛ 1 5 ⎞ ⎛ 1 5 ⎞H⎜– , − ⎟ and H⎜ , ⎟⎝ 6 36 ⎠ ⎝ 6 36 ⎠7. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )Neither an even nor an odd function.1/3y-intercept: 10; x-intercept: 1–11 ≈ –1.22 2f′ ( x) = 3 x –6x+ 3= 3( x–1) ; f′( x) = 0 whenx = 1.Critical point: 1f′ ( x) > 0 for all x ≠ 1.f(x) is increasing on (– ∞ , ∞ ) and decreasingnowhere.No local maxima or minimaf′′ ( x) = 6 x–6= 6( x–1); f′′( x) > 0 when x > 1.f(x) is concave up on (1, ∞ ) and concave downon (– ∞ , 1); inflection point (1, 11)192 Section 3.5 Instructor’s Resource Manual

⎡ 16 ⎞8. Domain: (– ∞ , ∞ ); range: ⎢– , ∞ ⎟⎣ 3 ⎠4 2 4 24(– s) – 8(– s) –12 4 s – 8 s –12F(– s ) = =3 3= F();s even function; symmetric with respectto the y-axisy-intercept: –4; s-intercepts: – 3, 316 3 16 16 2F′ () s = s – s = s( s –1); F′() s = 03 3 3when s = –1, 0, 1.Critical points: –1, 0, 1F′ () s > 0when –1 < x < 0 or x > 1.F(s) is increasing on [–1, 0] ∪ [1, ∞ ) anddecreasing on (– ∞ , –1] ∪ [0, 1]16 16Global minima F( − 1) =− , F(1) =− ; local3 3maximum F(0) = –42 16 2 1F′′ () s 16s 16 ⎛ ⎞= − = ⎜s − ⎟; F′′() s > 03 ⎝ 3⎠1 1when s < – or s >3 3⎛ 1 ⎞ ⎛ 1 ⎞F(s) is concave up on ⎜−∞, − ⎟∪ ⎜ , ∞⎟⎝ 3⎠ ⎝ 3 ⎠⎛ 1 1 ⎞and concave down on ⎜– , ⎟;⎝ 3 3⎠inflection points⎛ 1 128⎞ ⎛ 1 128⎞F⎜– , − ⎟, F⎜ , − ⎟⎝ 3 27 ⎠ ⎝ 3 27 ⎠down on (–1, ∞ ); no inflection points (–1 is notin the domain of g).x 1lim = lim = 1;x→∞x + 1 x→∞1+1limxx1= lim = 1;x→–∞ + 1 x→–∞1+1xhorizontal asymptote: y = 1As– –x→ –1 , x+ 1→ 0 so+ +as x→ –1 , x+ 1→ 0 sovertical asymptote: x = –1xxlim = ∞;x + 1x→–1–xlim = – ∞;x + 1x→–1+10. Domain: (– ∞ , 0) ∪ (0, ∞ );range: (– ∞ , –4π ] ∪ [0, ∞ )Neither an even nor an odd functionNo y-intercept; s-intercept: π2 2s – πg′ () s = ; g′() s = 0when s = –π , π2sCritical points: − π , πg′ () s > 0when s < –π or s > πg(s) is increasing on ( −∞, −π] ∪ [ π , ∞ ) and9. Domain: (– ∞ , –1) ∪ (–1, ∞ );range: (– ∞ , 1) ∪ (1, ∞ )Neither an even nor an odd functiony-intercept: 0; x-intercept: 01g′ ( x) = ; g′( x)is never 0.2( x + 1)No critical pointsg′ ( x) > 0 for all x ≠ –1.g(x) is increasing on (– ∞ , –1) ∪ (–1, ∞ ).No local minima or maxima2g′′ ( x) = – ; g′′( x) > 0 when x < –1.3( x + 1)g(x) is concave up on (– ∞ , –1) and concavedecreasing on [–π , 0) ∪ (0, π ].Local minimum g(π ) = 0;local maximum g(–π ) = –4π22πg′′ () s = ; g′′() s > 0when s > 03sg(s) is concave up on (0, ∞ ) and concave downon (– ∞ , 0); no inflection points (0 is not in thedomain of g(s)).2πgs () = s–2 π+ ; y= s–2π is an obliquesasymptote.– 2 2As s →0 ,( s– π) →π , solim gs ( ) = – ∞ ;s→0–Instructor’s Resource Manual Section 3.5 193

+2 2as s →0 ,( s– π) →π , sos = 0 is a vertical asymptote.lim gs ( ) =∞ ;s→0+No vertical asymptotes⎡ 1 1⎤11. Domain: (– ∞ , ∞ ); range: ⎢– ,4 4⎥⎣ ⎦– x xf (– x) = = – = – f( x);odd2 2(– x) + 4 x + 4function; symmetric with respect to the origin.y-intercept: 0; x-intercept: 024– xf′ ( x) = ; f′( x) = 0 when x = –2, 22 2( x + 4)Critical points: –2, 2f′ ( x) > 0 for –2 < x < 2f(x) is increasing on [–2, 2] anddecreasing on (– ∞ , –2] ∪ [2, ∞ ).1Global minimum f (–2) = – ; global maximum41f (2) =422 xx ( –12)f′′ ( x) = ; f′′( x) > 0 when2 3( x + 4)–2 3 < x < 0 or x > 2 3f(x) is concave up on (–2 3, 0) ∪(2 3, ∞ ) andconcave down on (– ∞, – 2 3) ∪ (0, 2 3);⎛ 3 ⎞inflection points⎜−2 3, −, (0, 0)8 ⎟,⎝ ⎠⎛ 3 ⎞⎜2 3, 8 ⎟⎝ ⎠1xxlim = lim = 0;x→∞214x + 4 x→∞+x 21xxlim = lim = 0;x→– ∞ 2 – 14x + 4 x→ ∞ +x2y = 0 is a horizontal asymptote.12. Domain: (– ∞ , ∞ ); range: [0, 1)2 2(– θ)θΛ (– θ ) = = =Λ( θ ); even2 2(– θ) + 1 θ + 1function; symmetric with respect to the y-axis.y-intercept: 0; θ -intercept: 02θΛ ′( θ) = ; Λ ′( θ) = 0 when θ = 02 2( θ + 1)Critical point: 0Λ ′( θ ) > 0 when θ > 0Λ(θ) is increasing on [0, ∞ ) anddecreasing on (– ∞ , 0].Global minimum Λ(0) = 0; no local maxima22(1– 3 θ )Λ ′′( θ) = ; Λ ′′( θ) > 0 when2 3( θ + 1)1 1– < θ

13. Domain: (– ∞ , 1) ∪ (1, ∞ );range (– ∞ , 1) ∪ (1, ∞ )Neither an even nor an odd functiony-intercept: 0; x-intercept: 01hx ( ) =− ; h′( x)is never 0.2( x −1)No critical pointsh′ ( x) < 0 for all x ≠ 1.h(x) is increasing nowhere anddecreasing on (– ∞ , 1) ∪ (1, ∞ ).No local maxima or minima2h′′ ( x) = ; h′′( x) > 0 when x > 13( x –1)hx ( ) is concave up on (1, ∞ ) and concave downon (– ∞ , 1); no inflection points (1 is not in thedomain of hx ( ))x 1lim = lim = 1;x→∞x –1 x→∞1–1xx 1lim = lim = 1;x→−∞x −1 x→−∞1−1xy = 1 is a horizontal asymptote.– –As x→1 , x–1→ 0 so lim = – ∞ ;x→1– x –1+ + xas x→1 , x–1→ 0 so lim = ∞ ;x→1+ x –1x = 1 is a vertical asymptote.14. Domain: ( −∞,∞ )Range: ( 0,1 ]Even function since1 1P( − x) = = = P( x)2 2( − x) + 1 x + 1so the function is symmetric with respect to they-axis.y-intercept: y = 1x-intercept: none−2xP'( x)= ; P'( x)is 0 when x = 0 .2 2( x + 1)critical point: x = 0P'( x ) > 0 when x < 0 so Px ( ) is increasing onx( −∞,0]and decreasing on [0, ∞ ) . Globalmaximum P (0) = 1; no local minima.26x− 2P''( x)=2 3( x + 1)P''( x ) > 0 on ( −∞, −1/ 3) ∪ (1/ 3, ∞ ) (concaveup) and P''( x ) < 0 on ( − 1/ 3,1/ 3) (concavedown).⎛ 1 3⎞Inflection points: ⎜±, ⎟⎝ 3 4 ⎠No vertical asymptotes.lim Px ( ) = 0; lim Px ( ) = 0x→∞y = 0x→−∞is a horizontal asymptote.15. Domain: (– ∞ , –1) ∪ (–1, 2) ∪ (2, ∞ );range: (– ∞ , ∞ )Neither an even nor an odd function3y-intercept: – ; x-intercepts: 1, 3223 x –10x+11f ′( x) =; f′( x)is never 0.2 2( x+1) ( x–2)No critical pointsf′ ( x) > 0 for all x ≠ –1, 2f(x) is increasing on(– ∞ , –1) ∪ (–1, 2) ∪ ( 2,∞ ) .No local minima or maxima3 2–6x + 30 x – 66x+42f′′ ( x) = ; f′′( x) > 0 when3 3( x+1) ( x–2)x < –1 or 1 < x < 2f(x) is concave up on (– ∞ , –1) ∪ (1, 2) andconcave down on (–1, 1) ∪ (2, ∞ );inflection point f(1) = 0( 2x –1)( x –3) xlimlim–4 x+=3x→∞( x+1)( x–2) x→∞2x – x–21– 4 + 3xx2= lim = 1;x→∞1– 1 – 2xx 21– 4 + 3( x–1)( x–3)xx2lim = lim = 1;x→– ∞( x+1)( x–2) x→–∞1– 1 – 2xx2y =1 is a horizontal asymptote.–As x→ –1 , x–1 → –2, x–3 → –4,–x– 2 → –3, and x+ 1 → 0 solim f( x) = ∞ ;x→–1–Instructor’s Resource Manual Section 3.5 195

+as x→ –1 , x–1 → –2, x–3 → –4,x – 2 → –3, and x + 1→ 0 , solim f( x) = – ∞x→–1+–x→2 , x–1→1, x–3 → –1, x+ 1→ 3, andAs–x – 2→ 0 , so lim f( x) =∞ ; asx→2–+x→2 , x–1→1, x–3 → –1, x+ 1→ 3, and+x– 2→ 0 , so lim f( x) = – ∞x→2+x = –1 and x = 2 are vertical asymptotes.+17. Domain: ( −∞,1) ∪( 1, ∞ )Range: ( −∞,∞ )Neither even nor odd function.y-intercept: y = 6 ; x-intercept: x =− 3, 22x − 2x+5g'( x)= ; g '( x ) is never zero. No2( x −1)critical points.g'( x ) > 0 over the entire domain so the functionis always increasing. No local extrema.−8f ''( x)= ; f ''( x ) > 0 when3( x −1)x < 1(concave up) and f ''( x ) < 0 whenx > 1 (concave down); no inflection points.No horizontal asymptote; x = 1 is a verticalasymptote; the line y = x+ 2 is an oblique (orslant) asymptote.16. Domain: ( −∞,0) ∪ ( 0, ∞ )Range: ( −∞, −2] ∪[2, ∞ )Odd function since2 2( − z) + 1 z + 1w( − z) = =− = −w( z); symmetric−zzwith respect to the origin.y-intercept: nonex-intercept: none1w'( z) = 1− ; w'( z ) = 0 when z =± 1 .2zcritical points: z =± 1 . w'( z ) > 0 on( −∞, −1) ∪ (1, ∞ ) so the function is increasing on( −∞, −1] ∪[1, ∞ ). The function is decreasing on[ −1,0) ∪ (0,1).local minimum w (1) = 2 and local maximumw( − 1) = − 2. No global extrema.2w''( z) = > 0 when z > 0 . Concave up on3z(0, ∞)and concave down on ( −∞ ,0).No horizontal asymptote; x = 0 is a verticalasymptote; the line y = z is an oblique (or slant)asymptote.No inflection points.18. Domain: (– ∞ , ∞ ); range [0, ∞ )3 3f (– x) = – x = x = f( x);even function;symmetric with respect to the y-axis.y-intercept: 0; x-intercept: 02 ⎛ x ⎞f′ ( x) = 3 x = 3 x x ; f′( x) = 0 when x = 0⎜ x ⎟⎝ ⎠Critical point: 0f′ ( x) > 0 when x > 0f(x) is increasing on [0, ∞ ) and decreasing on(– ∞ , 0].Global minimum f(0) = 0; no local maxima23xf′′ ( x) = 3 x + = 6 x asx2x =2x ;f′′ ( x) > 0 when x ≠ 0f(x) is concave up on (– ∞ , 0) ∪ (0, ∞ ); noinflection points196 Section 3.5 Instructor’s Resource Manual

19. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )R(– z) = – z – z = – z z = – R( z);odd function;symmetric with respect to the origin.y-intercept: 0; z-intercept: 02zR′ 2 2( z) = z + = 2 z since z = z for all z;zR′ ( z) = 0 when z = 0Critical point: 0R′ ( z) > 0 when z ≠ 0R(z) is increasing on (– ∞ , ∞ ) and decreasingnowhere.No local minima or maxima2zR′′ ( z) = ; R′′( z) > 0 when z > 0.zR(z) is concave up on (0, ∞ ) and concave downon (– ∞ , 0); inflection point (0, 0).21. Domain: (– ∞ , ∞ ); range: [0, ∞ )Neither an even nor an odd function.Note that for x ≤ 0, x = – x so x + x = 0, whilex + xfor x > 0, x = x so = x.2⎧⎪ 0 if x ≤ 0gx ( ) = ⎨ 2⎪⎩ 3x + 2x if x > 0y-intercept: 0; x-intercepts: ( − ∞ , 0]⎧0 if x ≤ 0g′ ( x)= ⎨⎩6x+ 2 if x > 0No critical points for x > 0.g(x) is increasing on [0, ∞ ) and decreasingnowhere.⎧0 if x ≤ 0g′′ ( x)= ⎨⎩6 if x > 0g(x) is concave up on (0, ∞ ); no inflection points20. Domain: (– ∞ , ∞ ); range: [0, ∞ )2 2H (– q) = (– q) – q = q q = H( q);evenfunction; symmetric with respect to the y-axis.y-intercept: 0; q-intercept: 03 3q 3q2H ′2( q) = 2q q + = = 3q q as q = qq qfor all q; H′ ( q) = 0 when q = 0Critical point: 0H′ ( q) > 0 when q > 0H(q) is increasing on [0, ∞ ) anddecreasing on (– ∞ , 0].Global minimum H(0) = 0; no local maxima23qH′′ ( q) = 3 q + = 6 q ; H′′( q) > 0 whenqq ≠ 0.H(q) is concave up on (– ∞ , 0) ∪ (0, ∞ ); noinflection points.22. Domain: (– ∞ , ∞ ); range: [0, ∞ )Neither an even nor an odd function. Note thatx – xfor x < 0, x = – x so = – x , while for2x − xx ≥ 0, x = x so = 0.23 2 ⎧− ⎪ x + x − 6x if x

points23. Domain: (– ∞ , ∞ ); range: [0, 1]f ( − x) = sin( − x) = − sin x = sin x = f( x);evenfunction; symmetric with respect to the y-axis.y-intercept: 0; x-intercepts: kπ where k is anyinteger.sin xπf′ ( x) = cos x; f′( x) = 0 when x = + kπsin x2and f ′( x)does not exist when x = kπ , where kis any integer.kπ πCritical points: and kπ + , where k is any2 2integer; f′ ( x) > 0 when sin x and cos x are eitherboth positive or both negative.⎡ π⎤f(x) is increasing on ⎢kπ,kπ+2 ⎥ and decreasing⎣ ⎦⎡ π ⎤on ⎢kπ+ , ( k + 1) π2 ⎥ where k is any integer.⎣⎦Global minima f(kπ ) = 0; global maxima⎛ π ⎞f ⎜kπ+ ⎟=1, where k is any integer.⎝ 2 ⎠2 2cos x sin xf′′ ( x)= −sin x sin x⎛1⎞ ⎛ sinx⎞+ sin x cos x⎜− ⎟ (cos x)2⎜ sin x ⎟⎜sin x ⎟⎝ ⎠ ⎝ ⎠2 2 2 2cos x sin x cos x sin x= − − =− = − sin xsin x sin x sin x sin xf′′ ( x) < 0 when x ≠ kπ , k any integerf(x) is never concave up and concave down on(kπ , (k + 1)π ) where k is any integer.No inflection points24. Domain: [2kπ , (2k + 1)π ] where k is anyinteger; range: [0, 1]Neither an even nor an odd functiony-intercept: 0; x-intercepts: kπ , where k is anyinteger.cos xπf′ ( x) = ; f′( x) = 0 when x = 2kπ+2 sinx2while f ′( x)does not exist when x = kπ , k anyinteger.πCritical points: kπ, 2kπ+ where k is any2integerπf′ ( x) > 0 when 2kπ < x< 2kπ+2⎡ π⎤f(x) is increasing on ⎢2 kπ, 2kπ+2 ⎥⎣⎦ and⎡ π ⎤decreasing on ⎢2 kπ + ,(2k+ 1) π ,2 ⎥ k any⎣⎦integer.Global minima f(kπ ) = 0; global maxima⎛ π ⎞f ⎜2kπ + ⎟=1, k any integer⎝ 2 ⎠2 2 2– cos x – 2sin x –1– sin xf′′ ( x)= =3/2 3/24sin x 4sin x21+sin x= – ;3/24sin xf′′ ( x) < 0 for all x.f(x) is concave down on (2kπ , (2k + 1)π );no inflection points25. Domain: ( −∞, ∞ )Range: [0,1]Even function since2 2h( − t) = cos ( − t) = cos t = h( t)so the function is symmetric with respect to they-axis.πy-intercept: y = 1; t-intercepts: x = + kπ2where k is any integer.kπh'( t)= − 2 cos tsint ; h'( t ) = 0 at t = .2kπCritical points: t =2198 Section 3.5 Instructor’s Resource Manual

πh'( t) > 0 when kπ+ < t < ( k+ 1) π . The2function is increasing on the intervals[ kπ + ( π /2),( k+ 1) π]and decreasing on thekπ, kπ + ( π /2) .intervals [ ]Global maxima h( kπ ) = 1⎛π⎞Global minima h⎜+ kπ⎟=0⎝ 2 ⎠2 2h''( t) = 2sin t− 2 cos t =− 2(cos 2 t)⎛ π π ⎞h''( t ) < 0 on ⎜kπ− , kπ+ ⎟ so h is concave⎝ 4 4⎠ ⎛ π 3π⎞down, and h''( t ) > 0 on ⎜kπ+ , kπ+ ⎟⎝ 4 4 ⎠ so his concave up.⎛kππ 1 ⎞Inflection points: ⎜ + , ⎟⎝ 2 4 2⎠No vertical asymptotes; no horizontalasymptotes.4 2cos t+sin t(3) cos tsintg'( t) = 26cos t2 2cos t+3sin t= 24cos t21+2sin t= 2 > 04cos tover the entire domain. Thus the function is⎛ π π ⎞concave up on ⎜kπ− , kπ+ ⎟; no inflection⎝ 2 2⎠ points.πNo horizontal asymptotes; t = + kπare2vertical asymptotes.27. Domain: ≈ (– ∞ , 0.44) ∪ (0.44, ∞ );range: (– ∞ , ∞ )π26. Domain: all reals except t = + kπ2Range: [0, ∞ )y-intercepts: y = 0 ; t-intercepts: t = kπ where kis any integer.Even function since2 2 2g( − t) = tan ( − t) = ( − tan t) = tan tso the function is symmetric with respect to they-axis.2 2sintg'( t) = 2sec ttant = ; g'( t ) = 0 when3cos tt = kπ .Critical points: kπ⎡ π ⎞g ( t)is increasing on ⎢kπ,kπ+ ⎟⎣ 2 ⎠ and⎛ π ⎤decreasing on ⎜ kπ− , kπ2 ⎥⎝ ⎦ .Global minima gkπ ( ) = 0; no local maximaNeither an even nor an odd functiony-intercept: 0; x-intercepts: 0, ≈ 0.243 274.6092 x – 58.2013x + 7.82109xf′ ( x) =;2(7.126 x – 3.141)f′ ( x) = 0 when x = 0, ≈ 0.17, ≈ 0.61Critical points: 0, ≈ 0.17, ≈ 0.61f′ ( x) > 0 when 0 < x < 0.17 or 0.61 < xf(x) is increasing on ≈ [0, 0.17] ∪ [0.61, ∞ )and decreasing on(– ∞ , 0] ∪ [0.17, 0.44) ∪ (0.44, 0.61]Local minima f(0) = 0, f(0.61) ≈ 0.60; localmaximum f(0.17) ≈ 0.013 2531.665 x – 703.043x + 309.887 x– 24.566f′′ ( x) =;3(7.126 x – 3.141)f′′ ( x) > 0 when x < 0.10 or x > 0.44Instructor’s Resource Manual Section 3.5 199

f(x) is concave up on (– ∞ , 0.10) ∪ (0.44, ∞ )30.and concave down on (0.10, 0.44);inflection point ≈ (0.10, 0.003)5.235 3 2 2x − 1.245 xlimlim5.235 x − 1.245 x= =∞x→∞7.126x− 3.141 x→∞7.126 − 3.141xso f(x) does not have a horizontal asymptote.As– 3 2x 0.44 , 5.235 x –1.245x0.20→ → while31.–7.126 x – 3.141 → 0 , solim f( x) = – ∞ ;x→0.44–as+ 3 2x 0.44 , 5.235 x –1.245x0.20→ → while7.126 x – 3.141 → 0 , so+lim f( x) =∞ ;x→0.44+x ≈ 0.44 is a vertical asymptote of f(x).32.33.28.34.29.200 Section 3.5 Instructor’s Resource Manual

35.36.37.38.4 3y′ = 5( x–1) ; y′′ = 20( x–1) ; y′′( x) > 0when x > 1; inflection point (1, 3)At x = 1, y′ = 0, so the linear approximation is ahorizontal line.3240. Let f ( x) = ax + bx+ c,then f ′( x) = 2ax+band f ′′( x) = 2 a.An inflection point occurswhere f ′′( x)changes from positive to negative,but 2a is either always positive or alwaysnegative, so f(x) does not have any inflectionpoints.( f ′′( x)= 0 only when a = 0, but then f(x) is not aquadratic curve.)3 241. Let f ( x) = ax + bx + cx+ d,then2f ′( x) = 3ax + 2bx+ c and f ′′( x) = 6ax+ 2 b.Aslong as a ≠ 0 , f ′′( x)will be positive on onebside of x = and negative on the other side.3abx = is the only inflection point.3a42. Let4 3 2f ( x) = ax + bx + cx + dx+ c,then3 2f ′( x) = 4ax + 3bx + 2cx+ d and2 2f ′′( x) = 12ax + 6bx+ 2c = 2(6ax + 3 bx+c)Inflection points can only occur when f ′′( x)changes sign from positive to negative andf′′ ( x) = 0. f ′′( x)has at most 2 zeros, thus f(x)has at most 2 inflection points.43. Since the c term is squared, the only differenceoccurs when c = 0. When c = 0,2 2 3y x x x= = which has domain (– ∞ , ∞ )2 2 2and range [0, ∞ ). When c ≠ 0, y = x x – chas domain (– ∞ , –|c|] ∪ [|c|, ∞ ) andrange [0, ∞ ).39.Instructor’s Resource Manual Section 3.5 201

44.The only extremum points are ± c . For c = 0 ,there is one minimum, for c ≠ 0 there are two.No maxima, independent of c. No inflectionpoints, independent of c.cx cxf( x)= =2 2 24 + ( cx) 4+c x2 2c(4 – c x )2f′ ( x) = ; f′( x) = 0 when x =±2 2 2(4 + c x )cunless c = 0, in which case f(x) = 0 andf′ ( x) = 0.⎡ 2 2⎤If c > 0, f(x) is increasing on ⎢– ,c c⎥⎣ ⎦ and⎛ 2⎤ ⎡2⎞decreasing on ⎜– ∞, – ∪ , ∞ ,c⎥⎢ ⎟ thus, f(x) has⎝ ⎦ ⎣c⎠⎛ 2⎞ 1a global minimum at f ⎜– ⎟ = – and a global⎝ c ⎠ 4⎛2⎞ 1maximum of f ⎜ ⎟ = .⎝c⎠ 4⎛ 2⎤ ⎡ 2 ⎞If c < 0, f(x) is increasing on ⎜– ∞, ∪ – , ∞c⎥⎢ ⎟⎝ ⎦ ⎣ c ⎠⎡2 2⎤and decreasing on ⎢ , – .c c⎥Thus, f(x) has a⎣ ⎦⎛ 2⎞ 1global minimum at f ⎜– ⎟ = – and a global⎝ c ⎠ 4⎛2⎞ 1maximum at f ⎜ ⎟ = .⎝c⎠ 43 2 22 cxcx ( –12)f′′ ( x) =, so f(x) has inflection2 2 3(4 + c x )45.12 2 2f( x) =, then( cx –4) + cx22 cx(7−2 cx )f′ ( x) =;2 2 2 2[( cx – 4) + cx ]If c > 0, f′ ( x) = 0 when x = 0, ±7 .2cIf c < 0, f′ ( x) = 0 when x = 0.Note that f(x) = 1 (a horizontal line) if c = 0.167If c > 0, f′ ( x) > 0 when x 016when x < 0, so f(x) is increasing on (– ∞ , 0] anddecreasing on [0, ∞ ). Thus, f(x) has a local1maximum f (0) = . Note that f(x) > 0 and has16horizontal asymptote y = 0.points at x = 0,2 3± , c ≠ 0c202 Section 3.5 Instructor’s Resource Manual

46.1f( x) = .2x + 4x+c2+ 4 + = 0 when –2 4– .x x cBy the quadratic formula,x = ± c Thus f(x)has vertical asymptote(s) at x = –2±4– c–2 x – 4when c ≤ 4. f′ ( x) = ; f′( x) = 02 2( x + 4 x+c)when x = –2, unless c = 4 since then x = –2 is avertical asymptote.For c ≠ 4, f′ ( x) > 0 when x < –2, so f(x) isincreasing on (– ∞ , –2] and decreasing on[–2, ∞ ) (with the asymptotes excluded). Thus1f(x) has a local maximum at f (–2) = .c – 4For2c = 4, f′ ( x) = – so f(x) is increasing on( 2)3x +(– ∞ , –2) and decreasing on (–2, ∞ ).If c < 0 :⎡(4k+ 1) π ( 4k−1)π ⎤f ( x ) is decreasing on ⎢ ,2c2c⎥⎣⎦⎡( 4k−1) π ( 4k−3)π ⎤f ( x ) is increasing on ⎢ ,2c2c⎥⎣⎦( 4k−1)f ( x ) has local minima at x = π and local2c( 4k− 3)πmaxima at x = where k is an integer.2cIf c = 0 , f ( x ) = 0 and there are no extrema.If c > 0 :⎡( 4k−3) π ( 4k−1)π ⎤f ( x ) is decreasing on ⎢ ,2c2c⎥⎣⎦⎡( 4k−1 ) π (4k+ 1) π ⎤f ( x ) is increasing on ⎢ ,2c2c⎥⎣⎦( 4k−1)f ( x ) has local minima at x = π and2c( 4k− 3)πlocal maxima at x = where k is an2cinteger.y−3π −2π −π π 2π 3πxcc = −2y47. f ( x) = c+ sin cx.Since c is constant for all x and sin cx is continuouseverywhere, the function f ( x ) is continuouseverywhere.f '( x)= c⋅coscxf '( x ) = 0 when cx = 1( k + 2 ) π or x = 1( k + π2 ) cwhere k is an integer.2f ''( x) =−c ⋅ sin cxf '' (( k 2 2+ 1 π12)) sin ( ( 2)) ( 1 )kc =− c ⋅ c ⋅ k + πc =− c ⋅ −In general, the graph of f will resemble the graph ofy = sin x. The period will decrease as c increasesand the graph will shift up or down depending onwhether c is positive or negative.If c = 0 , then f ( x ) = 0 .−3π −2π −π π 2π 3πcxc = −1Instructor’s Resource Manual Section 3.5 203

−3π −2π −π π 2π 3πyyc = 0xJustification:f(1) = g(1) = 14 4f ( − x) = g(( − x) ) = g( x ) = f( x)f is an even function; symmetric with respect tothe y-axis.4 3f '( x) = g'( x )4xf '( x) > 0 for xon (0,1) ∪(1, ∞)f '( x) < 0 for xon ( −∞, −1) ∪( −1,0)f '( x) = 0 for x = −1,0,1 since f ' is continuous.4 6 2f ''( x) = g''( x )16 x + g'( x)12x−3π −2π −π π 2π 3πcc = 1xf x xf x x xf x x x''( ) = 0 for =−1,0,1''( ) > 0 for on (0, 0) ∪(1, ∞)''( ) < 0 for on ( 0,1)Where x 0 is a root of f ''( x ) = 0 (assume thatthere is only one root on (0, 1)).−3π −2π −π π 2π 3πcyc = 248. Since we have f ''' ( c ) > 0 , we know that f '( x )49.is concave up in a neighborhood around x = c .Since f '( c ) = 0, we then know that the graph off '( x ) must be positive in that neighborhood.This means that the graph of f must be increasingto the left of c and increasing to the right of c.Therefore, there is a point of inflection at c.x50. Suppose H ′′′(1) < 0, then H ′′( x)is decreasing ina neighborhood around x = 1. Thus, H ′′( x)> 0to the left of 1 and H ′′( x)< 0 to the right of 1, soH(x) is concave up to the left of 1 and concavedown to the right of 1. Suppose H ′′′ (1) > 0, thenH ′′ ( x)is increasing in a neighborhood aroundx = 1. Thus, H ′′ ( x)< 0 to the left of 1 andH ′′ ( x)> 0 to the right of 1, so H(x) is concaveup to the right of 1 and concave down to the leftof 1. In either case, H(x) has a point of inflectionat x = 1 and not a local max or min.51. a. Not possible; F′ ( x) > 0 means that F(x) isincreasing. F′′ ( x) > 0 means that the rate atwhich F(x) is increasing never slows down.Thus the values of F must eventuallybecome positive.b. Not possible; If F(x) is concave down for allx, then F(x) cannot always be positive.c.204 Section 3.5 Instructor’s Resource Manual

52. a.53. a.b.c.d.No global extrema; inflection point at (0, 0)No global maximum; global minimum at(0, 0); no inflection pointsGlobal minimumf(–π ) = –2π + sin (–π ) = –2π ≈ –6.3;global maximumf ( π ) = 2π + sinπ = 2π≈ 6.3;inflection point at (0, 0)Global minimumsin(– π)f (– π ) = – π – = – π≈ 3.1; global2sin πmaximum f ( π ) =π+ = π≈ 3.1;2inflection point at (0, 0).b.f ′( x) = 2cos x–2cosxsinx= 2cos x(1–sin x);π πf′ ( x) = 0 when x = – ,2 22 2f ′′( x) = –2sin x– 2cos x+2sin x24sin –2sin –2; ′′( ) 0= x x f x = when1sin x = – or sin x = 1 which occur when2π 5π πx = – , – ,6 6 2⎛ π ⎞Global minimum f ⎜– ⎟ = –2; global⎝ 2 ⎠⎛π⎞maximum f ⎜ ⎟ = 2; inflection points⎝2⎠⎛ π⎞ 1 ⎛ 5π⎞1f ⎜– ⎟= – , f ⎜– ⎟=–⎝ 6⎠ 4 ⎝ 6 ⎠ 4f ′( x) = 2cos x+2sin xcosx= 2cos x(1 + sin x);f′ ( x) = 0 whenπ πx = – ,2 22 2f ′′( x) = –2sinx+2cos x–2sinx2–4sin x–2sin x 2; f′′( x) 0sin x = –1 or sin12π π 5πx = – , ,2 6 6⎛ π ⎞Global minimum f ⎜– ⎟ = –1;⎝ 2 ⎠⎛π⎞maximum f ⎜ ⎟ = 3;⎝2⎠⎛π⎞ 5 ⎛5π⎞5f ⎜ ⎟= , f ⎜ ⎟=.⎝6⎠ 4 ⎝ 6 ⎠ 4= + = whenx = which occur whenglobalinflection pointsInstructor’s Resource Manual Section 3.5 205

c.f ′( x) = –2sin2x+2sinx= –4sinx cosx+ 2sin x = 2sin x(1–2cos x);π πf′ ( x) = 0 when x = – π, – , 0, , π3 3f ′′( x) = –4cos 2x+ 2cos x;f′′ ( x) = 0 whenx ≈ –2.206, –0.568, 0.568, 2.206⎛ π⎞ ⎛π⎞Global minimum f ⎜– ⎟= f ⎜ ⎟=–1.5;⎝ 3⎠ ⎝3⎠Global maximum f(–π ) = f(π ) = 3;− ,Inflection points: ≈ ( 2.206,0.890)( −0.568, − 1.265), ( 0.568, − 1.265),( 2.206,0.890 )54.f ′( x) = 2cos2x+3sin3xUsing the graphs, f(x) has a global minimumat f(2.17) ≈ –1.9 and a global maximum atf(0.97) ≈ 1.9f′′ ( x) = –4sin 2x+ 9cos3 x; f′′( x) = 0 whenπ πx = – , and when2 2x ≈ –2.469, –0.673, 0.413, 2.729.⎛ π ⎞Inflection points: ⎜−,0⎟⎝ 2 ⎠ , ⎛π2,0 ⎞⎜ ⎟⎝ ⎠ ,≈ ( − 2.469,0.542), ( 0.673, 0.542)( 0.413,0.408 ) , ( 2.729, − 0.408)− − ,d.55.y5e.f′ ( x) = 3cos3 x– cos x; f′( x) = 0 when3cos 3x = cos x which occurs whenπ πx = – , and when2 2x ≈ –2.7, –0.4, 0.4, 2.7f ′′( x) = –9sin 3x+ sin x which occurs whenx = –π , 0, π and whenx ≈ –2.126, –1.016, 1.016, 2.126⎛π⎞Global minimum f ⎜ ⎟ = –2;⎝2⎠⎛ π ⎞global maximum f ⎜– ⎟ = 2;⎝ 2 ⎠Inflection points: ≈ ( − 2.126,0.755),1.016, − 0.755 ,( − 1.016,0.755), ( )( 2.126, − 0.755)0,0 , ( )−5−5a. f is increasing on the intervals ( −∞, − 3]and [ − 1, 0].f is decreasing on the intervals [ −3, − 1]and [ 0,∞ ) .b. f is concave down on the intervals, 2 2,∞ .5( −∞ − ) and ( )f is concave up on the intervals ( − 2,0)and ( 0, 2 ) .c. f attains a local maximum at x =− 3 andx = 0 .f attains a local minimum at x =− 1.xd. f has a point of inflection at x =− 2 andx = 2 .206 Section 3.5 Instructor’s Resource Manual

56.5y59. a.−55x−5a. f is increasing on the interval [ −1,∞ ) .f is decreasing on the interval ( −∞, − 1]b. f is concave up on the intervals ( − 2,0)and ( 2,∞ ).f is concave down on the interval ( 0, 2 ) .22 x –9x+40f ′( x) =; f′( x)is never 0,2x –6x+40and always positive, so f(x) is increasing forall x. Thus, on [–1, 7], the global minimum isf(–1) ≈ –6.9 and the global maximum iff(7) ≈ 48.0.3 22x − 18x + 147 x– 240f′′ ( x) = ; f′′( x) = 02 3/2( x – 6x+40)when x ≈ 2.02; inflection pointf(2.02) ≈ 11.4c. f does not have any local maxima.f attains a local minimum at x = − 1.b.d. f has inflection points at x = 0 andx = 2 .57.Global minimum f(0) = 0; global maximumf(7) ≈ 124.4; inflection point at x ≈ 2.34,f(2.34) ≈ 48.09c.58.No global minimum or maximum; noinflection pointsd.Global minimum f(3) ≈ –0.9;global maximum f(–1) ≈ 1.0 or f(7) ≈ 1.0;Inflection points at x ≈ 0.05 and x ≈ 5.9,f(0.05) ≈ 0.3, f(5.9) ≈ 0.3.Instructor’s Resource Manual Section 3.5 207

60. a.3.6 Concepts Reviewb.2f′ ( x) = 3 x –16x+ 5; f′( x) = 0 when1x = , 5.3Global minimum f(5) = –46;1global maximum f ⎛⎜⎞ ⎟≈4.8⎝3⎠f′′ ( x) = 6 x–16; f′′( x) = 0 wheninflection point: 8( , − 20.6 3 )8x = ;31. continuous; (a, b); f ( b)– f( a) = f′( c)( b– a)2. f ′(0)does not exist.3. F(x) = G(x) + C4.4x+ CProblem Set 3.6x1. f′ ( x)=xf(2) – f(1) 2 –1= = 12–1 1c= 1 for all c > 0, hence for all c in (1, 2)cc.Global minimum when x ≈ –0.5 andx ≈ 1.2, f(–0.5) ≈ 0, f(1.2) ≈ 0;global maximum f(5) = 46Inflection point: ( 0.5,0)8( ,20.6 3 )− , ( )1.2,0 ,2. The Mean Value Theorem does not applybecause g′ (0) does not exist.d.No global minimum or maximum;inflection point atx ≈−0.26, f( −0.26) ≈− 1.73. f ′( x) = 2x+1f(2) – f(–2) 6 – 2= = 12 – (–2) 42c + 1 = 1 when c = 0No global minimum, global maximum whenx ≈ 0.26, f(0.26) ≈ 4.7Inflection points when x ≈ 0.75 andx ≈ 3.15, f(0.75) ≈ 2.6, f(3.15) ≈ –0.88208 Section 3.6 Instructor’s Resource Manual

4.g′ ( x) = 3( x+1)g(1) – g(–1) 8 – 0= = 41–(–1) 222 23( c+ 1) = 4 when c = –1+ ≈ 0.1537.1 2 2 1f′ ( z) = (3z + 1) = z +3 3f(2) – f(–1) 2 – (–2) 4= =2–(–1) 3 32 1 4c + = when c = –1, 1, but −1is not in3 3( −1,2)so c = 1 is the only solution.5. H ′() s = 2s+3H(1) – H(–3) 3 – (–1)= = 11– (–3) 1– (–3)2c + 3 = 1 when c = –18. The Mean Value Theorem does not applybecause F(t) is not continuous at t = 1.6.F′ ( x)= x2( )8 – –83 3F(2) – F(–2) 4= =2 – (–2) 4 32 4c = when32c =± ≈± 1.1539.3h′ ( x) = –( –3)2xh(2) – h(0) –2 – 0= = –12–0 23– = –1 when c = 3 ± 3,2( c –3)c = 3– 3 ≈ 1.27 (3 + 3 is not in (0, 2).)Instructor’s Resource Manual Section 3.6 209

10. The Mean Value Theorem does not applybecause f(x) is not continuous at x = 3.14.5 2/3g′ ( x)= x3g(1) – g(–1) 1 – (–1)= = 11–(–1) 23/25 2/313 c = when ⎛3⎞c =± ⎜ ⎟ ≈± 0.46⎝5⎠11.h′ () t =21/33t2/3h(2) – h(0) 2 − 0 –1/3= = 22–0 22 –1/3 16= 2 when c = ≈ 0.591/33c2715. S′ ( θ ) = cosθS( π)– S(– π ) 0–0 = = 0π–(– π) 2ππcos c = 0 when c = ± .212. The Mean Value Theorem does not applybecause h′ (0) does not exist.16. The Mean Value Theorem does not applyC θ is not continuous at θ =− π,0,π .because ( )13.5 2/3g′ ( x)= x3g(1) – g(0) 1 – 0= = 11–0 13/25 2/313 c = when ⎛3⎞c =± ⎜ ⎟ ,⎝5⎠3/2⎛3⎞ ⎛ 3/2⎛3⎞⎞c = ⎜ ⎟ ≈0.46,– is not in (0, 1).⎝5⎠ ⎜ ⎜ ⎟⎝5⎠⎟⎝⎠17. The Mean Value Theorem does not applyπbecause T ( θ ) is not continuous at θ = .2210 Section 3.6 Instructor’s Resource Manual

18. The Mean Value Theorem does not applybecause f(x) is not continuous at x = 0.22. By the Mean Value Theoremf( b) − f( a)= f′() c for some c in (a, b).b−a0Since f(b) = f(a), = f′ (); c f′() c = 0.b−a19.12f′ ( x) = 1–x5f(2) – f(1) 2 –2 1= =2–1 1 21 11– when c2c = 22, 2 1.41( c = – 2 is not in (1, 2).)f(8) − f(0) 123.= −8−0 4There are three values for c such that1f′ () c = − .4They are approximately 1.5, 3.75, and 7.24. f′ ( x) = 2αx+βf( b)– f( a) 1 [ (2 –2b a ) ( b – a )]b– a = b–a α + β= ( a+ b)+αβa+b2 αc+ β = α( a+ b)+ β when c = which is2the midpoint of [a, b].20. The Mean Value Theorem does not applybecause f(x) is not continuous at x = 2.25. By the Monotonicity Theorem, f is increasing onthe intervals ( a, x0) and ( x0, b ).To show that f( x0) > f( x)for x in ( a, x 0),consider f on the interval ( a, x 0].f satisfies the conditions of the Mean ValueTheorem on the interval [ x, x 0]for x in ( a, x 0).So for some c in ( x, x 0),f( x0) − f( x) = f′( c)( x0− x).Becausef′ () c > 0 and x − x > 0, f( x ) − f() x > 0,0 0so f( x0) > f( x).Similar reasoning shows thatf( x) > f( x0) for x in ( x0, b).Therefore, f is increasing on (a, b).21. The Mean Value Theorem does not applybecause f is not differentiable at 0 x = .26. a.b.c.2f ( x) 3x0′ = > except at x = 0 in (– ∞ , ∞ ).3xf( x)= is increasing on (– ∞ , ∞ ) byProblem 25.4f ( x) 5x0′ = > except at x = 0 in (– ∞ , ∞ ).5xf( x)= is increasing on (– ∞ , ∞ ) byProblem 25.⎧ 2⎪3xx≤0f′ ( x)= ⎨⎪⎩ 1 x > 0(– ∞ , ∞ ).⎧ 3⎪xx≤0f( x)= ⎨⎪⎩ x x > 0(– ∞ , ∞ ) by Problem 25.> 0 except at x = 0 inis increasing onInstructor’s Resource Manual Section 3.6 211

27. s(t) is defined in any interval not containing t = 0.1s′ () c = – < 0for all c ≠ 0. For any a, b with2ca 0 while s′ () c < 0, hences(b) – s(a) < 0, or s(b) < s(a).Thus, s(t) is decreasing on any interval notcontaining t = 0.28.2s () c – 03c′ = < for all c > 0. If 0 < a < b, theMean Value Theorem sayss( b)– s( a) = s′( c)( b– a)for some c in (a, b).Since a < b, b – a > 0 while s′ () c < 0, hences(b) – s(a) < 0, or s(b) < s(a). Thus, s(t) isdecreasing on any interval to the right of theorigin.29. F′ ( x) = 0 and G( x) = 0; G′( x) = 0.By Theorem B,F(x) = G(x) + C, so F(x) = 0 + C = C.30.2 2 2 2F( x) = cos x+ sin x; F(0) = 1 + 0 = 1F′ ( x) = 2cos x( − sin x) + 2sin x(cos x) = 0By Problem 29, F(x) = C for all x.2 2Since F(0) = 1, C = 1, so sin x+ cos x = 1 forall x.31. Let Gx ( ) = DxF ; ′( x) = Dand G′( x)= D.By Theorem B, F(x) = G(x) + C; F(x) = Dx + C.32. F′ ( x) = 5; F(0) = 4F(x) = 5x + C by Problem 31.F(0) = 4 so C = 4.F(x) = 5x + 433. Since f(a) and f(b) have opposite signs, 0 isbetween f(a) and f(b). f(x) is continuous on [a, b],since it has a derivative. Thus, by theIntermediate Value Theorem, there is at least onepoint c,a < c < b with f(c) = 0.Suppose there are two points, c and c′ , c < c′in(a, b) with f() c = f( c′ ) = 0. Then by Rolle’sTheorem, there is at least one number d in (, cc′ )with f′ ( d) = 0. This contradicts the giveninformation that f′ ( x) ≠ 0 for all x in [a, b], thusthere cannot be more than one x in [a, b] wheref(x) = 0.34.2f′ ( x) 6 x –18x 6 x( x–3); f′( x) 0= = = whenx = 0 or x = 3.f(–1) = –10, f(0) = 1 so, by Problem 33, f(x) = 0has exactly one solution on (–1, 0).f(0) = 1, f(1) = –6 so, by Problem 33, f(x) = 0 hasexactly one solution on (0, 1).f(4) = –15, f(5) = 26 so, by Problem 33, f(x) = 0has exactly one solution on (4, 5).35. Suppose there is more than one zero betweensuccessive distinct zeros of f ′ . That is, there area and b such that f(a) = f(b) = 0 with a and bbetween successive distinct zeros of f ′ . Then byRolle’s Theorem, there is a c between a and bsuch that f′ () c = 0. This contradicts thesupposition that a and b lie between successivedistinct zeros.36. Let x 1 , x 2 , and x 3 be the three values such thatgx ( ) = gx ( ) = gx ( ) = 0 and1 2 3a x1 x2 x3b≤ < < ≤ . By applying Rolle’sTheorem (see Problem 22) there is at least onenumber x 4 in ( x1, x 2)and one number x 5 in( x2, x 3)such that g′ ( x4) = g′( x5) = 0. Then byapplying Rolle’s Theorem to g′ ( x), there is atleast one number x 6 in ( x4, x 5)such thatg′′ ( x ) = 0.637. f(x) is a polynomial function so it is continuouson [0, 4] and f′′ ( x)exists for all x on (0, 4).f(1) = f(2) = f(3) = 0, so by Problem 36, there areat least two values of x in [0, 4] where f′ ( x) = 0and at least one value of x in [0, 4] wheref′′ ( x) = 0.38. By applying the Mean Value Theorem and takingthe absolute value of both sides,f( x2) − f( x1)= f ′()c , for some c in ( x1, x 2).x2 − x1Since f ′( x)≤ M for all x in (a, b),f( x2) − f( x1)x2 − x1≤ M ; f( x2) − f( x1) ≤ M x2 −x1.39. f′ ( x) = 2cos2 x; f′( x) ≤ 2f ( x 2) − f ( x 1) ( 2) ( 1)( ) ; f x −= f′xf x ≤ 2x2 −x1 x2 −x1f ( x2) − f( x1) ≤ 2 x2 − x1;sin 2x2 − sin 2x1 ≤ 2 x2 − x1212 Section 3.6 Instructor’s Resource Manual

40. a.b.41. Suppose f′ ( x) ≥ 0 . Let a and b lie in the interiorof I such that b > a. By the Mean Value Theorem,there is a point c between a and b such thatf( b) − f( a) f( b) − f( a)f′ () c = ; ≥0.b−a b−aSince a < b, f(b) ≥ f(a), so f is nondecreasing.Suppose f′ ( x) ≤ 0. Let a and b lie in the interiorof I such that b > a. By the Mean Value Theorem,there is a point c between a and b such thatf ( b) − f( a)f( b) − f( a)f′ () c =;≤ 0 . Sinceb−a b−aa < b, f(a) ≥ f(b), so f is nonincreasing.42.2[ f ( x)] ′ = 2 f( x) f′( x)Because f(x) ≥ 0 and f′ ( x) ≥0 on I, [ f ( x)] ′ ≥ 0on I.As a consequence of the Mean Value Theorem,2 2f ( x2) f ( x1) 0− ≥ for all x2 > x1on I.2Therefore f is nondecreasing.43. Let f(x) = h(x) – g(x).f′ ( x) = h′ ( x) − g′ ( x); f′ ( x) ≥ 0 for all x in(a, b) since g′ ( x) ≤ h′( x)for all x in (a, b), so f isnondecreasing on (a, b) by Problem 41. Thusx < x ⇒ f( x ) ≤ f( x );1 2 1 2hx ( 1) −gx ( 1) ≤ hx ( 2) − gx ( 2);g( x2) g( x1) h( x2) h( x1)− ≤ − for all x 1 and x 2in (a, b).2144. Let f( x) = x so f′( x)= . Apply the Mean2 xValue Theorem to f on the interval [x, x + 2] forx > 0.1 1Thus x+ 2 − x = (2) = for some c in2 c c1 1 1(x, x + 2). Observe < < .x + 2 c x1Thus as x →∞, → 0.c1Therefore lim ( x+ 2 − x)= lim = 0 .x→∞x→∞c45. Let f(x) = sin x. f′ ( x)= cos x, sof′ ( x) = cosx≤ 1 for all x.By the Mean Value Theorem,f( x) − f( y)= f′() c for some c in (x, y).x−yThus,f( x) − f( y)x−ysin x − sin y ≤ x−y .= f′ () c ≤1;46. Let d be the difference in distance between horseA and horse B as a function of time t.Then d′ is the difference in speeds.Let t 0 and t 1 and be the start and finish times ofthe race.dt ( 0) = dt ( 1) = 0By the Mean Value Theorem,dt ( 1) − dt ( 0)= d′() ct1−t0Therefore d () c 0for some c in ( t0, t 1).′ = for some c in ( t0, t 1).47. Let s be the difference in speeds between horse Aand horse B as function of time t.Then s′ is the difference in accelerations.Let t 2 be the time in Problem 46 at which thehorses had the same speeds and let t 1 be thefinish time of the race.st ( 2) = st ( 1) = 0By the Mean Value Theorem,st ( 1) − st ( 2)= s′() ct1−t2Therefore s () c 0for some c in ( t2, t 1).′ = for some c in ( t2, t 1).Instructor’s Resource Manual Section 3.6 213

48. Suppose x > c. Then by the Mean ValueTheorem,f( x) − f( c) = f′( a)( x− c)for some a in (, cx. )Since f is concave up, f ′′ > 0 and by theMonotonicity Theorem f ′ is increasing.Therefore f′ ( a) > f′( c)andf( x) − f( c) = f′ ( a)( x− c) > f′( c)( x−c)f() x > f() c + f′()( c x− c),x > cSuppose x < c. Then by the Mean ValueTheorem,f( c) − f( x) = f′( a)( c− x) for some a in ( x, c).Since f is concave up, f ′′ > 0 , and by theMonotonicity Theorem f ′ is increasing.Therefore, f′ () c > f′() a andf() c − f() x = f′ ()( a c− x) < f′()( c c− x).− f( x) < − f( c) + f′( c)( c−x)f( x) > f( c) − f′( c)( c−x)f() x > f() c + f′()( c x− c),x f() c + f′()( c x−c),x ≠ c.49. Fix an arbitrary x.f ( y) − f ( x)f '( x)= lim = 0, sincey→xy−x50.( ) − ( )f y f x≤ M y−x.y−xSo, f ' ≡0→ f = constant.f ( x)1/3x= on [0, a] or [–a, 0] where a is anypositive number. f ′(0)does not exist, but f(x)has a vertical tangent line at x = 0.51. Let f(t) be the distance traveled at time t.f(2) − f(0) 112 −0= = 562−0 2By the Mean Value Theorem, there is a time csuch that f′ ( c) = 56.At some time during the trip, Johnny must havegone 56 miles per hour.52. s is differentiable with s (0) = 0 and s (18) = 20 sowe can apply the Mean Value Theorem. Thereexists a c in the interval ( 0,18)such that(20 − 0)vc ( ) = s'( c) = ≈1.11 miles per minute( 18 − 0)≈ 66.67 miles per hour53. Since the car is stationary at t = 0 , and since v is1continuous, there exists a δ such that vt ()

254. Given the position function s () t = at + bt+ c ,the car’s instantaneous velocity is given by thefunction s '()t = 2at+ b.A + BThe midpoint of the interval [ A,B ] is .2Thus, the car’s instantaneous velocity at themidpoint of the interval is given by⎛ A+ B⎞ ⎛ A+B⎞s ' ⎜ ⎟= 2a⎜ ⎟+b⎝ 2 ⎠ ⎝ 2 ⎠= a A+ B + b( )The car’s average velocity will be its change inposition divided by the length of the interval.That is,2 2( ) − s( A) ( aB + bB + c) − ( aA + bA + c)s B=B−A B−A2 2aB − aA + bB −bA=B−A2 2a( B − A ) + b( B−A)=B−Aa B A B A b B A=B−A= a B+ A + b= a A+ B + b( − )( + ) + ( − )( )( )This is the same result as the instantaneousvelocity at the midpoint.3.7 Concepts Review1. slowness of convergence2. root; Intermediate Value3. algorithms4. fixed pointProblem Set 3.71. Let f( x) = x + 2 x–6.f(1) = –3, f(2) = 6nh n3m f ( m )1 0.5 1.5 0.3752 0.25 1.25 –1.5468753 0.125 1.375 –0.6503914 0.0625 1.4375 –0.1545415 0.03125 1.46875 0.1059276 0.015625 1.45312 –0.02537167 0.0078125 1.46094 0.040018 0.00390625 1.45703 0.007256709 0.00195312 1.45508 –0.00907617r ≈ 1.464 32. Let f( x) = x + 5x+ 1.f(–1) = –3, f(0) = 1n h n m n f ( m n )1 0.5 –0.5 0.43752 0.25 –0.75 –0.7929693 0.125 –0.625 –0.06811524 0.0625 –0.5625 0.210225 0.03125 –0.59375 0.07768346 0.015625 –0.609375 0.006471697 0.0078125 –0.617187 –0.03039628 0.00390625 –0.613281 –0.0118549 0.00195312 –0.611328 –0.00266589r ≈ –0.613. Let f ( x) = 2cosx− sinx.f ( 1)≈ 0.23913 ; ( )nf 2 ≈− 1.74159( )n h m f mn n n1 0.5 1.5 −0.8560212 0.25 1.25 −0.3183403 0.125 1.125 −0.0399154 0.0625 1.0625 0.9980445 0.03125 1.09375 0.0299606 0.01563 1.109375 −0.004978r ≈ 1.11nInstructor’s Resource Manual Section 3.7 215

4. Let f ( x) = x− 2+2cosxf () 1 = 1− 2 + 2cos()1 ≈ 0.080605f ( 2) = 2 − 2 + 2cos( 2)≈ − 0.832294( )n h m f mn n n1 0.5 1.5 −0.3585262 0.25 1.25 −0.1193553 0.125 1.125 −0.0126474 0.0625 1.0625 0.0358795 0.03125 1.09375 0.0120656 0.01563 1.109375 −0.000183r ≈ 1.11nx n1 12 0.86363643 0.84126704 0.84069985 0.84069946 0.8406994r ≈ 0.840707. Let ( ) 2 2cosf x = x− + x.y5. Let3 2f( x) = x + 6x + 9x+ 1= 0.5−55x( )−5f ' x = 1−2sinx6. Let2f′ ( x) = 3x + 12x+9nx n1 02 –0.11111113 –0.12054844 –0.12061485 –0.1206148r ≈ –0.120613f ( x) = 7 x + x–5n x n1 42 3.7244153 3.6984294 3.6981545 3.698154r ≈ 3.698158. Let ( ) 2cos sinf x = x− x.5y−55x2f′ ( x) = 21x+ 1( )−5f ' x =−2sinx−cosxn x n1 0.52 1.19468333 1.10692444 1.10714875 1.1071487r ≈ 1.10715216 Section 3.7 Instructor’s Resource Manual

9. Let f(x) = cos x – 2x.3 2f ( x) 4 x –24x 44 x–24′ = +Note that f(2) = 0.nx n1 0.52 0.5753 0.5855864 0.585786f′ ( x) = –sin x–2nr ≈ 0.45018x n1 0.52 0.45062673 0.45018364 0.450183610. Let f ( x) 2x sinx1= − − .y12. Letnx n1 3.52 3.4253 3.4144144 3.4142145 3.414214r = 2, r ≈ 0.58579, r ≈ 3.414214 3 2f( x) = x + 6x + 2x + 24 x–8.5−55x11. Let( )−5f ' x = 2−cosxn x n1 12 0.8913963 0.8878664 0.8878625 0.887862r ≈ 0.887864 3 2f( x) = x –8x + 22 x –24x+8.3 2f′ ( x) = 4x + 18x + 4x+24nx n1 –6.52 –6.32996323 –6.31670224 –6.31662485 –6.3166248nx n1 0.52 0.32862903 0.31666944 0.31662485 0.3166248r ≈ –6.31662, r ≈ 0.31662Instructor’s Resource Manual Section 3.7 217

13. Let ( )2f x = 2x − sinx.y16. Let4f( x) = x –47.3f ′( x) = 4x2nx n1 2.5−2−22x2 2.6273 2.6183734 2.6183305 2.618330( )f ' x = 4x−cosx4 47 2.61833 ≈n x n1 0.52 0.4816703 0.4809474 0.480946r ≈ 0.4809514. Let ( ) 2cot15. Letf x = x− x .2−2 2−2y( )2f ' x =−2csc x−1n x n1 12 1.0743053 1.0768714 1.076874r ≈ 1.076873f( x) = x –6.2f ′( x) = 3xx4 3 217. f ( x) = x + x + x + x is continuous on thegiven interval.From the graph of f, we see that the maximumvalue of the function on the interval occurs at theright endpoint. The minimum occurs at astationary point within the interval. To findf ' x = 0where the minimum occurs, we solve ( )on the interval [ − 1,1].3 2f '( x) = 4x + 3x + 2x+ 1=g( x)Using Newton’s Method to solve g( x ) = 0 , weget:n x n1 02 −0.53 −0.6254 −0.606385 −0.605836 −0.60583Minimum: f ( −0.60583)≈ − 0.32645Maximum: f ( 1)= 4n3 6 1.81712 ≈x n1 1.52 1.8888893 1.8198134 1.8171255 1.8171216 1.817121218 Section 3.7 Instructor’s Resource Manual

3x + 1= is continuous on the given4x + 1interval.52n x nn x n18. f ( x)1 4.712389 1 7.853982π .2 4.479179 2 7.7223913 4.793365 3 7.7252514 4.493409 4 7.7252525 4.493409 5 7.725252Minimum: f ( 4.493409)≈− 0.21723From the graph of f, we see that the maximumMaximum: f ( 7.725252)≈ 0.128375and minimum will both occur at stationary pointswithin the interval. The minimum appears to20. ( ) 2 xf x = x sin is continuous on the givenoccur at about x =− 1.5 while the maximum2appears to occur at about x = 0.8 . To find theinterval.stationary points, we solve f '( x ) = 0.2 4− x ( x + 4x−3)f '( x)= = g( x)42( x + 1)Using Newton’s method to solve g( x ) = 0 onthe interval, we use the starting values of − 1.5and 0.8 .find these points, we need to solve ( )n x n n x nthe interval.1 −1.51 0.8x xx cos + 4xsin2 −1.6807342 0.694908f '( x) = 2 2 = g( x)3 −1.7666423 0.69251224 −1.7837664 0.692505Using Newton’s method to solve ( ) 05 −1.7843575 0.6925056 −1.78435813π 7 −1.784358.4Maximum: f ( 0.692505)≈ 1.083021 4.712389 1 10.210176nMinimum: f ( −1.78436)≈ − 0.42032x n n x n19.sin x2 4.583037 2 10.174197f ( x)= is continuous on the given interval.x3 4.577868 3 10.1739704 4.577859 4 10.1739705 4.577859Minimum: f ( 10.173970)≈− 96.331841Maximum:From the graph of f, we see that the minimumf ( 4.577859)≈ 15.78121value and maximum value on the interval willoccur at stationary points within the interval. Tofind these points, we need to solve f '( x ) = 0 onthe interval.xcosx−sin xf '( x) = = g( x)2xUsing Newton’s method to solve g( x ) = 0 onthe interval, we use the starting values of 3 π and2From the graph of f, we see that the minimumvalue and maximum value on the interval willoccur at stationary points within the interval. Tof ' x = 0 ong x = onthe interval, we use the starting values of 3 π and2Instructor’s Resource Manual Section 3.7 219

21. Graph y = x and y = 0.8 + 0.2 sin x.22.xn+ 1 = 0.8 + 0.2sin xnLet x 1 = 1.nx n1 12 0.968293 0.964784 0.964395 0.964346 0.964337 0.96433x ≈ 0.96431/3f( xn)xnxn+ 1 = xn – = xn–f′( x ) 1 –2/3n x3 n= xn – 3 xn = –2xnThus, every iteration of Newton’s Method getsn+1further from zero. Note that x (–2) x .n+ 1 = 0Newton’s Method is based on approximating f byits tangent line near the root. This function has avertical tangent at the root.23. a. For Tom’s car, P = 2000, R = 100, andk = 24, thus100 ⎡ 1 ⎤2000 = ⎢1 − ⎥ ori (1 )24⎢⎣+ i ⎥⎦120i= 1 − , which is equivalent to(1 + )24i24 24i + i − + i + = .20 (1 ) (1 ) 1 0b. Let24 24f i = i + i − + i +() 20(1 ) (1 ) 124= (1 + i) (20i− 1) + 1 .Thenf ′( i) = 20(1 + i) + 480 i(1 + i) − 24(1 + i)23= (1 + i) (500i− 4), so24 23 2324f in + in in− +in+1 = in − = in−f′ i 23n + in in−( ) (1 ) (20 1) 1( ) (1 ) (500 4)⎡ 2 −2320in + 19in − 1 + (1 + in)⎤= in−⎢ ⎥ .⎢⎣500in− 4 ⎥⎦c. n i n1 0.0122 0.01652973 0.01526514 0.01513235 0.01513086 0.0151308i = 0.0151308r = 18.157%f ( xn)24. From Newton’s algorithm, xn+ 1 – xn=− .f ′ ( x )nlim ( xn+ 1 – xn) = lim xn+1 – lim xnxn→x xn→x xn→x= x – x = 0f ( xn)lim exists if f and f ′ are continuous atxn→ x f ′( xn)f′ x ≠x and ( ) 0.f( xn) f( x)Thus, lim = = 0,xn→x f′ ( xn) f′( x)x is a solution of f(x) = 0.25.xn+ 1.5cos xnx n + 1 =2n x n n x n1 1 5 0.9148642 0.905227 6 0.9148563 0.915744 7 0.9148574 0.914773x ≈ 0.91486so f( x ) = 0.220 Section 3.7 Instructor’s Resource Manual

26. xn+ 1 = 2 − sinx29. a.n x n1 22 1.090703 1.113054 1.10295x ≈ 1.10606n x n5 1.107466 1.105437 1.106348 1.10612n x n9 1.1060310 1.1060711 1.1060612 1.1060627. xn+ 1 = 2.7 + xnx ≈ 2.2175628. xn+ 1 = 3.2 + xnx n1 12 1.9235383 2.1502414 2.2023265 2.2141206 2.2167817 2.2173828 2.2175179 2.21754810 2.21755411 2.21755612 2.217556nx ≈ 2.35742nx n1 472 7.0851963 3.2070544 2.5312165 2.3939966 2.3651637 2.3590608 2.3577669 2.35749110 2.35743311 2.35742112 2.35741813 2.357418b.c.30. a.b.x ≈ 0.52xn+ 1 = 2( xn – xn)nx n1 0.72 0.423 0.48724 0.49967235 0.49999986 0.57 0.52x = 2( x– x )22 x – x = 0x(2x – 1) = 01x = 0, x =2x ≈ 0.82xn+ 1 = 5( xn – xn)nx n1 0.72 1.053 –0.26254 –1.6570315 –22.013926 –2533.133Instructor’s Resource Manual Section 3.7 221

c.2x = 5( x– x )25 x –4x = 0x(5x – 4) = 04x = 0, x =531. a. x 1 = 0x 2 = 1=1b.x 3 = 1+ 1 = 2 ≈ 1.4142136x 4 = 1+ 1+ 1 ≈ 1.553774x 5 = 1+ 1+ 1+ 1 ≈ 1.5980532x = 1+x2x = 1+x2x −x− 1=01± 1+ 4⋅1⋅ 1 1±5x = =2 2Taking the minus sign gives a negativesolution for x, violating the requirement thatx ≥ 0 . Hence,1+5x = ≈ 1.618034 .2c. Let x = 1+ 1+ 1 +… . Then x satisfiesthe equation x = 1 + x.From part (b) weknow that x must equal( 1+ 5 ) / 2 ≈ 1.618034 .32. a. x 1 = 0x 2 = 5 ≈ 2.236068x 3 = 5 + 5 ≈ 2.689994x 4 = 5 + 5 + 5 ≈ 2.7730839x 5 = 5+ 5+ 5+ 5 ≈ 2.7880251b. x = 5 + x , and x must satisfy x ≥ 02x = 5 + x2x −x− 5=01± 1+ 415 ⋅ ⋅ 1±21x = =2 2Taking the minus sign gives a negativesolution for x, violating the requirement thatx ≥ 0 . Hence,x = ( 1+ 21 ) / 2 ≈ 2.791287833. a. x 1 = 11x 2 = 1+ = 21x 3 = 1 31 + 1.51+1 = 2=1x 4 = 1 51 + 1.66666671 = 13≈+1+111 8x 5 = 1+ = = 1.611 +511 +1 +111b. x = 1+x2x = x+12x −x− 1=01± 1+ 4⋅1⋅ 1 1±5x = =2 2Taking the minus sign gives a negativesolution for x, violating the requirement that1+5x ≥ 0 . Hence, x = ≈ 1.618034 .2c. Let1x = 1+.11+ 1+1Then x satisfies the equation x = 1+ .xFrom part (b) we know that x must equal( 1+ 5 ) / 2 ≈ 1.618034 .f () r34. a. Suppose r is a root. Then r = r – .f ′()rf()r0,f′() r = so f(r) = 0.f()rSuppose f(r) = 0. Then r – r –0 r,f′() r = =f ( x)so r is a root of x = x– .f ′(x)c. Let x = 5+ 5+ 5 +… . Then x satisfiesthe equation x = 5 + x.From part (b) we know that x must equal( 1+ 21 ) / 2 ≈ 2.7912878222 Section 3.7 Instructor’s Resource Manual

. If we want to solve f(x) = 0 and f′ ( x) ≠ 0 inf( x)[a, b], then 0f′( x )= orf( x)x = x– = g( x).f′( x)f′( x) f( x)g′ ( x) = 1– + f′′( x)f′ ( x) 2[ f′( x)]f ( x) f′′( x)=2[ f′( x)]f() r f′′() rand g'( r) = = 0.2[ f′( r)]35. a. The algorithm computes the root of1 – a = 0 for x 1 close to 1 .xab. Let1f ( x) = – a.x12f′ ( x) = –xf( x)2– x axf′( x )= +The recursion formula isf( xn)2xn+ 1 = xn – 2 xn – axn.f′( x )=36. We can start by drawing a diagram:2yn( x, y)⎛ π ⎞On the interval ⎜0, ⎟, there is only one⎝ 2 ⎠stationary point (check graphically). We will useNewton’s Method to find the stationary point,πstarting with x = ≈ 0.785398 .4n x n1 π 0.7853984 ≈2 0.8624433 0.8603354 0.8603345 0.860334x ≈ 0.860334 will maximize the area of therectangle in quadrant I, and subsequently thelarger rectangle as well.y = cos x = cos( 0.860334)≈ 0.652184The maximum area of the larger rectangle isAL= ( 2x) y ≈ 2( 0.860334)( 0.652184)≈ 1.122192 square units37. The rod that barely fits around the corner willtouch the outside walls as well as the insidecorner.DEaθFBθbC6.2 feetπ− 0 π2x2xA8.6 feetFrom symmetry, maximizing the area of theentire rectangle is equivalent to maximizing thearea of the rectangle in quadrant I. The area ofthe rectangle in quadrant I is given byA=xy= x cos xTo find the maximum area, we first need the⎛ π ⎞stationary points on the interval ⎜0, ⎟⎝ 2 ⎠ .A' ( x)= cosx−xsinxTherefore, we need to solveA' ( x)= 0cos x − xsin x = 0As suggested in the diagram, let a and b representthe lengths of the segments AB and BC, and let θdenote the angles ∠ DBA and ∠ FCB . Considerthe two similar triangles Δ ADB and Δ BFC ;these have hypotenuses a and b respectively. Alittle trigonometry applied to these angles gives8.66.2a = = 8.6secθand b 6.2csccosθ= θsinθ=Note that the angle θ determines the position ofthe rod. The total length of the rod is thenL = a+ b = 8.6secθ + 6.2cscθ⎛ π ⎞The domain for θ is the open interval ⎜0, ⎟⎝ 2 ⎠ .Instructor’s Resource Manual Section 3.7 223

As suggested in the diagram, let a and b representThe derivative of L is105 o3 38.6sin θ − 6.2cos θthe lengths of the segments AB and BC, and let θL '( θ ) =2 2denote the angle ∠ ABD . Consider the two rightsin θ ⋅cosθtriangles Δ ADB and Δ CEB ; these haveThus, L '( θ ) = 0 providedhypotenuses a and b respectively. A little3 38.6sin θ − 6.2cos θ = 0trigonometry applied to these angles gives3 388.6sin θ = 6.2cos θa = = 8cscθandsinθ3sin θ 6.2=83cos θ 8.6b = = 8csc( 75−θ)sin ( 75 −θ)3 6.2tan θ =Note that the angle θ determines the position of8.6the rod. The total length of the rod is then6.2tanθ= 3L = a+ b = 8cscθ + 8csc( 75−θ )8.6The domain for θ is the open interval ( 0,75 ).⎛ π ⎞On the interval ⎜0, ⎟, there will only be oneA graph of of L indicates there is only one⎝ 2 ⎠ extremum (a minimum) on the interval.solution to this equation. We will use Newton’smethod to solve tanθ − 6.2 3 = 0 starting with8.6πθ 1 = .4The derivative of L isn θn1 π2 2≈ 0.785408sin4( θ ⋅cos( θ −75) −cosθ ⋅sin ( θ −75))L '( θ ) =2 22 0.73373sin θ ⋅sin ( θ −75)3 0.73098We will use Newton’s method to solve4 0.73097L '( θ ) = 0 starting with θ 1 = 40 .5 0.73097n θ nNote that θ ≈ 0.73097 minimizes the length of1 40the rod that does not fit around the corner, which2 37.54338in turn maximizes the length of the rod that willfit around the corner (verify by using the Second3 37.50000Derivative Test).4 37.5L ( 0.73097) = 8.6sec( 0.73097) + 6.2csc( 0.73097)Note that θ = 37.5° minimizes the length of the≈ 20.84rod that does not fit around the corner, which inThus, the length of the longest rod that will fitturn maximizes the length of the rod that will fitaround the corner is about 20.84 feet.around the corner (verify by using the SecondDerivative Test).38. The rod that barely fits around the corner willL ( 37.5) = 8csc( 37.5) + 8csc( 75 −37.5)touch the outside walls as well as the insidecorner.= 16csc( 37.5)C≈ 26.28Thus, the length of the longest rod that will fit8 feetaround the corner is about 26.28 feet.bE 75 − θBaθA8 feetD224 Section 3.7 Instructor’s Resource Manual

2x39. We can solve the equation − + x + 42 = 0 to25find the value for x when the object hits theground. We want the value to be positive, so weuse the quadratic formula, keeping only thepositive solution.2−1− 1 −4( −0.08)( 42)x = = 302( −0.08)We are interested in the global extrema for thedistance of the object from the observer. Weobtain the same extrema by considering thesquared distance2 2 2Dx ( ) = ( x− 3) + (42 + x−.08 x )A graph of D will help us identify a starting pointfor our numeric approach.23.8 Concepts Review1.rxr−r+11 x; + C, r ≠−1r + 1r−12. [ ] [ ]3.r f( x) f′ ( x); f( x) f′( x)4 2 3u = x + 3x + 1, du = (4x + 6 x)dx∫ ∫4 2 8 3 8( x + 3x + 1) (4x + 6 x)dx = u du9 4 2 9u ( x + 3x+ 1)= + C = + C9 9∫ ∫4. c1 f( x) dx+c2g( x)dxProblem Set 3.8r∫1. 5dx = 5x + CFrom the graph, it appears that D (and thus thedistance from the observer) is maximized atabout x = 7 feet and minimized just before theobject hits the ground at about x = 28 feet.The first derivative is given by16 3 12 2 236D'( x) = x − x − x+ 78 .625 25 25a. We will use Newton’s method to find thestationary point that yields the minimumdistance, starting with x 1 = 28 .n x n1 282 28.02803 28.02794 28.0279x ≈ 28.0279; y ≈ 7.1828The object is closest to the observer when itis at the point ( 28.0279,7.1828 ).b. We will use Newton’s method to find thestationary point that yields the maximumdistance, starting with x 1 = 7 .n x n1 72 6.77263 6.77284 6.7728x ≈6.7728; y ≈ 45.1031The object is closest to the observer when itis at the point ( 6.7728,45.1031 ).∫ ∫ ∫2. ( x − 4) dx = xdx −4 1dx3.2x= − 4x+ C232 2x( x + π ) dx = x dx +π 1dx = +π x + C3∫ ∫ ∫∫ ∫ ∫2 23x + 3 dx = 3 x dx + 3 1dx4. ( )5.6.7.8.9.3x3= 3 + 3x + C = x + 3x+C3∫9/45/4 x 4 9/4x dx = + C = x + C9 94⎛ 5/32/3 2/33 3 3 x ⎞x dx = x dx = ⎜ + C5 1 ⎟⎜ ⎟⎝ 3 ⎠∫ ∫= 9 5/35 x + C1−2/3 1/3 3dx = x dx = 3x + C = 3 x + C3 2x∫ ∫∫ ∫−3/4 −3/4 1/47x dx = 7 x dx = 7(4 x + C )1/428x= + C3 22 2x x( x − xdx ) = xdx− xdx= − + C3 2∫ ∫ ∫1Instructor’s Resource Manual Section 3.8 225

10.∫ ∫ ∫2 2(3 x −π xdx ) = 3 xdx−πxdx⎛ 3 2x ⎞ ⎛ x ⎞= 3 + C1 −π + C2⎜ 3 ⎟ ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠23 πx= x − + C217.6 4x + x dx3= (4 x + 3 x ) dx34 3∫ ∫x3∫ x dx ∫24 3x= 4 + 3 x dx= x + + C211.∫ ∫ ∫5 3 5 3(4 x − x ) dx = 4 x dx−x dx⎛ 6 4x ⎞ ⎛ x ⎞= 4 + C1 − + C2⎜ 6 ⎟ ⎜ 4 ⎟⎝ ⎠ ⎝ ⎠6 42xx= − + C3 418.6x − x dx3 2( x x−= − ) dx3∫ ∫x4 −13 −2x x= ∫x dx − ∫ x dx = − + C4 −14x 1= + + C4 x12.13.∫ ∫ ∫100 99 100 99( x + x ) dx = x dx+x dx101 100x x= + + C101 100∫7 5 3(27x + 3x − 45x + 2 x)dx∫ ∫ ∫ ∫7 5 3= 27 x dx + 3 x dx − 45 x dx + 2 x dx8 6 4 227x x 45x 2 x= + − + + C8 2 4 214. ⎡ 2 3 2( 5 3 3 ⎤∫ ⎢+ − +⎣) ⎥⎦5 4 3 2= ( + 5 − 3 + 3 )15.∫x x x x dxx x x x dx∫ ∫ ∫ ∫5 4 3 2= x dx + 5 x dx − 3 x dx + 3 x dx6 4 3x 5 3x 3x= + x − + + C6 4 3∫⎛ 3 2 ⎞−2 −3⎜ − dx = (3x −2 x ) dx2 3 ⎟ ∫⎝ x x ⎠−2 −3= 3 x dx −2x dx∫ ∫−1 −23x2x= − + C−1 −2=− 3 1Cx+ 2x+⎛ 2x3 ⎞−1/2 −516. ⎜ + ⎟ = ( 2 + 3 )∫ ∫⎜dx x x dxx 5x⎟⎝ ⎠1/2 −42x3x= + + C1 −423= 2 2x− + C44x19.3 22 2x x( x + xdx ) = x dx+ xdx= + + C3 2∫ ∫ ∫∫ ∫ ∫x 3 + x dx = x 3 dx+x 1/2 dx20. ( )4 3/2 4 3x x x 2 x= + + C = + + C4 3 4 3221. Let u = x + 1; then du = dx.3 32 2 u ( x+1)( x + 1) dx = u du = + C = + C3 3∫ ∫2z z dz z⎤dz⎣ ⎦222. ( + 2 ) = ⎡( 1+2)23.24.∫ ∫( 1 2)( ) 2 31+2 z22= + ∫ z dz = + C32 2 4 2( z + 1) z + 2z+ 1∫ dz = ∫dzzz7/2 3/2 −1/2= z dz+ 2 z + z dz∫ ∫ ∫2 9/2 4 5/2 1/2= z + z + 2z + C9 52 3 2s( s+ 1) s + 2s + sds =dsss∫ ∫∫ ∫ ∫5/2 3/2 1/2= s ds + 2 s ds + s ds7/2 5/2 3/22s 4s 2s= + + + C7 5 3∫ ∫ ∫25. (sinθ − cos θ) dθ = sinθ dθ − cosθ dθ= −cosθ− sinθ+ C226 Section 3.8 Instructor’s Resource Manual

26.∫ ∫ ∫2 2( t − 2cos t) dt = t dt −2 cost dt3t= − 2sint+C327. Let g( x) = 2 x+ 1; then g′ ( x) = 2 .28. Let( + ) = [ ]3 32 x 1 2 dx g( x) g′( x)dx∫ ∫[ gx ( )] 4 ( 2 x + 1 ) 4= + C = + C4 43gx ( ) x 1=π + ; then∫ π 3 + 4 π 2 = ∫ [ ] 4[ gx ( )] 5 3 5( π x + 1)2g′ ( x) = 3π x .( x 1) 3 x dx g( x) g′( x)dx= + C = + C5 5329. Let u = 5x + 3x− 8; then du = (15x + 3) dx .∫2 3 6(5x + 1)(5x + 3x−8)dx1 (152 3)(53 3 8)6= ∫ x + x + x−dx371 6 1⎛u⎞= u du C13∫= +3⎜7 ⎟⎝ ⎠3 7(5x+ 3x−8)= + C213 230. Let u = 5x + 3x− 2; then du = (15x + 3) dx.∫2 3(5x + 1) 5x + 3x−2dx1 (152 3) 53= ∫ x + x + 3 x−2 dx31 1/2 1⎛23/2 ⎞= u du u C13∫= ⎜ + ⎟3⎝3⎠2 (53 3 2)3/2= x + x− + C92 (53 3 2)3= x + x− + C9231. Let u = 2t − 11; then du = 4tdt.3 2 3 2 1/3∫3t 2t − 11 dt = ∫ (4 t)(2t −11)dt43 1/3 3⎛34/3 ⎞= u du u C14∫= ⎜ + ⎟4⎝4⎠9 (2 2 11) 4/3= t − + C169 3 2 4= (2t− 11) + C162232. Let u = 2y + 5; then du = 4ydy3y3 (4 )(22 5)−1/2dy = y y + dy22y+ 5y4∫ ∫3 −1/2 3 (2 1/2= u du u C1)4∫= +43 2 2= y + 5 + C2333. Let u = x + 4 ; then du = 3x dx .2 3 1 2 3∫ x x + 4dx = ∫ 3x x + 4dx31 1 1/2= ∫ udu=∫ u du3 31⎛23/2 ⎞= ⎜ u + C1⎟3⎝3⎠2 33/2= ( x + 4)+ C94 234. Let u = x + 2x; then( 4 3 4 3) 4 ( )3 4 2( x + x)x + 2x dxdu = x + x dx = x + x dx .∫( )1 3 4 2= ∫ ⋅ 4 x + x x + 2 x dx41 1 1/2= ∫ udu=∫ u du4 41⎛23/2 ⎞= ⎜ u + C1⎟4⎝3⎠1 4 23/2= ( x + 2x ) + C635. Let u = 1+ cosx; then du =− sin x dx .∫( + ) =−∫− ( + )24 4sin x 1 cos x dx sin x 1 cos x dx24 ⎛15 ⎞=− ∫ u du =− ⎜ u + C1⎟⎝5⎠1 5=− ( 1 + cos x)+ C536. Let u = 1+ sin x ; then du = 2sin x cosx dx .∫sinxcosx 1+sin xdx1 2= ∫ ⋅ 2sin x cos x 1 + sin xdx21 1 1/2= ∫ udu=∫ u du2 21⎛23/2 ⎞= ⎜ u + C1⎟2⎝3⎠1 23/2= ( 1 + sin x)+ C32Instructor’s Resource Manual Section 3.8 227

37.38.39.40.41.42.3 2f ′( x) = ∫ (3x+ 1) dx = x + x+C123 2f ( x)⎛⎞= ∫ ⎜ x + x+C1⎟dx⎝2⎠1 3 1 2= x + x + C1x+C22 2∫∫2f ′( x) = ( − 2x+ 3) dx =− x + 3x+C12f ( x) = ( − x + 3 x+C1) dx1 3 3 2=− x + x + C x+C3 21 21/2 2 3/2f ′( x)= ∫ x dx = x + C32 3/2f ( x)⎛ ⎞= ∫ ⎜ x + C1⎟dx⎝3⎠= 4 5/21 215 x + Cx+C4/3 3 7/3f ′( x)= ∫ x dx = x + C173 7/3 9 10/3f ( x)⎛ ⎞= ∫ ⎜ x + C1⎟dx = x + C1x+C2⎝7 ⎠ 703f ′′( x)= x+x −2 −23 x xf ′−( x) = ∫ ( x+ x ) dx = − + C12 2⎛1 2 1 −2⎞f ( x)= ∫ ⎜ x − x + C1⎟dx⎝2 2 ⎠1 3 1 −1= x + x + C1x+C26 21 3 1= x + + C1x+C26 2x1/3 3 4/3f ′( x) = 2 ∫ ( x+ 1) dx = ( x+ 1) + C123 4/3f ( x) ⎡⎤= ∫ ⎢ ( x+ 1) + C1dx2⎥⎣⎦= 9 ( 1)7/31 214 x + + Cx+C43. The Product Rule for derivatives saysd [ f ( xgx ) ( ) + C ] = f ( xg ) ′( x ) + f ′( xgx ) ( ).dxThus,∫ [ f ( xg ) ′ ( x) + f ′ ( xgx ) ( )] dx= f( xgx ) ( ) + C.144. The Quotient Rule for derivatives saysd ⎡ f( x) ⎤ g( x) f′ ( x) − f( x) g′( x)Cdx⎢ +g( x) ⎥ =.2⎣ ⎦ g ( x)g( x) f′ ( x) − f( x) g′( x) f( x)Thus, ∫ dx = + C .2g ( x)gx ( )45. Let46. Let2f ( x) = x , g( x) = x− 1.1f′ ( x) = 2 x, g′( x)=2 x −1⎡ 2x⎤∫ ⎢ + 2x x−1⎥dx⎢⎣2 x −1⎥⎦= f ( xg ) ′( x) + f′( xgx ) ( ) dx= f( xgx ) ( ) + C∫ [ ]2= x x− + C13 −1/2f( x) = x , g( x) = (2x+ 5) .2 3/2f′ −( x) = 3 x , g′( x) =− (2x+5)1=−3/2(2x+ 5)⎡ 3 2−x3x⎤∫ ⎢ + ⎥ dx3/2⎢⎣(2x+ 5) 2x+ 5⎥⎦= f ( xg ) ′( x) + gx ( ) f′( x)dx∫ [ ]3 −1/2= f ( xgx ) ( ) + C= x(2x+ 5) + C3x= + C2x+ 547. ∫ f ′′( xdx ) = ∫d f′ ( xdx ) = f′( x)+ Cdx3 33 3x5x+ 2f′ ( x) = x + 1+ = so3 32 x + 1 2 x + 135x+ 2∫ f ′′( xdx ) = + C.32 x + 1d ⎛48.dx ⎜⎝f( x)⎞+ Cgx ( ) ⎟⎠1 1/2g( x) f′ −( x) − f( x) 2[ g( x)] g′( x)=gx ( )2 g( x) f′ ( x) − f( x) g′( x)=3/22[ gx ( )]Thus,2 gxf ( ) '( x) − f( xg ) '( x) f( x)∫= + C2 gx ( )gx ( )[ ] 3/2228 Section 3.8 Instructor’s Resource Manual

49. The Product Rule for derivatives says thatd [ f m( xg n) ( x ) + C ]dx= f m ( x)[ g n ( x)] ′ + [ f m ( x)] ′ g n ( x)m n−1 m−1n= f ( x)[ ng ( x) g′ ( x)] + [ mf ( x) f′( x)] g ( x)m−1 n− 1= f ( xg ) ( x)[ nf( xg ) ′( x) + mgx ( ) f′( x)].Thus,m−1 n− 1f ( xg ) ( x)[ nf( xg ) ′( x) + mgx ( ) f′( x)]dx∫m= f ( xg ) ( x)+ C.50. Let u = sin[( x + 1) ];n2 42 4 2 3then du = cos ⎡( x + 1) ⎤4( x + 1) (2 x)dx⎣ ⎦2 4 2 3du = 8x cos ⎡( x + 1) ⎤( x + 1) dx⎣ ⎦∫3 2 4 2 4 2 3sin ⎡( x + 1) ⎤cos ⎡( x + 1) ⎤( x + 1) xdx⎣ ⎦ ⎣ ⎦43 1 1 3 1⎛u⎞= ∫u ⋅ du = u du C18 8∫= +8⎜4 ⎟⎝ ⎠4 2 4sin ⎡( x + 1) ⎤=⎣ ⎦+ C321 251. If x ≥ 0, then x = x and ∫ x dx = x + C .21 2If x < 0, then x =− x and ∫ x dx = − x + C .2⎧12x + C if x≥0⎪2∫ xdx= ⎨⎪− 1 2x + C if x < 0⎪⎩ 252. Using 2 u 1−cosusin = ,2 22 1−cos2x1 1sin x dx = dx = x − sin 2x + C2 2 4∫ ∫ .53. Different software may produce different, butequivalent answers. These answers wereproduced by Mathematica.a. ∫ 6sin ( 3( 2) ) 2cos( 3( 2) )b.∫x − dx = − x − + C3 ⎛ x⎞ 1 ⎛ x⎞ 9 ⎛ x⎞sin ⎜ ⎟dx= cos⎜ ⎟− cos⎜ ⎟+C⎝6⎠ 2 ⎝2⎠ 2 ⎝6⎠.∫∫54. a. F1( x) = ( xsin x) dx = sin x− xcosx+C1b.F2( x) = (sinx− xcos x+C1)dx= −2cosx − xsinx+ C1x+C2∫F3( x) = ( −2cosx− xsin x+ C1x+C2)dx1 2= x cos x− 3sin x+ C1x + C2x+C321 2F4( x) = ∫ ( xcosx− 3sin x+ C1x + C2x+C3)dx21 16 23 2= xsinx+ 4cos x+ C1x + C2x + C3x+C416 16−nCnxF16( x) = xsinx+ 16cosx+∑n=1(16 − n)!3.9 Concepts Review1. differential equation2. function3. separate variables4.2− 32 t+ v0; − 16t + v0t+s0Problem Set 3.91.2.dy −2x −x= =dx 2 22 1−x 1−xdy x −x xdx + y = + = 02 21−x 1−xdyCdx = dy− x + y = − Cx + Cx = 0dxdy3. C1cos x C2sin x;dx = −2d y=−C 2 1sinx−C2cosxdx2d yy2dx += ( −C1sin x− C2cos x) + ( C1sin x+ C2cos x) = 0c.∫2 x sin 2x( x cos2x+ xsin2 x)dx = + C22Instructor’s Resource Manual Section 3.9 229

dy4. For y = sin(x + C), cos( x C)dx = +5.6.⎛ 2dy 2 2 2⎜⎞ ⎟ + y = cos ( x + C) + sin ( x + C) = 1⎝dx⎠dyFor y = ±1, 0dx = .⎛ 2dy 2 2 2⎜⎞ ⎟ + y = 0 + ( ± 1) = 1⎝dx⎠dy 2x 1dx = +2dy = ( x + 1) dx∫ ∫2dy = ( x + 1) dx3xy+ C1 = + x+C233xy = + x+C3At x = 1, y = 1:1 11= + 1 + CC ; =−3 33x 1y = + x−3 3dy −3= x + 2dx−3dy = ( x + 2) dx∫ ∫−3dy = ( x + 2) dx−2xy+ C1 =− + 2x+C221y =− + 2x+C22xAt x = 1, y = 3:1 33=− + 2 + CC ; =2 2y =− 1 + 2x+322x27.8.9.dy x=dx yydy=∫ ∫xdx2 2x+ C1 = + C2y2 22 2y = x + C2y = ± x + CAt x = 1, y = 1:1= ± 1 + C; C = 0 and the square root ispositive.2y = x or y = xdy=dxxy∫ ydy=∫ xdx2 3/2 3/2y + C 21 = x + C23 33/2 3/2y = x + C3/2 2/3y = ( x + C)At x = 1, y = 4:2/34 = (1 + C) ; C = 73/2 2/3y = ( x + 7)dz 2 2t zdt =∫ ∫−2 2z dz = t dt3−1t− z + C1 = + C233 31 t C−t=− + C3=z 3 33z =3C−t1At t = 1, z = :31 3= ; 1 9; 103 C 1 C − = C =−3z =310 − t230 Section 3.9 Instructor’s Resource Manual

10.11.12.dy 4ydt =−4∫ = ∫y dy dt1− + C3 1 = t+C2y31y =−33t+CAt t = 0, y = 1:C = –11y =−33t−1ds 216t4t1dt = + −∫ ∫2ds = (16t + 4t −1)dt16 3 2s + C1 = t + 2t − t+C2316 3 2s = t + 2t − t+C3At t = 0, s = 100:C = 1001633 2s = t + t − t+du 3 3u t tdt = −2 100( )∫ ∫−3 3u du = ( t −t)dt4 21 t t− + C2 1 = − + C22u4 242 2 tu − = t − + C24−1/2⎛2 t ⎞u = t − + C⎜ 2 ⎟⎝ ⎠At t = 0, u = 4:−1/2 1= C C =4 ;164−1/2⎛2 t 1 ⎞u = t − +⎜ 2 16⎟⎝ ⎠13.14.dy(2x1)dx = +44 1 4y = ∫(2x+ 1) dx = (2x+1) 2dx2∫5 51 (2x+ 1) (2x+1)= + C = + C2 5 10At x = 0, y = 6:1 596 = + CC ; =10 105 5(2x+ 1) 59 (2x+ 1) + 59y = + =10 10 10dy 2 2 4y x( x 2)dx =− +−2 1 2 (2 2)4− ∫y dy = x x + dx2∫2 51 1 ( x + 2)+ C1 = + C2y 2 52 51 ( x + 2) + C=y 1010y =2 5( x + 2) + CAt x = 0, y = 1:101 = ; C = 10 − 32 = −2232 + C10y =2 5( x + 2) −22dy15. 3xdx = 3 2y = ∫ 3xdx = x + C2At (1, 2):32 = 2 + C1C =223 2 1 3x+ 1y = x + =2 2 2Instructor’s Resource Manual Section 3.9 231

16.17.18.dy3dx =2y−2∫ y dy = 3∫dx1− + C1 = 3x+C2y13x Cy =− +1y =C−3xAt (1, 2):12 =C − 37C =21 2y = =7 − 3x7 − 6x22tv = ∫ tdt = + v022tv = + 32⎛ 2 3t ⎞ ts = ∫+ 3 dt = + 3t + s⎜02 ⎟⎝ ⎠63 3t ts = + 3t+ 0= + 3t6 6At t = 2:v = 5 cm/s22s = cm3∫−4v = (1 + t)dt =− + C3+ t13(1 )1 1v0 = 0: 0 = − + C;C =33(1 + 0) 31 1v =− +33(1 + t)3⎛ 1 1⎞1 1s = ∫− + dt = + t + C⎜3 3 23(1 t) ⎟⎝ + ⎠ 6(1 + t)31 1 59s0 = 10 : 10 = + (0) + C;C =26(1 + 0) 3 61 1 59s = + t+26(1 + t)3 6At t = 2:1 1 26v =− + = cm/s81 3 8119.20.1/3 1 1/3v = ∫(2t+ 1) dt = (2t+1) 2dt2∫3 4/3= (2 t+ 1) + C183 3v0 = 0: 0 = + C1;C1=−8 83 4/3 3v = (2t+ 1) −8 83 4/3 3s = (2t 1) dt 1dt8∫ + −8∫3 43 3= (2t 1) 2dt 1dt16∫ + −8∫9 73 3= (2t+ 1) − t+C2112 89 1111s0 = 10 : 10 = + C2;C2=112 1129 73 3 1111s = (2t+ 1) − t+112 8 1123 43 3At t = 2: v = (5) − ≈ 2.838 89 73 6 1111s = (5) − + ≈ 12.6112 8 1123 1 3v (3t 1) −−= ∫ + dt = (3t+1) 3dt3∫1 (3 1)216 t −= − + + C1 25v0 = 4: 4 =− + C1;C1=6 61 −225v =− (3t+ 1) +6 61 −225s =− (3t 1) dt dt6∫ + + ∫61 −225=− (3t 1) 3dt dt18∫ + +6∫1 −125= (3t+ 1) + t+C218 61 1s0 = 0: 0 = + C2;C2=−18 181 −125 1s = (3t+ 1) + t−18 6 181 −225At t = 2: v =− (7) + ≈ 4.166 61 −125 1s = (7) + − ≈ 8.2918 3 181 2 59 284s = + + = cm54 3 6 27232 Section 3.9 Instructor’s Resource Manual

21. v = –32t + 96,22.2 2s =− 16t + 96t+ s0=− 16t + 96tv = 0 at t = 3At t = 3,2s =− 16(3 ) + 96(3) = 144 ftdva = = kdtdsv = ∫ kdt = kt+ v0 = ;dtk 2 k 2s = ∫ ( kt+ v0)dt= t + vt 0 + s0 = t + vt 02 2vv = 0 when t =− 0. Thenk2 2 2k ⎛ v0 ⎞ ⎛ v ⎞0 v0s =2⎜− k⎟ + − = − .⎝ ⎠ ⎜ k ⎟⎝ ⎠2kdv23. 5.28dt =−dv =− 5.28dt∫ ∫dsv 5.28t v0dt= = − + = –5.28t + 56∫ ∫ds = ( − 5.28t + 56) dt2 2s =− 2.64t + 56t+ s0=− 2.64t + 56t+1000When t = 4.5, v = 32.24 ft/s and s = 1198.54 ft−5624. v = 0 when t = ≈10.6061. Then−5.282s ≈− 2.64(10.6061) + 56(10.6061) + 1000≈ 1296.97 ft25.dV=− kSdt4 3 2Since V = π r and S = 4 π r ,32 24 π drdrr k4 r so kdt=− π dt= − .∫dr=−∫k dtr = –kt + C2 = –k(0) + C and 0.5 = –k(10) + C, so33C = 2 and k = . Then, r =− t+ 2 .202027. vesc = 2gRFor the Moon, vesc ≈ 2(0.165)(32)(1080 ⋅ 5280)≈ 7760 ft/s ≈ 1.470 mi/s.For Venus, vesc ≈ 2(0.85)(32)(3800⋅5280)≈ 33,038 ft/s ≈ 6.257 mi/s.For Jupiter, vesc≈ 194,369 ft/s ≈ 36.812 mi/s.For the Sun, vesc≈ 2,021,752 ft/s≈ 382.908 mi/s.28.v 0 = 60 mi/h = 88 ft/sv = 0 = –11t + 88; t = 8 sec11 2s () t =− t + 88t211 2s ( 8) =− ( 8) + 88( 8)= 352 feet2The shortest distance in which the car can bebraked to a halt is 352 feet.29.30.dv Δv60 −45 a = = = = 1.5 mi/h/s = 2.2 ft/s 2dt Δt108 275 = (3.75) + v0(3.75) + 0; v0= 5 ft/s2dv231. For the first 10 s, a = = 6, t v = 3t, anddt3s = t . So v(10) = 300 and s(10) = 1000. Afterdv10 s, a = =− 10 , v = –10(t – 10) + 300, anddt2s =−5( t− 10) + 300( t− 10) + 1000. v = 0 att = 40, at which time s = 5500 m.32. a. After accelerating for 8 seconds, the velocityis 8 · 3 = 24 m/s.b. Since acceleration and deceleration areconstant, the average velocity during thosetimes is24= 12 m/s . Solve 0 = –4t + 24 to get the224time spent decelerating. t = = 6 s;4d = (12)(8) + (24)(100) + (12)(6) = 2568 m.1726. Solving v = –136 = –32t yields t = .42⎛17 ⎞ ⎛17⎞Then s = 0=− 16 ⎜ ⎟ + (0) ⎜ ⎟+s0, so⎝ 4 ⎠ ⎝ 4 ⎠s 0 = 289 ft.Instructor’s Resource Manual Section 3.9 233

3.10 Chapter ReviewConcepts Test1. True: Max-Min Existence Theorem2. True: Since c is an interior point and f isdifferentiable ( f ′()c exists), by theCritical Point Theorem, c is astationary point ( f ′()c = 0).3. True: For example, let f(x) = sin x.15. True:12 1+x + 12lim =xlimx→∞2 11– x x→∞2–1x1= = –1 and–112 1+x + 12lim =xlimx→– ∞ 2 – 11– x x→ ∞x21= = –1.–1–14. False:5. True:f ( x)1/3x= is continuous andincreasing for all x, but f '( )not exist at x = 0.′ 5 3= + +f ( x) 18x 16x 4 x;4 2x doesf′′ ( x) = 90x + 48x+ 4 , which isgreater than zero for all x.6. False: For example, f ( x)= x is increasingon [–1, 1] but f ′(0) = 0.7. True: When f ′( x) > 0, f( x)is increasing.8. False: If f′′ () c = 0, c is a candidate, but notnecessarily an inflection point. For4example, if f( x) = x , P′′ (0) = 0 butx = 0 is not an inflection point.9. True:2( ) = + + ;′ = + ′′ =f x ax bx cf ( x) 2 ax b; f ( x) 2a10. True: If f(x) is increasing for all x in [a, b],the maximum occurs at b.11. False:12. True:13. True:2tan x has a minimum value of 0.This occurs whenever x = kπ where kis an integer.3lim (2 x + x)=∞ whilex→∞3lim (2 x + x)=−∞x→−∞−x→ π23lim (2x + x+ tan x)=∞ while+x→− π23lim (2x + x+ tan x) =−∞ .14. False: At x = 3 there is a removablediscontinuity.316. True:23x + 2x+sinx sinx– (3x+ 2) = ;xxsin xsin xlim = 0 and lim = 0 .x→∞xx→–∞ x17. True: The function is differentiable on(0, 2).x18. False: f′ ( x)= so f ′(0)does not exist.x19. False: There are two points:3 3x =− , .3 320. True: Let g(x) = D where D is any number.Then g′ ( x) = 0 and so, by Theorem Bof Section 3.6,f(x) = g(x) + C = D + C, which is aconstant, for all x in (a, b).21. False: For example if f ( x) = x ,f′ (0) = f′′(0) = 0 but f has aminimum at x = 0.22. True:2dy d y= cos x; =− sin x;–sin x = 0dx2dxhas infinitely many solutions.23. False: The rectangle will have minimumperimeter if it is a square.KA = xy = K; y =x22K dP 2K d P 4KP = 2 x+ ; = 2 − ; =x dx 2 2 3x dx x2dP0dx = and d P> 02dxwhen x = K,y = K .24. True: By the Mean Value Theorem, thederivative must be zero between eachpair of distinct x-intercepts.4Instructor's Resource Manual Section 3.10 235

= + − are a counter-is an inflection point while if25. True: If f ( x 1 ) < f( x 2 ) and g( x1) < g( x2)b2 – 3ac = 0 the only critical pointfor x1 < x2,f ( x1) + g( x1) < f( x2) + g( x2),sof + g is increasing.26. False: Let f(x) = g(x) = 2x, f′ ( x) > 0 andg′ ( x) > 0 for all x, but2f ( xgx ) ( ) = 4xis decreasing on(– ∞ , 0).34. True: f ( x) a 027. True: Since f ′′( x) > 0, f′( x)is increasingfor x ≥ 0. Therefore, f′ ( x) > 0 for xin [0, ∞ ), so f(x) is increasing.35. True: Intermediate Value Theorem28. False: If f(3) = 4, the Mean Value Theoremrequires that at some point c in [0, 3],f(3) – f(0) 4 –1f′ () c = = = 1which3–0 3–0f( xn)37. False: xn+ 1 xn – –2 xn.does not contradict that f′ ( x) ≤ 2 forf′( x )nall x in [0, 3].29. True: If the function is nondecreasing,f ′( x)must be greater than or equal tozero, and if f′ ( x) ≥ 0, f isnondecreasing. This can be seen usingthe Mean Value Theorem.continuous on [ ab , ] and30. True: However, if the constant is 0, thefunctions are the same.a g( x)b31. False:xFor example, let f ( x) = e .on [ , ]limxe = 0, so y = 0 is a horizontalthese criteria.x→−∞asymptote.32. True: If f(c) is a global maximum then f(c)is the maximum value of f on(a, b) ↔ S where (a, b) is any intervalcontaining c and S is the domain of f.Hence, f(c) is a local maximum value.41. True: Theorem 3.8.C33. True:2f ′( x) = 3ax + 2 bx+ c;f′ ( x) = 02– b±b –3acwhen x = by the3a43. True:2 2( − sin ) = sin2= 1−cosQuadratic Formula. f ′′( x) = 6ax+2b44. True: If F( x) f( x) dx, f( x)so⎛ 2– b±b –3ac⎞derivative of F(x).2f ′′ ⎜⎟=±2 b –3 ac.⎜ 3a⎟2⎝⎠45. False: f ( x) = x + 2x+ 1 andThus, if b 2 – 3ac > 0, one criticalgx ( )2x 7x5point is a local maximum and theexample.other is a local minimum.(Ifb 2 – 3ac< 0 there are no criticalpoints.)On an open interval, no local maximacan come from endpoints, so there canbe at most one local maximum in anopen interval.′ = ≠ so f(x) has no localminima or maxima. On an openinterval, no local minima or maximacan come from endpoints, so f(x) hasno local minima.36. False: The Bisection Method can be veryslow to converge.38. False: Newton’s method can fail to exist forseveral reasons (e.g. if f’(x) is 0 at ornear r). It may be possible to achieveconvergence by selecting a differentstarting value.39. True: From the Fixed-point Theorem, if g is≤ ≤ whenever a≤ x≤ b ,then there is at least one fixed pointab . The given conditions satisfy40. True: The Bisection Method alwaysconverges as long as the function iscontinuous and the values of thefunction at the endpoints are ofopposite sign.42. True: Obtained by integrating both sides ofthe Product Rulex x x= ∫ is a236 Section 3.10 Instructor’s Resource Manual

46. False: The two sides will in general differ bya constant term.47. True: At any given height, speed on thedownward trip is the negative ofspeed on the upward.Sample Test Problems1. f′ ( x) = 2 x–2;2x – 2 = 0 when x = 1.Critical points: 0, 1, 4f(0) = 0, f(1) = –1, f(4) = 8Global minimum f(1) = –1;global maximum f(4) = 82.1 1f′ () t = – ;–is never 0.2 2t tCritical points: 1, 41f(1) = 1, f (4) =41Global minimum f (4) = ;4global maximum f(1) = 1.2 23. f′ ( z) = – ;–is never 0.3 3z z1Critical points: –2, – 2f 1 1(–2) = , – 44 f ⎛ ⎜⎞⎟=⎝ 2⎠1Global minimum f (–2) = ;41global maximum f ⎛ ⎜– ⎞ ⎟ 4.⎝ 2 ⎠=4.2 2f′ ( x) = – ;–is never 0.3 3x xCritical point: –21f (–2) =4f′ ( x) > 0 for x < 0, so f is increasing.1Global minimum f (–2) = ; no global4maximum.5. f ′( x) = x ; f′( x)does not exist at x = 0.x1Critical points: – , 0, 12f ⎛ 1⎞ 1⎜– ⎟ = , (0) 0, (1) 12 2f = f =⎝ ⎠Global minimum f(0) = 0;global maximum f(1) = 16. f ′() s = 1 + s ; f′() s does not exist when s = 0.sFor s < 0, s = – s so f(s) = s – s = 0 and7.8.9.f′ () s = 1–1=0.Critical points: 1 and all s in [–1, 0]f(1) = 2, f(s) = 0 for s in [–1, 0]Global minimum f(s) = 0, –1 ≤ s ≤ 0;global maximum f(1) = 2.3 2 2f′ ( x) = 12 x –12x = 12 x ( x–1); f′( x) = 0when x = 0, 1Critical points: –2, 0, 1, 3f(–2) = 80, f(0) = 0, f(1) = –1, f(3) = 135Global minimum f(1) = –1;global maximum f(3) = 135u(7 u–12)f′ ( u) ; f′( u) 02/33( u – 2)= = whenf ′(2)does not exist.12Critical points: –1, 0, , 2, 373f(–1) = –3 ≈ –1.44, f(0) = 0,f ⎛ 12 144 23⎜⎞ ⎟ – –1.94,⎝ 7 ⎠= 49 7≈12f ⎛⎜⎞ ⎟≈–1.94;⎝ 7 ⎠Global minimumglobal maximum f(3) = 912u = 0, 7f(2) = 0, f(3) = 94 3 3f′ ( x) = 10 x –20x = 10 x ( x–2);f′ ( x) = 0 when x = 0, 2Critical points: –1, 0, 2, 3f(–1) = 0, f(0) = 7, f(2) = –9, f(3) = 88Global minimum f(2) = –9;global maximum f(3) = 88Instructor's Resource Manual Section 3.10 237

2 2 310. f′ ( x) = 3( x–1) ( x+ 2) + 2( x–1) ( x+2)2= ( x–1) ( x+ 2)(5x+ 4); f′( x) = 0 when4x = –2, – , 154Critical points: –2, – , 1 , 254 26,244f(–2) = 0, f ⎜⎛ – ⎟⎞ – –8.40,⎝ 5⎠= 3125≈f(1) = 0, f(2) = 164Global minimum f ⎛⎜– ⎞ ⎟≈–8.40;⎝ 5 ⎠global maximum f(2) = 16π11. f′ ( θ ) = cos θ; f′( θ) = 0 when θ = in2⎡π4π⎤⎢ ,4 3 ⎥⎣ ⎦4Critical points:π , π π,4 2 3⎛π⎞ 1 ⎛π⎞f ⎜ ⎟= ≈ 0.71, f ⎜ ⎟=1,⎝4⎠ 2 ⎝2⎠⎛4π⎞ 3f ⎜ ⎟ = – ≈ –0.87⎝ 3 ⎠ 2⎛4π⎞Global minimum f ⎜ ⎟ ≈ –0.87;⎝ 3 ⎠⎛π⎞global maximum f ⎜ ⎟ = 1⎝2⎠12. f ′( θ ) = 2sinθ cos θ – cosθ = cos θ(2sin θ –1);π π 5πf ′( θ ) = 0 when θ = , , in [0, π ]6 2 6π π 5π Critical points: 0, , , , π6 2 6⎛π⎞ 1 ⎛π⎞f(0) = 0, f ⎜ ⎟= – , f ⎜ ⎟ = 0,⎝6⎠ 4 ⎝2⎠⎛5π⎞ 1f ⎜ ⎟ = – , f(π ) = 0⎝ 6 ⎠ 4⎛π⎞ 1 ⎛5π⎞1Global minimum f ⎜ ⎟= – or f ⎜ ⎟=– ;⎝6⎠ 4 ⎝ 6 ⎠ 4⎛π⎞global maximum f(0) = 0, f ⎜ ⎟ = 0, or⎝2⎠f(π ) = 0313. f′ ( x) = 3–2 x; f′( x) > 0 when x < .2f′′ ( x) = –2; f′′( x)is always negative.⎛ 3⎤f(x) is increasing on ⎜ – ∞, 2 ⎥ and concave down⎝ ⎦on (– ∞ , ∞ ).14.15.16.17.18.8f′ ( x) 9 x ; f′( x) 0= > for all x ≠ 0.7f′′ ( x) = 72 x ; f′′( x) < 0 when x < 0.f(x) is increasing on (– ∞ , ∞ ) and concave downon (– ∞ , 0).2 2f′ ( x) 3 x –3 3( x –1); f′( x) 0= = > whenx < –1 or x > 1.f′′ ( x) = 6 x; f′′( x) < 0 when x < 0.f(x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) andconcave down on (– ∞ , 0).2f′ ( x) = −6x − 6x+ 12 = –6(x + 2)(x – 1);f′ ( x) > 0 when –2 < x < 1.f′′ ( x) = −12x− 6 = –6(2x + 1); f′′ ( x) < 0 when1x >− .2f(x) is increasing on [–2, 1] and concave down on⎛ 1 ⎞⎜−, ∞ ⎟⎝ 2 ⎠ .3 4 3f′ ( x) = 4 x –20x = 4 x (1–5 x); f′( x) > 010 < x < .52 3 2f′′ ( x) = 12 x –80x = 4 x (3–20 x); f′′( x) < 0when3when x > .20f(x) is increasing on ⎡0, 1⎣ 5⎛ 3 ⎞⎜ , ∞ ⎟ .⎝20⎠⎤ and concave down on⎦2 4 2 2f′ ( x) = 3 x –6x = 3 x (1–2 x ); f′( x) > 011– < x < 0 and 0 < x < .223 2f′′ ( x) 6 x–24x 6 x(1–4 x ); f′′( x) 0when= = < when11– < x < 0 or x > .22⎡ 1 1 ⎤f(x) is increasing on ⎢– , ⎥ and concave⎣ 2 2⎦ 1 1down on⎛ ⎜– , 0 ⎞ ⎟∪⎛ ⎜ , ∞⎞⎟.⎝ 2 ⎠ ⎝2⎠238 Section 3.10 Instructor’s Resource Manual

19.2 3 2f′ ( x) 3 x –4 x x (3–4 x); f′( x) 0= = > when3x < .4f′′ ( x) 26 x–12x 6 x(1–2 x); f′′( x) 0= = < when1x < 0 or x > .2⎛ 3⎤f(x) is increasing on ⎜ – ∞, 4 ⎥ and concave down⎝ ⎦⎛1⎞on (– ∞, 0) ∪⎜, ∞⎟.⎝2⎠f′′ ( x) = 6 x–8; f′′( x) > 0 when4x > .3⎛4 ⎞f(x) is concave up on ⎜ , ∞⎟and concave down⎝3⎠⎛ 4 ⎞⎛4 128⎞on ⎜– ∞, ⎟;inflection point ⎜ , − ⎟⎝ 3 ⎠⎝3 27 ⎠20.g′ () t2 13 t – ; g′() t2t01t < –31/4or= > when 3t4 1t > , so3g′ () t is increasing on1t > .1/43212t> or⎛ 1 ⎤ ⎡ 1 ⎞⎜– ∞, – ,1/4 ⎥∪⎢∞1/4 ⎟⎝ 3 ⎦ ⎣3⎠⎡ 1 ⎞ ⎛ 1 ⎤and decreasing on ⎢– , 0 ∪ 0,1/4 1/43 ⎟ ⎜ ⎥⎣ ⎠ ⎝ 3 ⎦ .1/4Local minimum g ⎛ 1 1⎜⎞ 3 1.75;1/4 ⎟ 3/4⎝3 ⎠= 3+ ≈local maximumg ⎛ 1 1 1/4⎜– ⎞ – – 3 –1.751/4 ⎟ 3/4⎝ 3 ⎠= 3≈2g′′ () t = 6 t+ ; g′′() t > 0 when t > 0. g(t) has no3tinflection point since g(0) does not exist.22.23.8xf′ ( x) = – ; f′( x) = 0 when x = 0.2 2( x + 1)28(3 x –1)f′′ ( x) = ; f′′(0) = –8, so f(0) = 6 is a2 3( x + 1)local maximum. f′ ( x) > 0 for x < 0 andf′ ( x) < 0 for x > 0 sof(0) = 6 is a global maximum value. f(x) has nominimum value.3f′ ( x) 4 x –2; f′( x) 0= = when2f′′ ( x) 12 x ; f′′( x) 0= = when x = 0.1x = .32⎛ 1 ⎞ 12f ′′ ⎜0,3 ⎟= > so2/3⎝ 2 ⎠ 21 1 2 3f ⎛ ⎜⎞ – –3 ⎟= = is a global4/3 1/3 4/3⎝ 2 ⎠ 2 2 2minimum.f′′ ( x) > 0 for all x ≠ 0; no inflection pointsNo horizontal or vertical asymptotes21.2 2f′ ( x) = 2 x( x–4) + x = 3 x –8 x = x(3 x–8);f′ ( x) > 0 when x < 0 or8x >3⎡8⎞f(x) is increasing on (– ∞, 0] ∪⎢, ∞ ⎟⎣3⎠ and⎡ 8⎤decreasing on ⎢0, 3 ⎥⎣ ⎦8 256Local minimum f ⎛ ⎜⎞ ⎟ – –9.48;⎝3⎠= 27≈local maximum f(0) = 0Instructor's Resource Manual Section 3.10 239

24.2 2 3f′ ( x) = 2( x –1)(2 x) = 4 x( x –1) = 4 x –4 x;f′ ( x) = 0 when x = –1, 0, 1.2 2f′′ ( x) 12 x – 4 4(3 x –1); f′′( x) 0= = = when1x =± .3f′′ (–1) = 8, f′′ (0) = –4, f′′(1) = 8Global minima f(–1) = 0, f(1) = 0;local maximum f(0) = 1⎛ 1 4⎞Inflection points ⎜±, ⎟⎝ 3 9 ⎠No horizontal or vertical asymptotes27.Vertical asymptote x = 33 2 2f′ ( x) 12 x –12x 12 x ( x–1); f′( x) 0= = =when x = 0, 1.2f′′ ( x) = 36 x –24x = 12 x(3 x–2); f′′( x) = 0when x = 0, 2 3 .f ′′ (1) = 12, so f(1) = –1 is a minimum.Global minimum f(1) = –1; no local maxima⎛2 16⎞Inflection points (0, 0), ⎜ , − ⎟⎝3 27⎠No horizontal or vertical asymptotes.25.26.3 x –6f′ ( x) = ; f′( x) = 0 when x = 2, but x = 22 x –3is not in the domain of f(x). f′ ( x)does not existwhen x = 3.3( x – 4)f′′ ( x) = ; f′′( x) = 0 when x = 4.3/24( x – 3)Global minimum f(3) = 0; no local maximaInflection point (4, 4)No horizontal or vertical asymptotes.1f′ ( x) – ; f′( x) 02( x –3)= < for all x ≠ 3.2f′′ ( x) = ; f′′( x) > 0 when x > 3.3( x –3)No local minima or maximaNo inflection points2x –2 1–xlim = lim = 1x→∞x –3 x→∞1–3xHorizontal asymptote y = 128.1f′ ( x) 1 ; f′( x) 02x2f′′ ( x) – ; f′′( x) 03xf ( x) 0= + > for all x ≠ 0.= > when x < 0 and′′ < when x > 0.No local minima or maximaNo inflection points1f( x) = x– , sox⎛ 1 ⎞lim [ f( x) − x] = lim ⎜− ⎟=0x→∞x→∞⎝x ⎠oblique asymptote.Vertical asymptote x = 0and y = x is an240 Section 3.10 Instructor’s Resource Manual

29.1f′ ( x) 3 ; f′( x) 02x2f′′ ( x) – ; f′′( x) 03xf ( x) 0= + > for all x ≠ 0.= > when x < 0 and′′ < when x > 0No local minima or maximaNo inflection points1f( x) = 3 x– , sox⎛ 1 ⎞lim [ f( x) − 3 x] = lim ⎜− ⎟=0x→∞x→∞⎝x ⎠oblique asymptote.Vertical asymptote x = 0and y = 3x is an31. f′ ( x) = –sin x–cos x; f′( x) = 0 whenπ 3πx = – , .4 4f′′ ( x) = –cosx+ sin x; f′′( x) = 0 when3π πx = – , .4 4π3πf′′ ⎛ ⎞ ⎛ ⎞⎜– ⎟ = – 2, f′′⎜ ⎟ = 2⎝ 4⎠ ⎝ 4 ⎠⎛3π⎞Global minimum f ⎜ ⎟ = – 2;⎝ 4 ⎠⎛ π ⎞global maximum f ⎜– ⎟ = 2⎝ 4 ⎠⎛ 3 π ⎞ ⎛π⎞Inflection points ⎜−, 0 ⎟ , ⎜ , 0 ⎟⎝ 4 ⎠ ⎝4⎠30.4f′ ( x) = – ; f′( x) > 03( x + 1)f ( x) 0′ < when x > –1.when x < − 1 and12f′′ ( x) = ; f′′( x) > 0 for all x ≠ − 1.4( x + 1)No local minima or maximaNo inflection pointslim f( x) = 0, lim f( x) = 0, so y = 0 is ax→∞x→–∞horizontal asymptote.Vertical asymptote x = –132.2f′ ( x) cos x–sec x; f′( x) 0= = when x = 02f′′ ( x) = –sin x–2sec xtanx3= – sin x(1 + 2sec x)f′′ ( x) = 0 when x = 0No local minima or maximaInflection point f(0) = 0π πVertical asymptotes x = – ,2 2Instructor's Resource Manual Section 3.10 241

33.2f′ ( x) xsec x tan x; f′( x) 0= + = when x = 02f′′ ( x) 2sec x(1 xtan x); f′′( x)= + is never 0 on⎛ π π⎞⎜– , ⎟⎝ 2 2⎠ .f ′′(0) = 2Global minimum f(0) = 034.2f′ ( x) 2 csc x; f′( x) 0= + > on (0, π )2f′′ ( x) –2cotxcsc x; f′′( x) 0= = whenπx = ; f′′ ⎛π⎞( x) > 0 on ⎜ , π⎟2⎝2⎠⎛π⎞Inflection point ⎜ , π⎟⎝2⎠36. f′ ( x) = –2sin x– 2cos x; f′( x) = 0 whenπ 3πx = – , .4 4f′′ ( x) = –2cosx+ 2sin x; f′′( x) = 0 when3π πx = – , .4 4π3πf′′ ⎛ ⎞ ⎛ ⎞⎜– ⎟ = –2 2, f′′⎜ ⎟ = 2 2⎝ 4⎠ ⎝ 4 ⎠⎛3π⎞Global minimum f ⎜ ⎟ = –2 2;⎝ 4 ⎠⎛ π ⎞global maximum f ⎜– ⎟ = 2 2⎝ 4 ⎠⎛ 3 π ⎞ ⎛π⎞Inflection points ⎜−, 0 ⎟ , ⎜ , 0 ⎟⎝ 4 ⎠ ⎝4⎠35. f′ ( x) = cos x– 2cos xsin x = cos x(1– 2sin x);π π π 5πf′ ( x) = 0 when x = – , , ,2 6 2 62 2f′′ ( x) = – sin x+ 2sin x– 2cos x; f′′( x) = 0when x ≈ –2.51, –0.63, 1.00, 2.143f′′ ⎛– π ⎞ ⎛3, f π ⎞ ⎛– , fπ ⎞⎜ ⎟ = ′′ ⎜ ⎟ = ′′ ⎜ ⎟ = 1,⎝ 2⎠ ⎝6⎠ 2 ⎝2⎠5π3f ′′⎛ ⎞⎜ ⎟=–⎝ 6 ⎠ 2⎛ π ⎞Global minimum f ⎜– ⎟ = –2,⎝ 2 ⎠⎛π⎞local minimum f ⎜ ⎟ = 0;⎝2⎠⎛π⎞ 1 ⎛5π⎞1global maxima f ⎜ ⎟= , f ⎜ ⎟ =⎝6⎠ 4 ⎝ 6 ⎠ 4Inflection points (–2.51, –0.94),(–0.63, –0.94), (1.00, 0.13), (2.14, 0.13)37.242 Section 3.10 Instructor’s Resource Manual

38.39.40. Let x be the length of a turned up side and let l bethe (fixed) length of the sheet of metal.2V = x(16 − 2 x) l = 16xl−2x ldV16l 4 xl; V 0 when x 4dx = − ′= =2dV=− 4; l 4 inches should be turned up for2dxeach side.41. Let p be the length of the plank and let x be thedistance from the fence to where the planktouches the ground.See the figure below.By properties of similar triangles,2p x + 64=x+1 x⎛ 1 ⎞ 2p = ⎜1+ ⎟ x + 64⎝ x ⎠Minimize p:dp 1 2 ⎛ 1⎞x=− x + 64 + 1dx 2 ⎜ + ⎟x⎝ x ⎠ 2x + 641 ⎛ 2 ⎛ 1⎞3⎞= ⎜–( x + 64) + ⎜1+⎟x⎟2 2x x + 64 ⎝⎝ x ⎠ ⎠3x − 64=2 2x x + 643x − 64= 0; x = 42 2x x + 64dpdp< 0 if x< 4, > 0 if x > 4dxdx⎛ 1 ⎞When x = 4, p = ⎜1+ ⎟ 16 + 64 ≈11.18ft.⎝ 4 ⎠42. Let x be the width and y the height of a page.A = xy. Because of the margins,27(y – 4)(x – 3) = 27 or y = 4x − 3+27xA = + 4; xx − 3dA ( x −3)(27) −27x81= + 4= − + 4dx 2 2( x−3) ( x−3)dA0dx = when 3 15x =− ,2 222d A 162 d A15= ; > 0 when x =2 3 2dx ( x − 3) dx243.15x = ; y = 1021 21282 π r h = π256h =2rLet S be the surface area of the trough.2 2 256πS =π r +π rh =π r +rdS 256π= 2πr−dr2r256π3 32πr− = 0; r = 128, r = 4 22rSince2d S2dr> 0 whenminimizes S.256 3h = = 8 2324 2( )3r = 4 2,3r = 4 2Instructor's Resource Manual Section 3.10 243

44.⎧ x 3 + if − 2 < x < 0⎪2 2f′ ( x)= ⎨⎪ x + 2 − if 0 < x < 2⎪⎩ 3x 3+ = 0; x =− 3 , which is not in the domain.2 2x + 2− = 0; x =− 2, which is not in the domain.3c.( x−1) − ( x+ 1) −2g′ ( x)= =( x−1) ( x−1)2 2g(3) −g(2) 2 −3= =−13−2 1−2= –1; c = 1 ± 22( c −1)Only c = 1+ 2 is in the interval (2, 3).45. a.Critical points: x = –2, 0, 2f(–2) = 0, f(0) = 2, f(2) = 0Minima f(–2) = 0, f(2) = 0, maximum f(0) = 2.⎧1if − 2 < x < 0⎪2f′′ ( x)= ⎨⎪− 1 if 0 < x < 2⎪⎩ 3Concave up on (–2, 0), concave down on (0, 2)2f ′( x)= xf(3) − f( − 3) 9 + 9= = 33 −( −3) 62c = 3; c =− 3, 346.47.dy 3 24x 18x 24x3dx = − + −2d y 2 2= 12x − 36x+ 24; 12( x − 3x+ 2) = 0 when2dxx = 1, 2Inflection points: x = 1, y = 5and x = 2, y = 11dySlope at x = 1: = 7dx x = 1Tangent line: y – 5 = 7(x – 1); y = 7x – 2dySlope at x = 2: = 5dx x = 2Tangent line: y – 11 = 5(x – 2); y = 5x + 1b. The Mean Value Theorem does not applybecause (0) F′ does not exist.48.244 Section 3.10 Instructor’s Resource Manual

49. Let f ( x) = 3x− cos2x; a 1 = 0 , b 1 = 1 .f ( 0)=− 1; f () 1 ≈ 3.4161468n h m f ( m )n n n1 0.5 0.5 0.95969772 0.25 0.25 −0.12758263 0.125 0.375 0.39331114 0.0625 0.3125 0.12653695 0.03125 0.28125 −0.00217456 0.015625 0.296875 0.06177657 0.0078125 0.2890625 0.02969888 0.0039063 0.2851563 0.01373649 0.0019532 0.2832031 0.005774510 0.0009766 0.2822266 0.001798411 0.0004883 0.2817383 −0.000188412 0.0002442 0.2819824 0.000804913 0.0001221 0.2818604 0.000308214 0.0000611 0.2817994 0.000060015 0.0000306 0.2817689 −0.000064116 0. 0000153 0.2817842 −0.000001817 0.0000077 0.2817918 0.000029318 0.0000039 0.2817880 0.000013819 0.0000020 0.2817861 0.000006120 0.0000010 0.2817852 0.000002221 0.0000005 0.2817847 0.000000422 0.0000003 0.2817845 −0.000000623 0.0000002 0.2817846 −0.0000000x ≈ 0.28178550. f(x) = 3x – cos 2x, f ′( x) = 3+2sin2xLet x 1 = 0.5 .n x n1 0.52 0.29506523 0.28185634 0.28178465 0.2817846x ≈ 0.281785cos 2xn51. x n + 1 =3n x n1 0.52 0.180103 0.3119424 0.2705395 0.2857186 0.2803757 0.2822858 0.2816069 0.28184810 0.28176211 0.28179312 0.28178213 0.28178614 0.28178415 0.28178516 0.281785x ≈ 0.281852. y = x and y = tan x11 πLet x1= .8f(x) = x – tan x,n12f ′( x) = 1–sec x.x n11π82 4.646617953 4.600910504 4.546622585 4.506580166 4.494224437 4.493412598 4.49340946x ≈ 4.4934Instructor's Resource Manual Section 3.10 245

3 253. ∫ ( − 3 + 3 )3 2 1/2= ( − 3 + 3 )54.55.∫x x x dxx x x dx1 4 3 2 3/2= x − x + 3⋅ x + C4 31 4 3 3/2= x − x + 2x + C44 22x− 3x+ 1∫dx2x= x − + x dx2 −2∫ ( 2 3 )2 3 −1= x −3x− x + C33 4 22x 1 2x −9x−3= −3 x − + C or+ C3 x3x3 −1y − 9ysin y+26yy∫2∫ ( 9sin 26)dy= y − y+dy1 3= y + 9cos y+ 26y+C3256. Let u = y − 4 ; then du = 2ydyor 1 221∫ y y − 4 dy = ∫ u⋅du21 1/2=2∫ u du1 2 3/2= ⋅ u + C2 31 23/2= ( y − 4)+ C32du = ydy .57. Let u = 2z− 3; then du = 4zdzor 1 du zdz4 = .21/31/3 1∫ z( 2z − 3)dz = ∫ u ⋅ du41 1/3=4∫ u du1 3 4/3= ⋅ u + C4 43 24/3= ( 2 z − 3 ) + C1658. Let u = cos x; then du = − sin x dx or59.60.− du = sin x dx .∫∫∫∫4 =4( ) 4cos xsin xdx cos x sin xdx= u ⋅−du=−4udu1 5=− u + C51 cos5=− x+C52 2 2u = tan(3x + 6 x), du = (6x+ 6)sec (3x + 6 x)∫2 2 2 2( + 1tan ) ( 3 + 6 ) sec ( 3 + 6 )x x x x x dx1 2 1 3=6∫u du = u + C181 tan3 2= ( 3 x + 6 x)+ C184 3u = t + 9, du = 4t dt13dut 4∫ dt =4∫t + 9 u1 −1/2=4∫ u du1 21/2= ⋅ u + C41 4= t + 9 + C2561. Let u = t + 5 ; then du = 5t dt or∫( 5)4154 52/3 1 2/3t t + dt = ∫ u du51 2/3=5∫ u du1 3 5/3= ⋅ u + C5 53 55/3= ( t + 5)+ C254du = t dt .246 Section 3.10 Instructor’s Resource Manual

262. Let u = x + 4 ; then du = 2x dx or 1 2x 1 du∫ dx =24 2∫x +u1 −1/2= u du2∫1 21/2= ⋅ u + C22= x + 4 + C363. Let u = x + 9 ; then du = 3x dx or∫2x 1 dudx =39 3∫x +u1 −1/2=3∫u du1 21/2= ⋅ u + C32 3= x + 9 + C364. Let u = y+ 1; then du = dy .1 1∫ dy = ∫ duy + 1 u( ) 2 2 −2=∫u−1du=− u + C1=− + Cy + 165. Let u = 2y− 1; then du = 2dy.2 du∫ dy =3 ∫ 32y−1u( )=∫u−3du1 −2=− u + C21=− + C22 2 1( y − )213du = xdx .2du = x dx .66. Let u = y − 3y; then67.3( 3 2 3 2) 3 ( 1 )du = y − dy = y − dy .y2∫3( y − 3y)−1 1dy =3∫du2 2u1 −2=3∫ u du1 −1= ⋅− u + C31 1=− ⋅ + C33 y − 3y1=− + C33y− 9y3 2 2u = 2y + 3y + 6 y, du = (6y + 6y+6) dy1 −1/5 5 (23 32 6 )4/5u du y y y C6∫= + + +24∫ ∫68. dy = sin x dx69.y = − cos x+Cy = –cos x + 3∫dy= ∫1dxx + 1y = 2 x+ 1+Cy = 2 x+ 1+14∫ ∫70. sin ydy=dx− cos y = x+Cx = –1 – cos y∫ ∫71. dy = 2t −1dt72.1 (2 1)32y = t− + C31 (2 1) 32y = t− − 13∫ ∫−4 2y dy = t dt31 t− = + C33y331 t 2− = −33y3 31y = 332 − tInstructor's Resource Manual Section 3.10 247

73.∫ ∫32 ydy= (6 x−x ) dx2 2 1 4y = 3x − x + C42 2 1 4y = 3x − x + 942 1 4y = 3x − x + 9474. ∫cos ydy=∫ xdx2xsin y = + C22−1⎛ xy = sin⎞⎜ 2 ⎟⎝ ⎠275. s()t =− 16t + 48t+ 448; s = 0 at t = 7;() ()v t = s' t = − 32t+48when t = 7, v = –32(7) + 48 = –176 ft/s7. A region = 0.5( 1+ 1.5 + 2 + 2.5)= 3.58. A region = 0.5( 1.5 + 2 + 2.5 + 3)= 4.59.10.11.12.1 1 2Aregion = Arect + Atri= 1x+ x⋅ x = x + x2 21 1 1 2Aregion= bh = x ⋅ xt = x t2 2 2y = 5 − x;A = A + A1= 2( 2) + ( 2)( 2)= 62A = A + Aregion rect triregion rect tri1= 11 () + ()( 1 7)= 4.52Review and Preview Problems1.1 1 3Aregion= bh = aa sin 60 = a2 2 4o 2⎛1 ⎞ ⎛1 ⎞⎛ 3 ⎞A = 6⎜ base× height ⎟= 6⎜ a ⎜ a⎝2 ⎠ ⎝2 ⎟⎜ 2 ⎟⎠⎝ ⎠3 3 2= a22. region3.Aregion5= a42⎛1⎞ a= 10⎜base× height ⎟=5 cot 36⎝2 ⎠ 4cot 362region = rect + tri = 17 8.51 ⎛ 8.5 ⎞+ 172⎜ o ⎟⎝tan 45 ⎠= 216.754. A A A ( )5. A = A + A = 3.6⋅ 5.8 + π ( 1.8) 2region rect semic.≈ 25.976. Aregion A#5 Atri12⎛1⎞= + 2 = 25.97 + 2⎜⋅1.2⎟5.8⎝2⎠= 32.93248 Review and Preview Instructor’s Resource Manual

33. a.b.∫ ∫−1/210 V dV = C1dt; 20 V = C1t+C2;V(0) = 1600: C 2 = 20⋅ 40 = 800;800V(40) = 0: C 1 = − =− 20401Vt () = ( − 20t+ 800) 2 = 40−t400( ) 2b. Since the trip that involves 1 min more traveltime at speed v m is 0.6 mi longer,v m = 0.6 mi/min= 36 mi/h.c. From part b, v m = 0.6 mi/min. Note that theaverage speed during acceleration andv mdeceleration is = 0.3 mi/min. Let t be the2time spent between stop C and stop D at theconstant speed v m,so0.6t + 0.3(4 – t)= 2 miles. Therefore,2t = 2 min and the time spent accelerating34−2 23 2is = min.2 30.6 − 02a = = 0.9 mi/min .23dh34. For the balloon, 4dt = , so ht () = 4t+ C1. Sett = 0 at the time when Victoria threw the ball, andheight 0 at the ground, then h(t) = 4t + 64. The2height of the ball is given by s() t =− 16t + v0t,since s 0 = 0 . The maximum height of the ball isv0when t = , since then s′ () t = 0. At this time32⎛v0 ⎞ ⎛v0 ⎞ ⎛v0⎞h(t) = s(t) or 4⎜ 64 16 v032⎟+ =− ⎜ +32⎟ ⎜32⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ .Solve this for v 0 to get v0 ≈ 68.125 feet persecond.236. a.c. V (10) = ( 40 − 10) 2 = 900 cm3dP 3= C 1 P , P(0) = 1000, P(10) = 1700dtwhere t is the number of years since 1980.b. −1/3 3 2/3∫P dP = ∫ C1dt; P = C1t+C223 2/3P(0) = 1000: C 2 = ⋅ 1000 = 15023 2/3⋅1700 −1502P(10) = 1700: C1=10≈ 6.36603/2P = (4.2440t+100)c.3/24000 = (4.2440t+ 100)2/34000 −100t = ≈ 35.8124.2440t ≈ 36 years, so the population will reach4000 by 2016.37. Initially, v = –32t and s = − 16t+ 16 . s = 0 whent = 1. Later, the ball falls 9 ft in a time given by20= − 16t+ 9, or 3 s, and on impact has a4⎛3⎞velocity of − 32⎜⎟=−24ft/s. By symmetry,⎝4⎠24 ft/s must be the velocity right after the firstbounce. Soa.32tfor 0 t 1vt () = ⎧− ⎨ ≤

CHAPTER 4The Definite Integral4.1 Concepts Review1.5(6)2⋅ = 30; 2(5) = 1022. 3(9) – 2(7) = 13; 9 + 4(10) = 493. inscribed; circumscribed4. 0+ 1+ 2+ 3=6Problem Set 4.11.2.3.4.6 6 6∑ ∑ ∑( k− 1) = k−1k= 1 k= 1 k=16∑ i 2i=17∑k=1 k6(7)= −6(1)2= 156(7)(13)= = 9161 1 1 1= + ++ 1 1+ 1 2+ 1 3+11 1 1 1+ + + +4+ 1 5+ 1 6+ 1 7+11 1 1 1 1 1 1= + + + + + +2 3 4 5 6 7 81443=840481=2808∑ ll=32 2 2 2 2 2 2( + 1) = 4 + 5 + 6 + 7 + 8 + 9 = 2715.6.8∑m=1m( −1) 2m−21 −1 2 0 3 1= ( − 1) 2 + ( − 1) 2 + ( −1) 24 2 5 3 6 4+ ( − 1) 2 + ( − 1) 2 + ( −1) 27 5 8 6+− ( 1) 2 + ( − 1) 21= − + 1 − 2 + 4 − 8 + 16 − 32 + 64285=27∑k=3kk( −1) 2( k + 1)3 3 4 4( −1) 2 ( −1) 2= +4 5( −1) 2 ( −1) 2 ( −1) 2+ + +6 7 81154=−1055 5 6 6 7 76 6ncos( nπ ) = −1 n ⋅nn= 1 n=17. ∑ ∑ ( )= –1 + 2 – 3 + 4 – 5 + 6= 36⎛kπ⎞8. ∑ k sin ⎜ ⎟k=−1⎝ 2 ⎠⎛ π⎞ ⎛π⎞= −sin ⎜− ⎟+ sin ⎜ ⎟+ 2sin( π)⎝ 2⎠ ⎝2⎠⎛3π⎞⎛5π⎞+ 3sin ⎜ ⎟+ 4sin(2 π)+ 5sin ⎜ ⎟+ 6sin(3 π)⎝ 2 ⎠⎝ 2 ⎠= 1 + 1 + 0 – 3 + 0 + 5 + 0= 49.411+ 2+ 3+ + 41=∑ii=110.11.252+ 4+ 6+ 8+ + 50=∑2ii=11001 1 1 11+ + + + 2 3 100=∑i=1 iInstructor’s Resource Manual Section 4.1 249

12.13.100 i+11 1 1 1 ( −1)1− + − + − 2 3 4 100=∑i=1 i50a1+ a3 + a5 + a7 + + a99 =∑ a2i− 1i=114. f ( w1) Δ x+ f( w2) Δ x+ + f( wn) Δx15.16.17.18.19.n= ∑ f ( wi) Δxi=110∑( ai+ bi)i=110 10= ∑ai+ ∑bii= 1 i=1= 40 + 50= 9010∑ (3an+ 2 bn)n=110 10= 3∑an+ 2∑bnn= 1 n=1= 3(40) + 2(50)= 2209∑ ( ap+ 1−bp+1)p=010 10= ∑ap−∑bpp= 1 p=1= 40 −50= –1010∑ ( aq−bq−q)q=110 10 10= ∑ q −∑ q −∑q= 1 q= 1 q=1a b q10(11)= 40 −50−2=−65100∑(3i− 2)i=1100 100∑∑= 3 i − 2i= 1 i=1= 3(5050) −2(100)= 14,9501020. ∑[ ( i− 1)(4i+3) ]21.22.23.24.i=110∑2= (4i−i−3)i=110210 10∑ ∑ ∑= 4 i − i−3i= 1 i= 1 i=1= 4(385) – 55 – 3(10)= 145510 10 103 2 3 2∑ ( k − k ) = ∑k −∑kk= 1 k= 1 k=1= 3025 −385= 264010 102 3 2∑ ∑5 k ( k + 4) = (5k + 20 k )k= 1 k=110 103 2∑ ∑= 5 k + 20k= 1 k=1k= 5(3025) + 20(385)= 22,8252 2∑ n n n ni − i+ = ∑i − ∑i+∑i= 1 i= 1 i= 1 i=1(2 3 1) 2 3 12 nn ( + 1)(2n+ 1) 3 nn ( + 1)= − + n6 23 2 22n + 3n + n 3n + 3n = − + n3 23 24n − 3n − n=6nn2 2∑ i ∑ i ii= 1 i=1n 2n n= 4∑i− 12∑i+∑ 9i= 1 i= 1 i=1nn n nn(2 − 3) = (4 − 12 + 9)4 ( + 1)(2 + 1) 12 ( + 1)= − + 9n6 23 24n − 12n + 11n=325. S = 1+ 2+ 3 + + ( n− 2) + ( n− 1) + n+ S = n+ ( n− 1) + ( n− 2) + + 3+ 2+12 S = ( n+ 1) + ( n+ 1) + ( n+ 1) + + ( n+ 1) + ( n+ 1) + ( n+1)2S = n(n + 1)nn ( + 1)S =2250 Section 4.1 Instructor’s Resource Manual

26.2 nS − rS = a + ar + ar + + ar2 1( n n− ar + ar + + ar + ar + )n 1= a−ar += S(1 − r);S =n+1a−ar1 r−27. a.b.( ) 1110 k 1 10⎛1⎞ 1−2 ⎛1⎞⎜ ⎟ = = 2 − ,1 ⎜ ⎟k= 0⎝2⎠ ⎝2⎠210 k 10∑ so⎛1 ⎞ ⎛1 ⎞ 1023∑ ⎜ ⎟ = 1− ⎜ ⎟ = .⎝2 ⎠ ⎝2 ⎠ 1024k= 11011k 1−2 112 = = 2 −1,k= 0 −110k 11∑ 2 = 2 − 2=2046.k= 1∑ so28. [ ] [ ][ ] [ ] S = a+ ( a+ d) + ( a+ 2 d) + + a+ ( n− 2) d + a+ ( n− 1) d + ( a+nd)+ S = ( a+ nd) + a+ ( n− 1) d + a+ ( n− 2) d + + ( a+ 2 d) + ( a+ d)+ a2 S = (2 a+ nd) + (2 a+ nd) + (2 a+ nd) + + (2 a+ nd) + (2 a+ nd) + (2 a+nd)2S = (n + 1)(2a + nd)( n+ 1)(2 a+nd)S =229. ( )n∑⎡⎣3 3 2i+ 1 − i = 3i + 3i+1n3 3 2( i+ 1) − i ⎤ = ( 3i + 3i+1)⎦∑i= 1 i=1n n n3 3 2( )3 2 2( + 1)( 2n+1)∑ ∑ ∑i= 1 i= 1 i=1n∑n( n+1)n+ 1 − 1 = 3 i + 3 i+1n + 3n + 3n= 3 i + 3 + n2i=1n3 2 2 22n + 6n + 6n= 6 i + 3n + 3n+2nn n3 22n 3n n6+ + =6=∑i=1n2∑i=1n2i=1i∑iInstructor’s Resource Manual Section 4.1 251

4 4 3 2i+ 1 − i = 4i + 6i + 4i+1nn4 4 3 2∑⎡( i+ 1) − i ⎤ = ( 4i + 6i + 4i+1⎣ ⎦ ∑)i= 1 i=1n n n n4 4 3 2( n+ 1) − 1 = 4∑i + 6∑i + 4∑i+∑1i= 1 i= 1 i= 1 i=1n4 3 2 3 nn ( + 1)(2n+ 1) nn ( + 1)n + 4n + 6n + 4n = 4∑i + 6 + 4 + ni=16 2n3Solving for ∑igivesi=1n3 4 3 2 3 2 24∑i = n + 4n + 6n + 4n− ( 2n + 3n + n) − ( 2n + 2n)−ni=1n3 4 3 24∑i = n + 2n + ni=1n 4 3 223 n + 2n + n ⎡n( n+1)⎤∑i= = ⎢ ⎥i=1 4 ⎣ 2 ⎦30. ( )5 5 4 3 2i+ 1 − i = 5i + 10i + 10i + 5i+1n ⎡ 5 5 4 3 2( 1)n n n n ni i ⎤∑⎢+ − = 5 i + 10 i + 10 i + 5 i+1⎣ ⎥⎦∑ ∑ ∑ ∑ ∑i= 1 i= 1 i= 1 i= 1 i= 1 i=1n 2 25 5 4 n ( n+ 1 ) nn ( + 1)(2n+ 1) nn ( + 1)( n+ 1)− 1 = 5∑i + 10 + 10 + 5 + ni=1 4 6 2n5 4 3 2 4 5 2 2 105n + 5n + 10n + 10n + 5n= 5∑i + n ( n+ 1) + n( n+ 1 )(2n+ 1) + n( n+ 1) + n2 6i=12n4Solving for ∑ i yieldsi=1n24 1 5 5 4 5 3 1 nn ( + 1)(2n+ 1)(3n + 3n−1)∑ i = ⎡n + n + n − n⎤=5⎣2 3 6 ⎦i=13031. ( )32. Suppose we have a ( n 1)+ × n grid. Shade inn + 1 – k boxes in the kth column. There are n columns, and the shaded area is 1+ 2+ + n . The shaded area isnn+ ( 1)nn ( + 1)also half the area of the grid or . Thus, 1+ 2+ + n = .22nn ( + 1)Suppose we have a square grid with sides of length 1+ 2+ + n = . From the diagram the area is2223 3 3 ⎡nn+( 1) ⎤1 + 2 + + n or ⎢ 2 ⎥⎣ ⎦ . Thus, 3 3 3 ⎡nn( + 1) ⎤1 + 2 + + n = ⎢ 2 ⎥ .⎣ ⎦33.1 55x = (2 + 5 + 7 + 8 + 9 + 10 + 14) = ≈ 7.867 72 2 222 1⎡ ⎛ 55 ⎞ ⎛ 55 ⎞ ⎛ 55 ⎞ ⎛ 55 ⎞s = ⎢⎜2− ⎟ + ⎜5− ⎟ + ⎜7− ⎟ + ⎜8−⎟7 7 7 7 7⎢⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠2 2 2⎛ 55 ⎞ ⎛ 55 ⎞ ⎛ 55 ⎞ ⎤ 608+ ⎜9− ⎟ + ⎜10− ⎟ + ⎜14− ⎟ ⎥ = ≈ 12.4⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎥ 49 ⎦252 Section 4.1 Instructor’s Resource Manual

34. a.2sx = 1, = 0b. x = 1001, s = 0c. x = 222 2 2 2 2 2 2=1⎡(1 − 2) + (2 − 2) + (3 − 2) ⎤ =1 1 2( − 1) + 0 + 1 = ( 2)=3 3 3 3s ⎡ ⎤⎣ ⎦ ⎢⎣ ⎥⎦d.35. a.b.x = 1, 000, 002⎡⎣⎤⎦2 1 2 2 2s = ( − 1) + 0 + 1 =23 3n n n∑ xix ∑xi∑ x nx nxi= 1 i= 1 i=1( − ) = − = − = 0nns 2 = 1 2 1 2 2∑ i i inx − x = ∑ x x x xi= 1 n− +i=11 n 2 2 n 1n= x2∑xixixn − ∑i 1 n + ∑= i= 1 n i=1n1 2 2x1 2= ∑ xi− ( nx) + ( nx )n i=1 n n⎛ nn1 x 2⎞ 2 2 1 2 2i 2 x x ⎛ x ⎞= − + = i −⎜ ∑ n ⎟ ⎜ ∑xi= 1 n ⎟i=1( ) ( 2 )⎝ ⎠ ⎝ ⎠36. The variance of n identical numbers is 0. Let c be the constant. Then2 2 22s =1 ⎡( c− c) + ( c− c)+ + ( c− c) ⎤ = 0n ⎢⎣⎥⎦37. Letn∑Sc () = ( x−c)i=1dS'( c)= x −cdcS''( c) = 2nin2. Then∑( i )i=1n∑ ii=1dcn∑ ii=1n∑ ii=1d= x − c( )22= 2( x −c)( −1)=− 2 x + 2ncSet S'( c ) = 0 and solve for c :n∑ xii=1n1n ∑ ii = 1− 2 + 2nc= 0c = x = xSince S''( x) = 2n> 0 we know that x minimizes Sc ().Instructor’s Resource Manual Section 4.1 253

38. a. The number of gifts given on the nth day isii ( + 1)The total number of gifts is ∑ = 364 .212i=1iii ( + 1)∑ m = .2m = 1ii ( + 1)b. For n days, the total number of gifts is ∑ .i=12n n 2 n n nii ( + 1) i i 1 2 1 1 ⎡nn ( + 1)(2n+ 1) ⎤ 1 ⎡nn( + 1) ⎤∑ = ∑ + ∑ = ∑i+ ∑i=2 2 2 2 2 2⎢ 6 ⎥+2⎢ 2 ⎥⎣ ⎦ ⎣ ⎦i= 1 i= 1 i=1ni= 1 i=11 ⎛2n+ 1 ⎞ 11= nn ( + 1) ⎜ + 1 ⎟= nn ( + 1)(2n+4) ( 1)( 2)4 ⎝ 3 ⎠ 12= 6 nn+ n+39. The bottom layer contains 10 · 16 = 160 oranges, the next layer contains 9 · 15 = 135 oranges, the third layercontains 8 · 14 = 112 oranges, and so on, up to the top layer, which contains 1 · 7 = 7 oranges. The stack contains1 · 7 + 2 · 8+ ... + 9· 15 + 10 · 16=10∑i=1i(6 + i)= 715 oranges.40. If the bottom layer is 50 oranges by 60 oranges, the stack contains50∑i=1i(10 + i) = 55,675.41. For a general stack whose base is m rows of n oranges with m≤n,the stack containsm m m∑ ∑ ∑in ( − m+ i) = ( n− m)i+ii= 1 i= 1 i=1mm ( + 1) mm ( + 1)(2m+1)= ( n− m)+2 6mm ( + 1)(3n− m+1)=6242.43.44.45.46.1 1 1 1+ + + +12 ⋅ 23 ⋅ 34 ⋅ nn ( + 1)⎛ 1 1⎞ ⎛1 1⎞ ⎛1 1⎞ ⎛1 1 ⎞= ⎜ − ⎟+ ⎜ − ⎟+ ⎜ − ⎟+ + ⎜ − ⎟⎝ 2 ⎠ ⎝ 2 3 ⎠ ⎝ 3 4 ⎠ ⎝ n n + 1 ⎠1= 1− n + 11⎡3 5⎤7A = 1 22⎢+ + + =2 2⎥⎣ ⎦ 21 ⎡ 5 3 7 9 5 11⎤15A = 1 24⎢+ + + + + + + =4 2 4 4 2 4 ⎥⎣⎦ 41⎡3 5 ⎤ 9A = 2 32⎢+ + + =2 2 ⎥⎣⎦ 21⎡5 3 7 9 5 11 ⎤ 17A = 2 34⎢+ + + + + + + =4 2 4 4 2 4 ⎥⎣⎦ 4254 Section 4.1 Instructor’s Resource Manual

47.21⎡⎛1 2 ⎞ ⎛1 ⎛1⎞ ⎞ ⎛12 ⎞ ⎛ 21 ⎛3⎞⎞⎤A = ⎜ ⋅ 0 + 1⎟+ ⋅ ⎜ ⎟ + 1 + ⎜ ⋅ 1 + 1⎟2⎢ + ⋅ + 1⎟⎥⎣⎝2 ⎠ ⎜2 ⎝2⎠ ⎟ ⎝2⎝ ⎠ ⎠ ⎜ ⎜ ⎟2 2⎝ ⎝ ⎠ ⎟⎥⎠⎦1⎛9 3 17⎞23= ⎜1+ + + ⎟=2⎝8 2 8 ⎠ 848.21⎡⎛1 ⎛1⎞ ⎞ ⎛12 ⎞A = ⎢⎜ ⋅ ⎜ ⎟ + 1 + ⎜ ⋅ 1 + 1⎟2⎢⎜2 ⎝2⎠ ⎟ ⎝2⎣⎝ ⎠ ⎠⎛ 21 ⎛3⎞ ⎞ ⎛12 ⎞⎤+ ⋅ + 1 + ⋅ 2 + 1 ⎥⎜ ⎜ ⎟ ⎜ ⎟2 2 ⎟ 2⎝ ⎝ ⎠ ⎠ ⎝ ⎠⎥⎦1⎛9 3 17 ⎞ 31= ⎜ + + + 3⎟=2⎝8 2 8 ⎠ 849.A = 1(1 + 2 + 3) = 650.1⎡⎛ 3 ⎞ ⎛ 5 ⎞1 ⎛7 13 ⎞ 23A = ⎜3⋅ − 1⎟+ ( 3⋅2− 1)+ ⎜3⋅ −1⎟2⎢ + (3 · 3 – 1)] = ⎜ + 5+ + 8⎟=⎣ ⎝ 2 ⎠ ⎝ 2 ⎠2⎝2 2 ⎠ 251.2 2 2 2 21⎡⎛⎛13⎞ ⎞ ⎛⎛7⎞ ⎞ ⎛⎛5⎞ ⎞ ⎛⎛8⎞ ⎞ ⎛⎛17⎞⎞ ⎤2A = ⎢⎜ ⎟ − 1 + ⎜ ⎟ − 1 + ⎜ ⎟ − 1 + ⎜ ⎟ − 1 + ⎜ ⎟ − 1 + (3 −1)⎥6⎢⎜ 6 ⎟ ⎜ 3 ⎟ ⎜ 2 ⎟ ⎜ 3 ⎟ ⎜ 6 ⎟⎣⎝⎝ ⎠ ⎠ ⎝⎝ ⎠ ⎠ ⎝⎝ ⎠ ⎠ ⎝⎝ ⎠ ⎠ ⎝⎝ ⎠ ⎠⎥⎦1 ⎛133 40 21 55 253 ⎞ 1243= ⎜ + + + + + 8⎟=6 ⎝ 36 9 4 9 36 ⎠ 216Instructor’s Resource Manual Section 4.1 255

52.2 2 21⎡ ⎛2 ⎛ 4⎞ ⎛ 4⎞ ⎞ ⎛ ⎛ 3⎞ ⎛ 3⎞ ⎞ ⎛ ⎛ 2⎞ ⎛ 2⎞⎞2A = ⎢(3( − 1) + ( − 1) + 1) + 3⎜− ⎟ + ⎜− ⎟+ 1 + 3⎜− ⎟ + ⎜− ⎟+ 1 + 3⎜− ⎟ + ⎜− ⎟+ 1 + (3(0) + 0 + 1)5⎢ ⎜ ⎝ 5⎠ ⎝ 5⎠ ⎟ ⎜ ⎝ 5⎠ ⎝ 5⎠ ⎟ ⎜ ⎝ 5⎠ ⎝ 5⎠⎟⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 2 2 2 2⎛1⎞ 1 ⎞ ⎛ ⎛2⎞ 2 ⎞ ⎛ ⎛3⎞ 3 ⎞⎛ ⎛4⎞4 ⎞⎤2+ 3 + + 1 + 3 + + 1 + 3 + + 1⎟⎜3 + + 1 + (3(1) + 1+1) ⎥⎜ ⎜ ⎟5 5 ⎟ ⎜ ⎜ ⎟5 5 ⎟ ⎜ ⎜ ⎟ ⎜ ⎟5 5 ⎟⎜ 5 5 ⎟⎝ ⎝ ⎠ ⎠ ⎝ ⎝ ⎠ ⎠ ⎝ ⎝ ⎠ ⎠⎝ ⎝ ⎠ ⎠⎥⎦1= [3 + 2.12 + 1.48 + 1.08 + 1 + 1.32 + 1.88 + 2.68 + 3.72 + 5] = 4.656553.1 iΔ x = , x i =n n⎛ i ⎞⎛1⎞i 2f( xi) Δ x = ⎜ + 2⎟⎜ ⎟= +⎝n ⎠⎝n⎠2n n⎡⎛ 1 2⎞ ⎛ 2 2⎞ ⎛ n 2⎞⎤AS ( n ) = ⎢⎜ +2 ⎟+ ⎜ +n 2 ⎟+ + ⎜ +n 2 ⎟⎥⎣⎝n ⎠ ⎝n ⎠ ⎝nn⎠⎦⎛ 1 5⎞5lim AS ( n ) = lim ⎜ + ⎟=n→∞n→∞⎝2 n 2 ⎠ 21= (1 + 2 + 3 + + n)+ 22nnn ( + 1)= + 222n1 5= +2n254.1 iΔ x = , x i =n n⎡ 2 21 ⎛ i ⎞ ⎤⎛1⎞i 1f( xi) Δ x = ⎢ ⋅ ⎜ ⎟ + 1⎥ ⎜ ⎟= +2 n n 3⎢ ⎝ ⎠ ⎥⎝ ⎠ 2nn⎣ ⎦⎡⎛ 2 2 21 1⎞ ⎛ 2 1⎞ ⎛ n 1⎞⎤2 2 2 2AS ( n ) = ⎢+ + + + + +⎥ = 1 (1 + 2 + 3 + + n ) + 1⎜ 3 3 3⎢ 2n n⎟ ⎜2n n⎟ ⎜2nn⎟3⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥⎦2n3 21 ⎡nn( + 1)(2n+1) ⎤ 1 ⎡2n + 3n + n⎤1 ⎡ 3 1 ⎤= + 13 ⎢2n6 ⎥ = ⎢⎥+1 = 2 1⎣⎦ 12 32⎢⎣n ⎥⎦12⎢ + + +n⎥⎣ n ⎦⎡ 1 ⎛ 3 1 ⎞ ⎤ 7lim AS ( n ) = lim ⎢ ⎜2 + + + 1 =→∞ n→∞12 n 2 ⎟ ⎥⎣ ⎝ n ⎠ ⎦ 6n256 Section 4.1 Instructor’s Resource Manual

55.2 2iΔ x = , xi=− 1+n n⎡ ⎛ 2i⎞ ⎤⎛2⎞8if( xi) Δ x = ⎢2⎜− 1+ ⎟+ 2 =n⎥⎜ ⎟n 2⎣ ⎝ ⎠ ⎦⎝ ⎠ n⎡⎛ 8 ⎞ ⎛16⎞ ⎛8n⎞⎤AS ( n ) = ⎢⎜ + +2 ⎟ ⎜ 2 ⎟ + ⎜ 2 ⎟⎥⎣⎝n⎠ ⎝n⎠ ⎝n⎠⎦8 8 ⎡nn( + 1) ⎤= (1 + 2 + 3 + + n)=22 ⎢nn 2 ⎥⎣ ⎦⎡ 24 n + n ⎤ 4= ⎢ ⎥ = 4 +2⎢⎣n ⎥⎦n⎛ 4 ⎞lim AS ( n ) = lim ⎜4 + ⎟=4n→∞n→∞⎝n ⎠58.1 iΔ x = , x i =n n⎡ 3 3⎛ i ⎞ i ⎤⎛1⎞ i if( xi) Δ x = ⎢ ⎜ ⎟ + ⎥ ⎜ ⎟= +n n n 4 2⎢⎣⎝ ⎠ ⎥⎦⎝ ⎠ n n1 3 3 3 1A( Sn) = (1 + 2 + + n ) + (1 + 2 + + n)4 2nn21 ⎡ nn ( + 1) ⎤ 1 ⎡nn( + 1) ⎤=4 ⎢ 2 ⎥ +2 ⎢nn 2 ⎥⎣ ⎦ ⎣ ⎦2 2 2n + 2n+ 1 n + n 3n + 4n+1 3 1 1= + = = + +2 2 2 4 24n 2n 4n n 4n3lim AS ( n ) =n→∞456. First, consider a = 0 and b = 2.2 2iΔ x = , x i =n n2 2⎛2i⎞ ⎛2⎞8if( xi) Δ x = ⎜ ⎟ ⎜ ⎟=⎝ n ⎠ ⎝n⎠3n⎡2 2⎛ 8 ⎞ ⎛8(2 ) ⎞ ⎛8n⎞⎤AS ( n ) = ⎢⎜ 3 ⎟+ +⋅⋅⋅+⎥3 3⎢ n ⎜ n ⎟ ⎜ n ⎟⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥⎦8 2 2 2 8 ⎡nn( + 1)(2n+1) ⎤= (1 + 2 +⋅⋅⋅+ n ) =3 3 ⎢nn 6 ⎥⎣⎦3 24⎡2n + 3n + n⎤8 4 4= ⎢⎥ = + +3 3 3 n 2⎢⎣n ⎥⎦3n⎛8 4 4 ⎞ 8lim AS ( n ) = lim ⎜ + + .n→∞n→∞3 n 2 ⎟=⎝ 3n⎠ 3⎛8⎞16By symmetry, A = 2⎜⎟=.⎝ 3 ⎠ 31 i57. Δ x = , x i =n n3 3⎛ i ⎞ ⎛1⎞ if( xi) Δ x = ⎜ ⎟ ⎜ ⎟=⎝n⎠ ⎝n⎠4n⎡ 1 3 1 3 1 3 ⎤A( Sn) = ⎢ (1 ) + (2 ) + + ( n )4 4 4 ⎥⎣n n n ⎦21 (13 23 3 1 ⎡nn( + 1) ⎤= + + + n ) =44 ⎢nn 2 ⎥⎣ ⎦4 3 21 ⎡n + 2n + n ⎤ 1⎡2 1 ⎤= ⎢ ⎥ = 14n ⎢⎣42⎥⎦4⎢ + +n⎥⎣ n ⎦1⎡2 1 ⎤ 1lim AS ( n ) = lim 1n 4⎢ + +→∞→∞ n 2 ⎥ =⎣ n ⎦ 4n59.60.⎡i⎤1 i 2f( ti) Δ t = ⎢ + 2n ⎥ = +⎣ ⎦n 2n nn n n⎛ i 2⎞1 2AS ( n ) = ∑ ⎜ +2 ⎟= ∑ i+∑2i= 1⎝nn⎠n i= 1 i=1n1 ⎡ nn ( + 1) ⎤= 22 ⎢ +n 2 ⎥⎣ ⎦⎡ 2n + n⎤= ⎢ ⎥+22⎢⎣2n⎥⎦⎛1 1 ⎞= ⎜ + ⎟+2⎝2 2n ⎠1 5lim AS ( n ) = + 2 =n→∞2 21The object traveled 2 ft. 2⎡ 2 21⎛i ⎞ ⎤1 i 1f( ti) Δ t =⎢ ⎜ ⎟ + 1⎥ = +2 n n 3⎢ ⎝ ⎠ ⎥ 2nn⎣ ⎦n ⎛ 2n n1i1⎞1 2 1AS ( n ) = ∑ + = i +⎜ 3 3i= 1 2nn⎟∑ ∑⎝ ⎠ 2ni= 1 i=1n1 ⎡ nn ( + 1)(2n+1) ⎤ 1 ⎡ 3 1 ⎤= + 13 ⎢2n6 ⎥ = 2 1⎣ ⎦ 12⎢ + + +n 2 ⎥⎣ n ⎦1 7lim AS ( n ) = (2) + 1 = ≈ 1.17n→∞12 6The object traveled about 1.17 feet.Instructor’s Resource Manual Section 4.1 257

2 32⎛ib ⎞ ⎛b ⎞ b if( xi) Δ x = ⎜ ⎟ ⎜ ⎟=⎝ n ⎠ ⎝n⎠3n3 n 3b b 2 ⎡A0 = i3 ∑ 3n i=1 n ⎣3b ⎡ 3 1 ⎤= 26⎢ + +n 2 ⎥⎣ n ⎦3 3b 2bblim A0= =n→∞6 32 3 261. a.65. a. A0( x ) = = 43+1b n( n+ 1)(2n+1) ⎤= 31 + 31 +⎢ 6 ⎥2 3 2 1 1 15⎦b. A1( x ) = − = 4− =3+ 1 3+1 4 45+ 1 5+12 5 2 1 32 1 63c. A1( x ) = − = − =5+ 1 5+1 3 6 621= = 10.52b a bb. Since a ≥ 0, A0 = A0 + A a , or9+13 32 9 2 1024b b a b ad. A0( x ) = = = 102.4Aa= A0 − A0 = − .9+1 103 366. Inscribed:3 35 5 3 98Consider an isosceles triangle formed by one side62. A 3 = − = ≈ 32.73 3 3of the polygon and the center of the circle. TheThe object traveled about 32.7 m.angle at the center is 2 . The length of the basenπ35 5 125π π63. a. A 0 = =is 2rsin . The height is r cos . Thus the area3 3nn2 1 2 23 34 4 1 63of the triangle is r sin π cos π = r sinπ .b. A 1 = − = = 21n n 2 n3 3 3⎛1 2 2π⎞ 1 2 2πAn= n⎜r sin ⎟=nr sin3 3⎝2 n ⎠ 2 n5 5 2 117c. A 2 = − = = 393 3 3Circumscribed:Consider an isosceles triangle formed by one sideb biof the polygon and the center of the circle. The64. a. Δ x = , x i =n nangle at the center is 2 . The length of the basem m+1 m⎛bi ⎞ ⎛b ⎞ b inπf( xi) Δ x = ⎜ ⎟ ⎜ ⎟=⎝ n ⎠ ⎝n⎠m+1πnis 2rtan . The height is r. Thus the area of them+1 n nbA( S )mn = ∑ im+12 πn itriangle is= 1r tan .nm+ 1 m+1b ⎡n⎤= ⎢ + C⎛ 2 π ⎞ 2 πm+1 n ⎥Bn= nn ⎢⎣m + 1⎜r tan ⎟=nr tan⎥⎦⎝ n⎠nm+1 m+1b b Csin 2πn1 2 2π⎛ ⎞2 n= +lim Am + 1 m+1n = lim nr sin = lim πr⎜ ⎟nn→∞ n→∞ 2 n n→∞⎜ 2π⎟⎝ n ⎠m+1b mb2A0 ( x ) = lim A( Sn) == π rn→∞m + 12 sin πC2 π πr⎛ ⎞nnlim 0n→∞ m 1n + = since C n is a polynomial in nlim Bn= lim nr tan = lim ⎜ ⎟n→∞ n→∞ n n→∞cos π ⎜ π ⎟n ⎝ n ⎠of degree m.2= π rb. Notice that A ( x ) = A0( x ) − A0( x ).Thus, using part a,+ + .b m b m a mam+ 1 m+1b m b aAa( x ) = −m 1 m 131 +258 Section 4.1 Instructor’s Resource Manual

4.2 Concepts Review1. Riemann sumb2. definite integral; ∫ f ( xdx )a3. Aup − Adown4.1 158 − = 2 2Problem Set 4.21. RP= f(2)(2.5 − 1) + f(3)(3.5 − 2.5) + f(4.5)(5 − 3.5) = 4(1.5) + 3(1) + (–2.25)(1.5) = 5.6252. R P = f(0.5)(0.7 – 0) + f(1.5)(1.7 – 0.7) + f(2)(2.7 – 1.7) + f(3.5)(4 – 2.7)= 1.25(0.7) + (–0.75)(1) + (–1)(1) + 1.25(1.3) = 0.753.4.5.6.5RP = ∑ f( xi)Δxi= f(3)(3.75 − 3) + f(4)(4.25 − 3.75) + f(4.75)(5.5 − 4.25) + f(6)(6 − 5.5) + f(6.5)(7 − 6)i=1= 2(0.75) + 3(0.5) + 3.75(1.25) + 5(0.5) + 5.5(1) = 15.68754RP = ∑ f( xi)Δxi= f( − 2)( − 1.3 + 3) + f( − 0.5)(0 + 1.3) + f(0)(0.9 − 0) + f(2)(2 − 0.9)i=1= 4(1.7) + 3.25(1.3) + 3(0.9) + 2(1.1) = 15.9258RP = f( xi)Δxii=1∑ = [ f − + f − + f − + f − + f + f + f + f ]( 1.75) ( 1.25) ( 0.75) ( 0.25) (0.25) (0.75) (1.25) (1.75) (0.5)= [–0.21875 – 0.46875 – 0.46875 – 0.21875 + 0.28125 + 1.03125 + 2.03125 + 3.28125](0.5) = 2.6256RP = f( xi)Δxii=1∑ = [ f + f + f + f + f + f ](0.5) (1) (1.5) (2) (2.5) (3) (0.5)= [1.5 + 5 + 14.5 + 33 + 63.5 + 109](0.5) = 113.257.8.9.10.3 3∫ 1 x dx2 30 ( x + 1) dx∫12xdx− 1 1+ x∫π 20 (sin x)dx∫11.2 2iΔ x = , x i =n n2if( xi) = xi+ 1= + 1nnn⎡ ⎛2⎞⎤2∑ f( xi) Δ x = ∑ ⎢ 1+ i ⎜ ⎟i= 1 i=1 n ⎥⎣ ⎝ ⎠⎦nn n2 4 2 4 ⎡ nn ( + 1) ⎤= ∑1+in 2 ∑ = ( n)+2i= 1 n i=1 n ⎢n 2 ⎥⎣ ⎦⎛ 1 ⎞= 2+ 2⎜1+⎟⎝ n ⎠∫20⎡ ⎛ 1 ⎞⎤( x+ 1) dx = lim ⎢2+ 2⎜1+ ⎟ = 4n→∞n⎥⎣ ⎝ ⎠⎦Instructor’s Resource Manual Section 4.2 259

12.13.2 2iΔ x = , x i =n n2 2⎛2i⎞4if( xi) = ⎜ ⎟ + 1= + 1⎝ n ⎠ 2nnn ⎡ 2 ⎛ 4 ⎞⎤2∑ f( xi) Δ x = ∑ ⎢1+i ⎜ 2 ⎟⎥i= 1 i=1⎣⎝n⎠⎦nn n2 8 2 2 8 ⎡nn( + 1)(2n+1) ⎤= ∑1+in 3 ∑ = ( n)+3i= 1 n i=1 n ⎢n 6 ⎥⎣⎦4⎛3 1 ⎞= 2+ ⎜2+ +3 n 2 ⎟⎝ n ⎠2 2⎡ 4⎛3 1 ⎞⎤14( x + 1) dx = lim 2 20⎢ + ⎜ + + =n→∞3 n 2 ⎟⎥⎣ ⎝ n ⎠⎦3∫3 3iΔ x = , xi=− 2 +n n⎛ 3i⎞6if( xi) = 2⎜− 2+ ⎟+π= π− 4+⎝ n ⎠nnn⎡ 6i⎤3∑ f( xi) Δ x = ∑ ⎢π− 4+i= 1 i=1 n ⎥⎣ ⎦nnn3 1818 ⎡nn( + 1) ⎤= ∑( π− 4) + in2 ∑ = 3( π− 4) +2 ⎢i= 1 n i=1n 2 ⎥⎣ ⎦⎛ 1 ⎞= 3π− 12+ 9⎜1+⎟⎝ n ⎠1⎡ ⎛ 1 ⎞⎤(2 x+π ) dx = lim 3 12 9 1−2⎢ π− + ⎜ + ⎟n→∞n⎥⎣ ⎝ ⎠⎦= 3π − 3∫14.15.3 3iΔ x = , xi= − 2 +n n2 2⎛ 3i⎞36i 27if( xi) = 3⎜− 2+ ⎟ + 2= 14− +⎝ n ⎠n 2nnn ⎡ ⎛36 ⎞ ⎛27 ⎞ 2 ⎤ 3∑ f( xi) Δ x = ∑ ⎢14− ⎜ ⎟i+⎜ i2 ⎟ ⎥i= 1 i=1⎣⎝ n ⎠ ⎝n⎠ ⎦ nn n n= 3 108 81 2∑14i in − 2 ∑ 3i 1 n + ∑= i= 1 n i=1108 ⎡ nn ( + 1) ⎤ 81 ⎡nn ( + 1)(2n+1) ⎤= 42 −2 ⎢ 2 ⎥+3 ⎢nn 6 ⎥⎣ ⎦ ⎣ ⎦⎛ 1⎞27⎛3 1 ⎞= 42 − 54⎜1+ ⎟+ ⎜2+ +n 2 n 2 ⎟⎝ ⎠ ⎝ n ⎠1 2∫ (3 2)2 x + dx−⎡ ⎛ 1⎞27⎛3 1 ⎞⎤= lim ⎢42 − 54⎜1+ ⎟+ ⎜2+ +n→∞n 2 n 2 ⎟⎥= 15⎣ ⎝ ⎠ ⎝ n ⎠⎦5 5iΔ x = , x i =n n5if( xi) = 1+nnn⎡ ⎛5⎞⎤5∑ f( xi) Δ x = ∑ ⎢ 1+ i ⎜ ⎟i= 1 i=1 n ⎥⎣ ⎝ ⎠⎦nn n5 25 25 ⎡nn( + 1) ⎤= ∑1+in 2 ∑ = 5 +2 ⎢i= 1 n i=1 n 2 ⎥⎣ ⎦25 ⎛ 1 ⎞= 5+ ⎜1+⎟2 ⎝ n ⎠∫50⎡ 25 ⎛ 1 ⎞⎤35( x+ 1) dx = lim ⎢5+ ⎜1+ ⎟ =n→∞2 n⎥⎣ ⎝ ⎠⎦216.20 20iΔ x = , xi= − 10 +nn22⎛ 20i⎞ ⎛ 20i⎞380i400if( xi) = ⎜− 10+ ⎟ + ⎜− 10+⎟ = 90 − +⎝ n ⎠ ⎝ n ⎠ n 2nnn ⎡ 380 2 400 ⎤nn⎛ ⎞ ⎛ ⎞ 20 20 7600 8000∑ f( xi) Δ x = ∑ ⎢90− i⎜⎟+i ⎜ 2 ⎟⎥= ∑90− ii= 1 i=1⎣⎝ n ⎠ ⎝ n ⎠⎦n n2 ∑n 2+3 ∑ ii= 1 n i=1 n i=17600 ⎡nn ( + 1) ⎤ 8000 ⎡nn ( + 1)(2n+1) ⎤⎛ 1⎞4000⎛3 1 ⎞= 1800 − +2 ⎢ 2 ⎥ 3 ⎢nn 6 ⎥ = 1800 − 3800⎜1+ ⎟+ ⎜2+ +⎣ ⎦ ⎣ ⎦ n 3 n 2 ⎟⎝ ⎠ ⎝ n ⎠10 2∫ 10 ( x +−xdx ) ⎡ ⎛ 1 ⎞ 4000 ⎛ 3 1 ⎞⎤2000= lim ⎢1800 − 3800⎜1+ ⎟+ ⎜2+ + =n→∞n 3 n 2 ⎟⎥⎣ ⎝ ⎠ ⎝ n ⎠⎦3260 Section 4.2 Instructor’s Resource Manual

17.21. The area under the curve is equal to the area of asemi-circle:AA 2 2 1 2− x dx =− A 2π A∫ .5∫ f ( xdx )01 1 27= (1)(2) + 1(2) + 3(2) + (3)(3) =2 2 222. The area under the curve is equal to the area of atriangle:18.y42∫4−4f⎛1⎞x dx = 2⎜⎟4⋅ 4=16⎝2⎠( )19.∫2024 x1 1 9f( x) dx = ()( 1 3) + ()( 1 2) + ()( 1 2)=2 2 24 1 ⎛ 4 ⎞ 24 = v t dt = 40⎜ ⎟=2 ⎝60⎠1523. s( ) ∫ ( )414 = = 4+ 4 9− 1 = 200224. s( ) ∫ v( t) dt ( )4 14 = = 2 1 + 2 1 = 30 225. s( ) ∫ v( t) dt () ()4 1 2s 4 = v t dt = π 2 + 0=π0 426. ( ) ∫ ( ) ( )20.2 1 2 1 1 πf( x) dx = ( π⋅ 1 ) + (1)(1) = +0 4 2 2 4∫27. t s(t)20 4040 8060 12080 160100 200120 2402 1 2 1∫ f( x) dx =− ( π⋅2 ) −(2)(2) − (2)(4)−24 2= − − 8πInstructor’s Resource Manual Section 4.2 261

28. t s(t)20 1040 4060 9080 160100 250120 36029. t s(t)20 2040 8060 16080 240100 320120 40030. t s(t)20 2040 6060 8080 60100 0120 -1003−331. a. ∫ b. ∫ xx dx = ( −3−2− 1+ 0+ 1+ 2)(1) = −33 2 dx = [( − 3) 2 + ( −2)2−33c. ∫ ( )−32+− ( 1) + 0+ 1+ 4](1) = 19⎡1⎤x− x dx= 6 ⎢ (1)(1) = 32 ⎥⎣ ⎦1x− x dx = 6 x dx = 6⋅ = 23d. ∫ ( )∫3 2 1 2−3 03e.f.32. a.3−3∫1 1xdx= (3)(3) + (3)(3) = 92 233 3( −3) (3)xxdx= + = 0−33 3∫2 x x dx 0 x dx 1 x dx 2 x dx−1 −1 0 11 1= − (1)(1) + 1(1) + (1)(1) = 12 2g. ∫ =− ∫ + 0∫ + ∫2 2 0 2 1 2x x dx x dx x dx−1 −1 0h. ∫ =− ∫ + 0∫b.c.+ 2 2∫ 1 x dx3 3 31 ⎛2 1 ⎞= − + − = 23 ⎜ 3 3 ⎟⎝ ⎠1∫ f( x) dx = 0 because this is an odd−1function.1−1∫1−1∫1−1d. ∫ [ gx ]e.f.gxdx ( ) = 3+ 3=6f( x) dx = 3+ 3=6− ( ) dx=− 3 + ( − 3) =−61∫ xg( x) dx = 0 because xg(x) is an odd−1function.1 3f ( x) g( x) dx = 0−1∫ becauseis an odd function.n133. RP = ∑ ( xi + xi−1)( xi −xi−1)2i=1n1 2 2= ∑ ( xi−xi−1)2 i=12 2 2 2 2 2= 1 ⎡ ( x1 − x0 ) + ( x2 − x1 ) + ( x3 −x2)2 ⎣2 2+ + ( xn−xn−1) ⎤1 2 2= ( xn− x 0 )2= 1 (2 2 )2 b − a1 2 2 1 2 2lim ( b − a ) = ( b − a )n→∞2 2⎦f3 ( x) g( x )262 Section 4.2 Instructor’s Resource Manual

⎡12 2 ⎤34. Note that ( ) 12x = x + x x + xi ⎢ i−1 i−1i i3⎥⎣⎦1/2⎡1 2 2 2 ⎤≥ ⎢ ( x 1 1 1 13 i − + x i − + x i − ⎥ = x i − and⎣⎦1/2⎡1 2 2⎤xi = ⎢ ( xi−1+ xi−1xi + xi3⎥⎣⎦1/2⎡1 (2 2 2 ⎤≤ ⎢ x )3 i + x i + x i ⎥ = x⎣⎦i .n2Rp = ∑ xi Δxii=1n1 2 2= ∑ ( xi + xi−1xi + xi−1 )( xi −xi−1)i=1 3n1 3 3= ∑ ( xi−xi−1)3 i=13 3 3 3 3 3= 1 ⎡ ( x1 − x0 ) + ( x2 − x1 ) + ( x3 −x2)3 ⎣3 3+ + ( xn−xn−1) ⎤⎦1 3 3 1 3 3= ( xn− x0) = ( b − a )3 3∫35. Left: 2 3( x + 1) dx = 5.24Right:02 3( x + 1) dx = 6.840∫Midpoint:2 3( x + 1) dx = 5.980∫∫37. Left: 1 0Right:cos xdx≈ 0.863810∫Midpoint:cos xdx≈ 0.817810∫cos xdx≈ 0.841838. Left: 3 ⎛1⎞∫ dx ≈ 1.16821⎜ ⎟⎝ x ⎠3 ⎛1⎞Right: ∫ dx ≈ 1.03491 ⎜ ⎟⎝ x ⎠3 ⎛1⎞Midpoint: ∫ dx ≈ 1.09711 ⎜ ⎟⎝ x ⎠39. Partition [0, 1] into n regular intervals, so1P = .ni 1If xi= , f ( xi) 1n+ 2n= .nn1lim ∑ f( xi) Δ xi= lim ∑ = 1P →0i= 1n→∞i=1ni 1xi= , f ( xi) 0n+ πn= .nnlim ∑ f( xi) Δ xi= lim ∑ 0 = 0P →0i= 1n→∞i=1IfThus f is not integrable on [0, 1].∫36. Left: 1 0Right:tan xdx≈ 0.539810∫Midpoint:tan xdx≈ 0.695510∫tan xdx≈ 0.6146Instructor’s Resource Manual Section 4.2 263

4.3 Concepts Review1. 4(4 – 2) = 8; 16(4 – 2) = 3221 ax= =2 25. Ax ( ) x( ax)2.3sin x3.4∫ f ( xdx ) ;152∫x dx4. 5Problem Set 4.3Ax ( ) = 1 ( x−2)( − 1 + x/2) = 1 x−2 , x≥22 46. ( ) 21. A( x) = 2x2. A( x)= ax7. ⎧2x0≤x≤1⎪ 2 + ( x − 1) 1 < x ≤ 2⎪Ax ( ) = ⎨3+ 2( x− 2) 2< x≤3⎪ 5 + ( x − 3) 3 < x ≤ 4⎪⎪⎩ etc.3.1( ) 2Ax ( ) = x−1 , x≥124. If 1 x 2≤ ≤ , then1( ) 2Ax ( ) = x− 1 .2If 2 ≤ x , then Ax ( ) = x−328. ⎧12x0≤ x≤1⎪ 2⎪ 1 + 1(3 − x)( x − 1) 1 < x ≤ 2⎪ 2 21 2⎪1 + ( x − 2) 2 < x ≤ 3Ax ( ) = 2⎨⎪ 3 + 1 (5 −x)( x− 3) 3 < x≤4⎪ 2 2⎪ 1 22+ ( x 4) 4 x 5⎪ 2− < ≤⎪⎩ etc.264 Section 4.3 Instructor’s Resource Manual

9.10.2 22 f ( x) dx = 2 f ( x) dx = 2(3) = 61 1∫ ∫2 22 f ( xdx ) = 2 f( xdx )0 0∫ ∫1 2= 2⎡f( x) dx f( x) dx⎤⎢+ = 2(2+ 3) = 10⎣∫0 ∫ 1 ⎥⎦2 2 22 f ( x) gx ( ) dx 2 f( xdx ) gxdx ( )0 0 011. ∫ [ + ] = ∫ + ∫12.13.1 2 2= 2⎡f ( xdx ) f( xdx )⎤⎢+ + gxdx ( )⎣∫0 ∫1 ⎥⎦∫ 0= 2(2 + 3) + 4 = 141 1 1[2 f () s + g()] s ds = 2 f() s ds+g()s ds0 0 0∫ ∫ ∫= 2(2) + (–1) = 31 2 2∫ [2 f () s + 5 g()] s ds = −2 f() s ds 5 g()s ds2 ∫ −1 ∫ 12 1=−2(3) −5 ⎡g( s) ds g( s)ds⎤⎢−⎣∫0 ∫ 0 ⎥⎦= –6 – 5[4 + 1] = –3113 f( x) + 2 g( x) dx = 0114. ∫ [ ]215. ∫ [ + ]16.0 3 f ( t) 2 g( t)dt1 2 2= 3⎡f ( t) dt f( t) dt⎤⎢+ + 2 g( t)dt⎣∫0 ∫1 ⎥⎦∫ 0= 3(2 + 3) + 2(4) = 2320∫⎡ 3 f ( t) 2 g( t)⎤⎣+ +π⎦dt= 3⎡f ( t) dt f( t) dt⎤⎢+ + 2 g( t)dt⎣∫ ∫ ∫1 2 20 1 ⎥⎦020 dt+π ∫= 3(2+ 3) + 2(4) + 2π= 5 3+ 4 2+ 2π21.22.23.24.π /4G′ ( x) = D⎡x ( s 2)cot(2 s)ds⎤⎢−⎣∫x⎥⎦x= D⎡x ( s 2)cot(2 s)ds⎤⎢− −⎣∫ π /4⎥⎦=−( x−2)cot(2 x)xxG′ ( x)= D⎡x xtdt⎤D⎡1x x tdt⎤⎢=⎣∫ ⎥⎦ ⎢⎣ ∫ 1 ⎥⎦⎡ 2x⎡ 2t ⎤ ⎤ x 1Dxx D ⎡ ⎛ − ⎞= ⎢x x⎤⎢ ⎥ ⎥ = ⎢ ⎥2 ⎜ 2 ⎟⎢ ⎣⎢⎥⎦ ⎢ ⎥⎣ 1 ⎥⎦⎣ ⎝ ⎠⎦⎛ 3x x⎞3 2 1= Dx− = x −⎜ 2 2⎟⎝ ⎠2 2x22G ( x) D ⎡ ⎤′ = x ⎢∫sin tdt⎥= 2xsin( x )1⎢⎣⎥⎦⎡ x2xG ( x) D +⎤′ = x ⎢∫2z+sinz dz⎥1⎢⎣⎥⎦2 2= (2x + 1) 2( x + x) + sin( x + x)25. x2tGx ( ) = ∫−x 1+t2 dt202 2t x t= ∫ 2 dt + dt−x2 ∫021+ t 1+t−x2 2x2t t=− ∫ dt + dt0 2 ∫021+ t 1+t22( −x)2xG'( x) =− ( − 2x)+22 21+ −x1 + x( )5 22xx= +4 21+ x 1+x17.18.xG′ ( x) = D⎡x 2tdt ⎤⎢= 2x⎣∫1 ⎥⎦⎡ 1 ⎤ ⎡ x ⎤x ⎢∫x⎥ x ⎢ ∫ 1 ⎥G′ ( x) = D 2tdt = D − 2tdt =−2x⎣ ⎦ ⎣ ⎦19.⎡ xx ( )G′ ( x) = D 2t 2 t dt⎤⎢+ = 2x 2 + x⎣∫0⎥⎦26.sin x 5Gx ( ) = D⎡x tdt⎤⎢⎣∫ cos x ⎥⎦sin x 5 0 5= D⎡x t dt t dt⎤⎢+⎣∫0 ∫ cosx⎥⎦sin x 5 cos x 5= D⎡x t dt t dt⎤⎢−⎣∫0 ∫ 0 ⎥⎦5 5= sin x cos x+cos xsinx20.x 3G′ ( x) = D⎡x cos (2 t)tan( t)dt⎤⎢⎣∫1⎥⎦3= cos (2 x) tan( x)x1= =21+ x2+ 127. f′ ( x) ; f′′( x)( x ) 3/2So, f(x) is increasing on [0, ) ∞ and concave upon (0, ∞ ).Instructor’s Resource Manual Section 4.3 265

28. f ( x)f1+x′ =21 + x+ − + + −′′ ( x)= = −2 22( x ) ( x)x 2x x2 2( x + 1) ( x + 1)1 1 2 2 1So, f(x) is increasing on [0, ∞ ) and concave up on( 0, − 1+ 2 ).29. ′( ) cos ; ′′( )f x = x f x = − sin x⎡ π ⎤ ⎡3π 5π⎤So, f(x) is increasing on ⎢0, , , ,...2⎥ ⎢2 2⎥ and⎣ ⎦ ⎣ ⎦concave up on ( π,2 π) ,( 3 π,4 π ),....30. ′( ) ′′( )f x = x+ sin x; f x = 1+cos xSo, f(x) is increasing on ( 0,∞ ) and concave upon ( 0,∞ ).35.36.4 2 4∫ ∫ ∫f ( xdx ) = (2 − xdx ) + ( x−2)dx0 0 2= 2+ 2=4′ 1 ′′1xxSo, f(x) is increasing on (0, ∞ ) and neverconcave up.31. f ( x) = ; f ( x) = −232. f(x) is increasing on x ≥ 0 and concave up on33.( 0,1 ),( 2,3 ),...( 3+ x−3)4∫dx03 4= ∫ ( 3+ x − 3) dx + ( 3 3)0 ∫ + x − dx33 4 27 7= ∫ ( 6− x)dx+ xdx = + = 170 ∫32 237. a. Local minima at 0, ≈ 3.8, ≈ 5.8, ≈ 7.9,≈ 9.9;local maxima at ≈ 3.1, ≈ 5, ≈ 7.1, ≈ 9, 10b. Absolute minimum at 0, absolute maximumat ≈ 9c. ≈ (0.7, 1.5), (2.5, 3.5), (4.5, 5.5), (6.5, 7.5),(8.5, 9.5)d.4 2 4f ( x) dx = 2dx + x dx = 4+ 6=100 0 2∫ ∫ ∫34.38. a. Local minima at 0, ≈ 1.8, ≈ 3.8, ≈ 5.8;local maxima at ≈ 1, ≈ 2.9, ≈ 5.2, ≈ 10b. Absolute minimum at 0, absolute maximumat 104 1 2 4∫ ∫ ∫ ∫f ( xdx ) = dx+ xdx+ (4 −xdx)0 0 1 2= 1+ 1.5+ 2.0=4.5c. (0.5, 1.5), (2.2, 3.2), (4.2,5.2), (6.2,7.2),(8.2, 9.2)266 Section 4.3 Instructor’s Resource Manual

d.f.39. a.b.( )0 4F(0) = ∫ t + 1 dt = 00y = F( x)dy4= F'( x) = x + 1dx4dy = ( x + 1)dx1 5y = x + x+C5c. Now apply the initial condition y (0) = 0 :d.40. a.1 50= 0 + 0+C5C = 05Thus y = F( x)= 1 x + x5( )1 4 1 5 6∫ x + 1 dx = F(1) = 1 + 1=.05 5∫∫xGx ( ) = sintdt0002π∫0G(0) = sin tdt = 0G(2 π ) = sintdt= 0b. Let y = G( x). ThendyG'( x)sin xdx = = .dy = sin x dxy =− cos x+Cc. Apply the initial condition0 = y(0) = − cos0+ C. Thus, C = 1,and hence y = G( x) = 1− cosx.d.πsin ( ) 1 cos 20 xdx = G π = − π =∫41. For t ≥ 1, t ≤ t . Since4 41≤ 1+ x ≤ 1+ x .1 1 4 1 4≤ + ≤ +0 0 01 41 ∫ 1 x0∫ ∫ ∫41+ x ≥ 1 for all x,dx 1 x dx (1 x ) dxBy problem 39d,6≤ + dx≤54 442. On the interval [0,1], 2≤ 4+ x ≤ 4+ x .Thus( )1 1 2 1 22 ≤ 4+ ≤ 4+0 0 01 2 212≤ ∫ 4+ x dx≤0∫ ∫ ∫dx x dx x dx5Here, we have used the result from problem 39:4 4( 4+ x ) dx = ∫ ( 3+ 1+x ) dx1 1 4= 3 + ( 1+)1 10 0∫43. 5 ≤ f( x) ≤ 69 so∫∫40(3x )( x )4 30∫ ∫0 06 215 5= 3 + =45 ⋅ ≤ 5+ dx≤ 469 ⋅20 ≤ 5 + dx≤27644. On [2,4], ( x ) 5dx x dx5 58 ≤ + 6 ≤ 10 . Thus,5 4 55∫ ( x )2∫ ( x )28 ⋅ ≤ + 6 dx≤ 210 ⋅4 5265,536 ≤ + 6 dx≤200,000e. G attains the maximum of 2 whenx = π ,3π.G attains the minimum of 0 whenx = 0, 2 π , 4πInflection points of G occur atπ 3π 5π 7πx = , , ,2 2 2 2Instructor’s Resource Manual Section 4.3 267

45. On [1,5],2 2 23+ ≤ 3+ ≤ 3+5 x 1⎛17 ⎞ 5⎛ 2 ⎞4⎜ ⎟≤ 3 dx 4 55∫1⎜ + ⎟ ≤ ⋅⎝ ⎠ ⎝ x ⎠68 5⎛2 ⎞≤ 3 dx 205∫ + ≤1 ⎜ ⎟⎝ x ⎠48. On [0.2,0.4],2 20.002 + 0.0001cos 0.4 ≤ 0.002 + 0.0001cos≤ 0.002 + 0.0001cos 0.22( +)0.4 2≤ ∫0.2( 0.002 + 0.0001cos )2≤ 0.2( 0.002 + 0.0001cos 0.2)0.2 0.002 0.0001cos 0.4Thus,∫( )0.4 20.000417 ≤ 0.002 + 0.0001cos0.2≤ 0.0004192x dxx dxx46. On [10, 20],5 5 5⎛ 1 ⎞ ⎛ 1⎞ ⎛ 1 ⎞⎜1+ ⎟ ≤ ⎜1+ ⎟ ≤ ⎜1+⎟⎝ 20 ⎠ ⎝ x ⎠ ⎝ 10 ⎠5 5 5⎛ 21 ⎞ 20⎛ 1 ⎞ ⎛11⎞10⎜ ⎟ ≤ 1 dx 1020∫10⎜ + ⎟ ≤ ⎜ ⎟⎝ ⎠ ⎝ x ⎠ ⎝10⎠54,084,101 20⎛1 ⎞ 161,051≤ 1 dx320,000∫10⎜ + ⎟ ≤⎝ x ⎠ 10,000520⎛1 ⎞12.7628 ≤ ∫ 1+ dx ≤16.105110 ⎜ ⎟⎝ x ⎠x 1+t49. Let F( x)= ∫ dt. Then0 2 + t1 x 1 + t F( x) −F(0)lim dt limx 0 x∫=→ 0 2+ t x→0x−01+0 1= F '(0) = =2+0 247. On [ 4 π ,8π]1 25≤ 5+ sin x ≤ 5+120 208π2( 4 π) (5) ≤ ∫ ( 5+ 1 sin x) dx≤ ( 4π)( 5+120 20 )4π8π1 2 10120π≤ ( 5 + sin x)dx≤π∫4π20550.51.lim1 x 1+t dt∫x→1x− 1 1 2+t⎡ x t 1= lim ⎢∫dt −x→10 ∫01 1+ 1+t ⎤dtx− 1 2 t 2 t ⎥⎣ + + ⎦F( x) − F(1)= limx→1x −11+1 2= F '(1) = =2+1 3∫xf () t dt = 2 x−21Differentiate both sides with respect to x:d x df() t dt ( 2x2)dx∫ = −1 dxf( x) = 2If such a function exists, it must satisfyf( x ) = 2, but both sides of the first equalitymay differ by a constant yet still have equalderivatives. When x = 1 the left side is11∫ f() t dt = 0and the right side is 21 ⋅ − 2= 0.Thus the function f( x ) = 2 satisfiesx∫ f () t dt = 2 x−2.1268 Section 4.3 Instructor’s Resource Manual

52.53.∫xf () t dt = x02Differentiate both sides with respect to x:d x d 2f () t dt xdx∫ =0 dxf ( x) = 2x∫x20f () t dt =13x3Differentiate both sides with respect to x:2d x d=dxdx2 2f x x = x∫ f () t dt0( )( 2 )2( x )xf =2xf( x)=21 3( x )54. No such function exists. When x = 0 the leftside is 0, whereas the right side is 155. True; by Theorem B (Comparison Property)56. False. a = –1, b = 2, f(x) = x is acounterexample.57. False. a = –1, b = 1, f(x) = x is acounterexample.58. False; A counterexample is f ( x ) = 0 for all x,2except f () 1 = 1. Thus, ∫ f ( x)dx = 0 , but f isnot identically zero.30⎧t⎪ 5 du, 0≤t≤100⎪∫0⎪ 100 t ⎛ u ⎞62. a. s () t = ⎨∫5du + 6 100 7000 ∫du t100⎜ −100⎟< ≤⎪⎝ ⎠⎪ 100 700⎛ u ⎞ t⎪∫ 5du + 6− du + ( − 1 ) du, t > 7000 ∫100 ⎜100⎟ ∫⎪ ⎝ ⎠700⎩⎧⎪⎪5 t, 0 ≤t≤100⎪t2⎪ ⎡ u ⎤= ⎨500 + ⎢6u− ⎥100 < t ≤700⎪ ⎢⎣ 200 ⎥⎦100⎪700⎪ ⎡ 2 ⎤u⎪500 + ⎢6u− ⎥ −( t− 700)t > 700⎪ 200⎩ ⎢⎣ ⎥⎦100⎧5, t0≤t≤100⎪2⎪ t= ⎨− 50 + 6 t− , 100 < t ≤700⎪ 200⎪2400 − t, t > 700⎩∫ ∫bb59. True. a f ( xdx ) − agxdx( )∫b= a [ f ( x) −g( x)]dx60. False. a = 0, b = 1, f(x) = 0, g(x) = –1 is acounterexample.( )⎧ ⎪2+ t−2 , t ≤ 261. vt () = ⎨⎪⎩ 2 − ( t − 2 ), t > 2⎧ t, t ≤ 2= ⎨⎩4 − t, t > 2() = ∫ ( )0⎧ t⎪∫02∫s ttv u duudu, 0≤t≤2= ⎨t ⎪ udu+ ( 4 − u)du, ⎩ 0 ∫2t > 2⎧ 2t⎪ , 0≤t≤ 2⎪ 2= ⎨2⎪⎡ t ⎤2 + 4 t − , t > 2⎪⎢ ⎥⎩ ⎢⎣2 ⎥⎦⎧ 2t,⎪ 20≤t≤ 2= ⎨2 ⎪− t + t − t >4 4 2⎪⎩ 22t− 4t+ 4= 0; t = 4+ 2 2 ≈ 6.832Instructor’s Resource Manual Section 4.3 269

. () 0vt > for 0 ≤ t < 600 and vt () < 0 fort > 600 . So, t = 600 is the point at whichthe object is farthest to the right of the origin.At t = 600 , s() t = 1750 .c. s() t = 0 = 2400 − t; t = 240063. − f ( x) ≤ f( x) ≤ f( x), sobaba∫ ∫babf x dxa− f ( x) dx≤ f( x)dx⇒∫ ∫f ( x) dx≥−( )and combining this withbb∫ f ( x) dx≥f( x) dx,a ∫ awe can conclude thatbb∫ f ( xdx ) ≤ f( x)dxa ∫ ax64. If x > a , ∫ f ′( x) dx≤ M( x−a)by theaBoundedness Property. If x < a ,ax∫ f ( x) dx =− f′( x) dx≥−M( x−a)x ∫ byathe Boundedness Property. Thusx∫ ( )a f ′ x dx≤ M x−a.xxFrom Problem 63, f ′ ( x) dx≥f ′ ( x)dxaa∫ ∫ .x∫ f ′ ( xdx ) f( x) fa ( ) f( x) fa ( )a = − ≥ −Therefore, f ( x) − f( a)≤ M x− a orf ( x) ≤ f( a)+ M x− a .2.3.4.5.6.7.8.9.10.522 ⎡4 x ⎤ 32 1 33x dx = ⎢ ⎥ = + =−1⎢⎣5 ⎥⎦5 5 5−1∫2 2 3 22(3x − 2x+ 3) dx = ⎡x − x + 3x⎤−1 ⎣ ⎦−1∫= (8 – 4 + 6) – (–1 –1 – 3) = 152 3 42(4x + 7) dx = ⎡x + 7x⎤1 ⎣ ⎦1∫= (16 + 14) – (1 + 7) = 2244 1 ⎡ 1⎤dw =1 2 ⎢ −w w ⎥⎣ ⎦1∫33 2 ⎡ 1 ⎤dt =1 3 ⎢ −2 ⎥t ⎣ t ⎦1∫44 ⎡23/2⎤tdt= t0 ⎢3 ⎥⎣ ⎦0∫88 3 ⎡ 3 4/3⎤wdw= w1 ⎢4 ⎥⎣ ⎦1∫⎛ 1⎞3= ⎜− ⎟−( − 1) =⎝ 4⎠4⎛ 1⎞8= ⎜− ⎟−( − 1) =⎝ 9⎠9⎛2 ⎞ 16= ⎜ ⋅8⎟− 0=⎝3 ⎠ 33−2−2 ⎛ 2 1 ⎞ ⎡ y 1 ⎤y + dy = −−4⎜3 3 2y⎟⎢ ⎥⎝ ⎠ ⎢⎣ 2y⎥⎦−4⎛ 8 1⎞ ⎛ 64 1 ⎞ 1783= ⎜− − ⎟−⎜− − ⎟=⎝ 3 8⎠ ⎝ 3 32⎠96∫3 3 45= ⎛ ⎜ ⋅16 ⎞ ⎟− ⎛ ⎜ ⋅ 1⎞⎟=⎝4 ⎠ ⎝4 ⎠ 44 344 s − 8 4 ⎡2 2 s 8⎤−ds = ( s − 8 s ) ds = ⎢ + ⎥1 2s 1⎢⎣3 s ⎥⎦1⎛64 ⎞ ⎛1⎞= ⎜ + 2⎟− ⎜ + 8⎟=15⎝ 3 ⎠ ⎝3⎠∫ ∫4.4 Concepts Review1. antiderivative; F(b) – F(a)2. F(b) – F(a)3. F( d ) − F( c)4.∫ 2 1 4u du13π /2 π /200∫ = 1 – 0 = 111. cos xdx=[ sin x]π /2 π∫ /2 = 0+ 3 = 3π /6 π /612. 2sin tdt = [ −2cost ]13.11 4 2 ⎡25 3 ⎤(2x − 3x + 5) dx = x x 5x0⎢ − +5⎥⎣⎦0⎛2 ⎞ 22= ⎜ − 1+ 5⎟− 0=⎝5 ⎠ 5∫Problem Set 4.41.422 ⎡3 x ⎤x dx = ⎢ ⎥ = 4− 0=40⎢⎣4 ⎥⎦0∫14.11 4/3 1/3 ⎡3 7/3 3 4/3⎤( x − 2 x ) dx = x x0⎢ −7 2⎥⎣⎦0⎛3 3⎞15= ⎜ − ⎟− 0 =−⎝7 2⎠14∫270 Section 4.4 Instructor’s Resource Manual

15. u = 3x + 2, du = 3 dx1 2 3/2 2 (3 2)3/2∫ u⋅ du = u + C = x+ + C3 9 916. u = 2x – 4, du = 2 dx1/3 1 3 4/3 3 (2 4)4/3∫ u ⋅ du = u + C = x− + C2 8 817. u = 3x + 2, du = 3 dx1 1 1∫ cos( u) ⋅ du = sin u+ C = sin(3x+ 2) + C3 3 318. u = 2x – 4, du = 2 dx1 1∫ sin u⋅ du =− cosu+C2 21=− cos(2 x − 4) + C219. u = 6x – 7, du = 6dx1 1∫ sin u⋅ du = − cosu+C6 61=− cos(6 x − 7) + C620. u =πv− 7, du = π dv21.22.23.24.∫1 1 1cosu⋅ du = sin u+ C = sin( πv− 7) + Cπ π π2u = x + 4, du = 2xdx∫1 1 3/2 1 (2 4)3/2u⋅ du = u + C = x + + C2 3 33 2u = x + 5, du = 3x dx∫9 1 1 10 1 (3 5)10u ⋅ du = u + C = x + + C3 30 302u = x + 3, du = 2xdx−12 / 7 1 7 −5/ 7∫ u ⋅ du =− u + C2 107 (2 3)5/710 x −=− + + C2u = 3 v +π , du = 2 3vdv7/8 1 4 15/8∫ u ⋅ du = u + C2 3 15 34 2= ( 3 v +π ) 15 / 8+ C15 325.26.27.28.29.30.31.32.2u = x + 4, du = 2xdx1 1∫ sin( u)⋅ du =− cosu+C2 21 cos( 2= − x + 4) + C23 2u = x + 5, du = 3x dx∫1 1 1 3cosu⋅ du = sin u+ C = sin( x + 5) + C3 3 32xu = x + 4, du = dx2x + 4∫sin udu= − cosu+ C = − cos x + 4 + C3 22zu = z + 3, du =dz23 23⎛z + 3⎞⎜ ⎟⎝ ⎠3 3 3 3 2∫ cosu⋅ du = sin u+ C = sin z + 3 + C2 2 23 9u = ( x + 5) ,3 8 2 2 3 8du = 9( x + 5) (3 x ) dx = 27 x ( x + 5) dx1 1∫ cosu⋅ du = sin u+C27 271 sin (3 5)9= ⎡ x + ⎤+C27 ⎣ ⎦7 9 6 7 8u = (7 x +π ) , du = 441 x (7 x +π ) dx1 1∫ sin u⋅ du =− cosu+C441 4411 cos(7 7 ) 9= − x +π + C4412 2u = sin( x + 4), du = 2x cos( x + 4) dx∫1 1 3/2u⋅ du = u + C2 37u = cos(3x+ 9)1 ⎡ sin( 2x 4)3/2= + ⎤ + C3 ⎣ ⎦6 7du =− 21x sin(3x + 9) dx3 1 1 4/3u ⎛ ⎞∫ ⋅⎜− ⎟du = − u + C⎝ 21⎠281 74/3= − ⎡ cos(3 x + 9) ⎤ + C28 ⎣ ⎦2Instructor’s Resource Manual Section 4.4 271

33.34.35.36.3 2 3u = cos( x + 5), du =− 3x sin( x + 5) dx9 1 1 10u ⎛ ⎞∫ ⋅⎜− ⎟du =− u + C⎝ 3⎠301 cos10 (3=− x + 5) + C303x −− 4 2 −3u = tan( + 1) , du =− 3x sec ( x + 1) dx5 1 5 6/5u ⎛ ⎞∫ ⋅⎜− ⎟du =− u + C⎝ 3⎠185 36/5tan( −=− ⎡ x + 1) ⎤ + C18 ⎣ ⎦2u = x + 1, du = 2xdx⎡ 11u ⎤x + x dx = u du = ⎢ ⎥⎢⎣11 ⎥⎦1 2 10 2 10( 1) (2 )0 1∫ ∫1 11 1 11= ⎡ ⎢ (2) ⎤ −⎡ (1)⎤11 ⎥ ⎢11⎥ =⎣ ⎦ ⎣ ⎦3 2u = x + 1, du = 3x dx20471110 3 2 1 ⎡23/2⎤∫ x + 1(3 x ) dx = udu u−1 ∫ =0 ⎢3 ⎥⎣ ⎦0⎛2 3/2 ⎞ ⎛2 ⎞ 2= ⎜ ⋅1 ⎟−⎜ ⋅ 0⎟=⎝3 ⎠ ⎝3 ⎠ 337. u = t + 2, du = dt1 ⎡ 1⎤dt = u du = ⎢ −( t 2)u ⎥+⎣ ⎦3 5 −2−1 2 1∫ ∫⎡ 1⎤4= ⎢− −[ − 1]=5⎥⎣ ⎦ 538. u = y – 1, du = dy910 9 ⎡23/2⎤∫ y− 1dy = udu u2 ∫ =1 ⎢3 ⎥⎣ ⎦12 2 52= ⎡ ⎢ (27) ⎤ ⎡ (1)⎤3 ⎥− ⎢ =3 ⎥⎣ ⎦ ⎣ ⎦ 339. u = 3x + 1, du = 3 dx8 1 8 1 25∫ 3x + 1dx = 3x 1 3dx u du5 3∫ + ⋅ =5 3∫1625⎡2 3/2⎤ ⎡2 ⎤ ⎡2 ⎤ 122= ⎢ u(125) (64)9⎥ = ⎢ − =9⎥ ⎢9⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 91640. u = 2x + 2, du = 2 dx7 1 1 7 2∫ dx =1 2∫ dx2x+ 2 1 2x+21 16 −1/216= u du = ⎡ u⎤= 4− 2=22∫4 ⎣ ⎦4512141.42.2u = 7+ 2 t , du = 4tdt3 2 3 2∫ 7+ 2 t (8 t) dt = 2 7 2t ( 4t ) dt−3 ∫ + ⋅−32525 ⎡43/2⎤= 2∫udu= u25 ⎢ 3 ⎥⎣ ⎦25⎡4 ⎤ ⎡4⎤= ⎢ (125) − (125) = 03 ⎥ ⎢3⎥⎣ ⎦ ⎣ ⎦3 2u = x + 3, x du = (3x + 3) dx3232x + 1 1 3x+ 3∫ dx =1 3 3∫dx1 3x + 3x x + 3x361 16 −1/2 ⎡21/2⎤= u du u3∫=4 ⎢ 3 ⎥⎣ ⎦42 2 8= ⎜ ⎛ ⋅6⎟ ⎞ −⎜ ⎛ ⋅ 2⎟⎞ =⎝3 ⎠ ⎝3 ⎠ 343. u = cos x, du =− sin xdxπ /2 2 π /2 2∫ cos xsin xdx=− ( )0 ∫ cos x −sinx dx0300 2⎡ u ⎤=− ∫ u du = ⎢−⎥1⎢⎣3 ⎥⎦1⎛ 1⎞1= 0 −⎜− ⎟=⎝ 3⎠344. u = sin 3 x, du = 3cos3xdxπ /2 2∫ sin 3xcos3xdx01 π /2 2 1 −12= sin 3 ( 3cos3 )3∫ x x dx =0 3∫u du03−1⎡u⎤ ⎛ 1⎞1= ⎢ ⎥ = ⎜− ⎟− 0 = −⎢⎣9 ⎥⎦⎝ 9⎠9045.46.2u = x + 2 x, du = (2x+ 2) dx = 2( x+1) dx1 2 2( + 1)( + 2 )01 2 2∫02333 2⎡u⎤∫ u du ⎢ ⎥0⎢⎣⎥⎦0∫x x x dx1= ( x + 2 x ) 2( x+1) dx1 9= = =2 6 21u = x − 1, du = dx2 x4343( x −1) ( x −1)∫ dx = 21 ∫ dxx1 2 x411 3⎡u⎤ 1= 2∫u du = 2⎢⎥ =0⎢⎣4 ⎥⎦20272 Section 4.4 Instructor’s Resource Manual

47. u = sin θ , du = cosθ dθ41/21/2 3⎡u⎤ 1 1udu= ⎢ ⎥ = − 0 =0⎢⎣4 ⎥⎦64 640∫48. u = cos θ , du =− sinθ dθ3/2 3 1 23/2 1 4 1u − du ⎡u− ⎤ ⎛ 1⎞1⎜ ⎟1∫− = = − =2⎣ ⎦ 2⎝3 ⎠ 649. u = 3x − 3, du = 3dx1 0 1 0 1cos [ sin ] (0 sin( 3))3∫udu= u−33 −= − −33sin 3=350. u = 2 π x, du = 2πdx51.52.1 π1 π 1sin udu=− [ cos u] 00=− ( −1−1)2π∫2π 2π1= πu x2 , du 2 xdx=π = πππ∫ udu [ u] 01 1 1sin =− cos =− ( −1−1)2π 02π 2π1= π5 4u = 2 x , du = 10x dx2π52π5cosudu=[ sin u]001 110∫101 5 1 5= (sin(2 π ) − 0) = sin(2 π )10 1053. u = 2 x, du = 2dx1 π/2 1 π/2cos sin2∫ udu+udu0 2∫01 π/2 1 π/2= [ sin u] −0[ cosu]2 201 1= (1 −0) − (0 − 1) = 12 254. u = 3, x du = 3 dx; v = 5, x dv = 5dx1 3 π/2 1 5 π/2cossin3∫ udu+vdv−π 3 /2 5∫−π 5 /21 3 π/2 1 5 π/2= [ sin u] −3 /2[ cos v]3−π5−π 5 /21 1 2= [( −1) −1] − [0 − 0] =−3 5 355. u = cos x, du =− sin xdx0 udu 0[ u ]1 1∫− sin = cos = 1 − cos156. u = π sin θ , du =π cosθ dθ57.58.1 π1 πcosudu= [ sin u]= 0π∫−π π −π2 2u = cos( x ), du = − 2xsin( x ) dx4cos141 cos1 3 1⎡u⎤ cos 1 1−2∫udu=− ⎢ ⎥ =− +12⎢⎣4 ⎥⎦8 8=41−cos 1813 2 3u = sin( x ), du = 3x cos( x ) dx3 sin(3sin( π / 8) 2 3π / 8)sin(3 u du = ⎡u⎤− π / 8) ⎣ ⎦ −sin( π3/ 8)3 ⎛ π3 ⎞⎜ 8 ⎟1 13∫92sin=⎝9⎠59. a. Between 0 and 3, f( x ) > 0. Thus,30∫ f( x) dx > 0.b. Since f is an antiderivative of f ' ,c.30∫30∫f '( x) dx = f(3) − f(0)= 0− 2=− 2

1V t =∫V t =∫− t dt = t− t + C2V ( 0)= C = 0 since no water has leaked out at2time t = 0 . Thus, V() t = 20t− 1 t , so2V ( 20) − V ( 10)= 200 − 150 = 50 gallons.1 2Time to drain: 20t− t = 200; t = 20 hours.2261. () ′() ( 20 ) 20121 ⎡ t ⎤ 219V − V = V t dt = ⎢t− ⎥ =0⎢⎣220 ⎥⎦220010201( 10 ) ( 9 ⎛) 1 t ⎞V − V = ∫ dt9 ⎜ − 110 ⎟ =⎝ ⎠ 2202T ⎛ t ⎞ T55 = V ( T) − V ( 0)= ∫ 1− dt = T −0 ⎜ 110⎟ 22062. () 1 ( 0) ∫′()T ≈ 110 hrs63. Use a midpoint Riemann sum with n = 12partitions.12∑i=1( )V = f x Δxii≈ 1(5.4 + 6.3 + 6.4 + 6.5 + 6.9 + 7.5 + 8.4+ 8.4 + 8.0 + 7.5 + 7.0 + 6.5)= 84.864. Use a midpoint Riemann sum with n = 10partitions.10∑i=1( )V = f x Δxii⎛6200 + 6300 + 6500 + 6500 + 6600 ⎞≈ 1⎜⎟⎝+ 6700 + 6800 + 7000 + 7200 + 7200⎠= 67,00065. Use a midpoint Riemann sum with n = 12partitions.12∑i=0( )E = P t Δtii≈ 2(3.0 + 3.0 + 3.8 + 5.8 + 7.8 + 6.9+ 6.5 + 6.3 + 7.2 + 8.2 + 8.7 + 5.4)= 145.2x= = 1+425mass = ∫ δ ( x) dx = m( 2)=0266. δ ( x) m′( x)⎝⎠bbnnn∫an ∫ an67. a. x dx = B ; y dy = Ab.c.Using Figure 3 of the text,nn( a)( a ) + A + B = ( b)( b ) orn nn+ 1 n+1Bn+ An= b − a . Thusbnn bn n+ 1 n+1∫ x dx +a ∫ an y dy = b −ab bnn n∫ x dx +a ∫ any dy1b⎡ n+bx ⎤ ⎡ n ny( n+1)/ n⎤= ⎢ ⎥ +n 1⎢n 1⎥⎢ + ⎣ + ⎦an⎣ ⎥⎦a⎛ n+ 1 n+1b a ⎞ ⎛ n n+ 1 n n+1⎞= − + b − a⎜ n 1 n 1⎟⎜ ⎟⎝+ +⎠ ⎝n+ 1 n+1 ⎠n+ 1 n+1( n+ 1) b − ( n+1) a n+ 1 n+1= = b −an + 1b n 1 n+1bBn= x dx ⎡x⎤∫ =a n + 1 ⎣ ⎦a1 (n+ 1 n+1b a )= −n + 1bnbnn ⎡ n ( n+1)/ n⎤An= ∫ an ydy = ⎢ yn 1 ⎥⎣ + ⎦ann n+ 1 n+1= ( b −a)n + 1n n+ 1 n+1nBn= ( b − a ) = An + 1x68. Let y = Gx ( ) =∫ ftdt ( ) . ThenadyG'( x) f( x)dx = =dy = f ( x)dxLet F be any antiderivative of f . ThenG x = F x + C. When x = a , we must have69.( ) ( )( ) 0G( x) = F( x) − F( a). Now choose x = b toG a = . Thus, C = −F( a)andobtainb∫ () ( ) ( ) ( )a f t dt = G b = F b −Fa333 2⎡x⎤x dx = ⎢ ⎥ = 9− 0=90⎢⎣3 ⎥⎦0∫nn70.422 3⎡x⎤x dx = ⎢ ⎥ = 4− 0=40⎢⎣4 ⎥⎦0∫274 Section 4.4 Instructor’s Resource Manual

ππsin = − cos = 1+ 1 = 2071. ∫ xdx [ x ] 072.22 2 ⎡ 1 2 1 3⎤∫ (1 + x+ x ) dx = x x x0⎢ + +2 3⎥⎣⎦0⎛ 8⎞20= ⎜2+ 2+ ⎟− 0=⎝ 3⎠373. The right-endpoint Riemann sum isn2n⎛ 1−0 ⎞ ⎛1⎞1 2⎜0 + i⎟ ⎜ ⎟=i ,3i= 1⎝ n ⎠ ⎝n⎠n i=1∑ ∑ which forn = 10 equals 77 0.385200 = .11 2 ⎡1 3 ⎤ 1∫ x dx = x 0.3330 ⎢ = =3⎥⎣ ⎦0374 ∫ ( 2 − 3 ) = 2∫ −3∫= 2( [ −2− 1+ 0+ 1+ 2+3)(1) ]75.4 x x dx 4 x dx 4 x dx−2 −2 −2⎡1 1 ⎤− 3 ⎢ (2)(2) + (4)(4)2 2 ⎥⎣⎦= –24d ⎛1 ⎞ 1 ⎛ x ⎞ x⎜ x x ⎟ = x⎜⎟ + = xdx ⎝2 ⎠ 2 ⎝ x ⎠ 2bb 1 1x dx ⎡ ⎤∫ = x x b b a aa ⎢ = −2⎥⎣ ⎦a2( )76. For b > 0, if b is an integer,b x dx = 0 + 1 + 2 +⋅⋅⋅+ ( b−1)∫ 0b−1( b−1) b= ∑ i = .i=1 2 If b is not an integer, let n = b . Thenb∫ x dx = 0 + 1 + 2 +⋅⋅⋅+ ( n − 1) + n ( b −n)0( n−1)n= + nb ( − n )2b−b b b b ( 1)= + ( − ).277. a. Let c be in ( , )ab . Then G'( c) = f( c)by theFirst Fundamental Theorem of Calculus.Since G is differentiable at c , G iscontinuous there. Now suppose c = a.xThen lim Gx ( ) = lim ∫ ftdt ( ) . Since f isx→c x→aacontinuous on [ ab , ] , there exist (by theMin-Max Existence Theorem) m and M suchthat f ( m) ≤ f( x) ≤ f( M)for all x in [ ab , ] .Thenx x x∫ f ( mdt ) ≤ ( ) ( )a ∫ ftdt≤a ∫ f M dta( x−a) f( m) ≤G( x) ≤( x−a) f( M)By the Squeeze Theoremlim ( x−a) f( m) ≤ lim G( x)x→a +x→a+≤ lim ( x −a) f( M)x→a+Thus,aalim Gx ( ) = 0 = ftdt ( ) = Ga ( )x→a+∫Therefore G is right-continuous at x = a .Now, suppose c = b . Thenblim Gx ( ) = lim ftdt ( )xx→b −x→b−∫As before,( b− x) f( m) ≤G( x) ≤( b− x) f( M)so wecan apply the Squeeze Theorem again toobtainlim ( b−x) f( m) ≤ lim G( x)x→b −x→b−≤ lim ( b−x) f( M)x→b−Thusbblim Gx ( ) = 0 = ftdt ( ) = Gb ( )x→b−∫Therefore, G is left-continuous at x= b .b. Let F be any antiderivative of f. Note that Gis also an antiderivative of f. Thus,F( x) = G( x)+ C . We know from part (a)that ( )ab , . ThusGx is continuous on [ ]F( x ) , being equal to Gx ( ) plus a constant,is also continuous on [ ab , ] .Instructor’s Resource Manual Section 4.4 275

⎧1, x > 0x78. Let f( x)= ⎨ and F( x) = f( t)dt⎩0, x ≤ 0∫ .−1If x < 0 , then F( x ) = 0. If x ≥ 0 , thenxF( x) = ∫ f( t)dt−10 x= ∫ 0dt+ 1dt−1 ∫ 0= 0 + x = xThus,⎧x, x≥0F( x)= ⎨⎩0, x < 0which is continuous everywhere even thoughf ( x ) is not continuous everywhere.4.5 Concepts Review1 bf a1. ( )b− a∫2. f ( c )20∫3. ( )x dx0; 2 f x dx4. f ( x+ p) = f ( x);periodProblem Set 4.51.2.3.1 3 3 1 434xdx= ⎡x⎤ = 403−1 ∫ 1 2⎣⎦141 4 2 1⎡53⎤5x dx = x = 354−1∫1 3⎢3⎥⎣ ⎦133x2dx⎡x 16⎤0 2 ⎢⎣⎥⎦01 1 1= + =3− 0∫x + 16 3 31x dx2+3 −31 0 2=⎛( x x)dx 2xdx⎞⎜ − + + ⎟5 ⎝∫−3 ∫0⎠1 22 4= ⎡x⎤ =5⎣⎦0526. ∫ ( x+)1 π 1 ππ 0 π01= [ sin π − sin 0 ] = 0π7. ∫ cos xdx=[ sin x]1 π 1π − 0 0 π1 2=− ( −1− 1)=π π8. ∫ sin x dx = ( −cosx)9.10.π01 π 2 1 ⎛12⎞∫ xcosx dx = sin xπ 0 0⎜ ⎟− π ⎝2⎠1= ( 0− 0)= 0π1 π /2 2sin x cos xdxπ /2−0∫0π /22⎡1 3 ⎤ 2= sin x =π⎢3 ⎥⎣ ⎦03π1 2 2 ⎡1211. y( 1+ y ) dy = ( 1+y )12.3 4⎤−∫⎢ ⎥⎣ ⎦2 1 18625 609= − 2 = = 76.1258 81 π /4 2 1 ⎡12 ⎤tan xsec x = tan xπ /4−1 ∫ 0π /4−1 ⎢2⎥⎣ ⎦2 2= ( 1− 0)=π −4 π −4210ππ /404.22x ⎡ 3dx x0 3⎢1 1 2 ⎤= + 162− 0∫16 2 3 ⎥x + ⎣ ⎦1 2= ( 24 − 4) = ( 6 −2)3 31x dx1+2 −21 0 1=⎡( 2 x) dx ( 2 x)dx⎤3 ⎢− + +⎣∫−2 ∫0⎥⎦15. ∫ ( 2 + )1 ⎧ 20211 1 ⎫= ⎨⎡2x− x ⎤ + ⎡2x+x ⎤3 2 2 2 ⎬⎩⎣ ⎦−⎣ ⎦0⎭1 2( 1 1 17= −2( − 2) + ( − 2) + 2 +2 2)=3 62013.14.π1 π /2sin z 4 dz ⎡ 2cos/4∫= −π /4 z π ⎣( π π )8= cos / 4 − cos / 2 ≈ 0.815π1 π /2 sinvcosv dvπ /2∫0 21+cos v2=⎡− 1+cosπ ⎢⎣( 1 2)2= − +π2v⎤⎥⎦π /20π /2z ⎤ ⎦π /4276 Section 4.5 Instructor’s Resource Manual

315. ∫ x+ 1dx = c+ 1( 3−0)03⎡23/2⎤⎢ ( x+ 1)3 13 ⎥ = c+⎣ ⎦011514 / 3 = 3 c+ 1; c = ≈1.42811 2 216. ∫ x dx = c ( 1−( −1))−113⎤2⎡1 3⎢ x = 2 c ; c =± ≈± 0.583 ⎥⎣ ⎦3∫−13 2 217. ( 1− x ) dx = ( 1− c )( 3+4)−433⎤2⎡ 1⎢x− x = 7−7c3 ⎥⎣ ⎦−439c =± ≈± 2.083118. ∫ x( 1− x) dx = c( 1−c)( 1−0)0⎡−x⎢⎢⎣2( 2x−3)63±3c = ≈0.21 or 0.79621⎤⎥⎥⎦19. ∫ ( )020= c−c⎡xx⎤xdx= c 2− 0 ; ⎢ ⎥ = 2 c; c=1⎣ 2 ⎦20. ∫ ( )−2π2⎡xx⎤xdx= c 2+ 2 ; ⎢ ⎥ = 4 c; c=−1,1⎣ 2 ⎦21. ∫ sin zdz= sin c( π + π)−ππ[ z]π c c202−2− cos = 2 sin ; = 0π−π22. ∫ cos 2ydy= ( cos 2c)( π −0)0π⎡sin 2y⎤π 3π⎢ π cos 2 c; c ,2 ⎥ = =⎣ ⎦4 4∫02 2 223. ( v − v) dv = ( c −c)( 2−0)023 2⎤2⎡1 1⎢ v − v = 2c −2c3 2 ⎥⎣ ⎦21 + 3c = ≈ 1.266022 3 3 ⎡14 ⎤x dx = c 2− 0 ; x 2c30⎢4⎥ =⎣ ⎦0324. ∫ ( )c = 2 ≈1.26414⎡a 2 ⎤⎢ x bx ⎥⎣ ⎦13 ac 3 b ; c25. ∫ ( ax + b) dx = ( ac + b)( 4−1)5+ = + =2 2b2 2 ⎡13⎤2b ydy = c b − 0;0⎢ y bc 3 ⎥ =⎣ ⎦0bc =326. ∫ ( )27.∫⎡⎢⎣BA( ax + )b dx= f cB−A( )Ba 2 ⎤ x + bx2 ⎥⎦A= ac + bB−Aa( B − A )( B + A ) + b ( B − A )2= ac + bB−Aa aB+ A+ b= ac+b;2 21 1c = B+ A= ( A+B)/22 2b2 2 ⎡13⎤2b ay dy = ac b − 0;0⎢ ay abc3 ⎥ =⎣ ⎦0b 3c =328. ∫( )29.. Using c = π yields30. Using 0.842 π(5) = 1250π≈ 39272c = yields ( )2 3 + sin 0.8 ≈ 7.19Instructor’s Resource Manual Section 4.5 277

31. Using c = 0.5 yields22 3.221+0.5 =38.3π2 3 3π2 3=− 3π0∫ ∫x cos( x ) dx 2 x cos( x ) dx( )2 2= = π33 π⎡3sin( x ) ⎤ sin 3 33⎣⎦0332. Using c = 15 yields5⎛16⎞⎜ ⎟ (20 −10) ≈13.8.⎝15⎠39.∫π−π2(sin x + cos x)dx∫π2 2= (sin x + 2sin xcos x+cos x)dx−ππ π π∫ ∫ ∫= (1 + 2 sin x cos xdx ) = dx+sin 2xdx−π −π −π∫π0π0= 2 dx + 0= 2[ x] = 2π33. A rectangle with height 25 and width 7 hasapproximately the same area as that under thecurve. Thus1 7() 257∫ Htdt≈034. a. A rectangle with height 28 and width 24 hasapproximately the same area as that underthe curve. Thus,1 24Ttdt≈ () 2824 − 0∫ 035.36.b. Yes. The Mean Value Theorem for Integralsguarantees the existence of a c such that1 24Ttdt () = Tc ( )24 − 0∫ 0The figure indicates that there are actuallytwo such values of c, roughly, c = 11 andc = 16 .π + = π +π−π−π0∫ ∫ ∫(sin x cos xdx ) sin xdx 2 cos xdx[ x] π0= 0+ 2 sin = 0∫1−1x32 4(1 + x )odd.dx = 0 , since the integrand is40.41.42.π /2 2 3 3z z z dz =−π /22 3 3∫sin ( )cos( ) 0 , since( −z)sin [( −z) ]cos[( − z) ]2 3 3=−zsin ( −z )cos( − z )3 2 3=−z[ − sin( z )] cos( z )2 3 3=− zsin ( z )cos( z )1 (1 2 3+ x + x + x ) dx−1 1 1 1 2 1 3= ∫ +−1 ∫ +−1 ∫ +−1 ∫ −1311 ⎡x⎤ 8= 2[ x]+ 0+ 2 00⎢ ⎥ + =⎢⎣3 ⎥⎦30∫dx x dx x dx x dx100 3 5−100 ∫( v + sin v + v cos v + sin v ) dv = 03 5since ( − v+ sin( −v) −vcos( − v) + sin ( − v))3 5= ( −v−sin v−vcos v−sin v)3 5=− ( v+ sinv+ vcosv+sin v)1 3 3 1 3 1 3x + x dx = 2 x dx+x dx−1 0 −141⎡x⎤ 1= 2⎢⎥ + 0=⎢⎣4 ⎥⎦20∫ ∫ ∫43. ( )π /4 244. ∫ ( 5)−π /4sincex sin x+ x tan x dx = 052−x sin ( − x) + −x tan( − x)5 2=−x sin x−x tan x37.∫π /2−π /2sin xdx = 0 , since the integrand is odd.1+cosx−ab45. f ( xdx ) = f( xdx )−ba∫ ∫ when f is even.−a−bbf ( xdx ) =− f( xdx )a∫ ∫ when f is odd.278 Section 4.5 Instructor’s Resource Manual

46. u =− x,du =− dx47.bb( ) ( )af x dx −∫ − = −∫−af u du−a−a= f ( u) du = f( x)dx−b−b∫ ∫ since the variableused in the integration is not important.4 ππ/2∫ x dx =0 ∫ 0[ x]cos 8 cos x dxπ /20= 8sin = 848. Since sin x is periodic with period 2π , sin 2x isperiodic with period π .49.4ππ 2x dx =0 0π 2∫ ∫sin 2 8 sin 2x dx⎡ 1 ⎤= 8 ⎢ − cos2x2 ⎥⎣ ⎦0= –4(–1 – 1) = 81 +π = π =π1 0 0= − cos x π = 2∫ ∫ ∫[ ] 0sin x dx sin x dx sin x dxb.k b 1 bku = u dx = ku dx = kub−a∫ab−a∫ac. Note thatu 1 b 1 a= u ( x ) dx = u ( x )b−a dxa a−bbwe can assume a< b.u = 1 b 1 bb−a∫ udx≤ vdx va b−a∫= a54. a. V = 0 by periodicity.b. V = 0 by periodicity.φ + 1∫ ∫ , so2c. ˆ 2 2rms = ∫ sin ( 120π+ φ)φ1ˆ 2 2= ∫ V sin ( 120πt)dtV V t dt0by periodicity.u = 120πt , du = 120πdt2 1 120π2 2V ˆrms = V sin udu120π∫ 0ˆ 2V ⎡ 1 1 ⎤= cosusinu u120π⎢− +2 2 ⎥⎣⎦1 ˆ 2= V2120π0d.Vˆ 2120 =2V ˆ = 120 2 ≈ 169.71 Volts50.51.2 +π/2 π/2∫ ∫sin 2 x dx = sin 2x dx2 01= [ − cos 2x] π /2 = 1201 + π π π /2∫ = ∫ = ∫π /2= 2sin [ x] = 21 ( − 0)= 2cos x dx cos x dx 2 cos x dx1 0 0052. The statement is true. Recall that1 bf = f( x)dxb a∫ .− ab b 1 b b∫ fdx = f ( )a ∫ dx = f x dx ⋅ dxa b−a∫a ∫ a= 1 bbf ( xdx ) ( b a) f( xdx )b− a∫ a⋅ − = ∫ a53. All the statements are true.1 b 1 ba. u + v =b a audx +−∫b−a∫avdx= 1 b( )b− a∫ a u+ v dx = u+v55. Since f is continuous on a closed interval [ ab , ]there exist (by the Min-Max Existence Theorem)ab , such thatan m and M in [ ]f ( m) ≤ f( x) ≤ f( M)for all x in [ , ]ab . Thusb b b∫ f ( m) dx≤( ) ( )a ∫ f x dx≤a ∫ f M dxab( b−a) f( m) ≤∫f( x) dx≤( b−a) f( M)a1 bf ( m) ≤ f( x) dx≤f( M)b−a∫aSince f is continuous, we can apply theIntermediate Value Theorem and say that f takeson every value between f ( m ) and f ( M ). Since1 b( )b− a∫ a f xdx is between f ( m ) and f ( M ),there exists a c in [ ab , ] such that1 bf () c = f()x dxb a∫ .− a2 2 2 0 002 256. a. ∫ ∫ [ ](sin x + cos x ) dx = dx = x = 2 πInstructor’s Resource Manual Section 4.5 279

.2π2 2π2∫0 ∫ 02π22π2x dx0 ∫ 02π2sin xdx0c. 2π= cos x dx + sin= 2∫ cos , thus∫57. a. Evenb. 2π= =πc. On [ ]0,π , sin x = sin x .u = cos x, du =− sin x dxx dxcos x dx∫ ( ) = ∫ sin ⋅sin ( cos )∫ sin cos= cos( cos x)+ CLikewise, on [ π ,2π ] ,∫f x dx x x dx=− udu= u+C( ) cos( cos )f x dx =− x + Cπ /2f ( x)dx0π /2 π /2= 2−π /2 0∫= 1−cos1≈0.46∫ f ( x) dx ∫ f ( x)dx= 2( 1−cos1)≈0.923 /2 3 /2∫ π ( ) = ∫ π ( ) + ∫π( )f x dx f x dx f x dx0 0= cos1−1 ≈−0.46πf x dx = 2 f x dx3 π /2 3 /2∫ ( ) ( )−3 π /2 ∫ 0( )2π∫ f ( x)dx = 004 π /3π /6∫π= 2 cos1−1 ≈−0.92⎛ 3⎞ ⎛1⎞f ( x)dx = 2cos1− cos ⎜+ cos2 ⎟ ⎜ ⎟⎝ ⎠ ⎝2⎠≈−0.44πf x dx = f x dx ≈−0.4410 π / 3 4 / 313 π / 6 π / 6∫ ( ) ∫ ( )58. a. Oddb. 2πc. This function cannot be integrated in closedform. We can only simplify the integralsusing symmetry and periodicity, andapproximate them numerically.aNote that ∫ f ( x ) dx = 0 since f is odd, and−aπ + a∫ f ( x ) dx = 0 sinceπ −a f π + x =−f π − x .( ) ( )π2π /2∫ f ( x) dx = J1() 1 ≈0.69(Bessel0function)π /2f ( x)dx = 0−π /23 π /2 π /20 03 π /22πf ( x)dx = 0−3 π /2 ∫ 013 π / 6 2ππ /6 04 π /3f ( x)dx ≈ 1.055π /610 π / 3 4 π / 313 π /6 π /6∫∫ f ( x) dx ∫ f ( x)dx∫ ; f ( )∫ f ( x) dx ∫ f ( x)dx= ≈0.69x dx = 0= = 0∫ (numeric integration)∫ ( ) ∫ ( )59. a. Written response.b.f x dx = f x dx ≈1.055a a a ⎛c⎞A = ∫ gxdx ( ) = f x dx0 ∫ 0 ⎜ ⎟c ⎝a⎠c2a0 ( ) a a c= ∫ f x dx = f2 0( x ) dxc c∫cb b b ⎛c⎞B = ∫ h( x)dx = f x dx0 ∫ 0 ⎜ ⎟c ⎝b⎠c2b0 ( ) b b c= ∫ f x dx = f2 0( x ) dxc c∫cabThus, ∫ g( x) dx+h( x)dx0 ∫ 02 2a c b c= ( ) ( )2 ∫ f x dx+f x dx0 2∫cc 02 2a + b cc= f( x) dx = f( x)dx2c 0 0∫ ∫ since2 2 2a + b = c from the triangle.60. If f is odd, then f( − x) =− f( x)and we canwrite0 0( ) [ ( )]af x dx =− ∫ −−af − x dx = 0∫ ( )a f u du∫a a =− ∫ f ( u ) du =− f ( xdx )0 ∫ 0On the second line, we have made thesubstitution u = − x.280 Section 4.5 Instructor’s Resource Manual

4.6 Concepts Review1. 1, 2, 2, 2, …, 2, 12. 1, 4, 2, 4, 2, …, 4, 13.4n4. largeProblem Set 4.61.f( x) = 1 ; h= 3–1 = 0.252x 8x 0 = 1.00f( x 0) = 1x 5 = 2.25f( x5) ≈ 0.1975x 1 = 1.25f( x 1) = 0.64x 6 = 2.50f( x 6) = 0.16x 2 = 1.50f( x2) ≈ 0.4444x 7 = 2.75f( x7) ≈ 0.1322x 3 = 1.75f( x3) ≈ 0.3265x 8 = 3.00f( x8) ≈ 0.1111x 4 = 2.00f( x 4) = 0.253 1Left Riemann Sum: ∫ dx ≈ 0.25[ f( x1 20) + f( x1) +…+ f( x7)] ≈0.7877x3 1dx ≈ 0.25[ f ( x1 21) + f x2 + ... + f ( x8)] ≈0.5655x3 1 0.25∫ dx ≈ [ f ( x1 20 ) + 2 f ( x1 ) +…+ 2 f ( x7 ) + f ( x8)] ≈ 0.6766x 23 1 0.25dx ≈ [ f ( x1 20 ) + 4 f ( x1 ) + 2 f ( x2 ) +…+ 4 f ( x7 ) + f ( x8)] ≈ 0.6671x 333 1 ⎡ 1⎤1 2∫ dx = – – 1 0.66671 2 ⎢x x⎥ = + = ≈⎣ ⎦13 3Right Riemann Sum: ∫( )Trapezoidal Rule:Parabolic Rule:∫Fundamental Theorem of Calculus:2.f( x) = 1 ; h = 3–1 = 0.253x 8x 0 = 1.00f( x 0) = 1x 5 = 2.25f( x5) ≈ 0.0878x 1 = 1.25f( x 1) = 0.5120x 6 = 2.50f( x 6) = 0.0640x 2 = 1.50f( x2) ≈ 0.2963x 7 = 2.75f( x7) ≈ 0.0481x 3 = 1.75f( x3) ≈ 0.1866x 8 = 3.00f( x8) ≈ 0.0370x 4 = 2.00f( x 4) = 0.12503 1∫ dx ≈ 0.25[ f( x1 30) + f( x1) +…+ f( x7)] ≈0.5799xLeft Riemann Sum:3 1dx ≈ 0.25[ f( x1 31) + f x2 + ... + f( x8)] ≈0.3392xRight Riemann Sum: ∫( )Trapezoidal Rule:Parabolic Rule:3 1 0.25dx ≈ [ f ( x1 30 ) + 2 f ( x1 ) +…+ 2 f ( x7 ) + f ( x8)] ≈ 0.4596x 2∫3 1 0.25dx ≈ [ f ( x1 30 ) + 4 f ( x1 ) + 2 f ( x2 ) +…+ 4 f ( x7 ) + f ( x8)] ≈ 0.4455x 3∫Fundamental Theorem of Calculus:33 1 ⎡ 1 ⎤ 4dx = 0.44441 3 ⎢− 2⎥= ≈x ⎣ 2x⎦19∫Instructor's Resource Manual Section 4.6 281

3.f( x) =2–0x; h = = 0.258x 0 = 0.00f( x 0) = 0x 5 = 1.25f( x5) ≈ 1.1180x 1 = 0.25f( x 1) = 0.5x 6 = 1.50f( x6) ≈ 1.2247x 2 = 0.50f( x2) ≈ 0.7071x 7 = 1.75f( x7) ≈ 1.3229x 3 = 0.75f( x3) ≈ 0.8660x 8 = 2.00f( x8) ≈ 1.4142x 4 = 1.00f( x 4) = 1Left Riemann Sum:∫20xdx≈ 0.25[ f( x0) + f( x1) +…+ f( x7)] ≈1.6847Right Riemann Sum: ∫( )Trapezoidal Rule:Parabolic Rule:2xdx≈ 0.25[ f( x01) + f x2 + ... + f( x8)] ≈ 2.03832 0.25∫ xdx ≈ [ f ( x00 ) + 2 f ( x1 ) +…+ 2 f ( x7 ) + f ( x8)] ≈ 1.861522 0.25xdx ≈ [ f ( x01 ) + 4 f ( x2 ) + 2 f ( x3 ) +…+ 4 f ( x7 ) + f ( x8)] ≈ 1.8755322 ⎡2 3/2⎤4 2∫ xdx = x1.88560 ⎢ = ≈3⎥⎣ ⎦03∫Fundamental Theorem of Calculus:4.f( x) = x2 3–1x + 1; h= = 0.258x 0 = 1.00f( x0) ≈ 1.4142x 1 = 1.25f( x1) ≈ 2.0010x 2 = 1.50f( x2) ≈ 2.7042x 3 = 1.75f( x3) ≈ 3.5272x 4 = 2.00f( x4) ≈ 4.4721x 5 = 2.25f( x5) ≈ 5.5400x 6 = 2.50f( x6) ≈ 6.7315x 7 = 2.75f( x7) ≈ 8.0470x 8 = 3.00f( x8) ≈ 9.4868Left Riemann Sum:3 2∫ x x + 1dx ≈ 0.25[ f( x10) + f( x1) + + f( x7)] ≈8.60933 2∫ x x + 1dx ≈ 0.25[ f( x11) + f x2 + ... + f( x8)] ≈10.62743 2 0.25∫ x x + 1 dx ≈ [ f( x10) + 2 f( x1) + + 2 f( x7) + f( x8)] ≈9.618423 2 0.25x x + 1 dx ≈ [ f( x10) + 4 f( x1) + 2 f( x2) + + 4 f( x7) + f( x8)] ≈9.598133 2 ⎡1 2 3/2⎤3 1∫ x x + 1 dx = ( x 1) 10 10 – 2 2 9.59811⎢ + = ≈3⎥⎣⎦13Right Riemann Sum: ( )Trapezoidal Rule:Parabolic Rule:∫ Fundamental Theorem of Calculus: ( )2 1−0f( x) = x x + 1 ; h= = 0.1258x 0 = 0.00f( x 0) = 0x 1 = 0.125f( x1) ≈ 0.1351x 2 = 0.250f( x2) ≈ 0.3385x 3 = 0.375f( x3) ≈ 0.7240x 4 = 0.500f( x4) ≈ 1.52595. ( ) 5x 5 = 0.625f( x5) ≈ 3.2504x 6 = 0.750f( x6) ≈ 6.9849x 7 = 0.875f( x7) ≈ 15.0414x 8 = 1.000f( x 8) = 32282 Section 4.6 Instructor’s Resource Manual

1 2Left Riemann Sum: ( ) 5∫ x x + 1 dx ≈ 0.125[ f( x00) + f( x1) + + f( x7)] ≈3.49661 25∫ x x + 1 dx ≈ 0.125[ f( x01) + f x2 + ... + f( x8)] ≈7.49661 2 0.125∫ x x + 1 dx ≈ [ f( x00) + 2 f( x1) + + 2 f( x7) + f( x8)] ≈5.49662Right Riemann Sum: ( ) ( )Trapezoidal Rule: ( ) 51 2Parabolic Rule: ( ) 50.125∫ x x + 1 dx ≈ [ f( x00) + 4 f( x1) + 2 f( x2) + + 4 f( x7) + f( x8)] ≈5.258035 61⎤+ 1 = + 1 5.2512⎥ =⎣ ⎦01 2 ⎡ 1 2Fundamental Theorem of Calculus: x( x ) dx ⎢ ( x )∫06. ( ) 3/2 4 −f x x h1( ) = + 1 ; =8= 0.375x 0 = 1.000f( x0) ≈ 2.8284x 1 = 1.375f( x1) ≈ 3.6601x 2 = 1.750f( x2) ≈ 4.5604x 3 = 2.125f( x3) ≈ 5.5243x 4 = 2.500f( x4) ≈ 6.5479x 5 = 2.875f( x5) ≈ 7.6279x 6 = 3.250f( x6) ≈ 8.7616x 7 = 3.625f( x7) ≈ 9.9464x 8 = 4.000f( x8) ≈ 11.1803Left Riemann Sum: ( )4 3/2∫ x+ 1 dx ≈ 0.375[ f( x10) + f( x1) + + f( x7)] ≈18.54644 3/2∫ x+ 1 dx ≈ 0.375[ f( x11) + f x2 + ... + f( x8)] ≈ 21.67844 3/2 0.375∫ x+ 1 dx ≈ [ f( x10) + 2 f( x1) + + 2 f( x7) + f( x8)] ≈ 20.112424 3/2 0.375x+ 1 dx ≈ [ f( x10) + 4 f( x1) + 2 f( x2) + + 4 f( x7) + f( x8)] ≈ 20.0979344 3/2 ⎡25/2⎤∫ x+ 1 dx = x 1 20.09791⎢ + ≈5⎥⎣ ⎦1Right Riemann Sum: ( ) ( )Trapezoidal Rule: ( )Parabolic Rule: ( )∫ Fundamental Theorem of Calculus: ( ) ( )7.LRS RRS MRS Trap Parabolicn = 4 0.5728 0.3728 0.4590 0.4728 0.4637n = 8 0.5159 0.4159 0.4625 0.4659 0.4636n = 16 0.4892 0.4392 0.4634 0.4642 0.46368.LRS RRS MRS Trap Parabolicn = 4 1.2833 0.9500 1.0898 1.1167 1.1000n = 8 1.1865 1.0199 1.0963 1.1032 1.0987n = 16 1.1414 1.0581 1.0980 1.0998 1.0986Instructor’s Resource Manual Section 4.6 283

9.LRS RRS MRS Trap Parabolicn = 4 2.6675 3.2855 2.9486 2.9765 2.9580n = 8 2.8080 3.1171 2.9556 2.9625 2.9579n = 16 2.8818 3.0363 2.9573 2.9591 2.957910.LRS RRS MRS Trap Parabolicn = 4 10.3726 17.6027 13.6601 13.9876 13.7687n = 8 12.0163 15.6314 13.7421 13.8239 13.7693n = 16 12.8792 14.6867 13.7625 13.7830 13.76931 2=− =2 3xx11. f′ ( x) ; f′′( x)The largest that f ′′ ( c)can be on [ ]( − )31, 3 occurs when 13 1 400( 2)≤0.01;n ≥ Round up: n = 12212n33 1 0.167∫ dx ≈ [ f ( x10 ) + 2 f ( x1 ) + + 2 f ( x11 ) + f ( x12)]x 2≈ 1.1007f′ x =− 1 ; f′′( x)=21+ 1+12. ( )( x)( x)2 3The largest that f ′′ ( c)can be on [ ]( ) 33−1 ⎛1⎞ 1000.01; n2 ⎜ ≤ ≥12n4⎟Round up: n = 5⎝ ⎠63 1 0.4∫ dx ≈ [ f ( x10 ) + 2 f ( x1 ) + + 2 f ( x4 ) + f ( x5)]1+x 2≈ 0.6956′ 1 ′′12 x4x13. f ( x) = ; f ( x) =−3/2The largest that f ′′ ( c)can be on [ ]( ) 3c = , and ( )f ′′ 1 = 211, 3 occurs when c = 1 , and f ′′ () 1 = .411, 4 occurs when c = 1 , and f ′′ () 1 = .44−1 ⎛1⎞ 9000.01; n2 ⎜ ≤ ≥12n4⎟Round up: n = 8⎝ ⎠164 0.375∫ x dx ≈ [ f ( x10 ) + 2 f ( x1 ) + +2 f ( x7 ) + f ( x8)]2≈ 4.6637284 Section 4.6 Instructor’s Resource Manual

1 1= = −2 x + 1 4 + 114. f′ ( x) ; f′′( x)( x ) 3/2The largest that f ′′ ( c)can be on [ ]( ) 311, 3 occurs when c = 1 , and f ′′ () 1 = .3/24 × 23−1 ⎛ 1 ⎞ 1000.01;2 ⎜ 3/2 ⎟ ≤ n ≥ Round up: n = 312n⎝4×2 ⎠12 23 0.667∫ x + 1 dx ≈ [ f( x10) + 2 f( x1) + 2 f( x2) + f( x3)]2≈ 3.443915. f′ ( x) f′′ ( x) f′′′( x)f( 4 ) ( x)1 2 6=− ; = ; =− ;2 3 4x x x24=5xThe largest that( 4f ) ( c ) can be on [ ]1, 3 occurs when c = 1 , and( 4f ) () 1 = 24.( 4−1)5( )24 ≤0.01; n ≈ 4.545 Round up to even: n = 64180n3 1 0.333∫ dx ≈ [ f ( x10 ) + 4 f ( x1 ) + ... + 4 f ( x5 ) + f ( x6)]x 3≈ 1.09891 1= ; = − ;3/22 x + 1 4 + 116. f′ ( x) f′′( x)( x )( x )3 ( 4( )) 15f′′′ x = ; f ( x)=−8 + 1 16 + 1( x )5/2 7/2The largest that( 4f ) ( c ) can be on [ ]4,8 occurs when c = 4 , and( ) ( 4)4 3f = .400 5( ) 58− 4 ⎛ 3 ⎞0.01; n 1.17534 ⎜ ⎟ ≤ ≈ Round up to even: n = 2180n⎝400 5 ⎠8 2x+ 1 dx ≈ ⎡f ( x0) + 4 f ( x1) + f ( x2)⎤ ≈10.54644 3⎣⎦∫Instructor’s Resource Manual Section 4.6 285

17.∫+ 2 ⎡a3 b 2 ⎤( ax + bx + c)dx = ⎢ x + x + cx3 2⎥⎣⎦m hm–hm+hm–ha 3 b 2 a 3 b 2= ( m+ h) + ( m+ h) + c( m+h)– ( m– h) – ( m– h) – c( m– h)3 2 3 2a 2 3 bh= (6mh+ 2 h) + (4 mh) + c(2 h)=2 2[ a (6 m + 2 h ) + b (6 m ) + 6 c ]3 23h[ f ( m− h ) + 4 f ( m ) + f ( m+h )]3h =2 2 2[ am ( – h ) + bm ( – h ) + c+ 4 am + 4 bm+ 4 c+ am ( + h ) + bm ( + h ) + c ]3h =2 2[ a (6 m + 2 h ) + b (6 m ) + 6 c ]318. a. To show that the Parabolic Rule is exact, examine it on the interval [m – h, m + h].Let∫m+h3 2f ( x) = ax + bx + cx+ d,thenf ( xdx )m−h a 4 4 b 3 3 c 2 2= ⎡( m+ h) −( m− h) ⎤+ ⎡( m+ h) −( m− h) ⎤+ ⎡( m+ h) −( m− h) ⎤+ d[( m+ h) −( m−h)]4⎣ ⎦ 3⎣ ⎦ 2⎣ ⎦a 3 3 b 2 3 c= (8mh+ 8 hm) + (6mh+ 2 h) + (4 mh) + d(2 h).4 3 2The Parabolic Rule with n = 2 givesm+h h3 3 2 2 3∫ f ( x) dx = [ f ( m − h) + 4 f ( m) + f ( m + h)] = 2am h + 2amh + 2bm h + bh + 2chm + 2dhm−h3 3a 3 3 b 2 3 c= (8mh+ 8 mh) + (6mh+ 2 h) + (4 mh) + d(2 h)4 3 2which agrees with the direct computation. Thus, the Parabolic Rule is exact for any cubic polynomial.( l−k)(4)b. The error in using the Parabolic Rule is given by En=− f ( m)for some m between l and k.4180n2 (3)However, f ′( x) = 3ax + 2 bx+ c, f′′( x) = 6ax+ 2 b, f ( x) = 6 a,and f (4) ( x ) = 0, so E = 0.19. The left Riemann sum will be smaller than ∫ f ( x ) dx .aIf the function is increasing, then ( ) < ( ) on the interval [ , ]bf xif x i + 15xix i + 1n. Therefore, the left Riemann sum willunderestimate the value of the definite integral. The following example illustrates this behavior:If f is increasing, then f′ ( c) > 0 for any c ( a b)∈ , . Thus, the errorpositive, then the Riemann sum must be less than the integral.( b−a)2( )En= f′c > 0 . Since the error is2n286 Section 4.6 Instructor’s Resource Manual

f a∫ x dx .20. The right Riemann sum will be larger than ( )If the function is increasing, then ( ) < ( ) on the interval [ , ]f xif x i + 1xix i + 1. Therefore, the right Riemann sumwill overestimate the value of the definite integral. The following example illustrates this behavior:If f is increasing, then f′ ( c) > 0 for any c ( a b)∈ , . Thus, the errornegative, then the Riemann sum must be greater than the integral.( b−a)2( )En= − f′c < 0 . Since the error is2nbf a∫ x dx .21. The midpoint Riemann sum will be larger than ( )If f is concave down, then f′′ ( c) < 0 for any c ( a b)∈ , . Thus, the erroris negative, then the Riemann sum must be greater than the integral.( b−a)3( )En= f′′c < 0 . Since the error224nbf a∫ x dx .22. The Trapezoidal Rule approximation will be smaller than ( )If f is concave down, then f′′ ( c) < 0 for any c ( a b)( b−a)∈ , . Thus, the error En= − f′′c > 0 . Since the212nerror is positive, then the Trapezoidal Rule approximation must be less than the integral.23. Let n = 2.kf ( x) = x ; h = ax0 = – af ( x0) = – akx 1 = 0f( x 1) = 0kx2= af ( x2) = aa k a k k∫ x dx ≈ [– a + 2⋅ 0 + a ] = 0– a 2aa k 1 k 1 1 [k 1 –(– )k 1 ]– ax dx ⎡1 x + ⎤k – a k 1a + a + 1∫ = ⎢ ⎥ =[k+ 1 –k+1= a a ] = 0⎣ + ⎦ +k + 1A corresponding argument works for all n.24. a. T ≈ 48.9414;3f ′ x = x( ) 43 3 2[4(3) – 4(1) ](0.25)T – ≈ 48.9414 – 0.5417 = 48.399712The correct value is 48.4 .3( )Instructor’s Resource Manual Section 4.6 287

. T ≈ 1.9886; f ′( x) = cosxπ( ) 2[cos π – cos 0]12T – ≈ 1.99998712The correct value is 2.25. The integrand is increasing and concave down. By problems 19-22,LRS < TRAP < MRS < RRS.26. The integrand is increasing and concave up. By problems 19-22,LRS < MRS < TRAP < RRS27.28.10A ≈ [75 + 2 ⋅ 71 + 2 ⋅ 60 + 2 ⋅ 45 + 2 ⋅ 45 + 2 ⋅ 52 + 2 ⋅ 57 + 2 ⋅ 60 + 59] = 4570 ft 223A ≈ [23 + 424 ⋅ + 223 ⋅ + 421 ⋅ + 218 ⋅ + 415 ⋅ + 212 ⋅ + 411 ⋅ + 210 ⋅ + 48 ⋅ + 0] = 465 ft 23V = A⋅6≈ 2790 ft 329.20A ≈ [0 + 4 ⋅ 7 + 2 ⋅ 12 + 4 ⋅ 18 + 2 ⋅ 20 + 4 ⋅ 20 + 2 ⋅ 17 + 4 ⋅ 10 + 0] = 2120 ft 234 mi/h = 21,120 ft/h(2120)(21,120)(24) = 1,074,585,600 ft 330. Using a right-Riemann sum,240Distance = ∫ vt ( ) dt≈8∑i=1vt ( ) Δt= ( 31+ 54 + 53+ 52 + 35 + 31+28) 360852= = 14.2 miles6031. Using a right-Riemann sum,Water Usage = ∫ F( t)dt10i=11200≈∑ Ft ( i ) Δ t= 12(71+ 68+ + 148)= 13,740gallons4.7 Chapter ReviewConcepts Test1. True: Theorem 4.3.D2. True: Obtained by integrating both sides ofthe Product Rule3. True: If F( x) = ∫ f( x) dx, f( x)is aderivative of F(x).i4. False:2f ( x) = x + 2x+ 1 and2gx ( ) = x + 7x− 5 are acounterexample.5. False: The two sides will in general differ bya constant term.6. True: At any given height, speed on thedownward trip is the negative ofspeed on the upward.7. True: a1+ a0 + a2 + a1+ a3 + a2+ + an−1+ an−2 + an + an−1= a0 + 2a1+ 2a2 + + 2a n −1+a n8. True:9. True:100 100 100∑ ∑ ∑(2i− 1) = 2 i−1i= 1 i= 1 i=12(100)(100 + 1)= − 100 = 10,000210 10 10 1002 2∑ ai + = ∑ai + ∑ai+ ∑i= 1 i= 1 i= 1 i=1( 1) 2 1= 100 + 2(20) + 10 = 15010. False: f must also be continuous except at afinite number of points on [a, b].11. True: The area of a vertical line segment is 0.288 Section 4.7 Instructor’s Resource Manual

12. False:1∫ − 1 x dx is a counterexample.28. False:f( x)3x= is a counterexample.13. False: A counterexample is⎧0, x ≠ 0f ( x)= ⎨⎩1, x = 01 2∫ f x ⎤⎦dx = 0 .x is continuous, thenwith ⎡ ( )−1⎣If f ( )22[ f( x)] ≥ 0, and if [ f( x )] is greaterthan 0 on [a, b], the integral will bealso.14. False: D ⎡ xx ( ) ( )af z dz ⎤⎢ =⎣∫⎥⎦f x29. False: f ( x)= x is a counterexample.30. True: All rectangles have height 4,regardless of x i .b31. True: F( b) − F( a) =∫ F′( x)dxa32. False:= b∫ ( ) ( ) ( )a G′x dx = G b −Gaa−aaf ( xdx ) = 2 f( xdx )0∫ ∫ because fis even.15. True: sin x + cos x has period 2π , sox+ 2 π (sin x + cos xdx )∫xis independent of x.33. False:34. False:zt ()2t= is a counterexample.b f ( x ) dx = F ( b ) −F(0)0∫16. True: lim kf ( x) = k lim f ( x)andx→a x→alim [ f ( x) + g( x)]x→a= lim f( x) + lim g( x)when all thex→a x→alimits exist.17. True:13sin18. True: Theorem 4.2.Bx is an odd function.19. False: The statement is not true if c > d.20. False:⎡ x21 ⎤ 2xDx⎢∫dt0 2 ⎥ =2⎢⎣1+ t ⎥⎦1+x21. True: Both sides equal 4.22. True: Both sides equal 4.23. True: If f is odd, then the accumulationxF x = ∫ f t dt is even,function ( ) ( )and so is F ( x)0+ C for any C.35. True: Odd-exponent terms cancelthemselves out over the interval, sincethey are odd.36. False: a = 0, b = 1, f(x) = –1, g(x) = 0 is acounterexample.37. False: a = 0, b = 1, f(x) = –1, g(x) = 0 is acounterexample.38. True: a1+ a2 + a3 + + an≤ a1 + a2 + a3 + + anbecauseany negative values of a i make theleft side smaller than the right side.39. True: Note that − f ( x) ≤ f ( x) ≤ f ( x)and use Theorem 4.3.B.40. True: Definition of Definite Integral41. True: Definition of Definite Integral242. False: Consider∫cos( x )dx24. False:25. False:26. False:27. False:f( x)f( x)f( x)2x= is a counterexample.2x= is a counterexample.2x= is a counterexample.2xf( x)= , v(x) = 2x + 1 is acounterexample.43. True. Right Riemann sum always bigger.44. True. Midpoint of x coordinate is midpointof y coordinate.45. False. Trapeziod rule overestimates integral.46. True. Parabolic Rule gives exact value forquadratic and cubic functions.Instructor’s Resource Manual Section 4.7 289

Sample Test Problems1.2.3.⎡⎢⎣⎡⎢⎣11 4 3 3/2⎤5x − x + 2x4⎥ =⎦0422 3 1⎤13x −3x− 3 x⎥ =⎦16π3⎡1 3 26⎤50 26 π⎢ y + 9cos y− 9cos13 y⎥ = − + −⎣⎦13 π 312.13.3 2 2u = 2y + 3y + 6 y, du = (6y + 6y + 6) dy( y2+ y+1)5 1 355 −1/5∫ dy =1 5 3 2112 3 6 6∫ u duy + y + y1⎡5 4/5⎤355 5 3554/5 114/5= u6⎢= −4 ⎥⎣ ⎦11244∑i=1⎡ 2⎛ i ⎞ ⎤⎛1⎞7⎢⎜ ⎟ −1⎥ ⎜ ⎟=⎢⎝2⎠ ⎥⎝2⎠4⎣ ⎦( )4.5.⎡⎢⎣91 2 3/2⎤77 77( y − 4) 8 33⎥ =− +⎦4323( )8⎡ 3 15 125 5(22 3)4/3⎤− − +⎢ z − =16 ⎥⎣⎦ 166.7.8.9.10.π /2⎡ 1 5 ⎤ 1⎢− cos x5⎥ =⎣ ⎦052 2 2u = tan(3x + 6 x), du = (6x+ 6)sec (3x + 6 x)1 2 1 36∫u du = u + C181 3 2π⎡1 3 2tan (3x + 6 x) ⎤ = tan (3π + 6 π )18 ⎣⎦0184 3u = t + 9, du = 4t dt1 25 −1/2 1 1/225u du ⎡u⎤ 14∫= =9 2⎣⎦921⎡3 5 5/3⎤ 3 5/3 5/3( t 5) ⎡37 6 ⎤ 46.95⎢ +5⎥ = − ≈⎣⎦125⎣ ⎦3⎡ 1 ⎤ 4⎢ ⎥ =3⎢⎣9y−3y⎥⎦272∫211. ( + 1)sin( + 2 + 3)x x x dx2( )( )1= sin 2 3 2 22∫ x + x+ x+dx1= sin2∫ udu1 cos2=− ( x + 2 x+ 3 ) + C214.15.16.17.18.19.1 1f′ ( x) = , f′(7) =x + 3 103 2(2 1)0 − x +∫dx( 5 4 1)3= ∫ x + − x+dx03⎡1 2 8 3/2⎤5= ⎢ x + 5 x− ( x+ 1)2 3⎥ =⎣⎦0651 5 2 3 1⎡23 3/2⎤3x x − 4 dx = ( x −4)5− 2∫23⎢3⎥⎣ ⎦2= 29444⎛ 1 ⎞ ⎡ 1⎤395 dx 5x2 ⎜ −2 ⎟ = ⎢ + =xx⎥⎝ ⎠ ⎣ ⎦24∫ni i−1∑ (3 − 3 )i=12 3 2 n n−1= (3 − 1) + (3 − 3) + (3 − 3 ) + + (3 −3 )n= 3 − 110 10 102 2∑(6i − 8 i) = 6∑i −8∑ii= 1 i= 1 i=1⎡10(11)(21) ⎤ ⎡10(11)⎤= 6 ⎢ − 8 = 18706 ⎥ ⎢ 2 ⎥⎣ ⎦ ⎣ ⎦290 Section 4.7 Instructor’s Resource Manual

20. a.1 1 1 13+ + =2 3 4 12e.−2 2f x dx f x dx0 0∫ ∫( − ) =− ( ) =−2b. 1 + 0 + (–1) + (–2) + (–3) + (–4) = –924. a.c.21. a.2 21+ + 0− − 1=02 2781∑n=2 nb.502n∑ nxn=14x0∫1 1− 1 dx = (1)(1) + (3)(3) = 52 222.b.23. a.b.c.n ⎡ 2⎛3i⎞ ⎤⎛3⎞A = lim ∑ ⎢16−⎜ ⎟ ⎥⎜ ⎟n→∞ i = 1 ⎢ ⎝ n ⎠ ⎥⎝n⎣ ⎦ ⎠⎧ n⎪ ⎡48 27= lim ⎨∑⎢ −n→∞ ⎪⎩i= 1 ⎣ n2i3n⎧⎪48 27= lim ⎨ 1−n→∞⎪⎩⎤⎫⎪⎥⎬⎦⎪⎭n n2∑ 3 ∑ in i= 1 n i=1⎫⎪⎬⎪⎭⎧ 27 ⎡nn( + 1)(2n+ 1) ⎤⎫= lim ⎨48−n→∞3 ⎢⎬n 6⎥⎩ ⎣⎦⎭⎧ 9⎡3 1 ⎤⎫= lim ⎨48 − 2n→∞2⎢ + +n 2 ⎥⎬⎩ ⎣ n ⎦⎭= 48 – 9 = 392 0 2f ( xdx ) = f( xdx ) + f( xdx )1 1 0∫ ∫ ∫= –4 + 2 = –20 1f x dx1 0∫ ∫( ) =− f( x) dx =−42 2f u du0 0∫ ∫23 ( ) = 3 f ( u) du = 3(2) = 6d. ∫ [ 2 ( ) − 3 ( )]0g x f x dx2 2g x0 0∫ ∫= 2 ( ) −3 f( x)dx= 2(–3) – 3(2) = –1225. a.∫40 x dx = 1+ 2+ 3=64 x x dx 4 xdx 4 x dx0 0 041 2c. ∫ ( − )= ∫ −∫ ⎡62 x ⎤⎢ ⎥ − = 8 – 6 = 2⎣ ⎦02 2f x dx−2 0∫ ∫( ) = 2 f( x) dx = 2( − 4) =−8b. Since f ( x) ≤ 0 , f ( x) f ( x)c.∫2 2( ) =−∫( )f x dx f x dx−2 −22∫2−2gxdx ( ) = 0d. ∫ [ ( ) + ( − )]∫0( )=− and=− 2 f x dx = 82f x f x dx−22 2= 2 ∫ f ( xdx ) + 2 f( xdx )0 ∫ 0= 4(–4) = –16Instructor’s Resource Manual Section 4.7 291

26.27.20e. ∫ [ 2 ( ) + 3 ( )]f.g x f x dx2 2= 2 ∫ g( xdx ) + 3 f( xdx )0 ∫ 0= 2(5) + 3(–4) = –20 2gxdx ( ) =− gxdx ( ) =−5−2 0∫ ∫100 3 5100 ( x +−sin x) dx = 0∫28. a.−1 2 2∫ 4 3x dx =−3 c ( − 1 + 4)⎡ 3−12x 9c⎣⎤ ⎦= −42c = 7c =− 7 ≈− 2.65b.c.1G′ ( x)=2x + 12xG′ ( x)=4x + 13x1G′ ( x)= −6 2x + 1 x + 12c.d.e.f.30. a.b.1 x 1G′ ( x) =− f( z) dz f( x)2 ∫ +x 0 xxG′ ( x) = ∫ f( t)dt0g( x) dg( u)g( x)Gx ( ) = ∫ du=[ gu ( )]00du= g( gx ( )) − g(0)G′ ( x) = g′ ( g( x)) g′( x)−xxGx ( ) = ∫ f( − tdt ) = fu ( )( −du)0 ∫ 0x =− ∫ f ( u ) du0G′ ( x) = − f( x)∫∫4 2 3/24 160 xdx= ⎡x⎤ =3⎣⎦033 2 1 33 261 x dx = ⎡x⎤ =3⎣⎦135x1 5x1 2x131. f ( x)= ∫ dt = dt dt2xt∫ −1 t∫ 1 t1 1f ′ () x = 5 2 05x⋅ − 2x⋅ =29. a.2G′ ( x) = sin xb. G′ ( x) = f( x+ 1) − f( x)32. Left Riemann Sum:Right Riemann Sum:Midpoint Riemann Sum:∫1211+x4dx ≈ 0.125[ f( x ) + f( x ) +…+ f( x )] ≈0.23190 1 72 1dx ≈ 0.125[ f( x1 41) + f( x2) +…+ f( x8)] ≈0.17671+x∫2 1dx ≈ 0.125[ f( x1 40.5) + f( x1.5) +…+ f( x7.5)] ≈0.20261+x∫33.2 1 0.125dx ≈ [ f ( x1 40 ) + 2 f ( x1 ) +…+ 2 f ( x7 ) + f ( x8)] ≈ 0.20431+x 2∫( − )4( )2 42 44 c (5c− 3) (4)(2 ) (5)(2 ) 3f ''( c) = ≤ = 1544 3 3(1 + c ) 1+13(2 −1) 1 154En=− f ''( c) = f ''( c) ≤ ≈ 0.20052(12)8 (12)(64) 768Remark: A plot off '' shows that in fact f ''( c ) < 1.5 , so E < 0.002 .n292 Section 4.7 Instructor’s Resource Manual

34.35.4 1 0.5dx ≈ [ f ( x00 ) + 4 f ( x1 ) + 2 f ( x2 ) +…+ 4 f ( x7 ) + f ( x8)] ≈ 1.10501+2x3∫f( 4 ) ( c)384= ≤384( 1+2c)5( − ) ( 4 ) ( )5 54 0 4 ⋅384 8En=− ⋅f c ≤ =4 4180⋅8 180⋅815( + )4( )2 42 44 c (5c− 3) (4)(2 ) (5)(2 ) 3f ''( c) = ≤ = 1664 3 3(1 + c ) 1+13(2 −1) 1 166En=− f ''( c) = f ''( c) ≤ < 0.00012 2 212n 12n 12n2 166n > ≈ 138,333 so n > 138,333 ≈ 371.9 Round up to n = 372 .(12)(0.0001)Remark: A plot of f '' shows that in fact f ''( c ) < 1.5 which leads to n = 36 .36.f( 4 ) ( c)384= ≤384( 1+2c)5( − ) ( 4 ) ( )5 54 0 4 ⋅384En=− ⋅f c ≤ < 0.00014 4180⋅n180⋅n54 4 ⋅384n > ≈ 21,845,333 , so n ≈ 68.4 . Round up to n = 69 .180 0.0001( )37. The integrand is decreasing and concave up. Therefore, we get:Midpoint Rule, Trapezoidal rule, Left Riemann SumReview and Preview Problems21 1 1 1 11.⎛ ⎞− ⎜ ⎟ = − =2 ⎝2⎠2 4 4( )2 2 2x+ h− x + x+ h −x6. ( ) ( )( 2 )2 2= h + xh+h222.x −2x3. the distance between ( )3 31, 4 and ( 4, 4) is 4 -14. the distance between⎛ y ⎞, y and 3( y , y ) is 3 y⎜ ⎟y −⎝ 4 ⎠45. the distance between (2,4) and (1,1) is2 2(2 − 1) + (4 − 1) = 107.8.9.10.2= ( ⋅ 2 )0.4=1.6V π π2 2= [ (4 − 1 )]1=15V π π2 2V = [ π ( r2 −r1)] Δ x2 2= [ (5 − 4.5 )]6 = 28.5V π πInstructor’s Resource Manual Review and Preview 293

5 42⎡xx ⎤x − 2x + 2 dx = ⎢ − + 2x⎥−1⎢⎣5 2 ⎥⎦−12 4 311. ( )∫12 ⎛ 27 ⎞ 51= −⎜− ⎟=5 ⎝ 10⎠1012.13.∫y dy = 3 ⋅ y = 3 ⋅ 3 ≈ 3.745 53 23 533 530 02 4 3 522⎛ 161x x ⎞ ⎡ x x ⎤− + dx = x − + =0 ⎜ 2 16 ⎟⎢ ⎥⎝ ⎠ ⎣⎢6 80⎦⎥150∫14. Let u = 1 +9x; then du =9dxand4 49 4 4 2 31+ x dx = u du = u 2 + C4 9 9 3∫ ∫3 8 ⎛ 9 2127 4 x ⎞= ⎜ + ⎟ +⎝ ⎠C344⎡2 ⎤9 8 9Thus, 1 xdx ⎢ ⎛ ⎞1 x ⎥∫ + = +1 4 ⎢⎜ ⎟27⎝4 ⎠ ⎥⎣⎦1328 ⎛32 13 ⎞= 10 − ≈7.6327 ⎜ 8 ⎟⎝⎠294 Review and Preview Instructor’s Resource Manual

CHAPTER5Applications of the Integral5.1 Concepts Reviewbb1. ∫ f ( xdx ) ; − f( xdx )a ∫ a2. slice, approximate, integrate3. g( x) − f( x); f( x) = g( x)dc4. ∫ [ ( ) − ( )]qy py dyProblem Set 5.11. Slice vertically.2ΔA ≈ ( x + 1) Δ x22 2 ⎡13 ⎤A = ( x + 1) dx = x x–1⎢ +3 ⎥⎣ ⎦−1∫ = 62. Slice vertically.3ΔA ≈( x − x+ 2) Δ x22 3 ⎡1 4 1 2 ⎤A= ( x − x+ 2) dx = x x 2x−1⎢ − +4 2 ⎥⎣⎦−1∫ = 3343. Slice vertically.2 2ΔA ≈ ⎡( x + 2) −( −x) ⎤Δ x = ( x + x+ 2) Δx⎣⎦22 2 ⎡1 3 1 2 ⎤A = ∫ ( x + x+ 2) dx = x x 2x−2⎢ + +3 2 ⎥⎣⎦−28 8 40= ⎛ ⎜ + 2+ 4 ⎞ ⎟− ⎛ ⎜− + 2− 4⎞⎟=⎝3 ⎠ ⎝ 3 ⎠ 34. Slice vertically.2 2ΔA ≈− ( x + 2x−3) Δ x = ( −x − 2x+ 3) Δ x11 2 ⎡ 1 3 2 ⎤A = ( −x − 2x+ 3) dx = x x 3x−3⎢− − +3⎥⎣⎦−3∫ = 32 35. To find the intersection points, solve222–x = x .x + x− 2=0(x + 2)(x – 1) = 0x = –2, 1Slice vertically.2 2ΔA ≈ ⎡(2 −x ) −x⎤Δ x = ( −x −x−2)Δx⎣ ⎦A =1 2∫ (– – 2)–2 x x+dx ⎡ 1 3 1 2 ⎤= ⎢– x – x + 2x3 2⎥⎣⎦1 1 8 9= ⎛ ⎜− − + 2 ⎞ ⎟− ⎛ ⎜ −2− 4⎞⎟=⎝ 3 2 ⎠ ⎝3 ⎠ 21−26. To find the intersection points, solve2x2− − 6=0x+ 4= − 2.x x(x + 2)(x – 3) = 0x = –2, 3Slice vertically.2 2Δ ≈ ⎡( + 4) −( −2) ⎤Δ = ( − + + 6) Δ⎣⎦33 2 ⎡ 1 3 1 2 ⎤A= ∫ ( − x + x+ 6) dx = x x 6x−2⎢− + +3 2 ⎥⎣⎦−2⎛ 9 8 125⎜− 9+ + 18 ⎞ ⎟− ⎛ ⎜ + 2− 12⎞⎟=⎝ 2 ⎠ ⎝3 ⎠ 6A x x x x x x7. Solve x 3 − x 2 − 6x= 0.2xx ( − x− 6) = 0x(x + 2)(x – 3) = 0x = –2, 0, 3Slice vertically.3 2ΔA1 ≈( x −x −6 x)Δ x3 2 3 2ΔA2 ≈−( x −x −6 x) Δ x = ( − x + x + 6 x)Δ xA = A1+A20 3 2 3 3 2= ∫ ( x −x − 6 x) dx+ ( − x + x + 6 x)dx−2 ∫ 003⎡1 4 1 3 2⎤⎡ 1 4 1 3 2⎤= ⎢ x − x −3x4 3 ⎥ + ⎢− x + x + 3x⎣⎦−24 3 ⎥⎣⎦0⎡ ⎛ 8 ⎞⎤⎡ 81 ⎤= ⎢0− ⎜4+ − 12⎟+ − + 9+ 27−03⎥ ⎢ 4⎥⎣ ⎝ ⎠⎦⎣ ⎦16 63 253= + =3 4 128. To find the intersection points, solve2− x + 2 = x .2x + x− 2=0(x + 2)(x – 1) = 0x = –2, 1Slice vertically.2 2ΔA ≈ ⎡( − x+ 2) −x ⎤Δ x = ( −x − x+ 2) Δx⎣⎦11 2 ⎡ 1 3 1 2 ⎤A = ∫ ( −x − x+ 2) dx = x x 2x−2⎢− − +3 2 ⎥⎣⎦−21 1 8 9= ⎛ ⎜− − + 2 ⎞ ⎟− ⎛ ⎜ −2− 4⎞⎟=⎝ 3 2 ⎠ ⎝3 ⎠ 2Instructor’s Resource Manual Section 5.1 295

9. To find the intersection points, solve2y+ 1= 3– y .2y + y− 2=0(y + 2)(y – 1) = 0y = –2, 1Slice horizontally.2 2ΔA ≈ ⎡(3 − y ) − ( y+ 1) ⎤Δ y = ( −y − y+ 2) Δy⎣⎦A = 1 2∫ (– – 2)–2 y y+dy ⎡ 1 3 1 2 ⎤= ⎢– y – y + 2y3 2 ⎥⎣⎦1 1 8 9= ⎛ ⎜− − + 2 ⎞ ⎟− ⎛ ⎜ −2− 4⎞⎟=⎝ 3 2 ⎠ ⎝3 ⎠ 210. To find the intersection point, solve11.12.2y + y− 6=0(y + 3)(y – 2) = 0y = –3, 2Slice horizontally.2 2ΔA ≈ ⎡(6 − y) − y ⎤Δ y = ( −y − y+ 6) Δy⎣ ⎦1−22y = 6 − y .22 2 1 3 1 2A ( y y 6) dy ⎡⎤= − − + = y y 6y0⎢− − +3 2 ⎥⎣⎦0∫ = 223⎛ 1 2 ⎞ΔA ≈⎜3 − x ⎟Δx⎝ 3 ⎠33 ⎛ 1 2⎞ ⎡ 1 3⎤A= ∫ 3 x dx 3x x 9 3 60 ⎜ − ⎟ = − = − =3 ⎢ 9 ⎥⎝ ⎠ ⎣ ⎦0Estimate the area to be (3)(2) = 6.13.14.15.2ΔA ≈−( x− 4)( x+ 2) Δ x = ( − x + 2x+ 8) Δ x33 2 1 3 2A ( x 2x 8) dx ⎡⎤= ∫ − + + = x x 8x0⎢− + +3⎥⎣⎦0= –9 + 9 + 24 = 24Estimate the area to be (3)(8) = 24.2 2ΔA ≈−( x −4x−5) Δ x = ( − x + 4x+ 5) Δ x4A 4 2= ∫ 1 ( − x + 4x + 5) dx ⎡ 1 3 22 5−3 x x x ⎤= ⎢− + + ⎥⎣⎦−1⎛ 64 ⎞ ⎛1 ⎞ 100= ⎜− + 32 + 20⎟− ⎜ + 2 − 5⎟= ≈33.33⎝ 3 ⎠ ⎝3 ⎠ 3⎛ 1⎞ 1Estimate the area to be (5) ⎜6 ⎟ = 32 .⎝ 2⎠22ΔA ≈(5 x−x ) Δ x33 2 ⎡5 2 1 3⎤A= ∫ (5 x− x ) dx = x x 11.331⎢ − ≈2 3⎥⎣ ⎦1⎛ 1 ⎞Estimate the area to be (2) ⎜5 ⎟ = 11 .⎝ 2 ⎠1 (2ΔA ≈− x − 7) Δ x422 1 2 1 1 3A ( x 7) dx ⎡ ⎤= ∫ − − = − x 7x0 4 4⎢ −3⎥⎣ ⎦01⎛8 ⎞ 17=− ⎜ − 14⎟= ≈2.834⎝3 ⎠ 6⎛ 1 ⎞Estimate the area to be (2) ⎜1 ⎟ = 3.⎝ 2 ⎠296 Section 5.1 Instructor’s Resource Manual

16.19.17.3ΔA1≈ −x Δ x3ΔA2≈ x Δ x0 3 3 3A = A1+ A2= ∫ − x dx+x dx−3 ∫ 00 3⎡ 1 4⎤ ⎡1 4⎤ ⎛81⎞ ⎛81⎞81= ⎢− x x4⎥ + ⎢−3 4⎥ = ⎜ ⎟+ ⎜ ⎟=⎣ ⎦ ⎣ ⎦0⎝ 4 ⎠ ⎝ 4 ⎠ 2= 40.5Estimate the area to be (3)(7) + (3)(7) = 42.[ ( 3)( 1) ]ΔA ≈ x− x− x− Δ x2 2= ⎡x −( x − 4x+ 3) ⎤Δ x = ( − x + 5x−3)Δx⎣⎦To find the intersection points, solvex = (x – 3)(x – 1).2x − 5x+ 3=05± 25−12x =25±13x =25+132 2A = ∫ ( − x + 5x−3)dx5−1323ΔA1≈ − xΔx3ΔA2≈ xΔx0 3 2 3A = A1+ A2= ∫ − xdx+xdx−2 ∫ 00 2 3 3⎡ 3 4/3⎤ ⎡3 4/3⎤⎛3 2 ⎞ ⎛3 2 ⎞= ⎢− xx4 ⎥ + ⎢ = +−2 4 ⎥⎣ ⎦ ⎣ ⎦ ⎜0 2 ⎟ ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠3= 3 2 ≈ 3.78Estimate the area to be (2)(1) + (2)(1) = 4.20.5+13⎡ 1 3 5 2 ⎤ 2 13 13= ⎢− x + x − 3x7.813 2 ⎥ = ≈⎣⎦5−13 62Estimate the area to be 1 (4)(4) = 8 .218.ΔA ≈ −( x −10) Δ x = (10 − x)Δ x99 2 32A (10 x) dx ⎡ ⎤= ∫ − = 10x x0⎢ −3⎥⎣ ⎦0= 90 – 18 = 72Estimate the area to be 9 · 8 =72.( )ΔA ≈ ⎡ x ( x 4) ⎤⎣− −⎦Δ x = x − x+ 4 ΔxTo find the intersection point, solvex = ( x− 4) .2x = ( x−4)2x − 9x+ 16=09± 81−64x =29±17x =2⎛ 9− 17 9+17 ⎞⎜x = is extraneous so x = .2 2 ⎟⎝⎠Instructor’s Resource Manual Section 5.1 297

21.9+17A = 2∫ ( x − x+4)dx09+17⎡2 3/2 1 2 ⎤ 2= ⎢ x − x + 4x3 2⎥⎣⎦03/2 22 ⎛9 + 17 ⎞ 1 ⎛9 + 17 ⎞ ⎛9 + 17 ⎞= 43⎜ − +2 ⎟ 2⎜ 2 ⎟ ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠3/22 ⎛9 + 17 ⎞ 23 17= 15.923⎜+ − ≈2 ⎟⎝ ⎠ 4 4Estimate the area to be 1 ⎛ 1 ⎞⎛5 5 1 ⎞⎜ ⎟⎜ ⎟ 151 = .2⎝ 2⎠⎝ 2⎠823.2x + 5x+ 6=0(x + 3)(x + 2) = 0x = –3, –2A −2 2= ∫ 3 ( − x − 5x − 6)−dx−2⎡ 1 3 5 2 ⎤= ⎢− x − x −6x3 2⎥⎣⎦−3⎛8 ⎞ ⎛ 45 ⎞ 1= ⎜ − 10 + 12⎟−⎜9 − + 18⎟= ≈0.17⎝3 ⎠ ⎝ 2 ⎠ 6Estimate the area to be 1 ⎛(1) 5 42 ⎞⎜ − ⎟=12 ⎝ 3⎠6.22.2 2 2ΔA ≈ ⎡−x −( x −2 x) ⎤Δ x = ( − 2x + 2 x)Δx⎣⎦To find the intersection points, solve2 2− x = x − 2x.22x− 2x= 02x(x – 1) = 0x = 0, x = 111 2 ⎡ 2 3 2⎤A= ∫ ( − 2x + 2 x)dx = x x0⎢− +3⎥⎣ ⎦02 1=− + 1= ≈ 0.333 3⎛1⎞⎛1⎞ 1Estimate the area to be ⎜ ⎟⎜ ⎟ = .⎝2⎠⎝2⎠424.2ΔA ≈(8 y− y ) Δ yTo find the intersection points, solve28y− y = 0.y(8 – y) = 0y = 0, 888 2 ⎡ 2 1 3⎤A= ∫ (8 y− y ) dy = 4y y0⎢ −3⎥⎣ ⎦0512 256= 256 − = ≈ 85.333 3Estimate the area to be (16)(5) = 80.2ΔA ≈ ⎡( x −9) −(2x− 1)( x+ 3) ⎤Δx⎣⎦2 2= ⎡( x −9) − (2x + 5x−3)⎤Δx⎣⎦2= ( −x −5x−6)Δ xTo find the intersection points, solve2(2x− 1)( x+ 3) = x − 9 .2ΔA ≈(3 − y)( y+ 1) Δ y = ( − y + 2y+ 3) Δ y33 2 ⎡ 1 3 2 ⎤A = ∫ ( − y + 2y+ 3) dy = y y 3y−1⎢− + +3⎥⎣⎦−1⎛1 ⎞ 32= ( − 9 + 9 + 9) − ⎜ + 1− 3⎟= ≈10.67⎝3 ⎠ 3⎛ 1 ⎞Estimate the area to be (4) ⎜2 ⎟ = 10 .⎝ 2 ⎠298 Section 5.1 Instructor’s Resource Manual

25.27.2ΔA ≈ ⎡( − 6y + 4 y) −(2−3 y)⎤Δy⎣⎦2= ( − 6y + 7y−2)Δ yTo find the intersection points, solve2− 6y + 4y = 2−3 y.26y− 7y+ 2=0(2y−1)(3y− 2) = 01 2y = ,2 32/3A = 2/3 2∫1/2 ( − 6y + 7y−2)dy ⎡ 3 7 2 ⎤= ⎢− 2y + y −2y2⎥⎣⎦1/2⎛ 16 14 4 ⎞ ⎛ 1 7 ⎞ 1= ⎜− + − ⎟−⎜− + − 1 ⎟ = ≈ 0.0046⎝ 27 9 3 ⎠ ⎝ 4 8 ⎠ 216Estimate the area to be1⎛1⎞⎛1⎞ 1⎛1⎞⎛1⎞1⎜ ⎟⎜ ⎟− ⎜ ⎟⎜ ⎟=.2⎝2⎠⎝5⎠ 2⎝2⎠⎝6⎠12028.2 2 2ΔA ≈ ⎡(3 − y ) −2 y ⎤Δ y = ( − 3y + 3) Δy⎣⎦To find the intersection points, solve2 22y= 3− y .23y − 3=03(y + 1)(y – 1) = 0y = –1, 11 2 31A = ( 3y 3) dy ⎡ y 3y⎤∫ − + = − +−1 ⎣ ⎦−1= (–1 + 3) – (1 – 3) = 4Estimate the value to be (2)(2) = 4.26.2 2ΔA ≈ ⎡( y+ 4) −( y −2 y) ⎤Δ y = ( − y + 3y+ 4) Δy⎣⎦To find the intersection points, solve2y − 2y = y+ 4.2y −3y− 4=0(y + 1)(y – 4) = 0y = –1, 44A 4 2= ∫ 1 ( − y + 3y + 4)−dy ⎡ 1 3 3 2 ⎤= ⎢− y + y + 4y3 2⎥⎣⎦−1⎛ 64 ⎞ ⎛1 3 ⎞ 125= ⎜− + 24 + 16⎟− ⎜ + − 4⎟= ≈ 20.83⎝ 3 ⎠ ⎝3 2 ⎠ 6Estimate the area to be (7)(3) = 21.4 4 4ΔA ≈ ⎡(8 −4 y ) −(4 y ) ⎤Δ y = (8 −8 y ) Δy⎣⎦To find the intersection points, solve4 44y= 8− 4y.48y = 84y = 1y = ±111 4 ⎡ 8 5⎤A= ∫ (8 − 8 y ) dy = 8y y−1⎢ −5⎥⎣ ⎦−1⎛ 8⎞ ⎛ 8⎞64= ⎜8− ⎟−⎜− 8+ ⎟= = 12.8⎝ 5⎠ ⎝ 5⎠5⎛ 1 ⎞Estimate the area to be (8) ⎜1 ⎟ = 12 .⎝ 2 ⎠Instructor’s Resource Manual Section 5.1 299

29.3y = xy = x + 630.2y+ x = 0Let R 1 be the region bounded by 2y + x = 0,y = x + 6, and x = 0.0 ⎡ ⎛ 1 ⎞⎤A( R1) = ∫ ( x+ 6) − − x dx−4⎢ ⎜ ⎟2 ⎥⎣ ⎝ ⎠⎦0 ⎛364 2 x ⎞= ∫ + dx−⎜ ⎟⎝ ⎠Let R 2 be the region bounded by y = x + 6,3y = x , and x = 0.∫2 32 3A( R2) = ⎡( x+ 6) −x ⎤dx0 ⎣ ⎦= 0 ( − x + x+6) dxA( R) = A( R1) + A( R2)0 ⎛3⎞ 2 3= ∫ x 6 dx ( x x 6) dx−4 ⎜ + ⎟ + − + +⎝2 ∫ ⎠ 00 2⎡3 2 ⎤ ⎡ 1 4 1 2 ⎤= ⎢ x + 6x x x 6x4⎥ + ⎢− + +−4 4 2⎥⎣ ⎦ ⎣ ⎦0= 12 + 10 = 22∫An equation of the line through (–1, 4) and (5, 1)1 7is y = − x+ . An equation of the line through2 2(–1, 4) and (2, –2) is y = –2x + 2. An equation ofthe line through (2, –2) and (5, 1) is y = x – 4.Two integrals must be used. The left-hand part ofthe triangle has area2 ⎡ 1 7 ⎤ 2 ⎛3 3⎞∫ − x + − ( − 2 x+2) dx−1⎢2 2 ⎥ = ∫ x dx⎣⎦ −1⎜ + ⎟⎝2 2⎠.The right-hand part of the triangle has area5⎡ 1 7 ⎤ 5 ⎛ 3 15⎞∫ x ( x 4) dx x dx2 ⎢− + − −2 2 ⎥ = ∫ − +2⎜ ⎟ .⎣ ⎦ ⎝ 2 2 ⎠The triangle has area2 ⎛3 3⎞ 5⎛ 3 15⎞∫ x dx x dx−1⎜ + ⎟ + − +2 2∫ 2⎜ ⎟⎝ ⎠ ⎝ 2 2 ⎠2 5⎡3 2 3 ⎤ ⎡ 3 2 15 ⎤= ⎢ x + x x x4 2⎥ + ⎢− +4 2⎥⎣ ⎦ ⎣ ⎦27 27 27= + = = 13.54 4 2−1 231.32.9 2 3 29(3t − 24t+ 36) dt = ⎡t − 12t + 36t⎤−1 ⎣⎦−12∫ = (729 – 972 + 324) – (–1 – 12 – 36) = 130The displacement is 130 ft. Solve 3t− 24t+ 36= 0.3(t – 2)(t – 6) = 0t = 2, 6⎧ 2⎪3t − 24t+ 36 t ≤ 2, t ≥6Vt () = ⎨2 ⎪ ⎩ − 3t + 24t − 36 2 < t < 69 2 2 23 − 24 + 36 = (3 24 36)−1 ∫ − +−13 223 26⎡t 12t 36t⎤ ⎡ t 12t 36t⎤⎣ ⎦−1 ⎣ ⎦2∫t t dt t t dt= − + + − + −The total distance traveled is 194 feet.∫3 /20π ⎛1 ⎞ ⎡1 1 ⎤⎜ + sin 2t⎟dt = t−cos 2t2⎢2 2⎥⎝ ⎠ ⎣ ⎦3 π /206 2 9 2∫2 ∫ 63 2912 36 ⎤⎦6+ ( − 3t + 24t− 36) dt+ (3t − 24t+36) dt+ ⎡t − t + t⎣⎛3π1⎞ ⎛ 1⎞3π= ⎜ + ⎟−⎜0− ⎟= + 1⎝ 4 2⎠ ⎝ 2⎠4= 81 + 32 + 81 = 194The displacement is 3π 1 3.36 feet1 3π+ ≈ . Solve + sin 2 t = 0 for 0 ≤t≤ .4221 7π11πsin 2t =− ⇒ 2 t = , ⇒ 7 π 11 πt = ,2 6 6 12 12300 Section 5.1 Instructor’s Resource Manual

⎧ 1 7π 11π 3π+ sin 2t 0 ≤ t ≤ , ≤ t ≤1 ⎪+ sin 2 t =2 12 12 2⎨2 ⎪ 1 7π11π− − sin 2t< t

37AC ( ) = AB ( ) = (by symmetry)60 2 22∫–2 ∫ 03 203 22−2 0A( D) = [( x+ 6) – x ] dx+ [(– x+6) – x ] dx⎡ 1 1 ⎤ ⎡ 1 1 ⎤= ⎢– x + x + 6 x – x – x 6x3 2⎥ + ⎢ +3 2⎥⎣ ⎦ ⎣ ⎦44=3A(A) + A(B) + A(C) + A(D) = 36( )3 2 1 3–3(9– ) ⎡AA B C D x dx 9 x – ⎤+ + + = ∫= ⎢ x 3 ⎥⎣ ⎦= 3636. Let f(x) be the width of region 1 at every x.bA f( x) x, so A f( x)dxΔ ≈ Δ =∫ .1 1Let g(x) be the width of region 2 at every x.ΔA ≈ gx ( ) Δ x, so A =∫ gxdx ( ) .2 2abaSince f(x) = g(x) at every x in [a, b],bb1 = ( ) = ( ) =aa2∫ ∫ .A f x dx g x dx A37. The height of the triangular region is given byfor 0≤x ≤ 1. We need only show that theheight of the second region is the same in orderto apply Cavalieri'’s Principle. The height of thesecond region is2 2h = ( x − 2x+ 1) −( x − 3x+1)22 2= x − 2x+ 1− x + 3x−1= x for 0 ≤ x≤1.Since h1 = h2over the same closed interval, wecan conclude that their areas are equal.38. Sketch the graph.1 17πSolve sin x = for 0 ≤ x≤ .2 6π 5π 13π 17πx = , , ,6 6 6 6The area of the trapped region isπ /6 ⎛1 ⎞ 5 π /6⎛1 ⎞∫ − sin x dx +0⎜ ⎟sin x dx⎝2∫ −⎠ π /6⎜ ⎟⎝ 2 ⎠13 π / 6⎛1 ⎞ 17 π / 6⎛1 ⎞+ ∫ − sin x dx5 π /6⎜ ⎟ + sin⎝2∫ x − dx⎠ 13 π / 6⎜ ⎟⎝ 2 ⎠3−3π /6 5 π /6⎡1 ⎤ ⎡ 1 ⎤= ⎢ x+ cos x + −cosx−x2⎥ ⎢2⎥⎣ ⎦ ⎣ ⎦π⎡1cos2 x x ⎤ ⎡ 1 ⎤+ ⎢ + ⎥ + ⎢−cosx−x⎣ ⎦2⎥⎣ ⎦0 π /613 / 617 π / 65 π /613 π / 6⎛ π 3 ⎞ ⎛ π⎞ ⎛ 2π⎞= ⎜+ − 1 + 3− + 3+12 2 ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠⎛ π⎞π 7 3+ ⎜ 3− ⎟= + −1≈ 5.32⎝ 3⎠12 25.2 Concepts Review1.2.3.4.2π r h2 2π( R − r ) hπx4Δ x2 2π [( x + 2) −4]Δ xProblem Set 5.21. Slice vertically.2 2 4 2ΔV ≈π ( x + 1) Δ x =π ( x + 2x + 1) Δ x∫2 4 2V =π 0 ( x + 2x + 1) dx2⎡1 5 2 3 ⎤ ⎛32 16 ⎞ 206π=π ⎢ x + x + x25 3⎥ =π ⎜ + + ⎟=⎣ ⎦0⎝ 5 3 ⎠ 15≈ 43.142. Slice vertically.2 2 4 3 2ΔV ≈π− ( x + 4 x) Δ x =π( x − 8x + 16 x ) Δ x∫3 4 3 2V =π 0 ( x − 8x + 16 x ) dx3⎡1 5 4 16 3⎤=π⎢x − 2x + x5 3 ⎥⎣⎦0⎛243 ⎞=π⎜− 162 + 144 ⎟⎝ 5⎠153π= ≈ 96.135302 Section 5.2 Instructor’s Resource Manual

3. a. Slice vertically.2 2ΔV ≈π(4 −x ) Δ x∫2 2 4V 0 (16 – 8 + x ) dx256π= ≈ 53.6215b. Slice horizontally.x = 4– yNote that when x = 0, y = 4.( ) 22 4=π(16 − 8 x + x ) Δ xΔV ≈π 4 − y Δ y = π(4 − y)Δ y4 ⎡ 1 2 ⎤V =π ∫ (4– y) dy =π 4 y– y0⎢ 2 ⎥⎣ ⎦=π(16 − 8) = 8π≈25.134. a. Slice vertically.2ΔV ≈π(4 −2 x)Δ x0≤ x ≤ 222 2 1 3V (4 2 x) dx ⎡⎤=π∫− =π (4 2 x)0⎢− −6⎥⎣⎦032π= ≈ 33.513b. Slice vertically.yx = 2 −22⎛ y ⎞ΔV≈ π⎜2− ⎟ Δy⎝ 2 ⎠0 ≤ y ≤ 42 344⎛ y⎞ ⎡ 2 ⎛ y⎞⎤V =π∫2 dy 20 ⎜ − ⎟ =π⎢− ⎜ − ⎟ ⎥⎝ 2⎠ ⎢ 3⎝ 2⎣ ⎠ ⎥⎦016π= ≈ 16.7635.406.7.8.3 2 6ΔV ≈π( x ) Δ x =πx Δ x33 6 ⎡1 7⎤2187πV =π ∫ x dx =π x981.520 ⎢ = ≈7⎥⎣ ⎦072⎛1⎞⎛ 1 ⎞ΔV ≈π⎜⎟ Δ x =π⎜Δxx2 ⎟⎝ ⎠ ⎝ x ⎠44 1 ⎡ 1⎤ ⎛ 1 1⎞πV = π ∫ dx = π2 2 ⎢− x x⎥ =π⎜− + ⎟=⎣ ⎦2⎝ 4 2⎠4≈ 0.793/2 2 3ΔV ≈π( x ) Δ x =πx Δ x33 3 ⎡1 4⎤ ⎛81 16⎞V =π ∫ x dx =π x2 ⎢4⎥ =π⎜− ⎟⎣ ⎦2⎝ 4 4 ⎠65π= ≈ 51.05422⎛4x ⎞ xΔV ≈π Δ x = Δx⎜ π ⎟⎝ ⎠π444 x 1 ⎡1 5 ⎤ 1024V = ∫ dx = x = ≈65.190 π π⎢5⎥⎣ ⎦05πInstructor’s Resource Manual Section 5.2 303

9.12.22 2ΔV ≈ π⎛9 x⎞⎜ − ⎟ Δ x =π(9 −x ) Δx⎝ ⎠33 2 ⎡ 1 3⎤V =π∫(9 − x ) dx =π 9x x−2⎢ −3⎥⎣ ⎦−2⎡ ⎛ 8 ⎞⎤ 100π=π⎢(27 −9) −⎜− 18+ ⎟ = ≈104.723⎥⎣ ⎝ ⎠⎦32⎛ 2⎞⎛ 1 ⎞ΔV ≈π⎜ ⎟ Δ y = 4π Δyy2⎝ ⎠⎜y⎟⎝ ⎠66 1 ⎡ 1⎤ ⎛ 1 1⎞V = 4π ∫ dy = 4π 42 2 ⎢− yy⎥ = π⎜− + ⎟⎣ ⎦2⎝ 6 2⎠4π= ≈ 4.19310.13.11.2/3 2 4/3ΔV ≈π( x ) Δ x =πx Δ x2727 4/3 ⎡3 7/3⎤ ⎛6561 3 ⎞V =π ∫ x dx =π x1⎢7⎥ =π⎜− ⎟⎣ ⎦1⎝ 7 7⎠6558π= ≈ 2943.22714.( ) 2ΔV ≈π 2 y Δ y = 4πyΔy44 ⎡12 ⎤V = 4π ∫ ydy = 4π y 32 100.530 ⎢ = π≈2⎥⎣ ⎦02 2 4ΔV ≈ π( y ) Δ y =πy Δ y33 4 ⎡1 5 ⎤ 243πV =π ∫ y dy =π y152.680 ⎢ = ≈5⎥⎣ ⎦052/3 2 4/3ΔV ≈π( y ) Δ y =πy Δ y2727 4/3 ⎡3 7/3⎤6561πV =π ∫ y dy =π y0⎢ =7⎥⎣ ⎦07≈ 2944.57304 Section 5.2 Instructor’s Resource Manual

15.18. Sketch the region.8y2y = 6xy = 6x−11x16.3/2 2 3ΔV ≈ π( y ) Δ y = πy Δ y99 3 ⎡1 4⎤6561πV =π ∫ y dy =π y5153.000 ⎢ = ≈4⎥⎣ ⎦0422 2ΔV ≈ π⎛4 y⎞⎜ − ⎟ Δ y = π(4 − y ) Δy⎝ ⎠22 2 ⎡ 1 3⎤V =π∫(4 − y ) dy =π 4y y−2⎢ −3⎥⎣ ⎦−2⎡⎛ 8⎞ ⎛ 8⎞⎤ 32π=π⎢⎜8− ⎟−⎜− 8+ ⎟ = ≈33.513 3⎥⎣⎝ ⎠ ⎝ ⎠⎦3−4To find the intersection points, solve 6x= 6x.26( x − x) = 06x(x – 1) = 0x = 0, 12 2 2 2 4ΔV ≈π⎡(6 x) −(6 x ) ⎤Δ x = 36 π( x −x ) Δx⎣⎦11 2 4 1 3 1 5V 36 ( x x ) dx 36⎡ ⎤= π∫− = π x x0⎢ −3 5⎥⎣ ⎦0⎛1 1⎞24π= 36π⎜− ⎟= ≈15.08⎝3 5⎠519. Sketch the region.217. The equation of the upper half of the ellipse is21 x b 2 2y = b − or y = a − x .2a aa2b 2 2V =π∫a 2 ( a −x ) dx− a2 3ab π ⎡2 x ⎤= ⎢a x−⎥2a ⎢⎣3 ⎥⎦−a2 3 3b π ⎡⎛ 3 a ⎞ ⎛3 a ⎞⎤4 2= ⎢a − − − a + ⎥ = ab π2a ⎢⎜ 3 ⎟ ⎜ 3 ⎟⎣⎝ ⎠ ⎝ ⎠⎥⎦3xTo find the intersection points, solve = 2 x .22x= 4x42x − 16x= 0x(x – 16) = 0x = 0, 16222xxV ⎡ ⎛ ⎞Δ ≈π⎢( 2 x)− ⎤ x ⎛ 4x ⎞⎜ ⎟ ⎥Δ =π − Δx⎢ ⎝2⎠⎥ ⎜ 4 ⎟⎣⎦ ⎝ ⎠2 31616 ⎛ x ⎞ ⎡2 x ⎤V =π∫4x− dx =π 2x−0 ⎜⎢ ⎥4 ⎟⎝ ⎠ ⎣⎢12⎦⎥0⎛ 1024 ⎞ 512π=π⎜512 − ⎟= ≈536.17⎝ 3 ⎠ 3Instructor’s Resource Manual Section 5.2 305

20. Sketch the region.22.3 2 1 2y = x + 3, y = x + 516 16Sketch the region.rr 2 2 ⎡ 2 1 3⎤V =π∫( r − x ) dx =π r x xr−h⎢ −3⎥⎣ ⎦r−h= 1π (3 − )23 h r h21. Sketch the region.−142y1y yTo find the intersection points, solve = . 4 22y y=16 42y − 4y= 0y(y – 4) = 0y = 0, 4⎡ 2⎛2y ⎞ 2⎛ y⎞⎤ y yΔV ≈π⎢− ⎥Δ y =π ⎛ − ⎞Δy⎢⎜ 2 ⎟ ⎜ ⎟⎝ 4⎠⎥ ⎜ 4 16 ⎟⎣⎝ ⎠ ⎦ ⎝ ⎠424⎛ y y ⎞ ⎡ 2 3y y ⎤V =π∫−dy =π0 ⎜ 4 16 ⎟⎢ − ⎥⎝ ⎠ ⎢⎣8 48⎥⎦2π= ≈ 2.094432x023.To find the intersection point, solve3 2 1 2x + 3= x + 5.16 161 22 08 x − =2x − 16 = 0(x + 4)(x – 4) = 0x = –4, 42 24 ⎡⎛ 1 2 ⎞ ⎛ 3 2 ⎞ ⎤V = π∫⎢ x 5 2 x 3 2 dx0⎜ + − ⎟ − ⎜ + − ⎟ ⎥⎢⎝16 ⎠ ⎝16⎣⎠ ⎥⎦4 ⎡⎛ 1 4 3 2 ⎞=π∫90 ⎢⎜x − x + ⎟⎣⎝256 8 ⎠⎛ 9 4 3 2 ⎞⎤−⎜x − x + 1 dx256 8⎟⎥⎝⎠⎦44⎛1 4 ⎞ ⎡ 1 5 ⎤=π∫8 − x dx0⎜ ⎟ =π 8xx⎝ 32 ⎢ −⎠160⎥⎣ ⎦⎛ 32 ⎞ 128π=π⎜32 − ⎟= ≈80.42⎝ 5 ⎠ 5The square at x has sides of lengthshown.22⎛ 2 ⎞ 2 2⎜ ⎟−2 −2∫ ∫022 4−xV = 2 4− x dx = 4(4 −x ) dx⎝ ⎠32⎡ x ⎤ ⎡ ⎤, as⎛ 8 ⎞ ⎛ 8 ⎞ 128= 4⎢4x− ⎥ = 4 8 83⎢⎜ − ⎟−⎜− + ⎟ =3 3⎥⎢⎣⎥⎦⎝ ⎠ ⎝ ⎠ 3−2⎣⎦≈ 42.67306 Section 5.2 Instructor’s Resource Manual

24. The area of each cross section perpendicular to2 2the x-axis is1 (4) ⎛ 2 4 x⎞⎜ − ⎟= 4 4 −x.2 ⎝ ⎠The area of a semicircle with radius 2 is∫ 4− = 2π.2 2x dx−22 2∫ −2 V = 4 4 − x dx = 4(2 π ) = 8π≈25.1325. The square at x has sides of length cos x .π /2 π /2−π /2 −π /2∫V = cos xdx = [sin x ] = 226. The area of each cross section perpendicular to2 4 2 8 6 4the x-axis is [(1 −x ) −(1 − x )] = x − 2 x + x .∫1 8 6 4V =1 ( x − 2 x + x ) dx−1⎡1 9 2 7 1 5⎤16= ⎢ x − x + x9 7 5⎥ = ≈ 0.051⎣⎦ 315−127. The square at x has sides of length1021− x .1 ⎡ 32 x ⎤ 2V = ∫ (1 − x ) dx = ⎢x− ⎥ =0⎢⎣3 ⎥⎦3≈ 0.6728. From Problem 27 we see that horizontal crosssections of one octant of the common region aresquares. The length of a side at height y is2 2r y− where r is the common radius of thecylinders. The volume of the “+” can be foundby adding the volumes of each cylinder andsubtracting off the volume of the common region(which is counted twice). The volume of oneoctant of the common region isr 2 2 2 1 2 r∫ ( r − y ) dy = r y−y |0033 1 3 2 3= r − r = r3 3Thus, the volume of the “+” isV = vol. of cylinders - vol. of common region2 ⎛23⎞=2( π r l) − 8 ⎜ r ⎟⎝ 3 ⎠2 ⎛2 3⎞128= 2 π(2 )(12) − 8 ⎜ (2) ⎟= 96π−⎝3 ⎠ 32≈ 258.93 in29. Using the result from Problem 28, the volume ofone octant of the common region in the “+” isr 2 2 2 1 2 r∫ ( r − y ) dy = r y−y |0033 1 3 2 3= r − r = r3 3Thus, the volume inside the “+” for two cylindersof radius r and length L isV = vol. of cylinders - vol. of common region2 ⎛23⎞= 2( π r L) −8 ⎜ r ⎟⎝ 3 ⎠2 16 3= 2πr L−r330. From Problem 28, the volume of one octant of2 3the common region is r . We can find the3volume of the “T” similarly. Since the “T” hasone-half the common region of the “+” inProblem 28, the volume of the “T” is given byV = vol. of cylinders - vol. of common region2 ⎛23⎞=( π r )( L1+ L2) −4 ⎜ r ⎟⎝ 3 ⎠With r = 2, L1 = 12, and L2= 8 (inches), thevolume of the “T” isV = vol. of cylinders - vol. of common region2 ⎛23⎞=( π r )( L1+ L2) −4 ⎜ r ⎟⎝ 3 ⎠2 ⎛23⎞= ( π 2 )(12+ 8) −4 ⎜ 2 ⎟⎝3⎠64 3= 80 π − in33≈ 229.99 in31. From Problem 30, the general form for thevolume of a “T” formed by two cylinders withthe same radius isV = vol. of cylinders - vol. of common region2 ⎛23⎞=( π r )( L1+ L2) −4 ⎜ r ⎟⎝ 3 ⎠2 8 3= π r ( L1+ L2)− r332. The area of each cross section perpendicular to1 ⎡12 ⎤the x-axis is ( ) 2π x −x2 ⎢2⎥⎣ ⎦π 4 5/2= ( 2 ).8 x − x + xV π 1 ( 4 2 5/2=8∫ x − x + x)dx01π ⎡1 5 4 7/2 1 2⎤9π= x x x8⎢ − +5 7 2⎥ =⎣⎦0560≈ 0.050Instructor’s Resource Manual Section 5.2 307

33. Sketch the region.a. Revolving about the line x = 4, the radius of2 2/3the disk at y is 4− y = 4− y .V =π 8 (4 y 2/3 ) 2∫ − dy0 =π 8 (16 8 2/3 4/3∫ − y + y ) dy08⎡ 24 5/3 3 7/3⎤=π⎢16y− y + y5 7⎥⎣⎦0⎛ 768 384 ⎞=π⎜128− + ⎟⎝ 5 7 ⎠1024π= ≈ 91.9135b. Revolving about the line y = 8, the inner33 3/2radius of the disk at x is 8− x = 8− x .V 4 8 2 (8 3/2 ) 2=π ⎡ x ⎤∫ − − dx0 ⎣⎦=π 4 (16 3/2 3∫ x −x) dx04⎡32 5/2 1 4⎤⎛1024 ⎞=π⎢x − x5 4⎥ =π⎜− 64 ⎟⎣⎦0⎝ 5 ⎠704π= ≈ 442.34534. Sketch the region.a. Revolving about the line x = 4, the inner2 2/3radius of the disk at y is 4− y = 4− y .( ) 2V8 ⎡ 2 2/30 4 4 y ⎤=π∫⎢ − − ⎥⎣⎦dy=π 8 2/3 4/3∫ 0 (8 y −y) dy38⎡24 5/3 3 7/3⎤=π⎢y − y5 7⎥⎣⎦0⎛768 384 ⎞ 3456π=π⎜− ⎟= ≈310.21⎝ 5 7 ⎠ 35b. Revolving about the line y = 8, the radius of3 3/2the disk at x is 8− x = 8− x .V =π 4 3/2 2∫ 0 (8 −x) dx=π 4 3/2 3∫ (64 16 )0 − x + x dx4⎡ 32 5/2 1 4⎤=π⎢64x− x + x5 4⎥⎣⎦0⎡ 1024 ⎤ 576π=π⎢256 − + 64 = ≈361.915 ⎥⎣⎦ 535. The area of a quarter circle with radius 2 is2 2∫ 4 − y dy =π.02 ⎡ 2 22 4 y 4 y⎤∫ − + − dy0 ⎢⎣⎥⎦2 2 2 2= 2∫ 4 − y dy + (4 − y ) dy0 ∫ 021 38= 2π+ ⎢⎡ 4y− y⎤ 2 ⎛ 8⎞3⎥ = π+ ⎜ − ⎟⎣ ⎦0⎝ 3⎠16= 2π+ ≈ 11.62336. Let the x-axis lie along the diameter at the baseperpendicular to the water level and sliceperpendicular to the x-axis. Let x = 0 be at the2 2center. The slice has base length 2 r − x andheight hxr .2hr 2 2V = x r x dxr∫ −02 2( ) 3/2 r2h⎡ 1 ⎤ 2h⎛1 3⎞2 2= r x r r hr⎢− −3⎥ = ⎜ ⎟=⎣⎦ r ⎝3 ⎠ 337. Let the x-axis lie on the base perpendicular to thediameter through the center of the base. The sliceat x is a rectangle with base of lengthand height x tanθ .r2 2V = ∫ 0 2 x tan θ r −x dxr⎡ 2 tan (2 2 )3/2⎤= ⎢− θ r −x3⎥⎣⎦0= 2 3tan3 r θ02 22 r − x308 Section 5.2 Instructor’s Resource Manual

38. a. x = 4 ykSlice horizontally.2⎛V 4 y ⎞ ⎛y y ⎞Δ ≈ π ⎜ Δ = π Δyk ⎟ ⎜ k ⎟⎝ ⎠ ⎝ ⎠If the depth of the tank is h, thenhh y π ⎡23/2⎤V =π ∫ dy = y0 k k⎢3⎥⎣ ⎦02π3/2= h .3 kThe volume as a function of the depth of the2π3/2tank is V( y)= y3 kdVb. It is given that =− m y.dtdV π 1/2 dyFrom part a, = y .dt k dtπ dyThus, y =− m y and dy −=m kk dtdt πwhich is constant.39. Let A lie on the xy-plane. Suppose Δ A = f( x)Δ xwhere f(x) is the length at x, so A = ∫ f( x)dx.Slice the general cone at height z parallel to A.The slice of the resulting region is A z and Δ Azis a region related to f(x) and Δ x by similartriangles:⎛ z⎞ ⎛ z⎞Δ Az= ⎜1 − ⎟ f( x) ⋅⎜1− ⎟Δx⎝ h⎠ ⎝ h⎠2⎛ z ⎞= ⎜1 − ⎟ f ( x)Δx⎝ h ⎠2 2⎛ z⎞ ⎛ z⎞Therefore, Az= ⎜1 − ⎟ f( x) dx = 1 − A.h∫ ⎜ ⎟⎝ ⎠ ⎝ h⎠22⎛ z ⎞h⎛ΔV ≈ AzΔ z = A⎜1− ⎟ Δz⎝ h ⎠0 1 z ⎞V = A∫⎜ − ⎟ dz⎝ h ⎠3h⎡ h⎛z⎞⎤ 1= A ⎢− ⎜1 − ⎟ ⎥ = Ah.⎢ 3⎝h ⎠ 3⎣ ⎥⎦0is1 3 3 2A = r⋅ r = r .2 2 4The center of an equilateral triangle is2 3 1⋅ r = r from a vertex. Then the3 2 3height of a regular tetrahedron is2h r 2 ⎛ 1 2 2 23 r ⎞= − ⎜ ⎟ =3r =⎝ ⎠3r1 2 3V = Ah = r3 1240. If two solids have the same cross sectional area atevery x in [a, b], then they have the same volume.41. First we examine the cross-sectional areas ofeach shape.Hemisphere: cross-sectional shape is a circle.2 2The radius of the circle at height y is r − y .Therefore, the cross-sectional area for thehemisphere is2 2 2 2 2Ah= π( r − y ) = π( r − y )Cylinder w/o cone: cross-sectional shape is awasher. The outer radius is a constant , r. Theinner radius at height y is equal to y. Therefore,the cross-sectional area is2 2 2 2A2 = πr − πy = π( r − y ) .Since both cross-sectional areas are the same, wecan apply Cavaleri’s Principle. The volume ofthe hemisphere of radius r isV = vol. of cylinder - vol. of cone2 1 2= πr h−πr h32 2= π r h3With the height of the cylinder and cone equal tor, the volume of the hemisphere is2 2 2 3V = π r () r = π r3 3.a.2A =π r1 1 2V = Ah = π r h3 3b. A face of a regular tetrahedron is anequilateral triangle. If the side of anequilateral triangle has length r, then the areaInstructor’s Resource Manual Section 5.2 309

5.3 Concepts Review3. a, b.1. 2 πx f( x)Δ x2.2 2 2 22 π∫ x dx; π (4 − y ) dy0 ∫ 03.22 π ∫ (1 +x)xdx04.22 π ∫ (1 + y)(2 − ydy )0c.3/2ΔV ≈ 2πx xΔ x = 2πx Δ xProblem Set 5.31. a, b.d, e.4. a,b.33 3/2 ⎡25/2 ⎤V = 2π ∫ x dx = 2π x0⎢5 ⎥⎣ ⎦036 3= π≈ 39.185c.⎛1⎞ΔV ≈ 2πx⎜⎟Δ x = 2πΔx⎝ x ⎠c.2 3ΔV ≈ 2 πx(9 −x ) Δ x = 2 π(9 x−x ) Δ x4 4= 2 π = 2 π =116 π≈ 18.85d,e. V ∫ dx [ x]2. a, b.d, e.33 3 9 2 1 4V 2 (9 x x ) dx 2⎡ ⎤= π∫− = π x x0⎢ −2 4⎥⎣ ⎦0⎛81 81⎞81π= 2π⎜− ⎟= ≈127.23⎝ 2 4 ⎠ 25. a, b.c.d, e.2 3ΔV ≈ 2 πx( x ) Δ x = 2πx Δ x11 3 ⎡14 ⎤ πV = 2π ∫ x dx = 2π x 1.570 ⎢ = ≈4⎥⎣ ⎦02c. ΔV ≈ 2 π(5 −x)xΔx1/2 3/2= 2 π(5 x −x ) Δ x5 1/2 3/2∫ 05⎡10 3/2 2 5/2⎤= 2π⎢x − x3 5⎥⎣⎦0⎛50 5 ⎞ 40 5= 2π ⎜− 10 5 = π≈93.663 ⎟⎝⎠ 3d, e. V = 2 π (5 x −x ) dx310 Section 5.3 Instructor’s Resource Manual

6. a, b.c.d, e.9. a, b.2 2 3ΔV ≈ 2 πx(3 x−x ) Δ x = 2 π(3 x −x ) Δ x33 2 3 3 1 4V 2 (3 x x ) dx 2⎡ ⎤= π∫− = π x x0⎢ −4⎥⎣ ⎦0⎛ 81⎞27π= 2π⎜27 − ⎟= ≈ 42.41⎝ 4 ⎠ 2c.2ΔV ≈ 2 π(3 −x)(9 −x ) Δ x2 3= 2 π(27−9x − 3 x + x ) Δ xd, e.7. a, b.3 2 3V = 2 π∫(27−9x− 3 x + x ) dx03⎡ 9 2 3 1 4⎤= 2π⎢27x− x − x + x2 4⎥⎣⎦0⎛ 81 81⎞135π= 2π⎜81− − 27+ ⎟= ≈ 212.06⎝ 2 4 ⎠ 2c.d, e.10. a, b.2 3ΔV ≈ 2 πy( y ) Δ y = 2πy Δ y11 3 ⎡14⎤πV = 2π ∫ y dy = 2π y 1.570 ⎢ = ≈4⎥⎣ ⎦02c.d, e.8. a, b.⎡⎛13 ⎞ ⎤ΔV ≈ 2π x⎢⎜x + 1 ⎟−(1 −x)Δx4⎥⎣⎝⎠ ⎦⎛14 224 x x ⎞= π ⎜ + ⎟Δ⎝ ⎠x1 1 4 2V 2⎛ ⎞= π ∫ x + x dx0 ⎜ ⎟⎝4⎠11 5 1 3 1 1= 2π ⎡ ⎢ x + x⎤ 2⎛ ⎞20 3⎥ = π ⎜ + ⎟⎣ ⎦0⎝20 3 ⎠23π= ≈ 2.4130d, e.c. ( )3/2ΔV ≈ 2π y y + 1 Δ y = 2 π ( y + y)Δ y4 3/2V = 2 π ∫ ( y + y)dy042 5/2 1 2 64= 2π ⎢⎡ y + y⎤ 2 ⎛ 8⎞5 2⎥ = π ⎜ + ⎟⎣ ⎦0⎝ 5 ⎠208π= ≈ 130.69511. a, b.Instructor’s Resource Manual Section 5.3 311

c.d, e.2 2 3ΔV ≈ 2 π(2 − y) y Δ y = 2 π(2 y − y ) Δ y22 2 3 2 3 1 4V 2 (2 y y ) dy 2⎡ ⎤= π∫− = π y y0⎢ −3 4⎥⎣ ⎦0⎛16 ⎞ 8π= 2π⎜− 4⎟= ≈8.38⎝ 3 ⎠ 315.12. a, b.a.A = ∫131x3dxb.3 ⎛ 1 ⎞ 3 1V = 2π ∫ x dx 2 dx1⎜ 3 ⎟ = π ∫ 1 2⎝ x ⎠ xc. ΔV ≈ 2 π(3 − y) ( 2y + 1)Δ y1/2 3/22 ( 3 3 2 2 )= π + y − y− y Δ yd, e. = 2π 2 ( 3+ 3 2 1/2 − − 23/2)13. a.∫V y y y dy02⎡3/2 1 2 2 2 5/2⎤= 2π ⎢3y+ 2 2y − y − y ⎥⎣2 5 ⎦0⎛ 16 ⎞ 88π= 2π ⎜6+ 8−2− ⎟= ≈55.29⎝ 5 ⎠ 5b 2 2π ⎡f ( x) g( x)⎤∫ − dxa ⎣⎦16.c.d.23 ⎡⎛1 ⎞ ⎤2V =π ∫⎢ 1 ( 1) ⎥ dx1 ⎜ + − −3 ⎟⎢⎣⎝x ⎠ ⎥⎦3 ⎛ 1 2 ⎞=π ∫dx1 ⎜ +6 3 ⎟⎝ x x ⎠3 ⎛ 1 ⎞V = 2 π∫(4 −x)dx1 ⎜ 3 ⎟⎝ x ⎠3 ⎛ 4 1 ⎞= 2π∫dx1 ⎜ −3 2 ⎟⎝ x x ⎠bb. 2 π ∫ [ ( ) − ( )]a x f x g x dxbc. 2 π∫( − )[ ( ) − ( )]ax a f x g x dxbd. 2 π ∫ ( − )[ ( ) − ( )]a b x f x g x dxa.A = 2 ( 3∫ x + 1) dx014. a.dπ∫c⎡ 2 2f ( y) −g( y)⎤dy⎣⎦b.2 3 2 4V = 2 π ∫ x( x + 1) dx = 2 π ( x + x)dx0 ∫ 0db. 2 π∫[ ( ) − ( )]cdy f y g y dyc. 2 π∫(3 − )[ ( ) − ( )]cy f y g y dyc.d.2 3 2 2V =π ⎡( x 2) ( 1) ⎤∫ + − −0 ⎣⎦=π 2 ( 6 4 3∫ x + x + 3) dx02 3V = 2 π∫(4 − x)( x + 1) dx02 4 3= 2 π∫( − x + 4x − x+4) dx0312 Section 5.3 Instructor’s Resource Manual

17. To find the intersection point, solve18.6yy =10246y − 1024y= 03yy = .325y( y − 1024) = 0y = 0, 44 ⎛ 3y ⎞V = 2π∫y y −dy0 ⎜ 32 ⎟⎝ ⎠4 ⎛ 43/2 y ⎞= 2π∫y −dy0 ⎜ 32 ⎟⎝ ⎠542 5/264 32 642 ⎡ yy⎤ ⎛ ⎞= π⎢ − ⎥ = 2π⎜ − ⎟=π⎢⎣5 160⎥⎦ ⎝ 5 5 ⎠ 50≈ 40.214 ⎛ 3y ⎞V = 2 π∫(4 − y)y −dy0 ⎜ 32 ⎟⎝ ⎠4 ⎛3 41/2 3/2 y y ⎞= 2π∫4y − y − +dy0 ⎜8 32 ⎟⎝⎠4 54⎡8 3/2 2 5/2 y y ⎤= 2π⎢y − y − + ⎥⎢⎣3 5 32 160⎥⎦0⎛64 64 32 ⎞ 208π= 2π⎜− − 8+ ⎟= ≈ 43.56⎝ 3 5 5 ⎠ 1519. Let R be the region bounded by20.2 2y = b − x ,2 2y =− b − x , and x = a. When R is revolvedabout the y-axis, it produces the desired solid.b 2 2 2 2V = 2π x ⎛ b x b x⎞∫ ⎜ − + − ⎟dxa ⎝⎠bb 2 2 ⎡ 1 2 2 3/2⎤= 4π∫x b − x dx = 4 π ( b x )a⎢− −3⎥⎣⎦a⎡1 2 2 3/2⎤4π2 2 3/2= 4 π⎢( b − a ) = ( b −a)3 ⎥⎣⎦ 32y =± a −x 2 , −a≤x≤aa2 2V = 2 π ( b x) ⎛2 a x⎞∫ − ⎜ − ⎟dx−a⎝ ⎠=a 2 2 a 2 24 πb ∫ a −x dx− 4 π x a x dx−a∫ −−aa⎛1 2⎞ ⎡ 1 2 2 3 2⎤2 2= 4πb⎜ πa ⎟−4 π ( a x ) 2 a b2⎢− − = π3⎥⎝ ⎠ ⎣ ⎦−a(Note that the area of a semicircle with radius a isa 2 2 1 2∫ a − x dx = πa.)−a221. To find the intersection point, solve22.2 2sin( x ) = cos( x ) .2tan( x ) = 12 πx =4x =π2/2 2 2V = 2π π x⎡cos( x ) sin( x ) ⎤∫− dx0 ⎣⎦π /2 2 2= 2π ⎡x cos( x ) xsin( x ) ⎤∫− dx0 ⎣⎦π /2⎡1 2 1 2 ⎤= 2π ⎢ sin( x ) + cos( x )2 2 ⎥⎣⎦0( )⎡⎛1 1 ⎞ 1⎤= 2π ⎢⎜+ ⎟− ⎥ = π 2−1 ≈1.30⎣⎝2 2 2 2 ⎠ 2⎦2V = 2 π ∫π x(2+sin x)dx02π= 2 π ∫ (2x + xsin x)dx02π2π= 2 π ∫ 2x dx + 2 π xsinx dx0 ∫ 022π2π= 2π ⎡x ⎤ + 2π[ sinx−xcosx⎣ ⎦]002 2= 2 π(4 π ) + 2 π( −2 π ) = 4 π (2π−1) ≈ 208.5723. a. The curves intersect when x = 0 and x = 1.b.1 2 2 2 1 2 4V = π∫ [ x − ( x ) ] dx = π ( x −x ) dx0 ∫ 01⎡1 3 1 5⎤ ⎛1 1⎞2π= π ⎢ x − x π0.423 5⎥ = ⎜ − ⎟= ≈⎣ ⎦0⎝3 5⎠151 2 1 2 3V = 2 π∫ x( x − x ) dx = 2 π ( x −x ) dx0 ∫ 01⎡1 3 1 4⎤ ⎛1 1⎞π= 2π⎢x − x 23 4⎥ = π⎜− ⎟=⎣ ⎦0⎝3 4⎠6≈ 0.52c. Slice perpendicular to the line y = x. At(a, a), the perpendicular line has equationy = −( x− a) + a = − x+ 2a. Substitute2y = –x + 2a into y = x and solve for x ≥ 0 .2x + x− 2a= 0− 1± 1+8ax =2− 1+ 1+8ax =2Instructor’s Resource Manual Section 5.3 313

Substitute into y = –x + 2a, so1+ 4a− 1+8ay = . Find an expression for22r , the square of the distance from (a, a) to⎛− 1+ 1+ 8a 1+ 4a− 1+8a⎞⎜,2 2 ⎟.⎝⎠22 1 1 8ar⎡ − + += ⎢a−⎤⎥⎣ 2 ⎦2⎡ 1+ 4a− 1+8a⎤+ ⎢a−⎥⎣ 2 ⎦2⎡2a+ 1− 1+8a⎤= ⎢ ⎥⎣ 2 ⎦2⎡ 2a+ 1− 1+8a⎤+ ⎢−⎥⎣ 2 ⎦2⎡2a+ 1− 1+8a⎤= 2 ⎢ ⎥⎣ 2 ⎦2= 2a + 6a+ 1− 2a 1+ 8a − 1+8a2ΔV ≈πr Δ a1 2V = π∫(2 a + 6 a+10− 2a 1+ 8a − 1+8 a)da12 3 2 1 3/2= π ⎡⎢ a + 3 a + a− (1+8 a)⎤3 12⎥⎣⎦0− π1∫ 2a1+8ada02 9 1= π ⎡⎢⎛ ⎜ + 3+ 1− ⎞ ⎟− ⎛ ⎜−⎞⎟⎤3 4 12⎥⎣⎝ ⎠ ⎝ ⎠⎦− π1∫ 2a1+8ada05π1 = − π 2a1 8ada2∫ +01To integrate ∫ 2a1+8ada, use the0substitution u = 1 + 8a.1 9 1 1∫ 2a 1+ 8 ada = ( u 1) u du0 ∫ −1 4 81 9 3/2 1/2= ( u u ) du32∫ −191 ⎡2 5/2 2 3/2⎤= u u32⎢ −5 3⎥⎣⎦11 ⎡⎛486 ⎞ ⎛2 2 ⎞⎤149= 1832⎢⎜ − ⎟−⎜ − ⎟ =5 5 3⎥⎣⎝ ⎠ ⎝ ⎠⎦605π 149π πV = − = ≈ 0.0522 60 6024.25.2ΔV ≈ 4πx Δ xrr 2 1 3 4 3V 4 x dx 4⎡ ⎤= π ∫ = π x r0 ⎢ = π3⎥⎣ ⎦032xΔV≈ SΔx2rrS r 2 S ⎡1 3⎤1V =2∫x dx = x rS02 ⎢ =rr 3⎥⎣ ⎦035.4 Concepts Review1. Circle2. x ;2 2 2 2x + y = 16cos t+ 16sin t = 16b2x + 12 2f t g t dt3. ∫ [ ′()] + [ ′()]a4. Mean Value Theorem (for derivatives)Problem Set 5.41.2.3/2 1/2f( x) = 4 x , f′( x) = 6x5 1/2 2 5L = ∫ 1 + (6 x ) dx = 1+36xdx1/3 ∫ 1/35⎡ 1 2 (1 36 )3/2⎤= ⎢ ⋅ + x36 3⎥⎣⎦1/31= ( 181 181 − 13 13 ) ≈ 44.23542 2 3/2 2 1/2f( x) = ( x + 1) , f′( x) = 2 x( x + 1)32 2 1/22L = 1 ⎡2 x( x 1) ⎤∫ + + dx1 ⎣⎦2 4 2 2 2= ∫ 4x + 4x + 1 dx = (2x + 1) dx1 ∫ 122 3 16 2 17= ⎡ ⎢ x + x⎤ 2 1 5.673⎥ = ⎛ ⎜ + ⎞ ⎟− ⎛ ⎜ + ⎞⎟= ≈⎣ ⎦1⎝ 3 ⎠ ⎝3 ⎠ 3314 Section 5.4 Instructor’s Resource Manual

3.2/3 3/2f( x) = (4 − x ) ,3 2/3 1/2 2 1/3f′ ( x) (4 x )⎛ x − ⎞= − ⎜ − ⎟2 ⎝ 3 ⎠−1/3 2/3 1/2=−x(4 − x )6.5y 1x = +30 32y5 4y 1 y 3g( y) = + , g′( y)= −30 3 6 42y2y4.8 1/3 2/3 1/21 ⎡ − (4 )1 ⎣∫L = + −x −x ⎤ dx⎦8 −2/3 8 −1/3= ∫ 4x dx = 2x dx1 ∫ 18⎡32/32 3(4 1) 92 x ⎤= ⎢ ⎥ = − =⎣ ⎦14 3x + 3 x 1f( x)= = +6x6 2x2x 1f′ ( x)= −2 22x2423 ⎛ y 3 ⎞L = ∫ 1+ −dy1 ⎜ 6 42y⎟⎝ ⎠83 y 1 9= ∫ + +1 36 2 4y83 ⎛ 4y 3dy = ∫+1 ⎜⎝6 2y4 533 ⎛ y 3 ⎞ ⎡y1 ⎤= ∫+ dy = −1 ⎜ 6 4 30 32y⎟⎢ ⎥⎝ ⎠ ⎣⎢2y⎦⎥1⎛81 1 ⎞ ⎛ 1 1 ⎞ 1154= ⎜ − ⎟−⎜ − ⎟= ≈8.55⎝10 54 ⎠ ⎝30 2 ⎠ 1352⎞dy4 ⎟⎠223 ⎛ x 1 ⎞L = ∫ 1+ −dx1 ⎜ 2 22x⎟⎝ ⎠7.5.4 223 x 1 1 3 ⎛ x 1 ⎞= ∫ + + dx = dx1 4 2 4 ∫+1 2 24x⎜ 2x⎟⎝ ⎠2 333 ⎛ x 1 ⎞ ⎡x1 ⎤= ∫+ dx = −1 ⎜ 2 22x⎟⎢ ⎥⎝ ⎠ ⎣⎢6 2x⎦⎥1⎛9 1⎞ ⎛1 1⎞14= ⎜ − ⎟−⎜ − ⎟= ≈ 4.67⎝2 6⎠ ⎝6 2⎠34 3y 1 y 1g( y) = + , g′( y)= −16 2 4 32yy32−2⎛ y 1 ⎞L = ∫ 1+ −dy−3⎜ 4 3y ⎟⎝ ⎠−26y 1 1 −2⎛ 3y 1= ∫ + + dy =−316 2 6 ∫+y−3⎜⎝4 y3 4−2−2⎛ y 1 ⎞ ⎡y1 ⎤= ∫ − + dy = − −−3⎜ 4 3 16 2y ⎟⎢ ⎥⎝ ⎠ ⎣⎢2y⎦⎥−3⎡⎛ 1⎞ ⎛81 1 ⎞⎤595=−⎢⎜1− ⎟−⎜ − ⎟ = ≈4.138 16 18⎥⎣⎝ ⎠ ⎝ ⎠⎦1442⎞dy3 ⎟⎠8.dx 2 dy= t , = tdt dt1 2 2 2 1 4 2L = ∫ ( t ) + ( t)dt = t + t dt0 ∫ 011 2 ⎡1 2 3/2⎤1= ∫ t t + 1 dt = ( t 1) 2 2 10⎢ + = −3⎥⎣ ⎦03≈ 0.61dx dy= 6, t = 6tdt dt2( )4 2 2 2 4 2 4L = ∫ (6 t) + (6 t ) dt = 36t + 36t dt1 ∫ 14 2 2 3/24= 6t 1 t dt ⎡2(1 t ) ⎤∫ + = +1 ⎣ ⎦1= 2( 17 17 −2 2 ) ≈ 134.53Instructor’s Resource Manual Section 5.4 315

9.12.3x = y+23g( y) = y+ , g′( y) = 123 2 3L = ∫ 1 + (1) = 2 dy = 2 21 ∫ 13 5At y = 1, x = 1+ = . 2 210.dx dy= 4cos t,=− 4sintdt dtπ2 2L = ∫ (4cos t) + ( −4sin t)dt0π 2 2 π= ∫ 16cos t+ 16sin t dt = 4dt0 ∫ 0= 4π≈12.57−3 −2 −1−1At y = 3,3 9x = 3 + = . 2 22⎛9 5⎞2d = ⎜ − ⎟ + (3 − 1) = 8 = 2 2⎝2 2⎠dx dy13. = 1, = 2tdt dt2 2 2 2 2L = ∫ 1 + (2 t) dt = 1+4t dt0 ∫ 02Let f( t) = 1+4 t . Using the Parabolic Rule−2−3dxdy= 2 5cos2 t, = − 2 5sin2tdtdtπ /40∫( 2 5cos2 ) ( 2 5sin2 )2 2L = t + − t dtπ/4 2 2 π/4= ∫ 20cos 2t+ 20sin 2t dt = 2 5 dt0 ∫ 05 π= ≈ 3.51211. f( x) = 2x+ 3, f′( x) = 23 23L = ∫ 1 + (2) dx = 5 dx = 2 51 ∫ 1At x = 1, y = 2(1) + 3 = 5.At x = 3, y = 2(3) + 3 = 9.2 2d = (3 − 1) + (9 − 5) = 20 = 2 514.with n = 8,2−0⎡ ⎛1⎞ ⎛1⎞ ⎛3⎞L ≈ f(0) + 4f ⎜ ⎟+ 2f ⎜ ⎟+4f⎜ ⎟3× 8⎢⎣ ⎝4⎠ ⎝2⎠ ⎝4⎠⎛5⎞ ⎛3⎞ ⎛7⎞+ 2 f(1) + 4f ⎜ ⎟+ 2f ⎜ ⎟+ 4 f ⎜ ⎟+f(2)]⎝4⎠ ⎝2⎠ ⎝4⎠1≈ [1 + 4 × 1.118 + 2 × 1.414212+ 4× 1.8028 + 2×2.2361+ 4× 2.6926 + 2×3.1623+ 4× 3.6401+ 4.1231] ≈ 4.6468dx dy 1= 2, t =dt dt 2 t2L 4 2 1 4 2 1(2 t ⎛ ⎞≈ ∫ ) + 41 ⎜ ⎟2 tdt = ∫ t +⎝ ⎠ 1 4tdt2 1f t = t +4twith n = 8,Let ( ) 4 . Using the Parabolic Rule4 1 11 14L ≈ − ⎡ f(1) + 4 f ⎛ ⎜ ⎞ ⎟+2 f⎛ ⎜ ⎞⎟3× 8⎢⎣ ⎝ 8 ⎠ ⎝ 8 ⎠⎛17 ⎞ ⎛20 ⎞ ⎛23 ⎞ ⎛26⎞+ 4f ⎜ ⎟+ 2f ⎜ ⎟+ 4f ⎜ ⎟+2f⎜ ⎟⎝ 8 ⎠ ⎝ 8 ⎠ ⎝ 8 ⎠ ⎝ 8 ⎠⎛ 29 ⎞ ⎤+ 4 f ⎜ ⎟+ f(4) ≈ 1 ( 2.0616 + 4×2.81188⎥⎝ ⎠ ⎦ 8+ 2× 3.562 + 4× 4.312 + 2× 5.0621+ 4×5.81222× 6.5622 + 4× 7.3122 + 8.0623)≈ 15.0467316 Section 5.4 Instructor’s Resource Manual

dx dy15. = cos t, = − 2sin 2tdt dtπL = 2 cost 2 + −2sin 2t 2 dt∫0( ) ( )π 2 2 2= ∫ t+0cos 4sin 2t dt2 2Let f( t) = cos t+4sin 2 t. Using theParabolic Rule with n = 8,π 2−0⎡ ⎛ π ⎞ ⎛2π⎞L ≈ f(0) + 4 f ⎜ ⎟+2 f ⎜ ⎟3× 8⎢⎣ ⎝16 ⎠ ⎝16⎠⎛3π ⎞ ⎛4π ⎞ ⎛5π ⎞ ⎛6π⎞+ 4f ⎜ ⎟+ 2f ⎜ ⎟+ 4f ⎜ ⎟+2f⎜ ⎟⎝16 ⎠ ⎝16 ⎠ ⎝16 ⎠ ⎝16⎠⎛7π ⎞ ⎛π ⎞⎤ π+ 4 f ⎜ ⎟+ f ⎜ ≈ [ 1+ 4×1.244116 2⎟⎥⎝ ⎠ ⎝ ⎠⎦ 48+ 2× 1.6892 + 4× 2.0262 + 2× 2.1213 + 4×1.9295+ 2× 1.4651+ 4× 0.7898 + 0) ≈ 2.3241π /2 2 2 4 2 2 49a cos tsin t 9a sin tcost dt0π /2 2 2 2 2 29a cos tsin t(sin t cos t)dt0π /2π /2 ⎡ 1 2 ⎤3acostsintdt 3a cos t0⎢2 ⎥ =⎣ ⎦0∫= +∫= +∫= = −(The integral can also be evaluated asπ /2⎡12 ⎤3a⎢sin t2⎥ with the same result.)⎣ ⎦0The total length is 6a.18. a. OT = length ( PT )= aθ3a2b. From Figure 18 of the text,PQ PQsin θ = PC= aand cosθ= QC QCPC= a.Therefore PQ = a sinθand QC = a cosθ.16.17.dx dy= 1, = secdt dt2t4 2 2 2 4 4L ππ= ∫ 1 + (sec t) dt = 1+sec t dt0 ∫ 04tLet f( t) = 1+sec . Using the Parabolicπ 4−0⎡ ⎛ π ⎞Rule with n = 8, L ≈ f(0) + 4 f ⎜ ⎟3× 8⎢⎣ ⎝32⎠⎛2π ⎞ ⎛3π ⎞ ⎛4π ⎞ ⎛5π⎞+ 2f ⎜ ⎟+ 4f ⎜ ⎟+ 2f ⎜ ⎟+4f⎜ ⎟⎝ 32 ⎠ ⎝32 ⎠ ⎝ 32 ⎠ ⎝ 32 ⎠⎛6π ⎞ ⎛7π ⎞ ⎛π ⎞⎤+ 2f ⎜ ⎟+ 4f ⎜ ⎟+f ⎜32 32 4⎟⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦π≈ [1.4142 + 4× 1.4211+ 2× 1.4425 + 4×1.480796+ 2× 1.5403 + 4× 1.6288 + 2×1.7585+ 4× 1.9495 + 2.2361] ≈ 1.278c. x = OT − PQ = aθ − asin θ = a( θ − sin θ)y = CT − CQ = a− acos θ = a(1 − cos θ )19. From Problem 18,x = a( θ − sin θ), y = a(1−cos θ)dxdy= a(1− cos θ ), = asinθsodθdθ2 2⎛ dx ⎞ ⎛ dy ⎞2 2⎜ ⎟ + ⎜ ⎟ = [ a(1− cos θ )] + [ asinθ]⎝dθ⎠ ⎝dθ⎠2 2 2 2 2 2= a − 2a cosθ + a cos θ + a sin θ2 2 2= 2a − 2a cosθ= 2 a (1−cos θ )21− cosθ2 2⎛θ⎞= 4a= 4asin ⎜ ⎟2 ⎝ 2⎠ .The length of one arch of the cycloid is⎛θ⎞ ⎛θ⎞θ⎝2⎠ ⎝2⎠2π2 22π4a sin dθ2asind0⎜ ⎟ =0⎜ ⎟∫ ∫2π⎡ θ ⎤a⎢a a2⎥⎣ ⎦0= 2 − 2cos = 2 (2+ 2) = 8dx2 dy2= 3acostsin t, =− 3asintcostdtdtThe first quadrant length is Lπ /2 2 2 2 2(3acostsin t) ( 3asin tcos t)dt0∫= + −Instructor’s Resource Manual Section 5.4 317

20. a. Using θ = ωt, the point P is at x = aωt−asin( ωt),y = a− acos( ωt)at time t.dx= aω − aωcos( ωt) = aω(1 − cos( ωt))dtdy= aωsin( ωt)dt2 2ds ⎡dy ⎤ ⎡dx⎤=dt⎢ +dt⎥ ⎢dt⎥⎣ ⎦ ⎣ ⎦2 2 2 2 2 2 2 2 2 2= a ω sin ( ωt) + a ω − 2a ω cos( ωt) + a ω cos ( ωt)1= 2 aω(1−cos( ωt))2ω22 t ωt= 2asin = 2asin 2ωω2 2 2 2= 2a ω − 2a ω cos( ωt)b. The speed is a maximum when sin ω tπ= 1, which occurs when t = (2 k + 1). The speed is a minimum when2ωω t2kπsin = 0, which occurs when t = .2ωc. From Problem 18a, the distance traveled by the wheel is aθ, so at time t, the wheel has gone aθ = aω t miles.Since the car is going 60 miles per hour, the wheel has gone 60t miles at time t. Thus, aω = 60 and themaximum speed of the bug on the wheel is 2aω = 2(60) = 120 miles per hour.dy21. a.dx =3x− 12 3 2 3/2L = ∫ 1+ x − 1dx = x dx1 ∫ 12⎡ 5/2⎤= ⎢ x ( )5⎥ = − ≈⎣ ⎦152 2 4 2 1 1.86b. f′ () t = 1− cos, t g′() t = sint4π 4π ⎛ t ⎞L = ∫ 2− 2cost dt = 2 sin dt0 ∫ 0⎜ ⎟⎝ 2 ⎠⎛ t ⎞sin ⎜ ⎟ is positive for 0 < t < 2π , and⎝ 2 ⎠by symmetry, we can double the integralfrom 0 to 2π .2π2π ⎛ t ⎞ ⎡ t ⎤L = 4∫sin dt 8cos0⎜ ⎟ = −2⎢2⎥⎝ ⎠ ⎣ ⎦0= 8 + 8 = 16dy22. a.dx =2 4x x64sin cos − 1π /3 2 4L = ∫ 1+ 64sin cos x−1dxπ /6⎡ 8= ∫= −⎣ 3π /3 2 38sin xcos xdx cos xπ /6⎢1=− + 3 ≈ 1.403⎤⎥⎦π /3π /6dxb. asin t asin t atcost atcostdt =− + + =dyacost acost atsin t atsintdt = − + =1 22 2 22 2L = ∫ a t cos t+a t sin tdt−11 1 0= ∫ at dt = at dt at dt−1 ∫ −0 ∫ −11 0⎡ a 2⎤ ⎡ a 2⎤= a a⎢ t t2⎥ − ⎢0 2⎥ = + = a⎣ ⎦ ⎣ ⎦−12 223. f( x) = 6 x, f′( x) = 624.1 1A = 2π ∫ 6x 1+ 36dx = 12 37πxdx0 ∫ 01⎡1212 372 x ⎤= π ⎢ ⎥ = 6 37π ≈ 114.66⎣ ⎦02xf( x) = 25 − x , f′( x)= −225 − x322 xA = 2π∫25− x 1+dx−2252− x3 2 22 ∫ 25 x x dx−23 3dx x−2−2= π − += 2π ∫ 5 = 10 π [ ] = 50π≈ 157.08318 Section 5.4 Instructor’s Resource Manual

25.26.3x2f x f′x x( ) = , ( ) =373x 4A = 2π ∫ 1+x dx1374 3/2⎤x ⎥⎦1( )⎡ 1π= 2 π ⎢ (1 + ) = 250 2−2 2⎣18 9248π2= ≈ 122.4396 4 3x + 2 x 1 x 1f( x) = = + , f′( x)= −2 8 2 2 3x x x8 4 23 ⎛ 4 3x 1 ⎞ ⎛ x 1 ⎞A = 2π ∫+ 1+ −1 ⎜ 8 2 2 34x⎟ ⎜ 2x⎟⎝ ⎠ ⎝ ⎠3 ⎛ 4 6x 1 ⎞ x 1 1= 2π ∫+ + + dx1 ⎜ 8 2 4 2 64x⎟⎝ ⎠ 4x3 ⎛ 4 3x 1 ⎞⎛ x 1 ⎞= 2π ∫+ ⎟⎜ +dx1 ⎜ 8 2 2 34x⎟⎜ 2x⎟⎝ ⎠⎝ ⎠3 ⎛ 7x 3x1 ⎞= 2π ∫+ +dx1 ⎜16 16 58x⎟⎝⎠8 23⎡ x 3x1 ⎤= 2π ⎢ + − ⎥128 32 4⎢⎣32x⎥⎦1⎡⎛6561 27 1 ⎞ ⎛ 1 3 1 ⎞⎤= 2π ⎢⎜ + − ⎟− ⎜ + − ⎟128 32 2592 128 32 32⎥⎣⎝ ⎠ ⎝ ⎠⎦8429π= ≈ 326.9281229.2 2y f x r x= ( ) = −2 2 −1/2f x x r x'( ) =− ( − )∫r2 2 2 2 −1/2A = 2πr − x 1 + ⎡−x( r −x ) ⎤ dx−r⎣⎦∫r2 2 2 2 2 −1= 2πr − x 1 + x ( r −x ) dx−rr∫2 2 2 2 2 −1( )( )= 2πr − x 1 + x ( r −x ) dx−rr 2 2 2r x x dx−rr 2 rr dx rdx−r∫−rrrx − r2r∫= 2π− +∫= 2π = 2π = 2 π | = 4π30. x = f() t = rcosty = g() t = rsintf '( t)= −rsintg '( t)= rcostπ2 2A = 2π∫ rsin t ( − rsin t) + ( rcos t)dt0π 2 2 2 2= 2π∫ rsint r sin t+r cos tdt0π 2= 2π∫ rsint r dt0π 2 2 π= 2π∫r sintdt =−2πr cos t|002 2=−2 r ( −1− 1) = 4πr31. a. The base circumference is equal to the arclength of the sector, so 2 π r = θl.Therefore,2πrθ = .l227.dx dy= 1, = 3dt dt2t1A = 2π t 31 9t 4∫ + dt01⎡ 1 4 3/2⎤⎢ t ⎥⎣⎦0( )π= 2 π (1+ 9 ) = 10 10−154 27≈ 3.56dx dy28. =− 2, t = 2dt dt1 2A = 2π ∫ 2t 4t + 4dt01⎡ 2 3 2⎤⎢ t ⎥⎣ ⎦01 2t t0∫= 8π + 1dt1 8π= 8 π ( + 1) = (2 2−1)≈ 15.323 3b. The area of the sector is equal to the lateralsurface area. Therefore, the lateral surface1 2 1 2⎛2πr⎞area is l θ = l ⎜ ⎟=πrl.2 2 ⎝ l ⎠c. Assume r 2 > r 1 . Let l 1 and l 2 be the slantheights for r 1 and r 2 , respectively. ThenA = πrl −π rl =π r( l + l)−π rl.2 2 11 2 1 112πr From part a,2 2πr2 2πrθ = = =1.l l + l lSolve for l 1 : l 1 r 2 = l 1 r 1 + lr 1l ( r − r ) = lr1 2 1 1lrl11 =r − r2 12 1 1⎛ lr1 ⎞ ⎛ lr1⎞A=π r2⎜ + l⎟−πr1⎜ ⎟⎝r2 −r1 ⎠ ⎝r2 −r1⎠⎡r1+r(2⎤=π lr1+ lr2) = 2π ⎢ l2 ⎥⎣ ⎦Instructor’s Resource Manual Section 5.4 319

32. Put the center of a circle of radius a at (a, 0).Revolving the portion of the circle from x = b tox = b + h about the x-axis results in the surface inquestion. (See figure.)dx dy34. =− asin t, = acostdt dtx = b, the surface area is2 2 2 2 2A = 2 π∫π( b− acos t) a sin t+a cos tdt02π= 2 πa∫( b−acos t)dt02π2= 2 πabt [ −asin t] 0 = 4πab35. a.The equation of the top half of the circle is2 2y = a −( x−a) .dy −( x −a)=b.dx 2 2a −( x−a)b+ h22 2 ( x−a)A = 2 π∫a −( x− a) 1+dxb2 2a −( x−a)b+h 2 2 2= 2 π∫a −( x− a) + ( x−a)dxbb hb h= 2 π ∫ + adx= 2 πa[ x]bb+c.= 2π ahA right circular cylinder of radius a and height hhas surface area 2π ah.dxdy33. a. = a(1 − cos t), = asintdtdt2πd.A= 2 π∫a(1−cos t)⋅02 2 2 2a (1 − cos t) + a sin t dt2π2 2= 2 πa∫(1 −cos t) 2a −2a cost dt02 2π3/2= 2 2 πa ∫ (1−cos t)dt0e.2 ⎛ t ⎞b. 1− cost= 2sin ⎜ ⎟,⎝2⎠ so2 2π 3/2 3⎛ t ⎞A = 2 2πa ∫ 2 sin dt0⎜ ⎟⎝2⎠2 2π ⎛ t ⎞ 2⎛ t ⎞= 8πa∫ sin sin dt0 ⎜ ⎟ ⎜ ⎟⎝2⎠ ⎝2⎠f.2 2π ⎛ t ⎞⎡ 2⎛ t ⎞⎤= 8πa∫ sin 1−cosdt0 ⎜ ⎟ ⎜ ⎟2⎢2⎥⎝ ⎠⎣⎝ ⎠⎦2π2 ⎡ ⎛ t ⎞ 2 3⎛ t ⎞⎤= 8πa⎢− 2cos⎜ ⎟+cos ⎜ ⎟2 3 2⎥⎣ ⎝ ⎠ ⎝ ⎠⎦02 2 2= 8πa ⎡⎢⎛ ⎜2− ⎞ ⎟− ⎛ ⎜− 2+⎞⎟⎤ 64 23 3⎥⎣⎝ ⎠ ⎝ ⎠⎦= 3 π aSince the circle is being revolved about the line320 Section 5.4 Instructor’s Resource Manual

36. a. f ′() t =− 3sin t, g′() t = 3cost37.2π2 2L = ∫ 9sin t+9cos tdt02π20 3dt3[ t]π= = 0 = 6 π∫ ≈ 18.850b. f ′() t =− 3sin t, g′() t = cost2π2 2L = ∫ 9sin t+ cos tdt ≈13.3650c. f ′() t = cost− tsin t, g′() t = tcost+sint6π2 2L = ∫ (cost− tsin t) + ( tcost+sin t)dt06π2= ∫ 1+ t dt ≈179.7180d. f ′() t =− sin t, g′() t = 2cos2t2π2 2L = ∫ sin t+ 4cos 2t dt ≈9.4290e. f ′() t =− 3sin3, t g′() t = 2cos2t2π2 2L = 9sin 3t+4cos 2t dt0∫ ≈ 15.289f. f ′() t =− sin t, g′() t =πcosπt40 2 2 2L = sin t+π cos πt dt0y = x, y′ = 1,∫ ≈ 86.5811L = ∫ 2dx = ⎡ 2 x⎤= 2 ≈1.414210 ⎣ ⎦02 1y x , y′2= = 2x,L = 0 1+4xdx∫ ≈ 1.478944 3 1y x , y′6= = 4 x , L = ∫ 1 16 1.600230 + x dx ≈10 9y = x , y′= 10 x ,L = 1 18∫ 1 100 1.754410 + dx ≈100 99y = x , y′= 100 x ,L = 1 198∫ 1 10,000 1.951670 + x dx ≈When n = 10,000 the length will be close to 2.5.5 Concepts Reviewba1. F ⋅( b−a); ∫ F( x)dx2. 30 · 10 = 3003. the depth of that part of the surface4. δ hAProblem Set 5.51.F ⎛1⎞ 1⎜ ⎟ = 6; 6, 122 k ⋅ =2k =⎝ ⎠F(x) = 12x1/2 21/2 3W = 12xdx ⎡6x⎤∫ = = = 1.5 ft-lb0 ⎣ ⎦022. From Problem 1, F(x) = 12x.2 22W = 12xdx = ⎡6x⎤0 ⎣ ⎦ 0∫ = 24 ft-lb3. F(0.01) = 0.6; k = 60F(x) = 60x0.02 20.02W = 60xdx ⎡30x⎤∫ = = 0.012 Joules0 ⎣ ⎦04. F(x) = kx and let l be the natural length of thespring.9−l9−l⎡12 ⎤W = ∫ kxdx = kx8−l⎢2 ⎥⎣ ⎦8−l12 2= (81 18 ) (64 16 )2 k ⎡ − l + l − − l + l ⎤⎣⎦= 1 (17 2 ) 0.052 k − l =0.1Thus, k = .17 − 2l10−l10−l⎡12 ⎤W = ∫ kxdx = kx9−l⎢2 ⎥⎣ ⎦9−l12 2= (100 20 ) (81 18 )2 k ⎡ − l + l − − l + l ⎤⎣⎦= 1 (19 2 ) 0.12 k − l =0.2Thus, k = .19 − 2l0.1 0.2 15Solving, l17 2l= 19 2l= .− − 2Thus k = 0.05, and the natural length is 7.5 cm.Instructor’s Resource Manual Section 5.5 321

dd 111. A slab of thickness Δ y at height y has width⎡ 2 ⎤5. W = ∫ kxdx = kx0 ⎢2 ⎥⎣ ⎦30y + 3 and length 10. The slab will be lifted a1 2 142= kd ( − 0) = kd⎛3⎞2 2distance 9 – y. ΔW ≈δ⋅10⋅ ⎜ y+ 3 ⎟Δy(9 − y)⎝4⎠115 26. F(8) = 2; k16 = 2, k == δ (36 + 5 y− y ) Δ y82272741 4/3 1 ⎡3 7/3⎤656115 2W = ∫ s ds = s =W =0 8 8⎢7⎥∫ δ (36 + 5 y−y ) dy0⎣ ⎦05624≈ 117.16 inch-pounds15 ⎡ 5 2 1 3⎤= (62.4) 362⎢ y+ y − y2 3⎥⎣⎦022 ⎡12 ⎤7. W = ∫ 9sds = 9 s = 180 ⎢2⎥⎣ ⎦ft-lb15 ⎛ 64 ⎞= (62.4) ⎜144 + 40 − ⎟02 ⎝ 3 ⎠= 76,128 ft-lb8. One spring will move from 2 feet beyond itsnatural length to 3 feet beyond its natural length. 12. A slab of thickness Δ y at height y has widthThe other will move from 2 feet beyond its2natural length to 1 foot beyond its natural length.2 6y − y and length 10. The slab will be lifted3 1 2321W = 6sds+ 6sds = ⎡3s ⎤ + ⎡3s⎤a distance 8 – y.∫2 ∫ 2 ⎣ ⎦2 ⎣ ⎦22ΔW ≈δ⋅10⋅2 6 y− y Δy(8 − y)= 3(9 – 4) + 3(1 – 4) = 6 ft-lb2= 20δ6 y− y (8 − y)Δ y9. A slab of thickness Δ y at height y has width4W = 3 2∫ 20 δ 6 y− y (8 − y ) dy04 − y and length 10. The slab will be lifted a5 3 2= 20δdistance 10 – y.∫ 6 y − y (3 − y)dy0⎛ 4 ⎞3 2ΔW ≈δ⋅10⋅⎜4 − y⎟Δy(10 − y)+ 20δ∫ 6 y − y (5) dy⎝ 50⎠32= 8 δ ( y − 15y+ 50) Δ y1 2 3/220 δ ⎡ (6 y y )⎤3 2= ⎢ −W = 53⎥ + 100δ∫ 6y − y dy2∫ 8 δ ( y − 15 y+50) dy⎣⎦ 000 3 2Notice that5∫ 6y−y dy is the area of a0⎡1 3 15 2 ⎤= 8(62.4) ⎢ y − y + 50y3 2⎥quarter of a circle with radius 3.⎣⎦0⎛125 375 ⎞W 1= 20 δ (9) + 100 δ ⎛ ⎜ π9⎞⎟= 8(62.4) ⎜ − + 250⎟= 52,000 ft-lb⎝4⎠⎝ 3 2 ⎠ = (62.4)(180 + 225π ) ≈ 55,340 ft-lb10. A slab of thickness Δ y at height y has width13. The volume of a disk with thickness Δ y is416πΔ y . If it is at height y, it will be lifted a4 − y and length 10. The slab will be lifted a3 distance 10 – y.distance 8 – y.ΔW ≈δ16 πΔy(10 − y) = 16 πδ(10 − y)Δ y⎛ 4 ⎞ΔW ≈δ⋅10⋅⎜4 − y⎟Δy(8 − y)10 ⎡ 1 2 ⎤⎝ 3 ⎠W = ∫ 16 πδ(10 − y) dy = 16 π(50) 0⎢10y−y2⎥⎣ ⎦040 (24 112= δ − y+ y ) Δ y= 16π (50)(100 – 50) ≈ 125,664 ft-lb33 40 (24 112W = ∫ δ − y+y ) dy0 3340 ⎡ 11 2 1 3⎤= (62.4) 24y− y + y3⎢2 3⎥⎣⎦040 ⎛ 99 ⎞= (62.4) ⎜72 − + 9⎟3 ⎝ 2 ⎠ = 26,208 ft-lb10322 Section 5.5 Instructor’s Resource Manual

14. The volume of a disk with thickness Δ x at height2x is π (4 + x)Δ x . It will be lifted a distance of10 – x.2ΔW ≈δπ (4 + x) Δx(10 − x)2 3=π δ (160 + 64x + 2 x −x ) Δ x10 2 3δ0∫W = π (160 + 64x+ 2 x −x ) dx⎡2 2 3 1 4⎤=π (50) ⎢160x+ 32x + x − x3 4⎥⎣⎦10⎛2000 ⎞=π (50) ⎜1600 + 3200 + −2500⎟⎝3 ⎠≈ 466,003 ft-lb15. The total force on the face of the piston is A · f(x)if the piston is x inches from the cylinder head.The work done by moving the piston from16.17.x to x is W = A ⋅ f ( x ) dx = A f ( x ) dx1 2x2 x2x1 x1∫ ∫ .This is the work done by the gas in moving thepiston. The work done by the piston to compressx1the gas is the opposite of this or A∫ f( x)dx .1.4c = 40(16)1.4A= 1; p( v)= cv −f( x)1.4= cx −16 2x1 = = 16, x2= = 21 116 −1.4 −0.416W = cx dx c⎡2.5x⎤∫ = −2 ⎣ ⎦ 21.4 −0.4 −0.440(16) ( 2.5)(16 2 )= − −≈ 2075.83 in.-lb1.4c = 40(16)1.4A= 2; p( v)= cv −1.4f( x) = c(2 x)−0x216 2x1 = = 8, x2= = 12 28 −1.4 −0.48W = 2 c(2 x) dx 2c⎡1.25(2 x)⎤∫= −1 ⎣⎦ 11.4 −0.4 −0.480(16) ( 1.25)(16 2 )= − −≈ 2075.83 in.-lb19. The total work is equal to the work W 1 to haul theload by itself and the work W 2 to haul the rope byitself.W 1 = 200⋅ 500 = 100,000 ft-lbLet y = 0 be the bottom of the shaft. When therope is at y, ΔW2 ≈ 2 Δy(500 − y).500⎡ 1 2 ⎤W2= ∫ 2(500 − y)dy = 2 500yy0⎢ −2⎥⎣ ⎦= 2(250,000 – 125,000) = 250,000 ft-lbW = W1+ W2 = 100,000 + 250,000= 350,000 ft-lb50020. The total work is equal to the work W 1 to lift themonkey plus the work W 2 to lift the chain.W 1 = 10⋅ 20 = 200 ft-lbLet y = 20 represent the top. As the monkeyclimbs the chain, the piece of chain at height y(0 ≤ y ≤ 10) will be lifted 20 – 2y ft.1ΔW2≈ Δy(20 − 2 y) = (10 − y)Δ y221.22.10 1 2W2(10 y) dy ⎡ ⎤= ∫ − = 10 y−y0⎢2⎥⎣ ⎦= 100 – 50 = 50 ft-lbW = W1+ W2= 250 ft-lbkf( x) = ; f(4000) = 50002xk5000240004200 80,000,000,000W = ∫dx4000 2x4200⎡ 1 ⎤= 80,000,000,000 ⎢ −x ⎥⎣ ⎦4000= , k = 80,000,000,00020,000,000= ≈ 952,381 mi-lb21kF( x)= where x is the distance between the2xkcharges. F(2) = 10; = 10, k = 40455 40 ⎡ 40⎤W = ∫ dx = 32 ergs1 2 ⎢− =x x⎥⎣ ⎦1100018. 80 lb/in. 2 = 11,520 lb/ft 21.4c =11,520(1) = 11,520−1.4ΔW ≈ p() v Δ v = 11,520v Δ v4 1.4 0.4411,520 − ⎡ −28,800 ⎤1 ⎣⎦ 1∫W = v dv = − v−0.4 −0.4=−28,800(4 − 1 ) ≈ 12,259 ft-lbInstructor’s Resource Manual Section 5.5 323

23. The relationship between the height of the bucket1and time is y = 2t, so t = y. When the bucket is2a height y, the sand has been leaking out of thebucket for 1 y seconds. The weight of the bucket2⎛1 ⎞ 3and sand is 100 + 500 − 3⎜y ⎟= 600 − y.⎝2 ⎠ 2⎛ 3 ⎞ΔW ≈⎜600 − y⎟Δy⎝ 2 ⎠8080⎛3 ⎞ ⎡ 3 2 ⎤W = ∫ 600 − y dy0 ⎜ ⎟ = 600yy⎝ 2 ⎢ −⎠4⎥⎣ ⎦0= 48,000 – 4800 = 43,200 ft-lb24. The total work is equal to the work W 1 needed tofill the pipe plus the work W 2 needed to fill thetank.2⎛1⎞ δπyΔ W1= δπ⎜⎟ Δ y( y)= Δy⎝2⎠4δ30( 62.4) π 130 πy⎡ 2 ⎤W1= ∫ dy = y0 4 4⎢2⎥⎣ ⎦0≈ 22,054 ft-lbThe cross sectional area at height y feet2(30 ≤ y ≤50)is π r where2 2 2r = 10 −(40 − y) = − y + 80y− 1500 .2 3 2Δ W2 = δπr Δ y y = δπ( − y + 80y −1500 y)Δ y50 3 2W2= ∫ δπ( − y + 80y −1500 y)dy3050⎡ 1 4 80 3 2⎤= (62.4) π⎢− y + y −750y4 3⎥⎣⎦30⎡⎛10,000,000⎞= (62.4) π⎢⎜− 1,562,500 + −1,875,000⎟⎣⎝3⎠−( − 202,500 + 720,000 −675,000)⎤ ⎦≈ 10,455,220 ft-lbW = W1+ W2 ≈ 10,477, 274 ft-lb25. Let y measure the height of a narrow rectanglewith 0 ≤ y ≤ 3. The force against this rectangleat depth 3 – y is ΔF ≈δ(3 − y)(6)Δ y. Thus,233yF = ∫ δ(3 − y)(6) dy = 6δ ⎡ ⎢3y−⎤⎥0⎢⎣2 ⎥⎦0= 6⋅ 62.4( 4.5)= 1684.8 pounds26. Let y measure the height of a narrow rectanglewith 0 ≤ y ≤ 3. The force against this rectangleat depth 5 – y is ΔF ≈δ(5 − y)(6)Δ y. Thus,233yF = ∫ δ(5 − y)(6) dy = 6δ ⎡ ⎢5y−⎤⎥0⎢⎣2 ⎥⎦0= 6⋅62.4⋅ 10.5 = 3931.2 pounds27. Place the equilateral triangle in the coordinatesystem such that the vertices are( −3,0),(3,0) and ( 0, − 3 3 ).The equation of the line in Quadrant I isyy= 3 ⋅x− 3 3 or x = + 3.3⎛ ⎛ y ⎞⎞ΔF ≈δ( − y) ⎜2⎜+ 3 ⎟⎟Δyand⎝ ⎝ 3 ⎠⎠0( ) ⎛2 ⎛ y3 ⎞⎞F = ∫ δ − y ⎜ + ⎟dy−3 3 ⎜ ⎟⎝ ⎝ 3 ⎠⎠0 ⎛ 2y ⎞=− 2δ∫+ 3y dy−3 3⎜3 ⎟⎝ ⎠3 20⎡ y 3y⎤=− 2δ⎢ + ⎥ =−2⋅62.4(0 −13.5)⎢⎣3 3 2 ⎥⎦−3 3= 1684.8 pounds28. Place the right triangle in the coordinate systemsuch that the vertices are (0,0), (3,0) and (0,-4).The equation of the line in Quadrant IV is4 3y = x− 4 or x = y+3.3 4⎛3⎞ΔF ≈δ(3 − y) ⎜ y+ 3 ⎟Δyand⎝4⎠0 ⎛ 3 3 2 ⎞F = ∫ δ 9 y y dy−4⎜ − − ⎟⎝ 4 4 ⎠2 30⎡ 3yy ⎤= δ ⎢9y− − ⎥ = 62.4⋅26⎢⎣8 4 ⎥⎦−4= 1622.4 pounds29. Δ ≈δ(1 − ) ( ) Δ ; = ∫1δ(1 − ) ( )01 12 32= δ ( y − y ) dyF y y y F y y dy∫0⎡2 32 2 52⎤ ⎛ 4 ⎞= δ ⎢ y − y 62.43 5 ⎥ = ⎜ ⎟⎣ ⎦0⎝15⎠= 16.64 pounds1324 Section 5.5 Instructor’s Resource Manual

30. Place the circle in the coordinate system so thatthe center is (0.0). The equation of the circle is2 2x + y = 16 and in Quadrants I and IV,2 2x = 16 − y . ΔF ≈δ(6 − y) ⎛2 16 y⎞⎜ − ⎟Δy⎝ ⎠F4 2= (6 ) 2 164 δ y ⎛y ⎞∫ − ⎜ − ⎟ dy− ⎝ ⎠Using a CAS, F ≈ 18,819 pounds.31. Place a rectangle in the coordinate system suchthat the vertices are (0,0), (0,b), (a,0) and (a,b).The equation of the diagonal from (0,0) to (a,b)is b ay = x or x = y. For the upper left triangle I,a b⎛a⎞ΔF ≈δ( b− y) ⎜ y⎟Δyand⎝b⎠b ⎛a⎞F = ∫ δ ( b− y)y dy0⎜ ⎟⎝b⎠b2 3⎛ a 2 ⎞ ⎡ay ay ⎤= δ∫y y dy δ0⎜ − ⎟ = ⎢ − ⎥⎝ b ⎠ ⎢⎣2 3b⎥⎦⎛ ab 2 ab 2 ⎞ 2= δ− = δab⎜ 2 3 ⎟⎝ ⎠6For the lower right triangle II,⎛ a ⎞ΔF ≈δ( b− y) ⎜a−y⎟dyand⎝ b ⎠b ⎛ a ⎞F = ∫ δ ( b− y)a−y dy0 ⎜ ⎟⎝ b ⎠b ⎛ a 2 ⎞= ∫ δ ab − 2ay + y dy0 ⎜⎟⎝ b ⎠3b⎡ 22 ay ⎤ ⎛2 2 ab ⎞= δ ⎢aby − ay + ⎥ = δab − ab +⎢3b⎜3 ⎟⎣ ⎥⎦0⎝ ⎠2ab= δ3The total force on one half of the dam is twice theabδtotal force on the other half since3abδ6b022= 2.32. Consider one side of the cube and place thevertices of this square on (0,0), (0,2), (2,0) and(2,2).2ΔF ≈δ(102 − y)(2) Δ y; F = ∫ 2 δ(102 − y)dy022y= 2δ ⎡⎢102y− ⎤⎥ = 2⋅62.4⋅ 202 = 25, 209.6⎢⎣2 ⎥⎦0The force on all six sides would be 6(25,209.6) =151,257.6 pounds.33. We can position the x-axis along the bottom ofthe pool as shown:204Δxh810From the diagram, we let h = the depth of anarbitrary slice along the width of the bottom ofthe pool.20Using the Pythagorean Theorem, we can find thatthe length of the bottom of the pool is2 220 + 4 = 416 = 4 26Next, we need to get h in terms of x. This can bedone by using similar triangles to set up aproportion.x 44 26h − 4 x 4x= → h = +4 4 26 26Δ F = δ ⋅h⋅ΔAx4h − 44 26⎛ x ⎞F =∫δ ⎜4+⎟( 10)dx0 ⎝ 26 ⎠4 26⎛ x ⎞=∫62.4⎜4 + ⎟( 10)dx0 ⎝ 26 ⎠4 26⎛x ⎞= 624∫⎜4+ ⎟dx0 ⎝ 26 ⎠24 26⎡ x ⎤= 624 ⎢4x+ ⎥⎣ 2 26⎦0= 624 16 26 + 8 26 = 624 24 26( ) ( )( )= 14,976 26 lb ≈76,362.92 lbInstructor’s Resource Manual Section 5.5 325

34. If we imagine unrolling the cylinder so we have aflat sheet, then we need to find the total forceagainst one side of a rectangular plate as if it hadbeen submerged in the oil. The rectangle wouldbe 2π( 5)= 10πfeet wide and 6 feet high.Thus, the total lateral force is given by6F = 50⋅y⋅10πdy∫0626ydy ⎡ πy⎤00∫= 500π= ⎣250 ⎦= 250 36 = 9000 π lbs ( ≈ 28,274.33 lb)π ( )35. Let W 1 be the work to lift V to the surface andW 2 be the work to lift V from the surface to 15feet above the surface. The volume displaced bythe buoy y feet above its original position is2 31 ⎛ a ⎞ 1 2 ⎛ y⎞π⎜a− y⎟ ( h− y) = πa h⎜1−⎟3 ⎝ h ⎠ 3 ⎝ h⎠ .2The weight displaced isδ yπa h⎛ ⎜1−⎞⎟3 ⎝ h ⎠ .δ 2Note by Archimede’s Principle m= π a h or32 3ma h = , so the displaced weight isδ π3⎛m 1 y ⎞⎜ − ⎟⎝ h ⎠ . 3 3⎛ ⎛ y⎞ ⎞ ⎛ ⎛ y⎞⎞ΔW1 ≈ m−m 1− Δ y = m 1− 1− Δy⎜ ⎜ ⎟ ⎜ ⎟⎝ h⎠ ⎟ ⎜ ⎝ h⎠⎟⎝ ⎠ ⎝ ⎠3h⎛10 1 ⎛1 y ⎞ ⎞W = m∫−⎜− ⎟dy⎜ ⎝ h ⎠ ⎟⎝ ⎠4h⎡ h⎛y⎞⎤ 3mh= m⎢ y+ ⎜1− ⎟ ⎥ =⎢ 4⎝h ⎠ 4⎣⎥⎦0W2 = m⋅ 15 = 15m3mhW = W1+ W2= + 15m4336. First calculate the work W 1 needed to lift thecontents of the bottom tank to 10 feet.ΔW1 ≈δ40 Δy(10 − y)4W1= ∫ δ 40(10 − y)dy04⎡ 1 2 ⎤= (62.4)(40) ⎢− (10 − y)2⎥⎣⎦0= (62.4)(40)(–18 + 50) = 79,872 ft-lbNext calculate the work W 2 needed to fill the toptank. Let y be the distance from the bottom of thetop tank.ΔW2 ≈δ(36 π)Δ y ySolve for the height of the top tank:160 4036π h = 160; h = =36π9π40 / 9πW2= ∫ δ 36πydy040 / 9π⎡12 ⎤= (62.4)(36 π) ⎢ y 2 ⎥⎣ ⎦0⎛ 800 ⎞= (62.4)(36 π) ⎜ 812 ⎟ ≈ 7062 ft-lbs⎝ π ⎠W = W1+ W2≈ 86,934 ft-lbs⎛1 2 ⎞22537. Since δ ⎜ π a ⎟(8) = 300, a = .⎝3 ⎠2πδWhen the buoy is at z feet (0 ≤ z ≤ 2) belowfloating position, the radius r at the water level is⎛8 z 225 8 zr+ ⎞ ⎛a+ ⎞= ⎜ ⎟ = ⎜ ⎟⎝ 8 ⎠ 2πδ⎝ 8 ⎠ .1 2F = δ ⎛⎜ π r ⎞⎟(8 + z) −300⎝3⎠75 (8 ) 3= + z − 3001282 ⎡ 75 (8 )3 ⎤W = ∫ + z − 300 dz0 ⎢128⎥⎣⎦2⎡ 75 (8 )4 ⎤= ⎢ + z − 300 z512⎥⎣⎦⎛46,875 ⎞= ⎜ − 600 ⎟− (600 − 0)⎝ 32 ⎠8475= ≈ 264.84 ft-lb320326 Section 5.5 Instructor’s Resource Manual

5.6 Concepts Review1. right; 41 ⋅ + 63 ⋅= 2.24+62. 2.5; right; x(1+x); 1 + x3. 1; 34.24 40;16 16The second lamina balances at x = 3, y = 1 .The first lamina has area 12 and the secondlamina has area 4.12 1 4 3 24 12 3 4 1 40x = ⋅ + ⋅ = , y = ⋅ + ⋅ =12 + 4 16 12 + 4 16Problem Set 5.61.25 ⋅ + (2)7 − ⋅ + 19 ⋅ 5x = =5+ 7+9 212. Let x measure the distance from the end whereJohn sits.180⋅ 0 + 80⋅ x + 110⋅12= 6180 + 80 + 11080x + 1320 = 6 · 37080x = 900x = 11.25Tom should be 11.25 feet from John, or,equivalently, 0.75 feet from Mary.3.7 72 5/2x xdx ⎡ x ⎤ 2∫0 ⎣5 ⎦05 ( 49 7 ) 21x = = = =73/2722xdx3 ( 7 7)5∫⎡ x ⎤0 ⎣3⎦06. M y = ( −3) ⋅ 5 + ( −2) ⋅ 6+ 3⋅ 2+ 4⋅ 7+ 7⋅ 1=14M x = 2⋅ 5 + ( −2) ⋅ 6+ 5⋅ 2+ 3⋅ 7 + ( −1) ⋅ 1=28m = 5 + 6 + 2 + 7 + 1 = 21M y 2 M x 4x = , ym= 3 = m= 37. Consider two regions R1and R2such that R1isbounded by f(x) and the x-axis, and R2isbounded by g(x) and the x-axis. Let R 3 be theregion formed by R1−R2. Make a regularpartition of the homogeneous region R 3 suchthat each sub-region is of width , Δxand let x bethe distance from the y-axis to the center of massof a sub-region. The heights of R1and R 2 at xare approximately f(x) and g(x) respectively. Themass of R3is approximatelyΔ m =Δm1−Δm2≈ δ f ( x) Δx−δg( x)Δx= δ[ f ( x) −g( x)]Δxwhere δ is the density. The moments for R 3 areapproximatelyM = M ( R ) −M ( R )x x 1 x 2δ 2 δ 2≈ [ f ( x)] Δx− [ g( x)]Δx2 2δ 2 2= ⎡( f ( x)) −( g( x))⎤Δx2 ⎣⎦My = My( R1) −My( R2)≈ xδf( x) Δx−xδg( x)Δx= xδ[ f( x) −g( x)]ΔxTaking the limit of the regular partition asΔx→ 0 yields the resulting integrals inFigure 10.4.7 2 573 1 1x(1 x ) dx ⎡ x x ⎤∫ ++0 ⎣2 5 ⎦0x = =7 3 47(1 x ) dx x 1∫ + ⎡ + x ⎤0 ⎣ 4 ⎦016,807( + )49 33,8592 5 10 9674= = = ≈5.582401 2429( 7 +4 )173548.5. M y = 1⋅ 2+ 7⋅ 3 + ( −2) ⋅ 4 + ( −1) ⋅ 6+ 4⋅ 2=17M x = 12 ⋅ + 13 ⋅ + ( −5)4 ⋅ + 06 ⋅ + 62 ⋅ =− 3m = 2 + 3 + 4 + 6 + 2 = 17M y M x 3x = 1, ym= = m=− 17f( x) = 2 − x; g( x) = 0x ==2x[(2 −x) −0]dx02∫ [(2 −x) −0]dx02 2[2 x−x ] dx02∫ [2 − xdx ]0∫∫Instructor’s Resource Manual Section 5.6 327

2⎛ 2 1 3⎞8⎜x− x ⎟34 −⎝ ⎠0= =32⎛ 1 4 22 ⎞ −⎜2x−x ⎟⎝ 2 ⎠02=310.9.1y =2=2 2 2∫ −x−02∫ [(2 −x) −0]02 2[4 − 4 x+x ] dx0∫[(2 ) 0 ] dx4dx2 123⎞83 ⎠03⎛⎜4x− 2x + x ⎟ 8− 8+⎝= =4 42=3x( )= = =y4 1 2 14 3∫ x x dx0 3 3 ∫ x dx0= =41 2 14 2∫ x dx x dx0 3 3∫0441⎡1 x ⎤ 643⎣4 ⎦0 3334 641⎡1x⎤ 93⎣3 ⎦024141 2 4 4 1 1 5( x 1) dx x dx ⎡ x ⎤2∫0 3 18 ∫0 18 ⎣5⎦0= = =41 264 64∫ x dx39 9051245 864 59= =11.x = 0 (by symmetry)y=12 2 22 ∫ (2 − x )− 2=2 2∫ (2 − x ) dx− 212 2 42 ∫ − +− 22⎡ 1 3 ⎤⎢2x−x⎣ 3 ⎥⎦−2dx(4 4 x x ) dx3 521⎡4x− 4x +1x ⎤ 32 22⎣3 5 ⎦−2 158 2 8 23 3= = =451 1 513 4 1xx ( ) dx x dx ⎡ x ⎤ 1∫0 ∫0 ⎣5⎦0 5 4x = = = = =1 3 41 1 1xdx 1 5∫⎡ x ⎤ 4 40 ⎣4⎦01 3 2 1 6 71( x ) dx x dx ⎡ 1x ⎤2∫0 2∫0 ⎣14⎦01 31 1∫ x dx 4 40114 21 1y = = == =1 7 4328 Section 5.6 Instructor’s Resource Manual

12.13.x = 0 (by symmetry)⎡( ( x −10)) 21 2 1 22∫dx−2⎢−2 ⎥y =⎣⎦2⎡ 1 22 2( x 10) ⎤∫ − − dx− ⎣ ⎦12 4 2−8 ∫ ( x − 20x + 100) dx−2=12 2− 2 ∫ ( x −10)dx−25 32−1⎡1x − 20 x + 100x⎤8⎣5 3 ⎦ − 574−215 287== = −3252− 1⎡1x−10x⎤13032⎣3 ⎦−2To find the intersection point, solve2x− 4= 2 x .x − 2 = x2x − 4x+ 4=x2x − 5x+ 4=0(x – 4)(x – 1) = 0x = 4 (x = 1 is extraneous.)4∫ x ⎡2 x (2x 4) ⎤1 ⎣− −⎦dxx =4∫⎡2 x −(2x−4)⎤dx1 ⎣⎦4 3/2 22 ∫ ( x − x + 2 x)dx=14 1/22 ∫ ( x − x+2) dx15/2 3 242 ⎡2 x −1x + x ⎤ 64⎣5 3 ⎦15= = =3/2 24 192⎡2 x −1x + 2x⎤3⎣3 2 ⎦1⎤1929514.y =⎡( )⎤x x⎥dx⎦x x dx214 22 ∫1⎢2 −(2 −4)⎣4∫⎡2 1 ⎣−( 2 −4)⎤⎦4 2− x +1x−dx1932 ∫ ( 5 4)=3 242⎡− 1 53x + 2x −4x⎤⎣⎦19 27= = =1919 193 3To find the intersection points,2x − x− 3=01±13x =21+13x ==( )2∫( 1−13)2( 1+13)2∫( 1−13)2( 1+13)2∫( 1−13)22x( x+ 3 −x ) dx2( x + 3 −x ) dx2 3( x + 3 x−x ) dx( 1+13)⎡1 2 1 3x + 3x−x ⎤ 2⎣2 3 ⎦ 1 132( − )( 1+13)2x = x + 3 .⎡1 3 3 2 4x + x − 1 x ⎤ 2⎣3 2 4 ⎦( 1−) 1313 3=2=1213 13 13 13=6 612y =( 1+13)2∫( 1−13)2( 1+13)2∫(1−13)2⎡ 2 2 2( x + 3) −( x ) ⎤dx⎣⎦2( x+ 3 −x ) dx12Instructor’s Resource Manual Section 5.6 329

15.16.==( 1+13)∫( − )1 22 1 132⎡2 4( x + 6x+ 9 −x)13 136( 1+13)( − )1 1 3 2 1 5x + 3x + 9x−x22⎣3 5 ⎦ 1 13213 136143 133013 136= =115To find the intersection points, solvey =±x2⎤2y = 2 .12 2 2 2 2⎡142 ( ) ⎤2∫−(4 − )− 2 ⎣ ⎦ 2∫−2= =2 2 32(2 y ) dy 1∫ −⎡22y−3y ⎤−⎣ ⎦−2521⎡4y−1y ⎤ 16 22⎣5 ⎦−2 5 68 2 8 2 53 3= = =y = 0 (by symmetry)y dy y dyTo find the intersection points, solve2y −3y− 4=− y.y2−2y− 4=02±20y =2y = 1±5x=11+5⎡ 2 2 2( y) ( y 3y 4) ⎤2 ∫ − − − − dy1−5 ⎣⎦=1+5⎡ 2( y) ( y 3y 4) ⎤∫ − − − − dy1−5 ⎣⎦11+5 4 32 ∫ ( − y + 6y −24y−16)dy1−51+5 2∫ ( − y + 2y+4) dy1−55 4 21+51⎡− 1y +3y −12y −16y⎤2⎣ 5 2 ⎦1−5 −3 21+5 20 5⎡− 1y + y + 4y⎤3⎣ 3 ⎦1−5= == − 3y =1+5⎡∫1−5 ⎣21+5⎡ 2∫1−5 ⎣1+51−5−3+2+20 53y ( −y) −( y −3y−4)⎤ dy⎦( −y) −( y −3y−4)⎤dy⎦( y 2y 4 y)dy∫ =4 3 21+5⎡− 1y +2y + 2y⎤ 20 5⎣ 4 3 ⎦1−5 3= = = 120 5 20 53 320 517. We let δ be the density of the regions and Aibethe area of region i.Region R 1 :1mR ( 1) = δ A1= δ(1/ 2)(1)(1) = δ211 1 3 1( ) x∫ xxdx |0 3 0 2x31 = = = =1 11 2 1xdx3∫0x |2 20Since R1is symmetric about the line y = 1− x ,the centroid must lie on this line. Therefore,2 1y1 = 1− x1= 1− = ; and we have3 31My( R1) = x2⋅ m( R1)= δ31Mx( R1) = y2⋅ m( R1)= δ6Region R 2 :mR ( 2) = δ A2= δ(2)(1) = 2δBy symmetry we get1x2= 2 and y 2 = .2Thus,My( R2) = x2⋅ m( R2) = 4δM ( R ) = y ⋅ m( R ) = δx2 2 2330 Section 5.6 Instructor’s Resource Manual

18. We can obtain the mass and moments for thewhole region by adding the individual regions.Using the results from Problem 17 we get that1 5m= m( R1) + m( R2) = δ + 2δ = δ2 21 13My = My( R1) + My( R2) = δ + 4δ = δ3 31 7Mx = Mx( R1) + Mx( R2)= δ + δ = δ6 6Therefore, the centroid is given by13M δy 3 26x = = =m 5δ1527δM x 6 7y = = =m 5δ152∫19. mR ( 1) = δ ( gx ( ) − f( x))dxbacmR ( 2) = δ ∫ ( gx ( ) − f( x))dxbδ b 2 2M x ( R1) = (( ( )) ( ( )) )2∫ g x − f x dxaδ c 2 2M x ( R2) = (( g( x)) −( f( x)) ) dx2∫bbM y ( R1) = δ ∫ x( g( x) − f( x))dxacM y ( R2) = δ ∫ x( g( x) − f( x))dxbNow,mR ( ) = δ ( gx ( ) − f( x))dx3= δcaba∫∫cb( g( x) − f( x)) dx+ δ ( g( x) − f( x))dx= mR ( 1) + mR ( 2)δ c 2 2M x ( R3) = (( ( )) ( ( )) )2∫ g x − f x dxaδ b 2 2= (( ( )) ( ( )) )2∫ g x − f x dxaδ c 2 2+ (( g( x)) ( f( x)) ) dx2∫ −b= M ( R ) + M ( R )3xc1 x 2M ( R ) = δ x( g( x) − f( x))dxy∫∫∫abacby= δ x( gx ( ) − f( x))dx+ δ x( gx ( ) − f( x))dx= M ( R ) + M ( R )1 y 2∫b∫ ab∫ a20. mR ( 1) = δ ( hx ( ) −gx ( )) dxmR ( ) = δ ( gx ( ) − f( x))dx2δ b 2 2M x ( R1) = (( ( )) ( ( )) )2∫ h x − g x dxaδ b 2 2M x ( R2) = (( g( x)) −( f( x)) ) dx2∫abM ( R ) = δ x( h( x) −g( x))dxy1M ( R ) = δ x( g( x) − f( x))dxyNow,32∫∫bababaabamR ( ) = δ ( hx ( ) − f( x))dx= δ∫∫∫( hx ( ) − gx ( ) + gx ( ) − f( x))dx= δ ( hx ( ) − gx ( )) dx+ δ ( gx ( ) − f( x))dx= mR ( 1) + mR ( 2)δ b 2 2Mx( R3) = (( ( )) ( ( )) )2∫ h x − f x dxaδ b 2 2 2 2= (( ( )) ( ( )) ( ( )) ( ( )) )2∫ hx − gx + gx − f x dxaδ b 2 2= (( ( )) ( ( )) )2∫ hx − gx dxaδ b 2 2+ (( gx ( )) ( f( x)) ) dx2∫ −a= M ( R ) + M ( R )3xb1 x 2M ( R ) = δ x( h( x) − f( x))dxy∫∫∫ababay1 y 2ba= δ xhx ( ( ) − gx ( ) + gx ( ) − f( x))dx∫= δ x( hx ( ) − gx ( )) dx+ δ xgx ( ( ) − f( x))dx= M ( R ) + M ( R )21. Let region 1 be the region bounded by x = –2,x = 2, y = 0, and y = 1, so m 1 = 41 ⋅ = 4.1By symmetry, x 1 = 0 and y 1 = . Therefore2M1y= x1m1 = 0 and M1x= y1m1 = 2 .Let region 2 be the region bounded by x = –2,x = 1, y = –1, and y = 0, so m 2 = 31 ⋅ = 3.11By symmetry, x 2 =− and y 2 = − . Therefore2233M2y= x2m2=− and M2x= y2m2=− .22M31y+ M2y−2 3x = = = −m + m 7 141 211x+ M2x2m1+m2M1y = = =7 14ba∫Instructor’s Resource Manual Section 5.6 331

22. Let region 1 be the region bounded by x = –3,x = 1, y = –1, and y = 4, so m 1 = 20 . By3symmetry, x =− 1 and y 1 = . Therefore,2M1y= x1m1 =− 20 and M1x= y1m1 = 30 . Letregion 2 be the region bounded by x = –3, x = –2,y = –3, and y = –1, so m 2 = 2 . By symmetry,5x 2 =− and y 2 =− 2 . Therefore,2M2y= x2m2 =− 5 and M2x= y2m2 =− 4 . Letregion 3 be the region bounded by x = 0, x = 1,y = –2, and y = –1, so m 3 = 1. By symmetry,12x 3 = and 3y =−32. Therefore,13M3y= x3m3= and M3x= y3m3=− .22M491y + M2y + M3y−2 49x = = =−m1+ m2 + m323 4649M1x + M2x + M3x2 49y = = =m1+ m2 + m323 4623. Let region 1 be the region bounded by x = –2,x = 2, y = 2, and y = 4, so m 1 = 4⋅ 2= 8. Bysymmetry, x1 = 0 and y1= 3. Therefore,M1y= x1m1 = 0 and M1x= y1m1 = 24 . Letregion 2 be the region bounded by x = –1,x = 2, y = 0, and y = 2, so m 2 = 32 ⋅ = 6. By1symmetry, x2 = and y2= 1 . Therefore,2M2y= x2m2 = 3 and M2x= y2m2 = 6 . Letregion 3 be the region bounded by x = 2, x = 4,y = 0, and y = 1, so m 3 = 21 ⋅ = 2. By symmetry,1x3 3 and y22M x = y m = .M1y + M2y + M3y9x = =m1+ m2 + m316M1x + M2x + M3x31y = =m1+ m2 + m316= = . Therefore, M3 = x3m3 = 6and 3 3 3 124. Let region 1 be the region bounded by x = –3,x = –1, y = –2, and y = 1, so m 1 = 6 . By1symmetry, x 1 =− 2 and y 1 =− . Therefore,2M1y= x1m1 = − 12 and M1x= y1m1 =− 3. Letregion 2 be the region bounded by x = –1, x = 0,y = –2, and y = 0, so m 2 = 2 . By symmetry,1x 2 =− and y 2 =− 1 . Therefore,2y25.M2y= x2m2 =− 1 and M2xy2m2 2region 3 be the remaining region, so m 3 = 22 .1By symmetry, x3 2 and y32M3y= x3m3 = 44 and M3x= y3m3 =− 11.M1y + M2y + M3yx = =m + m + m= =− . Let= =− . Therefore,31301 2 3M1x + M2x + M3x16 8y = =− =−m1+ m2 + m330 1511 3 ⎡1 4 ⎤ 1A= ∫ x dx = x0 ⎢ =4⎥⎣ ⎦044From Problem 11, x = .51⎛4⎞2πV = A(2 π x) = ⎜2π⋅ ⎟=4⎝5⎠5Using cylindrical shells:11 3 1 4 ⎡1 5⎤2V = 2π∫x⋅ x dx = 2π x dx 2 x0 ∫ = π0 ⎢ =5⎥⎣ ⎦0526. The area of the region is π a . The centroid is thecenter (0, 0) of the circle. It travels a distance of2 32π (2a) = 4π a. V = 4πa27. The volume of a sphere of radius a is24 3.3 aπ If2 2the semicircle y = a − x is revolved aboutthe x-axis the result is a sphere of radius a. Thecentroid of the region travels a distance of 2 π y.1 2The area of the region is .2 π a Pappus'sTheorem says that⎛1 2⎞2 2 4 3(2 πy)⎜ π a ⎟=π a y = πa.⎝2 ⎠ 34ay = , x = 0 (by symmetry)3 π28. Consider a slice at x rotated about the y-axis.bΔ V = 2 π xh( x)Δ x, so V = 2 π∫ xh( x)dx.abΔm≈h( x)Δ x, so m= ∫ h( x)dx = A.abΔM y ≈ xh( x)Δ x so M ( ),y = ∫ xh x dx .abM ∫ y( )a xhxdxx = =m AThe distance traveled by the centroid is 2π x .b(2 π x) A= 2 π∫ xh( x)dxaTherefore, V = 2π xA.π332 Section 5.6 Instructor’s Resource Manual

29. a. ΔV ≈ 2 π( K − y) w( y)Δ y30. a.dV = 2 π∫( K − y) w( y)dycdb. Δm≈ w( y)Δ y, so m= ∫ w( y)dy = A.cb.dΔM x ≈ yw( y)Δ y , so M x = ∫ yw( y)dy .c∫dyw( y)dycy =AThe distance traveled by the centroid is2 π( K − y).2 π( K − y) A= 2 π( KA−M x )dd= 2 π⎛Kw( ydy ) ywydy ( )⎞⎜−⎟⎝∫c∫ c ⎠d= 2 π∫( K − ywydy ) ( )cTherefore, V = 2 π( K − y)A.1m=bh2bThe length of a segment at y is b− y.h⎛ b ⎞ ⎛ b 2 ⎞ΔM x ≈ y⎜b− y⎟Δ y = ⎜by− y ⎟Δy⎝ h ⎠ ⎝ h ⎠h⎛b 2 ⎞M x = ∫ by − y dy0 ⎜ ⎟⎝ h ⎠h⎡1 2 b 3⎤1 2= ⎢ by − y bh2 3h⎥ =⎣⎦06M x hy = =m 31A = bh ; the distance traveled by the2⎛ h ⎞centroid is 2π⎜k− ⎟3⎝ ⎠ .1 πbh( )⎛ h⎞⎛ ⎞V = 2π⎜k− ⎟⎜ bh⎟= 3k−h⎝ 3⎠⎝2 ⎠ 331. a. The area of a regular polygon P of 2n sides2 π πis 2r nsin cos . (To find this consider2n2nthe isosceles triangles with one vertex at thecenter of the polygon and the other verticeson adjacent corners of the polygon. Eachπsuch triangle has base of length 2rsin 2 nb.πr cos , so the centroid travels a distance2nπof 2π r cos . 2 nThus, by Pappus's Theorem, the volume ofthe resulting solid is⎛ 22 rcos π ⎞⎛2r nsin π cosπ ⎞⎜ π ⎟⎜ ⎟⎝ 2n⎠⎝ 2n 2n⎠3 π 2 π= 4π rnsin cos .2n2n3 π 2 πlim 4πrnsin cosn→∞2n2nsinπ2n2 3 2 πlim 2πr cosn→∞π2 n2n2 3= 2πrAs n →∞, the regular polygon approachesa circle. Using Pappus's Theorem on the2circle of area π r whose centroid (= center)travels a distance of 2πr, the volume of the2 2 3solid is ( πr )(2 π r) = 2π r which agreeswith the results from the polygon.32. a. The graph of f (sin x ) on [0, π ] isπsymmetric about the line x = since2πf (sin x) = f ( sin( π− x)) . Thus x = .2π∫ xf(sin x)dx0πx = =πf(sin x)dx2∫0Thereforeππ π∫ x f(sin x) dx = f(sin x)dx0 2∫ 0b.4 2 2sin x cos x sin x(1 sin x)2 2f ( x) = x(1 − x ) .= − , soπ4 π π 4∫ xsin xcos xdx = sin xcosxdx0 2∫ 0ππ ⎡ 1 cos5 ⎤ π= x2⎢− 5⎥ =⎣ ⎦05π ⎞and height r cos . ⎟⎠ Since P is a regular2npolygon the centroid is at its center. Thedistance from the centroid to any side isInstructor’s Resource Manual Section 5.6 333

33. Consider the region S – R.yS−R11⎡ 2 22 ∫ −0 ⎣g ( x) f ( x)⎤dx=⎦≥ yS − R11 22 ∫ f ( xdx )0=R1 1 2 2 1 1 2R ⎡g ( x) f ( x) ⎤ ( ) ( )2∫ − dx≥ S −R f x dx0⎣⎦ 2∫ 01 1 2 2 1 1 2R ⎡g ( x) f ( x) ⎤( )2∫ − dx+R f x dx0⎣⎦ 2∫ 01 1 2 1 1 2≥ ( S − R) ( ) ( )2∫ f x dx+R f x dx0 2∫ 01 1 2 1 1 2R ( ) ( )2∫ g x dx≥S f x dx0 2∫ 011 2 11 22∫g ( xdx ) f ( xdx )0 2∫0≥SRyS≥ yR34. To approximate the centroid, we can lay thefigure on the x-axis (flat side down) and put theshortest side against the y-axis. Next we can usethe eight regions between measurements toapproximate the centroid. We will let h i ,theheight of the ith region, be approximated by theheight at the right end of the interval. Eachinterval is of width Δ x = 5 cm. The centroid canbe approximated as8∑ xh i ii=1 (5)(6.5) + (10)(8) + + (35)(10) + (40)(8)x ≈ =86.5 + 8 + + 10 + 8∑ hii=11695= ≈ 23.3872.581 2∑( hi)2 2 2 2 2i=1 (1/ 2)(6.5 + 8 + + 10 + 8 )y ≈ =8(6.5 + 8 + + 10 + 8)∑ hii=1335.875= ≈ 4.6372.5R35. First we place the lamina so that the origin iscentered inside the hole. We then recompute thecentroid of Problem 34 (in this position) as8∑ xh i ii=1x ≈8∑ hii=1( − 25)(6.5) + ( − 15)(8) + + (5)(10) + (10)(8)=6.5 + 8 + + 10 + 8−480= ≈−6.6272.5812 2∑(( hi−4) −( −4) )2i=1y ≈8∑ hii=12 2 2 2(1/ 2)((2.5 −( − 4) ) + + (4 −( −4) ))=6.5 + 8 + + 10 + 845.875= ≈0.63372.5A quick computation will show that these valuesagree with those in Problem 34 (using a differentreference point).Now consider the whole lamina as R 3 , thecircular hole as R 2 , and the remaining lamina asR 1 . We can find the centroid of R1by notingthatM x( R1) = Mx( R3) − Mx( R2)and similarly for M y ( R 1).From symmetry, we know that the centroid of acircle is at the center. Therefore, bothM x ( R2) and M y ( R2) must be zero in our case.This leads to the following equationsM y( R3) − My( R2)x =mR ( 3) − mR ( 2)δΔx( −480)=2δ Δ x(72.5) − δπ(2.5)−2400= ≈ −7342.87M x( R3) − Mx( R2)y =mR ( 3) − mR ( 2)δΔx(45.875)=2δ Δ x(72.5) − δπ(2.5)229.375= ≈0.669342.87Thus, the centroid is 7 cm above the center of thehole and 0.669 cm to the right of the center of thehole.334 Section 5.6 Instructor’s Resource Manual

36. This problem is much like Problem 34 except wedon’t have one side that is completely flat. Inthis problem, it will be necessary, in someregions, to find the value of g(x) instead of justf(x) – g(x). We will use the 19 regions in thefigure to approximate the centroid. Again wechoose the height of a region to be approximatelythe value at the right end of that region. Eachregion has a width of 20 miles. We will place thenorth-east corner of the state at the origin.The centroid is approximately19∑ xi( f( xi) − g( xi))i=1x ≈19∑( f( xi) − g( xi))i=1(20)(145 − 13) + (40)(149 − 10) + (380)(85 −85)=(145 − 13) + (149 − 19) + (85 −85)482,860= ≈173.6927801912 2∑[( f( xi)) − ( g( xi)) ]2 i=1y ≈19∑( f( xi) − g( xi))i=12 2 2 2 2 21 ⎡ (145 − 13 ) + (149 − 10 ) + + (85 − 85 ) ⎤2 ⎣⎦=(145 − 13) + (149 − 19) + + (85 −85)230,805= ≈83.022780This would put the geographic center of Illinoisjust south-east of Lincoln, IL.5.7 Concepts Review1. discrete, continuous2. sum, integral3.50∫f ( xdx )2. a. PX ( ≥ 2) = P(2) + P(3) + P(4)= 0.05 + 0.05 + 0.05 = 0.15b.EX ( ) =5∑ ii=1xpi= 0(0.7) + 1(0.15) + 2(0.05)+ 3(0.05) + 4(0.5)= 0.63. a. PX ( ≥ 2) = P(2) = 0.2b. EX ( ) = − 2(0.2) + ( − 1)(0.2) + 0(0.2)+ 1(0.2) + 2(0.2)= 04. a. PX ( ≥ 2) = P(2) = 0.1b. E( X ) = − 2(0.1) + ( − 1)(0.2) + 0(0.4)+ 1(0.2) + 2(0.1)= 05. a. PX ( ≥ 2) = P(2) + P(3) + P(4)= 0.2 + 0.2 + 0.2= 0.6b. EX ( ) = 1(0.4) + 2(0.2) + 3(0.2) + 4(0.2)= 2.26. a. PX ( ≥ 2) = P(100) + P(1000)= 0.018 + 0.002 = 0.02b. E( X ) = − 0.1(0.98) + 100(0.018)+ 1000(0.002)= 3.7027. a. PX ( ≥ 2) = P(2) + P(3) + P(4)3 2 1 6= + + = = 0.610 10 10 10b. E( X ) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 24. cumulative distribution functionProblem Set 5.71. a. PX ( ≥ 2) = P(2) + P(3) = 0.05 + 0.05 = 0.1b.4EX ( ) = ∑ xp i ii=1= 0(0.8) + 1(0.1) + 2(0.05) + 3(0.05)= 0.35Instructor’s Resource Manual Section 5.7 335

8. a. PX ( ≥ 2) = P(2) + P(3) + P(4)9. a.2 2 20 ( −1) ( −2) 5= + + = = 0.510 10 10 10b. E( X ) = 0(0.4) + 1(0.1) + 2(0) + 3(0.1) + 4(0.4)= 2b.10. a.20 1 1PX ( ≥ 2) = ∫ dx= ⋅ 18=0.92 20 2022020 1 ⎡x⎤E( X) = ∫ x⋅ dx = ⎢ ⎥ = 100 20 ⎢⎣40 ⎥⎦0c. For x between 0 and 20,b.11. a.x 1 1 xF( x)= ∫ dt = ⋅ x =0 20 20 2020 1 1PX ( ≥ 2) = ∫ dx= ⋅ 18=0.452 40 4022020 1 ⎡x⎤E( X) = ∫ x⋅ dx = ⎢ ⎥ = 5− 5=0−2040 ⎢⎣80 ⎥⎦−20c. For −20 ≤ x ≤ 20 ,x 1 1 1 1F( x) = ∫ dt = ( x+ 20) = x+−2040 40 40 2b.12. a.8 3PX ( ≥ 2) = ∫ x(8 −xdx)2 256383 ⎡2 x ⎤ 3 27= ⎢4x− ⎥ = ⋅ 72=256 ⎢⎣3 ⎥⎦256 328 3EX ( ) = ∫ x⋅ x(8 −xdx)0 2563= 8 ( 82 30)256∫ x −x dx3 483 ⎡8xx ⎤= ⎢ − ⎥ = 4256 ⎢⎣3 4 ⎥⎦c. For 0≤ x ≤ 80x33 3 ⎡2 t ⎤F( x) = ∫ t(8 − t) dt = ⎢4t− ⎥0 256 256 ⎢⎣3 ⎥⎦3 1= x − x64 2562 320 3PX ( ≥ 2) = ∫ x(20 −xdx)2 40003203 ⎡2 x ⎤= ⎢10x− ⎥ = 0.9724000 ⎢⎣3 ⎥⎦22x0b.13. a.20 3EX ( ) = ∫ x⋅ x(20 −xdx)0 40003= 20 ( 20x 2 −x 30) dx4000∫203 43 ⎡20xx ⎤= ⎢ − ⎥ = 104000 ⎢⎣3 4 ⎥⎦c. For 0≤ x ≤ 20x 3F( x) = ∫ t(20 −t)dt0 40003 ⎡3 1= ⎢10t − ⎥ = x − x4000 ⎢⎣3 ⎥⎦400 4000b.14. a.03x2 t ⎤2 34 3 2PX ( ≥ 2) = ∫ x(4 −xdx)2 643 43 ⎡4xx ⎤= ⎢ − ⎥ = 0.687564 ⎢⎣3 4 ⎥⎦4 3 2EX ( ) = ∫ x⋅ x(4 −xdx)0 6442( )0453 3 4 3 ⎡4 x ⎤= 4x x dx x 2.464∫ − = ⎢ − ⎥ =064 ⎢⎣5 ⎥⎦c. For 0≤ x ≤ 4b.x3 43 2 3 ⎡4t t ⎤F( x) = ∫ t (4 − t)dt = ⎢ − ⎥0 64 64 ⎢⎣3 4 ⎥⎦1 3= x − x16 2563 48 1PX ( ≥ 2) = ∫ (8 −xdx)2 3221 ⎡ x ⎤ 9= ⎢8x− ⎥ =32 ⎢⎣2 ⎥⎦168 1EX ( ) = ∫ x⋅ (8 −xdx)0 32381 ⎡2 x ⎤ 8= ⎢4x− ⎥ =32 ⎢⎣3 ⎥⎦3c. For 0≤ x ≤ 8820x21 1 ⎡ t ⎤F( x) = ∫ (8 − t) dt = ⎢8t−⎥0 32 32 ⎢⎣2 ⎥⎦1 1= x − x4 642x040x0336 Section 5.7 Instructor’s Resource Manual

15. a.b.16. a.4 π ⎛πx⎞PX ( ≥ 2) = ∫ sin dx2 ⎜ ⎟8 ⎝ 4 ⎠π 4⎡ 4 1 1cos πx ⎤= ( 1 0)8⎢− π 4⎥ =− − − =⎣ ⎦22 24 π ⎛πx⎞EX ( ) = ∫ x⋅ sin dx0 ⎜ ⎟8 ⎝ 4 ⎠Using integration by parts or a CAS,E( X ) = 2 .c. For 0≤ x ≤ 4xx π ⎛πt⎞ π ⎡−4πt⎤F( x) = ∫ sin dt cos0 ⎜ ⎟ =8 4 8⎢ π 4⎥⎝ ⎠ ⎣ ⎦01⎛πx⎞ 1 πx1=− ⎜cos − 1⎟=− cos +2⎝4 ⎠ 2 4 2b.17. a.π4 ⎛ x ⎞PX ( ≥ 2) = ∫ cos dx2 ⎜ ⎟8 ⎝ 8 ⎠4⎡ ⎛πx⎞⎤π π 1= ⎢sin ⎜ ⎟ sin sin 18⎥ = − = −⎣ ⎝ ⎠⎦22 4 2π4 π ⎛πx⎞E( X) = ∫ x⋅ cos dx0 ⎜ ⎟8 ⎝ 8 ⎠Using a CAS, E( X) ≈ 1.4535c. For 0≤ x ≤ 4xx π ⎛πt⎞ ⎡ ⎛πt⎞⎤F( x) = ∫ cos dt sin0 ⎜ ⎟ =8 8⎢ ⎜ ⎟8⎥⎝ ⎠ ⎣ ⎝ ⎠⎦0⎛π x ⎞= sin ⎜ ⎟⎝ 8 ⎠b.18. a.44 4 ⎡ 4 ⎤ 1PX ( ≥ 2) = ∫ dx= 2 2 ⎢− =3x3x⎥⎣ ⎦2344 4 ⎡4⎤EX ( ) = ∫ x⋅ dx= ln1 2 ⎢ x3x3 ⎥⎣ ⎦14= ln 4 ≈ 1.853c. For 1≤ x ≤ 41x 4 ⎡ 4⎤−4 4F( x)= ∫ dt =1 2 ⎢− 3t3t⎥ = +⎣ ⎦13x34x− 4=3x99 81 ⎡ 81 ⎤PX ( ≥ 2) = ∫ dx= 2 3 ⎢ −2 ⎥40x⎣ 80x⎦277= ≈ 0.24320b.99 81 ⎡ 81 ⎤EX ( ) = ∫ x⋅ dx= 1.81 3 ⎢− =40x40x⎥⎣ ⎦1c. For 1≤ x ≤ 9xx 81 ⎡ 81 ⎤F( x)= ∫ dt =1 3 ⎢ −2 ⎥40t⎣ 80t⎦1281 81 81x− 81=− + =2 80 280x80x19. Proof of F′ ( x) = f( x) :By definition, F( x) = ∫ f( t) dt. By the FirstAFundamental Theorem of Calculus,F′ ( x) = f( x).Proof of F( A) = 0 and F( B) = 1:∫∫AABAF( A) = f( x) dx = 0;F( B) = f( x) dx = 1Proof of Pa ( ≤ X≤ b) = Fb ( ) − Fa ( ) :ba∫xPa ( ≤ X≤ b) = f( xdx ) = Fb ( ) −Fa( ) due tothe Second Fundamental Theorem of Calculus.20. a.a+bThe midpoint of the interval [a,b] is .2⎛ a+ b⎞ ⎛ a+b⎞P⎜X< ⎟= P⎜X≤ ⎟⎝ 2 ⎠ ⎝ 2 ⎠a+b1 1 ⎛a+b ⎞= 2∫ dx = aa⎜ − ⎟b−a b−a⎝ 2 ⎠1 b−a 1= ⋅ =b−a 2 2b.c.2bb 1 1 ⎡ x ⎤E( X)= ∫ x⋅ dx = ⎢ ⎥a b−a b−a⎢⎣2 ⎥⎦a2 2b − a a+b= =2( b−a) 2x 1 1 x − aF( x) = ∫ dt = ( x− a)=a b−a b−a b−a21. The median will be the solution to thex0 1equation ∫ dx = 0.5 .a b−a1( x0− a)= 0.5b−ab−ax0− a =2a+bx0=2Instructor’s Resource Manual Section 5.7 337

15 2 2f ( x) = x (4 − x)is512symmetric about the line x = 2. Consequently,PX≤ ( 2) = 0.5 and 2 must be the median of X.22. The graph of23. Since the PDF must integrate to one, solve24.5∫ kx(5 − x) dx = 1.02 35⎡5kxkx ⎤⎢ − ⎥ = 1⎢⎣2 3 ⎥⎦0125k125k− = 12 3375k− 250k= 66k =1255 2 2Solve ∫ kx (5 − x) dx = 10( )5 2 3 4k∫25x − 10x + x dx = 103 4 55⎡25x 5x x ⎤k ⎢ − + ⎥ = 1⎢⎣3 2 5 ⎥⎦0625k = 166k =625425. a. ∫ ( )− − =Solve k 2 x 2 dx 10Due to the symmetry about the line x = 2, thesolution can be found by solving22∫kx dx = 1022k⋅ x = 104k= 11k =41P X x dx3 441 1 4= ∫ (2 −( x − 2)) dx = (4 )3 4 4∫−x dx3241⎡x ⎤ 1= ⎢4x− ⎥ =4⎢⎣2 ⎥⎦834b. (3 ≤ ≤ 4) = ∫ ( 2 − −2)c.d.4 1EX ( ) = ∫ x⋅ (2− x−2)dx0 42 1 4 1= ∫ x ⋅ (2 + ( x− 2)) dx+ (2 ( 2))0 4∫ x⋅ − x−dx2 41 2 2 1 4 2= (4 )4∫ x dx+ 0 4∫x−x dx2341 32 1⎡2 x ⎤ 2 4= x + ⎢2x− ⎥ = + = 212 0 4 ⎢⎣3 ⎥⎦3 322xx21 ⎡t ⎤ xIf 0 ≤ x ≤ 2, F( x)= ∫ tdt = ⎢ ⎥ =0 4 ⎢⎣8 ⎥⎦802 1 x 1If 2 < x ≤ 4, F( x) = ∫ xdx+ (4 )0 4∫ −t dt2 4222x2x 1⎡ t ⎤ 1 ⎛ x 3⎞= + ⎢4t− ⎥ = + x− −8 4 2 2 ⎜ 8 2⎟0 ⎣⎢⎥⎦2⎝ ⎠2x=− + x −18F( x)⎧ 0⎪if x < 02⎪ xif 0≤ x ≤ 2⎪ 8= ⎨2 ⎪− x + − < ≤⎪ x 1 if 2 x 48⎪⎪⎩ 1 if x > 4e. Using a similar procedure as shown in part(a), the PDF for Y is1f( y) = ( 120 − y−120 )14,400y 1If 0 ≤ y < 120, F( y)= ∫ tdt0 14,4002y2⎡ t ⎤ y= ⎢ ⎥ =⎢⎣28,800⎥⎦28,8000If 120 < y ≤ 240,1 y 1F( y) = + (240 −t)dt∫2 12014,40021 1 ⎡ t ⎤= + ⎢240t− ⎥2 14,400⎢⎣2 ⎥⎦1202 21 y y 3 y y= + − − = − + −12 60 28,800 2 28,800 60⎧ 0⎪if y < 02⎪ yif 0≤ y ≤120⎪ 28,800F( x)= ⎨2 ⎪− y y⎪ + − 1 if 120 < y ≤ 24028,800 60⎪⎪⎩ 1 if y > 240y338 Section 5.7 Instructor’s Resource Manual

26. a.180 2Solve ∫ kx (180 − x) dx = 1.04180⎡3 x ⎤k⎢60x− ⎥ = 1⎢⎣4 ⎥⎦01k =87,480,000b. P(100 ≤ X ≤ 150)= 150 1 2∫(180 )100 87,480,000 x −xdxc.27. a.41501 ⎡3 x ⎤= ⎢60x− ⎥ ≈0.46887,480,000 ⎢⎣4 ⎥⎦100180 1 2EX ( ) = ∫ x⋅ x(180 −xdx)0 87, 480,00051801 ⎡4 x ⎤= ⎢45x− ⎥ = 10887,480,000 ⎢⎣5 ⎥⎦00.6 6 8Solve ∫ kx (0.6 − x) dx = 1.00.6k 6(0.6 ) 8∫ x − x dx = 10Using a CAS, k ≈ 95,802,719b. The probability that a unit is scrapped is1 −P(0.35 ≤ X ≤ 0.45)c.d.0.45= 1 −k 6(0.6 ) 8∫ x −x dx0.35≈ 0.884 using a CAS0.6 6 8E( X) = ∫ x⋅kx (0.6 −x)dx00.6= k x 7(0.6 x)8∫ − dx0≈ 0.26256 8F( x) = ∫x95,802,719 t (0.6 −t)dt0Using a CAS,7 8 7F( x) ≈6,386,850 x ( x −5.14286x6 5 4+ 11.6308x − 15.12x + 12.3709x3 2− 6.53184x+ 2.17728x− 0.419904x+ 0.36)e. If X = measurement in mm, and Y =measurement in inches, then Y = X /25.4.Thus,FY( y) = P( Y ≤ y) = P( X /25.4≤y)= P X ≤ 25.4y = F 25.4y( ) ( )where F ( x ) is given in part (d).28. a.Alternatively, we can proceed as follows:3 127 6Solve ∫ ⎛ 3 ⎞k⋅y − y dy = 10 ⎜ ⎟⎝127⎠using aCAS.29k ≈ 1.132096857×108y 6 ⎛ 3 ⎞FY( y)= ∫ k⋅t 0 ⎜ −t⎟dt⎝127⎠Using a CAS,27 7 8 7FY( y) ≈ (7.54731× 10 ) y ( y −0.202475y6 5+ 0.01802y−0.000923y4 −7 3+ 0.00003 y − (6.17827×10 ) y−9 2 −11+ (8.108× 10 ) y − (6.156×10 ) y−13+ 2.07746×10 )200 2 8Solve ∫ kx (200 − x) dx = 1.0−23Using a CAS, k ≈ 2.417×10b. The probability that a batch is not accepted isc.d.e.200PX ( ≥ 100) = k 2(200 ) 8∫ x −x dx100≈ 0.0327 using a CAS.200 2 8EX ( ) = k∫xx ⋅ (200 −x)dx0= 50 using a CASx−23 2 8F( x) = ∫ (2.417× 10 ) t (200 −t)dx0−24 3Using a CAS, F(x) ≈ (2.19727× 10 ) x ⋅8 7 6 8 5( x − 1760x + 136889 x − (6.16×10 ) x11 4 13 3+ (1.76× 10 ) x − (3.2853×10 ) x15 2 17+ (3.942× 10 ) x − (2.816×10 ) x18+ 9.39×10 )100 2 8Solve ∫ kx (100 − x) dx. Using a CAS,0−20k = 4.95×10−20 2( ) ( )y8F( x) = ∫ 4.95× 10 t 100 −t dt0Using a CAS,−21 3F( x) ≈ (4.5×10 ) x8 7 6 7 5( x − 880x + 342, 222 x − (7.7×10 ) x10 4 12 3+ (1.1× 10 ) x − (1.027×10 ) x13 2 15+ (6.16× 10 ) x − (2.2×10 ) x16+ 3.667×10 )8Instructor’s Resource Manual Section 5.7 339

29. The PDF for the random variable X is⎧1 if 0≤x ≤1f( x)= ⎨⎩ 0 otherwiseFrom Problem 20, the CDF for X is F ( x)= xY is the distance from ( 1, X ) to the origin, so( ) ( )2 2 2Y = 1− 0 + X − 0 = 1+XHere we have a one-to-one transformation fromx:0 x 1 y:1≤ y ≤ 2 . Forthe set { }≤ ≤ to { }every 1< a < b< 2, the event a < Y < b willoccur when, and only when,2 2a − 1< X < b − 1.If we let a = 1 and b = y , we can obtain theCDF for Y.2 2( 1≤ ≤ ) = ( 1 −1≤ ≤ −1)2= P( 0≤ X ≤ y −1)2 2= F( y − 1)= y −1P Y y P X yTo find the PDF, we differentiate the CDF withrespect to y.d 2 1 1yPDF = dy y − 1= ⋅ ⋅ 2y=2 y 2 2−1 y −1Therefore, for 0≤ y ≤ 2 the PDF and CDF arerespectivelyy2g( y)= and G( y)= y − 1 .2y −133.34.⎧ 0 if x < 0⎪0.8 if 0 ≤ x < 1⎪F( x) = ⎨ 0.9 if 1 ≤ x < 2⎪0.95 if 2 ≤ x < 3⎪⎪⎩ 1 if x > 3F()x1.00.80.60.40.201 2 3⎧ 0 if x < 0⎪⎪0.7 if 0 ≤ x < 1⎪0.85 if 1 ≤ x < 2F( x)= ⎨⎪ 0.9 if 2 ≤ x < 3⎪0.95 if 3 ≤ x < 4⎪⎪⎩ 1 if x ≥ 4Fx ()1.00.80.60.40.2x30. PX ( = x) = ∫x ftdt ( ) = 0. Consequently,xPX ( < c) = PX ( ≤c). As a result, all fourexpressions, Pa ( < X< b), Pa ( ≤ X≤b),Pa ( < X≤b)and Pa ( ≤ X

36. a. PZ ( > 1) = 1 −PZ ( ≤ 1) = 1 − F(1)1 8= 1− = 9 95.8 Chapter ReviewConcepts Testb. P(1 < Z < 2) = P(1 ≤ Z ≤ 2) = F(2) − F(1)4 1 1= − =9 9 31. False:π∫ 0 cos xdx = 0 because half of the arealies above the x-axis and half below the x-axis.c.2zf( z) = F′( z) = ,0≤ z ≤ 392. True: The integral represents the area of theregion in the first quadrant if the center ofthe circle is at the origin.37.38.39.3332z⎡2z⎤d. EZ ( ) = ∫ z⋅ dz= ⎢ ⎥ = 20 9 ⎢⎣27 ⎥⎦04 15 2 2EX ( ) = ∫ x⋅ x(4 − x) dx=20 5122 4 2 15 2 2and E(X ) = ∫ x ⋅ x (4 −x)dx0 51232= ≈ 4.57 using a CAS72 8 2 3EX ( ) = ∫ x ⋅ x(8 − xdx ) = 19.2 and0 2563 8 3 3E( X ) = ∫ x ⋅ x(8 − x) dx = 102.40 256using a CAS2V( X) = E⎡( X − μ) ⎤, where μ = E( X) = 2⎣ ⎦4 2 15 2 2 4V( X) = ∫ ( x−2) ⋅ x (4 − x)dx =0 512 78 3μ = E X = x⋅ x 8− x dx = 40 2568 2 3 16V( X) = ∫ ( x−4) ⋅ x(8 − x)dx =0 256 540. ( ) ∫ ( )E⎡X μ ⎤⎢−⎥= E( X − 2 Xμ+μ )⎣ ⎦41. ( ) 2 2 2( μ)2 2= EX ( ) − E2 X + E( μ )2 2= EX ( ) −2 μ⋅ EX ( ) + μ2 2 2= EX ( ) − 2 μ + μ since EX ( ) = μ2 2= EX ( ) −μ2 2For Problem 37, V( X) = E( X ) −μ and32 2 4using previous results, V( X) = − 2 =7 73. False: The statement would be true if eitherf(x) ≥ g(x) or g(x) ≥ f(x) fora ≤ x ≤ b. Consider Problem 1 with f(x)= cos x and g(x) = 0.4. True: The area of a cross section of a cylinderwill be the same in any plane parallel tothe base.5. True: Since the cross sections in all planesparallel to the bases have the same area,the integrals used to compute the volumeswill be equal.6. False: The volume of a right circular cone of1 2radius r and height h is3 π r h . If theradius is doubled and the height halved2 2the volume is3 π r h.7. False: Using the method of shells,1 2= π∫− +0V 2 x( x x)dx . To use themethod of washers we need to solve2y = − x + x for x in terms of y.8. True: The bounded region is symmetric about1the line x = . Thus the solids obtained2by revolving about the linesx = 0 and x = 1 have the same volume.9. False: Consider the curve given bysin ty = , 2 ≤ t

11. False: If the cone-shaped tank is placed with thepoint downward, then the amount ofwater that needs to be pumped from nearthe bottom of the tank is much less thanthe amount that needs to be pumped fromnear the bottom of the cylindrical tank.12. False: The force depends on the depth, but theforce is the same at all points on a surfaceas long as they are at the same depth.13. True: This is the definition of the center ofmass.14. True: The region is symmetric about the point(π , 0).15. True: By symmetry, the centroid is on the lineπx = , so the centroid travels a distance2⎛π⎞ 2of 2 π ⎜ ⎟= π .⎝2⎠16. True: At slice y, ΔA ≈(9 − y ) Δ y.17. True: Since the density is proportional to thesquare of the distance from the midpoint,equal masses are on either side of themidpoint.18. True: See Problem 30 in Section 5.6.19. True: A discrete random variable takes on afinite number of possible values, or aninfinite set of possible outcomes providedthat these outcomes can be put in a listsuch as {x 1 , x 2 , …}.20. True: The computation of E(X) would be thesame as the computation for the center ofmass of the wire.21. True: EX ( ) = 5⋅ 1=5x22. True: If F( x) = ∫ f( t)dt, then F′ ( x) = f( x)Aby the First Fundamental Theorem ofCalculus.23. True:1= = ≤ ≤ = ∫ =1PX ( 1) P(1 X 1) f( xdx ) 0Sample Test Problems1.11 2 ⎡ 2 3⎤= ∫ ( − ) = − =0⎢ ⎥⎣ ⎦01 1 1A x x dx x x2 3 622.3.4.5.6.=π 1 − 2 2∫V 0 ( x x ) dx=π 1 2 − 3 + 4∫0 ( x 2 x x ) dx1⎡1 3 1 4 1 5⎤=π⎢x − x + x ⎥ =⎣⎦0π3 2 5 301 2 1 2 3= π∫ − = π −0 ∫ 01⎡1 3 1 4⎤π= 2π⎢x − x3 4⎥ =⎣ ⎦06V 2 x( x x ) dx 2 ( x x ) dx1 2 2 2=π ⎡( 2) (2) ⎤∫ − + −0 ⎣⎦V x x dx=π 1 4 − 3 − 2+∫0 ( x 2x 3x 4 x)dx1⎡ 5 4 3 2⎤=π⎢x − x − x + 2x⎥ =⎣⎦01 1 7π5 2 101 2= π∫− −0V 2 (3 x)( x x ) dx1 3 2= 2 π∫( x − 4x + 3 x)dx01⎡ 4 3 2⎤= 2π⎢x − x + x ⎥ =⎣⎦01 4 3 5π4 3 2 61 3 412 1 1xx ( x ) dx ⎡ x x ⎤∫ −−0 ⎣3 4 ⎦01x = = =1 2 2 31( x x ) dx 1 1 2∫ − ⎡ x − x ⎤0 ⎣2 3 ⎦01 3 4 511 2 2 1 1 1 1( x x ) dx ⎡ x x x ⎤2 ∫ −− +0 2⎣3 2 5 ⎦0y = =1 2 2 31( x x ) dx 1 x 1∫ −⎡ − x ⎤0 ⎣2 3 ⎦01=1017. From Problem 1, A = .61 1From Problem 6, x = and y = .2 10⎛ 1 ⎞⎛1⎞πV( S1) = 2π ⎜ ⎟⎜ ⎟=⎝10 ⎠⎝6 ⎠ 30⎛1⎞⎛1⎞πV( S2) = 2π ⎜ ⎟⎜ ⎟=⎝2⎠⎝6⎠6⎛ 1 ⎞⎛1⎞7πV( S3) = 2π ⎜ + 2⎟⎜ ⎟=⎝10 ⎠⎝6 ⎠ 10⎛ 1⎞⎛1⎞5πV( S4) = 2π⎜3− ⎟⎜ ⎟=⎝ 2 ⎠⎝ 6 ⎠ 6342 Section 5.8 Instructor’s Resource Manual

8. 8 = F(8) = 8k, k = 19.a.b.88 ⎡1 2 ⎤ 1W = ∫ xdx = x (64 4)2 ⎢ = −2⎥⎣ ⎦22= 30 in.-lb4−4 ⎡12 ⎤W = ∫ xdx = x = 8 in.-lb0 ⎢2 ⎥⎣ ⎦0W = 6 2∫ (62.4)(5 ) π (10 − y ) dy066 ⎡ 1 2 ⎤= 1560 π∫(10 − ydy ) = 1560π 10y y0⎢ −2⎥⎣ ⎦0= 65,520π≈ 205,837 ft-lb10. The total work is equal to the work W 1 to pull upthe object to the top without the cable and thework W 2 to pull up the cable.W 1 = 200⋅ 100 = 20,000 ft-lbThe cable weighs 120 = 6 lb/ft.100 56 6Δ W2= Δy⋅ y = yΔy5 5100100 6 6⎡12 ⎤W2= ∫ ydy = y0 5 5 ⎢2 ⎥⎣ ⎦0= 6000 ft-lbW = W + W = 26,000 ft-lb1 211. a. To find the intersection points, solve24x = x .2x − 4x= 0x(x – 4) = 0x = 0, 444 2 2 1 3(4 ) ⎡ ⎤∫20⎢3⎥⎣ ⎦0A= x− x dx = x − x⎛ 64 ⎞ 32= ⎜32− ⎟=⎝ 3 ⎠ 3b. To find the intersection points, solvey= y .42y= y162y − 16y= 0y(y – 16) = 0y = 0, 161616 ⎛ y ⎞ ⎡2 3/2 1 2⎤∫ 0⎜ ⎟4⎢3 8⎥⎝ ⎠ ⎣ ⎦0A= y − dy = y − y⎛128 ⎞ 32= ⎜ − 32⎟=⎝ 3 ⎠ 312.13.4 2 4 2 3∫ x(4 x−x ) dx (4 )0 ∫ x −x dx0x = =4 232∫ (4 x−x ) dx303 44⎡4 x − 1 x ⎤ 64⎣3 4 ⎦0 3= = = 232 323 314⎡ 2 2 22(4 x) ( x ) ⎤∫ − dx0y =⎣⎦4 2∫ (4 x−x ) dx01∫ 4 2 4(16 x − x ) dx2 0=3233 541⎡16x − 1 x ⎤ 10242⎣3 5 ⎦0 15 32= = =32 32 53 34 2 2 2V =π ⎡(4 x) ( x ) ⎤∫ − dx0 ⎣⎦=π 4 (16 2 4∫ x −x) dx04⎡16 3 1 5⎤2048π=π⎢x − x3 5⎥ =⎣ ⎦015Using Pappus’s Theorem:32From Problem 11, A = .332From Problem 12, y = .5⎛32 ⎞⎛32 ⎞ 2048πV = 2πy⋅ A= 2π ⎜ ⎟⎜ ⎟=⎝ 5 ⎠⎝ 3 ⎠ 1514. a. (See example 4, section 5.5). Think ofcutting the barrel vertically and opening thelateral surface into a rectangle as shown inthe sketch below.At depth 3 – y, a narrow rectangle has width16π , so the total force on the lateral surfaceis (δ = density of water = 62.4 lbs3)ft3 3δ π πδ0 ∫023∫(3 − y)(16 ) dy = 16 (3 − y)dy⎡ y ⎤= 16πδ⎢3y− ⎥ = 16 πδ (4.5) ≈14,114.55 lbs.⎢⎣2 ⎥⎦0Instructor’s Resource Manual Section 5.8 343

16.15.b. All points on the bottom of the barrel are atthe same depth; thus the total force on thebottom is simply the weight of the column ofwater in the barrel, namely2F = π(8 )(3) δ ≈ 37,638.8 lbs.dy 2= x −dx 4x123 ⎛ 2 1 ⎞L = ∫ 1+ x dx1⎜ −2 ⎟⎝ 4x⎠3 4 1 1 3 ⎛ 2 1 ⎞= ∫ x + + dx = x dx1 2 4 ∫ 1 ⎜ +2 ⎟16x⎝ 4x⎠31 3 1 1 1 1 53= ⎡ ⎢ x − ⎤ 93 4x⎥ = ⎛ ⎜ − ⎞ ⎟− ⎛ ⎜ − ⎞⎟=⎣ ⎦1⎝ 12⎠ ⎝3 4⎠62Lbaf x dxbL2 = ∫ a1 + [ g′( x)] 2 dxL3 = f( a) − g( a)L4 = f( b) − g( b)= L + L + L + L22. 1 = ∫ 1 + [ ′( )] 2Total length 1 2 3 4bA f x f x dxabA2 = 2 π ∫ gx ( ) 1 + [ g′( x)] 2 dxa23. 1 = 2 π ∫ ( ) 1 + [ ′( )] 22 2A3 =π⎡f ( a) −g ( a)⎤⎣⎦2 2A4 =π⎡f ( b) −g ( b)⎤⎣⎦Total surface area = A1+ A2 + A3 + A4.24. a. PX ( ≥ 1) = P(1≤ X≤2)2241 3 1 ⎡ x ⎤= ∫1( 8− x ) dx = ⎢8x−⎥12 12 ⎢⎣4 ⎥⎦117= ≈0.3544817.The loop is − 3 ≤t≤ 3. By symmetry, we candouble the length of the loop from t = 0 todx dy 2t = 3 , = 2; t = t − 1dt dt3 4 2 3 2L = 2∫ t + 2t + 1dt = 2 ( t + 1) dt0 ∫ 03⎡132 4 33 t t ⎤= ⎢ + ⎥ =⎣ ⎦0322 3 2V =⎛9 x⎞∫ ⎜ − ⎟ dx = (9 −x ) dx−3 ⎝ ⎠∫ −33⎡ 1 3 ⎤= ⎢9 x− x (27 9) ( 27 9) 363⎥ = − − − + =⎣ ⎦−3bA f x g x dxa18. = ∫ [ ( ) − ( )]19.b 2 2V =π ⎡f ( x) g ( x)⎤∫ − dxa ⎣⎦bV x a f x g x dxa20. = 2 π∫( − )[ ( ) − ( )]bM x f x g x dxaδ b 2 2M x = ⎡f ( x) g ( x)⎤dx2∫ −a ⎣⎦21. y = δ ∫ [ ( ) − ( )]25. a.b. P(0 ≤ X < 0.5) = P(0 ≤ X ≤ 0.5)20.541 3 1 ⎡ x ⎤= ∫0( 8− x ) dx = ⎢8x−⎥12 12 ⎢⎣4 ⎥⎦185= ≈0.332256522 1 3 1 ⎡2 x ⎤E( X) = ∫ x⋅ 8− x dx = ⎢4x− ⎥0 12 12 ⎢⎣5 ⎥⎦0= 0.8c. ( )4x1 1 ⎡ t ⎤F( x) = ∫ 8− t dt = ⎢8t−⎥0 12 12 ⎢⎣4 ⎥⎦04 41 ⎛ x ⎞ 2 x= 8x− = x−12 ⎜ 4 ⎟⎝ ⎠3 48x 3d. ( )b.c.2(6 − 3) 3PX ( ≤ 3) = F(3) = 1− =36 41 6−xf( x) = F′( x) = 2 ⋅ (6 − x)= ,36 180≤ x ≤ 66 ⎛6− x ⎞E( X)= ∫ x⋅ dx0 ⎜ ⎟⎝ 18 ⎠361 ⎡32 x ⎤= ⎢ x − ⎥ = 218 ⎢⎣3 ⎥⎦0344 Section 5.8 Instructor’s Resource Manual

Review and Preview Problems1. By the Power Rule− 2+1 −11 −2x x 1dx = x dx = = = − + C2x− 2+1 −1x∫ ∫2. By the Power Rule− 15 . + 1 −05.1 −15. x x 2dx = x dx = = =− + C15 .x− 15 . + 1 −05. x∫ ∫3. By the Power Rule∫− 101 . + 1 −001.1 −101. x x 100101 . ∫001 .xdx = x dx = = =− + C− 101 . + 1 −001. x10. a.b.c.d.11. a.1 1 1( 1+ ) = 2 = 21101 10 11( 1 ) ⎛ ⎞+ = 2.59374210⎜ ≈10⎟⎝ ⎠1001 100 101( 1 ) ⎛ ⎞+ = 2.704814100⎜ ≈100⎟⎝ ⎠10001 1000 ⎛1001⎞( 1+ ) = ≈ 2 71692391000⎜1000⎟ .⎝ ⎠1 2 13( 1 ) ⎛ ⎞+ = 2.252⎜ =2⎟⎝ ⎠24. By the Power Rule− 099 . + 1 001 .1−099 . x x001 .∫ dx = x dx = = = 100x + C099 . ∫x− 099 . + 1 001 .5.11F()1 = ∫ dt = 01t6. By the First Fundamental Theorem of Calculusd x1 1F′ ( x)= ∫ dt =1dx t x27. Let g( x)= x ; then by the Chain Rule andproblem 6,2DFxx( ) = DFgxx( ( )) = F′ ( gx ( )) g′( x)⎛ 1 ⎞= ⎜ ( 2x)22 ⎟ =⎝ x ⎠ x38. Let hx ( ) = x ; then by the Chain Rule andproblem 6,3x 1Dx∫dt = D ( ( )) ( ( )) ( )1xF h x = F′ h x h′xt⎛ 1 ⎞ 2 3= ⎜ ( 3x) =3 ⎟⎝ x ⎠ xb.c.d.e.12. a.21 1 10 105 1 1115⎛ ⎞ ⎛ ⎞+ = 1+ = ≈ 2 593742( )2⎜ .10⎟ ⎜10⎟⎝ ⎠ ⎝ ⎠21 1 20 2010 1 21110⎛ ⎞ ⎛ ⎞+ = 1+ = ≈ 2 6533( )2⎜ .20⎟ ⎜20⎟⎝ ⎠ ⎝ ⎠12 1 100 10050 1 101150⎛ ⎞ ⎛ ⎞( + ) = 12⎜ +100⎟ = ⎜100⎟⎝ ⎠ ⎝ ⎠≈ 2.7048112 1 200 200100 1 2011100⎛ ⎞ ⎛ ⎞( + ) = 12⎜ +200⎟ = ⎜200⎟⎝ ⎠ ⎝ ⎠≈ 2.711522 1( 1+ ) 2= 3 ≈ 1.73205112 10 5( 1+ ) = 1. 2 ≈ 2.4883210b.2( )2 100 50( 1+ ) = 1. 02 ≈ 2.691588100c.2( )2 1000 500( 1+ ) = 1. 002 ≈ 2.7155691000d.2( )9. a.b.c.d.e.11( 1+ 1)= 2 = 211 1 51 615⎛ ⎞+ = = 2 48832( )5⎜ .5⎟⎝ ⎠1 1 101 11110⎛ ⎞+ = ≈ 2 593742( )10⎜ .10⎟⎝ ⎠1 1 501 51150⎛ ⎞+ = ≈ 2 691588( )50⎜ .50⎟⎝ ⎠1 1 1001 1011100⎛ ⎞+ = ≈2 704814( )100⎜ .100⎟⎝ ⎠13. We know from trigonometry that, for any x andany integer k , sin( x + 2 kπ) = sin( x). Since⎛π⎞ 1 ⎛5π⎞ 1sin ⎜ ⎟ = and sin ⎜ ⎟ = ,⎝ 6 ⎠ 2 ⎝ 6 ⎠ 21 π 12k+ 1sin( x) = if x = + 2kπ= π2 6 65π12k+ 5or x = + 2kπ= π6 6where k is any integer.Instructor’s Resource Manual Review and Preview 345

14. We know from trigonometry that, for any x andany integer k , cos( x + 2 kπ) = cos( x). Sincecos( π ) = -1, cos( x) =− 1if x = π + 2 kπ = (2k+ 1) π where k is any integer.15. We know from trigonometry that, for any x andany integer k , tan( x + kπ) = tan( x). Since⎛π⎞ π 4k+ 1tan ⎜ ⎟ = 1, tan( x) = 1 if x = + kπ= π⎝ 4⎠4 4where k is any integer.116. Since sec( x)= , sec( x)is never 0 .cos( x)17. In the triangle, relative to θ ,2opp = x − 1, adj = 1, hypot = x so that2x −1 12sinθ = cosθ = tanθ= x −1xx1xcotθ = secθ = x cscθ=2 2x −1 x −118. In the triangle, relative to θ ,opp = x , adj = 1 − x , hypot = 1 so that2xsinθ = x cosθ = 1− x tanθ=21−x21−x1 1cotθ = secθ = cscθ=x21−x x19. In the triangle, relative to θ ,opp = 1, adj = x , hypot = 1+ x so that1 x1sinθ = cosθ = tanθ=2 21+ x1+x x21+x2cotθ = x secθ = cscθ= 1+xx20. In the triangle, relative to θ ,2opp = 1 − x , adj = x , hypot = 1 so that221−xsinθ = 1− x cosθ = x tanθ=xx1 1cotθ = secθ = cscθ=2 21−xx 1−x2221.22.2 2y'= xy → dy = xy dx1dy = xdx2ydy∫ = xdx2 ∫y1 1 2− = x + Cy 2When x = 0 and y = 1 we get C =− 1. Thus,21 1 2 x − 2− = x − 1 =y 2 22y =−2x − 2cos x cos xy'= → dy = dxyyydy=cos xdx∫ ydy = ∫ cos x dx1 2y = sin x+C2When x = 0 and y = 4 we get C = 8 . Thus,1y2y22= sin x+8= 2sin x+16346 Review and Preview Instructor’s Resource Manual

CHAPTER 6TranscendentalFunctions6.1 Concepts Review1.2.3.∫x1dt ; (0, ∞ ); (– ∞ , ∞ )t11x1 ; ln x + Cx4. ln x + ln y; ln x – ln y; r ln xProblem Set 6.11. a. ln 6 = ln (2 · 3) = ln 2 + ln 3= 0.693 + 1.099 = 1.792b.c.d.e.f.⎛3⎞ln1.5 = ln ⎜ ⎟= ln 3 − ln 2 = 0.406⎝2⎠4ln 81 = ln 3 = 4ln 3 = 4(1.099) = 4.3961/2 1 1ln 2 = ln 2 = ln 2 = (0.693) = 0.34652 2⎛ 1 ⎞ 2 2ln ⎜ ⎟ = – ln 36 = – ln(2 ⋅ 3 )⎝36⎠=−2ln2− 2ln3=−3.5844ln 48 = ln(2 ⋅ 3) = 4ln 2 + ln 3 = 3.8712. a. 1.792 b. 0.4053.4.5.c. 4.394 d. 0.3466e. –3.584 f. 3.8712Dxln( x + 3 x+π)12= ⋅ D ( 3 )2x x + x+π=x + 3x+πx2x+ 32+ 3x+π3 13Dxln(3x + 2 x) = D (3 2 )3 x x + x3x+ 2x29x+ 2=33x+ 2x3D ln( x– 4) = D 3ln( x– 4)x1 3= 3 ⋅ Dx( x–4)=x– 4 x–4x6.7.8.9.10.11.12.13.14.1Dxln 3 x–2 = Dxln(3 x–2)21 1 3= ⋅ Dx(3 x– 2) =2 3 x– 2 2(3 x– 2)dy 1 3= 3⋅ =dx x xdy 2 1= x ⋅ + 2x⋅ lnx= x(1 + 2 ln x)dx x2 2 3z = x ln x + (ln x)2 3= x ⋅ 2ln x+(ln x)dz 2 2 2 1= x ⋅ + 2x⋅ 2ln x+ 3(ln x)⋅dx x x3 2= 2x + 4xln x+(ln x)x3ln x ⎛ 1 ⎞r = + ln2 2 ⎜ ⎟x ln x ⎝ x ⎠= 1 –2 3– (ln )2 x xdr –2 –3 2 1= x –3(ln x)⋅dx 2xln x3= + (– ln x)2x ⋅ 2lnx21 3(ln x)=− −3x xg 1 1′ 2 –1/2( x ⎡) 1 ( 1) 221 2x x ⎤= ⎢ + + ⋅ ⎥x+ x + ⎣⎦1=2x + 1h 1 1′ 2 –1/2( x ⎡) 1 ( –1) 22–1 2x x ⎤= ⎢ + ⋅ ⎥x+x ⎣⎦1=2x − 13 1f ( x) = ln x = lnx31 1 1f′ ( x)= ⋅ =3 x 3x1 1f ′(81)= =3⋅81 2431f ′( x) = (–sin x) = –tanxcos xπ πf ′⎛ ⎞ ⎛ ⎞⎜ ⎟= − tan ⎜ ⎟= −1.⎝4⎠ ⎝4⎠Instructor’s Resource Manual Section 6.1 347

15. Let u = 2x + 1 so du = 2 dx.1 1 1∫ dx = du2x+ 1 2∫u1 1= ln u + C = ln 2x+ 1 + C2 216. Let u = 1 – 2x so du = –2dx.1 1 1∫ dx = – du1–2x2∫u1 1= – ln u + C = – ln1–2x + C2 2217. Let u = 3v + 9vso du = 6v + 9.6v+ 9 1∫ dv = du ln u C2 ∫ = +3v+ 9vu2= ln 3v + 9v + C218. Let u = 2z+ 8 so du = 4z dz.z 1 1∫ dz = du22z+ 8 4∫u1 1 2= ln u + C = ln ( 2z + 8)+ C4 4119. Let u = ln x so du = dxx2lnx ∫ dx = 2 udux∫2 2= u + C = (ln x)+ C120. Let u = ln x, so du = dx .x–1–2∫ dx = – u du2 ∫x(ln x)= 1 1C Cu+ = ln x+521. Let u = 2x+π so du = 10x dx .4x 1 1∫ dx = du52x+π 10∫u1 1 5= ln u + C = ln 2x +π + C10 10433 x ⎡ 1 5 ⎤dx = ln 2 x0 5 ⎢ +π2x10⎥+π ⎣⎦0∫1= [ln(486 +π ) – ln π ] = 10 486 +πln ≈ 0.504810π4222. Let u = 2t + 4t+ 3 so du = (4t + 4) dt .t + 1 1 1dt = du22t+ 4t+3 4 u1 1 2ln u C ln 2t 4t34 411 t + 1 ⎡1 ln 22 ⎤dt = 4 30 2⎢ t + t +2t4t3 4⎥+ + ⎣⎦0∫ ∫= + = + + + C∫1 1 94 4ln 9 – ln 3 ln ln 34 4 3= = = = 1 ln 3423. By long division,so222x1= x + 1+x −1 x −1x1dx = x dx + 1dx + dxx−1 x−1∫ ∫ ∫ ∫24. By long division,2x= + x+ ln x− 1 + C222x + x x 3 3= + +2x−1 2 4 4(2x−1)x + x x 3 3dx = dx + dx + dx2x−1 2 4 4(2x−1)∫ ∫ ∫ ∫x 3 3 1= + x+∫ dx4 4 4 2x−1Let u = 2x− 1; then du = 2dx. Hence1 1 1 1∫ dx = ∫ du = ln u + C2x−1 2 u 21= ln 2 x− 1 + C2and∫2 225. By long division,4x + x x 3 3dx = + x + ln 2 x − 1 + C2x−1 4 4 8x 3 2 256= x − 4x + 16x− 64+x+ 4 x+4∫4xdx =x + 43 21x dx − 4x dx + 16xdx − 64dx + 256 dxx + 44 3so∫ ∫ ∫ ∫ ∫x 4x2= − + 8x − 64x+ 256ln x+ 4 + C4 33 2x + x 2 426. By long division, = x − x+ 2 − sox+ 2 x+23 2x + x 21dx = x dx − xdx + 2 dx −4dx∫ ∫ ∫ ∫ ∫x+ 2 x+23 2x x= − + 2x− 4ln x+ 2 + C3 2so348 Section 6.1 Instructor’s Resource Manual

27.28.2 ( 1)2ln( x + 1) – ln x = ln( x+1) – ln x ln x +=x1 1ln( x – 9) + ln x = ln x– 9 – ln x2 2x– 9 x–9= ln = lnx x29. ln(x – 2) – ln(x + 2) + 2 ln x30.= ln( x – 2) – ln( x+ 2) + ln x2ln( x – 9) – 2ln( x– 3) – ln( x+ 3)2 2= ln( x −9) −ln( x−3) − ln( x+3)2ln x –9 1== ln ( –3) 2x ( x+3) x – 32x2 ( x–2)= lnx + 2231.1 3ln y = ln( x+11) – ln( x – 4)21 dy 1 1 1= ⋅1– ⋅ ⋅3xydx x+11 2 3x –41 3x–= x + 11 32( x – 4)dy ⎡ 1 3x2 ⎤= y ⋅⎢ – ⎥dx x 11 3⎢⎣+ 2( x – 4) ⎥⎦2x+11 ⎡ 1 3x⎤= ⎢ – ⎥3 11 3x –4⎢⎣x + 2( x – 4) ⎥⎦3 2– x + 33x+ 8=2( 3 – 4) 3/2x2232.33.34.35.2 2ln y = ln( x + 3 x) + ln( x– 2) + ln( x + 1)1 dy 2x + 3 1 2x= + +ydx 2 –2 2x + 3x x x + 1dy 2 2 ⎛ 2x + 3 1 2x⎞= ( x + 3 x)( x–2)( x + 1) ⎜ + +dx2 –2 2 ⎟⎝ x + 3x x x + 1⎠1 1ln y = ln( x+ 13) – ln( x– 4) – ln(2x+1)2 31 dy 1 1 2= – –ydx 2( x+ 13) x– 4 3(2x+1)4 3 2= 5x + 4 x –15x + 2 x–62dy x + 13 ⎡ 1 1 2 ⎤ 10x+ 219 x–118= – –dx 3 ⎢( x–4) 2x1 2( x + 13) x – 4 3(2x+ 1)⎥ = –2 1/2 4/3+ ⎣ ⎦ 6( x – 4) ( x+ 13) (2 x+1)2 21ln y = ln( x + 3) + 2ln(3x+ 2) – ln( x+1)3 21 dy 2 2x2⋅3 1= ⋅ + –ydx 3 2x + 3 3x+ 2 2( x+1)2 2/3 2dy ( x 3) (3x 2) ⎡ 4x6 1 ⎤3 2+ +(3x+ 2)(51x + 70x + 97x+90)= ⎢ + – ⎥ =dx x + 1 2⎢⎣3( x + 3) 3x + 2 2( x + 1)2 1/3 3/2⎥⎦6( x + 3) ( x+1)36.y = ln x is reflected across the y-axis.The y-values of y = ln x are multiplied by 1 ,21since ln x = ln x.2Instructor’s Resource Manual Section 6.1 349

37.38.39.y = ln x is reflected across the x-axis since⎛1⎞ln ⎜ ⎟ = – ln x.⎝ x ⎠y = ln x is shifted two units to the right.42. Let r(x) = rate of transmission2 1 2= kx ln = − kx ln x.x2 1r′ ⎛ ⎞( x) = −2kxln x− kx ⎜ ⎟=− kx(2lnx+1)⎝ x ⎠11r′ ( x) = 0 if ln x = − , or − ln x = , so221 1ln .x = 21 1ln1.65 ≈ , so x ≈ ≈ 0.606.2 1.651r′′ ⎛ ⎞( x) = − k(2lnx+ 1) −kx⎜2⋅⎟ = –k(2 ln x + 3)⎝ x ⎠r′′ (0.606) ≈ − 2k< 0 since k > 0, sox ≈ 0.606 gives the maximum rate oftransmission.43. ln 4 > 1mso ln 4 = mln 4 > m⋅ 1 = mmThus x > 4 ⇒ lnx > mso lim ln x = ∞x→∞1+44. Let z = so z →∞as x→0x⎛1⎞Then lim ln x = lim ln ⎜ ⎟=lim (– ln z)x→0+ z→∞ ⎝ z ⎠ z→∞= – lim ln z = – ∞z→∞y = ln cos x+ln sec x1= ln cos x + ln cos x⎛ π π⎞= ln cos x− ln cos x = 0 on ⎜−, ⎟⎝ 2 2⎠40. Since ln is continuous,sin x sin xlim ln = ln lim = ln1 = 0x→0 x x→0x41. The domain is ( 0,∞ ).2 1f′ ⎛ ⎞( x) = 4xlnx+ 2x ⎜ ⎟− 2x = 4xlnxxf '( x ) = 0 if ln x = 0 , or x = 1 .f '( x ) < 0 for x < 1 and ( )⎝f ' x > 0 for x > 1so f(1) = –1 is a minimum.⎠x 1 x145. ∫ dt = 2 dt1/3t∫ 1 t1 1 x1 x1∫ dt + dt 2 dt1/3t ∫ =1 t∫ 1 t1 1 x1∫ dt = dt1/3t∫ 1 t1/31 x1– ∫ dt = dt1 t∫ 1 t1– ln = ln x3ln 3 = ln xx = 346. a.1 1< for t > 1,t tx1 x 1 x –1/2so ln x = ∫ dt < dt t dt1 t∫ =1 ∫t 1x= ⎡2 t⎤⎣ ⎦= 2( x –1)1so ln x

47.48.49. a.b. If x > 1, 0< lnx< 2( x –1),ln 2( –1)so 0 x x< < .x xln x 2( x + 1)Hence 0≤ lim ≤ lim = 0x→∞x x→∞xln xand lim = 0.x→∞x⎡ 1 1 1 ⎤lim ⎢ + +⋅⋅⋅+n n 1 n 2 2n⎥→∞ ⎣ + + ⎦lim⎡ 1 1 1 ⎤ 1= ⎢ + +⋅⋅⋅+ ⎥⋅n→∞⎢ 1 1 1 2 1 n n⎣+ + + ⎥n n n ⎦n ⎛ 1 ⎞ 1= lim ∑ ⎜ ⎟⋅n→∞ 1 1ii =⎜ + ⎟ n⎝ n ⎠1,000,00072,382ln1,000,000 ≈b.2 1= ∫ dx = ln 2 ≈0.6931 xc⎛ax – b ⎞ ⎛ax − b ⎞f( x) = ln⎜ ⎟ = cln⎜ ⎟⎝ax + b ⎠ ⎝ax + b ⎠2 2a – b= [ln( ax – b) – ln( ax + b)]2ab2 2a – b ⎡ a a ⎤f′ ( x) = –2 ab ⎢ax – b ax + b ⎥⎣⎦2 2 2 2a − b ⎡ 2 ab ⎤ a – b=2 ab⎢( ax b)( ax b) ⎥ =⎣ − + ⎦ a x – b2 2a − bf ′(1) = = 12 2a − b2 duf′ ( x) = cos u⋅dx2 2 2x+ 1= cos [ln( x + x–1)]⋅2x + x –1(1) cos 2 [ln(1 21–1)] 21 ⋅ + 1f ′ = + ⋅1 2+ 1–12= 3cos (0) = 32 2 251. From Ex 10,π3π4∫sec xcsc xdx =⎡− ⎣ ln cos x + ln sin x ⎤⎦π3π4=⎡⎣ln tan x ⎤ ⎦ = ln tan −ln tanππ3 4= ln( 3) − ln1 = 0.5493 − 0 = 0.549352. Let u = 1+ sinx; then du = cos x dx so that53.54.cos x 1∫ dx = ∫ du = ln u + C1+sinxu= ln 1+ sin x + C = ln(1 + sin x)+ C(since 1+ sinx≥ 0for all x ).4 4 2πxV = 2 π ∫ xf( x)dx = dx1 ∫ 1 2x + 42Let u = x + 4 so du = 2x dx.2πx 1∫ dx = π du ln u C2 ∫ = π +x + 4 u2= π ln x + 4 + C4 2πx 4dx2= ⎡π ln x + 4 ⎤1 2x + 4⎢⎣⎥⎦1∫= π ln 20 − π ln 5 = π ln 4 ≈ 4.3552 2x x 1y = – ln x = – ln x4 4 2dy 2 1 1 1= x – ⋅ =x –dx 4 2 x 2 2xL =π3π42 22 ⎛dy⎞ 2 ⎛ x 1 ⎞1+ dx 1 – dx1 ⎜ ⎟ = +dx 1 ⎜ ⎟⎝ ⎠ ⎝2 2x⎠∫ ∫22 ⎛ x 1 ⎞ 2⎛ x 1 ⎞= ∫ dxdx1 ⎜ + ⎟ =2 2x∫ +1 ⎜ ⎟⎝ ⎠ ⎝2 2x⎠221 ⎡ x1ln x⎤ 1⎡ ⎛ ⎞= ⎢ + ⎥ = 2 + ln 2 − + ln1⎤2 2 2⎢ ⎜ ⎟2⎥⎢⎣⎥⎦⎝ ⎠1 ⎣⎦3 1= + ln 2 ≈ 1.0974 250. From Ex 9,π 3 tan xdx=⎡−ln cos x π 30 ⎣ ⎤⎦0∫= ln cos 0 −ln cosπ3⎛ 1 ⎞= ln(1) − ln(0.5) = ln ⎜ ⎟⎝ 0.5 ⎠= ln 2 ≈0.69315Instructor’s Resource Manual Section 6.1 351

55.b.1+1.5sinxf′′ ( x)=−2(1.5 + sin x)On [0,3 π ], f ( x) 0′′ = when x ≈ 3.871,5.553.Inflection points are (3.871, –0.182),(5.553, –0.182).56.1 1 1+ +⋅⋅⋅+ = the lower approximate area2 3 n1 11+ +⋅⋅⋅+ = the upper approximate area2 n –1ln n = the exact area under the curveThus,1 1 1 1 1 1ln 1 .2 + 3 +⋅⋅⋅+ nn< < + 2 + 3 +⋅⋅⋅+ n − 1y1 x1 y1dt – dt dtln y– ln x ∫1 t∫1t∫x= =ty – x y– x y–x= the average value of 1 on [x, y].tSince 1 is decreasing on the interval [x, y], thetaverage value is between the minimum value of1y and the maximum value of 1 .xc.58. a.59.3πln(1.5 + sin xdx ) ≈ 4.0420∫sin(ln x)f′ ( x)=−xOn [0.1, 20], f′ ( x) = 0 when x = 1.Critical points: 0.1, 1, 20f(0.1) ≈ –0.668, f(1) = 1, f(20) ≈ –0.989On [0.1, 20], the maximum value point is(1, 1) and minimum value point is(20, –0.989).b. On [0.01, 0.1], f′ ( x) = 0 when x ≈ 0.043.f(0.01) ≈ –0.107, f(0.043) ≈ –1On [0.01, 20], the maximum value point is(1, 1) and the minimum value point is(0.043, –1).c.20∫ cos(ln xdx≈− ) 8.370.157. a.1 cosxf′ ( x) = ⋅ cosx=1.5 + sin x 1.5 + sin xf′ ( x) = 0 when cos x = 0.π 3π 5π Critical points: 0, , , , 3 π2 2 2f(0) ≈ 0.405,⎛π⎞ ⎛3π⎞f ⎜ ⎟≈0.916, f ⎜ ⎟≈ −0.693,⎝2⎠ ⎝ 2 ⎠5f⎛ π⎜⎞ ⎟ ≈ 0.916, f(3 π ) ≈ 0.405.⎝ 2 ⎠On [0,3 π ], the maximum value points are⎛π⎞ ⎛5π⎞⎜ ,0.916 ⎟, ⎜ ,0.916⎟and the minimum⎝2 ⎠ ⎝ 2 ⎠⎛3 π ⎞value point is ⎜ , − 0.693 ⎟ .⎝ 2 ⎠60.a.1⎡⎛1⎞ 2 ⎛1⎞⎤5xln x ln dx 0.1390 ⎢ ⎜ ⎟− ⎜ ⎟ = ≈x x⎥⎣ ⎝ ⎠ ⎝ ⎠⎦36∫b. Maximum of ≈ 0.260 at x ≈ 0.236a.∫107[ xln x− xln x] dx = ≈0.19436b. Maximum of ≈ 0.521 at x ≈ 0.0555352 Section 6.1 Instructor’s Resource Manual

6.2 Concepts Review1. f ( x1) ≠ f( x2)2. x;f–1 ( y )3. monotonic; strictly increasing; strictly decreasing4.–1 1( f )( ′ y)=f ′ ( x)Problem Set 6.21. f(x) is one-to-one, so it has an inverse.1Since f(4) = 2, f − (2) = 4 .2. f(x) is one-to-one, so it has an inverse.Since f(1) = 2, f−1 (2) = 1 .3. f(x) is not one-to-one, so it does not have aninverse.4. f(x) is not one-to-one, so it does not have aninverse.5. f(x) is one-to-one, so it has an inverse.Since f(–1.3) ≈ 2, f−1 (2) ≈− 1.3 .6. f(x) is one-to-one, so it has an inverse. Since⎛1⎞ 1 1f ⎜ ⎟ = 2, f − (2) = .⎝2⎠27.8.4 2 4 2f ( x) –5 x – 3 x –(5x 3 x ) 0′ = = + < for allx ≠ 0. f(x) is strictly decreasing at x = 0 becausef(x) > 0 for x < 0 and f(x) < 0 for x > 0. Thereforef(x) is strictly decreasing for x and so it has aninverse.6 4f ( x) 7x 5x0′ = + > for all x ≠ 0.f(x) is strictly increasing at x = 0 because f(x) > 0for x > 0 and f(x) < 0 for x < 0. Therefore f(x) isstrictly increasing for all x and so it has aninverse.9. f ′( θ ) = –sinθ< 0 for 0 < θ < πf (θ) is decreasing at θ = 0 because f(0) = 1 andf(θ) < 1 for 0 < θ < π . f(θ) is decreasing atθ = π because f(π ) = –1 and f(θ) > –1 for0 < θ < π . Therefore f(θ) is strictly decreasingon 0 ≤ θ ≤ π and so it has an inverse.10.2πf′ ( x) = –csc x< 0 for 0 < x 0 for z > 1f(z) is increasing at z = 1 because f(1) = 0 andf(z) > 0 for z > 1. Therefore, f(z) is strictlyincreasing on z ≥ 1 and so it has an inverse.12. f′ ( x) = 2x+ 1> 0 for x ≥ 2 . f(x) is strictlyincreasing on x ≥ 2 and so it has an inverse.13.14.4 2f ( x) x x 10 0′ = + + > for all real x. f(x) isstrictly increasing and so it has an inverse.() 1 cos 4 – r 4f r = ∫ tdt =r ∫ 1cos tdt4πf′ ( r) = – cos r < 0 for all r ≠ kπ+ , k any2integer.πf(r) is decreasing at r = kπ+ since f′ () r < 02on the deleted neighborhood⎛ π π ⎞⎜kπ+ −ε, kπ+ + ε ⎟.Therefore, f(r) is⎝ 2 2 ⎠strictly decreasing for all r and so it has aninverse.15. Step 1:y = x + 1x = y – 1–1Step 2: f ( y) = y–1–1Step 3: f ( x) = x–1Check:f–1 ( f( x)) = ( x+ 1)–1=x–1f( f ( x)) = ( x–1) + 1=x16. Step 1:xy = – + 13x– = y –13x = –3(y – 1) = 3 – 3y–1Step 2: f ( y) = 3–3y–1Step 3: f ( x) = 3–3xCheck:–1 ⎛ x ⎞f ( f ( x )) = 3–3 ⎜ – + 1 ⎟ = 3 + ( x –3) = x⎝ 3 ⎠–1 –(3 – 3 x)f( f ( x )) = + 1= (–1 + x) + 1 = x3Instructor’s Resource Manual Section 6.2 353

17. Step 1:y = x + 1 (note that y ≥ 0 )2x + 1 = yx = y2 –1, y ≥ 0–1 2Step 2: f ( y) = y –1, y ≥ 0–1 2Step 3: f ( x) = x –1, x≥0Check:–1 2f ( f( x)) = ( x+ 1) –1 = ( x+ 1)–1=x–1 2 2f ( f ( x)) = ( x –1) + 1= x = x = x18. Step 1:y = – 1– x (note that y ≤ 0 )1– x = – y2 21– x = (– y)= y2x = 1– y , y ≤ 0–1 2f ( y) = 1– y , y ≤ 0–1 2f ( x) = 1– x , x≤0Step 2:Step 3:Check:–1 2f ( f( x)) = 1–(– 1– x) = 1–(1– x)= x–1 2 2f ( f ( x)) = – 1–(1– x ) = – x = – x= –(–x) = x19. Step 1:– 1y =x – 31x – 3 = –y1x = 3–y–1 1Step 2: f ( y) = 3–y–1 1Step 3: f ( x) = 3–xCheck:–11f ( f( x)) = 3– = 3 + ( x–3)= x– 1x–3–11 1f ( f ( x )) = – =11( )– = x3– –3 –xx20. Step 1:1y = x – 2(note that y > 0)2 1y =x – 2x –2=12y1x = 2 + , y > 02yStep 2:Step 3:Check:–1 1f ( y) = 2 + , y > 02y–1 1f ( x) = 2 + , x > 02x–11 1f ( f( x )) = 2+ = 2+= 2 + (x – 2) = x2 1( ) ( )1 xx–2–2–1 1 1 2f ( f ( x))= = = x⎛2 1 ⎞–2⎛ 1 ⎞⎜ +x2 ⎟ ⎜x2 ⎟⎝ ⎠ ⎝ ⎠= x = x21. Step 1:2y 4 x , x 0= ≤ (note that y ≥ 0 )2x =y4y yx = – =− , negative since x ≤ 04 2Step 2:–1 yf ( y ) =−2Step 3:–1 xf ( x ) =−2Check:2–1 4x2f ( f( x)) = – = – x = – x = –(– x)= x22–1x xf ( f ( x)) 4 ⎛ ⎞= ⎜– = 4⋅ = x2 ⎟⎝ ⎠ 422. Step 1:2y = ( x–3) , x≥ 3 (note that y ≥ 0 )x –3=yx = 3 + yStep 2: f –1 ( y) = 3+yStep 3: f –1 ( x) = 3+xCheck:–1f ( f( x)) = 3 +2( x–3) = 3 + x–3= 3 + ( x –3) = x–1 2 2f ( f ( x)) = [(3 + x) – 3] = ( x)= x354 Section 6.2 Instructor’s Resource Manual

23. Step 1:y = ( x–1)x –1=x = 1+Step 2:Step 3:Check:33yy3–1 3f ( y) = 1+y–1 3f ( x) = 1+x–1 3 3f ( f( x)) = 1 + ( x–1) = 1 + ( x–1)= x–1 3 3 3 3f ( f ( x)) = [(1 + x) –1] = ( x)= x24. Step 1:y = x 5/2 , x≥02/5x = y–1 2/5Step 2: f ( y)= y–1 2/5Step 3: f ( x)= xCheck:–1 5/ 2 2 / 5f ( f( x)) = ( x ) = x–1 2 / 5 5/ 2f ( f ( x)) = ( x ) = x25. Step 1:x –1y =x + 1xy+ y = x–1x – xy = 1 + y1+yx =1– y–1 1+yStep 2: f ( y)=1 − y–1 1+xStep 3: f ( x)=1– xCheck:1+ x+ 1 + x–1 2xf f x = = = = xx+ 1– x+1 2x–1–1 x+1( ( ))1–x–1x+11+x–1 1– x( ( ))1+x+ 11– x–1 1 + x–1+x 2xf f x = = = = x1+ x+1– x 226. Step 1:⎛ x –1⎞y = ⎜ ⎟⎝ x + 1⎠1/3 x –1y =x + 11/3 1/3xy + y = x–131/3 1/3x – xy = 1+y1/31+yx =1/31– yStep 2: f–11+y( y)=1– y1/31/31/3–1 1+xStep 3: f ( x)=1/31– xCheck:1/3⎡ 31 x–1⎤+ ( 1)–1–1 ⎢ x+⎥ 1 + xx+1f ( f( x))=⎣ ⎦=31/3 1– x–1⎡1– x–1 ⎤ x+1⎢( x+1)⎣⎥⎦x+ 1 + x–1 2x= = = xx+ 1– x+1 21/33⎛ 1+x –1⎞1/3 1/33–1 ⎜ 1– x1/3 ⎟ ⎛1 + x –1+x ⎞f( f ( x))= ⎜1+x1/3 ⎟ =⎜ ⎜ 1/3 1/31 1 x 1– x ⎟⎜+⎝ + + ⎠1– x1/3 ⎟⎝ ⎠1/33⎛2x⎞1/3 3= = ( x ) = x⎜ 2 ⎟⎝ ⎠27. Step 1:3x + 2y =3x + 13 3x y+ y = x + 23 3x y– x = 2– y3 2– yx =y –1⎛2–y ⎞x = ⎜ ⎟⎝ y –1⎠Step 2: fStep 3: fCheck:f1/3⎛ y ⎞( y)= ⎜ ⎟⎝ y –1⎠–1 2–x ⎞( x)= ⎜ ⎟⎝ x –1⎠–1 ⎛2–1/31/331/3⎛2–x + 2 ⎞3 31/3–1⎜x3+ 1⎟ ⎛2x+ 2– x –2⎞( f( x))= ⎜x3⎟ =⎜ + 2 ⎜ 3 3–1 x 2– x –1 ⎟⎜+x3 ⎟ ⎝ ⎠+ 11/3⎝⎛ 3x ⎞= = x⎜ 1 ⎟⎝ ⎠3⎡ 2– x1/3⎤ ( –1 ) 2x + 2– x–1 x–1=⎣ ⎦=1/33 2– x⎡ 2– x ⎤ –1( –1 ) 1xx +f( f ( x))⎢⎣⎥⎦2– x+2 x–2x= = = x2– x+x–1 1⎠⎢ ⎥ + 2+ 1Instructor’s Resource Manual Section 6.2 355

28. Step 1:35⎛ x + 2 ⎞y = ⎜ 3x + 1 ⎟⎝ ⎠31/5 x + 2y =3x + 13 1/5 1/5 3x y + y = x + 23 1/5 3 1/5x y – x = 2– y1/53 2– yx =1/5y –11/51/3⎛2–y ⎞x = ⎜ 1/5y –1⎟⎝ ⎠1/3⎛ 1/5–1 2– y ⎞Step 2: f ( y)= ⎜ 1/5y –1⎟⎝ ⎠1/3⎛ 1/5–1 2– x ⎞Step 3: f ( x)= ⎜ 1/5x –1⎟⎝ ⎠Check:1/51/3⎧ ⎡ 35 ⎤ ⎫⎪2–⎛ x + 2 ⎞⎢⎪⎜ 3 ⎟ ⎥x 1–1 ⎪⎢⎝+ ⎠ ⎥ ⎪f ( f( x))=⎣ ⎦⎨51/5 ⎬⎪⎡⎛x3+ 2 ⎞ ⎤ ⎪⎪⎢⎜–1x3 ⎟ ⎥ ⎪⎪ ⎝ +⎩⎢⎣1 ⎠ ⎥⎦⎪⎭31/3⎛2– x + 2 ⎞3 31/3⎜ x3+ 1 ⎟ ⎛2x+ 2– x –2⎞= ⎜x3 ⎟ =⎜ + 2 ⎜ 3 3–1 x 2– x –1 ⎟⎜+x3 ⎟ ⎝ ⎠⎝ + 1 ⎠31/3⎛ x ⎞= = x⎜ 1 ⎟⎝ ⎠35⎧⎡1/51/3⎤⎫⎪ ⎛ 2– x ⎞⎢2⎪⎜ 1/5 ⎟ ⎥ +x –1–1 ⎪⎢⎝⎠ ⎥⎪f( f ( x))=⎣ ⎦⎨1/33 ⎬⎪⎡⎛2– x1/5 ⎞ ⎤ ⎪⎪⎢⎜+ 1x1/5 ⎟ ⎥ ⎪⎪⎩⎢⎣⎝–1 ⎠ ⎥⎦⎪⎭1/55⎛ 2– x 2⎞1/5 1/55⎜+x1/5 –1 ⎟ ⎛2– x + 2 x –2⎞= ⎜2– x1/5 ⎟ =⎜ ⎜ 1/5 1/51 2– x x –1 ⎟⎜ +⎝ + ⎠x1/5 ⎟⎝ –1 ⎠1/55⎛ x ⎞= = x⎜ 1 ⎟⎝ ⎠29. By similar triangles,This gives( 4 h2 /9)r 4h = 6. Thus, hr =232 3π r h π h4πhV = = =3 3 273 27Vh =4πVh = 3 34π30. v = v0 − 32tv = 0 when v0 = 32t, that is, whenv0t = . The position function is322s() t v0t 16tof= − . The ball then reaches a height2 2v0 v0 v0H = s( v0 / 32) = v0 − 16 =32 232 642v0= 64Hv0= 8 H31. f′ ( x) = 4x+ 1; f′( x) > 0 whenf′ ( x) < 0 when1x < − .41x >− and4⎛ 1 ⎤The function is decreasing on ⎜ −∞,−4⎥ and⎝ ⎦⎡ 1 ⎞increasing on ⎢−, ∞ ⎟ . Restrict the domain to⎣ 4 ⎠⎛ 1 ⎤⎡ 1 ⎞⎜ −∞,−4⎥ or restrict it to ⎢ − ,⎝ ⎦4∞ ⎟⎣ ⎠ .−1 1Then f ( x) = ( −1− 8x+ 33) or4−1 1f ( x) = ( − 1+ 8x+33).4356 Section 6.2 Instructor’s Resource Manual

32.33.3f′ ( x) = 2x− 3; f′( x) > 0 when x >23and f′ ( x) < 0 when x

−42. Find f1 ( x):1y = ,x−1 1f ( y)=y1x =y−1 1f ( x)=x1Find g− ( x):y = 3x + 2y − 2x =3−1 y − 2g ( y)=3−1 x − 2g ( x)=31hx ( ) = f( gx ( )) = f(3x+ 2) =3x+ 2(11 1 1 1 1 ) 2− − − − ⎛ ⎞ −xh ( x) = g ( f ( x))= g ⎜ ⎟=⎝ x ⎠ 3−1 − 1⎛1 ⎞ (3x+ 2) −2 3xh ( h( x))= h ⎜ ⎟= = = x⎝3x+ 2⎠3 3⎛ 1( ) − 2 ⎞−1x1 1hh ( ( x))= h⎜⎟= = = x⎜ 3 ⎟ ⎡ 11( ) 2⎤⎝ ⎠− + 2⎣ x ⎦( x )43. f has an inverse because it is monotonic(increasing):2f′ ( x) = 1+ cos x > 0a.−11 1( f )( ′ A) = 1f ′ π( ) = 22 1 cos π( 2 )=+44. a.ax + by =cx + dcxy + dy = ax + b(cy – a)x = b – dyb−dy dy−bx = = −cy − a cy − a−1 dy − bf ( y ) =−cy − a−1 dx − bf ( x ) =−cx − ab. If bc – ad = 0, then f(x) is either a constantfunction or undefined.c. If f = f −1 , then for all x in the domain wehave:ax + b dx −b+ = 0cx + d cx −a(ax + b)(cx – a) + (dx – b)(cx + d) = 02 2 2acx + ( bc−a ) x− ab+dcx2+ ( d −bc) x− bd = 02 2 2( ac + dc) x + ( d − a ) x + ( −ab − bd) = 0Setting the coefficients equal to 0 gives threerequirements:(1) a = –d or c = 0(2) a = ±d(3) a = –d or b = 0If a = d, then f = f −1 requires b = 0 andaxc = 0, so f ( x)= = x. If a = –d, there aredno requirements on b and c (other thanbc − ad ≠ 0 ). Therefore, f = f −1 if a = –dor if f is the identity function.b.c.−11 1 1( f )( ′ B)= f ′ 5π( ) = 2 5 76 1 cos π( 6 )=+42=71 1 1f ′(0) 1+cos (0) 2− 1′ = = =2( f )(0)45.1 −1f ( y)dy =0∫ (Area of region B)= 1 – (Area of region A)12 3= 1 − ∫ f0( x ) dx = 1 − = 5 5358 Section 6.2 Instructor’s Resource Manual

46.a∫ f ( xdx= ) the area bounded by y = f(x), y = 0,0and x = a [the area under the curve].b –1∫ f ( ydy= ) the area bounded by x = f–1 ( y)0x = 0, and y = b.ab = the area of the rectangle bounded by x = 0,x = a, y = 0, and y = b.Case 1: b > f(a)The area above the curve is greater than the areaof the part of the rectangle above the curve, sothe total area represented by the sum of the twointegrals is greater than the area ab of therectangle.Case 2: b = f(a)47. Given p > 1, q > 1, 1 + 1 = 1, andp qp–1f( x) = x ,solving 1 1 q+ = 1 for p gives p = , sop qq –11 1 1 q –1= = = = q –1.p –1 q q–( q–1)–1 ⎡ ⎤ 1q–1 ⎢⎣q–1⎥⎦p–1Thus, if y = x then1p–1q–1x = y = y , so–1 q–1f ( y) = y .p 1By Problem 44, since f( x)= x − is strictlya p–1 b q–10 0∫ ∫increasing for p > 1, ab ≤ x dx + y dya b⎡ p qx ⎤ ⎡y⎤ab ≤ ⎢ ⎥ + ⎢ ⎥⎢⎣ p ⎥⎦ q0 ⎢⎣ ⎥⎦0p qa bab ≤ +p q6.3 Concepts Review1. increasing; exp2. ln e = 1; 2.723. x; xThe area represented by the sum of the twointegrals = the area ab of the rectangle.Case 3: b < f(a)xx4. e ; e + CProblem Set 6.31. a. 20.086b. 8.1662c.e2 1.41e≈ ≈ 4.1d.2. a.cos(ln 4) 0.18e ≈e≈ 1.203ln2 ln(23) ln8e = e = e = 8The area below the curve is greater than the areaof the part of the rectangle which is below thecurve, so the total area represented by the sum ofthe two integrals is greater than the area ab of therectangle.ab −1ab ≤ f ( x) dx + f ( y)dy0 0∫ ∫ with equalityholding when b = f(a).3.4.b.ln 64 2 ln(641/2) ln 8= = = 8e e e3ln x ln x33e = e = x–2ln x ln x−2−212e = e = x =xInstructor’s Resource Manual Section 6.3 359

5.6.7.8.ln ecos x= cos x–2 x–3ln e = –2 x– 33 –3x3 –3xln( x e ) = ln x + ln e = 3ln x– 3xex–lnxxxe e= =ln xe x19.x2x2x212 x2x[ + ] = x( ) + xD e e D e D e1= ( e ) D e + e D x21 212222(x )− x x x= e e Dxx + e ⋅2xx2−12 x2x22xx1 2122(x ) 2x x= e x+ e ⋅2x29.ln 3+2ln x ln 3 2ln x ln x 22e = e ⋅ e = 3⋅ e = 3x= x e +22 xx xex10.11.12.eln22x 2 2ln x – yln x e x x 2– y= = = = xylnxln xy yex+ 2 x+ 2 x+2De x = e Dx( x+ 2) = e2 x2– x 2 x2– x 2x = x2 x2– x= e (4 x –1)De e D(2 x – x)ex20.⎡ 1 ⎤D e D e D e⎢⎣ e⎥⎦1 x2x−2 −x2x ⎢ + ⎥ =2 x + xxx−2 −2 −x22xxx−2 −3−x2= e D x + e D [ − x ]= e ⋅( − 2 x ) + e ⋅( − 2 x)1 x22e2x=− −x e3x213.14.15.x+2x+ 2 x+2eDe x = e Dxx+ 2 =2 x + 2– 1 – 1x2x2 ⎛ 1 ⎞De x = e Dx⎜–2 ⎟⎝ x ⎠– 1– 1 22 –3 2 xxe= e ⋅ 2x=3x2ln xxDe= Deln x22x = Dx x = 2x1x x x (ln x) ⋅1–x ⋅16. ln x ln xxDe ln xx = e Dxx = e ⋅ln x2(ln x)e=xln x(ln x–1)(ln x)2xy21. D [ e + xy] = D [2]xxye ( xD y+ y) + ( xD y+ y) = 0xyxxyxe D y + ye + xD y + y = 0xyxxe D y+ xD y = – ye – yxxyxx− ye – y y( e + 1) yDx y = = – = –xy xyxe + x x( e + 1) xxxxyx+y22. D [ e ] = D [4 + x+y]23. a.xxx+ye (1 + Dxy) = 1+Dxyx+ y x+yx xx+ y x+ye Dxy– Dxy = 1– ex+yxye + e D y = 1+D y1– eDx y = = –1x+ye –117.3 x 3 x x 3x = x + x3 x x 2 2 xD ( x e ) x D e e D ( x )= xe + e ⋅ 3 x = xe( x+3)18.x3ln x x3ln x 3x = xx3 ln x⎛3 2De e D( x ln x)1= e ⎜x ⋅ + ln x⋅3 x⎝ xx3 ln x 2 2= e ( x + 3x ln x)⎞⎟⎠The graph ofx-axis.xy = e is reflected across thex3x2= x eln (1+3ln x)360 Section 6.3 Instructor’s Resource Manual

24.25.b.xThe graph of y = e is reflected across they-axis.– a – ba< b⇒ – a > – b⇒ e > e , sinceincreasing function.2xf ( x)= e Domain = ( −∞, ∞ )2x2xxe is anf ′( x) = 2 e , f′′( x) = 4eSince f′ ( x) > 0for all x, f is increasing on( −∞, ∞ ).Since f′′ ( x) > 0for all x, f is concave upward on( −∞, ∞ ).Since f and f ′ are both monotonic, there are noextreme values or points of inflection.8yxe −26. f( x)=2Domain = ( −∞, ∞ )1 x1 xf ( x) e −−′ =−2, f′′( x)= e22 4Since f′ ( x) < 0for all x, f is decreasing on( −∞, ∞ ).Since f′′ ( x) > 0for all x, f is concave upward on( −∞, ∞ ).Since f and f ′ are both monotonic, there are noextreme values or points of inflection.−584x27. f ( x)= xe − Domain = ( −∞, ∞ )y− x5f ′( x) = (1 − x) e , f′′( x) = ( x−2) ex ( −∞,1) 1 (1,2) 2 (2, ∞)f ′ + 0 − − −f ′′ − − − 0 +x− xf is increasing on ( −∞,1]and decreasing on−242x[1, ∞ ) . f has a maximum at (1, 1 )ef is concave up on (2, ∞)and concave down on( −∞ ,2). f has a point of inflection at (2, 22)ey5−38x−5Instructor’s Resource Manual Section 6.3 361

x28. f ( x)= e + x Domain = ( −∞, ∞ )xf ′( x) = e + 1, f′′( x)= eSince f′ ( x) > 0for all x, f is increasing on( −∞, ∞ ).Since f′′ ( x) > 0for all x, f is concave upward on( −∞, ∞ ).Since f and f ′ are both monotonic, there are noextreme values or points of inflection.5yx30. f( x) = ln(2x− 1) . Since 2x − 1> 0 if and only if11x > , domain = ( , ∞ )222 −4f′ ( x) = , f′′( x)=22x−1 (2x−1)Since f′ ( x) > 0for all domain values, f is1increasing on ( , ∞ ).2Since f′′ ( x) < 0for all domain values, f is1concave downward on ( , ∞ ).2Since f and f ′ are both monotonic, there are noextreme values or points of inflection.y−5 5x5−58x29.2f( x) = ln( x + 1) Since x + 1> 0for all x,domain = ( −∞, ∞)22x−2( x −1)f′ ( x) = , f′′( x)=2 2 2x + 1 ( x + 1)x ( −∞, −1) −1 ( −1,0) 0 (0,1) 1 (1, ∞)f ′ − − − 0 + + +f ′′ − 0 + + + 0 −f is increasing on (0, ∞)and decreasing on( −∞ ,0). f has a minimum at (0,0)f is concave up on ( −1,1)and concave down on( −∞, −1) ∪ (1, ∞ ). f has points of inflection at( − 1,ln2) and (1, ln 2)5y2−5x31. f ( x) = ln(1 + e ) Since 1+ e > 0for all x,domain = ( −∞, ∞)xxeef′ ( x) = , f′′( x)=xx 21 + e(1 + e )Since f′ ( x) > 0for all x, f is increasing on( −∞, ∞ ).Since f′′ ( x) > 0for all x, f is concave upward on( −∞, ∞ ).Since f and f ′ are both monotonic, there are noextreme values or points of inflection.5yx−5 5x−5 5x−5−5362 Section 6.3 Instructor’s Resource Manual

32.1 xe −2f( x)= Domain = ( −∞, ∞ )1 − x2 1−xf′ ( x) =− 2 xe , f′′( x) = (4x −2)e2 23yx2 2 2 2 2 2( −∞, − ) − ( − ,0) 0 (0, ) ( , ∞)2 2 2 2 2 2f ′ + + + 0 − − −f ′′ + 0 − − − 0 +−1 2 4xf is increasing on ( −∞,0]and decreasing on[0, ∞ ) . f has a maximum at (0, e )2 2 , )2 2f is concave up on ( −∞, − ) ∪ ( ∞ and2 22 2concave down on ( − , ). f has points of22inflection at ( − , e)and3−3 3y2( , e )2x−3x x34. f ( x)= e − e − Domain = ( −∞, ∞ )x − x x − xf ′( x) = e + e , f′′( x)= e − ex ( −∞,0) 0 (0, ∞)f ′ + + +f ′′ − 0 +f is increasing on ( −∞, ∞)and so has no extremevalues. f is concave up on (0, ∞)and concavedown on ( −∞ ,0). f has a point of inflection at(0,0)3y−333.( x 2)2f( x)= e − − Domain = ( −∞, ∞ )−( x−2)2f′ ( x) = (4−2 x) e ,−3 3x2 −( x−2)f′′ ( x) = (4x − 16x+14) e2Note that 4x− 16x+ 14= 0 when4±2x = ≈ 2±0.70722−3x ( −∞,1.293) ≈1.293 (1.293,2) 2 (2,2.707) ≈2.707 (2.707, ∞)f ′ + + + 0 − − −f ′′ + 0 − − − 0 +f is increasing on ( −∞,2]and decreasing on[2, ∞ ) . f has a maximum at (2,1)4− 2 4+2f is concave up on ( , ) ( , ∞ )2 24− 2 4+2concave down on ( , )2 2. f has points4−2 1 4+2 1of inflection at ( , ) and ( , )2 e 2.e−∞ ∪ andInstructor’s Resource Manual Section 6.3 363

35.36.x0− t2f ( x)= ∫ e dt Domain = ( −∞, ∞ )− xf ′( x) = e , f′′( x) =− 2xe− x2 2x ( −∞,0) 0 (0, ∞)f ′ + + +f ′′ + 0 −f is increasing on ( −∞, ∞)and so has no extremevalues. f is concave up on ( −∞,0)and concavedown on (0, ∞ ) . f has a point of inflection at(0,0)3−3 3x0−3− tyf ( x)= ∫ te dt Domain = ( −∞, ∞ )− xf ′( x) = xe , f′′( x) = (1 − x)ex− xx ( −∞,0) 0 (0,1) 1 (1, ∞)f ′ − 0 + + +f ′′ + + + 0 −f is increasing on [0, ∞)and decreasing on( −∞ ,0]. f has a minimum at (0,0)f is concave up on ( −∞,1)and concave down on(1, ∞ ) . f has a point of inflection at1te−t∫ dt .0(1, )Note: It can be shown with techniques in1 − t 2Chapter 7 that ∫ te dt = 1− ≈0.2640e4y37. Let u = 3x + 1, so du = 3dx.3 x + 1 1 3 x + 1 1 1∫e dx = e 3dx e u du e u C3∫ = = +3∫31 3x+1= e + C3238. Let u = x − 3, so du = 2x dx.x 2−3 1 x 2−31∫ xe dx = e 2xdx = e u du2∫ 2∫1 u 1 x2−= e + C = e 3+ C2 2239. Let u = x + 6x, so du = (2x + 6)dx.x2+ 6 x 1 u 1 u∫( x + 3) e dx = e du = e + C2∫21 x2+ 6 x= e + C2xx40. Let u = e − 1, so du = e dx .xe 1x∫ dx = du ln u C ln e 1 Cx ∫ = + = − +e −1u1 141. Let u = − , so du = dx .x2x42.−1/xeu u −1/xdx = e du = e + C = e + C2x∫ ∫x+exx exe dx = e ⋅e dx∫ ∫Let xxu = e , so du = e dx.x exu u exe ⋅ e dx = e du = e + C = e + C∫ ∫43. Let u = 2x + 3, so du = 2dx2x+ 3 1 u 1 u 1 2x+3∫e dx = e du e C e C2∫= + = +2 211 2 x+ 3 ⎡1 2 x+3 ⎤ 1 5 1– 3∫ e dx = e e e0 ⎢2⎥ =⎣ ⎦02 2= 1 3 2( 1) 64.22 e e − ≈−3−29x3344. Let u = , so du =− dx.x2x3/ xe 1 u 1 u∫ dx = – e du = – e + C2x 3∫31 3/ x= – e + C 33/22xe ⎡ 1 3/ x ⎤ 1 3/2 1 3dx = – –1 2 ⎢ e e ex 3⎥ = +⎣ ⎦13 3∫ ≈ 5.2364 Section 6.3 Instructor’s Resource Manual

45.46.ln 3 x 2 ln 3 2xV =π ∫ ( e ) dx =π e dx0 ∫ 0ln 3⎡1 2x⎤ ⎛1 2ln3 1 0⎞=π ⎢ e e e 4π12.572⎥ =π⎜− ⎟= ≈⎣ ⎦0⎝2 2 ⎠1 20 2 −xV = πxe dx∫ .2Let u =− x , so du = –2x dx.−x 2−x 2u2 π xe dx = −π e ( − 2 x)dx = −π e du∫ ∫ ∫u−x2=−π e + C =−π e + C2 211 − 1 02 x ⎡ −x⎤ −∫ π xe dx =−π e = – π( e −e)0⎢ ⎥⎣ ⎦01=π(1 − e − ) ≈ 1.990.3 ⎧⎡⎛0.3 ⎞0.3 ⎤0.3⎫≈ ⎨⎢⎜+ 1⎟+ 1 + 1⎬0.3 + 14 3⎥⎩⎣⎝⎠ ⎦ 2 ⎭= 1.34983750.3e ≈ 1.3498588 by direct calculation50. e( )tt t51. x = e sin t,so dx = ( e sint + e cos t)dtty = e cos t,so dy = ( e cost − e sin t)dtds =2 2dx + dyt= e2 2(sin t+ cos t) + (cost−sin t)dtt 2 2t= e 2sin t+ 2cos tdt = 2e dtThe length of the curve isπ2 ttππedt= 2 ⎡e⎤= 2( e −1) ≈31.3120 ⎣ ⎦0∫tt⎛ 1 ⎞47. The line through (0, 1) and ⎜1, ⎟ has slope⎝ e ⎠1 −1 e 1 1−e 1−e= − 1= ⇒ y− 1 = ( x−0);1−0 e e e1−ey = x+1e48.11⎡⎛1−e⎞ −x⎤ ⎡1−e2 −x⎤x 1 e dx x x e0 ⎢⎜+ ⎟− = + +e⎥ ⎢2e⎥⎣⎝ ⎠ ⎦ ⎣ ⎦0∫1−e1 3−e= + 1+ − 1 = ≈ 0.0522e e 2exx( e –1)(1)– x( e ) 1 – xf′ ( x) =– (– e )(–1)x 2 – x( e –1) 1– ex xe −1−xe1 ⎛ 1 ⎞= −x 2 −x ⎜ x ⎟( e −1) 1−e ⎝e⎠x xx x xe −1−xe1 e −1 −xe −( e −1)= − =x 2 xx 2( e −1) e −1( e −1)xxe=−x 2( e −1)When x > 0, f ( x) 0,′ < so f(x) is decreasing forx > 0.49. a. Exact:10! = 10987654321⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅b.= 3,628,800Approximate:10⎛10⎞10! ≈ 20π⎜⎟ ≈3,598,696⎝ e ⎠60⎛60⎞8160! ≈ 120π⎜⎟ ≈ 8.31×10⎝ e ⎠52. Use x = 30, n = 8, and k = 0.25.n −kx8 −0.2530⋅( kx) e (0.25⋅30)ePn( x)= = ≈ 0.14n! 8!53. a.b.ln xlim is of the form2x→0+ 1 + (ln x)∞∞1Dxln xx= lim= lim2 1x→0 + Dx[1 + (ln x) ] x→0+ 2ln x ⋅x1= lim = 0x→0+ 2ln xln x1lim = lim = 0x→∞21 + (ln x)x→∞2lnx2[1 + (ln x) ] ⋅ 1 – ln x⋅2 ln x⋅1xxf′ ( x)=2 2[1 + (ln x) ]21–(ln x)=2 2x[1 + (ln x) ]1f ′( x) = 0 when lnx =± 1 so x = e = e–1 1or x = e =eln e 1 1f()e = = =2 21 + (ln e) 1+1 21⎛1 ⎞ lne –1 1f ⎜ ⎟ = = = –⎝e⎠ 2 21+1( ln )1 + (–1) 2eMaximum value of 1 at x = e; minimum21value of − at x = e −1 .2.Instructor’s Resource Manual Section 6.3 365

c. F( x)=x2ln tdt121 + (ln t)∫2ln xF′ ( x) = ⋅2x2 21 + (ln x )F′ (2ln( e) 1e) = ⋅ 2 e = ⋅22 2 21 + [ln( e) ] 1+1e= e ≈ 1.65x0xe 0 –0x0 x0 x= f′( x 00) = e ⇒ e = x0e ⇒ x0= 1x0–0xy e x54. Let ( x , e 0 ) be the point of tangency. Thenso the line is =0or y = ex.58. a.b.y-axis so the area is⎧ 3⎪ 2 22 −xx 22 ⎨∫[ e −2 e (2x −1)]dx0⎪⎩3 −x22 −x2 ⎫⎪+ ∫ 3 [2 e (2x −1) −e ] dx⎬ 2⎪ ⎭≈ 4.2614p – xlim x e = 0x→∞p – x – x p–1f′ ( x) = x e (–1) + e ⋅ pxp–1 – x= x e ( p– x)f′ ( x) = 0 when x = pa.b.211 x x exA= ∫ ( e – ex) dx = ⎡⎢e– ⎤⎥0⎢⎣2 ⎥⎦0e 0 ee− −( e − 0) = –1≈0.362 2=1 2 20 [( xV =π∫ e ) – ( ex ) ] dx11 2 2 20 ( x=π∫ e – e x )⎡ 2 31 2xe x ⎤dx =π⎢ e – ⎥⎢⎣2 3 ⎥⎦0⎡ 21 2 e ⎛10⎞⎤π 2=π⎢e − −⎜e ⎟⎥= ( – 3) 2.30⎢⎣2 3 ⎝2⎠⎥⎦6 e ≈59.60.x→–∞2 – xlim ln( x + e ) =∞(behaves like − x )2 – xlim ln( x + e ) =∞ (behaves like 2lnx )x→∞x–1–2 x–1–2f ′( x) = –(1 + e ) ⋅ e (– x )1/ xe=x (1 + e )2 1/ x 255. a.3 ⎛ 1 ⎞ 3 ⎛ 1 ⎞exp dx 2 exp dx 3.11−3 ⎜− 2 ⎟ =0 ⎜− ≈2 ⎟⎝ x ⎠ ⎝ x ⎠∫ ∫b.8π 0.1e − x sin xdx0∫ ≈ 0.91056. a.1lim (1 + x) x = e ≈ 2.72x→ 057.b.−1x 1lim (1 + x)= ≈ 0.368ex→0f( x)= e −x22f′ ( x) =− 2xe −xx22 x2x22f′′ − − −( x) =− 2e + 4x e = 2 e (2x− 1)y = f(x) and y = f′′( x)intersect when−x2−x22 2e = 2 e (2x − 1); 1= 4x−2;2 34x− 3= 0, x = ±2Both graphs are symmetric with respect to thea.b.c.d.lim f( x) = 0x→0+lim f( x) = 1x→0–x1lim f( x)=→±∞ 2lim f′ ( x) = 0x→0e. f has no minimum or maximum values.366 Section 6.3 Instructor’s Resource Manual

6.4 Concepts Reviewx + 3 = 5x3x =3lnπ 1. e ;x ln ae42. eln129. log512 = ≈ 1.544ln 5ln x3.ln aln 0.1110. log70.11 = ≈ –1.1343ln 7a 14. ax − x; a ln a1/5 1ln8.1211. log 11(8.12) = ≈ 0.17475 ln117 ln8.5712. log 10(8.57) = 7 ≈ 6.5309ln10x 31. 2 = 8 = 2 ; x = 313. x ln 2 = ln 1722. x = 5 = 25ln17x = ≈ 4.08746ln 23/23. x = 4 = 814. x ln 5 = ln 1344. x = 64x = ln13 ≈ 1.5937ln 5x = 4 64 = 2 215. (2s – 3) ln 5 = ln 4⎛ x ⎞ 1ln 45. log9⎜ ⎟ =2 s –3=⎝3⎠2ln 51⎛ln4⎞s =x⎜3 + ⎟≈1.9307=1/22⎝ln5⎠9 = 33116. ln12 = ln 4x = 9θ –1ln123 1= θ –16. 4 =ln 42xln121 1θ = 1+ ≈ 2.7925x = =ln 4324 ⋅ 1282x 2x 2x17. Dx= ⋅ Dxx = ⋅7. log 2( x+ 3) – log2x = 2x + 3log2= 22 x2–3x 2 x2–3x2x18. x = ⋅ xx + 3 =22 x2–3x2 = 4= (4 x – 3) ⋅ 3 ln3xx + 3 = 4xx 1 xx = 119. Dxlog3e = ⋅ Dxexe ln 3x8. log 5( x+ 3) – log5x = 1e 1= = ≈ 0.9102xx + 3e ln 3 ln 3log5= 1xAlternate method:+ x3 = 1D = x log 3e = Dx( xlog 3e) = log3eProblem Set 6.4xx5 5(6 ) 6 ln 6 (2 ) 2 6 ln 6D (3 ) 3 ln 3 D (2 x – 3 x)ln e 1= = ≈ 0.9102ln 3 ln 3Instructor’s Resource Manual Section 6.4 367

3 1320. Dxlog 10( x + 9) = ⋅ D ( 9)3x x +( x + 9)ln1023x=3( x + 9)ln1021. zD [3 ln( z+5)]22.zz 1z= 3 ⋅ (1) + ln( z + 5) ⋅3 ln 3z + 5z ⎡ 1⎤= 3 ⎢ + ln( z + 5)ln3z + 5⎥⎣⎦θ2 – θ210 = D10Dθ log (3 ) θ ( θ – θ)log 32( θ – θ)ln3 ln3 2= Dθ= ⋅ Dθθln10 ln10ln 3 1 (2 – )–1/2 (2 –1)= ln10 ⋅ 2 θ θ θ2 θ –1 ln3=22 θ – θ ln10223. Let u = x so du = 2xdx.x2 1 u 1 2x ⋅ 2 dx = 2 du = ⋅ + C2 2 ln2∫ ∫x2x2 –12 2= + C = + C2ln2 ln224. Let u = 5x – 1, so du = 5 dx.∫ ∫5 x–1 1 u 1 1010 dx = 10 du = ⋅ + C5 5 ln105 x–110= + C5ln10125. Let u = x, so du = dx.2 xx5 u 5dx = 2 5 du = 2⋅ + Cxln 5∫ ∫x25 ⋅= + Cln 5∫14x5 ⎡ x5 ⎤⎛ 25 5 ⎞dx = 2⎢⎥ = 2⎜− ⎟x ⎢ln 5 ⎥ ⎝ln 5 ln 5⎣ ⎦⎠40= ≈ 24.85ln 541uuu– θ26.27.28.29.1 3 –3 1 3 1 –3(10 x + 10 x ) = 10 x + 10x0 0 0∫ ∫ ∫Let u = 3x, so du = 3dx.∫ ∫dx dx dx3x1 u 1 1010 dx = 10 du = ⋅ + C3 3 ln103x10= + C3ln10Now let u = –3x, so du = –3dx.–3x1 u 1 10∫10 dx = – 10 du – C3∫= ⋅ +3 ln10–3x10= – + C 3ln10Thus,3 –310 –10(10 10 x ⎡ x x ⎤+ ) dx = ⎢ ⎥⎢⎣3ln10 ⎥⎦1 3 x –30∫1 ⎛ 1 ⎞ 999,999= ⎜1000 – ⎟=3ln10 ⎝ 1000 ⎠ 3000ln10≈ 144.76d ( x 2) ( x 2) d 2 ( x 2x )= = x10 10 ln10 10 2 ln10dxdxd 210 d 20 19( x ) = x = 20xdx dxdy d ( x2) 2 10= [10 + ( x ) ]dx dx( x2) 19= 10 2xln10 + 20xd 2 dsin x = 2sin x sin x = 2sin xcosxdxdxd sin x sin x d sin x2 = 2 ln2 sinx= 2 ln2cosxdxdxdy d 2 sinx= (sin x + 2 )dx dxsin x= 2sin xcos x+2 cos xln 2d xπ+ 1( 1) xπ= π+dxd( 1) x xπ + = ( π+ 1) ln( π+ 1)dxdy d 1[ xπ+ x= + ( π+ 1) ]dx dx= ( π+ 1) + ( π+ 1) ln( π+ 1)x πxuu10368 Section 6.4 Instructor’s Resource Manual

30.31.d ( e ) ( ) ( )2 x e2 x x eln 2 2 x x= d e = e ln 2dxdxd e x e x e e x(2) = (2) ln2 = (2) eln2dxdy d ( e )[2 x e x= + (2 ) ]dx dx( e x ) x e x= 2 e ln2 + (2 ) eln22 ln x (ln x)ln( x2+ 1)y = ( x + 1) = edy (ln x)ln( x2+ 1) d2= e [(ln x)ln( x + 1)]dxdx(ln x)ln( x21) 1 2 2x⎢ ln( 1) ln2+ ⎡ ⎤= e x + + xx⎥⎣x + 1⎦⎛ 22 lnx ln( x + 1) 2xlnx⎞= ( x + 1)+⎜ x 2x + 1 ⎟⎝⎠35.− (ln 2)( )f( x) 2 x −= = ex Domain = ( −∞, ∞ )− xf′ ( x) = ( − ln2)2 , f′′( x) = (ln2) 2Since f′ ( x) < 0for all x, f is decreasing on2− x( −∞, ∞ ).Since f′′ ( x) > 0for all x, f is concave upward on( −∞, ∞ ).Since f and f ′ are both monotonic, there are noextreme values or points of inflection.−34y9x32.33.2 2 + 3 (2 + 3)ln(ln2)y = (ln x ) x = ex xdy (2x+3)ln(ln x2) d2= e [(2x+3)ln(ln x )]dxdx(2x+ 3)ln(ln x2) ⎡ 21 1 ⎤= e ⎢2ln(ln x ) + (2x+3) (2 x)2 2 ⎥⎣ln x x ⎦2x+ 3 ⎢⎡ 2x+ 3⎥⎤= (2ln x) 2ln (2ln x)+ ⎢ x ln x ⎥ln x2 ⎢⎣ ln x2 ⎥⎦sin x sin xlnxf( x)= x = esin xlnx df ′( x) = e (sinxln x)dxsin xln x ⎡ ⎛1⎞⎤= e ⎢(sin x) ⎜ ⎟+(cos x)(ln x)x⎥⎣ ⎝ ⎠⎦sin x ⎛sinx ⎞= x ⎜ + cos xlnx⎟⎝ x⎠sin1 sin1f ′⎛⎞(1) = 1 ⎜ + cos1ln1⎟= sin1 ≈0.8415⎝ 1 ⎠−236. f( x) = x2 − x Domain = ( −∞, ∞ )− xf′ ( x) = [1 −(ln2) x]2 ,− xf′′ ( x) = (ln 2)[(ln 2) x−2]21 1 1 2 2 2x ( −∞, ) ( , ) ( , ∞)ln2 ln2 ln2 ln2 ln2 ln2f ′ + 0 − − −f ′′ − − − 0 +⎛ 1 ⎤f is increasing on ⎜ −∞,⎥ and decreasing on⎝ ln 2⎦⎡ 1 ⎞1⎢ , ∞⎟ . f has a maximum at ( , 1 )⎣ln 2( ln2)⎠ln 2 e2f is concave up on ( , ∞)and concave down onln 22( −∞ , ). f has a point of inflection atln 2e34. f() e =π ≈ 22.46ge () = e π ≈ 23.14g(e) is larger than f(e).d x xf′ ( x) = π = π lnπdxef′ () e =π lnπ≈25.71d1g′ π π−( x)= x = π xdx1g′ () e =πe π− ≈ 26.74g′ () e is larger than f ′()e .2( , 22 )ln 2 ( e ln2)y3−2−38xInstructor’s Resource Manual Section 6.4 369

37.2 ln( x + 1)f( x) = log 2 ( x + 1) = . Sinceln 22x + 1> 0 for all x, domain = ( −∞, ∞ )25y2⎛ 2 ⎞⎛x ⎞ ⎛ 2 ⎞⎛1−x ⎞f′ ( x) = ⎜ ⎟⎜ , f ( x)2 ⎟ ′′ = ⎜ ⎟2 2ln 2 ⎝ x + 1 ⎠ ln 2 ⎜( x + 1) ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠x ( −∞, −1) −1 ( −1,0) 0 (0,1) 1 (1, ∞)f ′ − − − 0 + + +f ′′ − 0 + + + 0 −f is increasing on [0, ∞)and decreasing on( −∞ ,0]. f has a minimum at (0,0)f is concave up on ( −1,1)and concave down on( −∞, −1) ∪ (1, ∞ ). f has points of inflection at( −1,1)and (1,1)−55y5x39.−51x−5− t2f ( x) = ∫ 2 dt Domain = ( −∞, ∞ )− x52 2f′ ( x) = 2 , f′′( x) = − 2(ln2) x2x ( −∞,0) 0 (0, ∞)f ′ + + +f ′′ + 0 −x− xf is increasing on ( −∞, ∞)and so has no extremevalues.f is concave up on ( −∞,0)and concave down on(0, ∞ ) . f has a point of inflection at01∫− t2(0, 2 dt) ≈(0, −0.81)y−5538.2 xln( x + 1)f( x) = xlog 3( x + 1) = . Sinceln 32x + 1> 0 for all x, domain = ( −∞, ∞ )2−55x1 ⎡ 2x 2 ⎤ 2 ⎡ 32x + 3x⎤f′ ( x) = ⎢ + ln( x + 1) ⎥ , f′′( x)= ⎢ ⎥ln 3 ⎢x21 ln 3 x2⎣ + ⎥⎦ ⎢⎣ + 1 ⎥⎦x ( −∞,0) 0 (0, ∞)f ′ + 0 +f ′′ − 0 +−5f is increasing on ( −∞, ∞)and so has no extremevalues. f is concave up on (0, ∞)and concavedown on ( −∞ ,0). f has a point of inflection at(0,0)370 Section 6.4 Instructor’s Resource Manual

40.∫x2f ( x) = log0 10( t + 1) dt . Since log 10( t + 1) hasdomain = ( −∞, ∞ ), f also has domain = ( −∞, ∞)2 ln( x + 1)f′ ( x) = log 10( x + 1) = ,ln10⎛ 1 ⎞⎛ 2x⎞f′′ ( x)= ⎜ ⎟⎜ 2 ⎟⎝ln10 ⎠⎝ x + 1⎠x ( −∞,0) 0 (0, ∞)f ′ + 0 +f ′′ − 0 +f is increasing on ( −∞, ∞)and so has no extremevalues.f is concave up on (0, ∞)and concave down on( −∞ ,0). f has a point of inflection at (0,0)5y2244. 115 = 20log 10(121.3 P)log 10(121.3 P ) = 5.755.7510P = ≈ 4636 lb/in. 2121.345. If r is the ratio between the frequencies of12successive notes, then the frequency of C = r(the frequency of C). Since C has twice the1/12frequency of C, r = 2 ≈ 1.05951/12 3 4Frequency of C = 440(2 ) = 440 2 ≈ 523.25p46. Assume log 2 3 = where p and q are integers,qp q p qq ≠ 0 . Then 2 = 3 or 2 = 3 . Butp2 = 2 ⋅ 2 … 2 (p times) and has only powers of 2qas factors and 3 = 3 ⋅ 3 … 3 (q times) and hasonly powers of 3 as factors.p q2 = 3 only for p = q = 0 which contradicts ourassumption, so log23 cannot be rational.−5−55xx47. If y = A⋅ b , then ln y = ln A + x ln b, so theln y vs. x plot will be linear.dIf y = C⋅ x , then ln y = ln C + d ln x, so theln y vs. ln x plot will be linear.ln x ln x41. log1/2 x = = = −log1 2 xln − ln 2 242.43. M = 0.67log 10(0.37 E) + 1.46M −1.46log 10(0.37 E)=0.67M −1.4610 0.67E =0.37Evaluating this expression for M = 7 and M = 88gives E ≈ 5.017× 10 kW-h and10E ≈ 1.560× 10 kW-h, respectively.48. WRONG 1:y = f( x) g x( )g( x) 1y′ −g( x) f( x) f′( x)=WRONG 2:y = f( x) g x( )g( x) g( x)y′ f( x) (ln f( x)) g′ ( x) f( x) g′( x)ln f( x)= ⋅ =RIGHT:( ) ( )ln ( )y = f( x) g x = eg x f xg( x)ln f ( x)dy′ = e [ g( x)ln f( x)]dxg( x) ⎡1 ⎤= f ( x) ⎢g′ ( x)ln f( x) + g( x) f′( x)⎥⎣f( x)⎦g( x) g( x) 1f ( x) g′ −( x)ln f( x) f( x) g( x) f′( x)= +Note that RIGHT = WRONG 2 + WRONG 1.Instructor’s Resource Manual Section 6.4 371

49.50.x x ( x2) ( xx)f ( x) = ( x ) = x ≠ x = g( x)( x2) x2lnxf( x)= x = ex2 ln x df′ ( x) = e ( x 2ln x)dxx2 ln x⎛2 1 ⎞= e ⎜2xlnx+ x ⋅ ⎟⎝x ⎠( x2)= x (2xln x+x)( xx) xxlnxgx ( ) = x = eUsing the result from Example 5⎛ d x x ⎞⎜ x = x (1+ln x ) ⎟:⎝dx⎠xxln x d xg′ ( x) = e ( x ln x)dxxxln x ⎡ x x 1 ⎤= e ⎢x (1 + ln x) ln x+ x ⋅x ⎥⎣⎦( xx) x ⎡1 ⎤= x x ⎢(1 + ln x) ln x+x ⎥⎣⎦xx+ x ⎡ 2 1 ⎤= x ⎢ln x+ (ln x)+x ⎥⎣⎦xa −1f( x)=xa + 1x x x x x( a + 1) a ln a−( a −1) a lna 2a lnaf′ ( x)= =x 2 x 2( a + 1) ( a + 1)xSince a is positive, a is always positive.x 2( a + 1) is also always positive, thus f′ ( x) > 0if ln a > 0 and f′ ( x) < 0 if ln a < 0. f(x) is eitheralways increasing or always decreasing,depending on a, so f(x) has an inverse.xa −1y =xa + 1x xya ( + 1) = a − 1xa ( y− 1) = −1−yx 1+ya =1 − y1+yxln a = ln 1 − y1+y1− y +logaln 1 yx = =ln a 1−y− 1 1+yf ( y) = loga1 − y1 1 xf− ( x) = log +a 1 − x⎛ ax ⎞51. a. Let g(x) = ln f(x) = ln = aln x − xlnax.⎜a⎟⎝ ⎠ag′ ⎛ ⎞( x) = ⎜ ⎟−lna⎝ x ⎠ag′ ( x) < 0 when x > , so as x →∞ g(x)ln aais decreasing. g′′ ( x)=− , so g(x) is2xconcave down. Thus, lim gx ( ) =−∞ , sox→∞g( x)lim f( x) = lim e = 0.x→∞x→∞b. Again let g(x) = ln f(x) = a ln x – x ln a.Since y = ln x is an increasing function, f(x)is maximized when g(x) is maximized.aag′ ⎛ ⎞ ⎛ ⎞( x) = ⎜ ⎟− ln a, so g′( x) > 0 on ⎜0,⎟⎝ x ⎠ ⎝ ln a ⎠aand g′ ⎛ ⎞( x) < 0 on ⎜ , ∞⎟.⎝lna ⎠Therefore, g(x) (and hence f(x)) isamaximized at x0 = .ln aaxxac. Note that = is equivalent to g(x) = 0.aBy part b., g(x) is maximized at x0 = .ln aIf a = e, then⎛ e ⎞gx ( 0) = g⎜⎟= ge ( ) = eln e− eln e=0.⎝lne ⎠Since gx ( ) < gx ( 0) = 0 for all x ≠ x0,thea xequation g(x) = 0 (and hence x = a ) hasjust one positive solution. If a ≠ e, then⎛ a ⎞ ⎛ a ⎞ ag( x0) = g⎜ ⎟= aln ⎜ ⎟−(ln a)⎝ln a⎠ ⎝ln a⎠ln a⎡ ⎛ a ⎞ ⎤= a ⎢ln ⎜ ⎟−1ln a⎥⎣ ⎝ ⎠ ⎦ .aNow > e (justified below), soln a⎡ a ⎤gx ( 0) = a⎢ln − 1 > a(lne− 1) = 0.ln a ⎥Since⎣ ⎦g′ ( x) > 0 on (0, x0), g( x0) > 0, andlim gx ( ) = −∞ , g(x) = 0 has exactly onex→0solution on (0, x 0).Since g′ ( x) < 0 on ( x0, ∞ ),gx ( 0) > 0, and lim gx ( ) = −∞ , g(x) = 0 hasx→∞exactly one solution on ( x0, ∞ ). Therefore,372 Section 6.4 Instructor’s Resource Manual

a xthe equation g(x) = 0 (and hence x = a )has exactly two positive solutions.aTo show that e ln a > when a ≠ e:xConsider the function hx ( ) = , for x>1.ln xln( x )(1) − x 1( x )( )ln x −h′ x = =12 2(ln x) (ln x)Note that h′ ( x) < 0 on (1, e) and h′ ( x) > 0on (e, ∞ ), so h(x) has its minimum at (e, e).xTherefore > e for all x eln x ≠ , x > 1.d. For the case a = e, part c. shows thatgx ( ) = eln x− xln e< 0 for x ≠ e .e xTherefore, when x ≠ e , ln x < ln e , whiche ximplies x < e . In particular, π e < e π .u x52. a. fu( x)= x e −u 1 x u x u 1 xfu′ − − − − −( x) = ux e − x e = ( u−x)x eSince fu′ ( x) > 0 on (0, u) and fu′ ( x) < 0 on(u, ∞ ), fu( x ) attains its maximum at x 0 = u.b. fu( u) > fu( u+ 1) meansu −u u − ( u+1)u e > ( u+ 1) e .u+1e⎛u+ 1⎞uMultiplying by gives e >u⎜ ⎟u⎝ u ⎠ .fu+ 1( u+ 1) > fu+1( u) means1 ( 1) 1( 1) u + −u e u +u u + −+ > eu .u+11e1 u +⎛u+ ⎞Multiplying by giveseu+1 ⎜ ⎟ > .u ⎝ u ⎠Combining the two inequalities,uu+1⎛u+ 1⎞ ⎛u+1⎞⎜ ⎟ < e < ⎜ ⎟ .⎝ u ⎠ ⎝ u ⎠11 u +c.⎛u+ ⎞From part b., e < ⎜ ⎟⎝ u ⎠.uMultiplying byu + 1uu ⎛u+1⎞e < ⎜ ⎟u+ 1 ⎝ u ⎠ .⎛u+ ⎞We showed ⎜ ⎟ < e in part b., so⎝ u ⎠uu ⎛u+1⎞e< ⎜ ⎟ < e.u+ 1 ⎝ u ⎠1 u53.54.55.56.uSince lim e = e , this implies thatu→∞ u + 1uu⎛u+ 1⎞ ⎛ 1⎞lim ⎜ ⎟ = e, i.e., lim ⎜1+ ⎟ = e.u→∞⎝ u ⎠ u→∞⎝ u⎠x xlnxf( x)= x = eLet g( x) = xln x.Using L’Hôpital’s Rule,ln xlim gx ( ) = lim1x→0 +x→0+x1x= lim = lim ( − x) = 01x→0 + −2 x→0+xx 0Therefore, lim x = e = 1.x→0+g′ ( x) = 1+lnxSince g ( x) 0( 1/ , )′ < on ( )0,1/ e and g′ ( x) > 0 one ∞ , g(x) has its minimum at x =1.e−1 −1/eTherefore, f(x) has its minimum at ( e , e ).Note: this point could also be written as⎛1 ( 1 ) 1 e ⎞,ee.⎜ ⎟⎝ ⎠(2.4781, 15.2171), (3, 27)4πsin xx dx ≈ 20.22590∫Instructor’s Resource Manual Section 6.4 373

57. a. In order of increasing slope, the graphsx xrepresent the curves y = 2 , y = 3 , andy =x4.b. ln y is linear with respect to x, and at x = 0,y = 1 since C = 1.c. The graph passes through the points (0.2, 4)0.20.6and (0.6, 8). Thus, 4 = Cb and 8 = Cb .Dividing the second equation by the first,0.4 5 2gets 2 = b so b = 2 .32Therefore C = 2 .58. The graph of the equation whose log-log plot hasnegative slope contains the points (2, 7) and (7,2).rrThus, 7 = C2 and 2 = C7 , so 7 2 r⎛ ⎞= ⎜ ⎟ .2 ⎝7⎠7 2 ln7−ln2ln = rln ⇒ r = = −1and C = 14.2 7 ln2−ln71Hence, one equation is y = 14 x − .The graph of one equation contains the points(7, 30) and (10, 70). Thus, 30 = C7 r andr70 = C10 , so 3 7 r⎛ ⎞= ⎜ ⎟7 ⎝10⎠3 7 ln3−ln7ln = rln ⇒ r = ≈ 2.38 and7 10 ln7−ln10−2.38C ≈30⋅7 ≈ 0.29 . Hence, another equation is2.38y = 0.29 x .The graph of another equation contains the points(1, 2) and (7, 5). Thus, 2 = C1 rrand 5= C7 , soC = 2 andln 5 − ln 2ln 5 − ln 2 = r ln 7 ⇒ r = ≈ 0.47.ln 70.47Hence, the last equation is y = 2x.The given answers are only approximate.Student answers may also vary.6.5 Concepts Review1. ky ; ky( L − y)2.32 = 83. half-life 4. ( 1+h) 1/Problem Set 6.51. k = − 6 ,2.3.6ty0 = 4, so y = 4e −6tk = 6, y0= 1, so y = e0.005tk = 0.005, so y = y0e0.005(10) 0.05y(10)= y0e = y0e2y(10) = 2 ⇒ y0 =0.05e2 0.005t 0.005t−0.05 0.005( t−10)y = e = 2e = 2e0.05e–0.003ty = y0e(–0.003)(–2) 0.006y(–2)= y0e = y0e3y( − 2) = 3⇒ y0 =0.006e3 –0.003 t –0.003 t–0.006 –0.003( t+2)y = e = 3e = 3e0.006e4. k = –0.003, so5. y 0 = 10,000, y(10) = 20,00020,000 = 10,000ek(10)10k2 = eln 2ln 2 = 10k; k =10((ln 2) /10) tt/10y = 10,000e= 10,000⋅22.5After 25 days, y = 10,000⋅2 ≈ 56,568.6. Since the growth is exponential and it doubles in10 days (from t = 0 to t = 10), it will alwaysdouble in 10 days.h7.((ln 2) /10) t3y0 = y0e((ln 2) /10) t3 = eln 2ln 3 = 10 t10ln 3t = ≈ 15.8 daysln 2374 Section 6.5 Instructor’s Resource Manual

8. Let P(t) = population (in millions) inyear 1790 + t.In 1960, t = 170.ktPt () = Pe 0170k178 = 3.9e170k45.64 = eln 45.64k = ≈ 0.02248170In 2000, t = 2100.02248⋅210P(210) ≈3.9e≈ 438The model predicts that the population will be about438 million. The actual number, 275 million, isquite a bit smaller because the rate of growth hasdeclined in recent decades.9. 1 year: (4.5 million) (1.032) ≈ 4.64 million22 years: (4.5 million) (1.032) ≈ 4.79 million1010 years: (4.5 million) (1.032) ≈ 6.17 million100100 years: (4.5 million) (1.032) ≈ 105 million13.14.1 k(700)= e and y0= 102–ln 2 = 700kln 2k =− ≈− 0.000997000.00099ty = 10e −0.00099 300At t = 300, y = 10e − ⋅ ≈ 7.43.After 300 years there will be about 7.43 g.k(2)0.85 = eln 0.85 = 2kln 0.85k = ≈− 0.081321 0.0813t= e −2– ln 2 = − 0.0813tln 2t = ≈ 8.530.0813The half-life is about 8.53 days.10. y = y0ektk(1)1.032A= Aek = ln1.032 ≈ 0.03150(0.03150)(100)At t = 100, y = 4.5e≈ 105 .After 100 years, the population will be about105 million.11. The formula to use is y = ye 0 , where y =population after t years, y 0 =population at time t =0, and k is the rate of growth. We are givenk (12)235,000 = ye 0 andk(5)164,000 = ye 0Dividing one equation by the other yields12k−5k 7k1.43293 = e = e orln(1.43293)k = ≈ 0.05138887235,000Thus y0 = = 126,839.12(0.0513888)e12. The formula to use is y = ye 0 , where y = mass tmonths after initial measurement, y 0 = mass at timeof initial measurement, and k is the rate of growth.We are given6.76 = 4ek (4) so that1 ⎛6.76 ⎞ 0.5247k = ln ⎜ ⎟= ≈0.13124 ⎝ 4 ⎠ 4Thus, 6 months before the initial measurement, the(0.1312)( −6)mass was y = 4e≈ 1.82 grams. Thetumor would have been detectable at that time.ktkt15. The basic formula is y = ye 0 . If t * denotes thehalf-life of the material, then (see Example 3)1 kt ln(0.5)*= e or k = . Thus2t*−0.693 −0.693kC= =− 0.0229 and kS= = −0.024130.22 28.8To find when 1% of each material will remain, wekt ln(0.01)use 0.01 y0 = y0e or t = . Thusk−4.6052tC= ≈ 201 years (2187) and−0.0229−4.6052tS= ≈191 years (2177)−0.024116. The basic formula is y = ye 0 . We are givenk(2) k(8)15.231 = y e and 9.086 = y ektkt0 0Dividing one equation by the other gives15.2319.086k(2) −k(8) k( −6)= e = e so k = − 0.086115.231= ≈ 18.093 grams.Thus y0 ( .0861)(2)e −To find the half-life:ln(0.5) −0.693t*= = ≈8daysk −0.0861Instructor’s Resource Manual Section 6.5 375

17.18.1 5730k= e2ln 1( 2 )−k = ≈ − 1.210×10 45730( − 1.210×10−4) t0.7y0 = y0eln 0.7t = ≈ 2950−4− 1.210×10The fort burned down about 2950 years ago.12= e5730k( )12 − 4lnk = ≈ − 1.210×1057300.51y( − 1.210×10−4) t0 = y0eln 0.51t = ≈5565−4− 1.210×10The body was buried about 5565 years ago.22. From Example 4, Tt () = T1+ ( T0 − T1) ekt . In thisproblem, 250 = T(15) = 40 + (350 − 40) ek (15) so⎛210⎞ln ⎜ ⎟ 310k =⎝ ⎠=− 0.026 ; the brownies will be150.026 te −110 F when 110 = 40 + (310) or⎛ 70 ⎞ln ⎜ ⎟ 310t =⎝ ⎠= 57.2 min.−0.02623. From Example 4, Tt () = T1+ ( T0 − T1) ekt .Let w = the time of death; thenk(10 −w)82 = T(10 − w) = 70 + (98.6 −70)e76 = T(11 − w) = 70 + (98.6 −70)eorDividing:12 = 28.6e6=28.6ek( −1)k(10 −w)k(11 −w)k(11 −w)2 = e or k = ln (0.5) =− 0.69319. From Example 4, Tt () = T1+ ( T0 − T1) ekt . In thisproblem, 200 = T(0.5) = 75 + (300 − 75) ek(0.5) so⎛125⎞ln ⎜ ⎟ 225k =⎝ ⎠=− 1.1756 and0.5( 1.1756)(3)e −T(3) = 75 + 225 = 81.6 F20. From Example 4, Tt () = T1+ ( T0 − T1) ekt . In thisproblem, 0 = T(5) = 24 + ( −20− 24) ek (5) so⎛−24⎞ln ⎜ ⎟−44k =⎝ ⎠=− 0.1212 ; the thermometer will5register 20 C when 20 = 24 + ( − 44) or0.1212 te −⎛ −4⎞ln ⎜ ⎟−44t =⎝ ⎠= 19.78 min.−0.121221. From Example 4, Tt () = T1+ ( T0 − T1) ekt . In thisproblem, 70 = T(5) = 90 + (26 − 90) ek(5) so⎛−20⎞ln ⎜ ⎟−64k =⎝ ⎠=− 0.2326 and5( 0.2326)(10)e −T(10) = 90 − 64 = 90 − 64(0.0977) = 83.7 CTo find w :⎛ 12 ⎞ln ⎜ ⎟−0.693(10 −w)28.612 = 28.6 e so 10 − w= ⎝ ⎠= 1.25−0.693Therefore w = 10 − 1.25 = 8.75 = 8 : 45pm .24. a. From example 4 of this section,dTkT ( T1)dt = − or dT∫ kdt or ln T(t)-T1kt CT −T = = +1This gives Tt () − T1= e e . Now, if T 0 isCthe temperature at t = 0, T0 − T1= e and theLaw of Cooling becomesktTt () − T1 = T0 − Te 1 . Note that Tt () isalways between T 0 and T 1 so thatTt () −T1 and T0 − T1always have the samesign; this simplifies the Law of Cooling toTt () − T1 = ( T0 − T1) ektorTt () = T+ ( T−T) ekt1 0 1b. Since Tt () is always between T 0 and T 1, itkt Tt () −T1follows that e = < 1 so that k < 0 .T0 −T1Hencektlim Tt ( ) = T+ ( T− T) lim e = T+ 0 = Tt→∞kt1 0 1 1 1t→∞C376 Section 6.5 Instructor’s Resource Manual

25. a.b.c.d.26. a.b.c.d.27. a.28.b.2($375)(1.035) ≈ $401.7124⎛ 0.035 ⎞($375) ⎜1 + ⎟ ≈$402.15⎝ 12 ⎠730⎛ 0.035 ⎞($375) ⎜1 + ⎟ ≈$402.19⎝ 365 ⎠0.035⋅2($375) e ≈ $402.192($375)(1.046) = $410.2924⎛ 0.046 ⎞($375) ⎜1 + ⎟ ≈$411.06⎝ 12 ⎠730⎛ 0.046 ⎞($375) ⎜1 + ⎟ ≈$411.13⎝ 365 ⎠0.046⋅2($375) e ≈ $411.1412t⎛ 0.06 ⎞⎜1+ ⎟ = 2⎝ 12 ⎠121.005 t = 2ln 2ln 212t = so t = ≈ 11.58ln1.005 12ln1.005It will take about 11.58 years or11 years, 6 months, 29 days.0.06tln 2e = 2 ⇒ t = ≈ 11.550.06It will take about 11.55 yearsor 11 years, 6 months, and 18 days.5$20,000(1.025) ≈ $22,628.1629. 1626 to 2000 is 374 years.0.06⋅374y = 24 e ≈ $133.6 billion30.969 18$100(1.04) ≈ $3.201×10dy34. ky( L – y)dt =35.1dy = kdtyL ( – y) ⎡ 1 1 ⎤⎢ + dy = kdtLy L( L – y)⎥⎣⎦1 1 1dy kdtL ∫ ⎛ ⎞⎜ + ⎟ =⎝ y L–y∫ ⎠1 [ln y – ln L − y]= kt + CLyln Lkt LC1L–y = +y + 1 1y= e = e ⋅ e , so = CeL–y L−y⎛0 Lk⋅0Note that: C = Ce = Ce ⎞⎜⎟⎜y(0)y0⎟⎜= = .L– y(0) L–y ⎟⎝0 ⎠Lkt Lkty = LCe – yCeLkty+ yCe = LCe1Lkt LC LC Lkt LktLktLktLCe LC LCy = = =Lkt1+ Ce C + e1+ C– LkteLktyL ⋅0L– y0 Ly0y0– Lkt– Lkt+ e y– 0 + ( L– y0)eL y0= =( )16 6.4y =−16(0.00186)t6.4 + (16 −6.4)e102.4=−0.02976t6.4 + 9.6e20y31.32.(0.05)(1)1000 e = $1051.27(0.05)(1)Ae 0 = 10000.05A0 = 1000 e − ≈ $951.231033. If t is the doubling time, thent⎛ p ⎞⎜1+ ⎟ = 2⎝ 100 ⎠⎛ p ⎞t ln ⎜1+ ⎟=ln 2⎝ 100 ⎠ln2 ln2 100ln2 70t = ≈ = ≈p pln 1+p p100( ) 100−50 150 tInstructor’s Resource Manual Section 6.5 377

36. a.b.c.d.e.37. a.38.b.c.d.1000 1000lim (1 + x) = 1 = 1x→ 01/ xlim1 = lim1=1x→0 x→01/ xnlim (1 + ε) = lim (1 + ε)=∞x→0+ n→∞1/ x 1lim (1 + ε ) = lim = 0nx→0−n→∞(1 + ε )1/lim (1 + xx)= ex→ 01/1 1lim (1 − xx) = limx→0 x→0[1 ( )]1( −x)+ −x= e131/ x ⎡ ⎤3 3lim (1 + 3 x) = lim ⎢(1 + 3 x)x ⎥ = ex→0 x→0⎢⎣⎥⎦nn⎛n+ 2⎞ ⎛ 2⎞lim ⎜ ⎟ = lim ⎜1+⎟n→∞⎝ n ⎠ n→∞⎝ n⎠1/ x= lim (1 + 2 x)x→0+12⎡2 2lim (1 2 ) x ⎤= ⎢ + x ⎥ = ex→0+ ⎢⎣⎥⎦2n2n⎛n−1⎞ ⎛ 1⎞lim ⎜ ⎟ = lim ⎜1−⎟n→∞⎝ n ⎠ n→∞⎝ n⎠2/ x= lim (1 − x)x→0+1−2⎡1lim (1 ) x ⎤= ⎢ − x − ⎥ =2x→0+ ⎢⎣⎥⎦edyay bdt = +dy∫ = adty + b ∫abln y+ = at+Cab at+C b aty+ = e ; y+ = Aeaaat by = Ae −abby0 = A− ⇒ A= y0+aa⎛ b⎞at by = ⎜y0+ ⎟e−⎝ a⎠a39. Let y = population in millions, t = 0 in 1985,a = 0.012, b = 0.06 , y 0 = 10dy0.012y0.06dt = +0.06 0.012t0.06 0.012ty ⎛ ⎞= ⎜10 + ⎟e – = 15 e – 5⎝ 0.012 ⎠ 0.012From 1985 to 2010 is 25 years. At t = 25,0.012⋅25y = 15e−5 ≈ 15.25. The population in 2010will be about 15.25 million.40. Let N(t) be the number of people who have hearddNthe news after t days. Then = kL ( − N ).dt1∫ dN = k dtL−N∫–ln(L – N) = kt + Ckt CL− N = e − −N = L−Ae −ktN(0) = 0, ⇒ A = LNt () = L(1 − e −kt ).LN (5) = ⇒21 5k= e −2L5k= L(1 − e − )2ln 1k = 2≈0.1386−50.1386tNt () = L(1 − e − )0.1386t0.99 L = L(1 − e − )0.1386t0.01 = e −ln 0.01t = ≈ 33−0.138699% of the people will have heard about the scandalafter 33 days.kt41. If f(t) = e ,42.thenktf′() t ke= = k .f()t kten n–1f( x)= anx + an–1x +⋅⋅⋅+ a1x+a0lim f′( x)x →∞ f ( x )n–1 n–2nanx + ( n –1) an–1x +⋅⋅⋅+ a1= limx→∞nanx + an–1x+⋅⋅⋅+ a1 x+a0nan( n–1) a –1 a+ n +⋅⋅⋅+ 1xx2xn= lim = 0x →∞ an–1 a1 aa0n + +⋅⋅⋅+xxn –1+xn378 Section 6.5 Instructor’s Resource Manual

43.44.f′( x)= k > 0 can be written asf( x)1 dy= k where y = f(x).ydxdykx= kdx has the solution y = Ce .ykxThus, the equation f ( x)= Ce representsexponential growth since k > 0.f′( x)= k < 0 can be written as 1 dy kf( x)ydx = wherey = f(x). dykx= kdx has the solution y = Ce .ykxThus, f ( x)= Ce which represents exponentialdecay since k < 0.e. The maximum population will occur whend2( 0.0132t− 0.0001t) = 0dt0.0132 = 0.0002tt = 0.0132 / 0.0002 = 66t = 66 , which is year 2070.The population will equal the 2004 value of26.4 billion when 0.0132t− 0.0001t= 0t = 0 or t = 132 .The model predicts that the population willreturn to the 2004 level in year 2136.47. a. k = 0.0132 − 0.0001tb. y ' = ( 0.0132 − 0.0001 )t y45. Maximum population:2 640 acres 1 person13,500,000 mi ⋅ ⋅2 11 mi acre210= 1.728×10 peopleLet t = 0 be in 2004.9 0.0132 10(6.4× 10 ) et = 1.728×10⎛ 101.728⋅10⎞ln⎜96.4⋅10⎟t =⎝ ⎠≈ 75.2 years from 2004, or0.0132sometime in the year 2079.46. a. k = 0.0132 − 0.0002tb. y ' = ( 0.0132 − 0.0002 )t ydyt ydt = −dy( 0.0132 0.0002t)dty = −c. ( 0.0132 0.0002 )d.2ln y = 0.0132t− 0.0001t + C00.0132t−0.0001t2y = C1eThe initial condition (0) 6.4C 1 = 6.4 . Thus105yy = implies that0.0132t−0.0001t2y = 6.4edyt ydt = −dy( 0.0132 0.0001t)dty = −c. ( 0.0132 0.0001 )d.2ln y = 0.0132t− 0.00005t + C00.0132t−0.00005t2y = C1eThe initial condition (0) 6.4C 1 = 6.4 . Thus2010y100 200y = implies that0.0132t−0.00005t2y = 6.4e300e. The maximum population will occur whend2( 0.0132t− 0.00005t) = 0dt0.0132 = 0.0001tt = 0.0132 / 0.0001 = 132t = 132 , which is year 2136The population will equal the 2004 value of26.4 billion when 0.0132t− 0.00005t= 0t = 0 or t = 264 .The model predicts that the population willreturn to the 2004 level in year 2268.t50 100 150tInstructor’s Resource Manual Section 6.5 379

48.49.Ex ( + h)– Ex ( )E′ ( x) = lim h →0hExEh ( ) ( )– Ex ( )= limh→0hEh ( )–1 Eh ( )–1= lim Ex ( ) ⋅ = Ex ( ) limh→0 hh→0hEx ( ) = Ex ( + 0) = Ex ( ) ⋅ E(0)so E (0) = 1.Eh ( )– E(0)Thus, E′ ( x) = E( x) lim h →0hE(0 + h) – E(0)= Ex ( ) lim = Ex ( ) ⋅ E′(0)h→0h= kE(x) where k = E′ (0) .kx kx kx kxHence, Ex ( ) = Ee 0 = E(0) e = 1⋅ e = e .ku ( + v)ku+kvCheck: Eu ( + v)= e = eku kv= e ⋅ e = E( u) ⋅ E( v)6.6 Concepts Review∫1. exp ( Pxdx ( ) )2. yexp ( P( x)dx)3.∫1 d ⎛ y⎞ ; 1;2⎜ ⎟ = x + Cxx dx⎝x⎠4. particularProblem Set 6.61. Integrating factor is e x .xDye ( ) = 1– xy = e ( x+C )2. The left-hand side is already an exact derivative.2Dyx [ ( + 1)] = x –13x –3 x+Cy =3( x + 1)50. a.Exponential growth:In 2010 (t = 6): 6.93 billionIn 2040 (t = 36): 10.29 billionIn 2090 (t = 86): 19.92 billionLogistic growth:In 2010 (t = 6): 7.13 billionIn 2040 (t = 36): 10.90 billionIn 2090 (t = 86): 15.15 billionb.1/lim (1 + xx)= ex→ 01/ xlim (1– x)x→01=e3.x axy′ + y =2 21– x 1– xIntegrating factor:x2 –1/2exp dx exp ⎡ln(1– x ) ⎤∫ =21– x ⎣⎦2 –1/2= (1 – x )2 –1/2 2 –3/2Dy [ (1– x ) ] = ax(1– x )Then2 –1/2 2 –1/2y(1 – x ) = a(1 – x ) + C,so21/2y = a+C(1 – x ) .4. Integrating factor is sec x.2Dy [ sec x] = sec xy = sin x + C cos x5. Integrating factor is 1 .x⎡y⎤ xD⎢= ex ⎥⎣ ⎦xy = xe + Cx380 Section 6.6 Instructor’s Resource Manual

6. y′ – ay = f( x)– adx –Integrating factor: e∫ = e– ax – axDye [ ] = e f( x)– ax – axThen ye = ∫ e f ( x) dx,soax– axy = e ∫ e f( x)dx.ax7. Integrating factor is x. D[yx] = 1;8. Integrating factor is2 5Dyx [ ( + 1) ] = ( x+1)2( x + 1) .⎛1 ⎞ ( 1)4 ( 1)–2y = ⎜ ⎟ x+ + C x+⎝6⎠9. y′ + f( x) y = f( x)f ( xdx )e ∫Integrating factor:f ( xdx ) f( xdx )D⎡ye∫⎤= f( x)e∫⎢⎣⎥⎦f ( x) dx f ( x)dxye∫= e∫+ C,soThen– f ( xdx )y = 1+ Ce ∫ .10. Integrating factor is 2 xe .2x2xDye [ ] = xe⎛1⎞ ⎛1⎞–2xy = ⎜ ⎟x–⎜ ⎟+Ce⎝2⎠ ⎝4⎠11. Integrating factor is 1 .x12.4y x 2x⎡y⎤ 2D⎢= 3 x ;x ⎥⎣ ⎦= + goes through (1, 3).2xy′ + 3y = edxIntegrating factor: e∫ = e3 3x3x5xDye [ ] = e5x3xeye C.55x3xe 4ye = + .5 52 x –3xe + 4ey5ThenTherefore,–1y = 1+Cx4y = x + Cx4= + x = 0, y = 1 ⇒ C = , so5solution through (0, 1).= is the particular13. Integrating factor: xexx– x –1d[ yxe ] = 1;y = e (1 + Cx );goes through (1, 0).– x –1y = e (1 – x )14. Integrating factor is sin 2 x .2 2Dy [ sin x] = 2sin xcosx2 2 3ysinx = sin x+C32 Cy = sin x+3 2sin x2 5 2y = sin x+csc x3 12⎛π⎞goes through ⎜ , 2 ⎟.⎝6⎠15. Let y denote the number of pounds of chemical Aafter t minutes.dy ⎛ lbs ⎞⎛ gal ⎞ ⎛ y lbs ⎞⎛3 gal ⎞= ⎜2 ⎟⎜3 ⎟–⎜ ⎟⎜ ⎟dt ⎝ gal ⎠⎝ min ⎠ ⎝20 gal ⎠⎝ min ⎠3y= 6– lb/min 203y′ + y = 620( 3/20)dt 3/20 tIntegrating factor: e∫ = e3/20 t 3/20 tDye [ ] = 6e3/20 t 3/20 tThen ye = 40 e + C.t = 0, y = 10⇒ C = –30.–3 t /20Therefore, yt () = 40–30 e , so–3y(20) 40 – 30e38.506= ≈ lb.dy ⎛ y ⎞ y16. = (2)(4) – ⎜ ⎟(4) or y′+ = 8dt ⎝200 ⎠ 50t /50Integrating factor is e .t/50 t/50Dye [ ] = 8e17.– t /50yt () = 400+Ce– /50yt () 400–350et= goes through (0, 50).–0.8y(40) 400 – 350e242.735= ≈ lb of saltdy ⎡ y ⎤ ⎡ 3 ⎤= 4– ⎢ ⎥(6) or y′+ ⎢ ⎥ y = 4dt ⎣(120 – 2 t) ⎦ ⎣(60 – t)⎦–3Integrating factor is (60 – t ) .–3 –3Dy [ (60– t) ] = 4(60– t)3yt () = 2(60– t) + C(60– t)⎛ 1 ⎞ 3yt ( ) = 2(60 – t) – ⎜ ⎟(60 − t)⎝1800⎠(0, 0).goes throughInstructor’s Resource Manual Section 6.6 381

18.19.dy –2y=dt 50 + tor 2y′ + y = 0.50 + tIntegrating factor:⎛ 2 ⎞ 2ln(50 + t) 2exp ⎜∫dt ⎟ = e = (50 + t)⎝ 50 + t ⎠2Dy [ (50 + t) ] = 02Then y(50 + t) = C.t = 0, y = 30 ⇒ C = 750002Thus, y(50 + t) = 75,000.2If y = 25, 25(50 + t) = 75,000, sot = 3000 – 50 ≈ 4.772 min.6I′ + 10 I = 1Integrating factor =6exp(10 t )6 6DI [ exp(10 t)] = exp(10 t)–6 6I( t) = 10 + Cexp(–10 t)–6 6I( t) 10 [1– exp(–10 t)]20. 3.5I′ = 120sin 377t= goes through (0, 0).240I′ ⎛ ⎞= ⎜ ⎟sin377 t⎝ 7 ⎠240I ⎛ ⎞= ⎜– ⎟cos377t+C⎝ 2639 ⎠⎛ 240 ⎞I() t = ⎜ ⎟ (1–cos377) t through (0, 0).⎝2639⎠21. 1000 I = 120 sin 377tI(t) = 0.12 sin 377t22.dx 2x–dt = 1001x′ ⎛ ⎞+ ⎜ ⎟x= 0⎝50⎠Integrating factor ist /50Dxe [ ] = 0– t /50x = Ce– /50xt () 50ett /50e .= satisfies t = 0, x = 50.– t /50dy ⎛50e ⎞ ⎛ y ⎞= 2 –2dt ⎜ ⎜ ⎟100 ⎟⎝ ⎠ ⎝200⎠1 – t /50y′ ⎛ ⎞+ ⎜ ⎟y = e⎝100⎠Integrating factor isttDye [ ] = e/100 – /100t /100e .– t/100 – t/100yt ( ) = e ( C–100 e )– t/100 – t/100yt ( ) e (250 –100 e )= satisfies t = 0,y = 150.23. Let y be the number of gallons of pure alcohol in thetank at time t.a.dy5y′ ⎛ ⎞= = 5(0.25) – ⎜ ⎟y = 1.25 – 0.05ydt⎝100⎠0.05tIntegrating factor is e .–0.05yt ( ) = 25 + Ce t ; y= 100, t= 0, C=75–0.05tyt ( ) 25 75 e ;= + y = 50, t = T,T = 20(ln 3) ≈ 21.97 minb. Let A be the number of gallons of pure alcoholdrained away.200(100 – A) + 0.25A = 50 ⇒ A =32003It took minutes for the draining and the5same amount of time to refill, so2 200( 3 ) 80T = = ≈ 26.67 min.5 3c. c would need to satisfy200 2003 3+ < 20(ln 3).5 c10c > ≈ 7.7170(3ln 3 – 2)d. y′ = 4(0.25) – 0.05y = 1– 0.05ySolving for y, as in part a, yields–0.05ty = 20 + 80 e . The drain is closed whent = 0.8 T.We require that−0.05⋅0.8T(20 + 80 e ) + 4⋅0.25⋅ 0.2T= 50,–0.04Tor 400e+ T = 150.24. a. v′ + av = – gatIntegrating factor: eat at d at ate ( v′ + av) = – ge ; ( ve ) = – gedtat at – g at – g – atve = ∫ – ge dt = e + C;v = + Ceaav = v0, t = 0v –g g0 = + C ⇒ C = v0+aa– g ⎛ g ⎞ – atTherefore, v = + ⎜v0+ ⎟e, soa ⎝ a ⎠– atvt () = v∞+ ( v0– v∞) e .382 Section 6.6 Instructor’s Resource Manual

.25. a.dy– at= v∞+ ( v0– v∞) e , sodt−at( v0− v∞) e= ∞ ⋅ − +y v t C.a–( v0– v∞)y = y0, t = 0 ⇒ y0= + Cav0– v∞⇒ C = y0+a– at( v0 – v∞) e ⎛ v0– v∞⎞y = v∞t –+ y0a⎜ +a⎟⎝⎠v0– v∞– at= y0+ v∞t+(1 – e )a32v ∞ = – = –6400.050.05tvt ( ) = [120 −( − 640)] e − + ( − 640) = 0 if⎛19⎞t = 20ln ⎜ ⎟.⎝16⎠yt ( ) = 0 + (–640) t⎛ 1 ⎞–0.05t+ ⎜ ⎟ [120 – (–640)](1– e )⎝0.05⎠–0.05t= –640t+15, 200(1– e )Therefore, the maximum altitude is19 19 45,600y ⎜⎛ 20ln ⎜ ⎛ ⎟ ⎞ ⎟⎞ =− 12,800ln ⎜ ⎛ ⎟⎞ +⎝ ⎝16 ⎠⎠⎝16 ⎠ 19≈ 200.32 ft–0.05Tb. –640T+ 15,200(1– e ) = 0;–0.05T95 – 4 T – 95e = 026. For t in [0, 15],–32v ∞ = = –320.0.1027. a.28.–0.1 t–0.1tvt ( ) = (0 + 320) e – 320 = 320( e –1);–1.5v(15) = 320( e –1) ≈ –248.6–0.1tyt ( ) = 8000 – 320t+10(320)(1– e );–1.5y(15) = 3200(2 – e ) ≈ 5686Let t be the number of seconds after the parachuteopens that it takes Megan to reach the ground.32For t in [15, 15+T], v ∞ = – = –20.1.60 = yT ( + 15)–1.5= [3200(2 – e )]–1.5 –1.6T–20 T + (0.625)[320( e –1) + 20](1– e )1.6T≈ 5543 – 20 T –142.9e − ≈ 5543 – 20T [since–1.6 T –35T > 50, so e < 10 (very small)]Therefore, T ≈ 277, so it takes Megan about292 s (4 min, 52 s) to reach the ground.− ln x+ C ⎛dyy ⎞ 2 − ln x+Ce ⎜ − ⎟=x e⎝dxx ⎠−ln x C ⎛dyy ⎞ 2 C −lnxe e ⎜ − ⎟=x e e⎝dxx ⎠1 C dy C 1 2 C 1e − ye = x ex dx 2x xd ⎛ C 1 ⎞ e y xeC⎜ ⎟ =dx ⎝ x ⎠b.C y Ce ex = ∫ xdx2y xC1x = 23xy = + C1x2Pxdx ( ) CdyPxdx ( ) Ce∫ + ++ P( x)e∫ydx= Qxe ( )∫ Pxdx ( ) + Cd ⎛ Pxdx ( ) + C ⎞Pxdx ( ) + Ce∫ y = Q ( x ) e∫dx ⎜⎟⎝⎠Pxdx ( ) CPxdx ( )ye∫ += Q( x)e∫e Cdx + C1∫− P( x) dxP( x)dxy = e∫Q( x)e∫dx∫−+ Ce∫2Pxdx ( )Instructor’s Resource Manual Section 6.6 383

6.7 Concepts Review5.1. slope field2. tangent line3. yn−1+hf( xn−1, yn−1)4. underestimateProblem Set 6.71.6.The oblique asymptote is y= x.lim yx ( ) = 12 and y(2) ≈ 10.5x→∞2.7.The oblique asymptote is y = 3 + x/2.3.4.lim yx ( )x→∞=∞and y(2) ≈ 16lim yx ( ) = 0 and y(2) ≈ 6x→∞dy 1 1y; y(0)dx = 2 = 2dy 1dxy = 2xln y = + C2y = C1ex /2To find C 1 , apply the initial condition:10= y(0)= C1e = C121 x /2y = e2lim yx ( )x→∞=∞ and y(2) ≈ 13384 Section 6.7 Instructor’s Resource Manual

8.9.10.dyy; y(0) 4dx =− =dydxy =−ln y =− x+Cy = C1e −xTo find C 1 , apply the initial condition:−04 = y(0)= C1e = C1y = 4e −xy' + y = x+21dx xThe integrating factor is e∫ = e .e x y' + ye x = e x ( x+2)x( ) = ( + 2)d e y x e xdxxxe y = ( x+2) e dx∫Integrate by parts: let u = x+2, dv = e dx .xThen du = dx and v = e . Thusx x xe y = ( x+ 2) e −∫ e dxx x xe y = ( x+ 2) e − e + Cy = x+ 2− 1+Ce −xTo find C , apply the initial condition:−04 = y(0) = 0+ 1+ Ce = 1 + C → C = 3Thus, y = x+ 1+ 3e .x3y' + y = 2x+23e x y' ye x ⎛ ⎞+ = ⎜2x+⎟ex⎝ 2 ⎠d x ⎛ 3 ⎞ x( e y ) = ⎜2x + ⎟ edx ⎝ 2 ⎠x 3 xe y ⎛ ⎞= ∫ ⎜2x+⎟e dx⎝ 2 ⎠3Integrate by parts: let u = 2 x+,2xxdv = e dx . Then du = 2dxand v = e .Thus,3e x y ⎛ ⎞= ⎜2x+ ⎟e x − 2e x dx⎝ 2∫⎠3e x y ⎛ ⎞= ⎜2x+ ⎟e x − 2e x + C⎝ 2 ⎠1y = 2x− + Ce −x2To find C, apply the initial condition:1 −013 = y(0) = 0− + Ce = C−2 27Thus C = , so the solution is21 7y = 2x− + e −x2 2Note: Solutions to Problems 22-28 are given along withthe corresponding solutions to 11-16.11., 22. x nEuler's Improved EulerMethod y n Method y n0.0 3.0 3.00.2 4.2 4.440.4 5.88 6.57120.6 8.232 9.725380.8 11.5248 14.393561.0 16.1347 21.3024612., 23. x nEuler's Improved EulerMethod y n Method y n0.0 2.0 2.00.2 1.6 1.640.4 1.28 1.34480.6 1.024 1.102740.8 0.8195 0.904241.0 0.65536 0.74148Instructor’s Resource Manual Section 6.7 385

13., 24. x nEuler's Improved EulerMethod y n Method y n0.0 0.0 0.00.2 0.0 0.020.4 0.04 0.080.6 0.12 0.180.8 0.24 0.321.0 0.40 0.5014., 25. x nEuler's Improved EulerMethod y n Method y n0.0 0.0 0.00.2 0.0 0.0040.4 0.008 0.0240.6 0.040 0.0760.8 0.112 0.1761.0 0.240 0.34015., 26. x nEuler's Improved EulerMethod y n Method y n1.0 1.0 1.01.2 1.2 1.2441.4 1.488 1.609241.6 1.90464 2.164101.8 2.51412 3.024552.0 3.41921 4.39176516., 27. x nEuler's Improved EulerMethod y n Method y n1.0 2.0 2.01.2 1.2 1.3121.4 0.624 0.806091.6 0.27456 0.466891.8 0.09884 0.256982.0 0.02768 0.1356817. a. y 0 = 1y1 = y0 + hf( x0, y0)= y0 + hy0 = (1 + h)y0y2 = y1+ hf( x1, y1)= y1+hy12= (1 + hy ) 1 = (1 + h)y0y3 = y2 + hf( x2, y2)= y2 + hy23= (1 + hy ) 2 = (1 + h)y0yn = yn−1+ hf( xn−1, yn−1)= yn−1+hyn−1nn= (1 + hy ) n−1 = (1 + h) y0= ( 1+h)b. Let N = 1/ h. Then y N is an approximationto the solution at x = Nh = (1/ h) h= 1 . Theexact solution is y(1)= e . Thus,N( 1 1/ N)ewe know that lim ( 1 1/ N)+ ≈ for large N. From Chapter 7,+ N= e .N →∞18. y0 = y( x0) = 0y1 = y0 + hf ( x0) = 0 + hf ( x0) = hf ( x0)y2 = y1+ hf( x1) = hf( x0) + hf( x1)= h( f( x0) + f( x1))y3 = y2 + hf( x2)= h f( x ) + f( x ) + hf( x )19. a.[ ]0 1 231 −h f( x ) f( x ) f( x ) h f( xi)i=0= [ 0 + 1 + 2 ] = ∑At the nth step of Euler's method,b.n−1yn = yn−1+ hf( xn−1) = h∑f( xi)i=0x1 x1 2∫ y'( x)dx = sinx ∫ x dx0 x0( )2yx ( ) − yx ( ) ≈ x−x sinx1 0 1 0 021 − = 021yx ( ) y(0) hsinxyx ( ) −0≈0.1sin0yx ( ) ≈ 01x2 x2 2y'( x)dx = sinx0 x0∫ ∫x dx( )2yx ( ) − yx ( ) ≈ x−x sinx2 0 1 0 02+ ( x2 −x1)sinx12 22 − = 0 + 12 22yx ( ) y(0) hsinx hsinxyx ( ) −0 ≈ 0.1sin 0 + 0.1sin 0.1yx ( ) ≈ 0.000999982386 Section 6.7 Instructor’s Resource Manual

c.20. a.x3 x3 2y'( x)dx = sinx0 x0∫ ∫x dx( )2yx ( ) − yx ( ) ≈ x−x sinx3 0 1 0 02 2+ ( x2 − x1)sin x1 + ( x3 −x2)sinx12 2 23 − = 0 + 1 + 22 232y( x ) y(0) hsin x hsin x hsinxyx ( ) −0 ≈ 0.1sin 0 + 0.1sin 0.1+ 0.1sin 0.2yx ( 3) ≈ 0.004999Continuing in this fashion, we havexnxn2∫ y'( x)dx = sinx ∫ x dx0 x0n−12yx ( n) − yx ( 0) ≈∑( xi+1−xi)sinxii=0n−1yx ( n) ≈ h∑f( xi−1)i=0When n = 10 , this becomesyx ( 10) = y(1) ≈ 0.269097n−1d. The result yx ( n) ≈ h∑ f( xi−1) is the same asi=0that given in Problem 18. Thus, when f ( xy , )depends only on x , then the two methods (1)Euler's method for approximating the solutionto y' = f( x)at x n , and (2) the left-endpointxnRiemann sum for approximating ∫ f ( xdx ) ,0are equivalent.b.x1 x1x0 x0∫ ∫( )y'( x) dx = x+1 dxyx ( 1) − yx ( 0) ≈ x1− x0 x0+ 1yx ( ) − y(0) = h x + 11 0yx ( 1) −0≈ 0.1 0+1yx ( ) ≈ 0.11x2 x2x0 x0∫ ∫y'( x) dx = x+1 dx( )yx ( ) − yx ( ) ≈ x− x x + 12 0 1 0 0+ ( x2 − x1) x1+1yx ( 2) − y(0) = h x0 + 1+ h x1+1yx ( 2) −0 ≈ 0.1 0 + 1 + 0.1 0.1+1yx ( ) ≈ 0.2048812c.x3 x3x0 x0∫ ∫y'( x) dx = x+1 dx( )yx ( ) − yx ( ) ≈ x− x x + 13 0 1 0 0+ ( x − x ) x + 1 + ( x − x ) x + 12 1 1 3 2 2yx ( 3) − y(0) = 0.1 0 + 1 + 0.1 0.1+1+ 0.1 0.2 + 1yx ( 3) ≈ 0.314425Continuing in this fashion, we havexnxny'( x) dx = x+1 dxx0 x0n−1n 0 ∑ i+ 1 i i−1i=0n−1( n) ≈ ∑ i−1+ 1i=0∫ ∫yx ( ) − yx ( ) ≈ ( x − x) x + 1yx h xWhen n = 10 , this becomesyx ( 10) = y(1) ≈ 1.198119Δ y 121. a. = [ f ( x0, y0) + f( x1 + yˆ1)]Δx2y1−y0Δy1b. = = [ f( x0, y0) + f( x1+ yˆ1)]⇒h Δx22( y1− y0) = h[ f( x0, y0) + f( x1+ yˆ1)]⇒hy1− y0 = [ f( x0, y0) + f( x1+ yˆ1)]⇒2hy1 = y0 + [ f( x0, y0) + f( x1+yˆ1)]2c. 1. xn−1+ h2. yn−1+hf( xn−1, yn−1)h3. yn−1+ [ f( xn−1, yn−1) + f( xn, yˆn)]222-27. See problems 11-1628.hError fromEuler'sMethodError fromImprovedEuler Method0.2 0.229962 0.0155740.1 0.124539 0.0042010.05 0.064984 0.0010910.01 0.013468 0.0000450.005 0.006765 0.000011For Euler's method, the error is halved as the stepsize h is halved. Thus, the error is proportional to h.For the improved Euler method, when h is halved,the error decreases to approximately one-fourth ofwhat is was. Hence, for the improved Eulermethod, the error is proportional to2hInstructor’s Resource Manual Section 6.7 387

6.8 Concepts Review⎡ π π⎤1. ⎢– , ;2 2⎥arcsin⎣ ⎦⎛ π π⎞2. ⎜– , ⎟;arctan⎝ 2 2⎠3. 14. πProblem Set 6.813.14.15.16.⎛ 1 ⎞cos(arccot 3.212) = cos⎜arctan ⎟⎝ 3.212 ⎠≈ cos 0.3018 ≈ 0.95481sec(arccos 0.5111) =cos(arccos 0.5111)1= ≈ 1.9570.5111⎞sec (–2.222) = cos ⎜ ⎟≈2.038⎝ –2.222 ⎠–1 –1 ⎛ 11tan − ( − 60.11) ≈ –1.5541.2.3.⎛ 2 ⎞ πarccos⎜=2 ⎟since⎝ ⎠ 4π 2cos = 4 2⎛ 3⎞ π ⎛ π⎞3arcsin ⎜– = – since sin – = –2 ⎟⎜ ⎟⎝ ⎠ 3 ⎝ 3⎠2–1⎛ 3⎞ π ⎛ π⎞3sin ⎜– = – since sin – = –2 ⎟⎜ ⎟⎝ ⎠ 3 ⎝ 3⎠217.18.19.20.−1cos(sin(tan 2.001))≈ 0.62592sin (ln(cos 0.5555)) ≈ 0.02632θ = sinθ = tan−1−1x8x64.–1⎛ 2 ⎞ π ⎛ π⎞2sin ⎜– = – since sin – = –2 ⎟⎜ ⎟⎝ ⎠ 4 ⎝ 4⎠221.θ = sin−1 5xπ ⎛π⎞5. arctan( 3) = since tan ⎜ ⎟=33 ⎝3⎠22.9θ = cos or θ = secx−1 −1x96.7.8.9.10.11.⎛1⎞ π ⎛π⎞1arcsec(2) = arccos ⎜ ⎟= since cos⎜ ⎟=, so⎝2⎠ 3 ⎝3⎠2⎛π⎞sec⎜⎟ = 2⎝3⎠⎛ 1⎞ π ⎛ π⎞1arcsin ⎜– ⎟= – since sin ⎜– ⎟=–⎝ 2⎠ 6 ⎝ 6⎠2–1⎛ 3⎞ π ⎛ π⎞3tan ⎜– = – since tan – = –3 ⎟⎜ ⎟⎝ ⎠ 6 ⎝ 6⎠3–1sin(sin 0.4567) = 0.4567 by definition–1 2 −1cos(sin 0.56) = 1−sin (sin 0.56)2= 1– (0.56) ≈ 0.828−1sin (0.1113)≈ 0.111512. arccos(0.6341) ≈ 0.884023. Let θ 1 be the angle opposite the side of length 3,3and θ2 = θ1 – θ,so θ = θ1 – θ2.Then tanθ 1 =x1 –1 3 –1 1and tan θ 2 = . θ = tan – tan .xx x24. Let θ 1 be the angle opposite the side of length 5,and θ2 = θ1− θ , and y the length of the unlabeled2side. Then θ = θ1− θ2and y = x − 25.5 5 2 2tan θ1 = = , tanθ2= = ,y 2 2x −25 y x −25−1 5 −12θ = tan−tan2 2x −25 x −2525.–1 2 2 –1 2cos ⎡ ⎛ ⎞2sin – ⎤ 1– 2sin ⎡ ⎛ ⎞⎢ ⎜ ⎟ = sin ⎜–⎟⎤3⎥ ⎢3⎥⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦2⎛ 2⎞1= 1–2 ⎜– ⎟ =⎝ 3 ⎠ 9388 Section 6.8 Instructor’s Resource Manual

26.27.28.⎡tan ⎢2 tan⎣1321( )3–12⋅3= =1–4( ) ⎤3( )–12tan⎡1⎛1⎞⎤tan⎣ ⎦⎜ ⎟3⎥ =⎝ ⎠⎦2 –11–tan ⎡tan1 ⎤⎣ 3 ⎦–1 3 –1 5 –1 3 –1 5 –1 3 –1 5sin ⎡ ⎛ ⎞ ⎛ ⎞cos cos ⎤ sin ⎡ ⎛ ⎞cos ⎤ cos ⎡ ⎛ ⎞cos ⎤ cos ⎡ ⎛ ⎞cos ⎤ sin ⎡ ⎛ ⎞⎢ ⎜ ⎟+ ⎜ ⎟ = +cos⎤5 13⎥ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟5⎥ ⎢13⎥ ⎢5⎥ ⎢13⎥⎣ ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦2 2⎛3⎞ 5 3 ⎛ 5 ⎞ 56= 1– ⎜ ⎟ ⋅ + 1– ⎜ ⎟ =⎝5⎠ 13 5 ⎝13⎠65–1 4 –1 12 –1 4 –1 12 –1 4 –1 12cos ⎡ ⎛ ⎞ ⎛ ⎞cos sin ⎤ cos ⎡ ⎛ ⎞cos ⎤ cos ⎡ ⎛ ⎞sin ⎤ – sin ⎡ ⎛ ⎞cos ⎤ sin ⎡ ⎛ ⎞⎢ ⎜ ⎟+ ⎜ ⎟ =sin⎤5 13⎥ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟5⎥ ⎢13⎥ ⎢5⎥ ⎢13⎥⎣ ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦2 24 ⎛12 ⎞ ⎛4 ⎞ 12 16= ⋅ 1– ⎜ ⎟ – 1– ⎜ ⎟ ⋅ = –5 ⎝13⎠ ⎝5⎠13 6529.30.31.32.33. a.b.–1–1 sin(sin x)xtan(sin x)= =cos(sin x) 1– x–1 2–1 1 1sin(tan x)= =–1csc(tan x) 2 –11+cot (tan x)1 1 x= = =1+ 1 1+ 1 2tan2(tan–1)2 x + 1x x–1 2 –1 2cos(2sin x) = 1– 2sin (sin x) = 1– 2x–1–1 2tan(tan x) 2xtan(2 tan x)= =2 –1 21– tan (tan x) 1– x–1lim tan xx→∞π= since lim2 θ π–1 πlim tan x –x →–∞ 2lim tanθ=−∞θ→−π/2+= sincetan→ /2−θ = ∞34. a.b.35. a. Let⎞lim sec x = lim cos ⎜ ⎟x→∞x→∞⎝ x ⎠–1 π= lim cos z =z→0+ 2–1 –1 ⎛1–1 –1 ⎛1⎞lim sec x = lim cos ⎜ ⎟x→– ∞x→–∞ ⎝ x ⎠–1 π= lim cos z =z→0– 2x→1−−1sin(sin x) = x,−1L = lim sin x . Since−1xx→1 −x→1−lim sin(sin ) = lim x = 1 .Thus, since sin is continuous, the CompositeLimit Theorem gives us−1lim sin(sin x) lim sin( L)x→1 −x→1−= ; hencesin L = 1 and since the range of1sin − is⎡ π ,π ⎤⎢−2 2⎥⎣ ⎦ , L = π .2Instructor’s Resource Manual Section 6.8 389

. Let−1L = lim sin x . Sincex→−1+−1sin(sin x) = x,−1lim sin(sin x) = lim x =− 1.x→−1 +x→−1+Thus, since sin is continuous, theComposite Limit Theorem gives us−1lim sin(sin x) = lim sin( L);x→−1 +x→−1+hencesin L =−1and since the range of⎡ π ,π ⎤⎢−2 2⎥⎣ ⎦ , πL =− .2−136. No. Since sin x37. Let38.−1lim sin xx→1+two-sided limit1sin − isis not defined on (1, ∞ ) ,does not exist so neither can the−1lim sin xx→1−1f( x) y sin x= = ; then the slope of thetangent line to the graph of y at c isf′ () c =121−c. Hence,.lim ′( c)=∞ sofc→1−that the tangent lines approach the vertical.39. y = ln(2 + sin x). Let u = 2+ sinx; theny = ln u so by the Chain Ruledy dy du ⎛1⎞du⎛ 1 ⎞= = ⎜ ⎟ = ⎜ ⎟⋅cosxdx du dx ⎝u ⎠dx ⎝2+sinx⎠cos x=2 + sinx40.41.42.43.44.45.46.d tan x tan x d tan x 2e = e tan x = e sec xdxdxd sec tan secln(sec tan )x x +x+ x =xdx sec x + tan x(sec x)(tan x+sec x)= = sec xsec x+tan xd – csc cot – csc[– ln(csc x+ cot x)] = –x x xdx csc x + cot xcsc x(cot x+csc x)= = csc xcot x+csc xd –1 2 1 4xsin (2 x ) = ⋅ 4x=dx 2 2 41–(2 x ) 1–4xxd x 1 x earccos( e ) = – ⋅ e = –dx x 2 2x1–( e ) 1– exd 3 –1 x 3 e 2 –1 x[ x tan ( e )] = x ⋅ + 3x tan ( e )dxx 21 + ( e )⎡ x2 xe⎤–1 x= x ⎢ + 3tan ( e ) ⎥2x⎢⎣1+e⎥⎦d x 2 x 2xx 2( e arcsin x ) = e ⋅ + e arcsin xdx2 21–( x )⎛x 2x= e+ arcsin x⎜ 4⎝ 1– x2⎞⎟⎠2247.d –1 3 –1 2 1(tan x) = 3(tan x)⋅dx1+x2–1 23(tan x)=21+x48.−1 2d –1 d sin(cos x) d 1– xtan(cos x)= =dx dx −1cos(cos x)dx x1 12x⋅ ⋅ (–2 x) – 1– x ⋅121– x2=2x2 2– x –(1– x ) 1= = –x 1– x x 1– x2 2 2 2390 Section 6.8 Instructor’s Resource Manual

49.50.51.52.d –1 3 12sec ( x ) = ⋅3x=dx 3 3 2x ( x ) –1 xd –1 3 –1 2 1(sec x) = 3(sec x)⋅dx 2x x –1–1 23(sec x)=2x x –1d –1 3 –1 2(1 + sin x) = 3(1+ sin x)⋅dx−1⎛ 1 ⎞y = sin ⎜ 2 ⎟⎝ x + 4 ⎠–1 23(1 + sin x)=21– x11– x236x –154.y = xarcsec( x + 1)2dy ⎡ d 2 ⎤ ⎛ d ⎞ 2= x arcsec( x 1) x arcsec( x 1)dx ⎢+dx ⎥+ ⎜ ⎟⋅ +⎣ ⎦ ⎝dx⎠⎡⎤⎢ 2x⎥2= x⎢1 arcsec( 1)2 2 2 ⎥+ ⋅ x +⎢( x + 1 ) ( x + 1) −1⎣⎥⎦⎡2 ⎤⎢ 2x⎥2=⎢arcsec( x 1)2 4 2 ⎥+ +⎢( x + 1)x + 2x⎣⎥⎦⎡2 ⎤⎢ 2x⎥2=⎢arcsec( x 1)2 2 ⎥+ +⎢( x + 1)⋅ x x + 2⎣⎥⎦⎡⎤⎢ 2 x ⎥2=⎢arcsec( x 12 2 ⎥+ + )⎢( x + 1)x + 2⎣⎥⎦12Let u =x + 4; then −1y = sin ( u( x)) so by theChain Rule:dy dy du 1 du= = ⋅ =dx du dx 21−u dx1 ⎛ −2x⎞⋅ =2 ⎜2 21 ( x 4)⎟⎛ ⎞ ⎝ + ⎠1− ⎜ 2 ⎟⎝ x + 4 ⎠⎛ 2( x + 4) ⎞ ⎛ −2x⎞⎟⋅ =⎜ 4 2 ⎟ ⎜ 2 2x 8x15 ( x 4)⎟⎝ + + ⎠ ⎝ + ⎠−2x2 4 2( x + 4) x + 8x+ 15−53. y = tan1 ( ln x2)Let2 −1u x , v lnuy = tan v( u( x))soby the Chain Rule:dy dy dv du 1 1= = ⋅ ⋅ 2 x =dx dv du dx 21+v u1 1⋅ ⋅ 2 x =2 2 21 + (ln x ) x22 2x[1 + (ln x ) ]= = ; then ( )55. ∫ cos3x dxLet u = 3, x du = 3dx; then1∫cos3xdx= cos3 (3 )3∫x dx =1 1 1cosudu= sin u+ C = sin 3x+C3∫3 3256. Let u = x , so du = 2x dx .2 1 2∫xsin( x ) dx = sin( x ) ⋅2xdx2∫1 1= sin udu cosu C2∫=− +21 cos( 2= − x ) + C257. Let u = sin 2x, so du = 2 cos 2x dx.1∫sin2x cos2xdx=sin2 x(2cos2 x)dx2∫= 12∫ udu2u 1 sin2= + C = 2 x+C4 458. Let u = cos x, so du =− sin x dx .sin x 1∫tan x dx = ∫ dx = − ( −sin x)dxcos x∫cos x1= − ∫ du =− ln u + C =− ln cos x + Cu= ln sec x + CInstructor’s Resource Manual Section 6.8 391

2x22 x59. Let u = e , so du = e dx .2x2x∫ e cos( e ) dx 1 cos(2x)(22x= e e ) dx2∫1= cos udu2∫1 1 2x= sin u+ C = sin( e ) + C2 2⎡1⎤cos( ) = ⎢ sin( )2 ⎥⎣ ⎦1 2 x 2 x 2 xe e dx e0∫⎡1 2 1 0 ⎤= ⎢ sin( e ) − sin( e )2 2 ⎥⎣⎦2sin e − sin1= ≈ 0.0262260. Let u = sin x, so du = cos x dx.61.62.63.103 32 2 u sin xsin x cos xdx= u du = + C = + C3 33π /2π /2 ⎡2sin x ⎤ 1 1sin xcos xdx = ⎢ ⎥ = − 0 =0⎢⎣3 ⎥⎦3 3∫ ∫∫12/2 2/2dx = [arcsin x]0 20∫1– x2π= arcsin – arcsin 0 =2 4dxdx2 2 −12= = ⎡secx⎤2 2 2 2 ⎣ ⎦ 2∫ ∫x x −1 x x −1−1 −1= sec 2 − sec 2− 1⎛1⎞− 1⎛ 2 ⎞ π π π= cos ⎜ ⎟− cos= − =⎝2⎠ ⎜ 2 ⎟⎝ ⎠ 3 4 121 1111 1dx ⎡ −tan x⎤− −∫ = = tan 1−tan ( −1)−1 21+x ⎣ ⎦−1π ⎛ π⎞π= −⎜− ⎟=4 ⎝ 4⎠264. Let u = cos θ , so du =− sin θ dθ.sin θ∫ dθ =− 1 ( sin )2 ∫ − θ dθ21+ cos θ 1+cos θ1−1=− ∫ du =− tan u + C21+u−1=− tan (cos θ ) + Csinθπ /2 −1π /2dθ= ⎡−tan (cos θ)⎤0 2 ⎣⎦ 0∫1+cos θ−1 −1π π=− tan 0 + tan 1 =− 0 + =4 4065. Let u = 2x, so du = 2 dx.∫ 1 dx =1 11 4 2∫+ x 1 + (2 x)2 22dx1 1 1= arctan2∫du = u + C21+u 21= arctan 2 x + C2xx66. Let u = e , so du = e dx.67.68.∫xxee1dx = dx = du2xx 2 21+ e 1 + ( e ) 1+u∫ ∫ ∫= arctan u + C = arctan e x + C1 1dx =2∫ dx⎛ 312 12 ⎞12 − 9x⎜ − x ⎟⎝ 4 ⎠1 1= ∫ dx2 32⎛ 3 ⎞1−⎜⎜x2 ⎟⎝ ⎠3 3Let u = x,du = dx ; then2 21 1 1 ⎛ 2 ⎞ 1∫ dx =2 3 2⎜ ⎟∫du⎛23 ⎞ 2 3⎝ 3⎠ 1−u1−⎜⎜x2 ⎟⎝ ⎠1 1 3= sin u+ C = sinx + C3 3 ⎜2 ⎟⎝ ⎠∫−1 −1⎛ ⎞x12 − 9x2dx . Let u = 12 − 9 x , du =− 18 xdx;thenx 1 1∫ dx =− ( 18 )2 18∫ − dx212 −9x12 −9x1 1 ⎛ 1 ⎞=− du = ⎜− ⎟ (2 u ) + C18∫u ⎝ 18 ⎠212 − 9x=− + C92392 Section 6.8 Instructor’s Resource Manual

69.70.71.1 1∫dx =2 ∫dx2x − 6x+ 13 ( x − 6x+ 9) + 41= ∫ dx2( x − 3) + 4Let u = x − 3, du = dx, a = 2; then1 1 1 tan−1⎛u⎞∫ dx =2 ∫ du =2 2⎜ ⎟+C( x− 3) + 4 u + a a ⎝a⎠1 −1⎛ x − 3⎞= tan ⎜ ⎟+C2 ⎝ 2 ⎠∫1 1= =∫dx dx2 2172x + 8x+ 25 2( x + 4x+ 4 + )21 12∫dx2( 2)2 ⎛ 17 ⎞x + + ⎜ ⎟⎝ 2 ⎠17Let u = x+ 2, du = dx, a = ; then21 1 1 12∫dx =2 172∫du =2 2( x + 2) + u + a21 1 −1⎛u⎞1 2 ⎛−1x+2⎞⋅ tan ⎜ ⎟+ C = ⋅ tan+ C2 a ⎝a⎠ 2 17 ⎜ 17 ⎟⎝ 2 ⎠34 −1⎡ 34 ⋅ ( x + 2) ⎤= tan ⎢⎥+C34 ⎣ 17 ⎦∫1dx . Let u = 2 x , du = 2 dx , a = 3 ;2x 4x − 9then1 1∫ dx =2∫(2 dx)=2x 4x −9 2x 4x−91 1 sec−1⎛ u ⎞∫ du = ⎜ ⎟+ C =2 2u u − a a ⎝ a ⎠1 1 2 x sec− ⎛ ⎞⎜ ⎟ + C3 ⎝ 3 ⎠73. The top of the picture is 7.6 ft above eye level,and the bottom of the picture is 2.6 ft above eyelevel. Let θ 1 be the angle between the viewer’sline of sight to the top of the picture and thehorizontal. Then call θ2 = θ1− θ , so θ = θ1− θ2.7.6 2.6tan θ1 = ; tan θ2= ;b b−17.6 −12.6θ = tan − tanb bIf b = 12.9, θ ≈ 0.3335 or 19.1°.⎡ π⎤74. a. Restrict 2x to[0, π ] , i.e., restrict x to ⎢0, .2 ⎥⎣ ⎦Then y = 3 cos 2xy= cos 2x3y2x = arccos 3–1 1 yx = f ( y) = arccos2 3–1 1 xf ( x ) = arccos2 3⎡ π π⎤b. Restrict 3x to ⎢– , ,2 2⎥i.e., restrict x to⎣ ⎦⎡ π π⎤⎢– ,6 6⎥⎣ ⎦Then y = 2 sin 3xy= sin 3x2y3x = arcsin 2–1 1 yx = f ( y) = arcsin3 2–1 1 xf ( x ) = arcsin3 2⎛ π π⎞c. Restrict x to ⎜– , ⎟⎝ 2 2⎠72.x+1 x1∫ dx =2∫ dx + ∫ dx4− 9x 4−9x 24−9x2These integrals are evaluated the same as those inproblems 67 and 68 (with a constant of 4 ratherthan 12). Thus1y = tan x22y = tan xx = f –1 ( y ) = arctan 2 yf –1 ( x) = arctan 2x∫x+ 1 1 2 1 −1⎛3x⎞dx =− 4− 9x + sin ⎜ ⎟+C24−9x9 3 ⎝ 2 ⎠Instructor’s Resource Manual Section 6.8 393

75.⎛ 2⎞ ⎛2⎞d. Restrict x to ⎜– ∞, – ⎟∪⎜ , ∞⎟⎝ π⎠ ⎝π⎠ so 1 x is⎛ π ⎞ ⎛ π⎞restricted to ⎜– , 0⎟∪⎜0,⎟⎝ 2 ⎠ ⎝ 2⎠1then y = sinx1arcsin yx =−1 1x = f ( y)=arcsin yf−1 1( x)=arcsin x⎡ –1 ⎛1⎞⎤2tan tan ⎜ ⎟⎡ –1 1⎢4⎥⎛ ⎞⎤tan 2 tan⎣ ⎝ ⎠⎦⎢ ⎜ ⎟4⎥ =⎣ ⎝ ⎠⎦2 ⎡ –1⎛1⎞⎤1–tan ⎢tan⎜ ⎟4⎥⎣ ⎝ ⎠⎦77.⎡ –1 ⎛1⎞⎤2tan⎢2tan⎜ ⎟55⎥⎝ ⎠ 2⋅12 120=⎣ ⎦= =522 ⎡ –1⎛1⎞⎤1–1191– tan ⎢2 tan ⎜ ⎟ ( 12 )5⎥⎣ ⎝ ⎠⎦⎡ –1 ⎛1⎞ –1 ⎛ 1 ⎞⎤tan ⎢4 tan ⎜ ⎟– tan ⎜ ⎟5 239⎥⎣ ⎝ ⎠ ⎝ ⎠⎦⎡ –1 ⎛1⎞⎤ ⎡ –1 ⎛ 1 ⎞⎤tan ⎢4 tan ⎜ ⎟ – tan tan5⎥ ⎢ ⎜ ⎟239⎥⎝ ⎠ ⎝ ⎠=⎣ ⎦ ⎣ ⎦⎡ –1 ⎛1⎞⎤ ⎡ –1 ⎛ 1 ⎞⎤1+ tan ⎢4 tan ⎜ ⎟ tan tan ⎜ ⎟5⎥ ⎢239⎥⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦120 – 1119 239 28,561 π= = = 1 = tan1+ 120 ⋅ 1 28,561 4119 239–1 ⎛1⎞ –1 ⎛ 1 ⎞ −1πThus, 4 tan ⎜ ⎟– tan ⎜ ⎟ = tan (1) =⎝5⎠ ⎝239⎠476.2⋅14 8= =121–15( 4 )–1 1 –1 1 –1 1tan ⎡ ⎛ ⎞3tan ⎤ tan ⎡ ⎛ ⎞ ⎛ ⎞⎢ ⎜ ⎟ = 2 tan + tan⎤4⎥ ⎢ ⎜ ⎟ ⎜ ⎟4 4⎥⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠ ⎝ ⎠⎦–1 –1tan ⎡2 tan 1( ) ⎤+tan ⎡tan14 ( 4)⎤=⎣ ⎦ ⎣ ⎦1 1 –11– tan ⎡ −2 tan 1( 4) ⎤tan ⎡tan( 4)⎤⎣ ⎦ ⎣ ⎦8 + 115 4 47= =1– 8 ⋅ 1 5215 4⎡ –1 ⎛1⎞ –1 ⎛ 5 ⎞⎤tan ⎢3tan⎜ ⎟+tan ⎜ ⎟4 99⎥⎣ ⎝ ⎠ ⎝ ⎠⎦–1 –1tan ⎡3tan5( 1 ) ⎤+tan ⎡tan( ) ⎤=⎣ 4 ⎦ ⎣( ) ( 99 ⎦–1 1 –11– tan ⎡3tan ⎤ 54tan ⎡tan99 ) ⎤⎣ ⎦ ⎣ ⎦47 + 552 99 4913 π= = = 1 = tan1– 47 ⋅ 5 4913 452 99–1 ⎛1⎞ –1 ⎛ 5 ⎞ –1 πThus, 3tan ⎜ ⎟+ tan ⎜ ⎟= tan (1) =⎝4⎠ ⎝99⎠4⎡ –1 ⎛1⎞⎤2tan tan ⎜ ⎟⎡ –1 1⎢5⎥⎛ ⎞⎤tan 2 tan⎣ ⎝ ⎠⎦⎢ ⎜ ⎟5⎥ =⎣ ⎝ ⎠⎦2 ⎡ –1⎛1⎞⎤1–tan ⎢tan⎜ ⎟5⎥⎣ ⎝ ⎠⎦2⋅15 5= =21– 1 12( 5 )–1 1 –1 1tan ⎡ ⎛ ⎞4 tan⎤ tan ⎡ ⎛ ⎞⎢ ⎜ ⎟ = 2⋅2 tan ⎜ ⎟⎤5⎥ ⎢5⎥⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦.θLet θ represent ∠ DAB,then ∠ CAB is .2–12cos .2a1 ⎛ –1 b ⎞ 2cos2 2 cos–1 b⎜ ⎟b= b .2⎝2a⎠2aSincebb θ 2 b= , cos = = and2 2 2Δ ABC is isosceles, AEabθ = Thus sector ADB has areaaLet φ representφ φ∠ DCB, then ∠ ACB is and ∠ ECA is , so2 4bφsin 2 b = = and–1 bφ = 4sin . Thus sector4 a 2 a2a1 ⎛ –1 b ⎞ 2 2 –1 bDCB has area ⎜ 4sin ⎟a= 2 a sin .2⎝2a⎠2aThese sectors overlap on the triangles ΔDAC andΔCAB, each of which has area2 2 21 1 2 b 1 4 a – bAB h b a –⎛ ⎞= ⎜ ⎟ = b2 2 ⎝2⎠2 2The large circle has area π b2 , hence the shadedregion has areaπ 2 2 –1 b 2 –1 b 1 2 2b – b cos –2a sin b 4 a – b2a2a+ 2.394 Section 6.8 Instructor’s Resource Manual

78.y = sin(arcsin x) is the line y = x, but onlydefined for −1≤ x ≤ 1.y = arcsin(sin x) is defined for all x, but only theπ πportion for – ≤ x ≤ is the line y = x.2 279.They have the same graph.Conjecture: arcsin x = arctanx21– xfor–1 < x < 1Proof: Let θ = arcsin x, so x = sin θ.Thenxsinθsinθtanθ2 21– x 1–sin θ cosθso θ = arctanx21– x.81.dx∫ ∫dx=2 2a – x 2 ⎡ 2a 1– ( x ⎤⎢ a )⎣⎥⎦= 1 dx 1 dxa⋅ 2 21– x( ) = a⋅1– xa( a)∫ ∫ since a > 0x 1Let u = , so du = dx.a a1 dx du −1∫ ⋅ = sin u Ca 2∫ = +21−u( )1– x a−1x= sin + Ca80.It is the same graph as y = arccos x.πConjecture: – arcsin x = arccos x2πProof: Let θ = – arcsin x2⎛π⎞Then x = sin ⎜ − θ ⎟=cosθ⎝2⎠so θ = arccos x.82.−1x 1 1Dxsin = ⋅a21– x a( a )1 1 a 1a2– x2 a 2 2a – x aa2a 1 ,2 2a – x a1= ⋅ = ⋅= ⋅ since a > 0=2 2a – xx 183. Let u = , so du = dxa a∫dx 1 1 12 2 1 x2 dxa + x = a∫+a( )1 1 1 tan−1= du u Ca∫= +21+u a1 tan−1⎛ x ⎞= ⎜ ⎟+Ca ⎝a⎠aInstructor’s Resource Manual Section 6.8 395

x84. Let u ,adx= so du ( )= 1/ a dx.Since a > 0,1 1 1=dxx x 2 − a2 a x x( ) ( )2 a−1a a∫ ∫=1 1dua∫u u2−11 −1 1 −1xsec u C sec Ca a a= + = +d 1 1sin− ⎛ x⎜⎞ ⎟ =dx a 2 2a x85. Note that(See⎝ ⎠ −Problem 67).d ⎡ x 22 2 a 1sinx ⎤−⎢ a − x + + C⎥dx ⎢⎣2 2 a ⎥⎦1 2 2 x 1= a − x + ( −2 x)2 2 22 2a − x2a 1+ + 02 2 2a − x2 21 2 2 1 − x + a 2 2= a − x + = a −x2 2 2 2a − x86.a−a2 2 a 2 2− = 2 −02a⎤2 2 −1∫ ∫a x dx a x dx⎡ x a x= 2⎢a − x + sin ⎥⎢⎣2 2 a ⎥⎦02−1 0 22−1⎡a a a ⎤= 2 ⎢ (0) + sin (1) − a − sin (0) ⎥⎢⎣2 2 2 2 ⎥⎦22 −1πa= a sin (1) =2This result is expected because the integralshould be half the area of a circle with radius a.87. Let θ be the angle subtended by viewer’s eye.−1⎛12 ⎞ −1⎛2⎞θ = tan ⎜ ⎟−tan⎜ ⎟⎝ b ⎠ ⎝b⎠d θ 1 12 1 2= ⎛ − ⎞ − ⎛ −⎞db 2 2 2 2112⎜ b ⎟ ⎜ ⎟+ ⎝ ⎠ 1+2 ⎝ b ⎠( ) ( )b2 12 10(24 − b )= − =2 22 2b + 4 b + 144 ( b + 4)( b + 144)dθSince > 0 for b in ⎡0, 2 6 )db⎣dand θ < 0 for b in ( 2 6, ∞ ),the angle isdbmaximized for b = 2 6 ≈ 4.899 .The ideal distance is about 4.9 ft from the wall.b288. a.b.−1⎛ x ⎞ −1⎛ x ⎞θ = cos ⎜ ⎟−cos⎜ ⎟⎝b⎠ ⎝a⎠⎛ ⎞ ⎛ ⎞dθ⎜ −1 ⎟⎛1⎞⎛dx⎞ ⎜ −1 ⎟⎛1⎞⎛dx⎞⎛ 1 1 ⎞dx= ⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟⎜ ⎟ = −dt 2 21x⎝b ⎠⎝ dt ⎠ ( ) 1x⎝a ⎠⎝ dt ⎠ ⎜ 2 2 2 2 ⎟ dt⎜ − ⎟ ⎜ −b( a)⎟⎝ a −x b −x⎠⎝ ⎠ ⎝ ⎠⎛−1 a+ x ⎞−1⎛ x⎞θ = tan−sin⎜ 2 2 ⎟ ⎜ ⎟⎝ b − x ⎠ ⎝b⎠⎛ ⎞⎜ ⎟ ⎛ 2 2 ( a+x)xb − x + ⎞ ⎛ ⎞dθ⎜ 1 ⎟⎜ b2x2 ⎟− ⎛dx⎞ ⎜ 1 ⎟⎛1⎞⎛dx⎞= ⎜dt 2 ⎟⎜⎟2 2 ⎜ ⎟−⎜ ⎟⎜ ⎟⎜ ⎟⎜b x dt 2⎛1 a x1 x b dt+ ⎞ ⎟− ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠+ ⎜⎟ ⎜ − ( b ) ⎟⎜ ⎜ ⎟ ⎟b2−x2 ⎟⎝ ⎠ ⎝ ⎠⎝ ⎝ ⎠ ⎠⎡⎛ 2 2 2b − x ⎞⎛ b + ax ⎞ 1 ⎤dx= ⎢⎟⎜ − ⎥⎜ 2 2 2 2 2 3/2b x ( a x) ⎟⎜( b x ) ⎟ 2 2⎣⎢⎝ − + + ⎠⎝ − ⎠ b − x ⎦⎥dt⎡2b + ax 1⎤dx= ⎢− ⎥⎢ 2 2 2 2 2 2( b + a + 2 ax)b −x b −x⎥ dt⎣⎦⎡2a + ax⎤dx= ⎢−⎥⎢ 2 2 2 2( b + a + 2 ax)b −x⎥ dt⎣⎦396 Section 6.8 Instructor’s Resource Manual

89. Let h(t) represent the height of the elevator (thenumber of feet above the spectator’s line of sight) tseconds after the line of sight passes horizontal, andlet θ () t denote the angle of elevation.−1⎛15t⎞ −1⎛ t ⎞Then h(t) = 15t, so θ () t = tan ⎜ ⎟=tan ⎜ ⎟⎝ 60 ⎠ ⎝4⎠ .dθ 1 ⎛1⎞4= ⎜ ⎟=dt+ ⎝4 ⎠ 16 + tAt t = 6,( ) 2 21 t4dθ 4 1= =dt 216 + 6 13about 4.41° per second.radians per second or90. Let x(t) be the horizontal distance from the observerto the plane, in miles, at time t., in minutes.Let t = 0 when the distance to the plane is 3 miles.Then2 2x(0) = 3 − 2 = 5 . The speed of the plane is 10miles per minute, so x() t = 5− 10. t The angle of−1⎛ 2 ⎞ −1⎛ 2 ⎞elevation is θ () t = tan ⎜ ⎟=tan ⎜ ⎟,⎝ xt () ⎠ ⎝ 5−10t⎠sodθ 1 ⎛ −2⎞= ⎜( −10)dt⎟1+ 2/ 5−10⎠When t = 0,( ( )) 2 ⎜t ⎝( 5−10 t)220=.2( 5− 10 t) + 4dθ 20 = ≈ 2.22 radians per minute.dt 991. Let x represent the position on the shoreline and letθ represent the angle of the beam (x = 0 and θ = 0when the light is pointed at P). Then−1⎛ x ⎞ dθ1 1 dx 2 dxθ = tan ⎜ ⎟, so = =⎝2⎠ dt 2 2 dt 2+4 + x dt( )1 x2When x = 1,dx dθ2= 5 π , so = (5 π ) = 2πThe beacondt dt 24+1revolves at a rate of 2π radians per minute or 1revolution per minute.92 Let x represent the length of the rope and let θrepresent the angle of depression of the rope.−1 ⎛8⎞Then θ = sin ⎜ ⎟, so⎝ x ⎠dθ 1 8 dx 8 dx= − = − .dt 2 28 x dt 21x x − 64 dt−( )xdxWhen x = 17 and =− 5 , we obtaindtd θ 8 8=− ( − 5) = .dt217 17 − 64 51The angle of depression is increasing at a rate of8/ 51 ≈ 0.16 radians per second.93. Let x represent the distance to the center of the earthand let θ represent the angle subtended by the−1 ⎛6376⎞earth. Then θ = 2sin ⎜ ⎟⎝ x ⎠ , sodθ 1 ⎛ 6376 ⎞dx= 22 ⎜ −dt2 ⎟1 6376 ⎝ x ⎠ dt− ( x )12,752 dx=−2 2x x − 6376 dtWhen she is 3000 km from the surfacedxx = 3000 + 6376 = 9376 and 2dt = − . Substitutingdθ−4these values, we obtain ≈ 3.96× 10 radiansdtper second.Instructor’s Resource Manual Section 6.8 397

6.9 Concepts Review1.2.– –e x – e x e x + ex;2 22 2cosh x − sinh x = 13. the graph of2 2x − y = 1, a hyperbola4. catenary; a hanging cable or chainProblem Set 6.91.2.3.4.x – x x – xe + e e – ecosh x+ sinh x = +2 2x2ex= = e22 x –2x 2 x –2xe + e e – ecosh 2x+ sinh 2x= +2 22x2e2x= = e2x – x x – xe + e e – ecosh x– sinh x = –2 2– x2e– x= = e22 x –2x 2 x –2xe + e e – ecosh 2 x– sinh 2 x = –2 2–2x2e–2x= = e25.6.x – x y – y x – x y – ye – e e + e e + e e – esinh xcosh y+ cosh xsinhy = ⋅ + ⋅2 2 2 2x+ y x– y – x+ y – x– y x+ y x– y – x+y – x–ye + e – e – e e – e + e – e= +4 4x+ y –( x+ y) x+ y –( x+y)2 e –2 e e – e= = = sinh( x + y)4 2x – x y – y x – x y – ye – e e + e e + e e – esinh xcosh y– cosh xsinh y = ⋅ – ⋅2 2 2 2x+ y x– y – x+ y – x– y x+ y x– y – x+y – x–ye + e – e – e e – e + e – e=–4 4x– y – x+y x– y –( x– y)2 e –2 e e – e= = = sinh( x – y)4 2398 Section 6.9 Instructor’s Resource Manual

7.8.x – x y – y x – x y – ye + e e + e e – e e – ecosh xcosh y+ sinh xsinhy = ⋅ + ⋅2 2 2 2x+ y x– y – x+ y – x– y x+ y x– y – x+y – x–ye + e + e + e e – e – e + e= +4 4x+ y – x– y x+ y –( x+y)2e + 2e e + e= = = cosh( x + y)4 2x – x y – y x – x y – ye + e e + e e – e e – ecosh xcosh y– sinh xsinh y = ⋅ – ⋅2 2 2 2x+ y x– y – x+ y – x– y x+ y x– y – x+y – x–ye + e + e + e e – e – e + e=–4 4x– y – x+y x– y –( x– y)2e + 2e e + e= = = cosh( x – y)4 29.sinh x sinh ytanh x+tanh y +cosh x cosh y=1+ tanh xtanh y sinh sinh y1+ x ⋅cosh x cosh ysinh x cosh y+ cosh xsinh y sinh( x+y)= =cosh x cosh y+ sinh xsinh y cosh( x+y)= tanh (x + y)16.3 2D cosh x = 3cosh xsinhxx17. D cosh(3x+ 1) = sinh(3x+ 1) ⋅ 3 = 3sinh(3x+1)18.x2 2D sinh( x + x) = cosh( x + x) ⋅ (2x+1)x= (2x + 1)cosh( x + x)210.sinh x sinh y–cosh x cosh ysinh x sinh ycosh x cosh ytanh x– tanh y1– tanh xtanh y = 1– ⋅sinh x cosh y– cosh xsinh y sinh( x– y)= =cosh x cosh y– sinh xsinh y cosh( x– y)= tanh(x – y)11. 2 sinh x cosh x = sinh x cosh x + cosh x sinh x= sinh (x + x) = sinh 2x12.13.14.2 2cosh x + sinh x = cosh xcosh x+sinh xsinhx= cosh( x + x) = cosh 2x2D sinh x = 2sinh xcosh x = sinh 2xx2D cosh x = 2cosh xsinh x = sinh 2xx19.20.21.22.1 coshxDxln(sinh x)= ⋅ cosh x =sinh x sinh x= coth x1 2Dxln(coth x) = (–csch x)coth xsinh x 1 1= – ⋅ = –cosh x 2sinh x sinh xcoshx= − csch x sech x2 2Dx ( x cosh x) = x ⋅ sinh x+ cosh x⋅2x= x 2 sinh x+2xcoshx–2 –2 –3Dx ( x sinh x) = x ⋅ cosh x+ sinh x⋅(–2 x )−2 −3= x coshx−2x sinhx15.2Dx (5sinh x) = 10sinh x⋅ cosh x = 5sinh 2x23. D (cosh 3xsinh x) = cosh 3x⋅ cosh x+ sinh x⋅sinh 3x⋅3= cosh 3x cosh x+3sinh 3xsinhxx24. D (sinh xcosh 4 x) = sinh x⋅sinh 4x⋅ 4 + cosh 4x⋅ cosh x = 4 sinh x sinh 4x + cosh x cosh 4xx25.Dx (tanh xsinh 2 x) = tanh x⋅cosh 2x⋅ 2 + sinh 2x⋅sech x2= 2 tanh x cosh 2x+sinh 2xsechx226.Dx (coth 4xsinh x) = coth 4x⋅ cosh x+ sinh x(–csch 4 x) ⋅ 42= cosh x coth 4 x– 4sinh x csch 4x2Instructor’s Resource Manual Section 6.9 399

27.28.29.–1 2 1 2xDxsinh ( x ) = ⋅ 2x=2 2 4( x ) + 1 x + 12–1 3 1 2 3x= ⋅ =3 2 6Dxcosh ( x ) 3x( x ) –1 x –1Dx–112 2tanh (2 x– 3) = ⋅2= =22 21–(2 x –3) 1–(4 x –12x+ 9) –4x + 12 x–8=1–2( x2 – 3 x + 2)30.31.104–1 5 −1⎛ 1 ⎞ 1 ⎛ 5 ⎞ x ⎛ 5 ⎞ 5xDxcoth ( x ) = Dxtanh ⎜ 5 ⎟ = ⋅2 ⎜− 6 ⎟= ⋅10 ⎜−6 ⎟ = –10⎝ x ⎠ 1 ⎝ x ⎠ x –1 ⎝ x ⎠1–⎛ ⎞x –1⎜x 5 ⎟⎝ ⎠–1 1–13x–1Dx[ xcosh (3 x)] = x⋅ ⋅ 3+ cosh (3 x) ⋅1= + cosh 3x22(3 x) –19 x –132.2 –1 5 2 1 4 –1 5Dx( x sinh x ) = x ⋅ ⋅ 5x + sinh x ⋅2x5 2( x ) + 165x= + 2xsinh10x + 1–1 5x33.Dx=–1 1 1ln(cosh x)= ⋅–1cosh x 2x –112 –1x –1cosh x38. Let u = 3x + 2, so du = 3 dx.1 1∫sinh(3x + 2) dx = sinh u du = cosh u + C3∫31= cosh(3 x + 2) + C334.35.36.–1cosh (cos x ) does not have a derivative, since−1Ducosh ucos 1is only defined for u > 1 whilex ≤ for all x.2 2Dxtanh(cot x) = sech (cot x) ⋅ (– csc x)2 2= – csc x sech (cot x)–1 –1 ⎛ 1 ⎞Dxcoth (tanh x) = Dxtanh ⎜ ⎟⎝tanhx ⎠–1= Dxtanh (coth x)1 22 –csch x= (–csch x) = = 12 21–(coth x)–csch x37. Area = ln 3⎡1⎤ln 3∫ cosh 2xdx= sinh 2x0⎢ 2 ⎥⎣ ⎦ 02ln3 –2ln3 0 –01 ⎛ e – e ––e e ⎞= 2⎜2 2 ⎟⎝⎠1 ln 9ln91 1⎛1⎞= ( e − e ) = ⎜9−⎟4 4⎝9⎠=209239. Let u =π x + 5, so du = 2π xdx .2 1∫x cosh( π x + 5) dx = cosh udu2 π∫1 1 2= sinh u+ C = sinh( π x + 5) + C2π2π140. Let u = z, so du = dz .2 zcosh z dz = 2 cosh u du = 2sinh u + Cz∫ ∫= 2sinh z + C1/4 1 –3/4 141. Let u = 2 z , so du = ⋅ 2 z dz = dz.4 4 32 z1/4sinh(2 z )∫ dz = 2 sinhu du = 2coshu + C4 3∫z1/4= 2cosh(2 z ) + C400 Section 6.9 Instructor's Resource Manual

x42. Let u = e , so du = e dx .∫ ∫x xe sinh e dx = sinh u du = cosh u + C= cosh ex + C43. Let u = sin x, so du = cos x dxcos xsinh(sin xdx ) = sinh udu= cosh u+C∫ ∫= cosh(sin x) + C44. Let u = ln(cosh x),so1du = ⋅ sinh x = tanh x dx .cosh x2u∫tanh x ln(cosh x)dx = ∫ u du = + C21 [ln(cosh )]2= x + C2245. Let u = ln(sinh x ) , so12 2du = ⋅cosh x ⋅ 2xdx = 2x coth x dx .2sinh xx22 2 1 1 u∫x coth x ln(sinh x ) dx = udu C2∫= ⋅ +2 21 [ln(sinh 2 )] 2= x + C4ln 5 ln 5∫ cosh 2x dx = 2 cosh 2x dx–ln5 ∫ 0ln 5⎡1⎤= 2 ⎢ sinh2x2 ⎥⎣ ⎦01 2ln5 2ln5= sinh(2ln 5) = ( )2 e − e−ln 11 ln 25 251⎛1 ⎞46. Area == ( e − e ) = ⎜25−⎟2 2⎝25⎠= 312 12.4825 =47. Note that the graphs of y = sinh x and y = 0intersect at the origin.ln 2 ln 2Area = ∫ sinh xdx=[cosh x]00ln 2 − ln 2 0 0e + e e + e 1⎛1⎞1= − = ⎜2+ ⎟− 1=2 2 2⎝2⎠448. tanh x = 0 when sinh x = 0, which is when x = 0.0 8Area = ∫ ( − tanh x) dx + tanhx dx−8 ∫ 08 8 sinh x= 2∫ tanhx dx = 2 dx0 ∫ 0 cosh xLet u = cosh x, so du = sinh xdx.sinh 12∫ x dx = 2 du = 2ln u + Ccosh x∫u8 sinh x 82∫dx =⎡2ln cosh x ⎤0 cosh x⎣ ⎦0= 2(ln cosh 8 − ln1) = 2ln(cosh 8) ≈ 14.6149. Volume = 1 2 π∫ π cosh x dx = 1 (1 + cosh 2 x)dx0 2∫ 01π ⎡ sinh 2x⎤= x2⎢ +2⎥⎣ ⎦0π ⎛ sinh 2 ⎞= ⎜1+ −0⎟2⎝2 ⎠π πsinh 2= + ≈ 4.422 4∫50. Volume = ln10 2πsinh xdx02ln10⎛x −xe − e ⎞=π∫dx0 ⎜ 2 ⎟⎝ ⎠ln102 x –2xe –2+ e π ln10 2 x –2x=π ∫dx = ( e –2 + e ) dx0 4 4∫0ln10π ⎡1 2 x 1 –2x⎤= e –2 x–e4⎢2 2⎥⎣⎦0π 2 x –2xln10= [ e –4 x– e ] 08π ⎛1 ⎞= ⎜100 – 4ln10 – ⎟≈35.658 ⎝100 ⎠2 251. Note that 1+ sinh x = cosh x and2 1+cosh2xcosh x =21Surface area = π y + ( ) 2∫2 1 dy0 dx= 1 2∫ 2 π cosh x 1 + sinh xdx0= 1∫ 2 π cosh x cosh xdx0dx1= ∫ π (1+cosh 2 x)dx01⎡ π ⎤ π= ⎢π x+ sinh 2xsinh 2 8.842⎥ =π+ ≈⎣⎦02Instructor’s Resource Manual Section 6.9 401

52. Surface area =212 1 ⎛dy⎞π y + dx0 ⎜ ⎟⎝dx⎠∫1 22 sinh 1 cosh0 x∫= π +Let u = cosh x, so du = sinh x dx2 2 ⎡u2 12 ⎤∫2π sinhx 1+ cosh xdx = 2π ∫ 1+u du = 2π ⎢ 1+ u + ln u+ 1+ u + C2 2⎥⎣⎦2 22=π cosh x 1+ cosh x +π ln cosh x+ 1+ cosh x + C (The integration of ∫ 1 + u du is shown in Formula 44 ofthe Tables in the back of the text, which is covered in Chapter 8.)11 2∫ 2 sinh 1 cosh0 π x + xdx ⎡2 2 ⎤=π⎢cosh x 1+ cosh x + ln cosh x+ 1+cosh x⎣⎥⎦0xdx( )π ⎡⎤⎣⎥⎦ ≈ 5.532 2=⎢cosh1 1+ cosh 1 + ln cosh1+ 1+ cos 1 − 2 + ln 1+2⎛53. y acosh x ⎞= ⎜ ⎟+C⎝a⎠dy ⎛ x ⎞= sinh ⎜ ⎟dx ⎝a⎠2d y 1 ⎛ x⎞= cosh2 ⎜ ⎟dx a ⎝a⎠22d y 1 ⎛dy⎞= 1 + .2 ⎜ ⎟dx a ⎝dx⎠2⎛ x ⎞ 2⎛ x ⎞ ⎛ x ⎞1+ sinh ⎜ ⎟=cosh ⎜ ⎟ and cosh ⎜ ⎟ > 0.⎝a⎠ ⎝a⎠ ⎝a⎠We need to show thatNote that21 ⎛dy ⎞ 1 2⎛ x ⎞ 1 2⎛ x ⎞1+ ⎜ ⎟ = 1+ sinh ⎜ ⎟ = cosh ⎜ ⎟a ⎝dx⎠ a ⎝a⎠ a ⎝a⎠Therefore,21 ⎛ x ⎞ d y= cosh ⎜ ⎟=a ⎝a⎠2dx⎛54. a. The graph of y b acosh x ⎞= − ⎜ ⎟ is symmetric about the y-axis, so if its width along the⎝a⎠⎛a⎞x-axis is 2a, its x-intercepts are (±a, 0). Therefore, ya ( ) = b− acosh⎜⎟=0, so b= acosh1 ≈ 1.54308 a.⎝a⎠55. a.b. The height is y(0) ≈1.54308a− acosh 0 = 0.54308a.c. If 2a = 48, the height is about 0.54308a = (0.54308)(24) ≈ 13 .b. Area under the curve is2424 ⎡ ⎛ x ⎞⎤ ⎡ ⎛ x ⎞⎤37 24cosh dx 37x576sinh 422−24⎢ − ⎜ ⎟ = − ≈24⎥ ⎢ ⎜ ⎟24⎥⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦−24∫Volume is about (422)(100) = 42,200 ft 3 .402 Section 6.9 Instructor’s Resource Manual

c. Length of the curve is224 ⎛dy ⎞ 24 2 ⎛ x ⎞ 24 ⎛ x ⎞1+ dx 1 sinh dx cosh dx−24 ⎜ ⎟ = +dx −24 ⎜ ⎟ =24 −24⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝24⎠2≈ (56.4)(100) = 5640 ft∫ ∫ ∫Surface area24⎡ ⎛ x ⎞⎤= ⎢24sinh ⎜ ⎟ = 48sinh1 ≈56.424⎥⎣ ⎝ ⎠⎦−24cosh t1 cosh tcosh sinh2 1 ⎡1 t t−x − 12 1 2 ⎤dx2∫= cosh tsinh t− x x 1 ln x x 112⎢ − − + −2 2⎥⎣⎦11 ⎡1 2 12 ⎤= cosh tsinh t− cosh t cosh t 1 ln cosh t cosh t 1 02 ⎢− − + − −2 2⎥⎣⎦1 1 1 1 t t= cosh tsinh t− cosh tsinh t+ ln cosh t+ sinh t = ln e =2 2 2 2 256. Area =57. a.b.x – x x – xr r⎛r e – e e + e ⎞ ⎛ x2e⎞rx(sinh x+ cosh x)= += = e⎜ 2 2 ⎟ ⎜⎝⎠2 ⎟⎝ ⎠rx – rx rx – rx rxe – e e + e 2erxsinh rx + cosh rx = + = = e2 2 2– –r⎛ x x x xr e e e – e ⎞ –r+⎛ x2e⎞– rx(cosh x– sinh x) = –= = e⎜ 2 2 ⎟ ⎜⎝⎠2 ⎟⎝ ⎠rx – rx rx – rx – rxe + e e – e 2e– rxcosh rx – sinh rx = – = = e2 2 2ix −ix ix −ixrrr ⎛e + e e −e⎞ ⎛ ix2e⎞irxx+ i x = + i= = e⎜ 2 2i⎟ ⎜⎝⎠2 ⎟⎝ ⎠irx −irx irx −irxirxe + e e −e2eirxcos rx + isinrx = + i = = e2 2i2c. ( cos sin )ix −ix ix −ixrrr ⎛e + e e −e⎞ ⎛ −ix2e⎞−irxx− i x = −i= = e⎜ 2 2i⎟ ⎜⎝⎠2 ⎟⎝ ⎠irx −irx irx −irx−irxe + e e −e2e−irxcos rx − i sin rx = −i= = e2 2i2d. ( cos sin )58. a.–1gd(– t) = tan [sinh(– t)]–1–1= tan (– sinh t)= – tan (sinh t) = − gd( t)so gd is odd.1 coshtDt[ gd( t)] = ⋅ cosht=2 21+sinh t cosh t= sech t > 0 for all t, so gd is increasing.2Dt[ gd ( t )] = Dt(sech t ) = − sech t tanh t2Dt[ gd ( t )] = 0 when tanh t = 0, sincesech t > 0 for all t. tanh t = 0 at t = 0 and2tanh t < 0 for t < 0, thus Dt[ gd ( t )] > 0 for2t < 0 and Dt[ gd ( t )] < 0 for t > 0. Hencegd(t) has an inflection point at−1(0, gd(0)) = (0, tan 0) = (0, 0).b. If–1y = tan (sinh t)then tan y = sinh t sotan y sinh tsin y = =2 2tan y+ 1 sinh t+1sinh t–1= = tanh t so y = sin (tanh t)cosh t1Also, Dy t = ⋅cosht21+sinh t= cosh t 1sech t,2cosh t= cosh t=tso y = ∫ sech udu by the Fundamental0Theorem of Calculus.Instructor’s Resource Manual Section 6.9 403

∫x59. Area = cosh tdt= [sinh t] 0 = sinh x0Arc length =x2 x 21 + [ D cosh ] 1 sinh0t t dt = + tdt0∫ ∫xx= ∫ cosh tdt= [ sinh t] 0 0= sinh x60. From Problem 54, the equation of an invertedxcatenary is y = b− acosh . Given theainformation about the Gateway Arch, the curvepasses through the points (±315, 0) and (0, 630).315Thus, b= acosh and 630 = b – a, soab = a + 630.315a+ 630 = acosh ⇒ a ≈ 128, so b ≈ 758 .axThe equation is y = 758 − 128cosh .128x61.62.The functions y = sinh x andare inverse functions.–1y gd( x) tan (sinh x)= =tan y = sinh x–1 –1x = gd ( y) = sinh (tan y)–1 –1y = gd ( x) = sinh (tan x)Thus,2y = ln( x+ x + 1)6.10 Chapter ReviewConcepts Test1. False: ln 0 is undefined.2. True:3. True:2d y 1=− < 0 for all x > 0.2 2dx x33ee 13dt =⎡ln t ln e ln1 31 t⎣ ⎤ ⎦ = − =1∫4. False: The graph is intersected at most onceby every horizontal line.5. True: The range of y = ln x is the set of allreal numbers.⎛6. False: ln x ln y ln x ⎞− = ⎜ ⎟⎝ y ⎠7. False:8. True:44lnx= ln( x )x+1x+1 x 2eln(2 e ) – ln(2 e ) = ln2x= ln e = 1e9. True:ln( x−4)f( g( x)) = 4+e= 4 + ( x− 4) = xandxxg( f( x)) = ln(4+ e − 4) = lne = x10. False: exp( x + y) = exp xexpy11. True: ln x is an increasing function.12. False: Only true for x > 1, or ln x > 0.z13. True: e > 0 for all z.14. True:15. True:16. True:xe is an increasing function.lim (ln sin x−ln x)x→0+⎛sinx ⎞= lim ln ⎜ ⎟= ln1 = 0x→0+ ⎝ x ⎠2 2lnππ = ed17. False: lnπ is a constant so ln π= 0.dxd18. True: (ln 3 x + C)dxd1= (ln x + ln 3 + C)=dxx404 Section 6.10 Instructor’s Resource Manual

19. True: e is a number.20. True: exp[ gx ( )] ≠ 0 because 0 is not in thexrange of the function y = e .21. False: D ( x ) = x (1+ln x)xx22. True: 2( tan x + sec x) ' − ( tan x+sec x) 2x2( x x x)= 2 sec + sec tan2 2−tan x −2 tan xsec x−secx2 2= sec x −tanx = 123. True: The integrating factor is( ) 44/ xdx 4lnx lnx4e∫= e = e = x24. True: The solution is ( )−4 2x−4 2xy x = e ⋅ e . Thus,slope = 2e⋅eand at x = 2 theslope is 2.25. False: The solution is y( x)( )22x= e , soy' x = 2ex . In general, Euler’smethod will underestimate thesolution if the slope of the solution isincreasing as it is in this case.26. False: sin ( arcsin(2) ) is undefined27. False: arcsin(sin 2 π ) = arcsin 0 = 028. True: sinh x is increasing.29. False: cosh x is not increasing.30. True:31. True:0cosh(0) = 1 = ex −x xIf x > 0, e > 1 while e < 1 < e so1 x −x 1 xcosh x = ( e + e ) < (2 e )2 2xx= e = e . If x < 0, –x > 0 andxe − x x> 1 while e < 1 < e − so1 x x 1 xcosh x ( e e −−= + ) < (2 e )2 2−x= e = ex.1 xsinh x ≤ e is equivalent to2x−xxe −e ≤ e . When x = 0,1 0 1sinh x = 0 < e = . If x > 0,2 232. False:33. False:34. False:35. True:x −x xe > 1 and e < 1 < e , thusx − x x −x xe − e = e − e < e = e−x x −xIf x < 0, e > 1 and e < 1 < e ,thusx x x xe e −−− = −( e − e )−x x −xx= e − e < e = e .−1⎛1⎞tan ⎜ ⎟ ≈ 0.4636⎝2⎠−1 sin 1( 2 ) 1but =−1 cos 1( 2 ) 2ln 3 ln 3e + e −cosh(ln 3) =21⎛1⎞5= ⎜3+ ⎟=2⎝3⎠3⎛sinx ⎞lim ln ⎜ ⎟ = ln1 = 0x→0⎝ x ⎠x−1πlim tan x =− , since→−∞ 2lim tan x =−∞.+x→− π236. False: cosh x > 1 for x ≠ 0 , while sinonly defined for −1≤u≤ 1.37. True:x.−1sinh xtanh x = ; sinh x is an oddcosh xfunction and cosh x is an evenfunction.38. False: Both functions satisfy y′′ − y = 0 .39. True:100ln 3 100ln 3 100 1= > ⋅ sinceln 3 > 1.40. False: ln(x – 3) is not defined for x < 3.41. True: y triples every time t increases by t 1 .42. False: x(0) = C; 1 C = Ce −kt when21e −kt 1= , so ln = − kt or22ln 12 −ln 2 ln 2t = = =−k −kkuisInstructor’s Resource Manual Section 6.10 405

43. True: ( yt () + zt ())′ = y′ () t + z′() t= ky() t + kz() t = k( y() t + z())t44. False: Only true if C = 0;( y () t + y ()) t ′ = y′ () t + y′() t1 2 1 2= ky1() t + C + ky2()t + C= k( y1( t) + y2( t)) + 2C.45. False: Use the substitution u = –h.–1/ h1 ulim (1 – h) = lim (1 + u)= e46. False:h→0 u→0by Theorem 6.5.A.120.05 ⎛ 0.06 ⎞e ≈ 1.051 < ⎜1+ ⎟ ≈1.062⎝ 12 ⎠47. True: If Dx( a x ) = a x ln a = ax , thenln a = 1, so a = e.Sample Test Problems8.9.10.11.d −11 d cosxtanh (sin x) = sin x =dx2 21−sin x dx 1−sinxcos x= = sec x2cos xd −11 dsinh (tan x) =tan xdx2tan x + 1 dx22sec x sec x== = sec x22tan x + 1 sec xd −12 d2sin 3x=3xdx2 dx1−3( x )2 3 3= =1−3x 2 3x 23x − 9xd −1xsec e =dx1 d xex x 2e ( e ) −1dx1.2.3.4xln = 4ln x − ln 224d x d4ln = (4ln x − ln 2) =dx 2 dx xd 2 3 3 d 3sin ( x ) = 2sin( x ) sin( x )dxdx3 3 d 3 2 3 3= 2sin( x )cos( x ) x = 6x sin( x )cos( x )dxd x2−4x x2−4xd 2e = e ( x − 4 x)dxdx2− 4= (2x−4) ex x12.13.xe 1= =x 2x 2xe e −1 e −1d 2⎛ x⎞ 1 d 2⎛ x⎞ln sin ⎜ ⎟=sindx 2 2sin x ⎜ ⎟⎝ ⎠ ( 2 ) dx ⎝2⎠1 ⎛ x ⎞ d ⎛ x⎞= 2sin sin2sin x ⎜ ⎟ ⎜ ⎟( 2 ) ⎝2⎠dx⎝2⎠1 ⎡ ⎛ x ⎞⎤1⎛ x⎞ ⎛ x⎞= 2sin cos cot2sin x ⎢ ⎜ ⎟ =( 2 ) 2⎥ ⎜ ⎟ ⎜ ⎟⎣ ⎝ ⎠⎦2 ⎝2⎠ ⎝2⎠5xd 5x3 5x15e3ln( e + 1) = (5 e ) =dx 5x5xe + 1 e + 14.d 5 1 d 5log 10( x 1) ( x 1)dx− = 5( x −1)ln10dx−45x=5( x −1)ln1014.d 3ln(2x− 4x+5)dx1 d 3= (2x− 4x+5)32x− 4x+5 dx26x− 4=32x− 4x+55.6.7.d x d2tan(ln e ) = tan x = sec xdxdxd ln cot x d2e = cot x =− csc xdx dxd 22 sech2 tanh x = 2sech x d x =xdx dx xd x x d x15. cos e =− sin e edxdxx x d= ( − sin e ) e xdxx xe sin e=−2 x406 Section 6.10 Instructor’s Resource Manual

16.17.18.19.20.d1 dln(tanh x)= tanh xdxtanh x dx1 2= sech x = csch xsechxtanh xd −1−22cos x =dx21 − ( x)−2 1 1= = −1−x 2 x 2x − xddxd⎡ 3x4 d x 44 + (3 x) ⎤ = (64 + 81 x )dx ⎣ ⎦ dxx3= 64 ln 64 + 324xd ln x d2csce= 2cscdxdxd=− 2csc x cot x xdxcsc=−x cotxxd2/3(log102 x)dx2 (log1/310 2 )− d= x (log 10 2 + log 10 x )3dx2 −1/3= (log 110 2 x)3 x ln102=3xln103log 2x10d21. 4 tan 5xsec5xdx222= 20sec 5xsec5x+20 tan 5xsec5xtan 5x2 2= 20sec5 x(sec 5x+tan 5 x)2= 20sec5 x(2sec 5x−1)2 2d −11tan⎛ x d x⎞ =⎛ ⎞dx ⎜ 2 ⎟ 22 dx ⎜ 2 ⎟⎝ ⎠ ⎛ x ⎞1⎝ ⎠⎜ 2 ⎟ +⎝ ⎠x 4x= =⎛ 4 4x ⎞1 x + 4⎜ 4 ⎟+⎝ ⎠2 2d ⎡1 x ⎤−−1x 4x⎢xtan ⎥ = (1) tan + ( x)dx ⎜ ⎟ ⎜ ⎟ 4⎣ ⎝ ⎠⎦⎝ ⎠ x +⎛ 2 2−1x ⎞ 4x= tan ⎜ ⎟+4⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ 2 ⎥ 2 ⎝ 4 ⎠xx⎜ 2 ⎟⎝ ⎠ x + 423.24.dxdxd= edx1 + x (1 + x)lnx(1 + x) ln x d= e [(1 + x)ln x]dx1+ x ⎡ ⎛1⎞⎤= x ⎢(1)(ln x) + (1 + x)⎜ ⎟x⎥⎣⎝ ⎠⎦= x+ ⎛ ⎜ln x+ 1+⎞⎟⎝ x ⎠1 x 1d 2 e 2 e−1 d 2(1 + x ) = e(1 + x ) (1 + x )dxdx2 12 xe(1 x ) e −= +25. Let u = 3x – 1, so du = 3 dx.3 x −1 1 3 x −11∫ e dx = e 3dx = e u du3∫ 3∫1 u 1 3x−1= e + C = e + C3 3Check:d ⎛1 x − ⎞ 1 x − de C e (3x 1) ex −⎜ + ⎟= − =dx ⎝3 ⎠ 3 dx26. Let u = sin 3x, so du = 3 cos 3x dx.1 1∫ 6cot3x dx = 2∫ 3cos3x dx = 2 dusin 3x∫u= 2ln u + C = 2ln sin3x + CCheck:d2 d(2ln sin3 x + C) = sin3xdxsin 3x dx2(3cos3 x)= = 6cot3xsin 3x3 1 3 1 3 1x27. Let u = e , so du = e dx .x x∫ ∫e sin e dx = sin udu =− cosu+C= − cos ex + CCheck:dd( − cos e + C) = (sin e ) edxdx= e sine2xx x x x x28. Let u = x + x− 5, so du = (2x + 1) dx .6x+ 3 1∫ dx = 3 (2x 1) dx2 ∫+2x + x− 5 x + x−512= 3∫du = 3ln u + C = 3ln x + x − 5 + CuCheck:d 2 3 d 2( 3ln x + x− 5 + C)= ( x + x−5)dx2x + x−5dx6x+ 3=2x + x−5Instructor’s Resource Manual Section 6.10 407

x+333. Let u = ln x, so du = dx .xCheck:⎛ π π ⎞Concave up on ⎜– ,–⎟d⎡ −11 dtan (sin x) C⎤⎝ 2 4⎠+ =sin xdx ⎣ ⎦ 21+sin x dx⎛ π π ⎞Concave down on ⎜– , ⎟cos x⎝ 4 2⎠=+2−∫ 1 dx =−x 31 ln( e2 ∫1 ⋅1 dx2x+ x(ln x) 1 + (ln x)x+ 1)= ln u + C = + C1ee−1 −1∫ 2Check:1+ux+3d ⎛ln( e + 1) ⎞Check:1 1 d+x+3C= ( e + 1)ddx ⎜ e ⎟ e x+3−11 d⎝⎠ e + 1 dx− + =−dx21 + (ln x)dxx+ 3 − 1 x+2e e e= =−1x+ 3 x+3=e + 1 e + 12x + x(ln x)u = x2 , so du = 2x dx.34. Let u = x – 3, so du = dx.2 2∫4x cos x dx = 2 ∫(cos x )2xdx = 2∫cosudu2∫sech ( 3)2∫ sech tanh2= 2sinu+ C = 2sin x + C= tanh( x − 3) + CCheck:Check:d 2 2 d 2 2d2 d(2sin x + C) = 2cos x x = 4xcosxx− = x− dxdxdxdxx−2= sech ( x − 3)4 1∫ dx = 2 2dx2∫35. f′ ( x) cos x–sin x; f′( x) 021−4x1 −(2 x)πx =14= 2∫du21−uf ( x) 0−1 −1π π= 2sin u+ C = 2sin 2x+C– ≤ x < .Check:2 4f′′ ( x) = –sin x–cos x; f′′( x) = 0 whend⎛−11⎞d(2sin 2 x + C) = 2⎜⎟ 2xπdx⎜ 21 − (2 x)⎟ dxtan x = –1, x = –⎝ ⎠44f ( x) 0=2π π1−4xwhen – ≤ x < – .2 4⎡ π π⎤Increasing on – ,cos x 1⎢−12 4⎥∫ dx = du = tan u + C⎣ ⎦2 ∫ 21+ sin x 1+u⎡ π π⎤Decreasing on ,−1⎢= tan (sin x)+ C4 2⎥⎣ ⎦x 329. Let u = e + + 1, so du = e dx .x+2e 1 1 x+3 1 1∫ dx = e dx dux+ 3 x 3e 1 e∫ =++ e + 1 e∫u30. Let31. Let u = 2x, so du = 2 dx.32. Let u = sin x, so du = cos x dx.1 sinx1= − du = − tan u + C =− tan (ln x)+ C[ tan (ln x) C] lnxx − dx = u du = u + C[tanh( 3)] sech ( 3) ( 3)= = when tan x = 1,′ > when cos x > sin x which occurs when′′ > when cos x < –sin x which occurs⎛ π ⎞Inflection point at ⎜– , 0⎟⎝ 4 ⎠⎛π⎞Global maximum at ⎜ , 2 ⎟⎝4⎠408 Section 6.10 Instructor’s Resource Manual

⎛ π ⎞Global minimum at ⎜– , –1⎟⎝ 2 ⎠37. a.4 2f′ ( x) = 5x + 6x+ 4≥ 4> 0 for all x, sof(x) is increasing.b. f(1) = 7, so g(7) =f − 1 (7) = 1.c.1 1g′ (7) = =f ′(1) 1536.2xxf( x)=ex 2 x2e (2 x) −x ( e ) 2x−xf′ ( x)= =x 2x( e ) ef is increasing on [0, 2] because f′ ( x) > 0 on(0, 2).f is decreasing on ( −∞, 0] ∪[2, ∞ ) becausef′ ( x) < 0 on ( −∞,0) ∪ (2, ∞ ).x2 x 2e (2 −2 x) −(2 x−x ) e x − 4x+2f′′ ( x)= =x 2x( e )eInflection points are at4± 16−4⋅2x = = 2± 2 .2The graph of f is concave up on( −∞, 2− 2) ∪ (2+ 2, ∞ ) because f′′ ( x) > 0on these intervals.The graph of f is concave down on(2 − 2, 2 + 2) because f′′ ( x) < 0 on thisinterval.The absolute minimum value is f(0) = 0.4The relative maximum value is f (2) = .2eThe inflection points are⎛ 6− 4 2 ⎞ ⎛ 6+4 2⎞2− 2, and 2+2, .⎜ 2− 2 ⎟ ⎜ 2+2 ⎟⎝ e ⎠ ⎝ e ⎠38.39.12= e10k( )12lnk =10≈− 0.06931y = 1001 = 1000.06931te −0.06931te −ln1( 100 )t = ≈66.44−0.06931It will take about 66.44 years.xny n1.0 2.01.2 2.41.4 2.9761.6 3.809281.8 5.028252.0 6.8384240. Let x be the horizontal distance from the airplanedxto the searchlight, 300.dt =500 −1500tan θ = , so θ = tan .xxdθ 1 ⎛ 500 ⎞dx= ⎜ − ⎟dt1+500 ⎝ x ⎠ dt=−x2( ) 2 2x500 dx+ 250,000 dt500When θ = 30°, x = = 500 3 andtan 30°dθ 500=− (300)dt2 2(500 3) + (500)300 3=− =− . The angle is decreasing at the2000 20rate of 0.15 rad/s ≈ 8.59°/s.Instructor’s Resource Manual Section 6.10 409

41.y = (cos x)= esin x sin x ln(cos x)dy sin x ln(cos x)d= e [sin xln(cos x)]dxdx⎡= ⎜ ⎟⎣⎝ ⎠sin x ln(cos x) ⎛ 1 ⎞e⎢cos xln(cos x) + (sin x) ( −sin x)cos x⎡2sin xsin x ⎤= (cos x) ⎢cos xln(cos x)− ⎥⎢⎣cos x ⎥⎦dy 0At x = 0, 1(1ln1 0) 0dx = − = .The tangent line has slope 0, so it is horizontal:y = 1.42. Let t represent the number of years since 1990.14,000 = 10,000e10kln(1.4)k = ≈ 0.0336510y = 10,000e0.03365t(0.03365)(20)y(20) = 10,000e≈ 19,601The population will be about 19,600.143. Integrating factor is x . D[ yx] = 0; y = Cx −44. Integrating factor is x2 .1Dyx [ ] x;y ⎛ ⎞= = ⎜ ⎟x + Cx⎝4⎠2 3 2 –2⎤⎥⎦45. (Linear first-order) y′ + 2xy = 2x2xdx xIntegrating factor: e∫ = e22 2 2 2Dye [ x ] = 2 xe x ; ye x = e x + C;– x2y = 1+CeIf x = 0, y = 3, then 3 = 1 + C, so C = 2.Therefore,– x2y = 1+2 e .46. Integrating factor is – axe .– axaxDye [ ] = 1; y= e ( x+C)47. Integrating factor is –2 xe .Dye [ ] = e ; y= – e + Ce–2 x – x x 2 x48. a. Q′ ( t) = 3 – 0.02Qb. Q′ ( t) + 0.02Q= 3Integrating factor is e0.02t0.02tDQe [ ] = 3e–0.02tQt () = 150+Ce–0.02tQt () 150–30e0.02t= goes through (0, 120).c. Q → 150 g, as t →∞.Review and Preview Problems1.2.3.4.1 1sin 2x dx = sin u du =− cosu + C =2 2∫ ∫u=2xdu = 2 dx1− cos 2 x+C23t 1 u 1 u 1 3te dt = e du = e + C = e + Cu=3t3 3 3du = 3dt∫ ∫2 1 1xsin x dx = sin udu = − cosu+ C =u=x2 2 2du = 2x dx∫ ∫1 cos2− x + C22 23x 1 u 1 u 1 3xxe dx = e du = e + C = e + C6 6 6∫ ∫u=3x2du = 6x dx5.6.7.8.sin t 1dt = − du =− ln u + C =costu∫ ∫u=costdu =−sint dt1 1ln + C = ln + C = ln sect + Cu costu=sin xdu = cos x dx3 32 2 u sin xsin x cos xdx= u du = + C = + C3 3∫ ∫32 1 1x x + 2 dx = udu = u 2 + C2 3∫ ∫u= x2+ 2du = 2x dx( x 2)32 21= + + C3x 1 1 1dx = du = ln u + C2x + 1 2 u 2u= x2+ 1du = 2x dx∫ ∫2= ln u + C = ln x + 1 + C410 Review and Preview Instructor’s Resource Manual

9.⎡ 1 ⎤f ′⎛ ⎞( x) = ⎢x⎜⎟+ (ln x)(1) − 1=lnxx⎥⎣ ⎝ ⎠ ⎦21.2 2 2 2( asec t) − a −= a (sec t− 1) =2a tan t = a ⋅ tan t10.11.⎡ x⎤ −2xf '( x)= ⎢ + (1)arcsin x⎥+⎢ 2 2⎣ 1−x⎥⎦2 1−x= arcsin x2f′ ( x) = ⎡( − 2 x)(cos x) + ( −x )( − sin x)⎤+⎣⎦[(2)(sin x) + (2 x)(cos x) ] + [ 2( −sin x)]2= x sin x22.23.a −x 1 xa 1e dx ⎡ −e ⎤∫ = ⇒ − = ⇒0 2 ⎣ ⎦02⎡ −a1 1 1− e + 1⎤= ⇒ = ⇒⎣ ⎦ 2 ae 2ae = 2⇒ a = ln21 1 x−(1 −x) 2x−1− = =1 − x x (1 −x) x x(1 −x)x12. ( )13.14.15.16.17.xf ′( x) = e cos x+ sin x + e (sin x−cos x)x= 2esinx22 1−cos2x= − x ; thus sin x =2cos 2x1 2sin2 1+cos2x= − ; thus cos x =22cos 2x2cos x 124 2 2 ⎛1+cos2x⎞cos x = (cos x)= ⎜ ⎟⎝ 2 ⎠sin( u+ v) + sin( u−v)sin ucosv = ⇒2sin(7 x) + sin( −x) sin 7x−sinxsin 3xcos 4x= =2 2cos( u+ v) + cos( u−v)cosucosv = ⇒2cos(8 x) + cos( −2 x)cos3xcos5x=2cos8x+cos 2x=224.25.26.7 8 7( x− 3) + 8( x+2)+ = =5( x+ 2) 5( x− 3) 5( x+ 2)( x−3)15x−5 5(3x−1)= =5( x+ 2)( x− 3) 5( x+ 2)( x−3)(3x−1)( x+ 2)( x−3)1 1 3− − +x 2( x+ 1) 2( x−3)− 2( x+ 1)( x−3) −xx ( − 3) + 3 xx ( + 1)=2 xx ( + 1)( x−3)2 2 2−2( x −2x−3) −( x − 3 x) + (3x + 3 x)=2 xx ( + 1)( x−3)10x+ 6 2(5x+3)= = =2 xx ( + 1)( x− 3) 2 xx ( + 1)( x−3)(5x+ 3)xx ( + 1)( x−3)1 1 (2000 − y) + y 2000+ = =y 2000 − y y(2000 − y) y(2000 − y)18.19.cos( u−v) − cos( u+v)sin usinv = ⇒2cos( −x) −cos(5 x)sin 2xsin 3x=2cos x−cos5x=22 2 2 2a − ( asin t) = a (1 − sin t)=2a cos t = a cost20.2 2 2 2a + ( atan t) = a (1 + tan t)=2a sec t = a sectInstructor’s Resource Manual Review and Preview 411

CHAPTER7Techniques of Integration7.1 Concepts Review1. elementary function2.3.4.5∫ uduxe2 3∫ 1 uduProblem Set 7.11.2.3.4.∫5 1 6( x –2) dx = ( x –2) + C61 2 3/23x dx = 3x ⋅ 3 dx = (3 x)+ C3 9∫ ∫2u = x + 1, du = 2xdxWhen x = 0, u = 1 and when x = 2, u = 5 .2 2 5 1 2 2 5∫ x( x + 1) dx = ( x + 1) (2 xdx)0 2∫ 01 5= u 5du2∫ 165⎡ 6 6u ⎤ 5 −1= ⎢ ⎥ =⎢⎣12 ⎥⎦12115624= = 1302122u = 1– x , du = –2xdxWhen x = 0, u = 1 and when x = 1, u = 0 .1 2 1 1 2∫ x 1– x dx = – 1 −x ( −2 xdx)0 2∫ 01 0=− 1/22∫ u du11 1= u 1/2du2∫01⎡1 3/2⎤1= ⎢ u =3⎥⎣ ⎦ 305.∫dx 1 tan–1 ⎛ x ⎞= C2⎜ ⎟+x + 4 2 ⎝2⎠x x6. u = 2 + e , du = e dxxe du∫ dx =x ∫2 + e u= ln u + C7.8.9.x= ln 2 + e + Cx= ln(2 + e ) + C2u = x + 4, du = 2x dxx 1 du∫ dx =2x + 4 2∫u1= ln u + C21 ln 2= x + 4 + C21 ln( 2= x + 4) + C22 22t2t+ 1−1∫ dt =2 ∫ dt22t+ 1 2t+ 11= ∫dt– ∫ dt2 2t + 1u = 2, t du = 2dt1 1 dut – ∫ dt = t –22 1 2∫ 12t + + u1 –1= t – tan ( 2 t)+ C22u = 4 + z , du = 2z dz∫6 4+ 2= 3∫3/2z z dz udu= 2u + C2 3/2= 2(4 + z ) + C412 Section 7.1 Instructor’s Resource Manual

10. u = 2t + 1, du = 2dt11.5 5dt =2t+1 2∫ ∫= 5 u + C= 5 2t+ 1+Ctanduuz dz2= tan z sec z dz2∫ ∫cos zu = tan z, du = sec22∫tansec = ∫= 1 2+z dzz zdz udu2 u C1 tan2= z+C212. u = cos z, du = –sin z dz∫ ∫u u=− ∫ =− +cos zcos zsin = – (– sin )e z dz e z dze du e Ccos ze= – + C16.17.18.1u = 1– x, du = – dx2 1– x3/4sin 1– x 1/2 dx = –2 sin u du0 1– x1∫ ∫1= 2∫sinudu1/21= [ −2cos u] 1/2⎛ 1 ⎞=−2⎜cos1−cos ⎟⎝ 2 ⎠≈ 0.674623 x + 2 1∫ x dx = (3 x –1) dx + dxx+ 1∫ ∫x+1= 3 2– ln 12 x x+ x+ + C3x + 7x 21∫ dx = ( x x 8) dx 8 dxx–1 ∫ + + + ∫x–11 3 1 2= x + x + 8x+ 8ln x–1+ C3 213.14.1u = t,du = dt2 tsin t∫ dt = 2∫sin u dut= –2 cos u + C= –2cos t + Cu = x 2 , du = 2x dx2x dx du∫ = ∫1– x 1– u–1= sin u+C4 2–1 2= sin ( x ) + C19.2 2u = ln 4 x , du = dxx2sin(ln 4 x ) 1∫ dx = sin u dux 2∫1= – cosu+C21 2= – cos(ln4 x ) + C2120. u = ln x, du = dxx2sec (ln x) 1 dx sec2= u du2x2∫ ∫15. u = sin x, du = cos x dxπ /4 cos x2/2 du∫ dx =0 ∫1+ sin x 0 1+u−1 2/2= [tan u]0−12= tan2≈ 0.61552 221.1= tan u+C21= tan(ln x ) + C2xu = e ,∫xdu = e dxx6edudx = 62x∫ du21−e1−u−1= 6sin u+C−1x= 6sin ( e ) + CInstructor’s Resource Manual Section 7.1 413

22.23.24.u = x 2 , du = 2x dxx 1 du∫ dx =x 4 2∫+ 4+u1 tan−1u= + C4 221 ⎛tan–1 x ⎞= + C4 ⎜ 2 ⎟⎝ ⎠4 22x2xu = 1– e , du = –2e dx2x3e3dx = –2x1– e 2∫ ∫= –3 u + C2x= –3 1– e + C3 3dux 1 4x∫ dx =4 4∫ dx4x + 4 x + 41 ln4= x + 4 + C41 ln(4= x + 4) + C4u27.sin x− cos x ⎛ cos x⎞∫ dx = 1 dxsin x∫ ⎜ − ⎟⎝ sin x ⎠u = sin x, du = cos x dxsin x − cos∫ x dx = x −dusin x∫u= x − ln u + C= x − ln sin x + C28. u = cos(4t – 1), du = –4 sin(4t – 1)dtsin(4t−1) sin(4t−1)∫ dt =2 ∫dt21−sin (4t−1) cos (4t−1)29.1 1=− du4∫ 2u1 −11= u + C = sec(4 t− 1) + C4 4xu = e , du = e dxx x∫ ∫= ln secu+ tan u + Ce sec e dx = secudux x= ln sece + tan e + Cx25.26.1t dt t dt21 2 1 2tt3 = 2 30 0∫ ∫⎡3 3 1= ⎢ ⎥ = –⎢2ln3⎥2ln3 2ln3⎣ ⎦2 1t ⎤01= ≈ 0.9102ln 3π/6 cos xπ/6cos x∫ x dx =0 ∫ 0cos/62 sin – 2 (–sin x dx)⎡ x2 ⎤π= ⎢– ⎥⎢⎣ln 2 ⎥⎦1 3/2= – (2 – 2)ln 2=≈2−2ln 20.25593/2030.31.32.xu = e , du = e dx∫ ∫xx 2 x2e sec ( e ) dx = sec u du = tan u + Cx= tan( e ) + C3 sinxsec x+e2 sinxdx = (sec x + e cos x)dxsec xsin xtan x e cos xdx∫ ∫= +∫u = sin x, du = cos x dxsin xtan x + cos tanue xdx = x+e du∫ ∫usin x= tan x + e + C = tan x+ e + C2u = 3t −t− 1,1 (32 1)−1/2du = t −t − (6 t − 1) dt22(6t−1)sin 3t −t−1∫ dt = 2 sinu du2∫3t−t−1= –2 cos u + C= −2cos 3t −t− 1+C2414 Section 7.1 Instructor’s Resource Manual

33.34.35.3u = t − 2 , du = 3t dt2 3t cos( t − 2) 1 cos u∫ dt =2 3 3∫ du2sin ( t − 2) sin uv = sin u, dv = cos u du1 cosu 1 2 1du v− dv v−13∫ = =− + C2sin u 3∫31=− + C3sinu1=− + C .33sin( t − 2)21+cos2x1 cos2xdx = dx + dx2 2 2sin 2x sin 2x sin 2x∫ ∫ ∫2∫ ∫= csc 2x dx + cot 2x csc 2x dx1 1=− cot 2x − csc 2x+C2 23u = t − 2 , du = 3t dt2 2 3 2t cos ( t − 2) 1 cos u∫ dt =2 3 3∫du2sin ( t − 2) sin u1 cot22= udu3∫= 1 (csc u –1) du3∫1= [ − cot u− u ] + C131 3 3= [ − cot( t − 2) − ( t − 2)] + C131 [cot(3 2)3=− t − + t ] + C336. u = 1 + cot 2t, du =− 2csc 2t37.38.2csc 2t1 1dt =− du1+cot2t2 u∫ ∫=− u + C=− 1+ cot2t + C−12u = tan 2t, du = dt21 + 4t−1tan 2te 1 u∫ dt = e du21+4t2∫1 1 −1utan 2t= e + C = e + C2 22u =−t −2t− 5,du = (–2t – 2)dt = –2(t + 1)dt22−t − 2 t−5 1 u∫ ∫( t+ 1) e = − e du21 u 1 2−t−=− e + C2 t−=− e 5+ C222239. u = 3y, du = 6y dyy 1 1dy =16 −9y6 4 −u∫ ∫4 2 21 sin−1⎛u⎞= ⎜ ⎟+C6 ⎝4⎠21 −13y= sin⎛⎞+ C6 ⎜ 4 ⎟⎝ ⎠40. u = 3x, du = 3dx∫cosh 3 xdx1 1= (cosh ) sinh3∫udu= u+C31= sinh 3 x+C3341. u = x , du = 3x dx2∫ ∫2 3 1x sinh xdx=sinh udu31= cosh u+C31 cosh3= x + C342. u = 2x, du = 2 dx5 5 1∫ dx =9 4x2∫−3 −u5 sin−1⎛u⎞= ⎜ ⎟+C2 ⎝3⎠5 −1⎛2x⎞= sin ⎜ ⎟+C2 ⎝ 3 ⎠3t43. u = e , du =3tdudu2 2 233e t dte 1 1∫ dt =4 e 3∫−2 −u1 sin−1⎛u⎞= ⎜ ⎟+C3 ⎝2⎠3t1 sin−1e= ⎛⎞+ C3 ⎜ 2 ⎟⎝ ⎠6t2 2du44. u = 2t, du = 2dtdt 1 1∫ =2 2∫ du22t 4t −1 u u −11 sec−1= ⎡ u ⎤+C2 ⎣ ⎦1 sec−1= 2 t + C2Instructor’s Resource Manual Section 7.1 415

45. u = cos x, du = –sin x dxπ /2 sin x0 1∫ dx =−0 2 ∫ du1 216 + cos x 16 + u1 1= ∫ du0 216 + u1⎡1 tan−1⎛u⎞⎤⎡1 −1⎛1⎞1 −1⎤= ⎢ ⎜ ⎟4 4⎥ = ⎢ tan ⎜ ⎟−tan 0⎣ ⎝ ⎠⎦4 4 4⎥⎣ ⎝ ⎠ ⎦46.47.48.49.01 −1⎛1⎞= tan ⎜ ⎟≈0.06124 ⎝4⎠2x2xu = e + e −2x−2x, du = (2e − 2 e ) dx2x−2x= 2( e − e ) dx2 22 21x − x−e − ee + e∫ dx =0 2x−2x∫ 21 1due + e 2 u12 −2e + e 1 ln2 2 1= ⎡ u ⎤2⎣ ⎦ = ln e + e − − ln 22 2 241 e + 1 1= ln − ln 22 2e 21 4 1 2 1= ln( e + 1) − ln( e ) − ln 22 2 241⎛⎛e+ 1⎞⎞= ln −2 ≈0.66252⎜⎜ 2 ⎟ ⎟⎝ ⎝ ⎠ ⎠∫ 1 dx =12 ∫dx2x + 2x+ 5 x + 2x+ 1+41= ∫( 1)2 2( x 1) 2 d x ++ +1 –1 ⎛ x + 1⎞= tan ⎜ ⎟+C2 ⎝ 2 ⎠∫ 1 dx =12 ∫dx2x – 4x+ 9 x –4x+ 4+51= ∫d( x–2)2 2( x –2) + ( 5)1 –1 ⎛ x –2⎞= tan ⎜ ⎟+C5 ⎝ 5 ⎠dxdx∫ =2 ∫ 29x + 18x+ 10 9x + 18x+ 9+1dx= ∫ 2 2(3x+ 3) + 1u = 3x + 3, du = 3 dxdx 1 du∫ =(3x3) 1 3∫+ + u + 12 2 2 21 tan–1 (3 3)= x + + C350.51.52.dxdx=2 216 + 6 x– x –( x – 6x+9 – 25)∫ ∫=dx∫ =2 2∫2 2–( x –3) + 5–1 ⎛ x –3⎞= sin ⎜ ⎟+C⎝ 5 ⎠dx5 –( x –3)x+ 1 1 18x+18∫ dx =2 18∫dx29x + 18x+ 10 9x + 18x+101 2= ln 9 x + 18 x+ 10 + C181 ln2= ( 9 x + 18 x+ 10 ) + C183– x 1 6–2xdx =16 + 6 x– x 2 16 + 6 x–x∫ ∫= 16 + 6 x – x + C53. u = 2, t du = 2dt54.2 2dt du=2 2 2t 2 t –9 u u –3∫ ∫⎛1 2t⎞sec–1= ⎜ ⎟+C3 ⎜ 3 ⎟⎝ ⎠2dxtan x cos x tan x∫ dx =2∫dxcos 2sec x– 4 x sec x– 4sin x= ∫dx21–4cos xu = 2 cos x, du = –2 sin x dx∫sin x 1 1dx =−21−4cos x 2∫ du1−u21 sin−11 –1= − u+C = – sin (2cos x)+ C2255. The length is given by2b ⎛dy⎞L = ∫ 1+ dxa ⎜ ⎟⎝dx⎠/4 1= ∫π 1 + ⎡ ( sin x)⎤ dx0 ⎢ −cos x⎥⎣⎦π /4 2 π /4 2= ∫ 1+tan x dx = sec x dx0∫ 0π /4/4= ∫ sec x dx =⎡ ⎣ln sec x+ tan x⎤⎦0π0= ln 2 + 1 −ln 1 = ln 2 + 1 ≈ 0.8812416 Section 7.1 Instructor’s Resource Manual

56.1 1+sinxsec x = =cos x cos x(1 + sin x)2 2 2sin x + sin x+ cos x sin x(1 + sin x) + cos x= =cos x(1 + sin x) cos x(1 + sin x)= sin x cos xcos x+ 1 + sin x⎛secsin x cos x ⎞∫ x = ∫ ⎜ + ⎟dx⎝cos x 1+sin x⎠sin x cos x= ∫ dx + dxcos x∫1+sin xFor the first integral use u = cos x, du = –sin x dx,and for the second integral use v = 1 + sin x,dv = cos x dx.sin cos∫ x dx + x dx –du dvcos x∫ = +1+sin x∫u∫v= –ln u + ln v + C= –ln cosx + ln1+ sinx + C1+sinx= ln + C cos x= ln sec x + tan x + C57. u = x – π , du = dx2πx sin x π ( u+π ) sin( u+π)∫ dx =0 2 ∫du– 21+ cos xπ1+ cos ( u+π)π ( u+π) sinu = ∫du– π 21+cos uπ u sin u π π sin u= ∫ du +du– π 2 ∫ – 21+ cos uπ1+cos uπ u sin u∫ du = 0 by symmetry.– π 21+cos uπ π sin uπ πsinu∫ du = 2du– π 2 ∫ 0 21+ cos u 1+cos uv = cos u, dv = –sin u du–1 π1 1− 2∫ dv = 2πdv1 2 ∫ –1 21+ v1+v−1 1 ⎡π ⎛ π⎞⎤= 2 π [tan v] − 1 = 2π ⎢ − ⎜ − ⎟4 4 ⎥⎣ ⎝ ⎠⎦⎛π⎞ 2= 2π ⎜ ⎟=π⎝2⎠58.3π4– π4V ⎛2 x π ⎞= π ∫ ⎜ + ⎟ sin –cos⎝ 4 ⎠x x dxu = x– π , du = dx4πV 22 ⎛ u π⎞ ⎛sin –cos– π2 u π⎞ ⎛4 u π⎞= π ∫ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟⎝ ⎠ ⎝ ⎠ ⎝ 4⎠duπ2– π22⎛ π ⎞ 2 2 2 2= 2π ∫ ⎜u+ ⎟ sin u+ cos u– cosu+sin u du⎝ 2⎠2 2 2 2π2 ⎛ π ⎞π 2 2πππ2−π⎜⎟π π2⎝2 ⎠− −2 2π2∫ u u du− πby symmetry. Therefore,2ππ2 2 2 22 2 2∫ 0sin 2 2 [ cos ] 02 2∫ ∫ ∫= 2π u + 2 sin u du = 2 2 u sin u du + 2 sin u du2 2π sin = 0V = π udu = π − u = πInstructor’s Resource Manual Section 7.1 417

7.2 Concepts Review1. uv –∫v du2. x; sin x dx3. 14. reductionProblem Set 7.21. u = xxdv = e dxdu = dxxv = ex xxe dx= xe −xedxx x= xe − e + C∫2. u = xdu = dx∫3xdv = e dx1 3xv = e33 1 3 1 3∫xe x dx = xe x − e x dx3∫31 3x1 3x= xe − e + C3 95t+π3. u = t dv = e dt1 5tdu = dt v = e +π55 t +π 1 5 t +π 1 5te dt te – e t +π∫ =dt5∫51 5t+π 1 5t+π= te – e + C5 252t+34. u = t + 7 dv = e dt1 2t3du = dt v = e +22t+ 3 1 2t+ 3 1 2t+3∫( t+ 7) e dt = ( t+7) e – e dt2∫21 2t+ 3 1 2t+3= ( t+ 7) e – e + C2 4t 2 3 13 et+ e2t+3= + + C2 45. u = x dv = cos x dxdu = dx v = sin x∫x cos xdx=xsin x– ∫ sin xdx= x sin x + cos x + C6. u = x dv = sin 2x dx1du = dx v = – cos 2x21 1∫xsin 2 xdx=– xcos 2 x– – cos 2xdx2∫21 1= – x cos2x+ sin2x+C2 47. u = t – 3 dv = cos (t – 3)dtdu = dt v = sin (t – 3)( t –3)cos( t –3) dt = ( t –3)sin( t –3)– sin( t –3) dt∫= (t – 3) sin (t – 3) + cos (t – 3) + C8. u = x – π dv = sin(x)dxdu = dx v = –cos x∫( x – π )sin( xdx ) = –( x– π )cos x+∫ cos xdx= (π – x) cos x + sin x + C9. u = t dv = t + 1 dt2du = dt ( 1)3/2v = t+32 3/2 2 3/2∫t t+ 1 dt = t( t+ 1) – ( t+1) dt3∫32 3/2 4 5/2= tt ( + 1) – ( t+ 1) + C3 1510. u = t dv = 3 2t + 7dt3du = dt(2 7)4/3v = t+83 3 4/3 3 4/3∫t 2t+ 7 dt = t(2t+ 7) – (2t+7) dt8∫83 4/3 9 7/3= t(2t+ 7) – (2t+ 7) + C8 11211. u = ln 3x dv = dx1du = dxxv = xln 3x dx = x ln 3x −1x dxx12.∫ ∫ ln 35u = ln(7 x ) dv = dx5du = dx v = xx∫ ∫5 5 5ln(7 x ) dx = x ln(7 x ) – x dxx5= x ln(7 x ) – 5x+C∫= x x− x+C418 Section 7.2 Instructor Solutions Manual

13. u = arctan x dv = dx1du = dx v = x21 + xx∫arctanx = xarctanx−∫dx21+x1 2x= x arctan x−dx2∫ 21+x1 2= x arctan x− ln(1 + x ) + C214. u = arctan 5x dv = dx5du = dx v = x21 + 25x5x∫arctan 5x dx = x arctan 5 x – ∫ dx1 252+ x1 50x dx= xarctan 5 x– 10∫ 21+25x12= x arctan 5 x– ln(1 + 25 x ) + C10dx15. u = ln xdv =2x11du = dx v = –xxln x ln x 1 ⎛1⎞∫ dx = – – – dx2x x∫ ⎜ ⎟x⎝x⎠= ln x 1– – Cx x+16.u = ln 2x55du = dxx12dv = dxx1v =−x533ln 2 ⎡ 1 5 ⎤ 3 1= ln 2 52 2 ⎢− x⎥ +2 2x ⎣ ⎦2x3x dx x dx∫ ∫⎡ 1 5 5⎤= ⎢− ln 2x−x x⎥⎣⎦2⎛ 1 5 5⎞ ⎛ 1 5 5⎞= ⎜− ln(2⋅3 ) − ⎟−⎜− ln(2⋅2 ) − ⎟⎝ 3 3⎠ ⎝ 2 2⎠1 5 5 5=− ln 2 − ln 3 − + 3ln 2 +3 3 3 28 5 5= ln 2 − ln 3 + ≈ 0.85073 3 617. u = ln t dv = t dt18.12 3/2du = dt v = tt3e∫ e ⎡2 3/2 ⎤ e21/2t ln tdt = t ln t – t dt1 ⎢3 ⎥ ∫⎣ ⎦ 11 3e2 3/2 2 ⎡43/2⎤= e ln e– ⋅1ln1− t3 3 ⎢9 ⎥⎣ ⎦2 3/2 4 3/2 4 2 3/2 4= e −0 − e + = e + ≈ 1.44043 9 9 9 93u = ln xdv = 2xdx31du = dx(2 )3/2v = xx355 3 ⎡1 32 3⎤5 32∫ 2xlnx dx = (2 x ) ln x 2 xdx1⎢−3⎥ ∫⎣ ⎦ 1⎡⎢⎣5/23/2 3 3/21 2= (2 x) ln x − x3 3⎤⎥⎦152 ⎛5232 3 3/2 32 31 2 1 2= (10) ln 5 − 5 − (2) ln1 −3 3 ⎜⎝3 34 2 32 4 2 32=− 5 + + 10 ln 5 ≈ 31.6993 319. u = ln z dv = z dz1 1 4du = dz v = zz 43 1 4 1 4 1∫z ln zdz = z ln z− z dz4∫ ⋅4 z1 4 1 3= z ln z− z dz4 4∫1 4 1 4= z ln z− z + C4 1620. u = arctan t dv = t dt1du = 1dt v = t221 + t221 2 1 t∫t arctan t dt = t arctan t – dt2 2∫21+t21 2 1 1+ t −1= t arctan t−dt2 2∫21+t1 2 1 1 1= t arctan t −2 2∫dt + dt2∫21+t1 2 1 1= t arctan t− t+ arctan t+C2 2 23151⎞⎟⎠Instructor's Resource Manual Section 7.2 419

21.⎛1⎞u = arctan ⎜ ⎟ dv = dt⎝t⎠1du = – dt v = t12+ t⎛1⎞⎛1⎞t∫arctan⎜ ⎟dt = t arctan ⎜ ⎟+dtt t∫⎝ ⎠ ⎝ ⎠21+t⎛1⎞1 2= tarctan ⎜ ⎟+ ln(1 + t ) + C⎝t⎠ 222.7u = ln( t )5dv = t dt7 1du = dt v = tt 65 7 1 6 7 7 5∫ t ln( t ) dt = t ln( t )– t dt6 6∫1 6 7 7 6= t ln( t ) – t + C6 36623. u = xdu = dx2dv = csc x dxv =− cot xπ /2 2π /2 π= [ − ] +/2π/6 π /6 ∫π/6∫x csc xdx x cot x cot xdx = ⎡− ⎣ xcotx+ ln sin x⎤⎦π ππ π 1 π=− ⋅ 0 + ln1+ 3 − ln = + ln 2 ≈ 1.602 6 2 2 324. u = xdu = dxdv = secv = tan x2x dxπ 4 sec 24[ tan π πx xdx x x]46 6tan4 2 3∫= − xdxππ ∫xtan x ln cos x π π ⎛ π ⎞=⎡ ln lnπ 6 ⎣ + ⎤ ⎦ = + − +π 6 4 2 ⎜ 6 3 2 ⎟⎝⎠π π 1 2= − + ln ≈ 0.284 6 3 2 32625.32 3dv = x x + 4dx23 3/2= 3x dx v = 2 ( x + 4) 95 3 2 3 3 3/2 2 2 3 3/2x x + 4 dx = x ( x + 4) – x ( x 4) dx9∫ +3u = xdu∫2 3 3 3/2 4 3 5/2= x ( x + 4) – ( x + 4) + C9 4526.u = x76 7dv = x x + 1 dx67 3/2du = 7x dx v = 2 ( x + 1) 21∫13 7 2 7 7 3/ 2 2 6 7 3/ 2 2 7 7 3/2 4 7 5/2x x + 1 dx = x ( x + 1) – x ( x + 1) dx21∫= x ( x + 1) – ( x + 1) + C321 10527.u= t43du = 4t dt∫3tdv =dt4 3/2(7 – 3 t )v =141/26(7 – 3 t )2– 37 4 34t t tt 1 41/2dt =4 3/2 4 1/2 ∫dt = + (7 – 3 t ) + C4 1/241/2(7 – 3 t ) 6(7 – 3 t ) (7 – 3 t ) 6(7 – 3 t ) 9420 Section 7.2 Instructor Solutions Manual

28.22u = xdv = x 4– x dx1 2 3/2du = 2x dx v = – (4 – x )3∫3 2 1 2 2 3/2 2 2 3/2x 4– x dx = – x (4– x ) + x(4– x ) dx3 3∫1 2 2 3/2 2 2 5/2= – x (4– x ) – (4– x ) + C3 1529.u = z43du = 4z dz∫3zdv = dz4 2(4 – z )1v =44(4 – z )7 4 3z z zdz = −4 2 4 ∫ dz4(4 – z ) 4(4 – z ) 4 – z4z 1 ln 4 –4= + z + C44(4 – z ) 430. u = x dv = cosh x dxdu = dxv = sinh xx cosh = x sinh – sinh ∫ ∫ = x sinh x – cosh x + C31. u = x dv = sinh x dxdu = dxv = cosh xx sinh = x cosh – cosh ∫ ∫ = x cosh x – sinh x + C32. u = ln x–1/ 2dv = x dx11/2du = dx v = 2xxln 1∫ x dx = 2 ln –2∫ = 2x1/2xx ln x–4x + C33. u = x49dv = (3x + 10) dx1du = dx(3 10)50v = x+15049 x50 150∫x(3x+ 10) dx = (3x+ 10) – (3x+10) dx150 150∫34. u tdv = t − 1 dt= ( ) 12du = dtv = 1 ( t−1) 131311 12 ⎡ t 13 ⎤ 1 1 13∫ t( t − 1) dt = ( ) ( )0 ⎢ t −1 113 ⎥ −00 13∫ t − dt⎣ ⎦1⎡ t 13 1 14 ⎤ 1= ⎢ ( t−1) − ( t− 1)=13 182⎥⎣⎦018235. u = x dv = 2 x dx1du = dx2xv =ln 2x x x 1 x∫x2 dx = 2 – 2 dxln 2 ln 2∫x x 1 x= 2 – 2 + Cln 2 2(ln 2)x50 151= (3x + 10) – (3x+ 10) + C150 22,950zdv = a dz36. u = z1du = dzv = aln az z z 1 z∫ za dz = a – a dzln a ln a∫= z 1 az– az Cln a2(ln a)+37.2u = xdu = 2x dx∫ ∫zxdv = e dxxv = e2 x 2 x xx edx= xe − 2xedxu = xdu = dx∫xdv = e dxv= ex( )2 2x e x dx = x e x −2xe x − e x dx2x x x= x e − 2xe + 2e + C∫Instructor's Resource Manual Section 7.2 421

38.39.u = x43du = 4x dxx2dv = xe dx1 x2v = e252 1 4232∫x e x dx = x e x – 2x e x dx2∫2x2u = xdv = 2xe dxx2du = 2x dx v = e52 1 4222 2x e x dx x e x – ⎛ x e x 2xe x ⎞∫ = − dx2⎜ ∫ ⎟⎝⎠1 4 x22 x2x2= x e – x e + e + C22u = ln z dv = dz2lnzdu = dzzv = z2 2ln zdz=zln z– 2 ln zdz∫ ∫u = ln zdv = dz1du = dzzv = z∫( )2 2ln zdz=zln z– 2 zln z–dz2= zln z– 2zln z+ 2z+C∫41.42.tu = edv = cost dttdu = e dt v = sin te t cost dt = e t sin t − e t sin t dt∫ ∫tu = edv = sin t dttdu = e dt v = –cos t∫ ∫t t t te costdt = e sin t−⎡− e cost+e costdt⎤⎣⎦e t costdt = e t sin t+ e t cost−e t costdt∫ ∫t2∫cos =tsin +tcos +∫e tdt e t e t Ct 1 te cos tdt = e (sin t+ cos t)+ C2atu = edv = sin t dtatdu = ae dt v = –cos te at sin tdt = – e at cost+a e at costdt∫ ∫atu = edv = cos t dtatdu = ae dt v = sin t∫∫( )e at sin t dt = – e at cost + a e at sin t – a e at sin t dt2e at sin t dt = – e at cost + ae at sin t – a e at sin t dt∫∫40.2 20u = ln x dv = dx2040ln xdu = dx v = xx2 20 2 20 20ln x dx = x ln x – 40 ln x dx∫ ∫20u = ln xdv = dx20du = dx v = xx∫( )ln 2 x 20 dx = x ln 2 x 20 – 40 x ln x 20 – 20 dx2 20 20= x ln x – 40xln x + 800x+C∫43.∫ + C2(1 + a ) e at sin tdt = – e at cos t+ae at sin t∫atatat – e cost ae sinte sin tdt = + + C2 2a + 1 a + 12u = x dv = cos x dxdu = 2x dx v = sin x∫ ∫2 2x cos xdx= x sin x−2xsinxdxu = 2x dv = sin x dxdu = 2dx v = − cos x2 2x cos xdx= x sin x− − 2xcos x+2cos xdx∫( )∫= x2 sin x+ 2xcos x− 2sin x+C44.2u = r dv = sin r drdu = 2r dr v = –cos r2 2r sin r dr = – r cos r + 2 r cos r dr∫ ∫u = rdu = dr∫dv = cos r drv = sin r( )2 2r sin rdr = – r cos r+2 rsin r – sin rdr∫2= – r cos r+ 2rsin r+ 2cos r+C422 Section 7.2 Instructor Solutions Manual

45. u = sin(ln x)dv = dx1du = cos(ln x)⋅ dxxv = x∫ ∫sin(ln x) dx = xsin(ln x) − cos(ln x)dxu = cos (ln x)dv = dx1du =−sin(ln x)⋅ dxxv = x∫ ∫∫ = − −∫2∫sin(ln ) = sin(ln ) − cos(ln ) +∫sin(ln x) dx = xsin(ln x) −⎡x cos(ln x) − −sin(ln x)dx⎤⎣⎦sin(ln x) dx xsin(ln x) x cos(ln x) sin(ln x)dxx dx x x x x Cxsin(ln x) dx = [sin(ln x) − cos(ln x)]+ C246. u = cos(ln x) dv = dx1du = –sin(ln x)dxxv = x∫ ∫cos(ln x) dx = x cos(ln x) + sin(ln x)dxu = sin(ln x)dv = dx1du = cos(ln x)dxv = xx∫ ∫2∫cos(ln x) dx = x[cos(ln x) + sin(ln x)]+ C∫cos(ln x) dx = x cos(ln x) + ⎡x sin(ln x) – cos(ln x)dx⎤⎣ ⎦xcos(ln x) dx = [cos(ln x) + sin(ln x)]+ C247.48.3u = (ln x)dv = dx23ln xdu = dx v = xx3 3 2(ln x)dx = x(ln x) –3 ln x dx∫ ∫3 2= x ln x– 3( xln x– 2xln x+ 2 x+C)3 2= + − +x ln x– 3xln x 6xln x 6x C4u = (ln x)dv = dx34ln xdu = dx v = xx4 4 34 3 2(ln x)dx = x(ln x) – 4 ln x dx = x ln x– 4( xln x– 3xln x+ 6xln x− 6 x+C)∫4 3 2= + − + +∫x ln x– 4xln x 12xln x 24xln x 24x C49. u = sin x dv = sin(3x)dxdu = cos x dx1v = – cos(3 x)31 1∫sin xsin(3 xdx ) = – sin xcos(3 x) + cos xcos(3 xdx )3 3∫u = cos x dv = cos(3x)dxdu = –sin x dx1v = sin(3 x )3Instructor's Resource Manual Section 7.2 423

1 1 1 1sin xsin(3 xdx ) – sin xcos(3 x) ⎡⎤∫ = + cos xsin(3 x) sin xsin(3 xdx )3 3⎢+3 3∫⎥⎣⎦1 1 1= – sin x cos(3 x) + cos xsin(3 x) + sin xsin(3 x)dx3 9 9∫8 1 1sin sin(3 ) – sin cos(3 ) cos sin(3 )9∫x xdx= x x + x x + C3 93 1∫ sin xsin(3 xdx ) = – sin xcos(3 x) + cos xsin(3 x)+ C8 850. u = cos (5x) dv = sin(7x)dx1du = –5 sin(5x)dx v = – cos(7 x)71 5∫cos(5 x)sin(7 xdx ) = – cos(5 x)cos(7 x) – sin(5 x)cos(7 xdx )7 7∫u = sin(5x) dv = cos(7x)dx1du = 5 cos(5x)dx v = sin(7 x )71 5⎡1 5⎤∫cos(5 x)sin(7 xdx ) = – cos(5 x)cos(7 x) – sin(5 x)sin(7 x) – cos(5 x)sin(7 xdx )7 7⎢7 7∫⎥⎣⎦1 5 25= – cos(5 x)cos(7 x) – sin(5 x)sin(7 x) + cos(5 x)sin(7 x)dx7 49 49∫24 1 5cos(5 )sin(7 ) – cos(5 )cos(7 ) – sin(5 )sin(7 )49∫ x xdx= x x x x + C7 497 5∫ cos(5 x)sin(7 xdx ) = – cos(5 x)cos(7 x) – sin(5 x)sin(7 x)+ C24 2451.u = e α zdv = sin βz dzα z1du = αe dz v = – cos β zβαz 1 αz α αze sin βzdz = – e cos βz+e cos βzdzββ∫ ∫u = e α zdv = cos βz dzα z1du = αe dz v = sin β zβαz 1 αz α ⎡1αz α αz⎤∫e sin βzdz =− e cos βz+ ⎢ e sin βz−e sin βzdz⎥β β ⎣β β∫⎦21 αz α αz α αz= – e cos βz+ e sin βz– e sin βzdzβ 2 2∫β β2 2+ αz 1 αz αze sin βzdz – e cos βz e sin βz C2 ∫= + +2β α αβββzαz – β αz α αze ( sin z– cos z)∫ e sin βzdz = e cos βz+ e sin βz+C = + C2 2 2 22 2+ ++α β α βα α β β βαβ424 Section 7.2 Instructor Solutions Manual

52.u = e α zdv = cos β z dzα zdu = αe dz1 v = sin β zβαz 1 αz α αze cos βzdz = e sin βz– e sin βzdzββ∫ ∫u = e α zdv = sin βz dzα z1du = αe dz v = – cos β zβαz 1 αz α ⎡ 1 αz α αz⎤∫e cos βzdz = e sin βz− e cos βz e cos βzdzβ β⎢− +β β∫⎥⎣⎦21 αz α αz α αz= e sin βz+ e cos βz– e cos βzdzβ 2 2∫β β2 2α + β αz α αz 1 αze cos βzdz e cos βz e sin βz C2 ∫= + +2βββ∫α zα z e ( αcosβz+βsin βz)e cos β zdz = + C2 2+αβα53. u = ln x dv = x dxα + 11xdu = dx v =xα + 1, α ≠ –11xα +α+ 1 α+1α1 α x x∫x ln xdx=ln x–x dxα + 1 α + 1∫= ln x– + C, α ≠ –1α + 1 2( α + 1)54.2αu = (ln x)dv = x dxα + 12lnxxdu = dx v =xα + 1, α ≠ –112 xα +1 1 1α2 2 α2∫x (ln x) dx = (ln x) – x ln xdxα + 1 α + 1∫x α+ 2 ⎡(ln )x α+ lnx α+⎤= x − ⎢ x− ⎥+Cα + 1 α + 1 α 1 2⎢⎣+ ( α + 1) ⎥⎦α+ 1 α+ 1 α+1x 2 x x= (ln x) – 2 ln x+ 2 + C, α ≠ –1α + 1 2 3( α + 1) ( α + 1)αProblem 53 was used for x ln xdx.∫55. u = x αβ xdv = e dxα –1du = x dxα1 xv = e ββxx x α βα β e α α–1βxx e dx = – x e dx∫ ∫56. u = x αdv = sin βx dxα –11du = α x dx v = – cosβxβββααx cos βxα α–1x sin xdx= – + x cos xdx∫ ββ β∫βInstructor's Resource Manual Section 7.2 425

57. u = x αdv = cos βx dxα –11du = α x dx v = sin β xβααx sin βxα α–1x cos xdx=– x sin xdx∫ ββ β∫58. u = (ln x)αdv = dxα –1α(ln x)dudxx–1(ln x) α dx = x(ln x) α – (ln x)α dx= v = x∫ α∫β59.60.61.2 2u ( a – x )α= dv = dx2 2 –1du –2 x( a – x )α dx= αv = x∫ α∫2 2 2 2 2 2 2 –1( a – x ) α dx = x( a – x ) α + 2 x ( a – x )α dxα –1u = cos xdv = cos x dx–2du = –( α –1)cos xsinx dx v = sin x∫ α ∫–1 –2 2cos α x dx = cos α xsin x + ( –1) cos α xsinx dxα ∫−∫ = + α − ∫α−1 α−2 2= cos xsin x+ ( −1) cos x(1 −cos x)dxα1 2cos α x dx cos α xsin x ( 1) cosα −x dxα –1α cos xsin x α –1cos x dx = +∫ ∫ααα–2cos x dxα –1u = cos β xdv = cos βx dxα–1 α–2αcos xsin x ( –1) cos xdx– ( –1) cos xdx= + α ∫ α ∫–2du = – ( –1)cos α xsinx dxβα β β1 v = sin β xβα –1α cos βxsinβxα–2 2cos x dx = + ( –1) cos xsinx dx∫ ∫β α β ββα −1cos βxsinβxα −2 2= + ( −1) cos x(1−cos xdx )βα –1cos βxsinβxβ∫α β βα–2α( α –1) cos βx dx –( α –1) cos βx dx= + ∫ ∫α −1α cos βxsinβxα−2α∫cos βx= + ( α −1) cos βxdxβ∫α –1α cos βxsin βxα –1cos βx dx = +αβ α∫ ∫α–2cosβx dx62.4 3 x 1 4 3 x 4 3 3–x 1 4 3 4 1 3 3 2 3∫x e dx = x e x e dx3 3∫x ⎡– x –x ⎤= x e x e x e dx3 3⎢3∫ ⎥⎣⎦1 4 3 x 4 3 3 4 1 2 3 2 3– x ⎡ x –x ⎤ 1 4 3= x e x e + x e xe dx3 9 3 ⎢ 3 3∫x 4 3 3 4 2 3 8 1 3 1 3⎥ – x x ⎡– x –x ⎤= x e x e + x e xe e dx⎣⎦ 3 9 9 9⎢3 3∫⎥⎣⎦1 4 3 4 3 3 4 2 3 8 3 8 3= x e x – x e x + x e x – xe x + e x + C3 9 9 27 81426 Section 7.2 Instructor Solutions Manual

63.64.4 1 4 4 31 4 4⎡1 3 2 ⎤∫x cos3xdx=x sin 3 x– x sin 3xdx3 3∫= x sin 3 x– – x cos3x x cos3xdx3 3⎢+3∫ ⎥⎣⎦1 4 4 3 4⎡1 2 2 ⎤= x sin 3x+ x cos3 x– x sin 3x xsin 3xdx3 9 3⎢−3 3∫⎥⎣⎦1 4 4 3 4 2 8⎡1 1 ⎤= x sin 3x+ x cos3 x– x sin 3 x+ – xcos3x cos3xdx3 9 9 9⎢+3 3∫⎥⎣⎦1 4 4 3 4 2 8 8= x sin3x+ x cos3 x– x sin3 x– xcos3x+ sin3x+C3 9 9 27 816 1 5 5 4cos 3x dx = cos 3xsin 3x + cos 3x dx18 61 5 5⎡1 3 3 2 ⎤∫ ∫ = cos 3xsin 3x+ cos 3xsin 3x cos 3xdx18 6 ⎢+12 4∫ ⎥⎣⎦1 5 5 3 5⎡1 1 ⎤= cos 3xsin 3x+ cos 3xsin 3x+ cos3xsin 3x dx18 72 8 ⎢+6 2∫ ⎥⎣⎦1 5 5 3 5 5x= cos 3xsin 3x+ cos 3xsin 3x+ cos3xsin 3x+ + C18 72 48 1665. First make a sketch.From the sketch, the area is given bye∫ 1 ln x dxu = ln xdv = dx1du = dx v = xxee e∫ ln x dx = [ x ln x] 1 1−∫ dx = [ x ln x− x] 11e = (e – e) – (1 · 0 – 1) = 166.e 2V = ∫ π(ln x)dx12u (ln x)= dv = dx2lnxdu = dxv = xxe 2 ⎛ 2e e ⎞π (ln x) dx ⎡x(ln x) ⎤∫ = π 2 ln x dx1 ⎜ −⎣ ⎦1∫1⎟⎝⎠=π[( e− 2e+ 2 e) −(0 − 0 + 2)] =π( e −2) ≈ 2.26(ln )22( ln ) ⎤e⎦1=π⎡x x − x x−x⎣2=π[ x(ln x) − 2xlnx+2 x] 1eInstructor's Resource Manual Section 7.2 427

67.9 – x/3 9 – x/3⎛1 ⎞3 e dx = –9 e – dx0 ∫0⎜ ⎟⎝ 3 ⎠∫– x /3 9 9= –9[ e ] 0 = – + 9 ≈ 8.553e68.69.70.9 – x/3 2 9 –2 x/3V = ∫ π (3 e ) dx = 9πe dx0 ∫ 03 9 –2 / 3 29 ⎛ – ⎞ xe⎛ – dx⎞ 27π –2 x / 3 9 27π 27π= π ⎜ ⎟2∫0⎜ ⎟ = – [ e ] 0 = – + ≈ 42.31⎝ ⎠ ⎝ 3 ⎠ 2 62e2π /4 π /4 π /4∫ ∫ ∫( x cos x– xsin x) dx=xcos xdx– xsinxdx0 0 0⎛ π 4 π 4 ⎞π 4 π 4⎜⎣xsinx⎦sin xdx0−∫ 0 ⎟ −⎛⎜[ − cos ] +0 ∫ 0= ⎡ ⎤⎝⎠⎝x x cos xdx⎞⎟⎠/4[ xsin x cos x xcos x– sin x] π 2 π= + +0 = –1 ≈ 0.114Use Problems 60 and 61 for ∫ x sin xdx and ∫ x cos xdx.2 x2 ∫ sin0 2V = ππ ⎛ ⎞x ⎜ ⎟dx⎝ ⎠u = xxdv = sin dx 2du = dxxv = –2cos 222 ⎛π⎡–2 cos x⎤2π2 02cos x ⎞V = π x + dx⎜⎢⎥ ∫⎣ ⎦02 ⎟⎝⎠⎛2π⎡ x ⎤ ⎞= 2π 4π+ 4sin = 8π⎜ ⎢2⎥ ⎟⎝ ⎣ ⎦0⎠271.e 2 ex dx =1 1∫ ∫ln 2 ln x dxu = ln x dv = dx1du = dxxv = xe( )2 e1 2 ⎛ e e ⎞∫ ln xdx= ⎜[ xln x] 1−∫dx1 ⎟= 2 e− [ x]1 =⎝⎠2e 2 e∫ x ln x dx = 2 xlnxdx1 ∫ 1u = ln x dv = x dx1 1 2du = dx v = xx 2ee ⎛⎡1 2 ⎤ e1⎞ ⎛e1 2 ⎡1 2⎤⎞ 1 22∫x lnxdx= 2 x ln x – xdx1 ⎜⎢ 2⎥ ∫ = 2 e – x = ( e + 1)⎣ ⎦ 11 2 ⎟ ⎜2 ⎢4⎥⎝⎠⎣ ⎦ ⎟1 2⎝⎠428 Section 7.2 Instructor Solutions Manual

12e1∫2(ln x)dx2u = (ln x)dv = dx2lnxdu = dx v = xx1 e 2 1 2 e e(ln x) dx⎛[ x(ln x) ]1 1 – 2 ln x dx⎞2∫= ⎜ ⎟2⎝∫1⎠1 2 22 ( e + 1) e + 1x = =2 412 ( e –2) e – 2y = =2 472. a. u = cot x2dv = cscdu = –csc x dx v = –cot x∫∫2x dx2 2 2cot x csc xdx=−cot x−cot xcscxdx2 22 cotxcsc xdx =− cot x+C∫2 1 2cot xcsc xdx =− cot x+C2b. u = csc x dv = cot x csc x dxdu = –cot x csc x dx v = –csc x∫∫2 2 2cot x csc xdx=−csc x−cot xcscxdx2 22 cotxcsc xdx =− csc x+C∫2 1 2cot xcsc xdx =− csc x+C2∫∫= 1 ( –2)2 ec.1 2 1 2– cot x = – (csc x–1)2 21 2 1= – csc x +2 273. a.b.3px ( ) = x − 2xxg( x)= exAll antiderivatives of g( x)= e∫3 x 3 x 2 x x x( x − 2 xe ) dx= ( x −2 xe ) −(3x − 2) e + 6xe − 6e + C2px ( ) = x − 3x+1g(x) = sin xG1( x) =− cosxG2( x) =− sinxG3( x) = cosx∫2 2( x − 3x+ 1)sin xdx = ( x − 3x+ 1)( −cos x) −(2x−3)( − sin x) + 2cosx+CInstructor's Resource Manual Section 7.2 429

74. a. We note that the nth arch extends from x = 2 π ( n−1)to x = π (2n− 1) , so the area of the nth arch isπ (2n−1)An ( ) = ∫ x sin xdx . Using integration by parts:2 π ( n−1)u = x dv = sin xdxdu = dx v =−cosxπ n−[ ] [ ] [ ]π(2n−1) π(2n−1) (2 1)π(2n−1) π(2n−1)An ( ) = ∫2 π( n−1) xsin xdx= −xcos x −2 ( 1) 2 ( n 1) cos xdx xcos x sin xπ n− ∫ π − − = − +2 π( n−1) 2 π( n−1)[ π(2n 1)cos( π(2n 1)) 2 π( n 1)cos(2 π( n 1)) ] [ sin( π(2n 1)) sin(2 π( n 1))]π(2n 1)( 1) 2 π( n 1)(1) 0 0 π[ (2n 1) (2n 2) ]. So A( n) (4n3) π= − − − + − − + − − −=− − − + − + − = − + − = −b.3 2V = 2π∫π x sinxdx2π2u = xdv = sin x dxdu = 2x dx v = –cos x2 ⎛– 23π3πV ⎡ x x⎤2 2x cos ⎞ ⎛ 2 2 3π⎞= π ⎜xdx⎣ ⎦⎟⎝π+∫= 2π 9 4 2xcosxdxπ⎜ π + π + ∫⎠2π⎟⎝⎠u = 2xdv = cos x dxdu = 2 dx v = sin x⎛ 2 3π 3V 2 13 [2xsin x] 2 – ⎞= π⎜π + π ∫ 2sin x2π⎟⎝⎠( x 2π π )2 3 2= 2π 13 π + [2cos ] = 2 π(13 π – 4) ≈ 78175. u = f(x) dv = sin nx dx1du = f ′( x)dx v =− cos nxnπ1 ⎡ ⎡ 1 ⎤ 1 π⎤an= − cos( nx) f( x) + cos( nx) f′( x)dxπ⎢ ⎢ n⎥ n∫−π⎥⎣ ⎣ ⎦−π ⎦Term 1Term 2Term 1 = 1 cos( nπ)( f( −π) − f( π )) =± 1 ( f( −π) − f( π ))nnπSince f ′( x)is continuous on [– ∞ , ∞ ], it is bounded. Thus, ∫ – cos( nx) f ′π( x)dx is bounded so1πlim an= lim⎡± ( f( −π) − f( π )) + cos( nx) f′( x) dx⎤= 0.n→∞n→∞πn⎢⎣∫ −π⎥⎦76.Gn[( n+ 1)( n+ 2) ⋅⋅⋅ ( n+n)]=nn 1 n[ n ]1 n⎛Gn⎞ 1 ⎡⎛ 1⎞⎛ 2⎞ ⎛ n ⎞⎤ln ⎜ ln 1 1 1n⎟= n⎢⎜ + ⎟⎜ + ⎟… ⎜ + ⎟n n n⎥⎝ ⎠ ⎣⎝ ⎠⎝ ⎠ ⎝ ⎠⎦1⎡⎛ 1⎞ ⎛ 2⎞ ⎛ln 1 ln 1 ln 1 n ⎞⎤=n⎢ ⎜ + ⎟+ ⎜ + ⎟+⋅⋅⋅+ ⎜ + ⎟n n n⎥⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦⎛Gn⎞ 2lim ln ⎜ = ln xdx = 2ln 2 –1→∞ n⎟ ∫⎝ ⎠1n⎛Gn⎞ 2ln2–1 –1 4lim ⎜ = e = 4e=→∞ n⎟⎝ ⎠en⎡⎛ 1⎞⎛ 2⎞ ⎛ n ⎞⎤= ⎢⎜1+ ⎟⎜1+ ⎟… ⎜1+⎟n n n⎥⎣⎝ ⎠⎝ ⎠ ⎝ ⎠⎦1/ n430 Section 7.2 Instructor Solutions Manual

77. The proof fails to consider the constants when integrating 1 t.1 tWe know that any two of theseThe symbol ∫ ( 1 t)dtis a family of functions, all of who whom have derivative .functions will differ by a constant, so it is perfectly correct (notationally) to write ∫( t) dt ∫ ( )1 = 1 t dt+178.d 5 5 5[ e x ( C 1cos7 x + C 2sin 7 x ) + C 3] = 5 e x ( C 1cos7 x + C 2sin 7 x ) + e x (–7 C 1sin 7 x + 7 C 2cos7 x )dx5x= e [(5C + 7 C )cos 7 x+(5 C – 7 C )sin 7 x]1 2 2 1Thus, 5C1+ 7C2= 4 and 5 C2 –7C 1 = 6.11 29Solving, C1 = – ; C2=37 3779. u = f(x) dv = dxdu = f ′( x)dx v = xbb∫ = [ ]a ∫af ( xdx ) xf ( x ) – xf ′( xdx )Starting with the same integral,u = f(x)dv = dxdu = f ′( x)dx v = x – abb( ) ( – ) ( ) – ( – ) ( )af x dx x a f x ax a f ′ x dxb∫ = [ ]a ∫ba80. u = f′( x)dv = dxdu = f ′′( x)dx v = x – abb bf ( b)– f( a) = ∫ f′( x)dx = [( – ) ( )]– ( – ) ( )a′ x a ∫ a′′ x dxb= f ′( b)( b– a)– ∫ ( x– a) f′′( x)dxaStarting with the same integral,u = f′( x)dv = dxdu = f ′′( x)dx v = x – bf ( b bb) − f ( a ) = ( ) ( – ) ( ) – ( – ) ( )af x dx = x b f x ax b f x dxbf a b a x b f x dxab∫′ [ ′ ]′′a ∫ = ′( )( − )– ∫ ( – ) ′′( )81. Use proof by induction.n = 1: f ( a tt t) + f ′( a )( t – a ) + ∫ ( – ) ( ) ( ) ( )( – ) [ ( )( – )] a ( )at x f ′′ x dx = f a + f ′ a t a + f ′ x t x + ∫ af ′ x dx= f ( a) + f′ ( a)( t– a)– f′( a)( t– a) + [ f( x)] t a = f( t)Thus, the statement is true for n = 1. Note that integration by parts was used with u = (t – x), dv = f ′′ ( x) dx.Suppose the statement is true for n.n () inf ( a) i t ( t – x)( n 1)f () t = f( a) + +∑ ( t – a) f ( x)dx1! ai=i+ ∫n!tn( t – x)( n+1)Integrate ∫ f ( xdx ) by parts.a n!( 1) nn+( t – x)u = f ( x)dv = dxn!( 2) n+1n+( t – x)du = f ( x)v = – ( n + 1)!n 1t⎡ n+ ⎤n 1( n 1) ( n 1) t++ + ( n+2)= ⎢⎥ +a⎢⎣+ ⎥⎦+at ( t – x) ( t – x) ( t – x)f ( xdx ) – f ( x) f ( xdx )a n! ( n 1)! ( n 1)!∫ ∫Instructor's Resource Manual Section 7.2 431

n+ 1 1( 1) tn+n+ ( n+2)∫ a( t – a) ( t – x)= f ( a) +f ( x)dx( n+ 1)! ( n+1)!Thusn () i n+ 1 1( ) ( – ) ( 1) tn+f a i t a n+ ( t – x)( n+2)∑1! ( 1)! ai=i n+ ∫( n+1)!() i( ) i tn+1( – ) ( n+2)af () t = f( a) + ( t – a) + f ( a) +f ( x)dxn+1f a t x= f ( a) + ∑ ( t – a) +f ( x)dxi! ∫( n+1)!i=1Thus, the statement is true for n + 1.82. a.b.1 α −1 β−10B( α, β) = ∫ x (1 −x)dx where α ≥1, β ≥ 1x = 1 – u, dx = –du1 1 1 0 1 1x α −(1 x) β −dx (1 u) α −( u) β −1 1 1∫ − = − ( −du)0 ∫10 (1 u) α−u β−= ∫ − du = B(β , α )Thus, B(α, β) = B(β, α).1 α −1 β−10∫B( α, β) = x (1 −x)dxu = x α −1( α 1)α −2du = − x dxβ − 1dv = (1 − x)dx1 v (1 x)β=− −β1⎡ 1 α−1 β ⎤ α −1 1 α−2 β α −11 α−2βB( α, β) = ⎢− x (1 − x) x (1 x) dx x (1 x)dxβ⎥ +0 00β∫ − =β∫ −⎣⎦α −1 = B ( α − 1, β + 1)Similarly,β1 1 1B( , ) x α − β−α β = ∫ (1 −x)dx01u = (1 − x)β −du =− −1(1 − x)dxα −1dv = x dx2( β )β −1v = x αα1⎡1 1 1 12B( , ) x α (1 x) β − ⎤ β −α βx α (1 x)β − β −1 1 α β−2β −1= ⎢ − + − dxα⎥⎣ ⎦ 00 α∫= x (1 x) dx B( α 1, β 1)α∫ − = + −0α(*)c. Assume that n ≤ m. Using part (b) n times,n −1( n−1( ) n−2)Bn ( , m) = Bn ( − 1, m+ 1) = Bn ( − 2, m+2)mm( m+1)( n ) n ( n )( )−1( −2) − 3…⋅21⋅=…= B(1, m+ n−1).mm ( + 1) m+ 2 …( m+ n−2)1 2 1 1 1 1B(1, m+ n− 1) = (1 xm nm n)+ −dx [(1 x) + −∫ − =− − ]0m n 1 0 =+ − m+ n−1( n−1 )( n−2) ( n− 3)…⋅2⋅1 ( n−1 )!( m−1 )! ( n−1 )!( m−1 )!Thus, Bn ( , m)= = =mm ( + 1) m+ 2 …( m+ n− 2) m+ n− 1 ( m+ n− 1)! ( n+ m−1)!If n > m, thenBn ( , m) = Bm ( , n)=( ) ( )( n−1! ) ( m−1!)( n+ m−1)!by the above reasoning.432 Section 7.2 Instructor Solutions Manual

83. u = f(t) dv = f ′′()t dtdu = f ′()t dtv = f′() tbb2∫′′() () = [ () ′()] – [ ′()]a ∫af t f t dt f t f t f t dtb 2= f ( b) f′ ( b) − f( a) f′ ( a) −∫ [ f′( t)]dt =− b 2[ f ′ ( t )] dta ∫ a2 2[ f′ ( t)] ≥0, so − b[ f′( t)] ≤0aba∫ .84.x⎛tf ( zdz )⎞∫ ⎜ ⎟ dt0 ⎝∫0 ⎠tu f( z)dz0= ∫ dv = dtdu = f(t)dtv = txx⎛0 ⎜ t t xf ( zdz )⎞dt=⎡t f( zdz )⎤– tf( tdt )⎝ 0 ⎟ ⎠ ⎣ ⎢ 0 ⎦ ⎥ 00∫ ∫ ∫ ∫ = ∫ x f z dz0 ∫ 0By letting z = t,xxx f( z) dz = x f( t) dt,0 0∫ ∫ sox⎛t x xf ( zdz )⎞⎜ ⎟dt=xf( tdt ) – tf( tdt ) =0 ⎝ 0 ⎠ 0 0 ∫xx∫ ∫ ∫ ∫ 0 ( – ) ( )x( ) – t f( t)dtt f t dtx85. Letx t1 tn−1I = ⋅⋅⋅ f ( t ) ...0 0 0n dtndt2dt1∫∫ ∫ be the iterated integral. Note that for i ≥ 2, the limits of integration of theintegral with respect to t i are 0 to ti− 1 so that any change of variables in an outer integral affects the limits, andhence the variables in all interior integrals. We use induction on n, noting that the case n = 2 is solved in theprevious problem.Assume we know the formula for n − 1, and we want to show it for n.x t1 t2 tn−1 t1 t2 tn−2I = ∫∫∫ ⋅⋅⋅ ( ) ...0 0 0 ∫ f t0n dtn dt3dt2dt1 = ∫∫ ⋅⋅⋅ (0 0 ∫ F t0n−1) dtn−1...dt3dt2dt1−1where ( ) ( )n 1t nF t − = ∫ f t dn.By induction,1 xn−2I = F( t )( x−t ) dtn − 2!∫ 0( )0n1 1 11= ( ) =∫ ( ) , dv = ( x − t )u F t f t dt1t0nnn−21= ( ) , v =− ( x−t)du f t dt1 11n −1n−1111 ⎧t = x⎪⎡1 n−1 t1⎤ 1 xn−1⎫⎪I = ⎨ − ( x− t1) f ( t ) ( 1)( 1)1( 2! ) 1 0n dtn+ f t x−t dt ⎬n− ⎢ n− ∫ ⎥0t1= 0 n−1∫.⎪⎩⎣⎦⎪⎭1 xn−1= f( t01)( x−t1)dt1( n −1)!∫(note: that the quantity in square brackets equals 0 when evaluated at the given limits)Instructor’s Resource Manual Section 7.3 433

86. Proof by induction.n = 1:xu = P1( x)dv = e dx1dudP ( x)x= dx v = edxx x x dP1( x)x dP1( x)x x x dP1( x)∫e P1( x) dx = e P1( x)–∫e dx = e P1( x)–e dxdxdx∫ = e P1( x)–edxdPNote that1( x)is a constant.dxSuppose the formula is true for n. By using integration by parts with u = P 1( x)and dv = e x dx,x x x dPn+ 1( x)e Pn+ 1( x) dx = e Pn+1( x)–e dxdxdPn+ 1 ( x )dx∫ ∫Note thatis a polynomial of degree n, so⎡ n j11( )n j+x x x j d ⎛dPn+ x ⎞⎤x x j d Pn+ 1 x∫ e Pn+ 1( x) dx = e Pn+ 1( x) −⎢e∑( − 1) e Pn1( x) e ( 1)j ⎜ ⎥ + j 1j 0 dx dx⎟ = − ∑ −⎢+⎣ = ⎝ ⎠⎥⎦j=0 dxn+1 jx x j d Pn1( x)n+1 j+ x j d Pn+ 1( x)= e Pn+ 1( x) + e∑( −1)= e ( 1)j ∑ −jj=1 dxj=0 dxn+( )87.∫4 j 4 24 2 x x j d (3x + 2 x )(3x + 2 x ) e dx = e ∑ (–1)jj=0 dxx 4 2 3 2= e [3x + 2 x –12 x – 4x+ 36x + 4 – 72x+72]x 4 3 2= e (3 x –12x + 38 x – 76x+76)7.3 Concepts Review1.2.3.∫∫∫1+cos2x dx22(1 – sin x) cos xdx2 2sin x(1– sin x)cosxdx124. cos mx cos nx = [ cos( m + n) x + cos( m − n)x]Problem Set 7.32 1–cos2x1. ∫sinx dx = ∫ dx21 1= dx – cos2x dx2∫ 2∫1 1= x – sin2x+C2 42. u = 6x, du = 6 dx1∫sin 6x dx = sin6∫4 4u du21 ⎛1–cos2u⎞= du6∫⎜ ⎟⎝ 2 ⎠12= (1– 2cos 2u+cos 2 u)du24∫1 1 1= – 2cos 2 (1 cos 4 )24∫du u du + + u du24∫48∫3 1 1= – 2cos2 4cos448∫du u du +u du24∫192∫3 1 1= (6 x) – sin12x+ sin 24x+C48 24 1923 1 1= x – sin12x+ sin24x+C8 24 192434 Section 7.2 Instructor Solutions Manual

3.4.∫ ∫= ∫ −∫3 2sin x dx = sin x(1 −cos x)dxsin x dx sin x cos x dx1 3=− cos x + cos x+C33∫ cos x dx == ∫ x −= ∫ −∫2cos (1 sin xdx )2cos x dx cos xsinx dx1 3= sin x − sin x+C325.π/2 5 π/22 2∫ cos θ dθ = (1– sin θ) cosθ dθ0 ∫ 0π /2 2 4= ∫ (1 – 2 sin + sin ) cos d0π /2⎡ 2 3 1 5 ⎤= ⎢sin θ – sin θ + sin θ3 5⎥⎣⎦0⎛ 2 1⎞8= ⎜1– + ⎟–0=⎝ 3 5⎠15θ θ θ θ6.7.3π/2 6 π/2⎛1–cos2θ⎞∫ sin θ dθ = dθ0 ∫ 0 ⎜ ⎟⎝ 2 ⎠1 π /2 2 3= (1– 3cos 2θ 3cos 2 θ – cos 2 θ)dθ8∫+01 /2 3 /2 3 /2 2 1 /2 3= – 2cos 2 cos 2 – cos 28∫ π dθ 0 16∫ π θ dθ +π θ πθ dθ0 8∫0 8∫01 π/2 3 π /2 3 π/2 ⎛1+cos4θ⎞ 1 π/22= [ θ ] 0 – [sin2 θ] 0 + dθ – (1–sin 2 θ)cos2θ dθ8 16 8 ∫ 0 ⎜ ⎟⎝ 2 ⎠ 8∫ 01 π 3 /2 3 /2 1 /2 1 /2 2= ⋅ + 4cos4 – 2cos2 sin 2 2cos28 2 16∫ π d + π d π d πd0 64∫0 16∫ + ⋅0 16∫ 0θ θ θ θ θ θ θ θπ 3 π= + + 3 [sin 4 θ] – 1 [sin 2 θ] +1 [sin 2 θ]16 32 64 16 48∫π/2 π/2 3 π/20 0 0∫5π=325 2 2 2 22 4 2sin 4x cos 4 xdx=(1– cos 4 x) cos 4xsin 4xdx= (1– 2cos 4x + cos 4 x)cos 4xsin 4xdx1 2 4 61 3 1 5 1 7= – (cos 4 x – 2cos 4x+cos 4 x)(–4sin 4 x)dx4∫= – cos 4x + cos 4 x– cos 4x+C12 10 28∫8.∫∫3 2 1/2(sin 2) t cos2 tdt = (1–cos 2)(cos2) t t sin2tdt1 3/2 1 7/2= – (cos 2 t) + (cos 2 t)+ C3 71 1/2 5/2= – [(cos 2 t) – (cos 2 t) ](–2sin 2 t)dt2∫9.10.3 –2 2 –21 2∫cos 3θ sin 3 θ dθ = ∫(1– sin 3 θ)sin 3θ cos3θ dθ(sin−= 3 θ − 1)3cos3 θ dθ3∫1 1= – csc3 θ – sin 3θ+ C3 3∫ ∫1/2 3 2 1/2sin 2zcos 2 zdz = (1– sin 2 z)sin 2zcos 2zdz1 (sin1/2 2 – sin5/21z 2 z )2cos 23/2 1 7/2= zdz2∫= sin 2 z– sin 2z+C3 7Instructor’s Resource Manual Section 7.3 435

11.12.2 24 4 ⎛1–cos6t⎞ ⎛1+cos6t⎞sin 3t cos 3t dt = ⎜ ⎟ ⎜ ⎟ dt2 2∫ ∫ ⎝ ⎠ ⎝ ⎠1 (1– 2cos 2 6 cos 4= t+6 t ) dt16∫1 12= – cos12 tdt+ (1 + 2cos12t+cos 12 t)dt1 ⎡12 ⎤= 1– (1 cos12 t) (1 cos12 t)dt16∫ ⎢ + + +4⎥⎣⎦ 16∫64∫1 1 1 1= – 12cos12 12cos12 (1 cos 24 )192∫ tdt+ dt tdt t dt64∫ + + +384∫128∫1 1 1 1 13 1 1= – sin12t+ t+ sin12t+ t+ sin 24t+C = t – sin12t+ sin24t+C192 64 384 128 3072128 384 307236 2 ⎛1+cos 2θ⎞ ⎛1– cos 2θ⎞ 1 3 4∫cosθ sin θ dθ = ∫ ⎜ ⎟ ⎜ ⎟ dθ= (1 + 2cos2 θ –2cos 2 θ –cos 2 θ ) dθ⎝ 2 ⎠ ⎝ 2 ⎠ 16∫1 1 1 2 12= 2cos2 – (1–sin 2 )cos2 – (1 cos4 )16∫dθ + θ dθ θ θ dθ θ dθ16∫ +8∫64∫1 1 1 1 2 12= 2cos 2 – 2cos 2 2sin 2 cos 2 – (1 2cos 4 cos 4 )16∫d + d d d d16∫ + + +16∫16∫64∫θ θ θ θ θ θ θ θ θ θ θ1 1 21 1 1= sin 2 2cos 2 – – 4cos 4 – (1 cos8 )16∫dθ + θ θ dθ dθ θ dθ θ dθ16∫ ⋅ +64∫128∫128∫1 1 3 1 1 1 1= θ + sin 2θ − θ − sin 4θ − θ − sin8θ+ C16 48 64 128 128 10245 1 3 1 1= θ + sin 2 θ – sin 4 θ – sin8θ+ C128 48 128 10241 1sin 4y cos5y dy = sin 9y + sin( − y) dy = (sin 9y −sin y)dy2 21⎛1 ⎞ 1 1= ⎜− cos9y + cos y⎟+ C = cos y− cos9y+C2⎝9 ⎠ 2 1813. ∫ ∫ [ ] ∫14.∫1cos y cos 4 y dy = [cos5y + cos( −3 y)]dy2∫1 1= sin 5y− sin( − 3 y)+ C10 61 1= sin 5y+ sin 3y+C10 615.24⎛ 2 1– cos 1 cossinw ⎞ ⎛cosw ⎞ ⎛dww ⎞ ⎛ + w ⎞ 1 2 3∫ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟dw⎝ 2 ⎠ ⎝ 2∫= (1 – cos w – cos w+cos w ) dw⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 8∫1 ⎡ 12 ⎤ 1 ⎡1 12 ⎤= 1– cos w– (1 + cos 2 w) + (1– sin w) cos w dw8∫⎢ 2⎥ = – cos 2 w– sin wcosw dw⎣⎦ 8∫⎢2 2⎥⎣⎦1 1 1 3= w– sin2 w– sin w+C16 32 241t t dt t t dt21=− ( cos 4 cos 2 )2∫ tdt −∫tdt1⎛1 1 ⎞=− ⎜ sin 4t− sin 2t⎟+C2⎝4 2 ⎠1 1=− sin4t+ sin2t+C8 416. ∫sin 3 sin = ∫− [ cos 4 −cos 2 ]436 Section 7.3 Instructor's Resource Manual

17.18.∫2xcosxsinxdxu = x du = 1dx2dv = cos xsinx dx2 1 3v =−∫(cos x) ( − sin x)dx =− cos xt=cos x3Thus∫2xcosxsinxdx =1 3 1 3x( − cos x) − (1)( cos )3∫ − x dx =31⎡ 3 3− xcosx+ cos xdx⎤=3 ⎣ ∫ ⎦1⎡ 3 2− x cos x+ cos x(1 − sin x)dx⎤=3 ⎣ ∫⎦1 ⎡ 3 2 ⎤⎢− x cos x+ (cos cos sin )3∫ x− x x dx⎥=⎣t=sin x ⎦1⎡3 1 3 ⎤− xcos x+ sin x− sin x + C3⎢3 ⎥⎣⎦∫3xsinxcosxdxu = x du = 1dx3dv = sin x cos x dx3 1 4v = ∫ (sin x) (cos x)dx = sin xt=sin x 4Thus∫3xsinxcosxdx =1 4 1 4x( sin x) − (1)( sin )4∫ x dx =41⎡ 4 2 2xsin x− (sin x)dx⎤=4 ⎣ ∫ ⎦1⎡4 12 ⎤xsin x− (1 − cos 2 x)dx =4⎢4∫⎥⎣⎦1⎡4 12 ⎤sin (1 2cos 2 cos 2 )4⎢x x− 4∫− x+ x dx⎥=⎣⎦1⎡4 1 1 1⎤sin sin 2 (1 cos 4 )4⎢x x− x+ x− x dx4 4 8∫+ ⎥ =⎣⎦1⎡4 3 1 1 ⎤xsin x x sin 2x sin 4x C4⎢− + − +8 4 32 ⎥⎣⎦4 2 219. ∫tan xdx=∫( tan x)( tan x)dx2 2= ∫ ( tan x)(sec x−1)dx2 2 2= ( tan sec −tan)∫∫x x x dx2 2 2= tan xsec xdx− (sec x−1)dx1 tan3= x− tan x+ x+C34 2 220. ∫cot = ∫( cot )( cot )2 2= ∫ ( cot x)(csc x−1)dx= ( cot 2 xcsc 2 x−cot2 x)dx∫∫x dx x x dx2 2 2= cot x csc xdx− (csc x−1)dx1 cot3=− x+ cot x+ x+C321. tan 3 2x = ∫ ( tan x)( tan x)dx2= ( tan )( sec −1)∫∫x x dx1 tan2 ln cos= x + x + C23 222. ∫cot 2 = ∫( cot 2 )( cot 2 )2= ( cot 2t)( csc 2t−1)dt∫∫tdt t t dt2∫= cot 2tcsc 2tdt−cot 2tdt1 2 1=− cot 2t− ln sin 2t + C4 2∫Instructor’s Resource Manual Section 7.3 437

23.5 ⎛θ⎞∫ tan ⎜ ⎟2 d θ⎝ ⎠⎛θ⎞ dθu = ⎜ ⎟;du =⎝ 2⎠25⎛θ⎞5∫tan ⎜ ⎟dθ= 2 tan u du⎝2∫⎠∫3 2( )( )= 2 tan u sec u−1du∫∫3 2 3= 2 tan usec u du−2 tan u du3 2 2( )= 2 tan u sec u du −2 tanu sec u −1du∫∫3 2 2∫ ∫ ∫1 tan4⎛θ ⎞ tan2⎛θ ⎞ θ 2ln cos C= 2 tan usec u du− 2 tanusec u du+2 tan u du= ⎜ ⎟− ⎜ ⎟− +2 ⎝ 2⎠ ⎝2⎠224.∫5cot 2 tdtu = 2; t du = 2dt∫5 1 5cot 2 tdt=cot2∫ udu1 3 2 1 3 2= ( cot )( cot ) ( cot )( csc 1)2∫u u du =2∫u − du1 3 2 1 3= ( cot )( csc ) cot2∫u u du−2∫u du1 3 2 12= ( cot )( csc ) ( cot )( csc 1 )2∫u u du −2∫u u − du1 3 2 1 2 1= ( cot )( csc ) ( cot )( csc ) cot2∫ u u du −2∫ u u du + u2∫1 4 1 2 1=− cot u+ cot u+ ln sin u + C8 4 21 4 1 2 1=− cot 2t+ cot 2t+ ln sin 2t + C8 4 2−3 4 −3 2 225. ∫tan xsec xdx = ∫( tan x)( sec x)( sec x)dx−3 2 2= ( tan )( 1+tan )( sec )∫∫x x x dx−3 2 −121 tan−2ln tan∫( )= tan xsec x dx + tan x sec x dx=− x+ x + C2−3/2 4 −3/2 2 226. ∫tan xsec x dx = ∫( tan x)( sec x)( sec x)−3/2 2 2= ( tan )( 1+tan )( sec )∫∫x x x dx−3/2 2 1/2 2= tan xsec xdx+tan xsecxdx∫−1/2 2 3/2=− 2tan x+ tan x+C3438 Section 7.3 Instructor’s Resource Manual

27.28.29.3 2∫ tan xsecxdx2Let u = tan x . Then du = sec x dx .3 2 3 1 4 1 4tan xsec xdx= u du= u + C= tan x+C4 4∫ ∫∫∫∫∫3 −1/2 2 −3/2tan xsec x dx = tan xsec x(sec xtan x)dx2 −3/2( sec 1)( sec )( sec tan )= x−x x x dx−( ) ∫ ( )1/2 3/2= sec x sec xtan x dx−sec x sec xtanx dx2 sec3/2 2sec−1/2= x+ x+C3π1 π1⎡1 1⎤∫ cos mx cos nx dx = (cos[( m + n) x] + cos[( m −n) x])dx– π2∫= sin[( m n) x] sin[( m n) x]– π2⎢+ + −m+ n m−n⎥⎣⎦= 0 for m ≠ n, since sin kπ = 0 for all integers k.π−ππ x π30. If we let u = then du = dx . Making the substitution and changing the limits as necessary, we getL LL mπx nπx L π∫ cos cos dx = cos mu cos nu du 0−LL L π∫ = (See Problem 29−π31.32.2∫ π ππ ( x +2 2sin x)dx =π ( 2 sin sin )0∫0 x + x x + x dxπ 2x dx 2 π π=π ∫ + π xsin x dx π(1 cos2 x)dx0 ∫ + −0 2∫ 0πππ π2 [ sin cos ] sin 200 0⎡1 3 ⎤ ⎡ 1 ⎤=π ⎢ x x x x x x3⎥ + π − + −2⎢2⎥⎣ ⎦ ⎣ ⎦Use Formula 40 with u = x for0π /2 2 2V 2 xsin ( x ) dx= π∫∫xsinxdxu = x2 , du = 2x dxπ /2π/2 /222π 1–cos2u⎡1 1 ⎤ πV =π ∫ sin udu =π du u– sin 2u2.46740 ∫=π0 2⎢= ≈2 4⎥⎣⎦041 4π1 4 5 2= π + 2 π (0+π− 0) + ( π−0−0)= π + π ≈ 57.14373 23 233. a.1 ππ ∫ 1 π ⎛ N ⎞N1 πf ( x)sin( mx)dx = a sin( ) sin( )−πn nx mx dxπ∫−π⎜∑= a⎟n sin( nx)sin( mx)dx⎝n=1 ⎠π∑ ∫ −πn=1From Example 6,π⎧0 if n≠m∫ sin( nx)sin( mx)dx = ⎨ so every term in the sum is 0 except for when n = m.−π⎩ π if n = mIf m > N, there is no term where n = m, while if m ≤ N, then n = m occurs. When n = mπan∫ sin( nx)sin( mx)dx = amπso when m ≤ N,−π1 π1f ( x)sin( mx) dx = ⋅am⋅π=a−πmπ∫ .πInstructor’s Resource Manual Section 7.3 439

.1 π 2f ( xdx )π ∫ 1 π ⎛ N ⎞⎛ N⎞ N N1= a sin( ) sin( )−πn nx ammx dxπ∫−π⎜∑⎟⎜∑⎟a n a π= m sin( nx )sin( mx ) dx⎝n= 1 ⎠⎝m=1 ⎠ π∑ ∑ ∫ −πn= 1 m=1From Example 6, the integral is 0 except when m = n. When m = n, we obtain1NN2an( anπ ) = ann= 1 n=1π ∑ ∑ .34. a. Proof by inductionx xn = 1: cos = cos2 2Assume true for k ≤ n.x x x x ⎡⎤ xcos cos cos cos cos cos cos cos2 4 2 2 ⎢⎣2 2 2 ⎥⎦2 2Note that⎛ k ⎞⎛ 1 ⎞ 1⎡ 2k 1 2 k –1cos x cos x cos +⎤⎜ cos ,n ⎟⎜ x xn+ 1 ⎟= 2⎢ +n+ 1 n+1 ⎥ so⎝ 2 ⎠⎝ 2 ⎠ ⎣ 2 2 ⎦n1 3 2 –1 1 ⋅ = ⎢ x+ x+ + x⎥n n+ 1 n n n n–1 n+1⎡ n11 3 2 –1 ⎤ n+⎛ 1 ⎞ 1 ⎡ 1 3 2 –1 ⎤ 1⎢cos x+ cos x+ + cos x⎥ cos x cos x cos x cos xn n n ⎜ n+ 1 ⎟ = ⎢ + + +⎥n–1 n+ 1 n+ 1 n+1 n⎢⎣ 2 2 2 ⎥⎦⎝2 ⎠2 ⎢⎣ 2 2 2 ⎥⎦2b.⎡ 1 3 2 –1 ⎤ 1 1 ⎡ 1 3 2 –1 ⎤lim ⎢cos cos cos ⎥ lim ⎢cos cos cos ⎥n→∞⎢⎣ 2 2 2 ⎥⎦2 x ⎢⎣ 2 2 2 ⎥⎦21 x= costdtx∫ 0nnxx+ x+ + x = x+ x+ + xn n n n–1 n→∞n n n n–11 1 sinc. cos tdt [sin t]00x∫ = =x xx x xx 1+cosx35. Using the half-angle identity cos = , we see that since2 2π2 2πcos = cos =4 2 2π 2π 1+2 2 2+2cos = cos = = ,8 4 2 2π 2+2π 1+2 2 2+ 2+2cos cos ,16 8 2 2= = = etc.2 2+ 2 2+ 2+2 ⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞2 2 2Thus, ⋅ ⋅ = cos cos cos2 2 2⎜ 2⎟ ⎜ 4⎟ ⎜ 8⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠π( )⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞ sin2 2 2 2 2= lim cos cos cosn→∞2 4 2 n= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2440 Section 7.3 Instructor’s Resource Manual

2 2π2 π 2 2∫ π( − sin ) = π ( − 2 sin + sin )0 ∫ 02k π dx 2k π sin xdx πππ2 π π π ⎡ 1 ⎤=π ∫ − π + (1−cos2 x)dx0 ∫0 2∫ =π k [ x] + 2kπ [ cosx]+ x sin2x00 0 2⎢ −2⎥⎣ ⎦022 2 2 2k 2 k ( 1 1) ππ=π + π− − + ( π− 0) =π k −4kπ+2 222 2 π22Let f( k) =π k −4 kπ+ , then f′ ( k) = 2π k −4π and f′ ( k) = 0 when k = .2π36. Since ( k − sin x) = (sin x− k) , the volume of S isThe critical points of f(k) on 0 ≤ k ≤ 1 are 0, 2 ,π 1.2 2 2π ⎛2⎞ π 2 πf(0) = ≈ 4.93, f ⎜ ⎟= 4 − 8 + ≈ 0.93, f(1) = π −4π+ ≈ 2.242 ⎝π⎠ 2 22a. S has minimum volume when k = . πb. S has maximum volume when k = 0.k x k k x x dx7.4 Concepts Review1. x – 32. 2 sin t3. 2 tan t4. 2 sec tProblem Set 7.44.24, 4,u = x+ u = x+ 2u du = dx2 2 2 2x + 3 x ( u –4) + 3( u –4)∫ dx = ∫2 u dux + 4u4 2 2 5 10 3= 2 ∫ ( u –5u + 4) du = – + 8 +5 3u u u C2 5/2 10 3/2 1/2= ( x + 4) – ( x+ 4) + 8( x+ 4) + C5 31.2= + 1, = + 1, 2 =u x u x udu dx∫ ∫x x+ 1 dx = ( u –1) u(2 udu)4 2= ∫ (2 – 2 )u u du22 5 2 3= u – u + C5 32 5/2 2 3/2= ( + 1) – ( + 1) +5 3x x C5.2, ,u = t u = t 2u du = dt2 dt 2 2u du 2u + e −e ∫ = 2du1 ∫ =t e 1 u+ e∫+1 u+e2 2 e= 2 ∫ du –2 du1 ∫1u+e22= 2[ u] 1 – 2e⎡ ⎣lnu+ e⎤⎦12.3.3 3 2= +π , = +π , 3 =u x u x u du dx33 2∫x x+π dx = ∫ ( u – π) u(3 u du)6 3 3 7 3π4= ∫ (3 u – 3 πu ) du = – +7 43 7/3 3π4/3= ( x +π ) – ( x+π ) + C7 42u 3t 4, u 3t4,u u C= + = + 2u du = 3 dt1 2 23 ( u − 4) udu3 2 ( 2u –4)tdt= =3t+ 4 u 92 3 8= u – u+C27 92 3/2 8 1/2= (3t+ 4) – (3t+ 4) + C27 9∫ ∫ ∫du6.= 2( 2 –1) – 2 e[ln( 2 + e) – ln(1 + e)]⎛ 2 + e ⎞= 2 2 –2–2eln⎜1+e ⎟⎝ ⎠2, ,u = t u = t 2u du = dttu1 1∫ dt = (2 u du)0t+ 1∫ 0 2u + 11212u u + 1−1= 2∫ du = 2 du0 2 ∫ 0 2u + 1 u + 11 1 11 –1 1= 2 ∫ du –2 du0 ∫ = 2[ u] 0 20 – 2[tan u]0u + 1–1 π= 2–2tan 1= 2– ≈ 0.42922Instructor’s Resource Manual Section 7.4 441

7.8.1/2 2= (3 + 2) , = 3 + 2, 2 = 3u t u t udu dt3/2 1 2 3⎛2⎞∫t(3t+ 2) dt = ∫ ( u – 2) u ⎜ udu⎟3 ⎝3⎠2 6 4 2 7 4 5= ( –2 )9∫u u du = u − u + C63 452 7/2 4 5/2= (3t+ 2) – (3t+ 2) + C63 451/3 3 2= (1 – ) , = 1 – , 3 = –u x u x u du dx∫ ∫2/3 3 2 2x(1 – x) dx = (1 – u ) u (–3 u ) du7 4 3 8 3 5= 3 ∫ ( u – u ) du = u − u + C8 53 8/3 3 5/3= (1 – x) – (1 – x)+ C8 59. x = 2 sin t, dx = 2 cos t dt24– x 2cos t∫ dx = (2cos t dt )x∫2sint21–sin t= 2∫ dt = 2 csc tdt– 2 sinsin t∫ ∫= 2ln csct− cot t + 2cost+C22−4– x2= 2ln + 4– +xxCtdt10. x = 4sin tdx , = 4costdt2 2x dx sin t cost ∫ = 162∫ dt16 – x cost2= 16∫ sin tdt = 8 ∫ (1–cos2 tdt )= 8 t – 4sin 2t+ C = 8t− 8sin tcost+C11.2–1 ⎛ x⎞x 16 – x= 8sin ⎜ ⎟–+⎝4⎠22= 2 tan , = 2sec2dx 2sec t dt 1= = cos2 3/2 2 3/2( + 4) (4sec ) 4x tdx tdt∫ ∫ ∫x1= sin t+C4x= +24 + 4xtCCtdt12. t = sec x, dt = sec x tan x dxπNote that 0 ≤ x < .2t2 –1= tanx = tanx–13 dt sec (3) sec x tan x∫ =2 2 2∫dxπ /3 2t t –1 sec x tan x–1sec (3)= ∫ cos x dxπ /3–1sec (3) −1π= [sin x] π /3= sin[sec (3)] − sin 3⎡ −1 ⎛1⎞⎤3 2 2 3= sin ⎢cos ⎜ ⎟ − = – ≈0.07683⎥⎣ ⎝ ⎠⎦2 3 213. t = sec x, dt = sec x tan x dxπNote that < x ≤π .2t2 –1 = tan x = – tan x2–1–3 tsec (–3)=–2 3 2 π / 3 3–1 –tanx∫ dt ∫sec x tan x dxtsec x–1 –1sec (–3) 2 sec (–3) 1 1–sin x dx ⎛cos2 –2 /3 2 /3 2 x ⎞= ∫=π∫ π ⎜ ⎟⎝ 2⎠dx–1sec (–3)⎡1 1 ⎤= ⎢ sin 2 x–x4 2 ⎥⎣⎦2 π /3–1sec (–3)⎡1 1 ⎤= ⎢ sin xcos x–x2 2 ⎥⎣⎦2 π /32 1 –1 3 π= – – sec (–3) + + ≈ 0.1512529 2 8 314. t = sin x, dt = cos x dxtdt = sin x dx21– t∫ ∫ = –cos x + C2= – 1–t + C15. z = sin t, dz = cos t dt2 z –3∫ dz = (2sin t – 3) dt2∫1– z= –2 cos t – 3t + C2 –1= –2 1– – 3sin +z z C442 Section 7.4 Instructor’s Resource Manual

16. x = π tan t, dx =π sec t dtπx–12∫ dx = ( π tan t –1)sect dt2 2∫x +π2 tan tsec tdt – sectdt17.18.=π ∫ ∫=π2 sec t – ln sect+ tan t + C2 2 1 2 2 x=π x +π − ln x +π + + Cπππ πx−1∫ dx0 2 2x +ππ⎡2 2 ⎤2 2 x +π x= ⎢π x +π –ln + ⎥⎢π π ⎥⎣⎦02 2= [ 2 π –ln( 2+ 1)]–[ π − ln1]2= ( 2 –1) π –ln( 2+ 1) ≈ 3.2072 2 2+ 2 + 5= + 2 + 1+ 4 = ( + 1) + 4x x x x xu = x + 1, du = dxdx du∫ =2∫2x + 2x+ 5 u + 42u = 2 tan t, du = 2sec t dtdu∫ = sectdt ln sect tan t C2∫ = + +u + 42u + 4 u= ln + +2 2C2x + 2x+ 5+ x+1= ln+22= ln + 2 + 5 + + 1 +12Cx x x C2 2 2+ 4 + 5= + 4 + 4+ 1 = ( + 2) + 1x x x x xu = x + 2, du = dxdx du∫ =2∫2x + 4x+ 5 u + 12u = tan t,du = sec tdtdu∫ = sectdt ln sect tan t C2∫ = + +u + 1dx2∫= ln u + 1 + u + C2x + 4x+52= ln + 4 + 5 + + 2 +x x x C1119.20.21.22.2 2 2+ 2 + 5= + 2 + 1+ 4 = ( + 1) + 4x x x x xu = x + 1, du = dx3x3 u–3∫ dx =2∫ du2x + 2x+ 5 u + 4udu= 3 ∫ du –32∫2u + 4 u + 4(Use the result of Problem 17.)2 2= 3 u + 4 –3ln u + 4+ u + C2 2= 3 + 2 + 5 –3ln + 2 + 5+ + 1 +x x x x x C2 2 2+ 4 + 5= + 4 + 4+ 1 = ( + 2) + 1x x x x xu = x + 2, du = dx2 x–1 2u−5∫ dx =2∫ du2x + 4x+ 5 u + 12ududu= ∫ –52∫2u + 1 u + 1(Use the result of Problem 18.)2 2= 2 u + 1–5ln u + 1+ u + C2 2= 2 + 4 + 5 –5ln + 4 + 5+ + 2 +x x x x x C2 2 25 4 9 (4 4 ) 9 ( 2)− x− x = − + x+ x = − x+u = x + 2, du = dx∫ ∫2 25–4 x – x dx = 9– u duu = 3 sin t, du = 3 cos t dt2 2 9∫ 9− u du = 9∫cos tdt = (1+cos2 t)dt2∫9⎛1 ⎞ 9= ⎜t+ sin 2 t⎟+C ( sin cos )2⎝2 ⎠= 2 t+ t t + C9 –1 ⎛u⎞ 1 2= sin ⎜ ⎟+ u 9 – u + C2 ⎝3⎠29 –1 ⎛ x+ 2⎞x+22= sin ⎜ ⎟+ 5 – 4 x – x + C2 ⎝ 3 ⎠ 22 2 216 + 6 x– x = 25 −(9 − 6 x+ x ) = 25 – ( x– 3)u = x – 3, du = dxdxdu∫ =2∫216 + 6 x – x 25 – uu = 5 sin t, du = 5 cos tdu–1 ⎛u⎞∫ = dt = t + C2∫ = sin ⎜ ⎟+C25 − u⎝5⎠–1 ⎛ x –3⎞= sin ⎜ ⎟+C⎝ 5 ⎠Instructor’s Resource Manual Section 7.4 443

23.24.25.26.27.2 2 24 x– x = 4 −(4− 4 x+ x ) = 4–( x–2)u = x – 2, du = dxdx du∫ =2∫24 x – x 4– uu = 2 sin t, du = 2 cos t dtdu–1 ⎛u⎞∫ = dt = t + C2∫ = sin ⎜ ⎟+C4 − u⎝2⎠–1 ⎛ x –2⎞= sin ⎜ ⎟+C⎝ 2 ⎠2 2 24 x– x = 4 −(4− 4 x+ x ) = 4–( x–2)u = x – 2, du = dxxu+2∫ dx =2∫ du24 x– x 4– u– udu du= – ∫ + 22∫24– u 4– u(Use the result of Problem 23.)2 –1⎛u⎞= – 4– u + 2sin ⎜ ⎟+C⎝2⎠2 –1 ⎛ x –2⎞= – 4 x – x + 2sin ⎜ ⎟+C⎝ 2 ⎠2 2 2x + 2x+ 2= x + 2x+ 1+ 1 = ( x+ 1) + 1u = x + 1, du = dx2x+1 2 u–1∫ dx =2 ∫ du2x + 2x+ 2 u + 12udu= ∫ du –2 ∫ 2u + 1 u + 12 –1= ln u + 1 – tan u+C( )2 −1= ln x + 2x+ 2 − tan ( x+ 1) + C2 2 2x – 6x+ 18 = x –6x+ 9+ 9 = ( x–3) + 9u = x – 3, du = dx2 x–1 2u+5∫ dx =2 ∫ du2x – 6x+ 18 u + 92ududu= ∫ + 52 ∫ 2u + 9 u + 92 5 −1⎛u⎞= ln ( u + 9)+ tan ⎜ ⎟+C3 ⎝3⎠2 5 −1⎛x − 3⎞= ln ( x − 6x+ 18)+ tan ⎜ ⎟+C3 ⎝ 3 ⎠21⎛1 ⎞V =π∫dx0 ⎜ 2 ⎟⎝ x + 2x+5⎠21⎡1 ⎤=π∫⎢ ⎥ dx0 2⎢⎣( x + 1) + 4⎥⎦28.29. a.2x + 1 = 2 tan t, dx = 2sect dt2π /4 ⎛ 1 ⎞ 2–12sectan (1/ 2) 2∫V =π ⎜ ⎟ tdt⎝4sect ⎠π π /4 1 π π /4 2= –1 dt8∫ =tan (1/ 2) 2–1sec t 8∫ cos tdttan (1/ 2)π π /4 ⎛1 1 ⎞= –1 cos 2 t dt8∫tan (1/ 2) ⎜ + ⎟⎝2 2 ⎠π /4π ⎡1 1 ⎤= sin 28 ⎢ t+t2 4 ⎥⎣⎦ –1tan (1/ 2)π /4π ⎡1 1 ⎤= sin cos8 ⎢ t+t t2 2 ⎥⎣⎦ –1tan (1/ 2)π ⎡⎛π1⎞ ⎛1 –1 1 1⎞⎤= – tan8⎢⎜ + ⎟ ⎜ + ⎟8 4 2 2 5⎥⎣⎝ ⎠ ⎝ ⎠⎦π ⎛ 1 π –1 1⎞= ⎜ + – tan ⎟≈0.08281116 ⎝10 4 2 ⎠1 1V = 2π∫xdx0 2x + 2x+51 x= 2π∫dx0 2( x + 1) + 41 x + 1 1 1= 2 π∫ dx –2π0 2 ∫dx0 2( x+ 1) + 4 ( x+ 1) + 411⎡1 2 ⎤ ⎡ 1 –1⎛ x + 1⎞⎤= 2π ⎢ ln[( x + 1) + 4] –2 tan2 ⎥ π⎢⎜ ⎟0 2 2⎥⎣ ⎦ ⎣ ⎝ ⎠⎦0⎡ –1 –1 1 ⎤=π[ln 8 – ln 5] – π ⎢ tan 1– tan2 ⎥⎣⎦⎛ 8 π –1 1⎞=π ⎜ln – + tan ⎟≈0.465751⎝ 5 4 2⎠2u = x + 9, du = 2xdxxdx 1 du 1∫ = ln u C2x 9 2∫= ++ u 21 2 1 2= ln x + 9 + C = ln ( x + 9)+ C2 2b. x = 3 tan t, dx = 3sec2t dtxdx∫ = tan tdt2 ∫ = –ln cost + Cx + 93 ⎛ 3 ⎞= − ln + C1 = − ln+ C12 ⎜ 2 ⎟x + 9 ⎝ x + 9 ⎠2= ln⎛x 9⎞⎜ + ⎟− ln 3 + C1⎝ ⎠2 1/2 1 2= ln (( x + 9) ) + C = ln ( x + 9 ) + C2444 Section 7.4 Instructor’s Resource Manual

30.31. a.2 2 2u 9 x , u 9 x ,= + = + 2u du = 2x dx33323 22xdx x u= x dx =0 2 0 2 3∫ ∫ ∫9+ x 9+x3 2− 9uduu3 2 ⎡ 32 u ⎤= ∫ ( u − 9) du = ⎢ –9u⎥= 18–9 23⎢⎣3 ⎥⎦3≈ 5.2722 2 2u = 4– x , u = 4– x , 2u du = –2x dx2 2 24– x 4−∫ dx = x x dx – u dux∫ =2 ∫ 4–2x u2− 4+4−1= ∫ u du =− 4 ∫ du + ∫ du2 24−u4−u1 u + 2=−4⋅ ln + u+C4 u − 224− x + 2 2=− ln + 4 − x + C24−x−2b. x = 2 sin t, dx = 2 cos t dt2 24– x cos tdx = 2 dtxsin t∫ ∫(1 – sin t)= 2∫dtsin t2 csc tdt– 2 sin tdt2= ∫ ∫= 2ln csct− cot t + 2cost+C2 4−x2= 2ln − + 4− x + Cx x222−4−x2= 2ln + 4− x + CxTo reconcile the answers, note that2 24− x + 2 4−x−2− ln= ln2 24−x−2 4− x + 22 2( 4−x−2)= ln2 2( 4− x + 2)( 4−x−2)2 2 2 2(2 − 4 −x) (2 −lnln4 −x= =)2 24−x−4−x2⎛2 22− 4−x⎞2− 4−x= ln ⎜ ⎟ = 2ln⎜ x ⎟x⎝ ⎠32. The equation of the circle with center (–a, 0) is2 2 2( x + a) + y = b , so2 2y b – ( x a)=± + . Bysymmetry, the area of the overlap is four timesthe area of the region bounded by x = 0, y = 0,2 2and y = b –( x+ a)dx .b– a 2 2A = 4 ∫ b –( x+a)dx0x + a = b sin t, dx = b cos t dt/2 2 2A = 4∫π–1 b cos tdtsin ( a/ b)2 π /2= 2 b ∫ –1 (1+cos2 t)dtsin ( a/ b)π /22 ⎡ 1 ⎤= 2b ⎢t+sin2t2 ⎥⎣ ⎦ –1sin ( a/ b)2 /2= 2 b [ t+sintcos t]π–1sin ( a/ b)⎡2 2 ⎤2 π⎛–1 a a b – a⎞⎛ ⎞= 2 b ⎢ – ⎜sin+⎟⎥⎢⎜ ⎟2 ⎜ ⎝b⎠b b ⎟⎥⎣ ⎝⎠⎦2 2 –1⎛a⎞ 2 2=πb –2b sin ⎜ ⎟–2 a b – a⎝b⎠33. a. The coordinate of C is (0, –a). The lower arcof the lune lies on the circle given by the2 2 2equation x + ( y+ a) = 2aor2 2y 2 a – x – a.=± The upper arc of thelune lies on the circle given by the equation2 2 2x + y = a or2 2y =± a – x .A a 2 2 a 2 2= – – 2 – –– aa x dx ⎛– aa x a ⎞∫ ∫ ⎜ ⎟⎝⎠dxa 2 2 a 2 2 2= ∫ a – x dx– 2 a – x dx 2a– a ∫+– aa 2 2Note that ∫ a – x dx is the area of a– asemicircle with radius a, soa22 2 πa∫ a – x dx = .– a2aFor2 2∫ 2 a – x dx,let– ax = 2asin t, dx = 2acostdta 2 2 π /4 2 2∫ 2 a – x dx = 2a cos tdt– a ∫ – π /4π /42 π /42 ⎡ 1 ⎤= a ∫ (1 + cos 2 t) dt = a t sin 2t– π /4⎢ +2 ⎥⎣ ⎦– π /42πa2= + a22 2πa⎛πa⎞2 2 2A = – + a + 2a = a2 ⎜ 2 ⎟⎝ ⎠Thus, the area of the lune is equal to the areaof the square.Instructor’s Resource Manual Section 7.4 445

. Without using calculus, consider thefollowing labels on the figure.7.5 Concepts Review1. proper2.5x –1+ x + 13. a = 2; b = 3; c = –1Area of the lune = Area of the semicircle ofradius a at O + Area (ΔABC) – Area of thesector ABC.1 2 2 1⎛π⎞ 2A = π a + a – ⎜ ⎟( 2 a)2 2⎝2⎠1 2 2 1 2 2= π a + a – π a = a2 2Note that since BC has length 2 a , theπmeasure of angle OCB is , so the measure4πof angle ACB is .234. Using reasoning similar to Problem 33 b, the area is1 2 1 2 2 1⎛–1 a ⎞ 2π a + (2 a) b – a – ⎜2sin⎟b2 2 2⎝b ⎠1 2 2 2 2 –1 a= π a + a b – a – b sin .2b35.2 22 2dy a – x–= – ; y = – a x dxdx x∫xx = a sin t, dx = a cos t dt2acost cos ty = ∫– acos tdt = – a dtasint∫sin t21–sin t= – a∫ dt = a (sin t – csc t)dtsin t∫(–cos ln csc cot )= a t− t− t + C2 2 2 2– –cos t = a x , csc t = a , cot t =a xa x x⎛ 2 2 2 2 ⎞a – x a a − xy = a⎜– −ln− ⎟+C⎜ a x x ⎟⎝⎠2 22 2ln a −a x a a −=− − − x + CxSince y = 0 when x = a,0 = 0 – a ln 1 + C, so C = 0.2 22 2 –– ln a −y = a −x − aa xx4.A B Cx+D+ +x –1 2 2( x–1) x + 1Problem Set 7.511.A B= +x( x+ 1) x x+11 = A(x + 1) + BxA = 1, B = –11 1 1∫ dx = dx – dxxx ( + 1)∫x∫x+12.3.= ln x – ln x+ 1 + C2 2 A B2x + 3x = xx ( + 3) = x + x+32 = A(x + 3) + Bx2 2A= , B = –3 32 2 1 2 B∫ dx = dx – dx23 3∫ x x x 3∫+x+32 2= ln x – ln x+ 3 + C3 33 3 A B= = +2x –1 ( x+ 1)( x–1) x+1 x–13 = A(x – 1) + B(x + 1)3 3A= – , B =2 23 3 1 3 1∫ –2–1 dx = 2∫ dx +xx+dx1 2∫x –13 3= – ln x + 1 + ln x–1+ C2 2446 Section 7.5 Instructor’s Resource Manual

4.5.6.5x5x5= =3 2 22x + 6x 2 x ( x+3) 2 xx ( + 3)= A Bx+ x + 35= A( x+ 3) + Bx25 5A= , B = –6 65x5 1 5 1∫ dx – dx3 22x 6x = 6∫ x 6∫+x+35 5= ln x – ln x+ 3 + C6 6x −11 x−11A B= = +2x + 3x−4( x + 4)( x− 1) x+ 4 x−1x – 11 = A(x – 1) + B(x + 4)A =3, B = –2x −11 1 1∫ dx = 3 22 ∫ dx − dxx 3x4 x+ 4∫+ −x−1= 3ln x + 4 −2ln x− 1 + Cx –7 x–7A B= = +2x – x–12( x–4)( x+ 3) x–4 x+3x – 7 = A(x + 3) + B(x – 4)3 10A= – , B =7 7x –7 3 1 10 1∫ dx = –2– –12 7∫ dx + dxx xx –4 7∫x+33 10= – ln x –4 + ln x+ 3 + C7 77.8.9.10.3x −13 3x−13A B= = +2x + 3x−10( x+ 5)( x− 2) x+ 5 x−23x− 13 = A( x− 2) + B( x+5)A = 4, B = –13x−131 1∫dx = 42 ∫ dx − dxx + 3x−10x+ 5∫x−2= 4ln x + 5 −ln x− 2 + Cx+π x+πA B== +2 2x –3π x+ 2π ( x–2 π)( x– π)x– 2 π x–πx+ π = A( x− π) + B( x−2 π)A = 3, B = –2x +π3 2∫ dx = –2 2 ∫ dx dxx –3 x 2 x–2 π∫π + πx–π= 3ln x – 2 π – 2ln x–π + C2x+ 21 2x+21=22x+ 9 x–5(2 x–1)( x+5)= A B2 x–1 + x + 52x + 21 = A(x + 5) + B(2x – 1)A = 4, B = –12x+ 21 4 1∫ dx = –2 ∫ dx dx2x9 x–52 x–1 ∫+x+5= 2ln 2 x –1 –ln x+ 5 + C2 22x − x− 20 2( x + x−6) −3x−8=2 2x + x− 6 x + x−63x+ 8= 2 −2x + x−63x+ 8 3x+8 A B== +2x + x−6( x+ 3)( x−2)x+ 3 x−23x + 8 = A(x – 2) + B(x + 3)1 14A= , B =5 522x−x−20∫dx2x + x−61 1 14 1= ∫2dx −5∫ x 3 dx −+ 5∫x−2dx1 14= 2x − ln x+ 3 − ln x− 2 + C5 511.17 x– 3 17 x– 3=23 x + x–2(3 x– 2)( x+1)= A B3 x –2 + x + 117x – 3 = A(x + 1) + B(3x – 2)A = 5, B = 417 x – 3 5 4 5∫ dx =2 ∫ dx + dx3 x + x–23 x–2 ∫ = ln 3 x – 2 + 4ln x+ 1 + Cx+1 3Instructor’s Resource Manual Section 7.5 447

12.5– x5– x A B== +2x – x( π+ 4) + 4π ( x– π)( x–4)x– π x–45 – x = A(x – 4) + B(x – π )5– π 1A= , B =π– 4 4– π5– x 5– π 1 1 1∫ dx = dx +dx2x x( 4) 4 π– 4∫− π+ + πx– π 4– π∫x–45– π1= ln x – π+ ln x– 4 + Cπ–4 4– π13.14.15.2 22x + x− 4 2x + x−4== A + B +C3 2x −x −2xxx ( + 1)( x−2)x x+ 1 x−222x + x− 4 = A( x+ 1)( x− 2) + Bx( x− 2) + Cx( x+1)A = 2, B = –1, C = 122x+ x−4 2 1 1dx = dx − dx + dx = x − x+ + x− + C3 2x −x −2xx x+ 1 x−2∫ ∫ ∫ ∫ 2ln ln 1 ln 227x + 2 x–3A B C= + +(2 x–1)(3x+ 2)( x–3) 2 x–1 3x+2 x–327x + 2 x–3 = A(3x+ 2)( x–3) + B(2 x–1)( x–3) + C(2 x–1)(3x+2)1 1 6A= , B = – , C =35 7 527x+ 2 x–3 1 1 1 1 6 1∫ dx = dx – dx + dx(2 x–1)(3x+ 2)( x–3) 35∫2 x–1 7∫3x+2 5∫x–31 1 6= ln 2 x –1 – ln 3x+ 2 + ln x– 3 + C70 21 52 26x + 22x− 23 6x + 22x−23== A + B +C2(2x− 1)( x + x−6)(2x− 1)( x+ 3)( x−2)2x− 1 x+ 3 x−226x + 22x− 23 = A( x+ 3)( x− 2) + B(2x−1)( x− 2) + C(2x− 1)( x+3)A = 2, B = –1, C = 326x+ 22x−23 2 1 3dx = dx − dx + dx = x − − x+ + x− + C2(2x− 1)( x + x−6)2x− 1 x+ 3 x−2∫ ∫ ∫ ∫ ln 2 1 ln 3 3ln 216.3 23 2x − 6x + 11x−61⎛x 6x 11x6⎞2− + − 1⎛x − 3x+2 ⎞= = 1+3 23 24x − 28x + 56x−324 ⎜ x − 7x + 14x−8⎟3 2⎝⎠4 ⎜ x − 7x + 14x−8⎟⎝⎠1 ⎛ ( x−1)( x−2)⎞ 1⎛1 ⎞= ⎜1+⎟ = ⎜1+⎟4 ⎝ ( x−1)( x−2)( x−4)⎠ 4⎝x − 4⎠3 2x –6x + 11 x–61 1 1 1 1∫dx =3 2 ∫ dx + dx4 x –28x + 56 x–324 4∫ = x + ln x – 4 + Cx –4 4 4448 Section 7.5 Instructor’s Resource Manual

17.3x3 x–2= x –1+2 2x + x– 2 x + x–23 x –2 3 x–2A B= = +2x + x–2( x + 2)( x–1) x+2 x–13x – 2 = A(x – 1) + B(x + 2)8 1A= , B =3 33x8 1 1 1∫ dx =2 ∫( 1)x + x– 2x − dx +3∫ x 2 dx ++ 3∫x−1dx 1 2 8 1= x – x+ ln x+ 2 + ln x–1+ C2 3 318.3 2x + x 14x+24= x –4 +2x + 5x+6 ( x+ 3)( x+2)14x + 24 A B= +( x 3)( x 2) x 3 x 2+ + + +14x + 24 = A(x + 2) + B(x + 3)A = 18, B = –4∫3 2x + xdx2x + 5x+618 4= ∫( x − 4) dx + ∫ –x+ 3 dx ∫x+2dx= 1 24 18ln 3 –4ln 22 x − x+ x+ x+ + C19.4 2 2x + 8x + 8 12x+ 8= x +3x − 4xxx ( + 2)( x–2)212x + 8 A B C= + +xx ( + 2)( x–2) x x+2 x–2212x + 8 = A( x+ 2)( x– 2) + Bx( x– 2) + Cx( x+2)A = –2, B = 7, C = 74 2x + 8x+ 8 1 1 1dx = x dx –2 dx + 7 dx + 7 dx3x –4xx x+2 x–2∫ ∫ ∫ ∫ ∫= 1 2–2ln 7ln 2 7ln –22 x x + x+ + x + C20.6 3 2x + 4x + 4 3 2272x+ 4= x + 4x + 16x+ 68+3 2 3 2x –4 x x –4x2272x + 4 A B C= + +2 2x ( x–4)x x x– 42 2272x + 4 = Ax( x –4) + B( x –4) + Cx1 1089A= – , B = –1, C =4 46 3x + 4x+ 4 3 21 1 1 1089 1∫dx ( 4 16 68) – –3 22x –4x= ∫ x + x + x + dx 4∫ xdx ∫xdx + 4∫x−4dx1 4 4 3 2 1 1 1089= x + x + 8x + 68 x– ln x + + ln x– 4 + C4 3 4 x 421.x+1 A B= +2 3 2( x−3) x − ( x−3)x + 1 = A(x – 3) + BA = 1, B = 4x + 1 1 4∫ dx =2 ∫ dx + dx32( x 3) x −∫−( x−3)4= ln x −3− + Cx − 3Instructor’s Resource Manual Section 7.5 449

22.5x+ 7 5x+7 A B= = +2 2 2 2x + 4x+ 4 ( x+ 2) x + ( x+2)5x + 7 = A(x + 2) + BA = 5, B = –35x+ 7 5 3∫ dx =2 ∫ dx − dx22x 4x 4 x +∫+ + ( x+2)3= 5ln x + 2 + + Cx + 223.24.25.26.27.3x+ 2 3x+2=3 2 3x + 3x + 3x+ 1 ( x+1)23x + 2 = Ax ( + 1) + Bx ( + 1) + CA B C= + +x + 1 2 3( x+ 1) ( x+1)A= 0, B = 3, C = –13x+ 2 3 1 3 1∫ dx =3 2 ∫ dx −2 ∫ dx =− + + C3 1 2x + 3x + 3x+ 1 ( x+ 1) ( x+ 1) x + 2( x+1)6x A B C D E F G2 5 –2 2 1–2 3 4 5( x –2) (1– x) = x+ ( x–2) + x+ (1– x) + (1– x) + (1– x) +(1– x)A = 128, B = –64, C = 129, D = –72, E = 30, F = –8, G = 16x⎡ 128 64 129 72 30 8 1∫ dx = ∫ ⎢ – + – + − +( x–2) (1– x) ⎢⎣x–2 ( x–2) 1– x (1– x) (1– x) (1– x) (1– x)64 72 15 8 1= 128ln x – 2 + –129ln 1– x + − + − + Cx–2 1– x 2 3 4(1– x) 3(1– x) 4(1– x)⎤⎥dx2 5 2 2 3 4 5⎥⎦2 23x − 21x+ 32 3x − 21x+32 A B C= = + +3 2 2 4 2x − 8x + 16 x x( x−4) x x−( x−4)2 23x − 21x+ 32 = A( x− 4) + Bx( x− 4) + CxA = 2, B = 1, C = –123x− 21x+32 2 1 11dx = dx + dx − dx = 2ln x + ln x− 4 + + C3 2 42x − 8x + 16 x x−( x−4)x − 4∫ ∫ ∫ ∫2 2x + 19x+ 10 x + 19x+10=4 3 32x + 5 x x (2x+5)A = –1, B = 3, C = 2, D = 2A B C D= + + +x 2 3x x 2x+52x + 19x+ 10 ⎛ 1 3 2 2 ⎞3 1dx = –dx4 3 ⎜ + + +2 3 ⎟ –ln x – – ln 2x 5 C2x + 5x ⎝ x x x 2x+5⎠= x x+ + +∫ ∫ 22 22 x + x–8 2 x + x–8A Bx+C= = +3 2 2x + 4 x x( x + 4) x x + 4A = –2, B = 4, C = 122 x + x–8 1 4x+1 1 2x1∫ dx = –2 dx dx3 ∫ +2x + 4x x∫ =− 2∫ dx + 2 dx dx2 2x + 4 x∫ + ∫x + 4 x + 42 1 –1⎛x ⎞= –2ln x + 2ln x + 4 + tan ⎜ ⎟+C2 ⎝2⎠450 Section 7.5 Instructor’s Resource Manual

28.3x+ 2 3x+2=2 2xx ( + 2) + 16 x xx ( + 4x+20)1 1 13A= , B = – , C =10 10 5∫A Bx+C= +x 2x + 4x+2013x+ 2 1 1 – x +10 5dx = dxdx22x( x+ 2) + 16x10∫ +x∫ = 1 1 dx +14 1 dx2x + 4x+20 10 x 5 ( x + 2) + 161 7 –1 ⎛ x + 2⎞= ln x + tan ⎜ ⎟10 10 ⎝ 4 ⎠131 2– ln x + 4x+ 20 + C20∫ ∫ − ∫ 21 2x+ 4dx20 x + 4x+2029.22 x –3 x–36A Bx+C= +2 2 –1 2(2x− 1)( x + 9) x x + 9A = –4, B = 3, C = 022 x –3 x–36 1 32 –4 xdx dx dx2 –1 2(2 x–1)( x + 9) = x+x + 9∫ ∫ ∫3 2= –2ln 2 x –1 + ln x + 9 + C230.31.32.1 =14 2x –16 ( x− 2)( x+ 2)( x + 4)A B Cx+D= + +x–2 x+ 2 2x + 41 1 1A= , B = – , C = 0, D = –32 32 81 1 1 1 1 1 1∫ –4 32 – 2 32 2 8 2–16 dx = ∫xdx ∫xxdx −+∫x + 4dx 1 1 1 –1 ⎛ x ⎞= ln x – 2 – ln x+ 2 – tan ⎜ ⎟+C32 32 16 ⎝2⎠1 A B C D2 2 –1 2 4 2( x–1) ( x+ 4) = x+ ( x–1) + x++( x+4)2 1 2 1A= – , B = , C = , D =125 25 125 251 2 1 1 1 2 1 1 1∫ –2 2 125 –1 25 2 125 4 25 2( x–1) ( x 4) dx = ∫xdx + ∫( x–1) dx + ∫+ x+dx + ∫( x+4)dx= 2 1 2 1– ln –1 – ln 4 –125 x 25( x–1) + 125 x+ 25( x+4)+ C3 2 2x – 8 x –1 –7x + 7 x–16= 1+2 2( x+ 3)( x –4x+ 5) ( x+ 3)( x − 4x+5)2–7x + 7 x–16A Bx+C= +2 3 2( x+ 3)( x –4x+ 5) x + x –4x+550 41 14A= – , B = – , C =13 13 133 241 14x –8 x –1 ⎡ 50⎛1 ⎞ – x + ⎤13 13dx = ⎢1– 2 ⎜ ⎟+⎥dx13 3 2( x+ 3)( x –4x+ 5) ⎢ x +⎣⎝ ⎠ x –4x+5⎥⎦∫ ∫50 1 68 1 41 2x− 4= ∫dx −13∫ dx − dx −dxx + 3 13∫ 2 26∫ 2( x− 2) + 1 x − 4x+550 68 –1 41 2= x – ln x+ 3 – tan ( x–2)– ln x –4x+ 5 + C13 13 26Instructor’s Resource Manual Section 7.5 451

33. x = sin t, dx = cos t dt3 2 3 2(sin t−8sin t−1)cos t x −8x−1dt =dx2 2(sin t+ 3)(sin t− 4sin t+ 5) ( x+ 3)( x − 4x+5)∫ ∫50 68 −1 41 2= x − ln x+ 3 − tan ( x−2) − ln x − 4x+ 5 + C13 13 26which is the result of Problem 32.∫3 2(sin t – 8sin t –1)cost50 68 –1 41 2dt = sin t – ln sin t + 3 – tan (sin t – 2) – ln sin t – 4sin t + 5 + C2(sin t+ 3)(sin t – 4sin t+5) 13 13 2634. x = sin t, dx = cos t dtcost1 1 1 1 −1⎛ x⎞∫ dt = ln 2 ln 2 tan4 ∫ dx = x − − x + − C4⎜ ⎟+sin t−16 x −1632 32 16 ⎝2⎠which is the result of Problem 30.cost1 1 1 –1 ⎛sint⎞∫ dt = ln sin t – 2 – ln sin t + 2 – tan4⎜ ⎟+Csin t –16 32 32 16 ⎝ 2 ⎠35.3x –4x Ax+ B Cx+D= +2 2 2 2 2( x + 1) x + 1 ( x + 1)A= 1, B = 0, C = –5, D = 03x –4x x xdx = dx −5dx2 2 2 2 2( x + 1) x + 1 ( x + 1)∫ ∫ ∫2= 1 ln x + 1 + 5 + C2 22( x + 1)36. x = cos t, dx = –sin t dt2 2(sin t)(4cos t –1) 4 x –1dt = –dx2 4 2 4(cos t)(1 + 2cos t+ cos t) x(1 + 2 x + x )∫ ∫2 24x −1 4x − 1 A Bx+ C Dx+E= = + +2 4 2 2 2 2 2x(1+ 2 x + x ) x( x + 1) x x + 1 ( x + 1)A = –1, B = 1, C = 0, D = 5, E = 0⎡ 1 x 5x⎤ 1 2 5−∫⎢− + + ⎥dx = ln x − ln x + 1 + + Cx 2 2 2 22⎢⎣x + 1 ( x + 1) ⎥⎦2( x + 1)1 25= ln cost − ln cos t+ 1 + + C2 22(cos t + 1)37.3 2 22x + 5x + 16 x x(2x + 5x+16)=5 3 4 2x + 8x + 16 x x( x + 8x+ 16)A = 0, B = 2, C = 5, D = 822x + 5x+ 16 Ax+ B Cx+D= = +( x + 4) x + 4 ( x + 4)2 2 2 2 23 22x + 5x + 16x 2 5x+8 2 5x8∫ dx = dx + dx5 3 ∫ 2 ∫ =2 2 ∫ +2 ∫ +2 2 ∫ 2 2x + 8x + 16x x + 4 ( x + 4)82To integrate ∫ dx,let x = 2 tan θ, dx = 2sec θ dθ.2 2( x + 4)∫8 dx = ∫16sec( x + 4) 16sec2 2 41 1 1 1= θ + sin 2θ + C = θ + sinθ cosθ+ C2 4 2 2∫2θd θ2 ⎛1 1 ⎞= ∫cos θ dθ = ∫ ⎜ + cos 2θ⎟dθθ⎝2 2 ⎠dx dx dxx + 4 ( x + 4) ( x + 4)1 –1 x x= tan + + C2 2 2x + 43 2+ +–1 –1dx tan – tan5 3 2 22x 5x 16x x 5 1 x x 3 –1 x 2 x–5= + + + C = tan + + Cx + 8x + 16x 2 2( x + 4) 2 22x + 4 2 2 2( x + 4)452 Section 7.5 Instructor’s Resource Manual

38.x –17 x–17A B= = +2x + x–12( x+ 4)( x–3) x+4 x–3A = 3, B = –26 x –17 6⎛3 2 ⎞6∫dx = – dx4 2 ∫x x–124⎜ ⎟ = ⎡ 3ln x 4 – 2ln x–3+⎝ x+4 x–3⎣ + ⎤⎦ = (3ln10–2ln3)–(3ln8–2ln1)⎠4= 3ln10– 2ln3–3ln8 ≈ –1.5339. u = sin θ, du = cos θ dθπ /4 cosθ1/ 2 1∫dθ=0 ∫(1– sin θ)(sin θ + 1) 0 (1– u )( u + 1)du =2 2 2 2 2 21 A B Cu+ D Eu+F= + + +(1 – u )( u + 1) 1– u 1+u u + 1 ( u + 1)2 2 2 2 2 21/ 2 1du02 2(1 – u)(1 + u)( u + 1)1 1 1 1A= , B = , C = 0, D = , E = 0, F =8 8 4 21/ 2 1 1 1/ 2 1 1 1/ 2 1 1 1/ 2 1 1 1/ 2 1∫0 2 2 2 8 0 1 8 0 1 4 0 2 2 0 2 2(1 u )( u 1) du = ∫− udu + ∫+ udu + ∫− + u + 1 du + ∫( u + 1)du1/ 2⎡ 1 1 1 –1 1⎛–1 u ⎞⎤= ⎢– ln1– u + ln1+ u + tan u+ ⎜tanu+8 8 4 4 2 ⎟⎥⎣⎝ u + 1⎠⎦01 2+1 1 –1 1 1= ln + tan + ≈0.658 2−1 2 2 6 21(To integrate ∫ du,let u = tan t.)2 2( u + 1)∫1/ 2⎡1 1+u 1 –1 u ⎤= ⎢ ln + tan u + ⎥8 1 u 2 2⎢⎣− 4( u + 1) ⎥⎦040.3x+ 13 3x+13 A B== +2x + 4x+ 3 ( x+ 3)( x+1) x + 3 x + 1A = –2, B = 55 3x+ 135∫ dx = ⎡ –2ln x 3 5ln x 11 2 ⎣ + + + ⎤⎦= –2 ln 8 + 5 ln 6 + 2 ln 4 – 5 ln 2 = 5ln3−2ln2≈ 4.111x + 4x+3dy41. y(1 y)dt = − so that 1∫ dy = 1dt = t + C1y(1 − y) ∫a. Using partial fractions:1 A B A(1 − y)+ By= + = ⇒y(1 − y) y 1 − y y(1 − y)1 1 1A+ ( B− A) y = 1+ 0y⇒ A= 1, B− A= 0⇒ A= 1, B = 1⇒ = +y(1 − y) y 1 − yThus: t C ⎛1 1 ⎞+ 1 = ∫ ⎜ + ⎟y 1 ydy =⎝ −ln ln(1 ) ln ⎛ y ⎞y− − y = ⎜ ⎟ so that⎠⎝ 1 − y ⎠tyeor yt () =+e tb.t+C1t= e = Ce1−yC( C=e 1 )Since1ey(0) = 0.5, 0.5 = or C = 1 ; thus yt () =+ 11 +e3ey(3) = ≈0.95331+e1C1CInstructor’s Resource Manual Section 7.5 453tt

dy 142. y (12 y )dt = 10− so that∫ 1 dy = 1 dt = 1 t + C1y(12 − y) ∫10 10a. Using partial fractions:1 A B A(12 − y)+ By= + = ⇒y(12 − y) y 12 − y y(12 − y)12 A+ ( B− A) y = 1+ 0y⇒ 12A= 1, B− A=01 1⇒ A= , B =12 121 1 1⇒ = +y(12 − y) 12y 12(12 − y)1 1 1Thus: 110 t C ⎛⎞+ = ∫ ⎜ + ⎟12y12(12 y)dy =⎝− ⎠1 [ ln ln(12 )]1 ⎛ y ⎞y− − y = ln ⎜ ⎟ so that12 12 ⎝12− y ⎠y 1.2t+12C11.2t= e = Ce or12 − y12C1b.12eyt () =+eSincethus1C1.2t1.2t( C=e )12y(0) = 2.0, 2.0 = or C = 0.2 ;+ 112eyt () =5 +e3.61.2t1.212ey(3) = ≈10.563.65 + etdy43. 0.0003 y(8000 y)dt = − so that 1∫ dy = 0.0003dt = 0.0003t + C1y(8000 − y) ∫a. Using partial fractions:1 A B A(8000 − y)+ By= + =y(8000 − y) y 8000 − y y(8000 − y)⇒ 8000 A+ ( B− A) y = 1+0y⇒ 8000A= 1, B− A=01 1⇒ A= , B =8000 80001 1 ⎡1 1 ⎤⇒ = +y(8000 − y) 8000⎢y (8000 − y)⎥⎣⎦Thus:1 1 10.0003 t C ⎛⎞+ 1 =8000∫ ⎜ + ⎟y (8000 y)dy =⎝ − ⎠1 [ ln ln(8000 )]1 ⎛ y ⎞y− − y = ln ⎜ ⎟8000 8000 ⎝8000− y ⎠1Cb.so thaty 2.4t+8000C12.4t= e = Ce8000 − yC8000eyt () =+ eSince1C2.4t2.4t8000( C=e 1 )1Cory(0) = 1000, 1000 = 8000 or C =1 ;+ 1 72.4t8000ethus yt () =7+e2.4t7.28000ey(3) = ≈7958.47.27 + edy44. 0.001 y(4000 y)dt = − so that 1∫ dy = 0.001dt = 0.001t + C1y(4000 − y) ∫a. Using partial fractions:1 A B= +y(4000 − y) y 4000 − yA(4000 − y)+ By=y(4000 − y)⇒ 4000 A+ ( B− A) y = 1+0y⇒ 4000A= 1, B− A=01 1⇒ A= , B =4000 40001 1 ⎡1 1 ⎤⇒ = +y(4000 − y) 4000⎢y (4000 − y)⎥⎣⎦Thus:1 1 10.001 t C ⎛⎞+ 1 =4000∫ ⎜ + ⎟y (4000 y)dy =⎝ − ⎠1 [ ln ln(4000 )]1 ⎛ y ⎞y− − y = ln ⎜ ⎟4000 4000 ⎝4000− y ⎠so thaty 4t+4000C14t= e = Ce or4000 − y4000C( C=e 1 )b.4000eyt () =+ eSincethus1C4t4ty(0) = 100, 100 = 4000 or C =1 ;+ 1 394t4000eyt () =39+e4t121C4000ey(3) = ≈3999.041239 + e454 Section 7.5 Instructor’s Resource Manual

dy45. ky( L y)dt = − so that 1∫ dy = k dt = kt + C1yL ( − y) ∫Using partial fractions:1 A B A( L− y)+ By= + = ⇒yL ( − y) y L− y yL ( − y)LA + ( B − A) y = 1+ 0y ⇒ LA = 1, B − A = 0⇒1 1 1 1 ⎡1 1 ⎤A= , B = ⇒ = +L L y( L−y)L⎢y L−y⎥⎣ ⎦1 ⎛1 1 ⎞Thus: kt + C1= dyL∫⎜ + ⎟ =⎝ y L−y⎠1 [ ln ln( )]1 ⎛ y ⎞y− L− y = ln ⎜ ⎟ so thatL L ⎝ L−y⎠y kLt+LC1kLtLe= e = Ce or yt () =L−yLC11+eIfy0L 1 L−y= y = =yfinal formula is(Note: if y0kLteu+ekLt( C=e )C(0) 01+ 1then C 0CkLtLeyt () =.⎛L−y0⎞ekLt⎜ ⎟+⎝ y0⎠L−y0< L , then u 0y0< 1; thus yt ()kLtkLt; so our= > and< Lfor all t)46. Since y′ (0) = ky0( L− y0)is negative if y0> L,the population would be decreasing at timet = 0. Further, sinceL Llim yt ( ) = lim = = Lt→∞t→∞⎛ L−y ⎞ 0+10 ⎟+ 1⎜ yekLt ⎟⎝ 0 ⎠(no matter how y 0 and Lcompare), and sinceL−y0y0ekLt is monotonic as t →∞ ,we concludethat the population would decrease toward alimiting value of L.47. If y 0 < L , then y′ (0) = ky0( L− y0) > 0 and thepopulation is increasing initially.48. The graph will be concave up for values of t thatmake y′′ () t > 0. Nowdy′dy′′ () t = = [ ky( L− y)] =dt dtk − yy′ + ( L− y) y′= k ky( L− y) L−2y[ ] [ ][ ]Thus if y0< L , then yt () < Lfor all positive t(see note at the end of problem 45 solution) andso the graph will be concave up as long asL− 2y> 0 ; that is, as long as the population isless than half the capacity.dy49. a. ky(16 – y)dt =b.c.dyy(16 – y )=dyy(16 – y )=kdt∫ ∫kdt1 ⎛1 1 ⎞dy kt C16∫ ⎜ + ⎟ = +⎝ y 16 – y⎠1( ln y – ln 16 – y ) = kt+C16yln 16ktC16 – y = +y 16ktCe16 – y =y(0) = 2:y(50) = 4:1 y 1 16kt= C;= e7 16– y 71 1 800ke ,3 7= so1( ln 7)y 1 te 50 316 – y = 7( 1 ln 7 7) t ( 1 ln )t7y = 16 e 50 3 – ye 50 31 7k = ln800 31( ln 7) t50 3( 1 ln 7 7) t –( 1 ln )16e16y = =t7+ e 50 3 1+7e50 316y(90) = ≈6.34billion– 1 7( ln ) 901+7e50 39 =16– 1 7( ln50 3 )1+7e1 7( ) t– ln50 3 167 e = –191 7( )– ln t50 3 1e =91 7( )t =1– 50 ln 3ln 9t⎛ln1 ⎞9t = –50⎜⎟≈129.66⎜ln7 ⎟⎝ 3 ⎠The population will be 9 billion in 2055.Instructor’s Resource Manual Section 7.5 455

dy50. a. ky(10 – y)dt =b.dyy(10 – y )=kdt1 ⎛1 1 ⎞⎜ + ⎟dy=10 ⎝ y 10 – y⎠∫ ∫yln 10ktC10 – y = +yCe10 – y =y(0) = 2:y(50) = 4:y 1e10 – y = 410ktkdt1 y 1= C;= e4 10– y 410kt2 1 500k1 8= e , k = ln3 4 500 31( ln 8)t50 3( 1 ln8 8) t ( 1 ln )t50 3 50 34y = 10 e – ye1( ln 8)t50 3( 1 ln 8 8) t –( 1 ln )10e10y = =t4+ e 50 3 1+4e50 310y(90) = ≈5.94billion– 1 8( ln ) 901+4e50 351. a. Separating variables, we obtaindx( a−x)( b−x )= kdt1 A B= +( a− x)( b−x)a−x b−x1 1A=− , B =a−b a−bdx∫( a−x)( b−x)1 ⎛ 1 1 ⎞= ⎜− + ⎟dx=a−b ⎝ a−x b−x⎠ln a−x −lnb−x= kt + Ca−b1 a−x ln = kt + Ca−b b−x∫ ∫a− x =( a−b)ktCeb−xk dtaSince x = 0 when t = 0, C = , so ba ( a−b)kta− x = ( b−x)eb( a−b) kt ⎛ a ( a−b)kt ⎞a( 1− e ) = x⎜1−e ⎟⎝ b ⎠( a−b)kt( a−b)kta(1 − e ) ab(1 − e )xt () = =( a b)kt1 − a −( a−b)kte b−aebc.9 =10– 1 8( ln50 3)1+4e1 8( ) t– ln50 3 104 e = –191 8( )– ln t50 3 1e =36⎛ 1 8⎞ 1– ⎜ ln ⎟t= ln⎝50 3 ⎠ 36⎛ln1 ⎞36t = –50⎜⎟≈182.68⎜ ln 8 ⎟⎝ 3 ⎠tThe population will be 9 billion in 2108.( a b)ktb. Since b > a and k > 0, e − → 0 ast →∞. Thus,ab(1)x → = a .b − 0c.−2kt8(1 − e )xt () =−2kt4−2e−40k−40kx(20) = 1, so 4− 2e= 8−8e40k6e − = 41 2k =− ln40 32 kt t / 20ln 2 / 3 ln(2 / 3)t /20 2− ⎛ ⎞e = e = e = ⎜ ⎟⎝3⎠⎛ t /202 ⎞4⎜1−( 3 ) ⎟xt () =⎝ ⎠t /202 −2( 3 )−2( 3 )32( )⎛ 3 ⎞4⎜1⎟38x(60) =⎝ ⎠= ≈1.65 grams2 −233t /20456 Section 7.5 Instructor’s Resource Manual

d. If a = b, the differential equation is, afterseparating variablesdx= kdt2( a−x)dx∫ = kdt2 ∫( a−x)1kt Ca−x = +1a xkt + C = −xt 1() = a − kt + CSince x = 0 when t = 0, C = 1 ,aso1 axt () = a−= a −kt +1 akt + 1a⎛ 1 ⎞ ⎛ akt ⎞= a⎜1−⎟ = a ⎜ ⎟ .⎝ akt + 1⎠⎝ akt + 1⎠52. Separating variables, we obtaindy( y−m)( M − y )= kdt .1 A B= +( y−m)( M − y)y−m M − y1 1A= , B =M −m M −mdy 1 ⎛ 1 1 ⎞∫= ⎜ + ⎟dy( y−m)( M − y)M −m∫⎝ y−m M − y⎠= ∫ kdtln y−m −lnM − yM − m= kt + C1ln y − m = kt + CM −m M − yy−m= CeM − y( M −mkt)y− m= ( M − y)Ce( M −mkt)( M − mkt ) ( M−mkt)y(1+ Ce = m+MCe( M −mkt)−( M−m)ktm+MCe me + MCy ==( M −mkt)−( M−m)kt1+Cee + C( M m)ktAs t →∞, e − − → 0 since M > m.MCThus y → = M as t →∞ .C53. Separating variables, we obtaindy( A− y)( B+y )= kdt1 C D= +( A − y)( B+ y)A− y B+y1 1C = , D =A + B A+Bdy 1 ⎛ 1 1 ⎞∫= ⎜ + ⎟dy( A − y)( B+y)A+ B∫⎝ A− y B+y⎠= ∫ kdt−ln( A− y) + ln( B+y)= kt + CA + B1ln B + y = kt + CA+ B A−yB+ y = CeA−y( A+Bkt )( A Bkt )B+ y = ( A−y)Ce +( A+ B) kt ( A+B)kty(1 + Ce ) = ACe − BACeyt () =1+Ce( A+B)kt− B( A+Bkt )54. u = sin x, du = cos x dxπ /2 cos x1 1∫ dx =π /6∫ 1sin x(sin x+ 1) u( u + 1)2 2 2 221 A Bu + C Du + E= + +uu ( + 1) u u + 1 ( u + 1)A = 1, B = –1, C = 0, D = –1, E = 01 1∫ 1 du2 2uu ( + 1)22 2 2 2 21 uudu du du1 1 1= ∫1 − 1 2 1 2 22u ∫ −∫2u + 1 2 ( u + 1)1⎡ 1 2 1 ⎤= ⎢ln u− ln( u + 1) + ⎥2 2⎢⎣2( u + 1) ⎥⎦121 1 ⎛ 1 1 5 2⎞= 0 − ln 2 + −⎜ln − ln + ⎟2 4 2 2 4 5du⎝ ⎠ ≈ 0.308Instructor’s Resource Manual Section 7.5 457

7.6 Concepts Review1. substitution2. 533. approximation4. 0Problem Set 7.6Note: Throughout this section, the notation Fxxxrefers to integration formula number xxx in theback of the book.1. Integration by parts.−5xu = x dv = e1 −5xdu = 1dx v = − e5−5x 1 −5x 1 −5x∫xe dx =− xe −5∫−e dx51 −5x1 −5x=− xe − e + C5 251 −5x⎛ 1⎞=− e ⎜x+ ⎟+C5 ⎝ 5⎠2. Substitutionx 1 12∫ dx = du = ln u + C = ln2( x + 9)+ Cx + 9 2∫u2u= x + 9du=2x dx3. Substitution21u=ln xdu = 1 dxxln 2 20 0ln 2 2( ln 2)ln x⎡u⎤dx = u du = ⎢ ⎥ = ≈0.2402x⎢⎣2 ⎥⎦2∫ ∫4. Partial fractionsxx∫ dx =2 ∫dxx − 5x+6 ( x−3)( x−2)x A B= + =( x−3)( x−2) ( x−3) ( x−2)Ax ( − 2) + Bx ( − 3) ( A+ Bx ) + ( −2A−3 B)= ⇒( x−3)( x−2) ( x−3)( x−2)A+ B = 1, −2A− 3B = 0 ⇒ A= 3, B =−2x3 2∫ dx =2 ∫ − dx =x − 5x+6 ( x−3) ( x−2)3( x − 3)3ln x −3 −2ln x− 2 = ln + C( 2)2x −5. Trig identitysubstitution.∫2 1+cos2ucos u = and224 ⎛1+cos4x⎞cos 2xdx= ∫⎜⎟ dx=⎝ 2 ⎠⎡⎤1 ⎢2 ⎥1 2cos 4 cos 44∫ ⎢+ x+ x dxu= 4x⎥=⎢⎣ du = 4 dx ⎥⎦⎡⎤⎢⎥1⎢1 ⎛1+cos8x⎞ ⎥sin 44⎢x+ x+ dx2∫ ⎜ ⎟2 ⎥=⎢⎝ ⎠⎥v= 8x⎢⎣dv = 8 dx ⎥⎦1⎡1 1 1 ⎤x sin 4x x sin8x C4⎢+ + +2 2 16 ⎥+ =⎣⎦1[ 24 x+ 8sin 4 x+ sin8 x]+ C646. Substitutionu sin xdu cos x dx4 43 3 u sin xsin xcosxdx= u du = + C = + C= 4 4=∫ ∫7. Partial fractions1 1∫ dx =2 ∫dxx + 6x+8 ( x+ 4)( x+2)1 A B= + =( x+ 4)( x+ 2) ( x+ 4) ( x+2)Ax ( + 2) + Bx ( + 4) ( A+ Bx ) + (2A+4 B)= ⇒( x+ 4)( x+ 2) ( x+ 4)( x+2)1 1A+ B = 0, 2A+ 4B = 1 ⇒ A=− , B =2 22 1 1 2⎛1 1 ⎞∫ =1 2 2∫1 ⎜ − ⎟ dxx + 6x+8 ⎝ x+ 2 x+4⎠21 2 1 ⎡ ( x + 2) ⎤= ⎡ ln x 2 ln x 4 ln2⎣ + − + ⎤ ⎦ =1 ⎢ ⎥2 ⎣ ( x + 4) ⎦11⎛4 3⎞1 10= ⎜ln − ln ⎟= ln ≈0.05272⎝6 5⎠2 9458 Section 7.6 Instructor’s Resource Manual

8. Partial fractions∫1 dt =12 ∫ dt1−t (1 − t)(1 + t)1 A B= + =(1 − t)(1 + t) (1 − t) (1 + t)A(1 + t) + B(1 −t) ( A− B) t+ ( A+B)= ⇒(1 − t)(1 + t) (1 − t)(1 + t)1 1A+ B = 1, A− B = 0 ⇒ A= , B =2 21 12 1 1 2⎛1 1 ⎞∫ dt =0 2 2∫0 ⎜ + ⎟ dx1−t⎝1− t 1+t⎠11⎡−ln 1 t ln 1 t 22⎣ − + + ⎤ ⎦ =011 ⎡ (1 + t)⎤ 2⎢ln ⎥ ≈ 0.54932 ⎣ (1 − t)⎦09. Substitution5 7 20 ∫ 2u= x+22u = x+ 22udu = dx∫∫x x+ 2 dx = ( u − 2)( u)2udu=5 377 ⎡4 2 u 2u⎤2u− 4u du = 2⎢− ⎥ =2⎢⎣5 3 ⎥⎦22 5 37⎡23u− 10u⎤ = ⎡77 7 + 8 2⎤≈ 28.6715 ⎣ ⎦ 2 15 ⎣ ⎦10. Substitution4 1 8 u8 1∫ dt = 23 ∫ du = du6 2 ∫ =t−2tu6 u − 2− u2u= 2, t u = 2tudu = dt8 8−22⎡⎣ln u −2⎤ ⎦ = 2ln ≈1.2236 6−211. Use of symmetry; this is an odd function, soπ22∫ cos xsin xdx = 0−π212. Use of symmetry; substitution2ππ4∫ sin 2x dx = 80 ∫ sin 2x dx =0 u= 2xdu = 2 dxππ24 2∫ sinudu= 4[ − cosu]= 40 0213. a. Formula 9623∫ x 3x+ 1dx = ( 9x− 2)( 3x+ 1)2 + C135F96a= 3, b=1b. Substitution; Formula 96∫ ∫x xe 3e + 1 dx = u 3u+ 1du=x xu= e , du=e dxF96a= 3, b=1x x( )( )32 9 e − 2 3 e + 1 2+ C13514. a. Formula 962(3 t − 4) t dt = 2 t(3− 4) t dt =∫ ∫F96a=−4,b=3⎡ 23 2⎤2 ⎢ ( −12t−6)(3 − 4 t)240⎥+ C =⎣⎦1 3− (2 t+ 1)(3 − 4 t ) 2 + C10b. Substitution; Formula 96∫cost 3 − 4costsin tdt =− u 3 − 4u du =u= cos t,du=−sintdtpart a.1 3(2cos t+ 1)(3 − 4cos t ) 2 + C2015. a. Substitution, Formula 18dx 1 du∫ =2 4∫ =29−16x9−uF18u= 4 x, du= 4dx a=31⎡ 1 u+ 3 ⎤ 1 4x+3lnln4⎢CC6 u 3⎥+ = +⎣ − ⎦ 24 4x−3b. Substitution, Formula 18xe1 du∫dx =2x4∫ =29−16e9−upart a.u= 4 ex, du=4e xdxx1 4e+ 3ln + C24 x4e− 316. a. Substitution, Formula 18dx dx 5 du= − = − =2 2 5 25x −11 11−5x 11−uF18u= 5 x, a=11du = 5 dx∫ ∫ ∫− 5 11 5x+ 11ln + C5 22 5x− 1155 5x− 11= ln + C110 5x+ 11∫Instructor’s Resource Manual Section 7.6 459

. Substitution, Formula 18∫xdx5 du=− =4 25x−11 10 11−uF18u= 5 x2, a=11du = 2 5x dx255 5x− 11ln+ C220 25x+ 1117. a. Substitution, Formula 572 2 2 2 2x 9− 2x dx = u 9− u du =u=2 x 4F57du = 2 dxa=3∫ ∫∫1 ⎛ 2 2 81 2 1 2xx(4x 9) 9 2x ⎞− ⎛ ⎞⎜ − − ⎟+ sin+ C16 ⎝ ⎠ 32 ⎜ 3 ⎟⎝ ⎠b. Substitution, Formula 57∫2 2 2 2 2sin x cos x 9 − 2sin xdx = 9u 2sinx4∫u −u du=du = 2cosx dx1 2 2=⎛ sin x (4sin x 9) 9 2sin x⎞⎜− − ⎟F5716 ⎝⎠a=381 2 −1⎛ 2 sin x ⎞+ sin+ C32 ⎜3 ⎟⎝ ⎠18. a. Substitution, Formula 552 216 −3t16 −udt = du =tu F55u=du =3t3 dta = 4∫ ∫22 4+ 16−3t16 −3t− 4ln+ C3tb. Substitution, Formula 55∫6 2 616 −3t t 16 −3tdt = dtt∫=3tu=3t3du = 3 3 t2dt21 16−u3∫du =u F55a = 4⎧6 ⎫1⎪6 4+ 16−3t⎪⎨ 16 −3t− 4ln+ C3 3 ⎬⎪3 t⎩⎪⎭19. a. Substitution, Formula 45dx 3 du= =2 3 2 F 455+ 3x5+uu=du=3x3 dxa=5∫ ∫3 ln 3 5 32x+ + x + C3b. Substitution, Formula 45x 3 dudx = =4 6 25+ 3x5+u∫ ∫F 45u=3x2 a=5du=2 3x dx3 ln 32 5 34x + + x + C620. a. Substitution; Formula 48∫2 2 5 2 2t 3+ 5t dt= u 3+ u du =25∫u=5 t F 48du = 5 dta = 3( )⎧⎛ 5 ⎞2 2t 10t 3⎛3 5t⎞⎫+ ⎜ + ⎟−5 ⎪⎜ 8 ⎟ ⎝ ⎠ ⎪⎨⎝ ⎠⎬+ C =25 ⎪ 9 ln 5 3 52 ⎪⎪ t+ + t8⎪⎩⎭1 2 2 2{ 5(10 t t + 3) 3+ 5t − 9 5ln 5t+ 3+ 5t 200} + Cb. Substitution; Formula 48∫8 6 6 6 2t 3+ 5t dt = t 3+ 5t t dt=575∫2 2∫u=5 t3du = 3 5t 2dtu 3 + u du =F 48a = 3( )⎧⎛ 5 3⎞ 6 6t 10t 3⎛3 5t⎞⎫+ ⎜ + ⎟−5 ⎪⎜ 8 ⎟ ⎝ ⎠ ⎪⎨⎝ ⎠⎬+ C =75 ⎪ 9 ln 53 3 56 ⎪⎪ t + + t8⎪⎩⎭⎧ 3 6 615 t (10t + 3) 3+ 5t−⎫⎪ ⎪⎨C600 3 6⎬ +⎪9 5ln 5t+ 3+5t⎪⎩⎭460 Section 7.6 Instructor’s Resource Manual

21. a. Complete the square; substitution;Formula 45.dt dt du∫ =2∫ =2∫ =2 Ft + 2t− 3 ( t+ 1) −4 u −4u=+t 1du = dt2ln ( t+ 1) + t + 2t− 3 + Cb. Complete the square; substitution;Formula 45.dtdt∫=2∫=3 2 29t + 3t− 5 ( t + ) −∫45a = 22 4u= t+3 2du = dtdu3 2= ln ( t+ ) + t + 3t− 5 + C2 29 F 452u −4 a = 29222. a. Complete the square; substitution;Formula 47.22x + 2x−3( x + 1) −4∫dx = dx =x+ 1∫x+1∫2u − 4du =u F 47a = 2u= x+1du = dx2 −1⎛x + 1⎞x + 2x−3− 2sec ⎜ ⎟+C⎝ 2 ⎠b. Complete the square; substitution;Formula 47.22x − 4x ( x −2) −4∫ dx = dx =x−2 ∫x−2∫2u − 4du =u F 47a = 2u= x−2du = dx2 −1⎛x − 2 ⎞x −4x − 2sec ⎜ ⎟+C⎝ 2 ⎠23. a. Formula 98y2∫ dy = (3 y − 10) 3 y + 5 + C3y+ 5 F9827a= 3, b=5b. Substitution, Formula 98sin tcost u∫ = ∫ du =3sint+ 5 3u+5 Fu=sin tdu = cost dt2 (3sin t− 10) 3sin t+ 5 + C2798a= 3, b=524. a. Formula 100adz 5 5−4z− 5∫ = ln+ Cz 5−4z F100a5 5− 4z+ 5a =−4b = 5b. Substitution, Formula 100asin xdu∫dx =− ∫=cos x 5 −4cos x u 5 −4uF100au= cos x a = −4du =−sinx dxb = 55 5−4cosx− 5− ln+ C =5 5− 4cosx+ 55 5− 4cosx+ 5ln5 5−4cosx− 5+ C25. Substitution; Formula 842 1 2∫sinh 3tdt= sinh udu =u=3t3∫Fdu = 3dt1⎛1 3 ⎞ 1⎜ sinh 6t− t⎟+ C = ( sinh 6t− 6t)+ C3⎝4 2 ⎠ 128426. Substitution; Formula 82sech x∫ dx = 2∫sechu du =xFu=x1du =2−12tandxxsinhx + C27. Substitution; Formula 98costsint u∫ dt =− ∫ du =2cost+ 1 2u+1 Fu=cos tdu =−sint dt821− (2cos t− 2) 2cos t+ 1 + C =61 (1 − cos t ) 2 cos t+ 1 + C328. Substitution; Formula 96∫u=costdu =−sint dt98a = 2b = 1costsin t 4cos t− 1 dt =− u 4u− 1 du =∫F96a = 4b = −11 3− (6cos t+ 1)(4cos t− 1) 2 + C60Instructor’s Resource Manual Section 7.6 461

29. Substitution; Formula 99, Formula 98∫2 2cos tsint udt =− ∫ du =cost+ 1 u+1 F99u=costn = 2du =−sint dta = 1b = 12 ⎡ 2u ⎤− 1 25⎢u u+ − ∫ du⎥=⎣u + 1 ⎦ F982⎡2 ⎛2⎞⎤− u u 1 2 ( u 2) u 1 C5⎢ + − ⎜ − + ⎟3⎥+ =⎣⎝⎠⎦2 ⎡ 2 4 ⎤− cost+ 1 cos t− (cost− 2) + C5 ⎢ 3 ⎥⎣⎦30. Formula 95, Formula 171∫ dx =2 3(9 + x ) F95n = 3a = 31 ⎡ xdx ⎤⎢ + 336 2 2 ∫ ⎥ =2 2⎢(9 ) (9 ) F95⎣ + x + x ⎥⎦n = 2a = 31 ⎡ x ⎡ 1 ⎛ x dx ⎞⎤⎤⎢ + 3⎢+ ⎥⎥36 2 2 18 2 2(9 x )⎜ ∫⎢ + ⎢ (9 + x ) 9 + x⎟⎣⎣ ⎝⎠⎥⎦⎥⎦1 ⎧⎪x x −1⎛ x⎞⎫⎪= ⎨ + + tan C36 2 2 2 ⎜ ⎟⎬+⎪⎩(9 + x ) 6 ⋅ (9 + x ) ⎝3⎠⎪⎭F17a = 331. Using a CAS, we obtain:∫π02cos x dx = π −2 ≈1.141591+sinx32. Using a CAS, we obtain:∫103sech xdx≈ 0.7680333. Using a CAS, we obtain:π /212 231π∫ sin xdx= ≈0.354350204834. Using a CAS, we obtain:π 4 x 3π∫ cos dx = ≈1.178100 2 835. Using a CAS, we obtain:∫14t1+t8dt ≈ 0.1108336. Using a CAS, we obtain:∫3 4 −x/2 − 3/20xe dx= 768 −3378e≈14.2663237. Using a CAS, we obtain:π /2 1∫dx ≈ 1.10577051+2cos x38. Using a CAS, we obtain:3π /4 x∫dx ≈−0.00921−π /44+tanx39. Using a CAS, we obtain:∫32xx22+ 2x−1dx = 4ln ( 2)+ 2 ≈ 4.77259− 2x+140. Using a CAS, we obtain:3 du−1π∫ = 2 tan ( 5)− ≈0.729731u 2u−1241.c 1c∫ dx = ⎡ ln x + 1 ⎤ = ln( c + 1)0 x + 1⎣ ⎦F3 0ln( c+ 1) = 1 ⇒ c+ 1 = e⇒c = e−1 ≈1.7182842. Formula 17c 21c1dx ⎡ −2tan x⎤−∫ = = 2tan c0 2x + 1 F17⎣ ⎦0−1 −112 tan c = 1 ⇒ tan c = ⇒21c = tan ≈0.5463243. Substitution; Formula 65ln( x+ 1) dx = ln udu =∫u= x+1du = dx[ x ]F 65( x+ 1) ln( + 1) −1 . Thus∫cc∫ ln( x+ 1) dx = ( x+ 1) [ ln( x+ 1) − 1]=00( c+ 1)ln( c+ 1) −cand( c+ 1)ln( c+ 1) − c = 1⇒ ln( c+ 1) = 1 ⇒c+ 1 = e⇒ c = e−1 ≈1.7182844. Substitution ; Formula 32c x 1 c + 1 1∫ dx = du0 2x 1 2∫ =+1 uu= x2+ 1du = 2x dx1 2c + 1 1 2[ ln u]= ln( c + 1)2121 ln(2 1) 12 12c + = ⇒ c + = e ⇒22c = e −1≈2.528462 Section 7.6 Instructor’s Resource Manual

45. There is no antiderivative that can be expressedin terms of elementary functions; anapproximation for the integral, as well as aprocess such as Newton’s Method, must be used.Several approaches are possible. c ≈ 0.5960146. Integration by parts; partial fractions; Formula 17a.b.∫3 3 xln( x + 1) dx = x ln( x + 1) − 3∫dx =3x + 1u= ln( x3+ 1)23xdu =x3+ 1dv = dx,v = x3 ⎛ 1 ⎞xln( x + 1) −3∫⎜1− dx =3 ⎟⎝ x + 1⎠3⎛ 1 ⎞x ln( x + 1) − 3x+ 3∫dx⎜2( x+ 1)( x − x+1)⎟⎝⎠1 A Bx+C= + =2 1 2( x+ 1)( x − x+ 1) x + x − x+12( A+ B) x + ( B+ C− A) x+ ( A+C)⇒2( x+ 1)( x − x+1)A+ C = 1 B+ C = A A=−B⇒1 1 2A= B =− C = .3 3 3Therefore3 1∫dx =( 2x+ 1)( x − x+1)1 x − 2∫ dx − dx =x + 1∫ 2x − x+1x − 2ln x+ 1 − ∫ dx =1 2 3( x − ) +2 41u= x−2du = dx3−2du2 3+ F174uln x+ 1 − ∫ =u1 2 −1ln x+ 1 − ln x − x+ 1 + 3 tan ( x−)2332 1( 2 )c. Summarizingc 3ln( x + 1) dx =∫0⎡ 3xln( x + 1) − 3x+ ln( x+ 1) − ⎤⎢⎥⎢1 ln(2 1) 3 tan−12 1x x ( ( x )⎥=⎢− + + −⎣ 23 2 ) ⎥⎦0⎧⎛3c + 1 ⎞ ⎫⎪c(ln( c + 1) − 3) + ln+ ⎪⎪⎜ 2 ⎟c c 1 ⎪⎨⎝ − + ⎠ ⎬⎪−12 1 3π⎪⎪ 3tan ( ( c− )3 2 ) +⎪⎩6 ⎭Using Newton’s Method , with⎧⎛3c + 1 ⎞ ⎫⎪c(ln( c + 1) − 3) + ln+ ⎪⎪⎜ 2 ⎟Gc ()c c 1 ⎪= ⎨ ⎝ − + ⎠ ⎬⎪−12 1 3π⎪⎪ 3tan ( ( c− )3 2 ) + −1⎪⎩6 ⎭3and G′ () c = ln( c + 1) we getn 1 2 3 4 5an2.0000 1.6976 1.6621 1.6615 1.6615Thereforec 3∫ ln( x + 1) dx = 1 ⇒0c ≈1.661547. There is no antiderivative that can be expressedin terms of elementary functions; anapproximation for the integral, as well as aprocess such as Newton’s Method, must be used.Several approaches are possible. c ≈ 0.1666848. There is no antiderivative that can be expressedin terms of elementary functions; anapproximation for the integral, as well as aprocess such as Newton’s Method, must be used.Several approaches are possible. c ≈ 0.250949. There is no antiderivative that can be expressedin terms of elementary functions; anapproximation for the integral, as well as aprocess such as Newton’s Method, must be used.Several approaches are possible. c ≈ 9.236550. There is no antiderivative that can be expressedin terms of elementary functions; anapproximation for the integral, as well as aprocess such as Newton’s Method, must be used.Several approaches are possible. c ≈ 1.96cInstructor’s Resource Manual Section 7.6 463

51.8f( x) = 8 − x g( x) = cx a = 0 b= c + 1a.b.c.b8c+1a082 ⎛c⎞ 3c+1∫ ∫x( f( x) − g( x)) dx = 8 x− ( c+ 1) x dx =⎡ + 1 ⎤ 256 512⎢4x− ⎜ ⎟x3⎥ = − =2 2⎣ ⎝ ⎠ ⎦0( c+ 1) 3( c+1)25623( c + 1)c+108c+10⎛c⎞82c+1∫ ∫( f ( x) − g( x)) dx = 8 − ( c+ 1) x dx =⎡ + 1 ⎤ 64 32⎢8x− ⎜ ⎟x2⎥ = − =⎣ ⎝ ⎠ ⎦0( c+ 1) ( c+1)32( c + 1)⎛ 256 ⎞ ⎛c+ 1⎞8x = =⎜ 2⎟⎜3( c 1)⎟ ⎟⎝ + ⎠⎝32 ⎠ 3( c + 1)8 1x = 2⇒ = 2⇒ c =3( c + 1) 352. f ( x) = c g( x) = x a = 0 b = ca.b.c.ba∫ ∫2 3c30c0x( f( x) − g( x))dx = cx− x dx =⎡cx x ⎤ c⎢ − ⎥ =⎢⎣2 3 ⎥⎦6ba∫ ∫( f ( x) − g( x)) dx = ( c− x)dx =2c2⎡ x ⎤ c⎢cx− ⎥ =⎢⎣2 ⎥⎦20⎛ 3c ⎞⎛ 2 ⎞⎜ ⎟⎜ 2 ⎟cx = =⎜ 6 ⎟⎝ ⎠ ⎝ c ⎠ 3x = 2⇒ c = 6c022− x53. 3( )a.f ( x) = 6e g x = 0 a = 0 b=cc − x30 u=x− xdv = e 3du = dx− xv=−3e3bx( f( x) − g( x)) dx = 6 xe dx =a∫∫c⎡ −x3⎤c −x6 3 18 3⎢− xe ⎥ + ∫ e dx =⎣ ⎦ 00xc⎡ −−c18 e 3 ⎤( x 3) 18 e 3⎢− + ⎥ =− ( c+ 3) + 54⎣⎦0ba⎛ − c18 e 3 ⎞− ⎜ −1⎟⎝ ⎠∫ ∫c0− xb. ( f ( x) − g( x)) dx = 6e 3 dx =c. For notational convenience, letu =− 18e − c 3 ; thenuc ( + 3) + 54 cu 3( u+18)x = = + =u+ 18 u+ 18 u+18cu+ 3u + 18cucx = 2⇒ = −1⇒ = −1⇒u + 18181+u1 1 − cc = −1⇒ = e 3− c3 c + 1eLet1 −c3 1 −c3 1hc () = − e , h′() c = e −c + 1 3 2( c + 1)and apply Newton’s Methodn1 2 3 4 5 62.0000 5.0000 5.6313 5.7103 5.7114 5.7114a n5.7114 c ≈⎛πx ⎞54. f( x) = csin ⎜ ⎟ g( x) = x a = 0 b = c⎝ 2c⎠(Note: the value for b is obtained by setting⎛πx ⎞csin⎜ ⎟ = x This requires that x be a zero for⎝ 2 c ⎠c⎛π⎞the function hu ( ) = u−sin⎜ u⎟⎝ 2 ⎠ . ApplyingNewton’s Method to h we discover that the zerosof h are -1, 0, and 1. Since we are dealing withpositive values, we conclude that x =1 or x = c.)c464 Section 7.6 Instructor’s Resource Manual

a.b.c.55. a.b.56. a.bc⎡⎛ x ⎞ 2 ⎤∫ x( f( x) − g( x)) dx = sina∫0⎢cx ⎜ ⎟−x dx2 c⎥⎣ ⎝ ⎠ ⎦3cc ⎛πx⎞⎡x⎤= ∫ cxsin dx0 ⎜ ⎟ −⎢ ⎥⎝ 2 c ⎠ ⎢⎣3 ⎥⎦π π0u= x,du=dx2c2cπ32 ⎛2c ⎞ ⎛2c⎞⎡c⎤= ∫ c u sin u du0 ⎜ ⎟ ⎜ ⎟ −⎢ ⎥⎝ π ⎠ ⎝ π ⎠ ⎢⎣3 ⎥⎦4c ⎡c ⎤ 4c c= − − = −π⎢⎣3 ⎥⎦π 3F 403 π 3 3 32[ sin u ucosu]2 0 ⎢ ⎥23 ⎛ 4 1⎞= c ⎜ −2 ⎟⎝π3 ⎠bc⎡⎛ x ⎞ ⎤∫ ( f( x) − g( x)) dx = sina∫ 0 ⎢c ⎜ ⎟− x dx 2 c⎥ =⎣ ⎝ ⎠ ⎦2 2c⎡2 22c ⎛πx⎞x ⎤ 2c c⎢− cos⎜⎟− ⎥ = − =⎢⎣π ⎝ 2c⎠ 2 ⎥⎦π 202 ⎛ 2 1⎞c ⎜ − ⎟⎝π2 ⎠23 ⎛12−π⎞c⎜ 23π⎟ ⎡ 22(12 −π) ⎤x =⎝ ⎠= c⎢ ⎥2 ⎛4 −π ⎞ 3 π (4 π )c⎢ − ⎥⎜ ⎟⎣ ⎦⎝ 2π⎠3 π(4 −π)x = 2 ⇒ c = ≈3.798212 −π2 x 2−terf ( x)= ∫ e dtπ 0d 2∴ erf ( x ) = edx πx sin tSi( x)= ∫ dt0 td sin x∴ Si( x)=dx x2−xx ⎛ 2πt⎞S( x) = ∫ sindt0 ⎜ 2 ⎟⎝ ⎠2d ⎛πx ⎞∴ Sx ( ) = sindx ⎜ 2 ⎟⎝ ⎠ππb.x ⎛ 2πt⎞Cx ( ) = ∫ cosdt0 ⎜ 2 ⎟⎝ ⎠2d ⎛πx ⎞∴ Cx ( ) = cosdx ⎜ 2 ⎟⎝ ⎠57. a. (See problem 55 a.) . Since erf ′( x)> 0 for allx , erf ( x ) is increasing on (0, ∞ ) .b.−4x2xerf ′′−( x)= e which is negative onπ(0, ∞ ) , so erf ( x)is not concave upanywhere on the interval.58. a. (See problem 56 a.) Sinceπ 2S′ ⎛ ⎞( x) = sin ⎜ x ⎟, S′( x) > 0 when⎝ 2 ⎠π 2 20 < x < π or 0 < x < 2; thus2( )0, 2 .S x is increasing on ( )π 2b. Since S′′ ⎛ ⎞( x) = π xcos ⎜ x ⎟, S′′( x) > 0⎝ 2 ⎠whenπ 2 π 3π π 20 < x < and < x < 2 π ,2 2 2 22 2or 0 < x < 1 and 3 < x < 4.Thus S( x ) is concave up on(0,1) ∪ ( 3,2) .59. a. (See problem 56 b.) Sinceπ 2C′ ⎛ ⎞( x) = cos ⎜ x ⎟, C′( x) > 0 when⎝ 2 ⎠π 2 π 3π π 20 < x < or < x < 2 π ; thus2 2 2 2Cx ( ) is increasing on (0,1) ∪ ( 3,2) .π 2b. Since C′′ ⎛ ⎞( x) =− π xsin ⎜ x ⎟, C′′( x) > 0⎝ 2 ⎠π 2when π < 2 .2 x < π Thus Cx ( ) is concaveup on ( 2,2).60. From problem 58 we know that S( xis ) concaveup on (0,1) and concave down on (1, 3) so thefirst point of inflection occurs at x = 1 . NowS1 ⎛π2(1) sin0t ⎞= ∫ ⎜ ⎟⎝dt . Since the integral cannot2 ⎠be integrated directly, we must use someapproximation method. Methods may vary butthe result will be S(1) ≈ 0.43826 . Thus the firstpoint of inflection is (1,0.43826)Instructor’s Resource Manual Section 7.6 465

7.7 Chapter ReviewConcepts Test1. True: The resulting integrand will be of theform sin u.17. False:18. True:2x1 1x2 =–11 + − 2( x − 1) 2( x + 1)2x + 2 2 3 3=− + +2xx ( −1)x 2( x+ 1) 2( x−1)2. True: The resulting integrand will be of the1form .2 2a + u3. False: Try the substitution4 3u = x , du = 4x dx4. False: Use the substitutiondu = (2x – 3)dx.u = x2 – 3x+5,5. True: The resulting integrand will be of the1form .2 2a + u6. True: The resulting integrand will be of the1form .2 2a − x7. True: This integral is most easily solvedwith a partial fraction decomposition.8. False: This improper fraction should bereduced first, then a partial fractiondecomposition can be used.9. True: Because both exponents are evenpositive integers, half-angle formulasare used.10. False: Use the substitutionx xu = 1 + e , du = e dx11. False: Use the substitution2u = – x –4 x, du = ( −2x−4) dx12. True: This substitution eliminates theradical.13. True: Then expand and use the substitutionu = sin x, du = cos x dx14. True: The trigonometric substitutionx = 3sin t will eliminate the radical.15. True: Let u = ln x1du = dxx16. False: Use a product identity.2dv = x dx1 3v = x319. True:20. False:2x + 2 2 – x= +2 2xx ( + 1) x x + 1x + 22 2x ( x −1)1 2 3 1=− − + −x 2x 2( x− 1) 2( x+1)21. False: To complete the square, add2b.4a22. False: Polynomials can be factored intoproducts of linear and quadraticpolynomials with real coefficients.23. True: Polynomials with the same values forall x will have identical coefficientsfor like degree terms.24. True: Let u = 2x; then du = 2dxand2 2 1 2 2∫x 25 − 4x dx = u 25 −u du8∫which can be evaluated usingFormula 57.25. False: It can, however, be solved by the2substitution u = 25 − 4x; thendu = −8x dx and2 1∫x 25 − 4x dx =−8∫udu =1 3(25 42− − x ) 2 + C1226. True: Since (see Section 7.6, prob 55 a.)2 2xerf ′−( x) = e > 0 for all x ,πerf ( x)is an increasing function.27. True: by the First Fundamental Theorem ofCalculus.28. False: Since (see Section 7.6, prob 55 b.)sin xSi′ ( x)= , which is negative on,xsay, ( π ,2 π ), Si( x)will be decreasingon that same interval.466 Section 7.7 Instructor’s Resource Manual

Sample Test Problems4 t21.dt⎡9 t⎤∫= + = 5− 3=20 2 ⎢ ⎥9 + t⎣ ⎦022 cos 2θ2. ∫cot (2 θ ) dθ = ∫ dθ2sin 2θ21−sin 2θd θ2= ∫=2 ∫ (csc 2θ−1)dθsin 2θ1=− cot 2 θ − θ + C23. ∫4. ∫(Use integration by parts with u = x,dv = sin 2x dx .)32 25.y + y ⎛⎞∫ dy = ⎜ y − y + 2 − ⎟ dyy+ 1∫⎝ 1+y⎠1 3 1 2= y − y + 2y− 2ln1+ y + C3 23 26. ∫sin (2 tdt ) = ∫ [1– cos (2 t)]sin(2 tdt )1 1 3= – cos(2 t) + cos (2 t)+ C2 6y–2 1 2 y–47. ∫ dy =2 2∫ dy2y – 4y+ 2 y –4y+21 ln2= y – 4 y+ 2 + C23/2 dy3/28. ∫ = ⎡ 2y+ 1⎤= 2− 1=10 2y+ 1 ⎣ ⎦0π /2 cos xcos xπ /2e sin xdx = ⎡− e ⎤ = e−1 ≈1.7180 ⎣ ⎦0π /4π /4⎡sin 2xx ⎤ 1xsin 2xdx = cos 2x0⎢ −4 2⎥ =⎣⎦04410.11.12.13.2sin x+cos x ⎛ cos x⎞∫ dx = cos x dxtan x∫+⎜ sin x ⎟⎝⎠⎛21−sin x ⎞= ∫cos x +dx⎜ sin x ⎟⎝⎠∫= (cos x + csc x−sin x)dx= sin x + ln csc x− cot x + cos x+C(Use Formula 15 for ∫ csc x dx .)∫dx16 + 4 x– 2(Complete the square.)∫1 1–1 ⎛ x − ⎞= sin ⎜ ⎟+C2x 2 ⎝ 3 ⎠2 x x2x edx= e − x+ x + C(2 2 )Use integration by parts twice.2 2tan , sec3 32y = t dy = tdt23dy sec t∫ =2∫ dt2+3y2sect1 1= ∫ sectdt= ln sect+ tan t + C3 322+23 yC12 23 31 y= ln + +32+23+ yC1231 y= ln+31 2 2= ln y + + y + C3 3yy +Note that tan t = , so sec t = .2312 23239.∫2te t tdt = e + 2ln e − 2 + Cte − 2(Use the substitution u = e − 2 ,tdu = e dtwhich gives the integral∫tu + 2du.⎞ ⎟⎠ u14.15.∫3w 1 2 1dw2= – w – ln1– w + C21– w 2 2Divide the numerator by the denominator.∫tan x dx = –ln ln cos x + Cln cos xUse the substitution u = ln cos x .Instructor’s Resource Manual Section 7.7 467

16.3dt1 t + 2∫ =3 ∫ dt − dt1 2t –1 t −∫t + t+11 1 2t+ 4= ∫ dt −dtt − 1 2∫2t + t+11 1 2t+ 1+3= ∫ dt −dtt − 1 2∫2t + t+11 1 2t+ 1 3 1= ∫ dt − dt −dtt − 1 2∫ t + t+ 2∫t + +( )2 211 32 41 2 1⎞= ln t−1 − ln t + t+ 1 − 3 tan ⎜ ⎟+C2 ⎝ 3 ⎠∫17. sinh x dx = cosh x + C118. u = ln y, du = dyy2 − 1⎛t +5(ln y) 5 1 (ln )6dy = u du = y + Cy6∫ ∫19. u = xdu = dx2dv = cot x dxv = –cot x – x∫ ∫2 2x cot xdx=– xcot x– x – (– cot x– x)dx1 2= – x cot x– x + ln sinx + C22 22cot x csc x 1 cot x dx.Use20. u = x ,21.= − for1 −1/2du = x dx2sin x dx = 2 sin u dux∫ ∫=− 2cos x + C2 2u = ln t , du = dtt∫2 2 2ln t [ln( t )]dt = + Ct 4∫22.23.24.25.26.27.28.2u = ln( y + 9) dv = dy2ydu = dy2y + 9v = y22 2 2yln( y + 9) dy = yln( y + 9) – dy2y + 9∫ ∫2 ⎛ 18 ⎞= y ln( y + 9) −∫2 − dy⎜2y + 9⎟⎝ ⎠2 –1⎛y ⎞= yln( y + 9) – 2y+ 6 tan ⎜ ⎟+C⎝ 3 ⎠∫t /3t /3 3 e (9cos3t sin3 t)e sin 3tdt −−= + C82Use integration by parts twice.t+ 9 1 – t+1∫ dt = dt dt3 ∫ +2t + 9t t∫t + 91 t 1= ∫ dt − dt dtt∫ +2 ∫ 2t + 9 t + 91 2 1 –1⎛t⎞= ln t – ln t + 9 + tan ⎜ ⎟+C2 3 ⎝3⎠3 cos cos2∫ sinx cosx dx = − x − x + C2 2 2 4Use a product identity.24 ⎛ x⎞ ⎛1+cosx⎞∫cos⎜ ⎟dx= ⎜ ⎟ dx⎝2 ∫⎠ ⎝ 2 ⎠1 1 1 2= 2cos cos4∫dx + x dx + x dx4∫ 4∫1 1 1= cos (1 cos 2 )4∫dx + x dx + + x dx2∫ 8∫3 1 1= x + sin x+ sin 2x+C8 2 163 1 2∫tan 2xsec 2 xdx=(sec 2 x–1) d(sec 2 x)2∫1 3 1= sec (2 x)– sec(2 x)+ C6 21u = x,du = dx2 x2x 2x⎛ 1 ⎞ u∫ dx = ∫ ⎜ dx ⎟ = 2∫du1+ x 1+ x ⎝2x ⎠ 1 + u( u+ 1)( u− 1) + 1 ⎛ 1 ⎞= 2∫ du = 2 ⎜u − 1+⎟duu+ 1∫⎝ u+1⎠⎛ 2u⎞= 2 − u+ ln u+ 1+ C⎜ 2⎟⎝⎠( )= x − 2 x + 2ln 1+ x + C468 Section 7.7 Instructor’s Resource Manual

29.∫∫∫ ∫3/2 4 3/2 2 23/2 2 7/2 2tan xsec xdx= tan x(1 + tan x)secxdx = tan xsec xdx+tan xsecxdx2 5/2 2 9/2= tan x + tan x+C5 930.1/6 6 5u = t + 1, ( u– 1) = t, 6( u− 1) du = dt5dt 6( u −1) du 6du= =1/6 6tt ( + 1) ( u−1)u uu ( –1)∫ ∫ ∫1 1= –6∫du + 6 duu∫u−11/6 1/6= –6ln t + 1 + 6ln t + C31.2 y2u = 9 − e , du =− 2e y dy∫2 ye1 −1/2dy =− u du =− u + C2 y9 − e 2∫2 y= − 9 − e + C32.∫∫5 2 2 1/2cos x sin xdx = (1– sin x) (sin x)cosxdx2 3/2 4 7/2 2 11/2= sin x – sin x+ sin x+C3 7 11∫ ∫ ∫1/2 5/2 9/2= sin x cos xdx– 2 sin xcos xdx+sin xcosxdx33.∫ ∫ln(3cos x)e dx = 3cosxdx = 3sin x+C34. y = 3 sin t, dy = 3 cos t dt29 − y 3costdy = ⋅3cost dty 3sint∫ ∫21−sin t= 3∫ = 3 (csct−sin t)dtsin t∫= 3⎡⎣ln csct− cott + cost⎤⎦+ C23 9 − y2= 3ln − + 9− y + Cy yy 3Note that sin t = , so csct= and3 y29– ycot t = .y4x 435. u = e , du = 4e x dx∫4xe 1 dudx =1 e 4∫+ 1+u8x21 tan−1 (4x= e ) + C436. x = a tan t,37.2 22dx = a sec t dtx + a asect2∫ dx = asect dt4 ∫ 4 4x a tan t31 sec t 1 cost=2∫ dt = dt4 2∫4a tan t a sin t1 ⎛ 1 1 ⎞ 1 3= – C2 ⎜ + 3 3 ⎟ = – csc t+C2a ⎝ sin t⎠3a2 2 3/21 ( x + a )= – + C32 3a x2 2xx + aNote that tan t = , so csc t = .ax2u = w+ 5, u = w+ 5, 2u du = dww 2 2dw 32 ( u –5) du u –10 u C∫ ∫= = +w + 53= 2 ( 5)3/2 –10( 5)1/23 w+ w+ + C38. u = 1 + cos t, du = –sin t dtsin tdt du∫ = – ∫ = –2 1+ cost+ C1+costu39.2u = cos y, du = –2cos y sin y dysin ycos y 1 dudy = –9+ cos y 2 9+u∫ ∫4 221 ⎛–1 cos y ⎞= – tan+ C6 ⎜ 3 ⎟⎝ ⎠Instructor’s Resource Manual Section 7.7 469

40.dx=dx1–6 x– x 10–( x+3)∫ ∫–1 ⎛ x + 3⎞= sin ⎜ ⎟+C⎝ 10 ⎠2 241.24x + 3x+ 6 A B Cx+D= + +2 2 2 2x ( x + 3) x x x + 3A = 1, B = 2, C = –1, D = 224x + 3x+ 6 1 1 – x+2∫ dx = dx 2 dx dx2 2 ∫ +2 2x ( x 3) x∫ + ∫+ x x + 31 1 1 2x1= ∫ dx + 2 dx dx 2 dxx∫ −2 2∫ +2 ∫ 2x x + 3 x + 32 1 2 2 –1⎛x ⎞= ln x – – ln x + 3 + tan ⎜ ⎟+Cx 2 3 ⎝ 3⎠242. x = 4 tan t, dx = 4secdxt dt1 116 16∫ = costdt= sin t+C2 3/2 ∫ ⎜ 2(16 + x )1 ⎛ x ⎞= + C16 ⎟⎝ x + 16 ⎠x= + C216 x + 1643. a.b.c.d.e.f.44. a.23–4x A B C= + +(2x+ 1) 2x+ 1 (2x+ 1) (2x+1)3 2 37 x–41A B C D E( x –1) (2 – x) = x–1 + ( x–1) + 2– x+ (2 – x) +(2 – x)2 3 2 2 33x+ 1 Ax+ B Cx+D= +( x + x+ 10) x + x+ 10 ( x + x+10)2 2 2 2 22( x+ 1)A B C D Ex+ F Gx+H= + + + + +( x – x+ 10) (1– x ) 1– x (1– x) 1+x (1 + x) x – x+ 10 ( x – x+10)2 2 2 2 2 2 2 2 25x A B C D Ex+ F Gx+H= + + + + +( x+ 3) ( x + 2x+ 10) x + 3 ( x+ 3) ( x+ 3) ( x+ 3) x + 2x+ 10 ( x + 2x+10)4 2 2 2 3 4 2 2 22 2(3x + 2 x–1)Ax+ B Cx+ D Ex+F= + +(2x + x+ 10) 2x + x+ 10 (2x + x+ 10) (2x + x+10)2 3 2 2 2 2 322⎡⎤ 21 1V =π ∫ ⎢ ⎥ dx =π1 2∫ dx1 2⎢⎣3 x–x ⎥⎦3 x–x1 A B23 x–x = x+ 3– x1 1A= , B =3 32 1⎛1 1 ⎞ πV =π ∫dx1⎜ + ⎟ = ⎡ln x – ln 3 – x ⎤3⎝x 3– x⎠3⎣⎦21π2π= (ln 2 + ln 2) = ln 2 ≈ 1.45173 3470 Section 7.7 Instructor’s Resource Manual

.V = 2π∫2x1 23 x–x22− 2x+ 3−3 2 3−2x2 1dx =−π ∫ dx = – π dx 3dx1 2∫ + π1 2∫ 1 23x−x 3x−x 3x−x⎡ 2 2 12 3x x⎤⎡ 2 −1⎛2x−3⎞⎤=−π − + 3πdx⎢⎣⎥⎦∫= ⎢−2π 3x− x + 3πsin⎜ ⎟3⎥⎣⎝ ⎠⎦( x − )119 32−4 211 1 12 2 3 sin − 2 2 3 sin− ⎛ ⎞ −=− π + π + π − π ⎜ − ⎟ = 6πsin ≈ 6.40583 ⎝ 3⎠31 12145.46.47.2x xy = , y′ =16 824 42⎛ x⎞x1 10⎜ ⎟0∫ ∫L = + dx = + dx⎝8⎠642x = 8 tan t, dx = 8sec ttan–11 tan–112 2 2 3sec 8sec 80 ∫ sec0∫L = t⋅ tdt =tdt = 4 ⎡ ⎣secttan t+ ln sect+ tan t ⎤⎦⎡⎛ 5⎞⎛1⎞1 5 ⎤ ⎛1+5⎞= 4⎢⎜ ⎟⎜ ⎟+ ln + ⎥2 ⎟= 5 + 4ln 4.1609⎢⎣⎝⎠ ⎝ 2 ⎠ 2 2⎜≈⎥⎦2 ⎟⎝ ⎠tan –1120Note: Use Formula 28 for3 13 1V =π∫dx =π0 2 2 ∫dx2 2( x + 5x+6)0( x+ 3) ( x+2)1 A B C D2 2 3 2 2 2( x+ 3) ( x+ 2) = x+ + ( x+ 3) + x++( x+2)A = 2, B = 1, C = –2, D = 13⎡2 1 2 1 ⎤ ⎡1 1 ⎤V =π ∫ ⎢ + – + ⎥dx=π0 x 3 2 2 2⎢⎣+ ( x+ 3) x+⎢2ln x+ 3 – –2ln x+2 –( x+2) ⎥⎦x+ 3 x+2⎥⎣⎦⎡⎛ 1 1⎞ ⎛ 1 1⎞⎤⎛ 7 4⎞=π ⎢⎜2ln6– –2ln5– ⎟ – ⎜ 2ln3– –2ln2– ⎟6 5 3 2 ⎥ =π ⎜ + 2ln ⎟≈0.06402⎣⎝ ⎠ ⎝ ⎠⎦⎝15 5 ⎠3 xV = 2π∫dx0 2x + 5x+6x A B= +2x + 5x+6 x+ 2 x+3A = –2, B = 33⎡2 3 ⎤V = 2 π ∫ –dx0 ⎢ +x+ 2 x+3⎥= 2 π [ –2ln( x+ 2) + 3ln( x+3) ] 3⎣⎦0⎛ 2⎞32= 2 π [(–2ln 5 + 3ln 6) – (–2ln 2 + 3ln 3)] = 2π ⎜3ln 2 + 2ln ⎟= 2πln ≈1.5511⎝ 5⎠2530∫3sec tdt.48.2 2V 2 4x 2– xdx= π∫0u = 2 – xx = 2 – udu = –dxdx = –du0 22 1/2 3/2 5/2 ⎡8 8 2V = 2π∫4(2– u) u(– du)= 8 π (4 u –4 u + u ) du2∫= 8 π u – u u0⎢+⎣3 5 7⎛16 2 32 2 16 2 ⎞ ⎛128 2 ⎞ 1024 2π= 8 π ⎜– +3 5 7 ⎟= 8π ⎜= ≈ 43.3287⎝⎠ 105 ⎟⎝ ⎠ 1053/2 5/2 7/2⎤⎥⎦20Instructor’s Resource Manual Section 7.7 471

49.ln 3V 2 2( e –1)(ln 3 – x)dx = 4 π [(ln3) e – xe –ln3 + x]dx= π∫0xln 3∫ 0x x x x x∫ ∫ by using integration by parts.Note that xe dx = xe – edx = xe – e + Cx x x 1 2V 4 ⎡⎤= π ⎢(ln3) e – xe + e –(ln3) x+x2⎥⎣⎦⎡ 1 2 ⎤= 4π⎢2 – ln 3 – (ln 3) ≈3.74372 ⎥⎣⎦ln 30xx⎡⎛2 1 2⎞⎤= 4π ⎢⎜3ln3–3ln3+ 3–(ln3) + (ln3) ⎟–(ln3+1)2⎥⎣⎝⎠ ⎦50.A =∫3 33 2 2x18dxx + 9x = 3 tan t, dx = 3sec t dt2π/3 18 2 π/3costA = ∫3sec tdt=2 dtπ/6 2 ∫ /6 227 tan tsect πsin tπ /3⎡ 1 ⎤ ⎛ 2 ⎞ ⎛ 1 ⎞= 2 ⎢– 2 – 2 4 1–sin t⎥ = ⎜ + ⎟=⎜ ⎟⎣ ⎦ ⎝ 3 ⎠ ⎝ 3⎠ ≈ 1.6906π /651.52.A = – ∫0–6t( t –1)2dtt A B= +2 ( –1) 2( t –1) t ( t –1)A = 1, B = 10 ⎡ 1 1 ⎤A = – ∫ ⎢ + ⎥dt–6 t –1 2⎢⎣( t –1) ⎥⎦– ⎡ln –1 – 1 ⎤= ⎢ tt –1 ⎥⎣⎦0–6⎡ ⎛ 1⎞⎤6= – ⎢(0 + 1) – ⎜ln 7 + ⎟ = ln 7 – ≈1.08887⎥⎣ ⎝ ⎠⎦72–1 –1⎛ 6 ⎞36V =π ∫ –3⎜ ⎟ dx =π ∫ dxx x 4 –3 2⎝ + ⎠ x ( x+4)36 A B C2 2x ( x+4)= x+ x+ x + 49 9A= – , B = 9, C =4 4–1⎡9 9 9V =π ∫ + +–3 ⎢⎣ x +– 4 x 2 4( x 4)9π⎡⎛ 4⎞⎤= (4 ln 3) – – ln 34⎢ + ⎜ + ⎟3⎥⎣ ⎝ ⎠⎦53. The length is given byπ /3 21 + [ ′( )]π /6∫π /3 cos xf x dx = ∫ 1+dx =π /6 sin2 ∫x/3x⎦π π /6=⎡ ⎣ln csc x − cot ⎤⎤ 9π –1⎛ 1 4 1 ⎞⎥dx= ––3 2⎦ 4∫ ⎜ + + ⎟⎝ x x x+4⎠dx 9π ⎡ 4 ⎤= –ln x – ln x 44⎢ + +x⎥⎣⎦9π⎛8 ⎞ 3π= ⎜ + 2ln 3 ⎟= (4 + 3ln 3) ≈34.38084 ⎝3 ⎠ 22π /3π /62 1= ln − −ln 2 − 33 3sin2 2x + cos2sin x–1–3x π/3 1 π/3dx = ∫ dx = csc x dxπ/6 sin x∫ π/6⎛ 1 ⎞⎛2 3+3⎞= ln ⎜ ⎟−ln(2 − 3) = ln⎝ 3 ⎠⎜ 3 ⎟≈ 0.768⎝ ⎠472 Section 7.7 Instructor’s Resource Manual

54. a. First substitute u = 2 x, du = 2dxto obtain∫2 22x81− 4x9 + 81−4dx = 81−4x − 9ln+ Cx2xb. First substitute u e , du e dx∫2 281−4x81−u∫ dx = dux∫ , then use Formula 55:u3 3x xx 2x2 2 2= = to obtain ( 9− ) = ( 9−)3 xe( ) ( )xx 2x 2 2x 2x243 1 ee 9− e dx = 45−2e 9− e + sin− ⎛⎞+ C8 8 ⎜ 3 ⎟⎝ ⎠55. a. First substitute u = sin x, du = cos xdx to obtain∫∫ e e dx ∫ u du , then use Formula 62:2 2∫cos x sin x + 4 dx = ∫ u + 4 du , then use Formula 44:2 sin x 2 2cos x sin x+ 4 dx = sin x+ 4 + 2ln sin x+ sin x+ 4 + C21 1 dub. First substitute u = 2x, du = 2dxto obtain1−4x2 1−u1 1 2x+ 1Then use Formula 18: ∫ dx = ln + C .21−4x4 2x−1∫ dx = ∫ .2 256. By the First Fundamental Theorem of Calculus,⎧sinx⎧ xcosx−sin x⎪ x ≠ 0⎪for x ≠ 0Si′ ( x) = xSi ( x)2⎨′′ = ⎨ x⎩⎪ 1 x = 0 ⎪⎩ 0 for x = 057. Using partial fractions (see Section 7.6, prob 46 b.):21 1 A Bx+ C ( A+ B) x + ( B+ C− A) x+ ( A+C)= = + = ⇒3 2 1 2 21 + x ( x+ 1)( x − x+ 1) x + x − x+ 1 ( x+ 1)( x − x+1)A+ C = 1 B+ C = A A=−B 1 1 2⇒ A= B =− C = .3 3 3Therefore:⎡⎤⎢⎥1 1⎡1 x−2 ⎤ 1⎢x−2⎥∫ dx = ln 13 3⎢∫ dx − dx = ⎢ x + −dx⎥1 2 31 2 31 x x +∫ ⎥ ∫+ ⎣ x − x+1 ⎦ ⎢ ( x − ) + ⎥⎢2 41 ⎥⎢u = x− , du = dx⎣⎥2 ⎦⎡ 3u − ⎤2 1⎡ x + 1⎤−12 1= ⎢ln x+ 1 − du⎥ = ⎢ln + 3 tan2 3 ( ( x−)3 2 ) ⎥⎢ ∫u + ⎥F173 ⎢ 24 ⎣ x − x+1⎥⎣⎦⎦+ ⎡ π ⎤⎤∫ = + − + ⎥0 1 x 3⎢23c c 16⎥.+ ⎣ − + ⎣⎦⎥⎦1⎡c + 1 ⎡ −12 1 π ⎤⎤Letting Gc () = ⎢ln + 3 tan ( ( c− )3 2 ) + ⎥−0.53⎢2⎢c c 16⎥and⎣ − + ⎣⎦⎥⎦c 1Method to find the value of c such that ∫ dx = 0.5 :0 31+xc 1 1⎡c 11 2 1so dx ⎢−ln 3 tan3⎢ ( ( c )2 )n 1 2 3 4 5 6a n 1.0000 0.3287 0.5090 0.5165 0.5165 0.5165Thus c ≈ 0.5165 .1G′ () c = we apply Newton’s31 + cInstructor’s Resource Manual Section 7.7 473

Review and Preview Problems1.2.3.2 2x + 1 2 + 1 5lim = =x→22 2x −1 2 −132x+ 1 2(3) + 1 7lim = =x→3x + 5 3+5 82x − 9 ( x+ 3)( x−3)lim = lim=x→3 x−3 x→3x−3lim ( x + 3) = 3 + 3 = 6x→3−114. Note that, ifθ= sec x,then1 −11secθ = x ⇒ cosθ = ⇒ θ = cos . Hencexx−1 −lim sec x = lim cos11= 1x→∞x→∞xx15. f ( x) = xe −2y4.5.6.2x − 5x+ 6 ( x−2)( x−3)lim= lim=x→2 x−2 x→2x−2lim ( x − 3) = 2 − 3 = −1x→2sin 2x 2sin xcosxlim = lim=x→0 x x→0x⎛sinx ⎞lim 2⎜⎟cos x = 2(1)(1) = 2x→0⎝ x ⎠tan3x⎛ sin3x⎞⎛ 3 ⎞= ⎜ ⎟⎜ ⎟=x ⎝ x⎠⎝ x⎠⎛sin 3x⎞⎛ 1 ⎞lim 3⎜ ⎟⎜ ⎟ = 3(1)(1) = 3⎝ 3x⎠⎝cos3x⎠lim limx→0 x→0cos3 3x→0−2510xWe would conjecture lim xe − = 0 .2 x16. f ( x) = x e −2yx→∞x7.12 1+x + 1 2 1 0lim limx += = = 1 or:x→∞2x −1 x→∞11−1−02x2x + 1 2lim = lim 1 + = 1 + 0 = 1x→∞2 2x −1 x→∞x −1−2We would conjecture510 x2xlim xe − = 0 .x→∞8.12 +2x+ 1 2 0lim limx += = = 2x→∞x + 5 x→∞51+1+0x3 x17. f ( x) = x e −5y9.10.11.− x 1lim e = lim = 0x→∞x→∞xe2− x 1lim e = lim = 02x→∞x→∞xelim ex→∞2 x=∞ (has no finite value)−5We would conjecture510 x3xlim xe − = 0 .x→∞12.limx→−∞e− 2x2u= lim e =∞( u=−x)u→∞(has no finite value)13.lim tanx→∞−1πx =2474 Review and Preview Instructor’s Resource Manual

4 x18. f ( x) = x e −19.2−2yWe would conjecture10 xy = x e −y480,000510 x10xlim x e − = 0 .x→∞23.∫ax1 22dx = ⎡ln(1 + x ) ⎤ = ln⎛1+a⎞⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎝ ⎠0 21+x 20u=x2du=2x dxaa 1 2 4 8 16⎛ln 12 ⎞⎜ + a ⎟⎝ ⎠0.3466 0.8047 1.4166 2.0872 2.774524. ∫ dx = [ ln(1 + x) ] = ln0( 1+a)25.011+xa 1 2 4 8 16( + a)aln 1 0.6931 1.0986 1.6094 2.1972 2.8332aa 1 ⎡ 1⎤1dx = 11 2 ⎢− x x⎥= −⎣ ⎦1a∫a 2 4 8 1611−a0.5 0.75 0.875 0.9375a240,00010 20 x2 xWe would conjecture lim xe − = 0 .x→∞20. Based on the results from problems 15-19, wewould conjecturelim n xxe − = 021.x→∞a a−x −x −ae dx = ⎡− e ⎤ = 1−e0 ⎣ ⎦0∫a 1 2 4 8 161− e−a0.632 0.865 0.982 0.9997 0.9999+26.27.∫1a1 ⎡ 1 ⎤ 1⎡ 1 ⎤dx = ⎢− ⎥ = ⎢1− ⎥x ⎣ x ⎦ ⎣ a ⎦3 2 22 12aa 2 4 8 161⎡1 ⎤⎢1−2 a2⎥⎣ ⎦0.375 0.46875 0.4921875 0.4980468754 14dx = ⎡2 x ⎤ 4 2 aa x⎣ ⎦= −a∫a 1 1214181164−2 a 2 2.58579 3 3.29289 3.54 1 4 4= = lna xa a28. ∫ dx [ ln x]a 1 1214181164lna1.38629 2.07944 2.77259 3.46574 4.1588822.2a 2 2 −a−x 1 ⎡ −x ⎤ exe dx = − e 10u=−x2 2⎢= −⎣ ⎥⎦2∫du=−2x dx1a 1−22ea1 0.816060282 0.932332364 0.9999999448 1 − (8.02×10−29)16 1Instructor’s Resource Manual Review and Preview 475

CHAPTER8Indeterminate Forms andImproper Integrals8.1 Concepts Review1. lim f ( x); lim g( x)x→a x→a2.3.f ′( x)g′( x)2sec x; 1; lim cos x ≠ 0x→04. Cauchy’s Mean ValueProblem Set 8.11. The limit is of the form 0 .02 x–sinx 2–cosxlim = lim = 1x→0 x x→012. The limit is of the form 0 .0cos x – sin xlim = lim = 1x→π/2 π /2– x x→π/2 –13. The limit is of the form 0 .0x–sin2x 1–2cos2x1–2lim = lim = = –1x→0 tan x x→02sec x 14. The limit is of the form 0 .03–1x21+9xx→0 –1 0 1sin x x→21– xtan 3 3lim = lim = = 315. The limit is of the form 0 .02x + 6x+ 8 2x+6lim = limx→–2 2x –3 x–10x→–22 x –32 2= = ––7 76. The limit is of the form 0 .03 2 2x – 3x + x 3x + 6x+1 1 1lim = lim = = –x→0 3 0 2x –2x x→3 x –2 –2 27. The limit is not of the form 0 .0– 2 2 –As x→ 1 , x –2x+ 2→1, and x –1→ 0 so2x –2x+ 2lim = – ∞– 2x→1x + 18. The limit is of the form 0 .012 2xln x2x 1lim = lim = lim = 1x→1 2 1 2 1 2x –1 x→ x x→x9. The limit is of the form 0 .013ln(sin x)3sin xlim = limx→π/2 x→π/2π /2– x–10= = 0–110. The limit is of the form 0 .0x – x x – xe – e e + e 2lim = lim = = 1x→0 2sin x x→02cosx211. The limit is of the form 0 .023sin x cos x21 3t – t–2 t2 t –2 3lim = lim = = –t→1 ln t t→11 1 2t12. The limit is of the form 0 .0xx7 ln7x7 –1 2 x 7 ln7lim = lim = lim+0 x+ x2 –1 0 2 ln2+x→ x→ x→0x2 ln22 xln 7= ≈ 2.81ln 213. The limit is of the form 0 . (Apply l’Hôpital’s0Rule twice.)–2sin 2xln cos 2 x cos 2x–2sin 2xlim = lim = limx→0 27xx→0 14x x→014xcos 2x–4cos2 x –4 2= lim = = –x→014cos 2 x– 28xsin 2x14 – 0 7476 Section 8.1 Instructor's Resource Manual

14. The limit is of the form 0 .03sin x 3cos xlim = lim– – 1x→0 – x x→0–2 – x= lim – 6 – xcos x = 0–x→015. The limit is of the form 0 . (Apply l’Hôpital’s0Rule three times.)2tan x– x sec x–1lim = limx→0sin 2 x– 2x x→02cos 2 x– 22 4 2 22sec x tan x 2sec x+4sec xtanx= lim= limx→0 –4sin 2 x x→0–8cos 2x2+0 1= = ––8 416. The limit is of the form 0 . (Apply l’Hôpital’s0Rule three times.)2sin x – tan x cos x– sec xlim= limx→0 2 02x sin x x→2xsin x+x cos x2–sin x –2sec xtanx= limx→02sin 4 cos –2x + x x x sin x4 2 2–cos x –2sec x–4sec xtanx= limx→026cos x– x cos x–6xsinx–1–2–0 1= = –6–0–0 217. The limit is of the form 0 . (Apply l’Hôpital’s0Rule twice.)2x2x2lim = lim = lim+ sin x – x + cos x–1 + −sinxx→0 x→0 x→0This limit is not of the form 0 .0 As+−x →0 , 2→ 2, and −sin x → 0 , so2lim =−∞ .+x→0sin x18. The limit is of the form 0 . (Apply l’Hôpital’s0Rule twice.)xxe –ln(1 + x)–1e –lim = limx→0 2 x→02xxe +12(1 + x)1+1= lim = = 1x→02 211+xx19. The limit is of the form 0 . (Apply l’Hôpital’s0Rule twice.)1–1–1tan x–x21+x= =x→0 3 x→0 2 x→0–2x2 2(1 + x )lim lim lim8x24x48 x1 1= lim – = –x→02 224(1 + x ) 2420. The limit is of the form 0 . (Apply l’Hôpital’s0Rule twice.)cosh x–1 sinh x cosh x 1lim = lim = lim =x→0 2`x x→0 2xx→02 221. The limit is of the form 0 . (Apply l’Hôpital’s0Rule twice.)1−cosx−xsinxlim+2x→02 − 2cos x−sin x−xcosx= lim+x→02sin x − 2cos x sin sxsinx– cos x= lim2 2x+→02cos x –2cos x+2sin xThis limit is not of the form 0 .0+As x → 0 , xsin x– cos x→− 1 and2 2 +2cos x–2cos x+ 2sin x→ 0 , soxsinx– cos xlim = – ∞+2 2x→02cos x–2cos x+2sin x22. The limit is of the form 0 .02sin x + tan x cos x+sec xlimlim– – – –x 0 x x =→ e + e –2 x→0e x – exThis limit is not of the form 0 .0– 2As x→ 0 , cos x+ sec x→ 2, and2x – x – cos x+sec xe – e → 0 , so lim = – ∞ .– x – xx→0e – e23. The limit is of the form 0 0 .x∫ 1+sintdt0x→0 x→0lim = lim 1+ sin x = 1xInstructor’s Resource Manual Section 8.1 477

24. The limit is of the form 0 .0x∫ t costdt 0x cos xlim= lim2x→0 + x x→0+ 2x= cos xlimx+→02 x= ∞25. It would not have helped us because we provedsin xlim = 1 in order to find the derivative ofx→0xsin x.26. Note that sin ( 1 0 ) is undefined (not zero), sol'Hôpital's Rule cannot be used.1 ⎛1⎞As x →0, →∞ and sin ⎜ ⎟ oscillates rapidlyx⎝ x ⎠between –1 and 1, sox2 sin ( 1 x ) x2lim ≤ lim .x→0 tan x x→0tanxx 2 2 cos=x xtan x sin x2x cos x ⎡⎛x ⎞ ⎤lim = lim xcos x 0x→0 sin x⎢⎜⎟ =x→0sin x⎥ .⎣⎝⎠ ⎦x2 sin ( 1x )Thus, lim = 0 .x→0tan xA table of values or graphing utility confirmsthis.27. a. OB = cos t, BC = sin t and AB = 1–cos t,so the area of triangle ABC is 1 sin t(1– cos t ).2The area of the sector COA is 1 2 t while the area of triangle COB is 1 costsin t , thus the area of the curved2region ABC is 1 ( –cos sin ).2 t t t 1area of triangle ABCsin t(1– cos t)2lim= lim+ 1t 0 area of curved region ABC +→t→0 2( t –costsin t)2 2sin t(1– cos t) cos t – cos t+sin t 4sin tcos t – sin t 4cos t –1 3= lim= lim= lim= lim =2 2t→0 + t –costsint t→0+ 1–cos t+ sin t+t 0 4costsint +→t→04cost4(L’Hôpital’s Rule was applied twice.)1 2b. The area of the sector BOD is cos ,2 t t so the area of the curved region BCD is 1 1 2costsin t – tcos t .2 21area of curved region BCD cos t(sin t – tcos t)lim= lim 2+ 1t 0 area of curved region ABC +→t→0( t –costsin t)22 2 2 2cos t(sin t – tcos t) sin t(2tcos t – sin t)2(cos t t –sin t) t(cos t –sin t)= lim= lim= lim= lim2 2t→0 + t –sintcost t→0+ 1–cos t+sin t+t 0 4costsint +→t→02costsint2 2= cos t – 4tcostsin t – sin t 1– 0 – 0 1lim2 2t 0 2cos t – 2sin t= +→2–0 = 2(L’Hôpital’s Rule was applied three times.)478 Section 8.1 Instructor’s Resource Manual

28. a. Note that ∠ DOE has measure t radians. Thus the coordinates of E are (cost, sint).Also, slope BC = slope CE . Thus,0−y sint−0=(1 −t) −0 cos t−(1 −t)(1 − t) sin t− y =cost+ t−1( t−1)sinty =cos t+t –1( t –1)sintlim y = lim+ +t→0 t→0cos t+t –1This limit is of the form 0 .0( t –1)sint sin t+( t –1)costlim= lim+ cos t+ t –1 + – sin t+1t→0 t→00 + (–1)(1)= = –1–0 + 1b. Slope AF = slope EF . Thus,t t−sin t=1−x1−costt(1−cos t)= 1−xt−sin tt(1+cos t)x = 1−t−sin ttcos t – sin tx =t –sinttcos t – sin tlim x = lim+ + t –sintt→0 t→0The limit is of the form 0 .0tcos t – sin t – tsintlim= lim+t 0 t – sin t +→t→01– cost– sin t – tcost tsin t – 2cost= lim= lim+t 0 sin t+→t→0cost(Apply l’Hôpital’s Rule three times.)0–2= = –2129. By l’Hộpital’s Rule 0 xx⎛ ⎞⎜ ⎟⎝0⎠ , we have e −1elim f( x) = lim = lim = 1and+ +x 0 x 0 x+→ → x→01xxe −1elim f( x) = lim = lim = 1 so we define f (0) = 1.− −x 0 x 0 x−→ → x→0130. By l’Hộpital’s Rule 0 1⎛ ⎞⎜ ⎟⎝0⎠ , we have ln xlim f( x) = lim limx1+ +x 1 x 1 x 1 +→ → −= x→11= and1ln xlim f( x) = lim limx1− −x 1 x 1 x 1 −→ → −= x→11= so we define f (1) = 1 .Instructor’s Resource Manual Section 8.1 479

231. A should approach 4 π b , the surface area of a sphere of radius b.⎡⎤22 222 2a – b a – b⎢ 2πabarcsin arcsin2 a ⎥a2alim 2 b 2 b 2 b lima b + ⎢ π +→2 2⎥ = π + πa→b+ 2 2⎢a – b ⎥a – b⎣⎦Focusing on the limit, we have2 22 2a – b 2 ⎛2 – 2 arcsinb ⎞a b a+ aa arcsina ⎜2 2⎟ ⎛2 2aa a – blimlim⎝ ⎠2 2 a – b⎞== lim ⎜2 a – b arcsin + b⎟=b.2 2aa→b + a – b a→b++a→b⎜a ⎟2 2⎝⎠Thus,2 2A = π b + π b b = π b+a→blim 2 2 ( ) 4 .a – b4 332. In order for l’Hôpital’s Rule to be of any use, a(1) + b(1) + 1 = 0, so b = –1 – a.Using l’Hôpital’s Rule,4 3 3 2ax + bx + 1 4ax + 3bxlim= limx→1( x –1)sinπx x→1sin π x+π( x–1)cosπxTo use l’Hôpital’s Rule here,3 24 a(1) 3 b(1) 0,+ = so 4a + 3b = 0, hence a = 3, b = –4.4 3 3 23 x –4x + 1 12 x –12xlim= limx→1 ( x – 1) sin πx x→1sin π x+π( x– 1) cos πx6a = 3, b = –4, c = – π236 x – 24x12= lim=x→122πcos πx– π ( x–1)sinπx–2π6= – π33. If f ′( a) and g′ ( a)both exist, then f and g areboth continuous at a. Thus, lim f ( x ) = 0 = f ( a )x→a and lim g ( x ) = 0 = g ( a ).x→a f ( x) f( x)– f( a)lim = limx→a g( x) x→ag( x)– g( a)f ( x)– f ( a)f ( x)– f ( a)limx–a x→ax–a f ′( a)lim = =x→ag( x)– g( a) g( x)– g( a)limg ′ ( a)x– ax→ax–a38.34.35.36.37.2cos x –1+x2 1lim=x→04x 24x 2 3e –1– x– x – x2 6 1lim=x→04x 2421–cos( x ) 1lim =x→03x sin x 2tan x−x sec x−1lim = lim = 2arcsin x−x−1x→0 x→0211−x2The slopes are approximately 0.02 / 0.01 = 2 and0.01/ 0.01 = 1. The ratio of the slopes istherefore 2/1= 2, indicating that the limit of theratio should be about 2. An application ofl'Hopital's Rule confirms this.480 Section 8.1 Instructor’s Resource Manual

39.41.The slopes are approximately 0.005/ 0.01 = 1/ 2and 0.01/ 0.01 = 1. The ratio of the slopes istherefore 1/2 , indicating that the limit of theratio should be about 1/ 2 . An application ofl'Hopital's Rule confirms this.The slopes are approximately 0.01/ 0.01 = 1 and− 0.01/ 0.01 = 1. The ratio of the slopes istherefore − 1/1=− 1, indicating that the limit ofthe ratio should be about − 1 . An application ofl'Hopital's Rule confirms this.40.42. If f and g are locally linear at zero, then, sincelim f( x) lim g( x) 0x→0 x→0g( x)qx p f= = , f ( x)≈ px and≈ , where = '(0) and q = g'(0).Then ( )/ ( ) / /f x g x px px p q≈ = when x isnear 0.The slopes are approximately 0.01/ 0.01 = 1 and0.02 / 0.01 = 2 . The ratio of the slopes istherefore 1/2 , indicating that the limit of theratio should be about 1/2 . An application ofl'Hopital's Rule confirms this.Instructor’s Resource Manual Section 8.1 481

8.2 Concepts Review1.2.f ′( x)g′( x)f ( x)g( x)lim or limx→a1 x→a1g( x)f ( x)3. ∞– ∞ , 0 ° , ∞° , 1 ∞4. ln xProblem Set 8.2∞1. The limit is of the form .∞1 99910001000 1000 xln xlim = limxx→∞x x→∞11000= lim = 0x→∞x∞2. The limit is of the form . (Apply l’Hôpital’s∞Rule twice.)3.1(ln x) 2 2(ln x)xlim = limx→∞xx2 x→∞2 ln21( )2lnx2x= lim = limx→∞xxx⋅ 2 ln2 x→∞2 ln2(1+xln2)2= lim = 0x→∞xx⋅ 2 ln2(1+xln2)10000xlim = 0 (See Example 2).x→∞xe∞4. The limit is of the form . (Apply l’Hôpital’s∞Rule three times.)3x3lim = = limx→∞xln(100 ) x 1xx + e →∞ (100 + e )x100x+exx300x + 3e 300 + 3e= lim= limx→∞xx100 + e x→∞ex3e= lim = 3x→∞xe∞5. The limit is of the form .∞3sec x +lim5 lim3sec x tan x=x→πtan x x→π2sec x2 23tanx= lim = lim 3sin x = 3sec xx→πx→π2 26. The limit is of the form – ∞.– ∞12ln sin x2sin xlim = lim+x 0 3lntan x +→x→03tan x2= 2cos x 2lim3 = 3+x→0∞7. The limit is of the form .∞1000ln(ln x )lim=x→∞ln x1000= lim = 0x→∞1000xlnx2sin x cosx2sec x1 11000 1000ln x xlimx→∞1x999( 1000x)8. The limit is of the form – ∞. (Apply l’Hôpital’s∞Rule twice.)122(4 – 8 x)(–8)ln(4 – 8 x)2(4–8 x)lim= lim1– tan πx1–2πsecπx( ) x→( )x→2 22–16cos πx 32πcos πxsinπx= lim= lim(4 – 8 ) –8π1 π x1( ) x→( )– –x→2 2= lim – 4cos πxsin π x = 01( ) –x→2∞9. The limit is of the form .∞cot x – csclim = lim–1x+0 –ln x x+→→02 x –lnx2 x –lnx= lim2x+→0sin x⎡ 2x⎤= lim csc x – ln x+ ⎢=∞x→0sin x⎥⎣⎦xsince lim = 1 while+x→0sin x+x→0lim – ln x = ∞ .+x→02xlim csc x =∞ and482 Section 8.2 Instructor’s Resource Manual

∞10. The limit is of the form ,∞be simplified.11.12.13.22csc x 2 2lim = lim = = 2cot x cos x 1x→0 2 x→02 210001000 ln xlim ( xln x ) = limx→0 x→01x∞The limit is of the form .∞1 99910001000xln x1000xlim = limx→0 1 x→0– 1x2x= lim –1000x= 0x→022 2 ⎛ x ⎞lim 3xcsc x = lim 3⎜⎟ = 3x→0 x→0⎝sinx ⎠xlim = 1x→0sin x22 2 1–cos xlim (csc x– cot x) = limx→0 x→0sin2x2sin x= lim = 1x→02sin xbut the fraction cansince15. The limit is of the formLet00.2x2y = (3 x) , then ln y = x ln 3x2ln 3xlim x ln 3x= lim+ + 1x→0 x→02x∞The limit is of the form .∞1 2ln 3x⋅33xx+ 1 + 2x 0 2 x 0 –+→ → 3 x→02xx2xln ylim (3 x) = lim e = 1+ +x→0 x→0lim = lim = lim – = 0∞16. The limit is of the form 1 .csc xLet y = (cos x) , then ln y = csc x(ln(cos x))ln(cos x)lim csc x(ln(cos x)) = lim x →0 x →0sin xThe limit is of the form 0 .01ln(cos x)(– sin x)cos xlim = limx→0 sin x x→0cos xsin x 0= lim – = – = 0x→02cos x 1csc x ln ylim (cos x) = lim e = 1x→0 x→014.sin x –1lim (tan x– sec x) = lim cos xx→πx→π2 2The limit is of the form 0 .0sin x–1 cos x 0lim lim 0x cos x= →πx→π– sin x= –1=2 217. The limit is of the form 0 ∞ , which is not antan xindeterminate form. lim (5cos x) = 0x→( π /2) –18.2 22⎛ 2 1 ⎞ ⎛ 1 1 ⎞ ⎛ 2 2x –sin x ⎞lim ⎜csc x – = lim –x→0 2 ⎟ ⎜x 0 2 2 ⎟ = lim⎝ x ⎠ →2 2⎝sinx x ⎠ x→0⎜x sin x ⎟⎝ ⎠2 2x –sin xConsider lim . The limit is of the form 0 . (Apply l’Hôpital’s Rule four times.)x→02 2x sin x02 2x – sin x 2 x– 2sin xcosxx–sinxcosxlim= lim= limx→0 2 2 0 2 2x sin x x→2 22xsin x+2x sin xcosx x→0x sin x+x sin x cos x2 21–cos x+sin x4sinxcosx= lim = limx→0sin2 4 sin cos2 cos2 –2 sin22 2 2x + x x x+x x x xx→06 cos + 6cos sin − 4 cos sin − 6 sin2 24cos x–4sinx4 1= lim = =x→012cos2 – 42 cos2 – 32 cos sin –12sin2 42 sin2x x x x x x x+x x 12 32⎛ 2 22x – sin x⎞ ⎛1 ⎞ 1Thus, lim= =x→0⎜ 2 2x sin x ⎟ ⎜ ⎟⎝ ⎠ ⎝3⎠9x x x x x x x x xInstructor's Resource Manual Section 8.2 483

∞19. The limit is of the form 1 .x /3 3/ x3 x /3Let y = ( x+ e ) , then ln y = ln( x+e ).xx /33 x /3 3ln( x+e )lim ln( x+ e ) = limx→0 xx→0xThe limit is of the form 0 .0x /3( 1+e )3 1x /33ln( x+e )x /3x+e 3lim= limx→0 x x→01x /33+e 4= lim = = 4x→0x /3x+e 1x/3 3/ x ln y 4lim ( x + e ) = lim e = ex→0 x→0020. The limit is of the form (–1) .The limit does not exist.21. The limit is of the form 1, 0 which is not anindeterminate form.cos xlim (sin x) = 1x→ π2∞22. The limit is of the form ∞ , which is not anindeterminate form.lim xx =∞x→∞23. The limit is of the form1/ x1y = x , then ln y = ln x.x∞0 . Let1 lnxlim ln x = limx→∞x x→∞xThe limit is of the form – ∞.∞1ln x x 1lim = lim = lim = 0x→∞ x x→∞ 1 x→∞x1/ x ln ylim x = lim e = 1x→∞x→∞∞24. The limit is of the form 1 .21/ x1Let y = (cos x) , then ln y = ln(cos x).2x1 ln(cos x)lim ln(cos x) = limx→0 2 x 0 2x→ xThe limit is of the form 0 .0(Apply l’Hôpital’s rule twice.)1ln(cos x)(– sin x)cos x− tanlim = lim = limx→0 2x x→0 2xx→02x2−sec x −1 1= lim = = −x→02 2 221/ x ln y −1/21lim (cos x)= lim e = e =x→0 x→0e25. The limit is of the form 0 ∞ , which is not anindeterminate form.2/ xlim (tan x) = 0+x→026. The limit is of the form ∞ + ∞, which is not anindeterminate form.–lim ( xxe – x) = lim ( e + x)=∞x→–∞x→∞027. The limit is of the form 0. Letxy = (sin x) , then ln y = xln(sin x).ln(sin x)lim xln(sin x) = lim+ + 1x→0 x→0xThe limit is of the form – ∞.∞1ln(sin x)cos xsin xlim = lim+ 1 + 1x→0 x 0 –x→ 2x⎡ x ⎤= lim (– xcos x) 1 0 0+ ⎢= ⋅ =x→0sin x ⎥⎣⎦x ln ylim (sin x) = lim e = 1+ +x→0 x→028. The limit is of the form 1 ∞ .Let1/ x1y = (cos x– sin x) , then ln y = ln(cos x– sin x).x1 ln(cosx− sin x)lim ln(cos x− sin x) = limx→0 xx→0x1( −sinx−cos x)cos x−sinx= limx→01−sinx−cosx= lim =−1x→0cos x−sin x1/ x ln y 1lim (cos x − sin x) = lim e = e −x→0 x→0x484 Section 8.2 Instructor's Resource Manual

29. The limit is of the form ∞ – ∞.⎛ 1⎞ ⎛ 1 1 ⎞ x –sinxlim ⎜csc x – ⎟= lim ⎜ – ⎟=limx→0⎝ x ⎠ x→0⎝sinx x⎠x→0xsinxThe limit is of the form 0 . (Apply l’Hôpital’s0Rule twice.)x –sinx 1–cosxlim = limx→0 xsin x x→0sin x+xcosxsin x 0= lim = = 0x→02cos x– xsinx2∞30. The limit is of the form 1 .x⎛ 1⎞ ⎛ 1⎞Let y = ⎜1 + ⎟ , then ln y = xln ⎜1+⎟⎝ x ⎠ ⎝ x ⎠ .1⎛ 1 ⎞ ln ( 1+x )lim x ln ⎜1+ ⎟=limx→∞⎝ x ⎠ x→∞1The limit is of the form 0 .01 111( ) (–+1+2)xln 1xxlim = limx→∞1 x→∞– 1x2x1= lim = 1x→∞1+1xx⎛ 1 ⎞ ln y 1lim ⎜1+ ⎟ = lim = =x→∞⎝x ⎠ x→∞xe e e31. The limit is of the form 3 ∞ ,indeterminate form.x 1/ xlim (1 + 2 e ) =∞+x→0which is not an32. The limit is of the form ∞ – ∞.2⎛ 1 x ⎞ ln x–x + xlim ⎜ – ⎟ = limx→1⎝x –1 ln x⎠x→1( x–1)lnxThe limit is of the form 0 .0Apply l’Hôpital’s Rule twice.21ln x− x + x − 2x+ 1xlim= limx→1 ( x− 1)lnx x→1lnx + x−1x21− 2x + x − 4x+ 1 −3 3= lim= lim = =−x→1xln x+ x− 1 x→1ln x+2 2 2∞33. The limit is of the form 1 .1/ x1Let y = (cos x) , then ln y = ln(cos x).x1 ln(cos x)lim ln(cos x) = limx→0 xx→0xThe limit is of the form 0 .01ln(cos x)(– sin x)cos xsin xlim = lim = lim – = 0x→0 x x→0 1 x→0cosx1/ x ln ylim (cos x) = lim e = 1x→0 x→034. The limit is of the form 0 ⋅– ∞ .1/2ln xlim ( x ln x) = lim+ + 1x→0 x→0xThe limit is of the form – ∞.∞1ln xlim1 = xlim lim – 2 010 0 –= x =+ + +x→ x→ 3/2 x→0x2x35. Since cos x oscillates between –1 and 1 asx →∞ , this limit is not of an indeterminate formpreviously seen.cos xLet y = e , then ln y = (cos x)ln e = cos xcos xlim cos x does not exist, so lim e does notx→∞x→∞exist.36. The limit is of the form ∞ – ∞.x + 1lim [ln( x+ 1) – ln( x–1)] = lim ln x →∞x →∞ x –11x+ 1 1+xx+1lim = lim = 1, so lim ln = 0x→∞ x–1 x→∞1–1 x→∞x–1x37. The limit is of the formindeterminate form.xlim = 0+ ln xx→00 ,–∞which is not an38. The limit is of the form – ∞⋅∞ , which is not anindeterminate form.lim (ln xcot x) = – ∞+x→0Instructor's Resource Manual Section 8.2 485

39. 1+ e −t > 1 for all t, sox t x1+ e − dt > dt = x−11 1The limit is of the form ∞ ∞ .∫ ∫ .x −t1 e dt −x∫ +11+elim = lim = 1x→∞x→∞140. This limit is of the form 0 .0xx∫ sin tdt1sin xlim = lim = sin(1)+x 1 x 1 +→ − x→11n141. a. Let y = a , then ln y = ln a.n1lim ln a = 0n→∞ nnln ylim a = lim e = 1n→∞n→∞b. The limit is of the form ∞n1Let y = n , then ln y = ln n .n1 lnnlim ln n = limn→∞n n→∞nThis limit is of the form ∞ ∞ .1ln n nlim = lim = 0n→∞n n→∞1nln ylim n = lim e = 1n→∞n→∞nc. lim n( a − 1) = lim1n→∞0 .na −1n→∞nThis limit is of the form 0 ,0nsince lim a = 1 by part a.n→∞1 nn− aa − 12nlim = limn→∞1 n→∞− 1n2nn= lim aln a = ln an→∞ln and. lim n( n − 1) = lim1n→∞nn −1n→∞nThis limit is of the form 0 0 ,nsince lim n = 1 by part b.n→∞n 1nn( 2 )(1 − ln n)n − 1nlim = limn→∞1 n→∞− 1n2nn= lim n(ln n− 1) =∞n→∞042. a. The limit is of the form 0.xLet y = x , then ln y = x ln x.ln xlim xln x = lim+ + 1x→0 x→0xThe limit is of the form – ∞.∞1ln xlim1 = xlim lim – 01x 0 x 0 –= x =+ + +→x→ 2 x→0xx ln ylim x = lim e = 1+ +x→0 x→0b. The limit is of the form 1, 0 sincexlim x = 1 by part a.+x→0x xLet y = ( x ) , then ln y = xln( x ).xlim xln( x ) = 0+x→0x xx+ +x→0 x→0ln ylim ( ) = lim e = 1Note that 0 1 is not an indeterminate form.c. The limit is of the formxlim x = 1 by part a.+x→0xx10, since( xLet) xy = x , then ln y = x ln xxlim x ln x = – ∞+x→0( xx ) ln yx e+ +x→0 x→0lim = lim = 0Note that10 is not an indeterminate form.486 Section 8.2 Instructor's Resource Manual

43.d. The limit is of the form 1, 0 sincex xlim ( x ) = 1 by part b.+x→0x x xx xLet y = (( x ) ) , then ln y = xln(( x ) ).x xlim xln(( x ) ) = 0+x→0x x xx+ +x→0 x→0ln ylim (( ) ) = lim e = 1Note that 0 1 is not an indeterminate form.e. The limit is of the form+x→0( xx )lim ( x ) = 0 by part c.00 , since( xx)x( x ) ( x )Let y = x , then ln y = x ln x.( x )lnlim xxx ln x = lim+ + 1x→0 x→0( xx)xThe limit is of the form – ∞.∞ln xlim = lim1x+ 1+x→0 0 x x– ( ) (ln 1)ln x x(x x→ ⎡ ⎤xxx )⎢x x+ x+x ⎥x⎣⎦( xx) 2( x )=lim+x→0– xx( x )x 2 x xx x(ln x)+ x xlnx+x0= = 010 ⋅ + 10 ⋅ + 1Note:2 (ln x)lim x(ln x) = lim+ +x→0 x→02ln xxx+ 1x 0 –+→ 2 x→0x( xx)( x ) ln yxe+ +x→0 x→01x= lim = lim – 2 ln x = 0lim = lim = 121ln x x1/ x ln yx ex→∞ x→∞ x→∞ x→∞lim = lim = 0, so lim = lim = 1x 11/ x1 ln xy = x = ex⎛ 1 lnx⎞1 ln xy′ = ex⎜ −2 2 ⎟⎝ x x ⎠y′ = 0 when x = e.y is maximum at x = e since y′ > 0 on (0, e) andy′ < 0 on (e, ∞). When x = e,1/ ey = e44. a. The limit is of the form (1+ 1) = 2 , whichis not an indeterminate form.x x 1/ xlim (1 + 2 ) =∞+x→0b. The limit is of the form (1+ 1) = 2 ,which is not an indeterminate form.1/lim (1 x + 2 x ) x = 0–x→00c. The limit is of the form ∞ .x x 1/ xLet y = (1 + 2 ) , then1 x xln y = ln(1 + 2 )x∞.∞– ∞ – ∞1 x x ln(1 + 2 )lim ln(1 + 2 ) = limx→∞xx→∞x∞The limit is of the form .∞(Applyl’Hôpital’s Rule twice.)1 x xx x(1 ln1 + 2 ln 2)ln(1 + 2 )x x1 + 2lim= limx→∞x x→∞1xx 22 ln2 2 (ln2)= = =x→∞x x x x1 + 2 x→∞1 ln1+2 ln2x x 1/ x ln y ln 2+ = e = e =x→∞x→∞lim lim ln 2lim (1 2 ) lim 2d. The limit is of the form 1, 0 xsince 1 = 1 forall x. This is not an indeterminate form.1/lim (1 x + 2 x ) x = 1x→−∞xxln xln y =xln x1/ x ln ylim , so lim x lim e 0+ x=−∞ = =+ +x→0 x→0 x→0Instructor's Resource Manual Section 8.2 487

45.46. Let47. a.k k k1 + 2 + +nlimn→∞k + 1nk k k1⎡⎛1⎞ ⎛2⎞ ⎛n⎞ ⎤= lim ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥n→∞n⎢⎣⎝n⎠ ⎝n⎠ ⎝n⎠⎥⎦nk1 ⎛ i ⎞= lim ∑ ⋅n→∞ ⎜ ⎟i = 1 n ⎝n⎠The summation has the form of a Reimann sumkfor f ( x) = x on the interval [ 0,1 ] using aregular partition and evaluating the function at1 ieach right endpoint. Thus, Δ x i = , xi= , andn nk⎛ i ⎞f ( xi) = ⎜ ⎟⎝n⎠ . Therefore,k k k n k1 + 2 + + n 1 ⎛ i ⎞lim= limnk + 1 ∑ ⋅n ⎜ ⎟→∞ n→∞ i 1 n n = ⎝ ⎠11 k ⎡ 1 k 1xdx x+ ⎤= ∫ =0 ⎢ k 1 ⎥⎣ + ⎦01=k + 11/ t⎛ nt⎞1 ⎛ nt⎞y = cixi,⎜∑ then ln y = ln⎟cixi⎝i=1 ⎠t ⎜∑ .⎟⎝i=1 ⎠ntlncxn1⎜∑i i⎛t⎞ ⎟i 1lim ln cx i i lim⎝ = ⎠=+t 0 t ⎜∑⎟ +→ ⎝i=1 ⎠ t→0tThe limit is of the form 0 n,0 since ∑ cii=1⎛ nt⎞lncx⎜∑i i⎟n⎝i=1 ⎠ 1 =tn ∑cxi i+t 0 t+→ t→ 0 t i=1∑cxi ii=1nnc= ln ln i∑ci xi = ∑ xii= 1 i=11/ t⎛ nt⎞ln ylimcx i i = lim e+ ⎜∑⎟+t→0 ⎝i=1 ⎠ t→0nc∑ ln x iini 1 c1 c2cn c= e = = x i1 x2… xn=∏ xii=1⎛⎞= 1.lim lim ln xb.+t→01/ t⎛1 t 1 t ⎞lim ⎜ 2 + 5 ⎟ = 2 5 ≈3.162⎝2 2 ⎠+t→01/ t⎛1 t 4 t ⎞ 5 5 4lim ⎜ 2 + 5 ⎟ = 2 ⋅ 5 ≈ 4.163⎝5 5 ⎠ic.48. a.b.c.+t→01/ t⎛ 1 t 9 t ⎞ 10 10 9lim ⎜ 2 + 5 ⎟ = 2 ⋅ 5 ≈ 4.562⎝10 10 ⎠2 −nxn xelimn→∞2n x=nxe2n x= , so the limit is of the form ∞ nxe∞ .2nxlim→∞ nxxenThis limit is of the form ∞ ∞ .2nx2xlim = lim = 0→∞ nx n 2 nxxe →∞ x en1 −x x x1xe dx ⎡ − −xe e ⎤0 ⎣ ⎦0∫2= − − = 1−e34 = −2 − = 1−2e− x − x − x1 49xe dx = ⎡−3xe − e ⎤ = 1−⎣⎦ 3e− x − x − x1 516xe dx = ⎡−4xe − e ⎤ = 1−⎣⎦ 4e− x − x − x1 625xe = ⎡−5xe − e ⎤ = 1−⎣⎦ 5e− x − x − x1 736e dx = ⎡−6xe − e ⎤ = 1−⎣⎦ 6e1 −2x 2x 2x1xe dx ⎡ − −xe e ⎤0 ⎣⎦0∫∫∫∫∫d. Guess:1 3 3 30 01 4 4 40 01 5 5 50 01 6 6 60 01 2 −nxlim ∫ n xe dx = 1n→∞01 2 nx nx nx1n xe − dx = ⎡−nxe − −e− ⎤0 ⎣⎦0− n n + 1( n 1) e 1 1n∫=− + + = −e1 2 −nx⎛ n + 1⎞lim ∫ n xe dx = lim 1n→∞0⎜ −n→∞n ⎟⎝ e ⎠n + 1= 1− lim if this last limit exists. Then→∞e nlimit is of the form ∞ ∞ .n + 1 1lim = lim = 0, so→∞ n n ne →∞ en1 2−nxlim ∫ n xe dx = 1 .n→∞0488 Section 8.2 Instructor's Resource Manual

49. Note f(x) > 0 on [0, ∞).⎛ 25 3 xx x ⎛2⎞ ⎞lim f( x) = lim + + = 0x→∞x→∞⎜x x ⎜ ⎟e e ⎝e⎠ ⎟⎝⎠Therefore there is no absolute minimum.24 2 x xf ′−( x) = (25x + 3x + 2 ln2) e25 3 x −x− ( x + x + 2 ) e25 24 3 2 x x x= ( − x + 25x − x + 3x − 2 + 2 ln2) e −Solve for x when f′ ( x) = 0. Using a numericalmethod, x ≈ 25.A graph using a computer algebra system verifiesthat an absolute maximum occurs at about x = 25.8.3 Concepts Review1. converge2.3.lim ∫ b→∞ 0bcosx dx0∞f xdx– ∞ 0∫ ∫4. p > 1( ) ; f( xdx )7.8.9.10.11.∫∞ dx ⎡ 1 ⎤= –1.00001 ⎢ 0.000010.00001 ⎥x ⎣ x ⎦1⎛ 1 ⎞ 1= 0 – ⎜– ⎟= = 100,000⎝ 0.00001⎠0.00001∞ x 1 2∞dx = ⎡ ln(1 + x ) ⎤10 21+x 2 ⎣ ⎦101= ∞ – ln 101 =∞2∫The integral diverges.0.00001∞∞ dx ⎡x⎤= ⎢ ⎥ =∞ –100,000 =∞1 0.99999x ⎢⎣0.00001⎥⎦1∫The integral diverges.∞∞ x ⎡ 1 ⎤dx = ⎢–⎥1 2 2 2(1 + x ) ⎢⎣2(1 + x ) ⎥⎦1⎛ 1⎞1= 0– ⎜– ⎟=⎝ 4 ⎠ 4∫∞ 1∞∫ dx = [ln(ln x)] e =∞ – 0 =∞e xlnxThe integral diverges.∞1Problem Set 8.3In this section and the chapter review, it is understoodthat [ gx ( )] ∞ means lim [ g ( x )] b and likewise fora b→∞asimilar expressions.1.2.3.4.5.6.∞ x x∞⎡ ⎤100 ⎣ ⎦100∫100edx= e =∞ – e =∞The integral diverges.∫∫∫15––5dx ⎡ 1 ⎤ 1 1∞ = – – –04 ⎢ 3⎥= =x ⎣ 3x⎦ 3(–125) 375– ∞2 2– x∞– x–1∞ 12 xe dx =⎡– e⎤⎢= 0–(– e ) =⎣ ⎥⎦e11 4x⎡1 4x⎤1 4 1 4e dx = e e –0 e– ∞ ⎢4⎥ = =⎣ ⎦–∞ 4 4∞∞xdx ⎡ 2x⎤9 2 ⎢⎣⎥⎦9∫= 1 + =∞ – 82 =∞1+xThe integral diverges.∞ dx ⎡ x ⎤ 2∫ = ⎢2 ⎥ =∞ – =∞1 πx⎣ π ⎦ π1The integral diverges.∞112.∞∞ ln x ⎡1 2 ⎤ 1dx = (ln x ) –e x⎢2⎥ =∞ =∞⎣ ⎦e2∫The integral diverges.1 1 113. Let u = ln x, du = dx, dv dx, v .x= 2x= − x∞ ln xb ln x∫ dx = lim dx2 2 b 2 2x →∞∫x14.15.∫1∞xe– xdxu = x, du = dxb⎡ ln x ⎤b 1= lim lim dxb→∞⎢− b 2 2x⎥ +→∞∫⎣ ⎦ x2⎡ ln x 1 ⎤ ln 2 + 1= limb→∞⎢− − =x x⎥⎣ ⎦ 2– x– xdv = e dx, v = – e∞ – x – x – xxe d – xe ∞ ∞= ⎡ ⎤ + e dx1 ⎣ ⎦11– x – x∞–1 –1∫ ∫2= ⎡– xe – e ⎤ = 0–0–(– e – e ) =⎣⎦1e∫1– ∞dx ⎡ 1 ⎤= ⎢–⎥3 2(2 x– 3) ⎢⎣4(2 x– 3) ⎥⎦1 1= – –(–0) =−4 4b21– ∞Instructor’s Resource Manual Section 8.3 489

16.17.18.19.20.∫∞dx( π − x)( ) 1/3 3π x ⎤= ⎡−3 − =∞+ 3 π − 4 =∞⎣⎦4 2/34The integral diverges.∞x 0 x ∞ x∞2 2∫ dx = dx dx– ∞ 2∫ +– ∞ 2∫=⎡x + 9⎤+⎡x + 9⎤= (3– ∞ ) + ( ∞–3)0 2 ⎢x + 9 x + 9 x + 9 ⎣ ⎥⎦ ⎢ ⎥– ∞ ⎣ ⎦00 x∞ xThe integral diverges since both ∫ dx and– ∞ 2∫ dx diverge.0 2x + 9x + 9∞ dx 0 dx ∞ dx= +– ∞ 2 2 – ∞ 2 2 0 2 2∫ ∫ ∫( x + 16) ( x + 16) ( x + 16)dx 1 –1 x x∫ = tan +by using the substitution x = 4 tan θ.2 2 128 4 2( x + 16) 32( x + 16)∫dx ⎡ 1 x x ⎤tan⎥( x + 16) ⎢⎣128 4 32( x + 16) ⎥⎦0 –1= ⎢ +– ∞ 2 2 2∞ dx ⎡ 1 –1 x x ⎤= ⎢ tan + ⎥0 2 2 128 4 2( x + 16) ⎢⎣32( x + 16) ⎥⎦∫∫∫∞ dx π π π= + =– ∞ 2 2( x + 16) 256 256 128∞– ∞ 2 – ∞ 20– ∞∞00⎡ 1 ⎛ π⎞⎤ π= 0– ⎢ ⎜– ⎟+ 0 =128 2⎥⎣ ⎝ ⎠ ⎦ 2561 ⎛π⎞π= ⎜ ⎟+ 0–(0) =128 ⎝2 ⎠ 2561 ∞ 10 1 ∞ 1dx = ∫dx = ∫ dx +dx2 2x + 2x+ 10 ( x+ 1) + 9– ∞ ∫( x+ 1) + 90( x+ 1) + 91 1 x + 1( x + 1) + 9 3 3–1∫ dx = tan by using the substitution x + 1 = 3 tan θ.2∫1 ⎡1 x + 1⎤= ⎢( x 1) 9 3 3 ⎥+ + ⎣ ⎦0 –1dx tan– ∞ 21 1 x 1⎤= ⎢( x 1) 9 3 3 ⎥+ + ⎣ ⎦∞ ⎡ –1 +dx tan0 2∫∫∞– ∞x20– ∞∞01 –1 1 1⎛ π ⎞ 1⎛ –1 1⎞= tan – ⎜– ⎟= ⎜π+2 tan ⎟3 3 3⎝ 2⎠ 6⎝ 3⎠1⎛π⎞ 1 –1 1 1⎛ –1 1⎞= ⎜ ⎟– tan = ⎜π–2tan⎟3⎝2⎠ 3 3 6⎝ 3⎠1 1 –1 1 1 –1 12tan–2tan2 10 dx ⎛ ⎞ ⎛ ⎞ π= ⎜π+ ⎟+ ⎜π ⎟=+ x+6⎝ 3⎠ 6⎝ 3⎠3∞ x 0 x ∞ x= +– ∞ 2 x – ∞ –2x0 2x∫ ∫ ∫dx dx dxee e0 x 0 2x2x1 2xFor ∫ dx = xe dx,– ∞ –2x∫ use u = x, du = dx, dv = e dx,v = e .e– ∞200 2 x ⎡1 2 x ⎤ 1 0 2= –x– ∞ ⎢2 ⎥ –– 2∫⎣ ⎦ ∞∞∞ x ∞dx–2x= xe dx ,0 2x0∫For⎡1 2x1 2x⎤1 1xe dx xe e dx = ⎢ xe – e 0– –(0) –2 4⎥ = =⎣⎦ 4 4∫ ∫ use u = x, du = dx,e∞∞ –2 x ⎡ 1 –2 x⎤1 ∞ –2x= –0 ⎢ 2⎥ +⎣ ⎦ 00 2∫∫∫∞– ∞0– ∞–2 x 1 –2xdv = e dx, v = – e . 2⎡ 1 –2 x 1 –2x⎤ ⎛ 1⎞1xe dx xe e dx = ⎢– xe – e 0– 0–2 4⎥ = ⎜ ⎟=⎣ ⎦ ⎝ 4⎠4ex2 x1 1 dx = – + = 04 4∞0∞490 Section 8.3 Instructor’s Resource Manual

21.22.23.24.∞∫ ∫ ∫0sech x dx = sech x dx = sech x dx– ∞– ∞0–1 0 –1[tan (sinh x)] – ∞ [tan (sinh x)]0∞= +⎡ ⎛ π⎞⎤ ⎡π⎤= ⎢0– ⎜– ⎟ + –0 =π2⎥ ⎢2⎥⎣ ⎝ ⎠⎦⎣ ⎦1 2csch x dx =sinh dx =x x – xe – edx∞ ∞ ∞∫ ∫ ∫1 1 1= ∫1∞x2edx2xe –1xLet u = e , du = e dx.1xx∞ 2e∞ 2 ∞⎛ 1 1 ⎞dx = du = – du2x 2e –1eu –1e⎜ ⎟⎝u–1 u+1⎠∫ ∫ ∫∞ ⎡ u –1⎤= [ln( u–1) – ln( u+ 1)] e = ⎢ lnu + 1 ⎥⎣ ⎦e –1= 0 – ln ≈0.7719e + 1⎛ b–1 b–1⎞⎜ lim ln = 0 since lim = 1⎟⎝b→∞b+ 1 b→∞b+1 ⎠∞ x1e − ⎡ ⎤cos xdx = (sin x cos x)0⎢ −x⎥⎣2e⎦∫1 1= 0 − (0− 1) =2 2(Use Formula 68 with a = –1 and b = 1.)∞ x1e − ⎡ ⎤sin xdx = (cos x sin x)0⎢− +x⎥⎣ 2e⎦∫1 1= 0 + (1+ 0) =2 2(Use Formula 67 with a = –1 and b = 1.)∞∞e∞0∞025. The area is given by∞ 2 1 1–1 24 1 dx ∞⎛ ⎞∫ = ∫ 1⎜ ⎟x − ⎝2 x–1 2x+1⎠dx1 ∞ 1⎡ 2x−1⎤= ⎡ln 2x1 ln 2x1 ln2 ⎣ − − + ⎤ ⎦ =1 2 ⎢2x+ 1 ⎥⎣ ⎦1⎛⎛1⎞⎞1= ⎜0− ln⎜⎟⎟=ln32⎝⎝3⎠⎠22x−1Note:. lim ln = = 0 sincex→∞2x+ 1⎛2x−1⎞ lim = 1 .⎜ ⎟x→∞⎝2x+ 1⎠26. The area is∞ 1 ∞⎛1 1 ⎞∫ dx = – dx1 2 ∫x x1⎜ ⎟+ ⎝ x x+1⎠∞ ⎡ x ⎤ 1=⎡⎣ln x − ln x+ 1 ⎤ ⎦ = ln 0 ln ln 21 ⎢ = − =x 1⎥⎣ + ⎦12.27. The integral would take the form∞ 1∞k∫dx = [ klnx] 3960 x3960=∞which would make it impossible to send anythingout of the earth's gravitational field.28. At x = 1080 mi, F = 165, so2 8k = 165(1080) ≈ 1.925× 10 . So the work donein mi-lb is8 ∞ 18 −1∞1.925× 10 dx 1.925 10 ⎡ x ⎤∫ = × −1080 2x⎣ ⎦ 108029.81.925×105= ≈ 1.782× 10 mi-lb.1080∞ −rt∞0 0∞1 100,000−0.08t⎤e ⎥⎦0∫ ∫FP = e f () t dt = 100,000e∞−0.08t⎡= ⎢ −= 1,250,000⎣ 0.08The present value is $1,250,000.∞130.∞ − 0.08t∫ 0−0.08t −0.08t −0.08t∞FP = e (100,000 + 1000 t)dt= ⎡−1,250,000e −12,500te −156, 250e⎣The present value is $1,406,250.⎤⎦0= 1,406,25031. a. ∞f 1( xdx a b) 0 dx dx ∞∫ = 0 dx−∞ ∫ +−∞ ∫ +a b−a∫ b= 10 + b[ x]0b−aa+ = 1( b a)b−a−Instructor’s Resource Manual Section 8.3 491

.∞μ = ∫ x f( x)dx−∞ax b 10 dx x dx ∞= ∫ ⋅ + x 0 dx−∞ ∫ + ⋅a b−a∫ b21 ⎡x⎤= 0+ ⎢ ⎥ + 0b− a⎢ ⎣2 ⎥⎦2 2b − a=2( b−a)( b+ a)( b−a)=2( b−a)a+b=2σba2 ∞ 2= ∫ ( x −μ)dx−∞a 2 b 2 1∫ ( μ) 0 ( μ)−∞ ∫a= x − ⋅ dx + x − dx + ( x −μ) ⋅0dxb−a b1⎡( x − μ ) 3 ⎤= 0+ ⎢ ⎥ + 0b− a⎢ 3 ⎥⎣ ⎦ba3 3( b−μ) −( a−μ)1=b − a 33 2 2 3 2 21 b − 3b μ + 3bμ − a + 3a μ −3aμ=b − a3Next, substitute μ = ( a+ b)/2to obtainσ 2 1 1 3 3 2 3 2 1 3= ⎡ b − b a+ ba − a ⎤3 ⎣4 4 4 4 ⎦( b−a)= 1 b a12−=( ) ( b−a)3( b−a) 212∫∞2c. 2PX ( 2) f( xdx )< =∫−∞0 20dx−∞ 0∫ ∫= +2 1= =10 51dx10 − 032. a.x( )∞ 0 ∞ β β −1β−( x / θ )( ) = 0 +−∞−∞ 0 θ θβ∫ ∫ ∫f xdx dx e dxIn the second integral, let u = ( x/ θ ) . Then,β −1du = ( β / θ)( t / θ)dt . When x = 0, u = 0 and whenx →∞,u →∞. Thus,x( )∞ ∞ β β −1β−( x / θ )( ) =−∞0 θ θ∞ ue −−u∞du = ⎡−e⎤ e0∫ ∫f xdx e dx0= ∫ =− 0+ = 1⎣ ⎦ 0492 Section 8.3 Instructor’s Resource Manual

33.b.β −1∞ 0 ∞ β ⎛ x ⎞ −( x / θ )( ) 0−∞−∞0⎜ ⎟∫ ∫ ∫μ = xf x dx = x⋅ dx+ x e dx∂θ ⎝θ⎠2 3∞22 −( x /3)= xe dx π3∫=02022 = ∞ 2 2 2 ∞ 2 −( x /9)∫ ( x − ) f ( x) dx = ( x ) 0 dx ( x ) xe dx−∞∫ − ⋅ + −∞9∫ − 03 3 3= π − μ = π − π = 02 2 2σ μ μ μc. The probability of being less than 2 isx( )2 0 2 β β −1β−( x / θ )f ( xdx ) = 0dx+e dx−∞−∞ 0 θ θ∫ ∫ ∫β2−(2/ θ ) −(2/3)= 1− e = 1−e≈ 0.359β 2−( x / θ )= 0 +⎡e⎤⎢−⎣ ⎥ ⎦0x – μf′ ( x) = – e3σ 2π2 2–( x– μ) / 2σ2 2 21 2 2–( x– μ) /2 σ ( x – μ)–( x– μ) /2σ3 5f′′ ( x) = – e + eσ 2πσ 2π⎛ 2( x – μ) 1 ⎞ 2 2–( x– μ) / 2σ= –e=⎜ 5 3σ 2 σ 2 ⎟⎝ π π ⎠12 22 2 –( x– μ) /2σ[( x−μ) −σ] e5σ 2π2 2( ) 0 when ( – )f′′ x = x μ = σ so x = μ ± σ and the distance from μ to each inflection point is σ.34. a.b.μk∞∞∞ CMk ⎡ 1 ⎤ k ⎛ 1 ⎞ Cf ( x) dx = dx = CM – CM 0– ∞M k+1 ⎢ k ⎥ = ⎜ +k ⎟=x ⎣ kx ⎦M⎝ kM ⎠ k∫ ∫ . Thus, 1∞ ∞kkMk ∞ 1 k ⎛ b 1 ⎞= ∫ xf( x) dx = x dx kM dx kM lim dx– ∞ ∫ =M k+1 ∫ =M k ⎜ ∫b M k ⎟x x ⎝ →∞ x ⎠This integral converges when k > 1.⎛bk⎡ 1 ⎤ ⎞k⎛ 1 ⎞ kMWhen k > 1, μ = kM ⎜ lim ⎢– ⎥⎟= kM–0 + =b→∞k–1 k–1⎜ ⎢ ( k –1) x ⎥⎟ ⎜( –1) k –1Mk M⎟⎝ ⎣ ⎦ ⎠ ⎝ ⎠The mean is finite only when k > 1.C= when C = k.kInstructor’s Resource Manual Section 8.3 493

c. Since the mean is finite only when k > 1, the variance is only defined when k > 1.22 ∞ 2– ( – ) ( ) ∞k2 2⎛M– kM ⎞ kMk ∞ ⎛2 2kMk M ⎞ 1σ = ∫ x μ f x dx = x dx∞ ∫ ⎜ ⎟ = kM –⎝ k –1⎠k+1 ∫x x +dx2 1xM ⎜ k –1 k+( k –1) ⎟⎝⎠ x∞2 k+ 1∞3 k+2∞k 1 2k M 1 k M 1= kM ∫ dx –dx dxM k–1 k –1∫ +M k 2 ∫ k 1x x ( k –1)M +xThe first integral converges only when k – 1 > 1 or k > 2. The second integral converges only when k > 1,which is taken care of by requiring k > 2.∞2 k+ 1∞3 k+2 ∞2 k⎡ 1 ⎤ 2k M ⎡ 1 ⎤ k M ⎡ 1 ⎤σ = kM ⎢– ⎥ – ⎢– ⎥ + ⎢–⎥⎢⎣ ( k –2) x ⎥⎦ ⎢⎣ ( k –1) x ⎥⎦( k –1) kx ⎦k–2 k –1k–1 2 kMM⎣2 k+ 1 3 k+2k⎛ 1 ⎞ 2k M ⎛ 1 ⎞ k M ⎛ 1 ⎞= kM–0 + – –0 + + –0 +⎜ k–2 –1k–1 2k( k –2) M⎟ k ⎜ ( k –1) M⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ( k –1) ⎝ kM ⎠2 2 2 2 22= kM –k M +k Mk –2 ( k –1) ( k –1)2 22 22⎛ 1 k ⎞ ⎛2 k –2k 1– k 2k⎞2+ + kM= kM–= kM=⎜k –2 2 2( k –1)⎟ ⎜ ( k –2)( k –1) ⎟⎝ ⎠ ⎝ ⎠ ( k –2)( k –1)2M35. We use the results from problem 34:a. To have a probability density function (34 a.)we need C = k ; so C = 3. Also,kMμ = (34 b.) and since, in our problem,k − 1μ = 20,000 and k =3, we have3 4×1020000 = M or M = .2 3436.u = Ar∞a∫dx( r + x )2 2 3/2∞A⎡ x ⎤ A⎛ a ⎞= ⎢ ⎥ = 1−r 2 2 r ⎜ 2 2 ⎟⎣⎢r + x ⎥⎦a⎝ r + a ⎠dxxNote that=2 2 3/2( r + x ) 2 2 2r r + x∫ by usingthe substitution x = rtan θ.2 kMb. By 34 c., σ =( k−2)( k −1)c.24282 3⎛4× 10 ⎞ 4×10σ = =4⎜3 ⎟⎝ ⎠32so that3∞ ⎛ 44×10 ⎞ t 3∫10 5 f ( xdx ) = lim105 dx=⎜ 3 ⎟ ∫ 4⎝ ⎠ t→∞x43⎛t4× 10 ⎞ ⎡ 1 ⎤−⎜ lim⎜ 3 ⎟ ⎢t→∞3 ⎥x 105⎝ ⎠ ⎣ ⎦43⎛4× 10 ⎞ ⎡ 1 1 ⎤ 64= lim − =⎜ 3 ⎟ ⎢t→∞15 3 ⎥ 3⎝ ⎠ ⎣10 t ⎦ 27×10≈ 0.002437. a.∞0∞∫ sin x dx = sin x dx sin x dx−∞ ∫ +−∞ ∫ 0a= lim [ − cos x] + lim [ − cos x] 0a→∞0a→−∞aBoth do not converge since –cos x isoscillating between –1 and 1, so the integraldiverges.aab. lim ∫ sin x dx = lim [ −cos x]a−aa→∞−a→∞= lim [ − cos a+ cos( − a)]a→∞= lim [ − cos a+ cos a]= lim 0 = 0a→∞a→∞38. a. The total mass of the wire is∞ 1 π∫ dx = from Example 4.0 21+x 2Thus 6 of one percent earn over25$100,000.b.∞∞ x ⎡1 ln 12 ⎤dx = x0 2 ⎢ +1 x 2⎥+ ⎣ ⎦0∫ whichdiverges. Thus, the wire does not have acenter of mass.494 Section 8.3 Instructor’s Resource Manual

139. For example, the region under the curve y =xto the right of x = 1.Rotated about the x-axis the volume is∞ 1π ∫ dx =π. Rotated about the y-axis, the1 2x1volume is 2π ∫ ∞x ⋅ dx which diverges.1 x40. a. Suppose lim f( x) = M ≠ 0, so the limitx→∞exists but is non-zero. Since lim f ( x) = M,x→∞there is some N > 0 such that when x ≥ N,Mf( x)– M ≤ , or2MMM – ≤ f( x)≤ M +2 2Since f(x) is nonnegative, M > 0, thusM > 0 2∞N∞∫ f ( xdx ) = f( xdx )0 ∫ +0 ∫ Nf( xdx )∞N∞ M N ⎡Mx⎤≥ ∫ f ( x) dx + dx f ( x)dx0 ∫ =N 2∫ + =∞0 ⎢2⎥⎣ ⎦Nso the integral diverges. Thus, if the limitexists, it must be 0.b. For example, let f(x) be given by⎧ 2 31⎪2 n x–2n + 1 if n– ≤ x≤n22⎪n⎪ 2 31f( x) = ⎨–2n x+ 2n + 1 if n< x≤ n+2⎪2n⎪0 otherwise⎪⎩for every positive integer n.⎛ 1 ⎞ 2⎛ 1 ⎞ 3f ⎜n– 2 – –2 12 ⎟= n ⎜n n2 ⎟ +⎝ 2n⎠ ⎝ 2n⎠3 3= 2 n –1–2n+ 1=02 3f( n) = 2 n ( n)–2n+ 1=12 3lim f ( n) = lim (–2n x+ 2n + 1) = 1 = f( n)+ +x→n x→n⎡ 1 ⎤ ⎡ 1 ⎤⎢nn , + and n 1 – , n 12⎥⎢ + + ⎥2⎣ 2n⎦ ⎢⎣2( n+1) ⎥⎦1 1will never overlap since22n ≤ 2and1 ≤1 .22( n + 1) 8The graph of f consists of a series of isoscelestriangles, each of height 1, vertices at⎛ 1 ⎞⎛ 1 ⎞⎜n– , 0 ,2 ⎟ (n, 1), and ⎜n+ , 0 ,2 ⎟⎝ 2n⎠⎝ 2n⎠based on the x-axis, and centered over eachinteger n.lim f ( x)does not exist, since f(x) will be 1x→∞at each integer, but 0 between the triangles.Each triangle has area1 1⎡1 ⎛ 1 ⎞⎤bh = ⎢n + – – (1)2 2 2 ⎜n2 ⎟⎥⎣ 2n⎝ 2n⎠⎦1⎛1 ⎞ 1= ⎜2 2 ⎟=2⎝n⎠ 2n∞∫ f ( xdx ) is the area in all of the triangles,0thus∞∞∞1 1 1∫ f( x)dx =0∑ =2 ∑ 2n= 12n2 n=1n∞1 1 1 1 1 ∞ 1= + ∑ ≤ + dx2 2 2 1 2n=2 n 2 2∫x∞1 1⎡1⎤1 1= + – (–0 1) 12 2⎢x⎥ = + + =⎣ ⎦12 2∞1(By viewing ∑ as a lower Riemann sum2n=2 n1for2x )∞Thus, ∫ f ( xdx ) converges, although0lim f ( x)does not exist.x→∞⎛ 1 ⎞ 2⎛ 1 ⎞ 3f ⎜n+ –2n n 2n12 ⎟= ⎜ +2 ⎟+ +⎝ 2n⎠ ⎝ 2n⎠3 3= –2 n –1+ 2n+ 1=0Thus, f is continuous at1 1n– , n, and n+.2 22n2nNote that the intervalsInstructor’s Resource Manual Section 8.3 495

41.42.43.100100 1 ⎡ 1⎤dx = – 0.991 2 ⎢ =x x⎥⎣ ⎦1100100 1 ⎡ 1 ⎤dx = – 3.691 1.1 ⎢ 0.1 ⎥ ≈x ⎣ 0.1x⎦1100100 1 ⎡ 1 ⎤dx = – 4.501 1.01 ⎢ 0.01 ⎥ ≈x ⎣ 0.01x⎦1∫∫∫1x100 100dx x11∫= [ln ] = ln100 ≈ 4.610.01100100 1 ⎡x⎤dx = ⎢ ⎥ ≈ 4.711 0.99x ⎢⎣0.01 ⎥⎦1∫10 1 110dx1 ⎡ tan −x⎤∫=0 2π (1 + x ) π ⎣ ⎦ 01.4711≈ ≈0.468π50 1 150dx1 ⎡ tan −x⎤∫=0 2π (1 + x ) π ⎣ ⎦01.5508≈ ≈0.494π100 1 1100dx1 ⎡ tan −x⎤∫=0 2π (1 + x ) π ⎣ ⎦ 01.5608≈ ≈0.497π1 1 exp(–0.52x ) dx ≈ 0.34130 2π2 1 exp(–0.52x ) dx ≈ 0.47720 2π3 1 exp(–0.52x ) dx ≈ 0.49870 2π4 1 exp(–0.52x ) dx ≈ 0.50000 2π∫∫∫∫8.4 Concepts Review1. unbounded2. 23.lim ∫b–b→404. p < 114–dxxProblem Set 8.42/333 dx ⎡3( x –1) ⎤∫ = lim ⎢ ⎥1 1/3 +( x –1) b→1⎢⎣2 ⎥⎦b2/3= 3 3 2 3( b –1) 3 32 – lim –02 3 3b 1 2 = +→2 =21.33 dx ⎡ 3 ⎤2. ∫ = lim ⎢–⎥1 4/3 + 1/3( x–1) b→1⎢⎣( x–1)⎥⎦b= 3 3 3– lim –3 1/3 32 + +b→1( x –1)= 2+∞The integral diverges.10 dx103. ∫ = lim ⎡2 x – 3⎤3 x –3 + ⎣ ⎦b→3b= 2 7 – lim 2 b –3 = 2 7+b→39 dxb4. ∫ = lim ⎡−2 9 – x ⎤0 9– x –b→9⎣ ⎦ 0= lim − 2 9 – b + 2 9 = 6–b→91 dx–1b5.lim ⎡sinx⎤∫ =0 2 – ⎣ ⎦101– x b→–1 –1 π π= lim sin b – sin 0 = –0=–b→12 2∞bx26.dx lim⎡1 x⎤∫= +100 21 xb→∞⎢⎣⎥+⎦1002= lim 1+ b + 10,001 =∞b→∞The integral diverges.3 1 b 1 3 17. ∫–1 3 dx = lim– ∫ lim–1 3 dx +3x dxb 0 x+ ∫→b→0b xb3⎡ 1 ⎤ ⎡ 1 ⎤= lim – lim ––⎢ 2⎥ ++⎢ 2⎥b→0 ⎣ 2x⎦–1b→0⎣ 2x⎦b⎛ 1 1⎞ ⎛ 1 1 ⎞= ⎜ lim – + – lim– 2 ⎟+ ⎜ +2b 0 2b2 18 +⎟⎝ →⎠ ⎝ b→02b⎠1 1= ⎜ ⎛ −∞ + ⎟ ⎞ + ⎜ ⎛ – + ∞⎟⎞⎝ 2⎠ ⎝ 8 ⎠The integral diverges.496 Section 8.4 Instructor's Resource Manual

8.–5 1 b 1 –5 1∫5 2/3 dx = limlim5 2/3 – 2/3x dx dx+ ∫ + ∫b→0 x b→0b x1/3b1/3–5= lim ⎡3x⎤ + lim ⎡3x⎤+ ⎣ ⎦5– ⎣ ⎦b→0 b→0b1/3 3 31/3= lim 3 b – 3 5 + 3 –5 – lim 3b+–b→0 b→03 3 3 3 3= 0–3 5+ 3 5–0= 3 −5− 3 5=−6 59.∫128x –5/ 7dx–1b –5/7 128 –5/7limlim– ∫b 0–1+ ∫→b→0bb128⎡7 2/7⎤ ⎡72/7⎤limlim– ⎢ xxb 0 2⎥ + ⎢→–1 b→02⎥⎣ ⎦ ⎣ ⎦b= x dx + x dx= +7 7 7 7= lim b – (–1) + (128) – lim b2 2 2 22/7 2/7 2/7 2/7–+b→0 b→07 7 21= 0– + (4)–0=2 2 210.1 xb x∫ dx = lim ∫1– x1– xb⎡ 3 2 2/3⎤= lim – (1– )– ⎢ xb→1 4⎥⎣⎦03 2 2/3 3= lim − (1– b ) +– 4 40 3 2 –b→10 3 2b→1dx3 3= –0 + = 4 411.dx dx dx= lim+ lim4 b4∫ ∫ ∫ –0 1/32– 0 1/3 1/32+(2 – 3 x) (2 – 3 )bb→x b→(2 – 3 x)3 31 1 1 1= lim − (2 – 3 b) + (2) – (–10) + lim (2 – 3 b)2 2 2 22/3 2/3 2/3 2/32–2+b→b→3 31 2/3 1 2/3 1 2/3 2/3= 0+ 2 − 10 + 0 = (2 − 10 )2 2 22 2+b→0 b→3 3b2/3 2/3⎡ 1 ⎤ ⎡ 1 ⎤= lim ⎢– (2 – 3 x) + lim – (2 – 3 x)2⎥ ⎢2⎥⎣ ⎦ ⎣ ⎦4b12.∫b8 x⎡ 3 21/3⎤dx = lim (16 2 x )5 2 2/3(16 2 x )− ⎢− −b→84⎥−⎣⎦ 5= 3 21/3 3 3 3 3lim − (16 2 b ) 6 64 − + 4 = 4b→−813.14.15.x x xdx = limdx + limdx–4 b–4∫ ∫ ∫0 2 + 0 2 –216 – 2x – 8 16 – 2bb→x b→– 8 16 – 2x+–b→– 8 0 b→– 8b2 2⎡ 1 ⎤ ⎡ 1 ⎤= lim ⎢– ln 16 – 2x+ lim – ln 16 – 2x4 ⎥ ⎢ 4 ⎥⎣ ⎦ ⎣ ⎦1 1 1 1= lim − ln16–2b+ ln16– ln16+lim ln16–2b4 4 4 42 2+–b→– 8 b→– 81 1= ⎡ ⎢–(– ∞ ) + ln16 ⎤ + ⎡ – ln16 + (– ∞)⎤4 ⎥ ⎢ 4 ⎥⎣ ⎦ ⎣ ⎦The integral diverges.3bx2dx lim⎡– 9 – x⎤∫ =0 2 – ⎢ ⎥b→3⎣ ⎦0–b→39– x2–4= lim − 9 – b + 9 = 3b∫–1 dx⎡ 3 ⎤3 3= lim ⎢–⎥–2 4/3 – 1/3( x+ 1) b→–1⎢⎣( x+1) ⎥⎦–2b→–1– ( b + 1) 1/3 (–1)1/3The integral diverges.b= lim – + = –(– ∞) – 3Instructor’s Resource Manual Section 8.4 497

dxdx ⎡ 1 1 ⎤16. Note that ∫ =x 2 ∫=+ x−2( x− 1)( x+∫ ⎢ − ⎥dx2) ⎣3( x− 1) 3( x+2)by using a partial fraction decomposition.⎦3 dx b dx 3 dx∫ = limlim0 2 – ∫ +0 2 2x x– 2 b 1 x x–2 + ∫+ → + b→1bx + x–2b⎡1 1 ⎤ ⎡1 1 ⎤= lim ⎢ ln x–1 – ln x+ 2 lim ln x–1 – ln x 23 3⎥ + ⎢ +3 3⎥⎣ ⎦ ⎣ ⎦–+b→1 0 b→1bb3⎡1 x–1 ⎤ ⎡1 x–1⎤ 1 b–1 1 1 1 2 1 b–1= lim ⎢ ln + lim ln3 x+ 2⎥ ⎢3 x+2⎥ = lim ln – ln + ln – lim ln–⎣ ⎦ ⎣ ⎦ b 1 3 b 2 3 2 3 5 +→ + b→13 b+2–+b→1 0 b→1b1 1 1 2= ⎛ ⎜– ∞ – ln ⎞ ⎟+ ⎛ ⎜ ln +∞⎞⎟⎝ 3 2⎠ ⎝3 5 ⎠The integral diverges.17. Note that 1 = 1 − 1 +1x 3−x 2− x+ 1 2( x−1) 2 4( x− 1) 4( x+1)dx dx dx= lim+ limx – x – x+ 1 x – x – x+ 1 x – x – x+13 b3∫ ∫ ∫0 3 2 – 0 3 2 + 3 2b→1 b→1bb⎡ 1 1 1 1 1 1lim – – ln 1 ln 1 lim – – ln 1 ln 12( –1) 4 x 4 x ⎤ ⎡x2( x–1) 4 x 4x ⎤= ⎢ − + + ⎥ + ⎢ − + + ⎥–+b→1 ⎣ ⎦0b→1⎣ ⎦b⎡⎛ 1 1 b+ 1 ⎞ ⎛ 1 ⎞⎤ ⎡ 1 1 ⎛ 1 1 b+1 ⎞⎤lim ⎢⎜– + ln ⎟+ – 0 lim – ln 2 – – ln–⎜ + ⎟⎥+ ⎢ + ⎜ + ⎟⎥b 1 2( b–1) 4 b 1 2 +→ ⎣⎝ − ⎠ ⎝ ⎠⎦ b→1⎣ 4 4 ⎝ 2( b–1) 4 b−1⎠⎦1 1 1= ⎛ ⎜∞+∞ – ⎞ ⎟+ ⎛ ⎜– + ln2 +∞–∞⎞⎟⎝ 2⎠ ⎝ 4 4 ⎠The integral diverges.18. Note that1/3x 1 9= +.2/3 1/3 1/3 2/3x −9 x x ( x −9)1/3b∫ 27 x⎡3 2/3 27 2/3 ⎤dx = lim ln – 90 2/3– ⎢ x + xx –9 b→27 2 2⎥⎣⎦0b→27–27 27= – ∞ – ln92 2The integral diverges.3⎛3 2/3 27 2/3 ⎞ ⎛ 27 ⎞= lim ⎜ b + ln b – 9 ⎟– ⎜0 + ln 9⎟⎝2 2 ⎠ ⎝ 2 ⎠319.∫π /40⎡ 1 ⎤tan 2xdx= lim ⎢– ln cos 2x2⎥⎣⎦–b→ π4–b→π41 1= lim − ln cos 2b+ ln1 = –(–∞) + 02 2The integral diverges.b020.π /2 π /2∫ csc xdx lim ln csc x – cot x0= ⎡ + ⎣⎤ ⎦bb→0= ln 1– 0 – lim ln csc b– cot b+b→01–cosb= 0– lim ln+b→0sin b1–cosbis of the form+blim 0 .→0sin b01–cosbsinb0lim = lim = = 0+b 0 sin b +→b→0cosb11–cosbThus, lim ln –+b→0sin bdiverges.= ∞ and the integral498 Section 8.4 Instructor's Resource Manual

21.22.23.24.π /2 sin xπ /2∫dx lim ln 1 – cos x0 1–cosx = ⎡ + ⎣ ⎤ ⎦bb→0= ln1– lim ln 1– cosb= 0 – (– ∞ )+b→0The integral diverges.π /2π /2 cos x ⎡32/3 ⎤∫ dx = lim sin x0 3sin x+ ⎢b→02⎥⎣ ⎦b3 2/3 3 2/3 3= (1) – (0) =2 2 2bπ /2 2 2 1 3tan0x sec xdx ⎡lim tan– 3x ⎤∫= ⎢ ⎥b→π ⎣ ⎦021 3 1 3= lim tan b – (0) =∞– 3 3b→ π2The integral diverges.2bπ /4 sec x⎡ 1 ⎤∫dx = lim –02– ⎢(tan x –1) b π tan x→ –1⎥⎣ ⎦04= 1 1lim − –(– ) –1– tan b –1 + 0 –1= ∞b→ π4The integral diverges.1−cosx2 x25. Since = sin ,2 21 1 csc2 x=− .cos x −1 2 226.27.ππ dx ⎡ x ⎤∫ = lim cot0 cos x –1 + ⎢b→02⎥⎣ ⎦bπ b= cot – lim cot = 0 – ∞2 +b→02The integral diverges.∫–1 dxb= lim ⎡2 ln(– x)⎤x x⎣ ⎦→–3 ln(– ) –b –1–3= lim 2 ln(– b ) –2 ln3 = 0–2 ln3–b→−1= –2 ln3∫ln 30xe dx=xe –1lim⎡ x2 e –1⎤⎢⎣⎥⎦+b→0ln 3= b2 3–1– lim 2 e –1 = 2 2 –0 = 2 2+b→0b28. Note that42 2 2 24x x 4 ( x 4x 4) 2 ( x 2) .− = − − + = − − (by completing the square)dx= limb dx4 x – x 4 x–x∫ ∫ lim–2 2 –b→42 2b⎡ –1 x – 2⎤–1 b –2 –1 π π= ⎢sinb→4 2⎥ = lim sin – sin 0 = –0=⎣ ⎦–2 b→422 229.30.31.e dxe= lim [ln(ln x)]1bxlnx +b→1∫The integral diverges.= ln(ln e) – lim ln(ln b)= ln 1 – ln 0 = 0 + ∞+b→110∫ 10 dx ⎡ 1 ⎤= lim –1 100 +⎢ 99 ⎥x ln x b→1⎣ 99ln x⎦= 1 1 1–b99 lim199 –99ln 10 + b 99ln b= +→99ln 99 10+∞The integral diverges.4c4cdx⎡ 2 2 ⎤2 2= lim ln x x 4c2c2 2 + ⎢+ −⎥= ln ⎡(4 2 3) c⎤lim ln 4x − 4cb→2c ⎣⎦ ⎣+⎦− b+ b − c+bb→2c∫= ln ⎡(4 2 3) c⎤⎣+⎦− ln 2c= ln(2 + 3)Instructor’s Resource Manual Section 8.4 499

32.( )( x+c) dx2( ) 2 0( )xdx xdx c dx= = −2c 2c 2c 2c∫ ∫ ∫ ∫c 2 2 c 2 c9 2 29 2 2x + xc–2 c 9 2x +c– c x+ c− c x+ c− c2 4 2 4 2 4⎡c c= lim x xc– 2c ln x x xc– 2c+⎢ + − + + +→ ⎣2 22 2 2 2b c bc 5c ⎡c c= 4c − ln + 4 c – lim b bc–2c ln b b bc–2c2 2 +⎢ + − + + +b→c⎣2 22 2 2 2 2 2c 9c ⎛ c 3c⎞ 9 3= 2c− ln – ⎜0− ln + 0 ⎟ = 2c− c ln c + c ln c = 2c−c ln32 2 ⎝ 2 2 ⎠ 2 2 2 2 2⎤⎥⎦2c⎤⎥⎦11 133. For 0 < c < 1, is continuous. Let u = , du = – dx .x(1 + x)1 + x2(1 + x)1dv = dx, v = 2 x .x∫11 ⎡ ⎤ 1c1 2 x xdxdx = ⎢ ⎥ + 2x(1 x) 1+x∫+ c⎣ ⎦ (1 + x)c22 2 c 1 xdx= – + 22 1 + c∫c(1 + x)1 1 ⎡ 2 c 1 xdx ⎤Thus, lim ∫ dx = lim ⎢1– + 2 ⎥c→0 c x(1 x) c→01+c∫ = 1–0+2+ c 2⎢⎣(1 + x)⎥⎦This last integral is a proper integral.1 134. Let u = , du = – dx1+ x3/22(1 + x)1dv = dx, v = 2 x .xFor 0 < c < 1,Thus,∫∫11 ⎡ ⎤ 1c1 10 c→0dx 2 x x= ⎢ ⎥ + ∫x(1 + x) 1 x c⎣ + ⎦ (1 + x)dxdx= lim ∫x(1 + x) c x(1 + x)This is a proper integral.c3/222 c 1 xdx= 1– + 21 + c∫ c(1 + x)∫10xdx(1 + x)2 1 2 c 1 xdx = – + ∫2 1 + c c(1 + x)⎡ 2 c x= +⎢⎣+ +1lim ⎢ 2 –0 1∫c→c c (1 )3/2x23/2⎤dx⎥= 2–0+⎥⎦2dx∫10x(1 + x)3/2dx35.36.x x x2 2dx = dx + dx = lim⎡– 9 – x⎤+ lim⎡– 9 – x⎤+ ⎢9– x 9– x 9– x b→–3 ⎣ ⎥⎦ ⎢ ⎥b b→3⎣ ⎦3 0 3∫ ∫ ∫ ––3 2 –3 2 0 22 2= – 9+ lim 9– b – lim 9– b + 9 = –3 + 0 – 0 + 3 = 0+–b→–3 b→33 x 0 x 3 x⎡ 1 2 ⎤ ⎡ 1 2 ⎤∫ dx = dx dx−3 2 ∫ +3 2 0 29−x − ∫ = lim ln 9 x lim ln 9 x9−x 9−x+ ⎢− −b 2⎥ +−⎢− −→b b→2⎥⎣ ⎦ ⎣ ⎦1 2 1 2=− ln 3 + lim ln 9 −b− lim ln 9 − b + ln 3 = ( −ln3 −∞ ) + ( ∞+ ln3)+b 3 2 −→−b→32The integral diverges.003 3 0b0b500 Section 8.4 Instructor’s Resource Manual

37.1 1 1⎡ 1 x+ 4 ⎤ ⎡ 1 x+ 4 ⎤dx = dx + dx = lim ln lim ln+⎢16 – x 16 – x 16 – x b→8 x–4 ⎥ + ⎢b→8 x–4⎥⎣ ⎦ ⎣ ⎦4 0 4∫ ∫ ∫ ––4 2 –4 2 0 20–4 b 4 0b1 1 b+ 4 1 b+4 1= ln1– lim ln + lim ln – ln1 = (0 + ∞) + (∞ – 0)8 +–b→–4 8 b–4 b→48 b–4 8The integral diverges.38.1 1 1 1 1dx dx dx dx dxx −ln x x −ln x x −ln x x −ln x x −lnx1 −1 2 0 1 2 1= + + +−1 −1 −1 2 0 1 2∫ ∫ ∫ ∫ ∫−12 b12= lim ⎡−2 − ln x ⎤ + lim ⎡−2 − ln x ⎤ + lim ⎡−2 − ln x ⎤ + lim ⎡−2 −lnx ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦+1 b−0 12 +b b b 0 b −→− → − → b→112= ( − 2 ln2+ 0) + ( −∞+ 2 ln2) + ( − 2 ln2 +∞ ) + (0+2 ln2)The integral diverges.b39.40.41.42.∞ 1 1 1 ∞ 1= +0 p 0 p 1 p∫ ∫ ∫dx dx dxx x x1 1If p > 1, ∫ 1 1 p0 p dx ⎡x p 1x − + ⎤= ⎢− + ⎥ diverges⎣⎦0− p+1since lim x =∞.+x→0If p < 1 and p ≠ 0,diverges since∞∞ 1 1 p 11 p dx ⎡x p 1x − + ⎤∫ = ⎢− + ⎥⎣⎦11lim x − p+=∞.x→∞∞If p = 0, ∫ 0 dx =∞.1 1 ∞ 1If p = 1, both ∫ dx and dx0 x∫ diverge.1 x∫∞f ( xdx )0lim b10( ) lim c ( ) lim b= ( )− ∫ ++ ∫ +b→b→1b ∫b→∞cwhere 1 < c 145.b1b∫ ln = lim ln ln0 − ∫ +c→0c ∫ 1= lim [ ln 1] [ ln bx x − xc+ x x − x]+1c→0x dx x dx x dx= − 1 − lim ( cln c − c) + blnb − b + 1+c→0= blnb−bThus, b ln b – b = 0 when b = e.1sin x∫ dx is not an improper integral since0 xsin xis bounded in the interval 0 ≤ x ≤ 1.x146. For x ≥ 1,41+x < 1 so∫1∞1

47. For x ≥ 1,e2x2– x – x≤ e .≥ x so2x– ≤ – x,thus1e dx e ee∞ – x– x b–1= lim [– ]11 = – lim +b→∞b→∞b∫1 1= –0 + = e eThus, by the Comparison Test,converges.∞1∫e2– x48. Since x+ 2−1≤ x+ 2 we know that1 1≥x+ 2− 1 x+ 2. Consider ∞ 1∫ 0 x + 2 dx∞ 1 b 1∫ dx = lim2 ∫ dxx+ 2 b→∞2 x+2∞= lim ⎡2 x 2⎤⎣+⎦= lim 2( b+ 2 − 2)=∞b→∞2b→∞Thus, by the Comparison Test of Problem 46, we∞ 1conclude that ∫ 0 x + 2 dx diverges.2 249. Since ln ( 1)x x+ ≥ x , we know that1 1∞ 1 ⎡ 1⎤x 2 ≤ln ( x+1)x2 . Since ∫ dx = 11 2 ⎢− =x x⎥⎣ ⎦1we can apply the Comparison Test of Problem 46∞ 1to conclude that ∫ 1 2x ln x+1 dx converges.( )50. If 0 ≤ f(x) ≤ g(x) on [a, b] and eitherlim f( x) = lim g( x)=∞ orx→a x→alim f( x) lim g( x) ,x→b x→bdx= =∞ then the convergencebof ∫ ( )a g xdx implies the convergence ofb∫ ( )a f xdxband the divergence of ∫ ( )a f xdxbimplies the divergence of ∫ ( ) .a g xdx∞ax51. a. From Example 2 of Section 8.2, lim = 0x→∞ xefor a any positive real number.n+1xThus lim = 0 for any positive realx→∞xenumber n, hence there is a number M suchn+1xthat 0< ≤ 1 for x ≥ M. Divide thexen–12x 1inequality by x to get that 0 < ≤x 2e xfor x ≥ M.52.b.53. a.b∞ 1 ⎡ 1⎤1 1∫ dx = lim – – lim1 2 ⎢x b→∞x⎥ = +⎣ ⎦1b→∞b 1= –0+ 1=1∞ n–1 – x M n–1 – x ∞ n–1 – x∫ x e dx = x e dx x e dx1 ∫ +1 ∫MM n –1 – x ∞ 1≤ ∫ x e dx+dx1 ∫ 1 2xM –1 –= 1 x ne x+∫ dx1by part a and Problem 46. The remainingintegral is finite, soconverges.∞ n–1 – xx e dx11 – x – x1 –1 1e dx ⎡– e ⎤∫ = = – e + 1=1– , so the0 ⎣ ⎦0eintegral converges when n = 1. For 0 ≤ x ≤ 1,n–1≤ x ≤ for n > 1. Thus,n–1n–1 – x – xx e exe1 n−1 – x00 1x= ≤ . By the comparison testfrom Problem 50, ∫ x e dx converges.(1)∞00x e−xdxΓ =∫ ⎣ ⎦0∫∞−x= ⎡− e ⎤ = 1∞ n −xb. Γ ( n+ 1) =∫ x e dx0Let n −xu = x , dv = e dx,n−1−xdu = nx dx,v = − e .n −x ∞ n 1 x( 1) [ ]∞ − −Γ + = − 0 +∫ 0∞ n 1 x= 0 + n x −e −∫ dx = nΓ( n)0c. From parts a and b,Γ (1) = 1, Γ (2) = 1 ⋅Γ (1) = 1,Γ (3) = 2 ⋅Γ (2) = 2⋅ 1= 2! .n x e nx e dxSuppose Γ ( n) = ( n− 1)!, then by part b,Γ ( n+ 1) = nΓ ( n) = n[( n− 1)!] = n!.502 Section 8.4 Instructor’s Resource Manual

54. n = 1,n = 2,n = 3,n = 4,n = 5,∞ – xe0∫∫∫∫∫∞0∞0∞0∞0xedx = 1= 0! = (1–1)!– x2 – xxe3 – xxe4 – xxedx = 1= 1! = (2–1)!dx= 2= 2! = (3–1)!dx= 6= 3! = (4–1)!dx= 24 = 4! = (5 –1)!55. a.b.c.∞∞( ) =– ∞0∫ ∫α–1 – βxf xdx Cx e dxLet y = βx, soy 1x = and dx = dy .β βα –1∞ α–1 – βx∞ ⎛ y ⎞ – y=0 ∫0⎜ ⎟1 C ∞ α–1 – y – α∫ Cx e dx C e dy = y e dy Cβ( α)⎝ β ⎠ β α ∫= Γβ0α– α1 βCβΓ ( α) = 1 when C = = .– αβ Γ( α)Γ ( α)∞∞ααβ α–1 – βxβμ = ∫ xf( x)dx = x x e dx– ∞ ∫=0 Γ( α)Γ ( α)∫Let y = βx, soyx = andβ1dx = dy.βα∞⎛y ⎞ – y 1 1 ∞ α – y0⎜ ⎟∫0∞0α – βxαβ1 1 αμ = e dy = y e dyΓ( α) ∫= Γ ( α + 1) = αΓ ( α)=⎝β ⎠ β βΓ( α)βΓ( α) βΓ( α)β(Recall that Γ(α + 1) = αΓ(α) for α > 0.)x e22 ∞ 2 ∞α⎛ α ⎞ β α–1 – βx∫ ( – μ) ( ) –– ∞ ∫0α2β ∞⎛2 2α α ⎞α–1 – βxσ = x f x dx = ⎜x ⎟ x e dx = x – x+x e dx⎝ β ⎠ Γ ( α)Γ( α)∫ 0 ⎜ β 2β ⎟⎝⎠α –1 2 –21 – x α2– –1 ––x αβ ∞ α+β αβ ∞ α β α β ∞ α β= x e dx x e dx+x e x dxΓ( α) ∫0 Γ( α) ∫0 Γ( α)∫ 0In all three integrals, let y = βx, soσy 1x = and dx dyβ= β.α α+1 –1 2 –2–12 – y α αy y – ––y ααβ ∞⎛ ⎞ αβ ∞⎛ ⎞ α β ∞⎛ y ⎞= ∫ y0⎜ ⎟ ∫ +0⎜ ⎟ ∫ 0⎜ ⎟1 2 1 1e dy e dy e dyΓ( α) ⎝β ⎠ β Γ( α) ⎝β ⎠ β Γ( α)⎝β ⎠ β21 ∞ α+1 – y 2α∞ α – –1 ––y α ∞ α y2 ∫0 2 ∫0 2 ∫ 0= y e dy y e dy +y e dyβ Γ( α) β Γ( α) β Γ( α)221 2αα1 2αα= Γ ( α + 2)– Γ ( α + 1) + Γ( α)= ( α + 1) αΓ( α)– αΓ ( α)+2 2 22 2 2β Γ( α) β Γ( α) β Γ( α)β Γ( α) β Γ( α)β2 2 2α + α 2α α α= – + =2 2 2 2β β β βdxInstructor’s Resource Manual Section 8.4 503

56. a.∞ α – st0αLt { }( s)= ∫ t e dtx 1Let t = , so dt dx,s= sthenα∫ ∞ α – st ∞⎛x ⎞ – x 1 ∞ 1 α – x Γ ( α + 1)t e dt = ∫ e dx x e dx .0 0⎜ ⎟ =s s ∫=⎝ ⎠0 α+ 1 α+1s sα −stIf s ≤ 0 , t e →∞as t →∞ , so the integral does not converge. Thus, the transform is defined only whens > 0.b.αt ∞ αt – st ∞ ( α– s)t= ∫ =0 ∫0Le { }( s)e e dt e dt∞( α– st ) ( α– sb )⎡ 1 ⎤ 1= e⎡lim e –1⎤⎢α– s⎥ =⎣⎦ α – s⎢⎣b→∞⎥⎦0( α – sb ) ⎧∞ if α > slim e = ⎨b→∞⎩0 if s > ααt−1 1Thus, Le { }( s)= = when s > α. (When s ≤ α , the integral does not converge.)α – s s − αc.– stL{sin( αt)}( s) = ∫ sin( αt)e dt∞∞0–Let0 sin( ) stI = ∫ αte dtand use integration by parts with u = sin(α t), du = α cos(α t)dt,– st1 – stdv = e dt, and v = – e .s∞– st⎤α ∞ – st⎥α00 s⎡ 1Then I = ⎢– sin( αte ) + cos( te ) dt⎣ s∫⎦Use integration by parts on this integral with– st1 – stu = cos(αt), du = –α sin(αt)dt, dv = e dt, and v = – e .s∞∞⎡ 1 – st ⎤ α ⎛⎡ 1 – st ⎤ α ∞⎞– stI = ⎢– sin( αte ) – cos( αte ) – sin( αte ) dts⎥ +00 s ⎜⎢ s⎥0 s∫⎣ ⎦ ⎣ ⎦⎟⎝⎠1 ⎡ – st ⎛ α ⎞⎤α= – e sin( αt) cos( αt) – Is⎢ ⎜ + ⎟s⎥ 2⎣ ⎝⎠⎦0sThus,2∞⎛ α ⎞ 1 ⎡ – st ⎛ α ⎞⎤1 + = – sin( α ) + cos( α )⎜ 2s ⎟ ⎜ ⎟s⎢s⎥⎣ ⎝ ⎠⎦0I e t t⎝ ⎠∞1 ⎡ – st ⎛ α ⎞⎤I = – e sin( t) cos( t)2 ⎢ ⎜ α + α ⎟s ( 1 αs⎥+ ⎣ ⎝⎠⎦2 )0s– sb ⎛ α ⎞ ⎧0 if s > 0lim e ⎜sin( αb) + cos( αb)⎟= ⎨b→∞⎝ s ⎠ ⎩∞ if s ≤0αThus, I = when s > 0.2 2s + α∞2s ⎡ – sb ⎛ α ⎞ α ⎤= – lim e sin( αb) cos( αb) –2 2 ⎢ ⎜ + ⎟s α b→∞ss⎥+ ⎣ ⎝⎠ ⎦504 Section 8.4 Instructor’s Resource Manual

57. a. The integral is the area between the curve2 1– xy = and the x-axis from x = 0 to x =1.x2 1– x ;2 1 – ; (2y = xy = x x y + 1) = 1x1x =2y + 11– xAs x→ 0, y = →∞ , whilex1–1when x = 1, y = = 0, thus the area is1∞ 1–1 b∫ dy = lim [tan y]0 20y + 1 b→∞–1 –1 π= lim tan b – tan 0 =b→∞2b. The integral is the area between the curve2 1+xy = and the x-axis from x = –1 to1– xx = 1.2 1+x 2 2 2 2y = ; y – xy = 1 + x; y – 1 = x( y + 1);1– x2y –1x =2y + 11 + (–1) 0When x = –1, y = = = 0, while1–(–1) 21+xas x→ 1, y = →∞ .1– xThe area in question is the area to the right of1+xthe curve y = and to the left of the1– xline x = 1. Thus, the area is∫∞⎛2y –1⎞∞ 21–dy = dy0 ⎜ 2 0 2y 1⎟∫⎝ + ⎠ y + 1–1b= lim ⎡2 tan y⎤b→∞⎣ ⎦0–1 –1 ⎛π⎞lim 2 tan b – 2 tan 0 = 2⎜⎟= πb →∞⎝ 2 ⎠p q58. For 0 < x < 1, x > x so 2x p > x p + xq and1 1 .q p> For 1 < x, x > x sop q px + x 2x2x q > x p + xq 1 1and > .p q qx + x 2x∞ 1 1 1 ∞ 1∫ dx =0 p q ∫ dx + dx0 p q ∫ 1 p qx + x x + x x + xBoth of these integrals must converge.1 1 1 1 1 1 1∫0 p q dx > ∫02p dx =2∫ 0 p dx whichx + x x xconverges if and only if p < 1.∞ 1 ∞ 1 1 ∞ 11 p q dx >12q dx =2 1 q dxx + x x xconverges if and only if q > 1. Thus, 0 < p < 1and 1 < q.∫ ∫ ∫ which8.5 Chapter ReviewConcepts Test1. True: See Example 2 of Section 8.2.2. True: Use l'Hôpital's Rule.3. False:4. False:41000x+ 1000 1000 6lim = = 10x→∞40.001x+ 1 0.0011/ xlim xe −x→∞x →∞ as x →∞.1/ x=∞ since e − → 1 and5. False: For example, if f(x) = x andxg( x) = e ,xlim = 0.x→∞ xe6. False: See Example 7 of Section 8.2.7. True: Take the inner limit first.8. True: Raising a small number to a largeexponent results in an even smallernumber.9. True: Since lim f ( x ) = –1 ≠ 0, it servesx→a only to affect the sign of the limit ofthe product.Instructor’s Resource Manual Section 8.5 505

10. False: Consider11. False: Consider2f ( x) ( x– a)1gx ( ) ,2( x – a)and lim gx ( ) ,x→a lim[ f ( x ) g ( x )] = 1.x→a = and= then lim f ( x ) = 0x→a =∞ while2f ( x) 3x= and2gx ( ) = x + 1, then2f ( xlim) lim3 x=x→∞gx ( ) x→∞2x + 13= lim = 3, butx→∞1+12xlim[ f( x)–3 g( x)]x→∞2 2= lim [3 x – 3( x + 1)]x→∞= lim [–3] = –3x→∞12. True: As x→a, f( x) → 2 while1gx ( )→∞ .13. True: See Example 7 of Section 8.2.14. True: Let1/ f ( x)y = [1 + f( x)] , then1ln y = ln[1 f( x)].f( x)+1 ln[1 + f ( x)]lim ln[1 + f( x)] = limx→a f( x) x→af( x)This limit is of the form 0 .01ln[1 + f( x)]f ′( x)1 + f ( x)lim= limx→a f( x) x→af′( x)1= lim = 1x→a1 + f( x)1/ f ( x) ln y 1lim [1 + f ( x)] = lim e = e = ex→a x→a15. True: Use repeated applications ofl'Hôpital's Rule.16. True:0e = 1 and p(0) is the constant term.17. False: Consider2f ( x) = 3x + x+ 1 and3gx ( ) = 4x + 2x+ 3; f ′( x) = 6x+12g ( x) 12x2,′ = + and sof′ ( x) 6x+1 1lim = lim =x→0 g′ ( x) x→0212x+ 2 22f( x) 3x + x+1 1lim = lim=x→0 gx ( ) x→034x + 2x+3 3while18. False: p > 1. See Example 4 of Section 8.4.19. True:∞ 1∫ 0 pxdx = 1 1 ∞ 1∫ dx + dx;0 p ∫ 1 px x1 1∫ 0 pxdx diverges for p ≥ 1 and∞ 1∫ 1 pxdx diverges for p ≤ 1.∞ 120. False: Consider ∫ 0 x + 1 dx .21. True:∞0∞f ( xdx ) = f( xdx ) + f( xdx )– ∞−∞0∫ ∫ ∫If f is an even function, thenf(–x) = f(x) so0∞f ( xdx ) = f( xdx ) .– ∞0∫ ∫Thus, both integrals making up∞∫ f ( xdx ) converge so their sum– ∞converges.22. False: See Problem 37 of Section 8.3.23. True:24. True:∞b∫ f ′( xdx ) = lim f( xdx )0 ∫′b→∞0= lim [ f ( x)] b0 = lim f( b) – f(0)b→∞b→∞= 0 – f(0) = –f(0).f(0) must exist and be finite sincef ′( x)is continuous on [0, ∞).∞∞ – x– x b∫ f( x) dx≤ e dx lim[– e ]0 ∫ =00b→∞– b∞= lim – e + 1 = 1, so ∫ f ( xdx )b→∞0must converge.25. False: The integrand is bounded on the⎡ π⎤interval ⎢0, .4 ⎥⎣ ⎦506 Section 8.5 Instructor’s Resource Manual

Sample Test Problems1. The limit is of the form 0 .04x4lim = lim = 4x→0tanx x→02sec x2. The limit is of the form 0 .02tan 2x2sec 2x2lim = lim =x→0 sin 3xx→03cos3x33. The limit is of the form 0 . (Apply l’Hôpital’s0Rule twice.)4.5.2sin x −tan x cos x−secxlim= limx→0 1 2 x 0 2x → x33−sin x−2sec x(sec xtan x)= lim = 0x→023cos xlimx→02xsince cos(0) = 1.)=∞ (L’Hôpital’s Rule does not apply2x cosxlim 2xcot x = limx →0 x →0sin xThe limit is of the form 0 .02x cos x 2cos x– 2xsinxlim = limsin xcos xx→0 x→02–0= = 21∞6. The limit is of the form .∞1ln(1 − x)−1−xlim = lim−2x 1 cot x −→ π x→1−πcscπx2sin πx=−xlim→1π (1 −x)The limit is of the form 0 .02sin πx 2πsin πxcosπxlim = lim = 0−x 1 (1 – x)−→ π x→1−π∞7. The limit is of the form .∞1ln t t 1lim = lim = lim = 0t→∞ 2 t 2tt 2t →∞ →∞ 2t∞8. The limit is of the form .∞3 22x6x3lim = lim = lim 6x=∞x→∞ ln x x→∞ 1 x→∞x9. As x →0, sin x→ 0 , and 1 →∞. A numberxless than 1, raised to a large power, is a very⎛ 32⎛1⎞⎞−10small number⎜ ⎟ = 2.328×10so⎜⎝2⎠⎟⎝⎠1/ xlim (sin x) = 0 .+x→010.ln xlim xln x = lim+ + 1x→0 x→0x∞The limit is of the form .∞1ln xx+ 1 + 1x 0 x 0 –+→x→ 2 x→0xlim = lim = lim – x = 0011. The limit is of the form 0.xLet y = x , then ln y = x ln x.ln xlim xln x = lim+ + 1x→0 x→0x∞The limit is of the form .∞1ln xx+ 1 + 1x 0 x 0 –+→x→ 2 x→0xx ln ylim x = lim e = 1+ +x→0 x→0lim = lim = lim – x = 0∞12. The limit is of the form 1 .2/ x2Let y = (1+ sin x) , then ln y = ln(1 + sin x).x2 2ln(1+sin x)lim ln(1 + sin x) = limx→0 xx→0xThe limit is of the form 0 .022ln(1+sin x)cos x1+sinxlim= limx→0 x x→012cosx2= lim = = 2x→01+sinx12/ x ln y 2lim (1 + sin x) = lim e = ex→0 x→0Instructor’s Resource Manual Section 8.5 507

13.ln xlim xln x = lim+ + 1x→0 x→0x∞The limit is of the form .∞1ln xx+ 1 + 1x 0 x 0 –+→ → 3/2 x→0x2xlim = lim = lim – 2 x = 014. The limit is of the form ∞1/ t1Let y = t , then ln y = ln t.t1 lntlim ln t = limt→∞t t→∞t∞The limit is of the form .∞1ln t t 1lim = lim = lim = 0t→∞ t t→∞ 1 t→∞t1/ t ln ylim t = lim e = 115.t→∞t→∞0 .⎛ 1 1 ⎞ x –sinxlim – lim+⎜ ⎟ =x 0 sin x x +→ ⎝ ⎠ x→0xsinxThe limit is of the form 0 .0Rule twice.)x –sinx 1–cosxlim = lim+ xsin x + sin x+xcosxx→0 x→0= sin x 0lim 0+x→02cos x– xsinx= 2=(Apply l’Hôpital’s∞16. The limit is of the form . (Apply l’Hôpital’s∞Rule three times.)tan 3 2xlim lim3sec 3 x=tan2x→πx x→πsec x2 22lim 3cos x cos xsinx== lim cos 2 x → π 3 x x→ π cos3 x sin 3 x222 2cos x − sin x 1= lim =−2 2x → π2 3(cos 3 x − sin 3 x ) 3(0 −1)1=3∞17. The limit is of the form 1 .tan xLet y = (sin x) , then ln y = tan xln(sin x).18.19.20.21.22.sin x ln(sin x)lim tan xln(sin x) = limcos xx→πx→π2 2The limit is of the form 0 .0cos ln(sin ) sin xsin xln(sin x)x x + cos xsin xlim= limx→πcos x x→πsin x2 2cos x(1 + ln(sin x)) 0= lim = = 0x→ π sin x 12tan x ln ylim (sin x) = lim e = 1x→πx→π2 2⎛ π ⎞ xsin x–lim ⎜xtan x– sec x⎟= lim⎝ 2 ⎠ cosxx→πx→π2 2The limit is of the form 0 .0xsin x– π2 sin x + xcosxlim = limx→πcos x x→πsin x2 2∞ dx ⎡ 1 ⎤= – 0 1 10 2 ⎢ = + =( x 1) x + 1⎥+ ⎣ ⎦∫∞ dx –1∞ π π= ⎡tan x⎤= – 0 =0 21+x ⎣ ⎦02 2∫∫∞0π21= = 1111 2x⎡1 2x⎤1 2 1 2e dx = e e –0 e– ∞ ⎢2⎥ = =⎣ ⎦–∞ 2 21 dxb∫ = lim[– ln(1– x)]–1–11– x b→1= – lim ln(1– b) + ln 2 = ∞b→1The integral diverges.∞ dx∞23. ∫ = [ln( x + 1)]00 =∞ – 0 =∞x + 1The integral diverges.508 Section 8.5 Instructor’s Resource Manual

2⎡5 4/5⎤ ⎡54/5⎤∫ 1/5 ∫ 1/5 ∫ = lim (ln ) lim (ln )1/5 – ⎢ xx12 2b 1 4⎥ ++ ⎢→b→14⎥⎣ ⎦ ⎣ ⎦b2⎛4/55 5 1 ⎞4/5⎛ ⎞ ⎛5 4/5 5 ⎞ 5 4/5 5⎛1⎞5 4/5 4/5= (0) – ⎜ln ⎟ + ⎜ (ln 2) – (0)⎜⎟ = (ln 2) – ln4 4⎝ 2⎠ ⎟⎜ ⎟ = [(ln 2) – (– ln 2) ]⎝4 4⎝⎠⎠ 4 4⎝2⎠45 [(ln 2)4/5 – (ln 2)4/5= ] = 042 dx b dx 2 dx24.1= lim–1+ lim+x(ln x) b→1 x(ln x) b→1b x(ln x)25.∞∞ dx ∞⎛ 1 1 ⎞ ⎡ 1 –1 ⎤ π –1 π π π= dx – –tan x1 2 4 ∫1⎜ − =2 2 ⎟ ⎢x x x 1 x x⎥ = 0– + 1+ tan 1= 1 + – = 1–+ ⎝ + ⎠ ⎣⎦12 4 2 4∫26.∫1 dx– ∞ 2(2 – x)1⎡ 1 ⎤ 1= ⎢0 12– x ⎥ = − =⎣ ⎦–∞ 127.28.29.30.31.32.b00 dx b dx 0 dx ⎡1 ⎤ ⎡1⎤= lim + lim = lim–2 2x+ 3 3– –2 2x+ 3– – 3+ b 2x+3⎢ ln 2x+ 3 lim ln 2 3b→b→– 3 2⎥ + x–2 – 3+ ⎢ +2⎥b→⎣ ⎦ b→⎣ ⎦b2 22 2∫ ∫ ∫ –⎛ ⎞ ⎛ ⎞⎜ 1 1 1 1lim ln 2b3 – (0) ⎟ ⎜ ln 3 – lim ln 2b3 ⎟ ⎛1⎞= + + + = (– ∞ ) + 3– 2 2 2– – 3+⎜ 2⎜ ln 3 +∞⎟b→⎟ ⎜ b→⎟ ⎝2⎠⎝ 2 ⎠ ⎝ 2⎠The integral diverges.4 dx4∫ = lim [2 x –1] 2 3– lim 2 –11b =x = 2 3–0=2 3x –1 + +b→1 b→1∞ dx ⎡ 1 ⎤ 1 1= – –02 2 ⎢x(ln x)ln x⎥ = + =⎣ ⎦ ln 2 ln 2∫∞∞ dx ⎡ 2 ⎤ 2= – –0 20 x/2 ⎢ x/2⎥= + =e ⎣ e ⎦01∫∞25 dx b dx 5 dx1/3b1/35∫ = limlim3 2/3 – ∫ +3 2/3 + ∫ = lim ⎡–3(4 – x) ⎤ + lim ⎡–3(4 – x)⎤2/3(4– x) b→4 (4– x) b→4b (4– x)– ⎣ ⎦b 4 3 + ⎣ ⎦→b→4b1/3 1/3 1/3 1/3= lim − 3(4 – b) + 3(1) – 3(–1) + lim 3(4 – b)= 0+ 3+ 3+ 0=6–+b→4 b→4∞∞ 2 2– x ⎡ 1 – x ⎤ 1 –4 1 –4xe dx = – e 0 e e2 ⎢2⎥ = + =⎣ ⎦22 2∫33.∞ x 0 x ∞ x= +– ∞ 2 – ∞ 2 0 2∫ ∫ ∫dx dx dxx + 1 x + 1 x + 11 20 1 2∞= ⎡ln( x + 1) ⎤ + ⎡ln( x + 1) ⎤ =2⎣ ⎦– ∞ 2⎣ ⎦0(0 +∞ ) + ( ∞– 0)The integral diverges.34.0∞∞ x 0 x ∞ x ⎡1 –1 2 ⎤ ⎡1–1 2 ⎤∫ dx = dx + dx– ∞ 4 ∫– 4 0 41+ x ∞ ∫ = ⎢ tan x tan x1+ x 1+x 2⎥ + ⎢– ∞ 2⎥⎣ ⎦ ⎣ ⎦01 –1 1⎛π⎞ 1⎛π⎞1 –1 π π= tan 0 – ⎜ ⎟+⎜ ⎟– tan 0 = 0– + –0=02 2⎝2⎠ 2⎝2⎠24 4Instructor’s Resource Manual Section 8.5 509

35.x xe e=2xx 2e + 1 ( e ) + 1xLet u = e , du = e dxx∞xe ∞ 1–1∞dx = du ⎡tanu⎤0 2x∫ =1 2e + 1 u + 1 ⎣ ⎦ 1∫–1= π –tan 1 = π –π =π2 2 4 43 236. Let u = x , du = 3x dx∞ 32 – x ∞ 1 – u∫ x e dx = e du– ∞ ∫–∞3The integral diverges.1 0 – u 1 ∞ – u=3∫e du+e du– ∞ 3∫01 – u0 1 – u= ⎡– e ⎤ + ⎡–e ⎤3⎣ ⎦ 3⎣ ⎦∞– ∞01 1= (–1 +∞ ) + (–0 + 1)3 337.3 xdx = 0−3 29 − x∫See Problem 35 in Section 8.4.138. let u = ln(cos x), then du = ⋅ –sin x dx = –tanx dxcos xπ2π31tan x – ∞ 1 ln 1dx = 1 – du = 2 du2 ln 2 – 2(ln cos x)2 u ∞ u∫ ∫ ∫1ln⎡ 1⎤2 1 1= ⎢– – 0u ⎥ = + =⎣ ⎦1– ∞ ln ln 22∞39. For p ≠ 1, p ≠ 0, ∫ ∞ 1 ⎡ 1 ⎤dx = ⎢– ⎥ = lim1 11 p ( –1)p–1x ⎢⎣p x ⎥⎦(1 – ) p–1+b→∞pb p –1111lim 0b→∞–1b p = when p – 1 > 0 or p > 1, and limb→∞p 1b − = ∞ when p < 1, p ≠ 0.∞ 1∞When p = 1, ∫ dx = [ln x] 11 =∞– 0 . The integral diverges.x∞When p = 0, 1 dx [ x]∞∫ =11 =∞ –1 . The integral diverges.∞ 1∫ 1 pxdx converges when p > 1 and diverges when p ≤ 1.40. For p ≠ 1, p ≠ 0,1limb→0p –1bWhen p = 1,1∫1 1 ⎡ 1 ⎤ 1 1dx = ⎢– ⎥ = + lim0 p ( –1)p–1x ⎢⎣p x ⎥⎦1– p b→0( p–1) bp–10converges when p – 1 < 0 or p < 1.1dx = x = 0 – lim ln b =∞ . The integral diverges.1 1[ln ]00x +b→01 dx = x 10 = =∫∫When p = 0, 1 [ ] 1–0 101 1∫ 0 pxdx converges when p < 1 and diverges when 1 ≤ p.41. For x ≥ 1,x6 + x > x6 , so6 6 3x + x > x = x and1 1< .6 3x + x xHence,∞ 1converges since 3 > 1 (see Problem 39). Thus ∫ dx converges.1 6x + x∞ 1 ∞ 1dx < dx1 61 3x + x x∫ ∫ which510 Section 8.5 Instructor’s Resource Manual

x42. For x > 1, ln x < e , so ln x< 1 andxeln x ln x 1= < .2x x 2 xe ( e ) eHence,∞lnx ∞ dx e– x dx– x[– e∞ –1 1∫ < = ]1 2x∫11 = –0 + e = .ee∞ ln xThus, ∫ dx converges.1 2xe43. For x > 3, ln x > 1, so ln x> 1 . Hence,x x∞ln x ∞1dx dx [ln x∞∫ > = ]3 33 =∞– ln 3.x∫x∞ ln xThe integral diverges, thus ∫ dx also3 xdiverges.44. For x ≥ 1, ln x < x, so ln x1x < and ln x 0 , then2x > 0 (AT)9.1 1 4 2 1 71+ + = + + =2 4 4 4 4 4Converse: IfContrapositive: If2. Original: If2x > 0 , then x > 02x ≤ 0 , then x ≤ 0 (AT)2x > 0 , then x > 0Converse: If x > 0 , thenContrapositive: If x ≤ 0 , then2x > 0 (AT)2x ≤ 03. Original:f differentiable at c⇒ f continuous at c (AT)Converse:f continuous at c⇒f differentiable at cContrapositive:f discontinuous at c⇒f non-differentiable at c(AT)10.11.12.1 1 1 1 11+ + + + + =2 4 8 16 3232 16 8 4 2 1 63+ + + + + =32 32 32 32 32 32 324∑i=1 i4∑k= 11 1 1 1 1 12+ 6+ 4+3 25= + + + = =1 2 3 4 12 12k( −1) −1 1 −1 1= + + + =k2 2 4 8 16− 8+ 4− 2+ 1 −5=16 16⎛∞⎞13. By L’Hopital’s Rule ⎜ ⎟⎝∞⎠ :x 1 1lim = lim =x→∞2x+ 1 x→∞2 2Instructor’s Resource Manual Review and Preview 511

⎛∞⎞14. By L’Hopital’s Rule ⎜ ⎟⎝∞⎠ twice:2lim n 2 2 12lim n= = =n→∞n→∞4 4 22n+ 1⎛∞⎞15. By L’Hopital’s Rule ⎜ ⎟⎝∞⎠ twice:2x 2x2lim = lim = lim = 0x→∞ x x x x xe →∞ e →∞ e⎛∞⎞16. By L’Hopital’s Rule ⎜ ⎟⎝∞⎠ twice:2n 2n2lim = lim = lim = 0n→∞ n n n n ne →∞ e →∞ e∞ 1 t 117. ∫ dx = lim dx1 x∫ =t→∞1 xtlim [ ln x] = lim1[ ln t]=∞t→∞t→∞Integral does not converge.∞ 1 t 118. ∫ dx = lim dx1 2 ∫ =t 1 2x →∞ xt⎡ 1⎤ ⎡ 1⎤lim ⎢− lim 1 1t→∞x⎥ = ⎢ − =t→∞t⎥⎣ ⎦ ⎣ ⎦1n20.∫∞xt xdx = lim dx =1 2 2x + 1 t→∞1 x + 1u= x2+ 1du = 2x dx2t + 11 1lim2∫ du =∞t→∞2 u∫Integral does not converge (see problem 17).∞ xt x 1 221. ∫ dx limln1 2 ∫ dx1 2 ( x + 1)22.= = =∞x + 1 t→∞x + 1 21Integral does not converge.∞ 1 t 1∫ dx = lim2 2 ∫ dx =2 2x(ln x) t→∞x(ln x)u=ln xdu = 1xdxln tln t 1 ⎡ 1⎤lim ∫ du = limt ln 2 2 ⎢− u t u ⎥ =→∞→∞ ⎣ ⎦ln 2⎡ 1 1 ⎤ 1lim ⎢ − = ≈1.443t→∞ln 2 ln t ⎥⎣ ⎦ ln 2Integral converges.∞Integral converges.19.∞ 1 t 1∫ dx = lim1 1.001 ∫ dx =1 1.001xt→∞xt⎡ 1000 ⎤ ⎡ 1000 ⎤lim ⎢− lim 1000 1000t→∞0.001 ⎥ = ⎢ − =0.001 ⎥⎣ x ⎦1t→∞⎣ t ⎦Integral converges.512 Review and Preview Instructor’s Resource Manual

CHAPTER 9Infinite Series9.1 Concepts Review1. a sequence2. lim a n exists (finite sense)n→∞3. bounded above4. –1; 1Problem Set 9.1= 1 = 2 = 3 = 4 =52 5 8 11 14n 1 1lim = lim = ;→∞ 3 n –1 n→∞3–1 31. a1 , a2 , a3 , a4 , a5nconverges2. a1 = , a2 = , a3 = , a4 = , a5=n5 8 11 14 172 3 4 5 62n1n3n+ 2 3 +lim = lim = 3;→∞ n + 1 n→∞1+nconverges= 6 = 2, = 18 = 2, =38 ,3 9 1766 22 102 34a4 = = , a5= =27 9 39 133. a1 a2 a3224n+ 24 +2n2 1 3 – 1+ 3 n–1n→∞+n 2nlim = lim = 4;n→∞nconverges4. a1 = 5, a 14 29 50 772 = , a3 = , a4 = , a5=3 5 7 9223n+ 2 3n+lim =nlim =∞ ;n→∞2 n –1 n→∞2–1ndiverges7 , 26 , 63 124 215a = a = a = , a4 = , a5=8 27 64 125 2163 2 3 2n + 3 n + 3 n n +limlim3 n + 3 n=n→∞3 3 2( n+ 1) n→∞n + 3n + 3n+15. 1 2 31+ 3 + 3n 2n= lim = 1n→∞1+ 3 + 3 + 1n 2 3n n6. a1 = 5 , a 14 292 = , a3= ,3 5 750 5 2 77a4 = = , a5=9 9 11223 +3n+ 2 2 3lim =nlim = ;n→∞2n+ 1 n→∞2 + 1 2nconvergesa = – 1 , a = 2 = 1 , a = – 3 , a = 4 =2 ,3 4 2 5 6 35a 5 = – 77. 1 2 3 4n 1lim = lim = 1,n→∞n + 2 n→∞1+2nbut since it alternatesbetween positive and negative, the sequencediverges.8. a1 = –1, a 2 3 4 52 = , a3 = – , a4 = , a5= –3 5 7 9⎧−1 for n oddcos( nπ ) = ⎨⎩ 1 for n evenn 1 1lim = lim = , but since cos(nπ )n→∞2 n –1 n→∞2– 1 2nalternates between 1 and –1, the sequencediverges.Instructor’s Resource Manual Section 9.1 513

= –1, = 1 , = – 1 , = 1 , = –12 3 4 5cos( nπ ) = ( − 1)n 1 cos( nπ) 1, so – ≤ ≤ .n n n1 1lim – = lim = 0, so by the Squeezen→∞n n→∞nTheorem, the sequence converges to 0.9. a1 a2 a3 a4 a510.–1 –2a1 = e sin1 ≈ 0.3096, a2= e sin 2 ≈ 0.1231,–3 –4a3 = e sin 3 ≈ 0.0070, a4= e sin 4 ≈ –0.0139,–5a5 = e sin 5 ≈ –0.0065–1 ≤ sin n ≤ 1 for all n, so– n – n – n– e ≤e sinn≤ e .– nlim – e– nlim e 0,n→∞n→∞= = so by the SqueezeTheorem, the sequence converges to 0.1 114. a1 = + 3 ≈ 1.9821, a2= + 3 = 3.0625,4 161 1a3 = + 3 3 ≈ 5.2118, a4= + 9 ≈9.0039,64 2561a 5 = + 9 3 ≈ 15.5891024n⎛1⎞1⎜ ⎟ converges to 0 since –1< < 1.⎝4⎠4n /23 = ( 3)ndiverges since 3 ≈ 1.732 > 1.Thus, the sum diverges.15. a1 = 2.99, a2 = 2.9801, a3≈ 2.9703,a4 ≈ 2.9606, a5≈ 2.9510(0.99) n converges to 0 since –1 < 0.99 < 1, thus2 + (0.99) n converges to 2.11.12.13.2 4eea1 = ≈ 2.4630, a2= ≈ 6.0665,3 96 8eea3 = ≈ 23.7311, a4= ≈ 110.4059,17 2710ea 5 = ≈ 564.781239Consider2x 2x 2xe 2e 4elim = lim = lim =∞x →∞ 2x + 3 x–1x →∞ 2x+ 3 x →∞ 2by using l’Hôpital’s Rule twice. The sequencediverges.2 4ee1 a2a = ≈ 1.8473, = ≈ 3.4124,4 166 8ee3 a4a = ≈ 6.3036, = ≈ 11.6444,64 25610ea 5 = ≈ 21.51010242n2n2e ⎛e ⎞ e= , > 1n4 ⎜ 4 ⎟⎝ ⎠4so the sequence diverges.2ππa1 = – ≈ –0.6283, a2= ≈ 0.3948,5 253 4ππa3 = – ≈ –0.2481, a4= ≈ 0.1559,125 6255πa5 = – ≈ –0.09793125n n(– π)⎛ π⎞π= – , –1 – 1,n ⎜ ⎟ <

18.19.1ln1 lna21 = = 0, a2= ≈ –0.3466,2 2ln 1 ln 13 4a3 = ≈ –0.4485, a4= ≈ –0.4901,6 2 2ln 1a 55 = ≈ –0.508910ln 1 1x − ln x −xConsider lim = lim = limx→∞ 2xx→∞ 2xx→∞12x2= lim − = 0 by using l’Hôpital’s Rule. Thus,x→∞xln 1nlim 0;n→∞2n= converges1/2⎛ 2 ⎞a1= ⎜1+ ⎟ = 3 ≈1.7321,⎝ 1 ⎠2/2⎛ 2 ⎞a2= ⎜1+ ⎟ = 2,⎝ 2 ⎠3/2 3/2⎛ 2⎞ ⎛5⎞a3= ⎜1+ ⎟ = ⎜ ⎟ ≈ 2.1517,⎝ 3⎠ ⎝3⎠4/2 2⎛ 2⎞ ⎛3⎞9a4= ⎜1 + ⎟ = ⎜ ⎟ = ,⎝ 4⎠ ⎝2⎠45/2 5/2⎛ 2⎞ ⎛7⎞a5= ⎜1+ ⎟ = ⎜ ⎟ ≈ 2.3191⎝ 5⎠ ⎝5⎠Let 2 = h,then as , 0n n →∞ h → andn /2⎛ 2 ⎞1/ hlim ⎜1+ ⎟ = lim (1 + h)= en→∞⎝n ⎠ h→0Theorem 6.5A; convergesby20.21.22.1/2 1/4 1/2a1 = 2 ≈ 1.4142, a2= 4 = 2 ≈ 1.4142,1/6 1/8 3/8a3 = 6 ≈ 1.3480, a4= 8 = 2 ≈ 1.2968,1/10a 5 = 10 ≈ 1.25891/2xConsider lim (2 x) .x→∞This limit is of the form0 1/2x1∞ . Let y = (2 x) , then ln y = ln 2 x.2x1 ln2xlim ln 2x= limx→∞2xx→∞2x∞This limit is of the form .∞1ln 2xx 1lim = lim = lim = 0x→∞ 2xx→∞ 2 x→∞2x1/2xln ylim (2 x) = lim e = 1x→∞x→∞1/2nThus lim (2 n) = 1; convergesn→∞n1an= or an= 1 − ;n+ 1 n+1⎛ 1 ⎞ 1lim ⎜1−⎟= 1−lim = 1; convergesn→∞⎝n+ 1⎠n→∞n+1an=2nn+1xx 1Consider .2 x Now, lim = lim = 0x→∞x x2 x→∞2 ln2nby l’Hôpital’s Rule. Thus, lim 0;n 12 n =→∞ +convergesn n n23. an= ( −1) ; lim2n−1 n→∞2n−11 1n= lim = , but due to ( − 1) , the terms ofn→∞2 −1 2nthe sequence alternate between positive andnegative, so the sequence diverges.24.25.1an= = n;1−n−1nlim n = ∞ ; divergesn→∞annn= ==2 2 2 2n – ( n–1) n –( n –2n+1)n 1 1lim = lim = ; convergesn→∞2 n –1 n→∞2– 21nn2 n –1;Instructor’s Resource Manual Section 9.1 515

26.27.28.29.30.2n n( n+ 1) n + nan= = = ;( n + 1) − 1 2 2n+1 ( n+ 1) − 1 n + 2n21n + n 1+nlim = lim = 1; convergesn→∞2 1 2n + 2nn→∞+n11 1 sinnannsin ; lim nsin lim 1n n→∞n n→∞1nsin xlim = 1; convergesx→0x= = = since2n na n = (–1) ;n32n 2n2lim = lim = lim = 0n→∞ n n n 23 n→∞ 3 ln3 n→∞3 (ln3)by using l’Hôpital’s Rule twice; convergesn2a n = ;2nn n n 22 2 ln2 2 (ln2)lim = lim = lim =∞ ;n→∞ 2n n→∞ 2nn→∞2diverges1 1 n+1– n 1an= – = = ;n n+ 1 nn ( + 1) nn ( + 1)1lim = 0; convergesn→∞nn ( + 1)a = 1 , a = 5 , a = 9 , a =132 4 8 16a n is positive for all n, and an+ 1 < anfor all4n− 7n ≥ 2 since an+ 1 − an = − , so { a }n+1 n2converges to a limit L ≥ 0.31. 1 2 3 4a = 1 ; a = 7 ; a = 17 ; a =312 6 12 2022n−1an= < 2 for all n, and a2n < a n + 1 for alln + n2n since an+ 1 − an= ,2n + 2nto a limit L ≤ 2.32. 1 2 3 43 ⎛3⎞⎛8⎞233. a2 = ; a3= ⎜ ⎟⎜ ⎟=;4 ⎝4⎠⎝9⎠3⎛3⎞⎛8⎞⎛15⎞5a4= ⎜ ⎟⎜ ⎟⎜ ⎟ = ;⎝4⎠⎝9⎠⎝16⎠8⎛3 ⎞⎛8 ⎞⎛15 ⎞⎛24 ⎞ 3a5= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ =⎝4⎠⎝9⎠⎝16⎠⎝25⎠5a n > 0 for all n and an+ 1 < ansince⎛ 1 ⎞an+1 = an1 −⎜( 1)2n +⎟⎝ ⎠{ n }a converges to a limit L ≥ 0.a = 1; a = 3 ; a = 5 ; a =412 3 24a < 2 for all n since34. 1 2 3 4so { a n}converges1and 1− < 1, so2( n + 1)n1 1 1 1 1+ + + ≤ + + +n 0 1 n+1∞ k⎛1⎞< ∑ ⎜ ⎟ = 2k=0⎝2⎠ana n +1 2! ! 2 2 2the sum never reaches 2. < 1 since eachterm is the previous term plus a positive quantity,a converges to a limit L ≤ 2.so { } n35. 1 3 1 ⎛ 3 ⎞a 71 = 1, a2 = 1 + (1) = , a3= 1 + ⎜ ⎟=2 2 2⎝2⎠41⎛7⎞15a4= 1+ ⎜ ⎟=2 ⎝ 4 ⎠ 8,Suppose that 1< a n < 2, then 1 < 1 a 1,2 2 n < so3 1 3< 1+ a 2,2 2 n < or < a n + 1 < 2. Thus, since21< a2< 2, every subsequent term is between 3 2and 2.a n < 2 thus 1 1a 1,2 n < so an < 1+ an = a n + 1 2and the sequence is nondecreasing, so { a n }converges to a limit L ≤ 2.516 Section 9.1 Instructor’s Resource Manual

1⎛2⎞336. a1 = 2, a2= ⎜2 + ⎟=,2⎝2⎠21 ⎛3 4 ⎞ 17 1 ⎛17 24 ⎞ 577a3 = ⎜ + ⎟= , a4= ⎜ + ⎟=2 ⎝2 3 ⎠ 12 2 ⎝12 17 ⎠ 408Suppose a n > 2 , and consider1⎛2 ⎞2⎜an+ ⎟> 2 ⇔ an+ > 2 2 ⇔2 ⎝ an⎠an2 2a + 2> 2 2a ⇔ a − 2 2a+ 2> 0⇔n n n n( ) 2na − 2 > 0, which is always true. Hence,a n > 2 for all n. Also,1⎛2 ⎞a ≤a ⇔ a + ≤ a⎜⎟⎝ ⎠1 1⇔ ≤ an⇔ 2 ≤ ana 2n+1 n n n2 annwhich is true. Hence, 2 < an+1 ≤ anand theseries converges to a limit L ≥ 2.37. n u n1 1.732052 2.175333 2.274934 2.296725 2.301466 2.302497 2.302718 2.302769 2.3027710 2.3027811 2.30278lim u ≈ 2.3028n→∞n138. Suppose that u n ( )0< < 1+ 213 , then13< 3+ u n < ( 7+ 213)and( ) ( )1 13 < 3+ un= u n + 1 < 7+ 13 = 1+132 2⎛ 1 ( ) 1⎜7+ 13 = ( 1+13)can be seen by⎝ 2 2squaring both sides of the equality and notingthat both sides are positive.) Hence, since10 < u1= 3 ≈ 1.73 < ( 1+ 13)≈ 2.3028,213 unif 23 + un > unor2u – u –3 < 0.nn( )2u – u –3 = 0 whenn1u n = 1 ± 13 , thus un+ 1 > unif21( 1– 13) < u 1n < ( 1+ 13 ),1 ( 1– 13 ) < 02 2 21and 0< u n < ( 1+ 13)for all n, as shown2above, so { u n } is increasing. Hence, by TheoremD, { u n } converges.39. If u = lim u , then u = 3 + u or un→∞n2 1u – u –3= 0 when u = ( 1 ± 13 ) so21u = ( 1 + 13 ) ≈ 2.3028 since u > 0 and21( 1– 13 ) < 0.2n2= 3 + u;1⎛2 ⎞40. If a = lim a n where an+1 = ⎜an+ ⎟ , thenn→∞2 ⎝ an⎠1⎛2⎞a = ⎜a+⎟2 ⎝ a ⎠ or 2 2 22a= a + 2; a = 2 whena =± 2, so a = 2, since a > 0.41. n u n1 02 13 1.14 1.110535 1.111656 1.111777 1.111788 1.11178lim u ≈ 1.1118n→∞na42. Since 1.1 > 1, 1.1 > 1.1 if a > b. Thus, sinceu3 = 1.1 > 1 = u2,u4 = 1.1 > 1.1 = u3.Suppose that un = since uN> uN–1by the induction hypothesis. Thus, u n isincreasing.1.1 u n < 2 if and only if u n ln1.1 < ln 2;ln 2u n < ≈ 7.3. Thus, unless u n > 7.3,ln1.1u nu n + 1 = 1.1 < 2. This means that { u n } isbounded above by 2, since u 1 = 0.Instructor’s Resource Manual Section 9.1 517

k143. As n →∞, → 0; using Δ x = , an equivalentnndefinite integral is1 1sin xdx=[– cos x ]0 0 = – cos1 + cos 0 = 1– cos1∫≈ 0.4597k144. As n →∞, → 0; using Δ x = , an equivalentnndefinite integral is1 1–1 1 –1 –1 π∫ dx = [tan x] 0 20 = tan 1– tan 0 =1+x445.n n− ( n+ 1) −1 1− 1 = = = ;n+ 1 n+ 1 n+ 1 n+11< ε is the same as 1 < n + 1. For any givenn + 1ε1ε > 0 , choose N > − 1 then εnn≥ N ⇒ − 1 0,same asn n= .2 2n + 1 n + 12n + 1 1 1 = n + > .n n ε1nnn< ε2n + 1is the1Since n+ > , it suffices to take n > . So forε1any given ε > 0 , choose N > , then εnn≥ N ⇒ 0 such thatnA– a n > ε for all n, since if A– aN≤ ε,A– a n ≤ ε for n≥ N since { a n } isnondecreasing and an≤ A for all n. However, ifεA– a n > ε for all n, an< A– < A for all n,2which contradicts A being the least upper boundfor the set S. For the second part of Theorem D,suppose that { a n } is a nonincreasing sequence,and L is a lower bound for { a n}.Then {– a n } isa nondecreasing sequence and –L is an upperbound for {– a n}.By what was just proven,{– a n } converges to a limit A ≤ –L, so { a n }converges to a limit B = –A ≥ L.49. If { b n } is bounded, there are numbers N and Mwith N ≤ bn≤ M for all n. ThenanN ≤ anbn ≤ anMlim a N = N lim a = 0 andn→∞nn→∞.nlim aM= M lim a = 0, so lim ab = 0n→∞nn→∞nn→∞n nby the Squeeze Theorem, and by Theorem C,lim ab = 0.n→∞n n50. Suppose { an+ bn}converges. Then, byTheorem Alim [( an + bn) – an] = lim ( an + bn) – lim an.n→∞ n→∞ n→∞But since ( an + bn)– an = bn,this would meanthat { b n } converges. Thus { an+ bn}diverges.51. No. Consider a n = ( − 1) n anda and b diverge, but{ } { }nnn n+1 n1b ( 1) n +n = − . Bothan+ bn= ( − 1) + ( − 1) = ( − 1) (1 + ( − 1)) = 0 so{ an+ bn}converges.52. a. f3 = 2, f4 = 3, f5 = 5, f6= 8,f = 13, f = 21, f = 34, f = 557 8 9 10b. Using the formula,1 ⎡1+ 5 1−5⎤ 1 ⎡2 5⎤f1= ⎢ − ⎥ = ⎢ ⎥ = 15 ⎣ 2 2 ⎦ 5 ⎣ 2 ⎦⎡ 2 21 ⎛ ⎤1+ 5⎞ ⎛1−5⎞f2= ⎢−⎥5 ⎢⎜ 2 ⎟ ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠ ⎥⎣⎦1 ⎡1+ 2 5+ 5 −(1− 2 5+5) ⎤= ⎢ ⎥5 ⎣ 4 ⎦1 ⎡4 5⎤= ⎢ ⎥ = 1.5 ⎣ 4 ⎦518 Section 9.1 Instructor’s Resource Manual

53.n+ 1 n+ 1 −n−1fn+1 φ −( −1)φlim = limn→∞f n n n −nn →∞ φ −( −1)φn+ 1 n+1n+1 ( −1) ( −1)φ −φ −n+ 1 2n+1φφ= lim= lim =nnn→∞n ( −1) n→∞( −1)φ −n1−2nφφ2− − 1= ⎢ 1+ 5 − 1+ 5 −12 ⎥⎣ ⎦ 2⎛3 5⎞ ⎛1 5⎞= ⎜+ − + − 1=02 2 ⎟ ⎜ 2 2 ⎟⎝ ⎠ ⎝ ⎠22 ⎡1 ⎤ 1c. φ φ ( ) ( )Therefore φ satisfies x −x− 1= 0.Using the Quadratic Formula on2x −x− 1= 0 yields1± 1+ 4 1±5x = = .2 21+5φ = ;21 2 2(1−5) 1−5− =− =− =φ 1+5 1−5 2From the figure shown, the sides of the trianglehave length n – 1 + 2x. The small right triangles3marked are 30-60-90 right triangles, so x = ;2thus the sides of the large triangle have lengths3n − 1+ 3 and B ( 1 3) 2n = n− +42( 2 3 2 2 3 4)= 34 n + n− n−+ while2nn ( + 1) ⎛1⎞π 2An= π ⎜ ⎟ = ( n + n)2 ⎝2⎠8π 2A( n + n)n8lim = limn→∞Bnn→∞3 2n n n4π ( 1+ 1n )π= lim=n→∞2 3 2 2 32 3 4 2 3( 1 + – – +n n 2 2n n)( + 2 3 –2 –2 3+4)φx⎛ 1 ⎞54. Let f( x) = ⎜1 + ⎟ .⎝ x ⎠x⎛ 1 ⎞1/ xlim ⎜1+ ⎟ = lim (1 + x) = e,sox→∞x+⎝ ⎠ x→0n⎛ 1 ⎞lim ⎜1+ ⎟ = e .n→∞⎝n ⎠x⎛ 1 ⎞55. Let f( x) = ⎜1 + ⎟ .⎝ 2x⎠x1/ x⎛ 1 ⎞ ⎛ x ⎞lim ⎜1+ ⎟ = lim 1x→∞2x +⎜ + ⎟⎝ ⎠ x→0⎝ 2⎠2/1/2⎡ x⎛ x ⎞ ⎤1/2= lim ⎢ 1 e ,+⎜ + ⎟ ⎥ = sox→0⎢⎝2⎣ ⎠ ⎥⎦n⎛ 1 ⎞ 1/2lim ⎜1 + ⎟ = e .n→∞⎝2n⎠56. Let12⎛ ⎞f( x) = ⎜1 + ⎟ .⎝ x ⎠x2⎛ 1 ⎞ ⎛ ⎛1⎞⎞lim⎜1+ lim 1x→∞2 ⎟ = ⎜ + ⎜ ⎟⎟⎝ x ⎠x→∞⎜⎝ x ⎠ ⎟⎝ ⎠Using the fact that lim ( )can write11/ xxx→∞⎛ 21 ⎞lim ⎜⎛ ⎞1+ ⎟ = lim 1++x→∞⎜⎜ ⎟⎝ x ⎠ ⎟ x→0⎝ ⎠to the indeterminate form 1 ∞ .2Let y ( 1 x ) 1/= + . Then,ln y = ln 1+1ln y = ln 1+xx1/ x2( x )2( x )2ln ( 1+x )11/ xf x 1lim f ⎛ ⎞=+ ⎜ ⎟x→0⎝ x ⎠ , we2( x )1/ x2x2lim ln y = lim = lim1+x+ + +x→0 x→0 x→01x2x= lim = 0+ 2x→01+xThis gives uslim ln y = 0+x→0ln⎛lim y⎞⎜ 0+⎟ =⎝ x→0⎠lim = = 1 or lim 1+ = 1( )1/ x0 2y e x+ +x→0 x→0⎛ 1 ⎞Thus, lim ⎜1+ = 1n→∞2 ⎟ .⎝ n ⎠nwhich leadsInstructor’s Resource Manual Section 9.1 519

57. Let f ( x)x⎛ x −1⎞= ⎜ ⎟⎝ x + 1⎠ .Using the fact that lim ( )can writex→∞f x 1lim f ⎛ ⎞=+ ⎜ ⎟x→0⎝ x ⎠ , wex1/ x1/ x11−x⎛ x −1⎞ ⎛ −1⎞ ⎛ ⎞xxlim ⎜ ⎟ = lim limx + →∞ 01=+01 xx 1 x→ 1 x→++ ⎜ + ⎟ ⎜ ⎟xx⎝ ⎠ ⎝ ⎠ ⎝ ⎠1/x⎛1−x ⎞= lim+ ⎜ ⎟ which leads to thex→0⎝ 1 + x ⎠indeterminate form 1 ∞ .⎛1−x ⎞Let y = ⎜ ⎟⎝1+x ⎠1/ x1/ x⎛1−x ⎞ln y = ln ⎜ ⎟⎝ 1 + x ⎠1/ x1 ⎛1−x ⎞ln y ln 1= ⎜ ⎟x ⎝ + x⎠. Then,1 ⎛1−x ⎞= ⎜ ⎟⎝ + x⎠lim ln y lim ln+ +x→0 x→0x 11−x( 1+x )lnln⎡lim y⎤⎢lim+=+⎣x→0 ⎥⎦x→0x−2= lim (l'Hopital's Rule)+ 2x→01−x−2This gives us,ln⎡lim y⎤⎢2+=−⎣x→0⎥⎦1/ x−2 ⎛1−x ⎞ −2lim y = e or lim e+ + ⎜ ⎟ =x→0 x→01 x⎛n−1⎞ Thus, lim ⎜ ⎟ = en→∞⎝n+ 1⎠n⎝−2.+⎠58. Let2x⎛2+ x ⎞f( x) = .⎜ 23 + x ⎟⎝ ⎠Using the fact that lim ( )can writexx→∞1⎛ 2 ⎛2⎞2 + x ⎞+2xlim = lim ⎜ ⎟x→∞⎜ 2 +0 313 x ⎟ x→⎜ ⎟⎝ + ⎠ +2⎝ x ⎠1/ x22x+ 1 22⎜ x ⎟x+ 23 12x 0 x+→ + x→0⎜ x2x⎝f x 1lim f ⎛ ⎞=+ ⎜ ⎟x→0⎝ x ⎠ , we1/ x⎛ ⎞ ⎛2 + 1⎞= lim = lim ⎜ ⎜ ⎟ 3 + 1 ⎟⎝ ⎠⎠to the indeterminate form 1 ∞ .⎛ 22x+ 1⎞Let y = ⎜23x+ 1⎟⎝ ⎠1/ x. Then,1/ x⎛ 22x+ 1⎞ln y = ln ⎜32x 1 ⎟⎝ + ⎠21 ⎛2x+ 1y+ +x→0 x→0⎜ 2⎞lim ln = lim ln ⎜ x 3 x + 1 ⎟⎝ ⎠lnln⎡lim y⎤⎢lim+=+⎣x→0 ⎥⎦x→02x( )2 + 123x+ 11/ xwhich leads⎡ 4x6x⎤= lim (l'Hopital's Rule)+ ⎢ −x 02 2 ⎥→ ⎣2x+ 1 3x+ 1⎦= 0This gives us,ln⎡lim y⎤⎢0+=⎣x→0⎥⎦x1/ x0⎛1−x ⎞lim y = e = 1 or lim 1+ + ⎜ ⎟ =x→0 x→0⎝1+x ⎠Thus,⎛ 22 + n ⎞lim = 1 .n→∞⎜23 + n ⎟⎝ ⎠n520 Section 9.1 Instructor’s Resource Manual

59. Let f ( x)⎛ 22 + x ⎞= ⎜23 + x ⎟⎝ ⎠2xUsing the fact that lim ( )can writex→∞f x 1lim f ⎛ ⎞=+ ⎜ ⎟x→0⎝ x ⎠ , we21/2x1x⎛⎛2⎞2 + x ⎞+2xlim = lim ⎜ ⎟x→∞⎜ 2 +0 3 13 x ⎟ x→⎜ ⎟⎝ + ⎠ +2⎝ x ⎠21/ x22x+ 1 22⎜ x ⎟x+ 23 12x 0 x+→ +x→0⎜ x2x⎝⎛ ⎞ ⎛2 + 1⎞= lim = lim ⎜ ⎜ ⎟ 3 + 1 ⎟⎝ ⎠⎠leads to the indeterminate form 1 ∞ .21/ x⎛ 22x+ 1⎞Let y = . Then,⎜ 23x+ 1⎟⎝ ⎠21/ x⎛ 22x+ 1⎞ln y = ln ⎜32x 1 ⎟⎝ + ⎠21 ⎛2x+ 1y+ + 2 2x→0 x→0⎜⎞lim ln = lim ln ⎜ x 3 x + 1 ⎟⎝ ⎠22x+ 1( 23x+ 1)221/ xwhichlnln⎡lim y⎤⎢lim+=+ 2⎣x→0 ⎥⎦x→0x⎡⎤−1= lim⎢⎥(l'Hopital's Rule)+x→0⎢ 2 2( 2x1)( 3x1⎥⎢+ +⎣) ⎥⎦=−1This gives us,ln⎡lim y⎤⎢1+=−⎣x→0⎥⎦21/ x−1 ⎛1−x ⎞ −1lim y = e or lim e+ + ⎜ ⎟ =x→0 x→0⎝ 1 + x ⎠Thus,2n⎛ 22 + n ⎞lim n→∞⎜32 ⎟⎝+ n⎠= e−19.2 Concepts Review1. an infinite series2. a1+ a2 +…+ ana3. r < 1; 1– r4. diverges.Problem Set 9.21.2.3.4.∞ k2⎛1⎞ 1 1 1 1⎛1⎞∑ ⎜ ⎟ = + ⋅ + ⎜ ⎟ +… ; a geometrick=1⎝7⎠ 7 7 7 7⎝7⎠1 11 1 7 7 1series with a = , r = ; S = = =7 7 1– 1 6 67 7∞ – k –2 –3 –4 –5⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞∑ ⎜– ⎟ = ⎜– ⎟ + ⎜– ⎟ + ⎜–⎟ +…k=1⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠3 4 5= (–4) + (–4) + (–4) +… ; a geometric serieswith3a = (–4) , r = –4; r = 4> 1 so the series diverges.∞ k2⎛1⎞ 1 ⎛1⎞2⎜ ⎟ = 2+ 2⋅ + 2 ⎜ ⎟ +… ;k=0 ⎝4⎠ 4 ⎝4⎠1 2 2 8series with a = 2, r = ; S = = = .4 1– 1 3 34 4∞ k2∑ a geometric⎛ 1⎞ 1 ⎛1⎞∑ 3 ⎜– ⎟ = 3–3⋅ + 3 ⎜ ⎟ −… ; a geometric⎝ 5⎠ 5 ⎝5⎠k=01series with a = 3, r = – ;53 3 5S = = =⎛ 1 ⎞ 6 21– ⎜– ⎟ 5⎝ 5 ⎠Thus, by Theorem B,∞ ⎡ k k⎛1⎞ ⎛ 1⎞⎤ 8 5 31∑ ⎢ 2⎜ ⎟ + 3 ⎜–⎟ ⎥ = + =k=0 ⎢ ⎝4⎠ ⎝ 5⎠⎥ 3 2 6⎣⎦∞k⎛1⎞ 5 5 1 5⎛1⎞∑ 5 ⎜ ⎟ = + ⋅ + ⎜ ⎟ +… ; a geometric⎝2⎠ 2 2 2 2⎝2⎠k=15 55 1series with a = , r = ; S = 2 = 2 = 5.2 2 1– 1 12 2∞ k+1 2⎛1⎞ 3 3 1 3 ⎛1⎞∑ 3 ⎜ ⎟ = + ⋅ + ⎜ ⎟ +… ; ak=1 ⎝7⎠ 49 49 7 49⎝7⎠3 1geometric series with a = , r = ;49 73 349 49 1S = = =1–1 6 147 7Thus, by Theorem B,∞ ⎡ k k+1⎛1⎞ ⎛1⎞⎤ 1 69∑ ⎢2 ⎜ ⎟ –3⎜ ⎟ ⎥ = 5– = .k=1⎢⎝4⎠ ⎝7⎠⎥ 14 14⎣⎦2Instructor’s Resource Manual Section 9.2 521

5.∞k –5 4 3 2 1 1 2∑ = – – – – + 0 + + +… ;k=1 k + 2 3 4 5 6 8 95k –5 1–klim = lim = 1 ≠ 0; the seriesk→∞k + 2 k→∞1+2kdiverges.6.∞⎛9⎞ 9 9 9 9⎛9⎞∑ ⎜ ⎟ = + ⋅ + ⎜ ⎟ +… ; a geometric⎝8⎠ 8 8 8 8⎝8⎠k=1k9 9 9series with a = , r = ; > 1, so the series8 8 8diverges.27.∞⎛1 1 ⎞ ⎛1 1⎞ ⎛1 1⎞ ⎛1 1⎞∑ ⎜ – ⎟= ⎜ – ⎟+ ⎜ – ⎟+ ⎜ – ⎟+…;k=2⎝k k –1⎠ ⎝2 1⎠ ⎝3 2⎠ ⎝4 3⎠⎛1 ⎞ ⎛1 1⎞ ⎛ 1 1 ⎞ ⎛1 1 ⎞S n = ⎜ –1 ⎟+ ⎜ – ⎟+…+ ⎜ – ⎟+⎜ – ⎟⎝2 ⎠ ⎝3 2 ⎠ ⎝n–1 n–2 ⎠ ⎝n n–1⎠∞1⎛1 1 ⎞lim Sn= lim –1 + = –1, so ∑ ⎜ – ⎟ = –1n→∞n→∞nk=2⎝k k –1⎠1= –1 + ;n8.∞ ∞3 1∑ = 3k= 1kk=1k∑ which diverges since∞1k=1 k∑ diverges.9.10.∞k! 1 2 6∑k 1100 = + + +…= 100 10,000 1,000,000n + 1 1Consider { a n},where an+1 = an, a1= .100 100n 0sequence is eventually an increasing sequence, hence lim ann→∞0.∞n! k!a n = , hencen ∑100k=1100 diverges.a > for all n, and for n >99, an+ 1 > an,so the≠ The sequence can also be described by∞∞2 2 2 2 ⎛1 1 ⎞ 1 1 1 1 1 1 1 1∑ = + + +…= ∑⎜ – ⎟ = ⎛ ⎜ – ⎞ ⎟+ ⎛ ⎜ – ⎞ ⎟+ ⎛ ⎜ – ⎞ ⎟+ ⎛ ⎜ –⎞⎟+…k= 1( k+ 2) k 3 8 15 k=1⎝k k+2⎠⎝1 3⎠ ⎝2 4⎠ ⎝3 5⎠ ⎝4 6⎠⎛ 1⎞ ⎛1 1⎞ ⎛1 1⎞ ⎛ 1 1 ⎞ ⎛1 1 ⎞S n = ⎜1– ⎟+ ⎜ – ⎟+ ⎜ – ⎟+…+ ⎜ – ⎟+⎜ – ⎟⎝ 3⎠ ⎝2 4⎠ ⎝3 5 ⎠ ⎝n–1 n+ 1⎠ ⎝n n+2⎠1 1 1 3 2n+ 3 3 2n+3= 1 + – – = – = –2 n+ 1 n+ 2 2 ( n+ 1)( n+ 2) 2 2n + 3n+22 3n ++n 2nSnn →∞ n →∞ 2 1 3 2n + 3n+2 n →∞ + +n 2n3 2 3 3 3 2 3lim = – lim = – lim = , so ∑ = .2 2 2 k 1 ( k+2) k 2∞=11.∞ k+1 2 2 2 2⎛ e⎞ ⎛ e⎞ ⎛ e⎞ e ⎛ e⎞ ⎛ e⎞⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ ⋅ + ⎜ ⎟ ⎜ ⎟ +… ;k=1⎝π⎠ ⎝π⎠ ⎝π⎠ π ⎝π⎠ ⎝π⎠e2e2( ) ( ) 2π π eS = = = ≈5.55621– e π–e ππ ( – e)π π∑ a geometric series with2⎛ e⎞ea = ⎜ ⎟ , r = < 1;⎝π⎠π12.∞k=1k+12∑ 4 16 4 ⎛4⎞= + 16⋅ + 16 ;k –1⎜ ⎟ +…7 1 7 ⎝7⎠a geometric series with a = 16,4r = < 1;716 16 112S = = =1– 4 3 37 7522 Section 9.2 Instructor’s Resource Manual

13.∞ ⎛ 3 3 ⎞ ⎛3 3⎞ ⎛3 3⎞ ⎛3 3 ⎞∑– = – + – + – +… ;⎜ 2 2k=2 ( k –1) k⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝⎠ ⎝1 4⎠ ⎝4 9⎠ ⎝9 16⎠⎛ 3⎞ ⎛3 1⎞ ⎛1 3 ⎞ ⎛ 3 3 ⎞ ⎛ 3 3 ⎞S n = ⎜3– ⎟+ ⎜ – ⎟+ ⎜ – ⎟+…+ – +–⎝ 4⎠ ⎝4 3⎠ ⎝3 16 ⎠ ⎜ 2 2 2 2( n–2) ( n–1) ⎟ ⎜( n–1)n⎟⎝ ⎠ ⎝ ⎠3 3= 3– ; lim S 3– lim 3,2 n = = sonn 2n →∞→∞ n∞ ⎛ 3 3 ⎞∑– = 3.⎜ 2 2k=2 ( k –1) k⎟⎝⎠14.15.16.17.18.19.∞2 2 2 2 2∑ = + + + +…k=6 k –5 1 2 3 4∞12k=1 k= ∑ which diverges since∞ k –12 ⎛ 1 ⎞0.22222…= ∑ ⎜ ⎟k=110 ⎝10⎠210 2= =1– 1 910∞1k=1 k∞k –121 ⎛ 1 ⎞0.21212121…= ∑ ⎜ ⎟k=1100 ⎝100⎠21100 21 7= = =1– 1 99 33100∞k –113 ⎛ 1 ⎞0.013013013... = ∑ ⎜ ⎟k=11000 ⎝1000⎠131000 13= =1−1 9991000∞k –1125 ⎛ 1 ⎞0.125125125... = ∑ ⎜ ⎟k=11000 ⎝1000⎠1251000 125= =1−1 9991000∞k –14 9 ⎛ 1 ⎞0.4999... = + ∑ ⎜ ⎟10 k=1100 ⎝10⎠94 100 1= + =10 1−1 210∑ diverges.20.36 71 ⎛ 1 ⎞0.36717171... = + ∑ ⎜ ⎟100 10,000 ⎝100⎠7110,0001−110036 727= + =100 1980∞k=1k= 0 k=0k –1 1−sk –121. Let s = 1 – r, so r = 1 – s. Since 0 < r < 2,–1 < 1 – r < 1, so∞∞kks < 1, and r(1 − r) = (1 −s)s22.k=1∑ ∑∞= ∑ (1 – ss ) = = 11−s∞ k k ∞ k ∞k –1∑ − x = ∑ − x = ∑ xk= 0 k= 0 k=1( 1) ( ) (– ) ;if –1 < x < 1 then–1 < –x < 1 so x < 1 ;∞(– x)∑k=1k –11 1= =1 −( − x) 1+xInstructor’s Resource Manual Section 9.2 523

k23. ln = ln k – ln( k+1)k + 1Sn= (ln1–ln2) + (ln2–ln3) + (ln3–ln4) +…+ (ln( n–1)–ln n) + (ln n–ln( n+ 1)) = ln1 – ln( n+ 1) = – ln( n+1)∞klim Sn= lim – ln( n+ 1) = – ∞ , thus ∑ ln diverges.n→∞n→∞k= 1k + 124.2⎛ 1 ⎞ k –1 2 2ln 1 – = ln =2⎜ln( k –1) – ln k2 ⎟= ln[( k+ 1)( k – 1)] – ln k = ln( k+ 1) + ln( k – 1) – 2 ln k2⎝ k ⎠ kS n = (ln 3 + ln1– 2ln 2) + (ln 4 + ln 2 – 2ln 3) + (ln 5 + ln 3 – 2ln 4) +…+ (ln n+ ln( n– 2) – 2ln( n–1)) + (ln( n+ 1) + ln( n–1) – 2ln n)1= –ln 2 + ln(n + 1) – ln n –ln2 ln n += +nn + 1lim Sn= – ln 2 + lim lnn→∞n→∞n⎛ n + 1⎞= – ln 2 + ln ⎜ lim ⎟= – ln 2 + ln1 = – ln 2⎝n→∞n ⎠2⎛2⎞25. The ball drops 100 feet, rebounds up 100⎜⎟⎝ 3 ⎠ feet, drops ⎛2⎞100⎜⎟⎝ 3 ⎠ feet, rebounds up ⎛2⎞100⎜⎟⎝ 3 ⎠2⎛2⎞100 ⎜ ⎟ , etc. The total distance it travels is⎝3⎠2 32 3⎛2⎞ ⎛2⎞ ⎛2⎞⎛2⎞ ⎛2⎞ ⎛2⎞100 + 200⎜ ⎟+ 200⎜ ⎟ + 200 ⎜ ⎟ + ... =− 100 + 200 + 200⎜ ⎟+ 200⎜ ⎟ + 200 ⎜ ⎟ + ...⎝3⎠ ⎝3⎠ ⎝3⎠⎝3⎠ ⎝3⎠ ⎝3⎠∞ k –1⎛2⎞200=− 100 + ∑ 200⎜⎟ =− 100 + = 5003 12k=⎝ ⎠−feet1 3feet, drops26. Each gets∞1 1 1 1⎛1 1⎞ 1⎛1⎞+ ⋅ + ⎜ ⋅ ⎟+ ... = ∑ ⎜ ⎟4 4 4 4⎝4 4⎠ 4⎝4⎠k=1k –114 114= =1−3(This can be seen intuitively, since the size of the leftover piece is approaching 0, and each person gets the sameamount.)27. $1 billion + 75% of $1 billion + 75% of 75% of $1 billion + ... =∞k –1 $1 billion($1 billion)0.75 = = $4 billion1−0.75∑k=128.∞$1 billion (0.90) k∑k=1–1$1 billion=1−0.90= $10 billion29. As the midpoints of the sides of a square are connected, a new square is formed. The new square has sides12times the sides of the old square. Thus, the new square has area 1 2the area of the old square. Then in the next step,18of each new square is shaded.Area =∞1 1 1 1 1 1⎛1⎞⋅ 1 + ⋅ + ⋅ + ... = ∑ ⎜ ⎟8 8 2 8 4 8⎝2⎠k=1k –118 112= =1−4The area will be 1 .4524 Section 9.2 Instructor’s Resource Manual

30.∞1 1⎛8⎞ 1⎛8 8⎞ 1⎛8⎞+ ⎜ ⎟+ ⎜ ⋅ ⎟+ = ∑ ⎜ ⎟9 9⎝9⎠ 9⎝9 9⎠ 9⎝9⎠k=1k –119−89= = 1;1the whole square will be painted.31.3 3⎛1 1⎞ 3⎛1 1⎞⎛1 1⎞ 3⎛ 1 ⎞+ ⎜ ⋅ ⎟+ ⎜ ⋅ ⎟⎜ ⋅ ⎟+ ... = ∑ ⎜ ⎟4 4⎝4 4⎠ 4⎝4 4⎠⎝4 4⎠ 4⎝16⎠∞k=1k –134116= =1−45The original does not need to be equilateral since each smaller triangle will have 1 4triangle.∞–1π π 3⎛1⎞4 3 π32. Ratio of inscribed circle to triangle is , so3 3∑ ⋅ ⎜ ⎟ = =3 3 4⎝4⎠1−3 3k=1kπ( )(This can be seen intuitively, since every small triangle has a circle inscribed in it.)14area of the previous larger33. a. We first note that, at each stage, the number of sides is four times the number in the previous stage and thelength of each side is one-third the length in the previous stage. Summarizing:length/ perimeterStage # of sidesside (in.) p n0 3 9 271( )1 3(4) 9 36 n31( n )n⎛4⎞⎟⎝ ⎠n3(4 ) 9 27⎜3 3The perimeter of the Koch snowflake isn⎛4⎞lim pn= lim 27⎜ ⎟n→∞n→∞⎝ 3 ⎠which is infinite since 4 13 > .b. We note the following:3 21. The area of an equilateral triangle of side s is4 s2. The number of new triangles added at each stage is equal to the number of sides the figure had at theprevious stage and3. the area of each new triangle at a given stage iscan summarize:3 (sidelength at that stage)2. Using results from part a. we4StageAdditional triangles Area of each new Δ(col 2, part a.) (see col 3, part a.)2 2( ) ( )2( )Additional area, A3 30 original 9 94 431 3 3 364 nnn−1( )2 n−23⎛ 9 ⎞ ⎛4⎞34 ⎜ 3 3n ⎟ ⎜ ⎟4 ⎝3⎠ ⎝9⎠Instructor’s Resource Manual Section 9.2 525

Thus the area of the Koch snowflake is∞∞n−181 3 27 3 ⎛4⎞∑An= + + ∑3 3⎜ ⎟n= 0 4 4 n=1 ⎝9⎠81 3 27 3 ⎛ 3 3 ⎞= + + 4 4 ⎜ 4( 19 ) ⎟⎝−⎠81 3 1 ⎛81 3 ⎞ 4 ⎛81 3 ⎞ 8 ⎛81 3 ⎞= + 4 3⎜ + =4 ⎟ 15⎜ 4 ⎟ 5⎜ 4 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠Note: By generalizing the above argument it can be shown that, no matter what the size of the originalequilateral triangle, the area of the Koch snowflake constructed from it will be 8 times the area of the original5triangle.34. We note the following:1. Each triangle contains the angles 90, θ,90− θ 2. The height of each triangle will be the hypotenuse of thesucceeding triangle. Summarizing:# triangle base height area A n1 hcosθ hsinθ 1h2sinθcosθ22 hsinθcosθ 2 1hsin θ h2sin3θcosθ2 n1 1h(sin n−θ )cosθ hsin nθ h2sin2n−1θcosθ2Thus the total area of the small triangles is∞ 2 ∞h ⎛cosθ⎞ 2 n−1A= ∑ An= ⎜ ⎟ ∑ (sin θ )n= 1 2 ⎝ sinθ⎠n=2∞2 n−11 1Now consider the infinite geometric series S = ∑ (sin θ ) = =2 2n=11−sin θ cos θ∞22 2 22 n−11 sin θh ⎛cosθ⎞ sin θ hthen: ∑ (sin θ ) = S − 1 = − 1 = Therefore:A = ⎛ ⎞tan2 2⎜ ⎟ = θ2n=2cos θ cos θ2 ⎝ sinθ⎠ ⎜ cos θ ⎟ ⎝ ⎠2In ABC , height = h and base = h tanθ ; thus the area of1 h ABC = ( h tan θ ) h = tan θ , the same as A .2 2235. Both Achilles and the tortoise will have moved.∞k –11 1 ⎛ 1 ⎞100 + 10 + 1 + + + ... = ∑ 100⎜ ⎟10 100 ⎝10⎠100 1= =1−110111 yards9k=1dAlso, one can see this by the following reasoning. In the time it takes the tortoise to run yards, Achilles will10run d yards. Solved 1000 1d = 100 + . d = = 111 yards10 9 9526 Section 9.2 Instructor’s Resource Manual

36. a. Say Trot and Tom start from the left, Joelfrom the right. Trot and Joel run towardseach other at 30 mph. Since they are 60miles apart they will meet in 2 hours. Trotwill have run 40 miles and Tom will haverun 20 miles, so they will be 20 miles apart.Trot and Tom will now be approaching eachother at 30 mph, so they will meet after2/3 hour. Trot will have run another40/3 miles and will be 80/3 miles from theleft. Joel will have run another 20/3 milesand will be at 100/3 miles from the left, sothey will be 20/3 miles apart. They will meetafter 2/9 hour, during which Trot will haverun 40/9 miles, etc. So Trot runs∞ k –140 40 ⎛1 ⎞ 4040 + + + = ∑ 40⎜⎟ =3 9 ⎝3⎠1−1= 60 miles .k=1 3b. Tom and Joel are approaching each other at20 mph. They are 60 miles apart, so they willmeet in 3 hours. Trot is running at 20 mphduring that entire time, so he runs 60 miles.37. Note that:1. If we let t n be the probability that Peter winson his nth flip, then the total probability that∞Peter wins is T = ∑ tnn=12. The probability that neither man wins in their2 2k4k⎛ ⎞ ⎛ ⎞first k flips is ⎜ ⋅ ⎟ = ⎜ ⎟ .⎝3 3⎠ ⎝9⎠3. The probability that Peter wins on his nth fliprequires that (i) he gets a head on the nthflip, and (ii) neither he nor Paul gets a headon their previous n-1 flips. Thus:n−∞ n−⎛1⎞⎛4⎞ ⎛1⎞⎛4⎞tn= ⎜ ⎟⎜ ⎟ and T = ∑ ⎜ ⎟⎜ ⎟ =⎝3⎠⎝9⎠ ⎝3⎠⎝9⎠1 1n=1⎛ 1 3⎞ 1 9 3= ⋅ =⎜( 1 4 9 ) ⎟⎝−⎠3 5 538. In this case (see problem 37),2n−1tn= p⎡( 1−p)⎤ so⎣ ⎦∞2n−1pT = ∑ p⎡( 1− p)⎤ =⎣ ⎦n=11−1−pp 1= =2 − p2( 2 p−p )2( ( ) )39. Let X = number of rolls needed to get first 6 ForX to equal n , two things must occur:1. Mary must get a non-6 (probability = 5 ) on 6each of her first n-1 rolls, and2. Mary rolls a 6 (probability = 1 ) on the nth6roll. Thus,n−1⎛5⎞ ⎛1⎞Pr( X = n) = ⎜ ⎟ ⎜ ⎟ and⎝6⎠ ⎝6⎠∞ n−11⎛5⎞ ⎛1⎞ 6∑ ⎜ ⎟ ⎜ ⎟ = = 1⎝6⎠ ⎝6⎠1−n=1( )1 ⎛5⎞EV X = n ⋅ X = n = ∑n⋅ ⋅ ⎜ ⎟6 ⎝6⎠5 640. ( ) ∑ Pr ( )∞n−1n= 1 n=1∞1⎛ 1 ⎞ n 1⎛6⎞ ⎛ 5/6 ⎞⎜ ⎟∑n p ⎜ ⎟ 2p n=1⎜ ⎟= ⋅ =6⎝ ⎠ 6⎝5⎠ ⎝( 1−5/6)⎠1⎛6⎞⎛5⎞⎛36⎞= ⎜ ⎟⎜ ⎟⎜ ⎟ = 66⎝5⎠⎝6⎠⎝ 1 ⎠41. (Proof by contradiction) Assume42.converges, and c ≠ 0. Then 1 c∞ ∞ ∞a 1 1k = ca k = ca kk= 1 k= 1cc k=1∞∞∑ cakk=1is defined, so∑ ∑ ∑ would alsoconverge, by Theorem B(i).∞∞1 1 1 1 1⎛1⎞1 1+ + + + = ∑ ⎜ ⎟=∑2 4 6 8 k= 12⎝k⎠ 2 k=1k∞1∑ diverges.k=1 ksince diverges43. a. The top block is supported exactly at itscenter of mass. The location of the center ofmass of the top n blocks is the average of thelocations of their individual centers of mass,so the nth block moves the center of massleft by 1 of the location of its center ofnmass, that is, 1 1 n ⋅ 2or 1to the left. But2nthis is exactly how far the (n + 1)st blockunderneath it is offset.b. Since∞1 1 1 1 1...2 4 6 2 k = 1 k+ + + = ∑ , whichdiverges, there is no limit to how far the topblock can protrude.Instructor’s Resource Manual Section 9.2 527

44. N = 31; S31 ≈ 4.0272 and S30≈ 3.9950.∞∑ ( ak+ bk)k=1∞converges. Since ∑ bkconverges, so wouldk=1∞ ∞ ∞∑ak = ∑( ak + bk) + ( −1) ∑ bk,byk= 1 k= 1 k=145. (Proof by contradiction) AssumeTheorem B(ii).∞ ∞1∑an=n= 1 n=1n∞ ∞1∑bn= ∑ ( −1)both diverge, butn= 1 n=1 n∞∞⎛1 1⎞( an+ bn)= ⎜ − ⎟n= 1 n=1⎝nn⎠46. (Answers may vary).∑ and∑ ∑ converges to 0.47. Taking vertical strips, the area isk –11 1 1 ⎛1⎞1⋅ 1+ 1⋅ + 1⋅ + 1⋅ + = ∑ ⎜ ⎟ .2 4 8k=1⎝2⎠Taking horizontal strips, the area is∞1 1 1 1k⋅ 1+ ⋅ 2+ ⋅ 3+ ⋅ 4+ =∑2 4 8 16 k=1 2 k .a.∞ ∞ k –1k ⎛1⎞1∑ = 2k ∑ ⎜ ⎟ = =1k= 1 2 k=1⎝2 ⎠ 1−2∞48. If49. a.∞kkrk=1∑ converges, so will rTheorem B.∞k∑ krk=1, by∞ ∞ ∞k k+1krS = r kr = kr = ( k –1) rk= 1 k= 1 k=2∞∞kkS = ∑ kr = r+∑ kr sok= 1 k=2∞ ∞kkS – rS = r+∑ kr – ∑ ( k –1) rk= 2 k=2∞[ –( –1)] k ∞ k ∞= r+ ∑ k k r = r+ ∑ r = ∑ rkk= 2 k= 2 k=1∞k rSince r < 1, ∑ r = , thusk=1 1– r∞1 k rS = ∑ r = .1– r2k=1 (1 – r)b.∑ ∑ ∑ while∞∞–11 n−nkt⎛ ⎞A= ∑Ce = ∑ C⎜ kt ⎟n= 0 n=1 ⎝ e ⎠ktC Ce= =1−kte −11kte1 −kt−6kln2 4= e = e ⇒ k = ⇒ A=C ;2 6 38if C = 2 mg, then A = mg.3b. The moment about x = 0 is∑∞ k⎛1⎞ ∞ k∞ k⎜ ⎟ ⋅ (1) k = ∑ = ∑ = 2.⎝2 ⎠k k2 2k= 0 k= 0 k=1moment 2x = = = 1area 250. Using partial fractions,k2 1 1= –k+ 1 k k k+1(2 –1)(2 –1) 2 –1 2 –11 1 1 1 1 1 1 1S n = ⎛ ⎜ – ⎞ – – –1 2 ⎟+ ⎛ ⎜ ⎞ 2 3 ⎟+…+ ⎛ ⎜ ⎞ n–1 n ⎟+⎛ ⎜ ⎞n n+ 1 ⎟⎝2 –1 2 –1⎠ ⎝2 –1 2 –1⎠ ⎝2 –1 2 –1⎠ ⎝2 –1 2 –1⎠1 1 1= – = 1–2–1 n+ 1 n+12 –1 2 –11lim Sn= 1– lim 1 – 0 1n→∞n→∞n+12 –1= =528 Section 9.2 Instructor’s Resource Manual

51.1 1 fk2 – fk1– = + =fk fk+ 1 fk+ 1fk+ 2 fk fk+ 1fk+ 2 fk fk+2since fk+ 2 = fk+1 + fk.Thus,∑∞∞1 ⎛ 1 1 ⎞= ∑ ⎜ – ⎟k= 1 fk fk+ 2 k=1 fk fk+ 1 fk+ 1fand⎝k+2 ⎠⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞Sn= ⎜ – ⎟+ ⎜ – ⎟ + + ⎜ – ⎟+⎜ – ⎟⎝ f1f2 f2f3 ⎠ ⎝ f2f3 f3f4 ⎠ ⎝ fn–1fn fn fn+ 1⎠ ⎝ fn fn+ 1 fn+ 1fn+2 ⎠= 1 1 1 1 1– – 1–f1f2 fn 1f = + n+ 2 11 ⋅ fn 1f =+ n+ 2 fn+ 1fn+2The terms of the Fibonacci sequence increase without bound, so1lim Sn= 1– lim 1– 0 1f f= =n→∞ n→∞ n+ 1 n+29.3 Concepts Review1. bounded above2. f(k); continuous; positive; nonincreasing3. convergence or divergence4. p > 1Problem Set 9.34.32is continuous, positive, and2x + 1nonincreasing on [1, ∞ ) .∞∞ 3 3 tan–1 21 22 1 dx ⎡x2x ⎤= ⎢ ⎥+ ⎣ ⎦1∫3 ⎛π−1⎞= ⎜ − tan 2 ⎟

7.8.9.10.11.7is continuous, positive, and4x + 2nonincreasingon.[2, ∞)∞∞ 7 ⎡7 ⎤ 7dx = ln 4x+ 2 =∞ – ln10 =∞2 4x+ 2⎢4⎥⎣ ⎦24∫The series diverges.2xxe[2, )is continuous, positive, and nonincreasing∞ . Using integration by parts twice, withi– xu = x , i = 1, 2 and dv = e dx,∞ 2 – x 2 – x ∞ ∞ – x∫ x e dx = [– x e ]22 + 2∫xe dx22 – x ∞ – x ∞ – x= [– x e ] 2 + 2⎛[– xe ]∞2 e dx⎞⎜ + ⎟⎝∫ 2 ⎠2 – x – x – x ∞= [– xe – 2 xe – 2 e ] 2–2 −2 2= 0+ 4e + 4e + 2e −The series converges.37/6(4 + 3 x)nonincreasing on [1, ∞ ) .–2= 10e

17.⎧1 k = 4j+1⎛kπ⎞ ⎪sin ⎜ ⎟ = ⎨–1 k = 4 j + 3,⎝ 2 ⎠ ⎪ ⎩ 0 k is evenwhere j is any nonnegative integer.⎛kπ⎞Thus lim sin ⎜ ⎟ does not exist, hencek→∞⎝ 2 ⎠⎛kπ⎞lim sin ⎜ ⎟ ≠ 0 and the series diverges.k→∞⎝ 2 ⎠1118. As k →∞, → 0. Let y = , thenkk1 1 sin ylim ksin = lim sin y = lim = 1 ≠ 0, sok→∞ k y→0 y y→0ythe series diverges.19.32 –xx e is continuous, positive, andnonincreasing on [1, ∞ ) .∞∞ 3 32 – x ⎡ 1 – x ⎤ 1 –1xe dx= – e 0 e ,1⎢3⎥ = +

27.28.29.30.31.∞1 ∞ 1 A 1En= ∑ < lim2 ∫ dx =n 2 ∫ dx =n 2k= n+1 k x A→∞x⎡1 1⎤1lim ⎢ −A→∞n A⎥=⎣ ⎦ n1< 0.0002 ⇒ n > 5000n∞1 ∞ 1 A 1En= ∑ < lim3 ∫ dx =n 3 ∫ dx =n 3k= n+1 k x A→∞x⎡ 1 1 ⎤ 1lim ⎢ −A 22 22⎥=→∞22⎣ n A ⎦ n1 1< 0.0002 ⇒ n > = 5022n0.0004En1 ∞ 1= ∑ < ∫1+ k 1+xk= n+1π2∞−12 n 2A 1−1 −1= lim dx lim ⎡tan A tan n⎤∫ = −A→∞n 21+x A→∞⎣⎦= −tannπ( )dxπ − tan < 0.0002 ⇒ tan > π −0.0002−1 −1nn2 2⇒ n > tan −0.0002 ≈50002∞k ∞ xEn= ∑ ln2n2e0.0004≈2.797 ∴ n > 2En∞k x 1 du= ∑ < ∫ = ∫1+ k 1+ x 1+uk= n+1∞Adx lim4 4 2nA→∞2 n 22u=xdu=2xdx( )( )1 −1 −1 2 1⎡π−2 2 ⎤= lim ⎡tan A− tan n ⎤ = −tann2 A→∞⎣ ⎦ 2⎢2⎥⎣ ⎦1 ⎡π−2 2 ⎤tan ( ) 0.00022⎢− n2⎥ 1.57039632⇒ n > tan 1.5703963 ≈5032.∞1 ∞ 1En= ∑ < dx =1( 1) nk= n+kk+ ∫xx ( + 1)A⎛1 1 ⎞lim ∫dxA n ⎜ − ⎟ =→∞ ⎝ x x+1⎠⎡ ⎛ A ⎞ ⎛ n ⎞⎤⎛ n ⎞lim ⎢ln ⎜ ⎟− ln ⎜ ⎟ = 0 − ln =A→∞A 1 n 1⎥ ⎜ ⎟⎣ ⎝ + ⎠ ⎝ + ⎠⎦⎝n+1⎠⎛n+ 1⎞ln ⎜ ⎟⎝ n ⎠⎛n+ 1⎞ 1 0.0002ln ⎜ ⎟ < 0.0002 ⇒ 1 + < e ≈ 1.0002⎝ n ⎠n1⇒ n > = 50000.0002∞ 133. Consider ∫ 2x(ln x) p dx. Let u = ln x,34.35.1du = dx.x∞ 1 ∞ 12(ln ) p dx =x xln2upp > 1.∫ ∫ du which converges for1is continuous, positive, andx ln xln(ln x)nonincreasing on [3, ∞ ) .∞ 1∫ 3 xlnxln(ln x) dx1Let u = ln(ln x), du = dx.xlnx1 1∫ ∞ dx = ∞ du = [ln u]3 ln(ln3)ln(ln ∞ 3)xln xln(ln x) ∫u= ∞ –ln(ln(ln3)) =∞The series diverges.The upper rectangles, which extend to n + 1 on1 1 1the right, have area 1 + + +…+ . These2 3 n1rectangles are above the curve y = from x = 1xto x = n + 1. Thus,n+ 1 1n+1∫ dx = [ln x] 11 = ln( n + 1) – ln1 = ln( n + 1)x532 Section 9.3 Instructor’s Resource Manual

1 1 1< 1 + + +…+ .2 3 nThe lower (shaded) rectangles have area1 1 1+ +…+ . These rectangles lie below the2 3 n1curve y = from x = 1 to x = n. Thusx1 1 1 1ln ,2 + ndx n3+…+ n< ∫ 1 x= so1 1 11 + 1 ln n.2 + 3+…+ n< +36. From Problem 35, B n is the area of the regionwithin the upper rectangles but above the curve1 y = . Each time n is incremented by 1, thexadded area is a positive amount, thusincreasing.From the inequalities in Problem 35,1 1 1B n is0< 1 + + +…+ –ln( n+ 1) < 1+ ln n–ln( n+1)2 3 nn= 1+lnn + 1n nSince < 1, ln < 0, thus B n < 1 for all n,n+ 1 n+1and B n is bounded by 1.37. { B n } is a nondecreasing sequence that isbounded above, thus by the Monotonic SequenceTheorem (Theorem D of Section 9.1), lim Bn→∞exists. The rationality of γ is a famous unsolvedproblem.138. From Problem 35, ln( n + 1) < ∑ 1 ln n,k=1k< + thus10,000,0001ln(10,000,001) ≈ 16.1181 < ∑k< 1+ ln(10,000,000) ≈ 17.1181nk=139. γ + ln( n+ 1) > 20 ⇒ ln( n+ 1) > 20 – γ ≈ 19.422819.4228⇒ n+ 1 > e ≈ 272,404,867⇒ n > 272,404,866nb. The leftmost rectangle has area1 ⋅ f (1) = f (1). If each shaded region to theright of x = 2 is shifted until it is in theleftmost rectangle, there will be no overlapof the shaded area, since the top of eachrectangle is at the bottom of the shadedregion to the left. Thus, the total shaded areais less than or equal to the area of theleftmost rectangle, or Bn≤ f(1).c. By parts a and b, { B n } is a nondecreasingsequence that is bounded above, so lim B nn→∞exists.1d. Let f( x) = , thenxn+ 1 n+11∫ f ( x) dx = dx ln( n 1)1 ∫ = + and1 xlim B n = γ as defined in Problem 37.n→∞41. Every time n is incremented by 1, a positiveamount of area is added, thus { A n } is anincreasing sequence. Each curved region hashorizontal width 1, and can be moved into theheavily outlined triangle without any overlap.This can be done by shifting the nth shadedregion, which goes from (n, f(n)) to(n + 1, f(n + 1)), as follows:shift (n + 1, f(n + 1)) to (2, f(2)) and (n, f(n)) to(1, f(2)–[f(n + 1) – f(n)]).The slope of the line forming the bottom of theshaded region between x = n and x = n + 1 isf( n+1)– f( n)= f( n+ 1)– f( n) > 0( n+1)– nsince f is increasing.By the Mean Value Theorem,f ( n+ 1)– f( n) = f′( c)for some c in (n, n + 1).Since f is concave down, n < c < n + 1 means thatf ′() c < f′() b for all b in [1, n]. Thus, the nthshaded region will not overlap any other shadedregion when shifted into the heavily outlinedtriangle. Thus, the area of all of the shadedregions is less than or equal to the area of theheavily outlined triangle, so lim A n exists.n→∞40. a. Each time n is incremented by 1, a positiveamount of area is added.Instructor’s Resource Manual Section 9.3 533

42. ln x is continuous, increasing, and concave down on [1, ∞ ) , so the conditions of Problem 41 are met.na. See the figure in the text for Problem 41. The area under the curve from x = 1 to x = n is ∫ ln x dx and the1area of the nth trapezoid is ln n+ ln( n+1) n ⎡ln1+ ln 2 ln( n–1) + ln( n)⎤, thus An= ln xdx–2∫ 1 ⎢ +…+2 2 ⎥⎣⎦.1Using integration by parts with u = ln x, du = dx, dv = dx, v = xxnn nnln x dx = [ x ln x] 11 – dx = [ x ln x – x]11∫ ∫ = nln n– n– (ln1–1) = nln n– n+1The sum of the areas of the n trapezoids isln1+ ln 2 ln 2 + ln 3 ln( n – 2) + ln( n –1) ln( n –1) + ln( n ) 2ln 2 + 2ln 3 +…+ 2ln( n –1) + ln+ +…+ + =n2 2 2 2 2ln nln n= ln 2 + ln 3 +…+ ln n – = ln(2⋅3 ⋅…⋅ n) – = ln n!– ln n2 2n nA = nln n– n+ 1– ln n!– ln n = nln n– n+ 1– ln n! + ln n = ln n – ln e + 1– ln n! + ln nThus, n( )nn⎛n⎞ n ⎡⎛n⎞n⎤= ln ⎜ ⎟ + 1+ ln = 1+ ln ⎢⎜ ⎟ ⎥⎝e⎠ n! ⎢⎝e⎠n!⎣ ⎥⎦⎡ ⎡ nn n⎤⎤b. By Problem 41, lim A n exists, hence part a says that lim ⎢⎛ ⎞1+ ln ⎢⎜⎟ ⎥⎥n→∞n→∞⎢ e n!⎣⎢⎣⎝ ⎠ ⎥⎦⎥⎦⎡ ⎡ n n nn n⎤⎤⎡ n n⎤ ⎡ n n⎤lim ⎢⎛ ⎞ ⎛ ⎞ ⎛ ⎞1+ ln ⎢⎜ ⎟ ⎥ ⎥ = 1+ lim ln ⎢⎜ ⎟ ⎥ = 1+ ln ⎢ lim ⎜ ⎟ ⎥→∞ ⎢ ⎢⎝e⎠ n! ⎥⎥→∞ ⎢⎝e⎠ n! ⎥ ⎢ →∞⎝e⎠n!⎣ ⎣ ⎦⎦⎣ ⎦ ⎣ ⎥⎦n n n⎛n⎞nSince the limit exists, lim ⎜ ⎟ = m.m cannot be 0 sincen→∞⎝e⎠n!nlim ln x = – ∞ .+x→0exists.Thus,n! 1 1 1lim = lim = = , i.e., the limit exists.n→∞ nn n→∞n n( ) ( e ) nlim n n mne( )n!n→∞e n!nn⎛n⎞⎛15⎞c. From part b, n! ≈ 2 πn ⎜ ⎟ , thus, 15! ≈ 30π⎜⎟ ≈ 1.3004×10⎝ e ⎠⎝ e ⎠The exact value is 15! = 1,307,674,368,000 .43. (Refer to fig 2 in the text). Let bTherefore∞t∑ a ≥ ∑ b k = ∫ f ( x ) dx so thatkk= n+ 1 k= n+1tn k kk= n+ 1t→∞k= n+1tkt1512k+1= ∫ f ( x ) dx; then from fig 2, it is clear that a ≥ b for k = 1,2, …, n,…n+1∑ ∑ ∫ ∫ .E = a = lim a ≥ lim f ( xdx ) = f ( xdx )tk∞t→∞n+ 1 n+1kk534 Section 9.3 Instructor’s Resource Manual

9.4 Concepts Review1. 0 ≤ ak ≤ bka2. lim kk→∞ bk3. ρ < 1; ρ > 1; ρ = 14. Ratio; Limit ComparisonProblem Set 9.41.2.3.4.ann 1= ; b2n =n + 2n+3 n2ann1n→∞ b n 212 3n →∞ n + 2n+3 n→∞+ +n 2nlim = lim = lim = 1;0 < 1 < ∞∞∞∑ bndiverges ⇒ ∑ andiverges.n=1an3n+ 1 1= ; b =n –4 nn=13 n 23 21a 3n 3n + n +nn→∞ b n 31–4n →∞ n –4 n→∞3nlim = lim = lim = 3;0 < 3 < ∞∞∞∑ bnconverges ⇒ ∑ anconvergesn=1a nn=11 1= =n n+ 1 n + n3 2;b n=n13/23/2 3ann nlim = lim = limn →∞ b n 3 2 n 3 2n →∞ →∞ n + n1nn+ n1= lim = 1; 0 < 1 < ∞n→∞1+∞∑ b converges ⇒ ∑ anconvergesnn=1∞n=12n+ 1 1an= ; b2 n =3/2n nann 2n+ 1 2n + nlim = lim = limn→∞ b n→∞ 2 n4n →∞ nn1n3/2 4 32 += lim = 2;0 < 2 1n→∞1+ 5+10+10+5+1The series diverges.n 2 3 4 5n n n n100 100an+1 ( n+1)! n nlim = lim = limn→∞ a n 100 99n →∞ ( n+ 1) n! n→∞( n+1)99n= lim =∞ sincen 1( ) 99n→∞ +nThe series diverges.5⎛n+ 1⎞lim ⎜ ⎟ = 1→∞⎝n ⎠n( )n( )n+11a ( n + 1)n+1 3 n + 1lim = lim = limn→∞ ann→∞ 1 n→∞n3 n31+1n 1= lim = < 1n→∞ 3 3The series converges.3an+1 ( n+1) (2 n)!lim = limn→∞a n (2 2)!3n →∞ n+n( 3 3 2n+ 1) n + 3 n +limlim3 n+= =1n→∞3 5 4 3(2n+ 2)(2n+ 1) n n→∞4n + 6n + 2n1 + 3 + 3 + 12 3 4 5n n n nn→∞4 + 6 + 2n 2n= lim = 0 < 1The series converges.n+1an+1 (3 + n+1) n!lim = limn→∞a n ( 1)!(3nn →∞ n+ + n )n+1lim 3 + n + 1=n→∞(3 n+ n )( n+1)+ n + 1n n3 321 n nn n3 32n+1lim 3 + n + 1=n→∞3 n 3 n 2n + + n + n3n= lim = 0

11.12.13.14.n1lim = lim = 1 ≠ 0n→∞n + 200 n→∞1+200nThe series diverges; nth-Term Testan+1 ( n+ 1)!(5 + n)lim = limn→∞ann→∞(6 + n ) n !2( n+ 1)(5 + n) n + 6n+5= lim= limn→∞6+ n n→∞6+nn + 6 + 5n= lim =∞ > 1n→∞6 + 1nThe series diverges; Ratio Testn + 3 1an= ; b2 n =3/2n n n5/2 3/23a 3 1n n + n+nlim = lim = lim = 1;n→∞ b n 5/2n →∞ nn→∞1∞∞0 < 1 < ∞ . ∑ bnconverges ⇒ ∑ ann=1n=1converges; Limit Comparison Testn + 1 1an= ; b2 n =3/2n + 1 n3/2 4 3ann n+ 1 n + nlim = lim = limn→∞ b n 2 2n →∞ n + 1 n→∞n + 117.18.19.an34n+ 3n1= ; b =n –4n + 1 n5 2 n 25 33a42n 4n + 3n+nn→∞ b n 5 2 1– 4 1n →∞ n –4n+ 1 n→∞+3 5n nlim = lim = lim = 4;0 < 4 < ∞∞bnn=1∞∑ converges ⇒ ∑ anconverges; LimitComparison Testn=12 nan+1 [( n + 1) + 1]3lim = limn→∞a n 3n+1 (2n →∞ n + 1)22 2n + 2n+21+ +n 2n 1= lim = lim = < 1n→∞2 3 33n+ 3 n→∞+ 32nThe series converges; Ratio Test1 1 1an= = ; b( 1) 2 n =nn+ 2n + n n2ann1lim = lim = lim = 1;n→∞ b n 2 n 1 1n →∞ n + n →∞ +n0 < 1 < ∞∞bnn=1∑ converges ⇒ ∑ anconverges;∞n=1Limit Comparison Test15.16.1n12n1+= lim = 1; 0 < 1

22.23.24.25.26.27.n 1an= ; b2 n =n + 1 n2ann1lim = lim = lim = 1; 0 < 1

33.34.n+1an+1 (4 + n+1) n!lim = limn→∞a n ( 1)!(4nn →∞ n+ + n )n+14 n 14 + n + 1+ +n n= lim= lim 4 4n→∞n( n+ 1)(4 + n)n→∞⎛ n ⎞( n + 1) ⎜1+4n ⎟⎝ ⎠4 + n + 1n n4 4= lim = 02n→∞1+ n + n + nn n4 42n nsince lim = 0, lim = 0 , andn→∞nn4 n→∞41lim 0.n→∞ 4 n= The series converges; Ratio Testnlim an+1lim ( n+ 1)(2+n5 )=n→∞a n [2 ( 1)5 n+1n →∞ + n+] n2n2n+ n 5 + 2+n5= limn→∞22 5n+ 1 5n+1n+ n + n2 2 1n15 25 1limn+ + nn+ n= = < 1n→∞2 5 5 5n+ +n5nThe series converge; Ratio Test35. Since ∑ anconverges, lim an= 0. Thus, theren→∞is some positive integer N such that 0< a n < 12for all n ≥ N. a < 1 ⇒ a < a , thus36.n n n∞ ∞∞22∑ an < ∑ an.Hence ann= N n=Nn=Nn∑ converges,2and ∑ a n also converges, since adding a finitenumber of terms does not affect the convergenceor divergence of a series.∞n!nn=1 nn!lim 0n→∞ nn∑ converges by Example 7, thus= by the nth-Term Test.an37. If lim = 0 then there is some positive integern→∞ bnanN such that 0≤ < 1 for all n ≥ N. Thus, forbnn ≥ N, an< bn. By the Comparison Test, since∞∞∑ bnconverges, ∑ analso converges. Thus,n=Nn=N∑ a n converges since adding a finite number ofterms will not affect the convergence ordivergence of a series.a38. If lim n = ∞ then there is some positiven→∞ bnaninteger N such that > 1 for all n ≥ N. Thus,b nfor n ≥ N , an > bnand by the Comparison∞∞Test, since ∑ bndiverges, ∑ analson=Nn=Ndiverges. Thus, ∑ andiverges since adding afinite number of terms will not affect theconvergence or divergence of a series.39. If lim nan= 1 then there is some positiven→∞integer N such that an≥ 0 for all n≥ N,Let1 ab n = ,nso lim = lim nan= 1 0 on (0, ∞ ).1+ x 1+xf(0) = 0 – ln 1 = 0, so since f(x) is increasing,f(x) > 0 on (0, ∞ ), i.e., x > ln(1 + x) for x > 0.Thus, since a n is a series of positive terms,∑ln(1 + an) < ∑ an,hence if ∑ anconverges,∑ ln(1 + a n ) also converges.1/ n41. Suppose that lim ( an) = R where a n > 0.n→∞If R < 1, there is some number r with R < r < 1and some positive integer N such that1/ n( an) – R r – R< for all n ≥ N. Thus,1/ nR– r < ( an) – R < r – R or1/– r ( an) n r 1. a n >1/ nn0 < ( an) < r and 0 anr∞ ∞nThus, an< rn= N n=N∞r < 1. Thus, ann=N< < < Since 0,< < for all n ≥ N.∑ ∑ , which converges since∑ converges so ∑ analsoconverges.If R > 1, there is some number r with 1 < r < Rand some positive integer N such that538 Section 9.4 Instructor’s Resource Manual

42. a.1/ n( an) – R R–r< for all n ≥ N. Thus,1/ nr – R < ( an) – R < R–r or1/ nr < ( an) < 2 R–r for all n ≥ N. Hence∞ ∞nnr a nr < ann= N n=N∞∞nsince ∑ r diverges (r > 1), ann=Nn=N∑ ∑ and< for all n ≥ N, so ,diverges, so∑ andiverges.∑ also1/ n⎡ n1/ n ⎛ 1 ⎞ ⎤ 1lim ( an) = lim ⎢⎜⎟ ⎥ = limn→∞ n→∞ ⎢⎝lnn⎠⎥ n→∞ln n⎣ ⎦= 0

e.∞ n⎛ e ⎞⎜n=2 ln n ⎟⎝ ⎠n⎛ e ⎞an= ⎜ ⎟⎝lnn ⎠associated series,∑ converges.With , the Root Test (Problem41) gives1/ n⎡ n1/ n ⎛ e ⎞ ⎤lim ( an) = lim ⎢⎜⎟ ⎥n→∞n→∞⎢ ln n⎣⎝⎠ ⎥⎦e= lim = 0 < 1n→∞ln nThus, ∑ ∞ n⎛ e ⎞⎜n=2 ln n ⎟⎝ ⎠converges, sou∞ ⎛ e ⎞duln 3⎜⎟⎝lnu ⎠∞1∑ converges.ln nn=3 (ln(ln n))∫ converges, whereby4an= 1/ n; bn= 1/(ln n)4an1/ n (ln n)lim = lim = limn→∞ b4n n→∞1/(ln n)n→∞n3 34(ln n) (1/ n) 4(ln n)= lim= limn→∞1 n→∞n.f.2 212(ln n) (1/ n) 12(ln n)= lim= limn→∞1 n→∞n24(ln n) 24(1/ n)= lim = limn→∞n n→∞124= lim = 0n→∞n∞∞1 1diverges ⇒n(ln n)∑ ∑n= 2 n=224diverges⎛ln x ⎞⎜ ⎟ is continuous, positive, and⎝ x ⎠nonincreasing on [3, ∞ ). Using integrationby parts twice,∫∞2 2⎛ln x ⎞ ⎡ (ln x) ⎤ ∞2lnx⎜ ⎟ dx = ⎢− ⎥ + dxx x∫ 2x3 ⎝ ⎠ 3⎢⎣⎥⎦32∞∞∞⎡ (ln x) ⎤ ⎡ 2ln x⎤2= ⎢− ⎥ + ⎢− ⎥ + ∫ dx⎢⎣x 3 2⎥⎦⎣ x ⎦33 x2∞⎡ (ln x) 2ln x 2⎤= ⎢− − − ⎥ ≈ 1.8

1 ⎛ 1 1 1 ⎞45. Let a n = 1p ⎜ + + +…+2p 3p p ⎟n ⎝ n ⎠ and 1 ab n .n ⎛ 1 1 1 ⎞ 1= Then lim = lim 1p⎜ + + +…+n n→∞b n 2p 3p p ⎟=∑ whichpn →∞⎝n ⎠ n=1 n∞∞converges if p > 1. Thus, by the Limit Comparison Test, if ∑ bnconverges for p > 1, so does ∑ an.Since∞ ∞bn=n= 1 n=1n1∑ ∑ converges for p > 1,pn=1n=1∑∞1 ⎛ 1 1 1 ⎞1p ⎜ + + +⋅⋅⋅+1 2p 3p p ⎟n=n ⎝n ⎠also converges. For p ≤ 1, since∞⎛⎞11 .p ⎜ + +…+ >p p ⎟ Hence, sincep ∑ diverges for p ≤ 1,p⎝ 2 ⎠n=1 n1 1 1 1 1 11+ 1,2 p +…+np >n n n∑∞1 ⎛ 1 1 ⎞1p ⎜ + +…+2p p ⎟n n ⎝n ⎠also diverges. The series converges for p > 1 and diverges for p ≤ 1.= 146. a. Leta n2 ⎛1⎞= sin ⎜ ⎟⎝n⎠ and 1b n = . Then2n1using the substitution u = . Since 0 < 1 < ∞ , bothn2an2 2⎛1⎞ ⎛1⎞2lim = lim n sin ⎜ ⎟=lim sinn→∞ bnn→∞ n+⎜ ⎟⎝ ⎠ u→0⎝u⎠∞ ∞∞ ∞1∑ bn= ∑ and a2nn= 1 n=1nn= 1 n=1∞⎛sinu ⎞u = lim 1+⎜ ⎟ =u→0⎝ u ⎠2 ⎛1⎞= sin ⎜ ⎟⎝n⎠∑ ∑ converge.1( n )1( )⎛1⎞b. Let a n = tan ⎜ ⎟⎝n⎠ and 1 asinb n = .n⎛1⎞ nThen lim = lim n tan ⎜ ⎟=limn n→∞ bnn→∞ ⎝n⎠n→∞cosn1( ) sin uu⎛sinu ⎞1= lim = lim ⎜ cosu⎟=1 using the substitution u = . Since 0 < 1 < ∞ , bothu→0 cosuu→0⎝u ⎠n∑∞ ∞⎛1⎞an= ∑ tan ⎜ ⎟⎝n⎠diverge.c.n= 1 n=12⎛ ⎞ ⎛ ⎞⎛ ⎞ n 1 − cos 1 2 1n n⎜ ⎟ ⎜ ⎟ = = sin n

9.5 Concepts Review1. lim a = 0n→∞n2. absolutely; conditionally3. the alternating harmonic series4. rearrangedProblem Set 9.51.2.3.2 2 2a n = ; > , so an> a n + 1;3n+ 1 3n+ 1 3n+42lim = 0. S9≈0.363.The error made byn→∞3n+ 1using S 9 is not more than a10 ≈ 0.065.1 1 1a n = ; > , so an> a n + 1;n n n+11lim = = 0. S9≈ 0.76695. The error maden→∞nby using S 9 is not more than a10 ≈ 0.31623.a n1 1 1= ; > , soln( n+ 1) ln( n+ 1) ln( n+2)an> a n + 1;1lim = 0. S9≈1.137.The error made byn→∞ln( n + 1)using S 9 is not more than a10 ≈ 0.417.n n n+14. an = ; > , so a2 2 2 n > an+1;n + 1 n + 1 ( n+ 1) + 1nlim = 0. S2 9 ≈0.32153.The error made byn→∞ n + 1using S 9 is not more than a10 ≈ 0.09901.7.8.9.10.11.12.un+1un( −3)4( −3)4n+13= = < 1, so the seriesn 4converges absolutely.∞ ∞ ∞1 1∑ u = ∑ = ∑ which converges3/2nnn= 1 n= 1n n n=1∞since 3 1,2 > thus 1∑ (–1) n convergesn=1 n nabsolutely.n+1n+1 2nn n2u n+1 n n+ 1 + 1 1= = ; lim = < 1, so theu 2n n→∞2n2series converges absolutely.unu+ 1n+1e += =2n2( n+1)nne2( n 1)en2( n + 1) 1lim = ≈ 0.36788 < 1, so the seriesn→∞2en econverges absolutely.2 2nn ( + 1) = n + n> n for all n > 0, thus21 1 1 1< , sonn ( + 1) 2 ∑ un= ∑ < ∑ 2n n= 1 n= 1nn( + 1)n=1nwhich converges since 2 > 1, thus∞n+1 1∑ (–1)n= 1nn+un+1unn+12( n+1)!( 1);∞ ∞ ∞converges absolutely.2 2= =n ; lim = 0 < 1, so the2 n + 1 n→∞n + 1!nseries converges absolutely.5.6.ln n ln n ln( n+1)an= ; > is equivalent ton n n+1n+1n+1nnln > 0 or > 1 which is true fornn( n + 1) ( n + 1)n > 2. S9 ≈− 0.041 . The error made by using S 9is not more than a10 ≈ 0.230.ln n ln n ln( n+1)an= ; > for n ≥ 7, so an> a n + 1n n n+1for n ≥ 7. S9 ≈ 0.17199 . The error made by usingS 9 is not more than a10 ≈ 0.72814 .13.14.∞∞ n+1n+1 1 1 ( −1)∑( − 1) = ∑ which convergesn= 15n5n=1n∞ + 1since∑n=1(–1) nnconverges. The series isconditionally convergent since∞ ∞ ∞n 1.1 1.1n= 1 n= 15n5n=1n∞1 1∑ diverges.5 nn= 11 1 1∑ u = ∑ = ∑ converges since1.1 > 1. The series is absolutely convergent.542 Section 9.5 Instructor’s Resource Manual

15.16.17.n 1lim = ≠ 0. Thus the sequence ofn→∞10n+ 1 10partial sums does not converge; the seriesdiverges.n n+1>, so a1.1 1.1 n > a n + 1;10n+ 1 10( n+ 1) + 1n1lim = lim = 0 . Then→∞1.1 0.110n+ 1 n→∞10n+1nalternating series converges.nLet an=1.110n+ 1and 1b n = . Then0.1n1.1ann 1 1lim = lim = ; 0 < isn→∞nln n nln n ( n+ 1)ln( n+1)1equivalent to ( 1) n +n+ > nn which is true for alln > 0 so an> a n + 1. The alternating seriesconverges.∞ ∞1 1∑ un= ∑ ; is continuous,n= 2 n=2nlnn xlnxpositive, and nonincreasing on [2, ∞ ).1Using u = ln x, du = dx,x1 1∫ ∞ dx = du ln .2 xlnx∫ ∞ = ⎡ u ⎤∞ =∞ln2u⎣ ⎦Thus,ln 2∞∞1n 1∑ diverges and ∑ (–1) isn=2nlnnn=2nlnnconditionally convergent.20.21.22.1 1 1a n = ; > ,2 2 2n –1 n –1 n + 2n1anan+1; lim 0,n→∞2n –1so> = hence thealternating series converges.1Let b n = , thennann1lim = lim = lim = 1;n→∞ bnn→∞ 2 1– 1n –1 n→∞2n0 < 1 < ∞ .∞ ∞1Thus, since ∑ bn= ∑ diverges,nn= 2 n=2∞ ∞1an=2n= 2 n=2 n –1∑ ∑ also diverges. The seriesconverges conditionally.n n n+1an= ; >2 2 2n + 1 n + 1 ( n+ 1) + 1is equivalent to2n + n–1> 0, which is true for n > 1, sonan> a n + 1; lim = 0,n→∞ 2n + 11series converges. Let b n = , thenn2annlim = lim = 1; 0 < 1 1. The series is2absolutely convergent.4+4n+1 ( n + 1)4 4( n 1)un+1= 2 = ;unn 2nn24( n + 1) 1lim = < 1.n→∞42n2The series is absolutely convergent.23.n 1 n+1cos nπ= ( − 1) = ( −1)( −1)∞ n+1( −1)−1 , −1n=1 nso the series is∑ times the alternatingharmonic series. The series is conditionallyconvergent.Instructor’s Resource Manual Section 9.5 543

nπ∞ sin2 1 1 124. ∑ = 1 − + − + , since2n=1 n 9 25 49nπ⎧⎪0 n evensin = ⎨ n−1.2 ⎪ ( − 1)2⎩ n odd∞ sin nπ∞2n+1 1Thus, = ( −1)2 2n= 1 n n=1 (2n−1)∞ ∞1∑ un= ∑ 2n= 1 n=1(2 n –1)2 2(2 n–1)> n for n > 1, thus∞∞1 1 1. The series is absolutely convergent.25. sin n ≤ 1 for all n, so26.27.∞ ∞ sin n ∞1∑ un= ≤3/2n=1 n= 1 n n n=1n∑ ∑ which convergessince 3 > 1. Thus the series is absolutely2convergent.1( )⎛1⎞ sinn1nsin⎜ ⎟ = . As n⎝n⎠ 1nnsin k⎛1⎞lim = 1, so lim nsin ⎜ ⎟=1.k→0kn→∞⎝n⎠diverges.a n =1;nn ( + 1)→∞, → 0 andThe series1 1>andnn ( + 1) ( n+ 1)( n+2)1lim = 0 so the alternating seriesn→∞nn ( + 1)converges.1Let b n = , thennann1lim = lim = lim = 1;n →∞ bnn →∞ 2 n 1 1n + n →∞ +n0 < 1 < ∞ .∞ ∞1Thus, since ∑bn= ∑ diverges,n= 1 n=1n∞ ∞1∑an= ∑ also diverges.n= 1 n=1 nn ( + 1)The series is conditionally convergent.28.29.1a n =;n+ 1 + n1 1>, so an> a n + 1;n+ 1+ n n+ 2+ n+11lim = 0. The alternating seriesn→∞n+ 1 + nconverges.1Let b n = , thennann1 1lim = lim = lim = ;n→∞ bnn→∞ n+ 1 + n n→∞1+ 1 + 1 2n∞ ∞110 < 0 andnnπ ππsin > sin , so an> a n + 1; lim sin = 0.n n+1n→∞nThe alternating series converges.sin πansinWe have lim limn nlim 1n b= n π= →∞ →∞ n→0n n n= .∞∞The series sin π∑ n and π∑ n either bothn=1n=1converge or both diverge. Since∞ ∞π 1n = π nn= 1 n=1∑ ∑ isdivergent, it follows that sin π∑ n is divergent.n=1The series is conditionally convergent.31. Suppose ∑ anconverges. Thus, ∑ 2 a nconverges, so ∑ ( an+ an)converges since32. Let0≤ an + an ≤ 2 an. By the linearity ofconvergent series, ∑ an = ∑( an + an)−∑anconverges, which is a contradiction.+ 1 1∑ an∑ ∑ bn. an∞= (–1) n =n∑ b n both converge, butdiverges.∑abn n=∑ and∑1n544 Section 9.5 Instructor’s Resource Manual

33. The positive-term series is∞1 1 1 11 + + + + ... =3 5 7 n=1 2n−1∞∞1 1 1∑ >n= 12n−1 2 n=1n∑ .harmonic series diverges.Thus,∞1n= 1 2n−1∑ which diverges since the∑ diverges.The negative-term series is∞1 1 1 1 1 1...2 4 6 8 2 n = 1 n− − − − − =− ∑ which diverges,since the harmonic series diverges.34. If the positive terms and negative terms bothformed convergent series then the series wouldbe absolutely convergent. If one series wasconvergent and the other was divergent, the sum,which is the original series, would be divergent35. a.b.c.11+ ≈ 1.3331 11+ − ≈ 0.8333 21 1 1 1 1 11+ − + + + + ≈ 1.383 2 5 7 9 111 1 1 1 1 1 11+ − + + + + − ≈ 1.133 2 5 7 9 11 436.1 1 1 1 1 1 1 1 1 11+ − + + + + − + + +3 2 5 7 9 11 4 13 15 17S20 ≈ 1.32651 1 1 1 1 1 1 1 1− + + + − + + + +6 19 21 23 8 25 27 29 3137. Written response. Consider the partial sum of thepositive terms of the series, and the partial sum ofthe negative terms. If both partial sums werebounded, the series would be absolutelyconvergent. Therefore, at least one of the partialsums must sum to ∞ (or −∞ ). If the series ofpositive terms summed to ∞ and the series ofnegative terms summed to a finite number, theoriginal series would not be convergent (similarlyfor the series of negative terms). Therefore, thepositive terms sum to ∞ and the negative termssum to −∞ . We can then rearrange the terms tomake the original series sum to any value wewish.38. Possible answer: take several positive terms, addone negative term, then add positive terms whosesum is at least one greater than the negative termpreviously added. Add another negative term,then add positive terms whose sum is at least onegreater that the negative term just added.Continue in this manner and the resulting serieswill diverge.1 1 1 139. Consider 1− 1 + − + − + ...2 4 3 9It is clear that lim an= 0. Pairing successiven→∞1 1 n −1terms, we obtain − = > 0 for n > 1.n 2 2n nn –1 1Let an= and b .2 n = Thennn2ann – nlim = lim = 1; 0 < 1 < ∞ .n→∞b n 2n →∞ n∞ ∞1Thus, since ∑ bn= ∑ diverges,n= 1 n=1n∑∞ ∞⎛1 1 ⎞an= ∑ ⎜ –2 ⎟n= 1 n=1⎝nn ⎠also diverges.1 1 240. − = , son − 1 n + 1 n − 1∞1 1 1 1 2− + − + = ∑2− 1 2+ 1 3− 1 3+1 n=2 n −1∞1= 2∑ which diverges.n=1 nInstructor’s Resource Manual Section 9.5 545

241. Note that ( a + b ) ≥ 0 and ( a – b ) ≥ 0 for all k. Thus, a ± 2a b + b ≥ 0, or a + b ≥± 2a b for all k,kkk k∞2akk=12∞2bkk=12 2k k k k∞2 2( ak+ bk)k=1∞∞1ab k k = 2 ab k kk= 1 2 k=12 2k k k k2 2and ak + bk ≥ 2 akbk. Since ∑ and ∑ both converge, ∑ also converges, and by the∞2 ab k kk=1Comparison Test, ∑ converges. Hence,absolutely.42.∑ ∑ converges, i.e.,∞ab k kk=1∑ converges∞ sin x∫ dx gives the area of the region above the x-axis minus the area of the region below.0 xNote that∫(2k+ 1) π ⎛sin x sin( x+π ) ⎞ (2k+ 1) π sin x (2k+ 1) πsin( x+π)dx dx dx2kπ ⎜ + ⎟ = +x x+π ∫ 2kπ x ∫⎝⎠2kπx+π(2k+ 1) π sin x (2k+ 2) πsinu (2k+ 2) πsinx= ∫dx +du2 kπ x∫= dx(2k+ 1) π u∫ 2kπxby using the substitution u = x + π , then changing the variable of integration back to x.∞∞sin x (2k+ 1) π∞⎛sin x sin( x+π)⎞(2k+ 1) π( x+π )sinx+ xsin( x+π)Thus, ∫ dx = + dx0∑x∫2kπ⎜⎟ = ∑ dxk=0 ⎝ x x+π∫⎠2kπk=0xx ( +π)∞(2k+ 1) π∞xsin x+πsin x– xsinx (2k+ 1) π πsinx= ∑ ∫dx =2kπ∑ dx .k=0xx ( +π)∫ 2kπk=0 xx ( +π)π π1For k > 0, on [2kπ , (2k + 1) π ] 0 ≤ sin x ≤ 1 while 0 < ≤ =.xx ( +π) 2 kπ(2 kπ+π ) 2(4k + 2 k)π(2k+ 1) π πsin x 1 (2k+ 1) π 1Thus, 0 ≤∫ dx ≤ dx = .2kπxx ( +π ) 2 ∫ 22(4k + 2 k) π kπ4k + 2k∞∞ sin x 1 1Hence, ∫ dx ≤2π∑ ≤x 2 ∑ which converges.24k + 2k 4k∞k= 1 k=12πsin x ∞ sin xAdding ∫ dx will not affect the convergence, so dx0 x∫ converges.0 x43. Consider the graph of sin x on the interval [kπ , (k + 1) π ].xπ 5 πNote that for kπ+ ≤ x≤kπ+ 1 1 1 ⎡⎛ 1⎞ ⎛ 5⎞⎤, ≤ sin x while ≤ . Thus on6 6 2⎛ 5 ⎞ x⎢⎜k+ ⎟π , ⎜k+ ⎟π6 6⎥⎜k+ ⎟π⎣⎝ ⎠ ⎝ ⎠ ⎦⎝ 6 ⎠1 1 sin x ( k+ 1) π sin x ( k+ 5 6) π sin x 1 ( k+ 5 6) π= ≤ , sodxdx2 5 5( k +6) π ( 2k+3)π x∫ ≥kπ x∫≥dx( k+ 1 6) π x5 ( 1 6)( 2k+ k+ π3 )∫ π1=3k + 5.2546 Section 9.5 Instructor’s Resource Manual

∞1a k =k=1 3k+2akk1 1 1lim = lim = lim = ;k→∞ b k 3 5 k 3 5k →∞ k + →∞ + 32 2k∞ sin xdiverges. Hence, dxπ x∞ sin x 1Hence, ∫ dx ≥ .π∑ Letx 3k+ 5521and b k = .k0 <

9.6 Concepts Review1. power series2. where it converges3. interval; endpoints4. (–1, 1)Problem Set 9.61.2.∞ nn+1x ( n−1)!x x; ρ = lim = lim = 01 ( 1)! nnn=n−→∞ nx ! n→∞n∑ . Series converges for all x .∞ n n n+1x 3 x x x; ρ = lim = lim = ;n n 1 nn 1 3 n +=→∞ 3 x n→∞3 3∑ convergence on ( − 3,3) .For x = 3, a n = 1 and the series diverges.For x =− 3, a ( 1) nn = − and the series diverges.Series converges on ( − 3,3)3.4.∑ ∞ n2 n+1x2 ; lim n x2lim xρ = = = xn2 1;n=1 n n→∞( n+1) x n→∞(1+ +convergence on ( − 1,1) .n 2n1For x = 1, an= (p-series, p=2) and the series converges.2nn( −1)For x =− 1, a n = ( alternating p-series, p=2) and the series converges. by the Absolute Convergence Test.2nSeries converges on [ − 1,1]∑∞1( 1) n +n n+ x ⎛ 1⎞nx ; ρ = lim = lim 1 x x ;n ⎜ + ⎟ =n=1n→∞nx n→∞⎝ n ⎠convergence on ( − 1,1) .For x = 1, an= nand the series diverges.For x 1, a ( 1) nn=− = − n and the series diverges ( lim ( −1) n ≠ 0 )Series converges on ( − 1,1)nn→∞5. This is the alternating series for problem 3; thus it converges on [ − 1,1] by the Absolute Convergence Test.6.∑ ∞ nn+1n x nx ⎛ n ⎞( − 1) ; ρ = lim = lim x x ;n ⎜ ⎟ =n=1 n n→∞( n+1) x n→∞⎝n+1⎠convergence on ( − 1,1) .n( −1)For x = 1, a n = (Alternating Harmonic Series) and the series converges.n1For x =− 1, an= (Harmonic Series) and the series diverges.nSeries converges on ( − 1,1]7. Let u = x− 2 ; then, from problem 6, the series converges when u ∈ ( − 1,1] ; that is when x ∈ (1, 3] .548 Section 9.6 Instructor’s Resource Manual

8.9.∞ nn+1( x+ 1) n!( x+ 1) x+1; ρ = lim = lim = 01 ! nnn=n→∞ ( n+ 1)!( x+1) n→∞n+1∑ . Series converges for all x.∞ n+ 1 n n+1n( −1)x x xn∑ ; ρ = lim÷ = lim x = xn=1 nn ( + 1) n→∞( n+ 1)( n+ 2) nn ( + 1) n→∞n + 2∞n+1 1When x = 1, the series is (–1)n= 1 nn ( + 1)∞n ∞n+ 1 (–1) 2n+1 1When x = –1, the series is ∑(–1) = ∑ (–1)n= 1 nn ( + 1) n=1 nn ( + 1)∞∞∞1 11= ∑(–1) = ( −1)∑ which converges sincen= 1 nn ( + 1) n=1nn( + 1)n= 1 nn ( + 1)The series converges on –1 ≤ x ≤ 1.∑ which converges absolutely by comparison with the series∑ converges.∞1∑ .2n=1 n10.∞ n n+1 nx x x∑ ; ρ = lim ÷n=0 n! n→∞( n+1)! n!The series converges for all x.1= lim x = 0n →∞ n + 111.∞ n 1 2n 1 2n 1 2n1( −1)x x x∑+ − + −; ρ = lim ÷n=1 (2n− 1)! n→∞(2n+ 1)! (2n−1)!The series converges for all x.2 1= lim x= 0n→∞2 n(2n+1)12.13.14.∞ n 2n 2n+2 2n( −1)x x x∑ ; ρ = lim ÷n=0 (2 n)! n→∞(2n+2)! (2 n)!2 1= lim x= 0n→∞(2n+ 2)(2n+1)The series converges for all x.∞n+1n ( n+1) x∑ nx ; ρ = limnn=1n→∞nxn + 1= lim x = xn→∞n∞When x = 1, the series is ∑ n which clearlyn=1diverges.∞nWhen x =− 1, the series is ∑ n(–1) ; an= n;n=1lim an≠ 0, thus the series diverges.n→∞The series converges on –1 < x < 1.∞2 n+12 n ( n+1) x∑ n x ; ρ = lim2 nn=1n→∞n x2( n + 1)= lim x = xn→∞2n15.∞2When x = 1, the series is ∑ n which clearlyn=1diverges.∞2 nWhen x = –1, the series is ∑ n (–1) ;n=12an= n ; lim an≠ 0, thus the series diverges.n→∞The series converges on –1 < x < 1.∞ n n n+1 n( −1)x x x1 + ∑ ; ρ = lim ÷n=1 n n→∞n+1 nn= lim x = xn→∞n + 1∞n 1When x = 1, the series is 1 (–1) ,n=1 n+ ∑ which is1 added to the alternating harmonic seriesmultiplied by –1, which converges.When x = –1, the series is∞n ∞n (–1) 11 + ∑ (–1) = 1+n= 1 n n=1n∑ , which diverges.The series converges on –1 < x ≤ 1.Instructor’s Resource Manual Section 9.6 549

16.17.∞ n n+1 nx x x1 + ∑ ; ρ = lim ÷n=1 n n→∞n+1 nn= lim x = xn→∞n + 11When x = 1, the series is 1+ ∑ whichn∞n=1diverges since 1 1.2 , so an> a n + 1 andn n n+11lim = 0, so the series converges.n→∞nThe series converges on –1 ≤ x < 1.∞ n n( −1)x1 + ∑ ;n=1 nn ( + 2)n+1nxxρ = lim ÷n→∞( n+ 1)( n+ 3) n ( n+2)19.converges by comparison withWhen x = –1, the series is∞1∑ .2n=1 n∞ n(–1)2n= 1 ( n + 1) –1∑ whichconverges absolutely by comparison with∞1∑ .2n=1 nThe series converges on –1 ≤ x ≤ 1.∞ n n n+1 n( −1)x x x x∑ ; ρ = lim ÷ = limn n 1 nn 0 2 n→∞+=2 2 n→∞2x x= ; < 1 when –2 < x < 2.2 2When x = 2, the series iswhich diverges.When x = –2, the series is∞ n ∞n 2n∑(–1) = (–1)n ∑n= 0 2 n=0∞ n ∞ ∞n (–2)n n(–1) = (–1) (–1) = 1nn= 0 2 n= 0 n=0∑ ∑ ∑ whichdiverges. The series converges on –2 < x < 2.2n + 2n= lim x= xn→∞2n + 4n+3∞n 1When x = 1 the series is 1 + ∑ (–1)n=1 nn ( + 2)which converges absolutely by comparison with∞1the series ∑ .2n=1 nWhen x = –1, the series is∞n ∞n (–1) 11 + ∑(–1) = 1+∑ whichn= 1 nn ( + 2) n=1nn( + 2)∞1converges by comparison with ∑ .2n=1 nThe series converges on –1 ≤ x ≤ 1.20.∞n+ 1 n+1n n 2 x∑ 2 x ; ρ = lim = lim 2x = 2 x ;n nn=0n→∞2 x n→∞1 12x < 1 when – < x < .2 2∞ n ∞1When x = ,n ⎛1⎞the series is ∑ 2 ⎜ ⎟ = ∑ 12n= 0 ⎝2⎠ n=0which diverges.1When x = – , the series is2∞n ∞n⎛1 ⎞n2 ⎜– ⎟ = (–1)⎝ 2 ⎠∑ ∑ which diverges.n= 0 n=0The series converges on1 1– < x < .2 218.∞ nx∑ ;2n= 1 ( n + 1) −1n+1nxxρ = lim ÷n→∞( 2)2 1 ( 1)2n+ − n+ − 12n + 2n= lim x= xn→∞2n + 4n+321.∞∑n=0n n n+ 1 n+1 n nx x x; ρ = lim ÷n n→∞n n2 2 2! ( + 1)! !1= lim 2x = 0 .n →∞ n + 1The series converges for all x.When x = 1, the series is∞ ∞ ∞1 1 1= =2 2n= 1( n+ 1) –1 n= 1n + 2nn=1nn( + 2)∑ ∑ ∑ which550 Section 9.6 Instructor’s Resource Manual

22.23.24.25.∞ n n+1 nnx ( n + 1) x nx∑ ; ρ = lim÷n=1 n+ n→∞n+ n+1 2 12n + 2n+1= lim x= xn→∞2n + 2n∞nWhen x = 1, the series is ∑ whichn= 1 n + 1ndiverges since lim = 1 ≠ 0.n→∞ n + 1∞ nn(–1)When x = –1, the series is ∑ whichn= 1n + 1ndiverges since lim = 1 ≠ 0.n→∞ n + 1The series converges on –1 < x < 1.∞∑n=1n n+1 n( x−1) ( x−1) ( x−1); ρ = lim ÷n n→∞n+1 nn= lim x− 1 = x−1; x –1 < 1 whenn→∞n + 10 < x < 2.∞(–1) nWhen x = 0, the series is ∑ whichn=1nconverges.∞1When x = 2, the series is ∑ which diverges.n=1nThe series converges on 0 ≤ x < 2.∞∑n=0n n+1 n( x+ 2) ( x+ 2) ( x+2); ρ = lim ÷n! n→∞( n+1)! n!1= lim x + 2 = 0n→∞n + 1The series converges for all x.∞∑n=0n n+1 n( x+ 1) ( x+ 1) ( x+1); ρ = lim ÷n n 1 n2 n→∞+2 2x+ 1 x+1= lim = ;n →∞ 2 2–3 < x < 1.When x = –3, the series iswhich diverges.When x = 1, the series isx + 1 < 1 when2∞n∞(–2)∑ = ∑ (–1)nn= 0 2 n=0∞∑2n∞nn= 02n=0diverges.The series converges on –3 < x < 1.= ∑ 1 whichn26.27.28.∞∑n=1n n+1 n( x−2) ( 2) ( 2)2 ; lim x− x−ρ = ÷n ( 1) 2 2n →∞ n+n2n= lim x− 2 = x−2; x − 2 < 1 whenn→∞2( n + 1)1 < x < 3.∞(–1)When x = 1, the series is ∑ nwhich2n=1 n∞1converges absolutely since ∑ converges.2n=1 n∞1When x = 3, the series is ∑ which2n=1 nconverges. The series converges on 1 ≤ x ≤ 3.∞ n n+1n( x+ 5) ( x+ 5) ( x+5)∑ ; ρ = lim÷n=1 nn ( + 1) n→∞( n+ 1)( n+ 2) nn ( + 1)n= lim x+ 5 = x+5 ; x + 5 < 1 whenn→∞n + 2–6 < x < –4.∞1When x = –4, the series isn= 1 nn ( + 1)∞1converges by comparison with ∑ .2n=1 n∞ n(–1)When x = –6, the series isn= 1 nn ( + 1)∞1converges absolutely since ∑n= 1 nn ( + 1)converges.The series converges on –6 ≤ x ≤ –4.∑ which∑ which∞n+1n+1 n ( n+ 1)( x+3)∑ ( − 1) nx ( + 3) ; ρ = limnn=1n→∞nx ( + 3)n + 1= lim x+ 3 = x+ 3 ; x + 3 < 1whenn→∞n–4 < x < –2.When x = –2, the series is∞1(–1) n +nn=1diverges since lim n 0.n→∞ ≠When x = –4, the series is∞∞n+1 n(–1) n(–1) = – nn= 1 n=1∑ which∑ ∑ , which diverges.The series converges on –4 < x < –2.29. If for somenot converge.nnx0 x0x0, lim 0, thenn→∞n! n!≠ ∑ couldInstructor’s Resource Manual Section 9.6 551

30. For any number k, sincek – n < k – n + 1 < … < k – 2 < k – 1 < k,( k –1)( k –2)…( k – n) < k , thusnn+1kk ( –1)( k–2) …( k– n)n klimx < lim xn→∞n! n→∞n!nk nn= k lim x . Since –1 < x < 1, lim x = 0,n→∞n!n→∞kand by Problem 21, lim = 0, hencen→∞n!kk ( –1)( k–2) …( k– n)nlim x = 0.→∞ n!n31. The Absolute Ratio Test givesn2 n + 3 2n+1( n+1)! x n!xρ = lim ÷n→∞135 ⋅ ⋅ (2 n+ 1) 135 ⋅ ⋅ (2 n−1)2 n+1 x= lim x =n→∞2n+ 1 22;2x2x < 2.The radius of convergence is 2 .< 1 whenn32. Using the Absolute Ratio Test,lim ( pn + p)! n+1 ( pn)!x ÷ xn→∞(( 1)!) p ( !) pn+n= lim xn→∞( pn + p)( pn + p − 1) …( pn + p −( p −1))( n + 1) p⎛ 1 ⎞⎛ 2 ⎞ ⎛ p −1⎞= lim x p⎜ p− ⎟⎜ p− ⎟…⎜ p−⎟n→∞⎝ n+ 1⎠⎝ n+ 1⎠ ⎝ n+1⎠=px pThe radius of convergence isnpp − .33. This is a geometric series, so it converges forx − 3 < 1, 2 < x < 4. For these values of x, the1 1series converges to =1 − ( x −3) 4− x.34.∞∑ an ( x–3) n converges on an interval of then=0form (3 – a, 3 + a), where a ≥ 0. If the seriesconverges at x = –1, then 3 – a ≤ –1, or a ≥ 4,since x = –1 could be an endpoint where theseries converges. If a ≥ 4, then 3 + a ≥ 7 so theseries will converge at x = 6. The series may notconverge at x = 7, since x = 7 may be an endpointof the convergence intervals, where the seriesmight or might not converge.35. a.n+1nlim (3x+ 1) (3x+1)n→∞( 1) 2 n+1 2 nn+ ⋅ n⋅ρ = ÷When x = –1, the series isWhen1x = , the series is3n 1 1= lim 3x+ 1 = 3x+1 ; 3 1 1n→∞2n+ 2 2 2 x + < when 1–1 < x < .3∞ n ∞(–2) n 1=nn= 1n⋅ 2 n=1n∞ n ∞∑ ∑ (–1) , which converges.∑2 1= ∑ , which diverges. The series converges onnn= 1n⋅ 2 n=1n1–1 ≤ x < .3b.n+ 1 n+1 n n(–1) (2 x– 3) (–1) (2 x– 3)ρ = limn→∞n+1n4 n+1 ÷4 nn 1= lim 2 x– 3 = 2 x– 3 ;n→∞4 n + 1 41 2 –3 14 x < when 1 7– < x < .2 2∞n ∞1n (–4) 1When x = – , the series is ∑(–1)=2n ∑ which diverges since 1 1.n= 1 4 n n=1 n2 ,n= 1 4 n n=1 n n n n+1∞11lim = 0, so (–1) n1 7∑ converges. The series converges on – < x ≤ .n→∞n n=1 n2 2so an> a n + 1;552 Section 9.6 Instructor’s Resource Manual

36. From Problem 52 of Section 9.1,⎡ nn1 ⎛ ⎤1+ 5⎞ ⎛1– 5⎞1 ⎡ nn⎤f n = ⎢– ⎥ = ( 1+5 ) –( 1– 5)5 ⎢⎜ 2 ⎟ ⎜ 2 ⎟ n ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎥ 2 5 ⎣ ⎦⎣⎦ρn+ 1 1 1 nx ⎡ n+ n+⎤ x ⎡ n n⎤= lim1( 1 5 ) – ( 1– 5) ( 1 5 ) – ( 1– 5nn)n→∞+ ⎢+ ÷ +2 5 ⎣⎥⎦⎢ ⎥2 5 ⎣ ⎦n( 1+5 ) –( 1– 5)n( + ) ( )+ 1 n+1lim x=n→∞2 1 5 – 1– 51+ 5 1+5= x ; x < 1 when2 2nn1– 5( ) ( )1 5n1– 5( )1+5 – 1– 5x += limn→∞21–2 2– < x < .1+ 5 1+5⎛n⎛1– 5 ⎞⎞⎜1– 5Note that lim = 0 since < 1. ⎟⎜n→∞⎜1 5⎟⎝ + ⎠1+5 ⎟⎝⎠2R = ≈0.6181+51+537. If an+ 3 = an,then a0 = a3 = a6 = a3n,a1 = a4 = a7 = a 3n+ 1, and a2 = a5 = a8 = a 3n+ 2. Thus,∞n2 3 4 52 3 6∑ anx = a0 + a1x+ a2x + a0x + a1x + a2x+ = ( a0 + a1x+ a2x )(1 + x + x + )n=0∞∞2 3n2 3= ( a0 + a1x+ a2x ) ∑ x = ( a0 + a1x+ a2x ) ∑ ( x ) n .n=0n=02a0 + a1x+ a2xis a polynomial, which will converge for all x.∞3( ) n3∑ x is a geometric series which, converges for x < 1 , or, equivalently, x < 1 .n=0∞23 n 1a0 + a1x+a2xSince ∑ ( x ) = for x < 1 , S( x)=for x < 1 .3n=0 1 −3x1−x38. If a n = a n+p , then a0 = ap = a2p = anp, a1 = ap+ 1 = a2p+ 1 = anp+1,etc. Thus,∞n p− 1 p p+ 1 2p−1∑ anx = a0 + a1x+ + ap−1x + a0x + a1x + + ap−1x+ n=01 2( 0 1 1p )(1 p p∞−p−1np= a + a x+ + ap−x + x + x + ) = ( a0 + a1x+ + ap−1x ) ∑ xn=01a0 + a1x+ + ap−1x p−is a polynomial, which will converge for all x.∞ ∞np p np∑ x = ∑ ( x ) is a geometric series which converges for 1n= 0 n=0∞p n 1p−1⎛ 1 ⎞Since ∑ ( x ) = for x < 1 , S( x) = ( ap0 + a1x+ + ap−1x) ⎜n=0 1 − x1p ⎟⎝ − x ⎠x < , or, equivalently, x < 1 . for x < 1 .Instructor’s Resource Manual Section 9.6 553

9.7 Concepts Review1. integrated; interior2.3.4.2 3 4 5x x x x– x – – – –2 3 4 54 62 x x1+ x + +2 62 3 43x x 3x1+ x + + +2 3 4Problem Set 9.711. From the geometric series for with x1−xreplaced by –x, we get12 3 4 51 x x x x x1+ x = − + − + − + ,radius of convergence 1.2.3.d ⎛ 1 ⎞ 1⎜ ⎟ =−dx ⎝1 + x ⎠ 2(1 + x)12 3 4 5= 1− x+ x − x + x − x + so1+ x12 3 4= 1− 2x+ 3x − 4x + 5x− ;2(1 + x)radius of convergence 1.d ⎛ 1 ⎞ 1 d ⎛ 1 ⎞ 2⎜ ⎟ = ; ,1 2 dx x 2 =⎝ − ⎠ 3(1 −x) dx ⎜(1 x) ⎟⎝ − ⎠ (1 −x)1so is 1 of the second derivative of(1 − x)3 211− x. Thus, 12 3= 1+ 3x+ 6x + 10x+ ;3(1 − x)radius of convergence 1.4. Using the result of Problem 2,x2 3 4 5= x− 2x + 3x − 4x + 5x− ;2(1 + x)radius of convergence 1.5. From the geometric series for6.replaced by 3 x , we get211−x2 31 1 3x 9x 27x2−3x = 2 + 4 + 8 + 16+ ;radius of convergence 2 3 .1 1⎛1 ⎞= . Since3 + 2x3 ⎜ 1 + 2 x ⎟⎝ 3 ⎠12 3 4 51 x x x x x ,1+ x = − + − + − +with x2 3 41⎛1 ⎞ 1 2x 4x 8x 16x= − + − + − ;3⎜1+2 x ⎟ 3 9 27 81 243⎝ 3 ⎠radius of convergence 3 2 .17. From the geometric series for with x1−x4replaced by x , we get2x 2 6 10 14x x x x41− x = + + + +;radius of convergence 1.3 3 ⎛ ⎞x x 1 x 3 x 6 x 9 x128. = ⎜ ⎟= + + + + 332 − x 2 ⎜1−x ⎟ 2 4 8 16⎝ 2 ⎠for3x23 3< 1 or – 2 < x < 2 .9. From the geometric series for ln(1+ x) with xreplaced by t, we get10.x2 3 4 5ln(1 ) x – x x – x+ tdt= + +…0 2 6 12 20∫ ;radius of convergence 1.x2 4 6 8tan –1 x – x x xtdt= + − +… ;0 2 12 30 56∫radius of convergence 1.554 Section 9.7 Instructor’s Resource Manual

11.2 3 4x x xln(1 + x) = x– + – +… , –1 < x≤12 3 42 3 4x x xln(1 – x) = – x– – – +… , –1 ≤ x 0. Thus, the naturalM + 17logarithm of any positive number can be found by using the series from Problem 11. For M = 8, x = , so93 5 7 9 11⎛7⎞ 2⎛7⎞ 2⎛7⎞ 2⎛7⎞ 2⎛7⎞ 2 ⎛7⎞ln 8 = 2⎜ ⎟+ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ +⋅⋅⋅⎝ 9 ⎠ 3 ⎝ 9 ⎠ 5 ⎝ 9 ⎠ 7 ⎝ 9 ⎠ 9 ⎝ 9 ⎠ 11 ⎝ 9 ⎠≈ 1.55556 + 0.31367 + 0.11385 + 0.04919 + 0.02315 + 0.01146 + 0.00586 + 0.00307 + 0.00164 + 0.00089+ 0.00049 + 0.00027 + 0.00015 + 0.00008 ≈ 2.07913. Substitute –x for x in the series for2 3 4 5−ex 1x x x x= − x + − + − + .2! 3! 4! 5!xe to get:14.x 2 ⎛ 4 621x x ⎞xe = x + x + + + ⎜ 2! 3! ⎟⎝⎠5 7 93 x x x= x+ x + + + +2! 3! 4!15. Add the result of Problem 13 to the series for2 4 62 2 2ex + e −x= 2 + x + x + x + .2! 4! 6!xe to get:16.2 32 x ⎛ 4 81 2 1 2 1 2x x ⎞e − − x =− − x+ + x+ + + ⎜ 2! 3! ⎟⎝⎠2 3 4 54x 8x 16x 32x= + + + +2! 3! 4! 5!17.18.2 31 ⎛x x x ⎞2 3 4 5−2 x x 3x 11xe ⋅ = 1 − x+ − + (1 + x+ x + )= 1+ + + + +1−x ⎜ 2! 3! ⎟⎝⎠2 3 8 30x ⎛ 2 3 3 5 71tan 1x x ⎞⎛ x x x ⎞3 4 5−e x = + x+ + + ⎟⎜x− + − +⎜ 2! 3! 2 x x 3x = x+ x + − + +⎟⎜ 3 5 7 ⎟⎝ ⎠⎝ ⎠6 6 4019.20.−1tan x⎛ 2 3 3 5 7−x−1= e tan x 1x x ⎞⎛ x x x ⎞= − x+ − + ⎟⎜x− + − +xe⎜ 2! 3! ⎟⎜ 3 5 7 ⎟⎝ ⎠⎝ ⎠2x3x2! 3!2x3x2 3xe 1+ x + + + 3 4 52 7x 47x 46x=1 + ln(1 + = 1+ x − + − +x) 1 + x − + − 6 24 153 4 52 x x 3x= x− x + + + +6 6 40Instructor’s Resource Manual Section 9.7 555

21.22.⎛ 3 5 71 2 4 x x x ⎞3 5 7−2 4 2x 13x 29x(tan x)(1 + x + x ) = x− + − + (1 + x + x ) = x + + − +⎜ 3 5 7 ⎟⎝⎠3 15 105tan3 5 7−1x x xx x − + − + 3 5 73 5 74x 8x 23x=x2 4 2 41+ x + x 1+ x + x= − + + −3 15 3523. The series representation ofx 2 3 4 5e x x 3x 11xis 1 + − + − + , so1+x 2 3 8 30xte 1 3 1 4 3 5∫ dt = x + x – x + x –…..01+t 6 12 4024. The series representation of−1tanxxis2 4 6x x x1 − + − +⋅⋅⋅ , so3 5 7∫0−1 3 5 7x tan t dt = x − x + x − x +⋅⋅⋅t9 25 4925. a.b.11+ x2 3 4 5= 1 − x+ x − x + x − x + , so2 3 4 5= 1 + x+ + + + + , soex x x x x2! 3! 4! 5!x12 3 4 5x x x x x+ x = − + − + − .x2 3e − (1 + x) 1 x x x= + + + + .2x 2! 3! 4! 5!c.2 3 4 5x x x x−ln(1 − x) = x+ + + + + , so2 3 4 52 3 44x 8x 16x−ln(1 − 2 x) = 2x+ + + +.2 3 426. a. Since12 3 41 x x x x1− x = + + + + + ,112 4 6 81 x x x x2− x = + + + + + .b. Again using12 3 41 x x x x1− x = + + + + + ,12 31 cosx cos x cos x1−cosx − = + + + .c.2 3 44 6 8x x x2 2 x x xln(1 − x)= −x− − − − , so ln(1 − x ) =−x− − − −, and2 3 42 3 42 4 6 81 2 1 x x x x− ln(1 − x ) = ln = + + + + .2 21−x 2 4 6 827. Differentiating the series for11−xx2 3 4= x+ 2x + 3x + 4 x + ,2(1 − x)28. Differentiating the series forgives2x(1 – x)31x –1yields hencetwice yields12 3= 1+ 2x+ 3x + 4x+2(1 − x)∞n xnx =2n=1 (1 − x)∑ for –1 < x < 1.23(1 – x)2 32 6x 12x 20x multiplying this series by x gives= + + + +⋅⋅⋅. Multiplying this series by x∞n 2xnn ( + 1) x =3n=1(1 – x)2 3 4= 2x+ 3⋅ 2x + 4⋅ 3x + 5⋅ 4 x +⋅⋅⋅ , hence ∑ for –1 < x < 1.29. a.3 51 ( 1) ( 1)tan − xxx x e e( e 1) ( e 1)− −3⎛− = − − + −x 2 x 3 ⎞ 21 ⎛ x ⎞= x+ + + − x+ + +3 5 ⎜ 2! 3! ⎟ 3⎜ 2! ⎟⎝ ⎠ ⎝ ⎠2 3x x= x + − −2 6556 Section 9.7 Instructor’s Resource Manual

.xx 2 x 3e –1 x ( e – 1) ( e – 1)e = 1 + ( e –1) + + +⋅⋅⋅2! 3!2 3 2223⎛ x x ⎞ 1 ⎛ x ⎞ 1 ⎛ x ⎞= 1+ x+ + + + x+ + + x+ +⎜ 2! 3! ⎟ 2! ⎜ 2! ⎟ 3! ⎜ 2! ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 2 3 3 4x x ⎞ 1 ⎛2 x ⎞ 1 ⎛3 x ⎞= 1+ x+ + + + x + 2 + + x + 3 +⎜ 2! 3! ⎟ 2! ⎜ 2! ⎟ 3! ⎜ 2! ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠32 5x= 1+ x+ x + +630.31.32.33.34.2 2f( x) = a0 + a1x+ a2x + = b0 + b1x+ b2x+ ;f (0) = a0 = b0,so a0 = b0.2 3f′ ( x) = a1+ 2a2x+ 3a3x + = b1+ 2b2x+ 3 b3x+ ;f ′(0) = a1 = b1,so a1 = b1.The nth derivative of f(x) is( n) ( n + 2)! 2( n + 2)! 2f ( x) = n! an + ( n+ 1)! an+ 1x+ an+2x+ = nb ! n + ( n+ 1)! bn+ 1x+ bn+2x+ ;22( n) f (0) = na ! n = nb ! n , so an= bn.xx 2 1= = −2x − 3x+ 2 ( x −2)( x−1) x−2 x−1=− 1 11 x + 1 − x22 3 ∞ n nx 3x 7 x (2 −1)x= + + + =∑2 4 8 nn=1 23 5x xy′′ =− x+ − + = − y, so y′′+ y = 03! 5!y y⎛ 2 3x x x ⎞2 3=− 1+ + + + +−( 1+ x+ x + x + ⎜)2 4 8 ⎟⎝ . It is clear that y(0) = 0 and y′ (0) = 1. Both the sine and cosinefunctions satisfy ′′ + = 0 , however, only the sine function satisfies the given initial conditions. Thus,y = sin x.2 2 3 2 3 2 3 4F( x)– xF( x)– x F( x) = ( f0 + f1x+ f2x + f3x + )–( f0x+ f1x + f2x + )–( f0x + f1x + f2x+ )2 3= f0 + ( f1 – f0) x+ ( f2 – f1 – f0) x + ( f3 – f2 – f1)x +∞0 ( 1 – 0 ) ( n – n–1 – n–2) n∞2= f + f f x+∑f f f x 0 ( 2 – 1 – ) n += + x+∑fn+ fn+fnxn=2n=02Since fn+ 2 = fn+ 1+ fn, fn+ 2 – fn+1– fn= 0 . Thus F( x)– xF( x)– x F( x) = x.xF( x)=21– x – xf0 f1 f2 2 f33 f44f1 f2 f32 f43y( x) = + x+ x + x + x + ; y′ ( x) = + x+ x + x + ;0! 1! 2! 3! 4!0! 1! 2! 3!f2 f3f42y"( x)= + x+ x +0! 1! 2!(Recall that 0! = 1.)⎛ f2 f3 f4 2 ⎞ ⎛ f1 f2 f3 2 ⎞ ⎛ f0f1 f22 ⎞y′′ ( x)– y′ ( x)– y( x) = ⎜ + x+ x + – x x – x x0! 1! 2!⎟ ⎜ + + + + + +0! 1! 2!⎟ ⎜ 0! 1! 2!⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠1 1 12= ( f2 – f1 – f0) + ( f3 – f2 – f1) x+ ( f4 – f3 – f2)x +0! 1! 2!∞1 ( 2 – 1 – )n= fn+ fn+fnx = 0n=0 n!∑ since fn+ 2 = fn+1+ fnfor all n ≥ 0.⎠Instructor’s Resource Manual Section 9.7 557

35.⎛1 1 1 1 1 ⎞ ⎛ 1 ⎞π≈16 ⎜ − + − + ⎟ −4⎜ ⎟⎝5 375 15,625 546,875 17,578,125 ⎠ ⎝239⎠≈ 3.14159n!36. For any positive integer k ≤ n, bothk and n!k!are positive integers. Thus, since q < n, n!pne ! = is a positiveqn! n! n!integer and M = n! e– n!– n!– – – – is also an integer. M is positive since2! 3! n!1 1 1 1e –1–1– – –= + + 2! n! ( n+ 1)! ( n+2)!1M < contradicts that M is a positive integer since for n ≥ 1, 1 ≤ 1 and there are no positive integersnnless than 1.9.8 Concepts Review1.( k) f (0)k!2. lim Rn( x) = 0n→∞3. –∞ ; ∞4.1 1 2 5 31 + x – x + x3 9 81Problem Set 9.81.2.3.3 5 7– x x – xsin x x + +… 3 53! 5! 7! x 2xtan x = == x + + +…2 4 6cos x 1– x + x – x +… 3 152! 4! 6!3 5x xsinh x x + + +… 3 53! 5! x 2xtanh x = == x – + +…2 4cosh x 1+ x + x +… 3 152! 4!2 3 4 3 5x x x x x xe sin x ⎛ 1 x ⎞⎛ x– –⎞3 52 x x= + + + + +… ⎟⎜ + …= x+ x + – –…⎜ 2! 3! 4! ⎟⎜ 3! 5! ⎟⎝ ⎠⎝ ⎠3 304.2 3 2 4– x x x x xe cos x = ⎛ 1 − x+ − +… ⎞⎛ ⎟⎜ ⎞1– + – …⎜ 2! 3! ⎟⎜ 2! 4! ⎟⎝ ⎠⎝ ⎠3 4 5x x x= 1– x + – + +…3 6 305.⎛ 2 4 2 3 4x x ⎞⎛ x x x ⎞ 2 3 5x x 3xcos xln(1 + x) = 1– + –… ⎟⎜x– + – +…= x – – + –…⎜ 2! 4! ⎟⎜ 2 3 4 ⎟⎝ ⎠⎝ ⎠2 6 40558 Section 9.7 Instructor’s Resource Manual

6.7.⎛ 3 5 2 3x x ⎞⎛ x x x ⎞(sin x) 1 + x = x– + –… ⎟⎜1 + – + –…⎜ 3! 5! ⎟⎜ 2 8 16 ⎟⎝ ⎠⎝ ⎠2 3 4 5x 7x x 19x= x + – – – +… , –1 < x < 12 24 48 19202 3 5x x x xe x sin x x ⎛ 1 x ⎞ ⎛ x– –⎞2 4 5x x 2x+ + = + + + +… + + …= 1+ 3x+ + + +…⎜ 2! ⎟ ⎜ 3! 5! ⎟⎝ ⎠ ⎝ ⎠2! 4! 5!8.2 4 6 8 10x x x x xcos x –1 + = – + – +…, so2 4! 6! 8! 10!2cos x –1 + x2 42 1 x x= – + – …4x 4! 6! 8!9.10.2 41 ⎛cosh (12 3 x x ⎞x = + x+ x + x +… ) 1 + + +…1– x⎜ 2! 4! ⎟⎝⎠2 3 4 5– ln(1 + x) – ln(1 + x)⎛ 2 3 4x x x ⎞2 3 4= = – x+ – + –… (1– x+ x – x + x –…)1+x 1–(– x)⎜ 2 3 4 ⎟⎝⎠2 3 4 53x 11x 25x 137x= – x + – + – +… , –1 < x < 12 6 12 603x 3x 37x 37x= 1 + x + + + + +… , –1 < x < 12 2 24 24∞1 1 1−x 1n= ⋅ = − = − = − + − +2 2 131+ x+ x 1+ x+ x − x 1−xn=0∑ , x < 13 3 411. ( 1 x) ( 1 x) x 1 x x x12.12 31 sin x (sin x) (sin x)1–sinx = + + + +…3 5 32333435⎛ x x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞= 1 + x– + –… + x– +… + x– +… + x– +… + x–+… + ⎜ 3! 5! ⎟ ⎜ 3! ⎟ ⎜ 3! ⎟ ⎜ 3! ⎟ ⎜ 3! ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 3 5 4 5x x ⎞ ⎛2 x ⎞ ⎛3 x ⎞4 5= 1 + x– + –… + x –2 +… + x –3 +… + ( x –…) + ( x –…)⎜ 3! 5! ⎟ ⎜ 3! ⎟ ⎜ 3! ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠3 4 52 5x 2x 61xπ= 1 + x+ x + + + +… , x < .6 3 120213.3 52⎛ 53 x x ⎞ ⎛ x x ⎞ ⎛ 4 3 52sin x = x– + –… x– + –…x ⎞⎛ x x ⎞= x –2 +… ⎟⎜x– + – …⎜ 3! 5! ⎟ ⎜ 3! 5! ⎟ ⎜⎝ ⎠ ⎝ ⎠3! ⎟⎜ 3! 5! ⎟⎝ ⎠⎝ ⎠53 x= x – +… 214.15.16.17.⎡⎛ 3 5 3 58x 32x ⎞ ⎛ 27x 243x⎞⎤x(sin 2x+ sin 3 x) = x⎢2 x– + –… + 3 x– + –…⎥⎢⎜ 3! 5! ⎟ ⎜ 3! 5! ⎟⎣⎝ ⎠ ⎝ ⎠⎥⎦3 52xx ⎛ x x ⎞xsec( x ) + sin x = + sin x = + x – + –…24 8cos( x ) 1–x x ⎜– 3! 5! ⎟+ … ⎝ ⎠⎛ 5 ⎞ ⎛ 3 5 ⎞2! 4!3 5x – x xx 61x= x+ +… + x + –…= 2 x – + +…⎜ 2 ⎟ ⎜ 3! 5! ⎟⎝ ⎠ ⎝ ⎠ 3! 120cos x(cos x)(1 x)1+ x = +–1/22 3 4 5⎛ 2 4 ⎞⎛ 2 3 ⎞x x x 3x 5x= 1– + –… ⎟⎜1– + – +…⎜ 2! 4! ⎟⎜ 2 8 16 ⎟⎝ ⎠⎝ ⎠x x x 49x 85x= 1– – – + – +… , –1 < x < 12 8 16 384 7682 3 4 53/2 3x 3x x 3x 3x(1 + x) = 1 + + – + – +… , –1 < x < 12 8 16 128 256⎛ 3 435x⎞2 35x= x 5 x– +… = 5 x – +…⎜ 3! ⎟⎝⎠3!Instructor’s Resource Manual Section 9.8 559

18.2 2/3 2 2/3(1 – x ) = [1 + (– x )]2 42 2 1 2 2 4 2 3 2xx= 1 + (– x )– (– x ) + (– x ) +… = 1– – –…,3 9 813 92x–1 < – < 1 or –1 < x < 119.20.21.( n ) xf ( x)= e for all n. f (1) = f′ (1) = f′′ (1) = f′′′(1) = e2 3e x ≈ e+ e( x– 1) + e ( x– 1) +e ( x– 1)2 6⎛ 1 3 1 3f π ⎞ ⎛; f π ⎞ ⎛; f π ⎞ ⎛; fπ ⎞⎜ ⎟= ′ ⎜ ⎟= ′′ ⎜ ⎟=− ′′′ ⎜ ⎟=−;⎝6⎠ 2 ⎝6⎠ 2 ⎝6⎠ 2 ⎝6⎠2⎛ 1 3 1 3f π ⎞ ⎛; f π ⎞ ⎛– ; f π ⎞ ⎛– ; fπ ⎞⎜ ⎟= ′ ⎜ ⎟= ′′ ⎜ ⎟= ′′′ ⎜ ⎟=;⎝3⎠ 2 ⎝3⎠ 2 ⎝3⎠ 2 ⎝3⎠22 31 3⎛ 1 3sin x x π ⎞ ⎛x π ⎞ ⎛xπ ⎞≈ + ⎜ − ⎟− ⎜ − ⎟ − ⎜ − ⎟2 2 ⎝ 6⎠ 4⎝ 6⎠ 12 ⎝ 6⎠2 31 3⎛ 1 3cos x – x– π ⎞ ⎛– x– π ⎞ ⎛x–π ⎞≈ ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟2 2 ⎝ 3⎠ 4⎝ 3⎠ 12 ⎝ 3⎠⎛22. f π ⎞ ⎛1; f π ⎞ ⎛2; f π ⎞ ⎛4; fπ ⎞⎜ ⎟= ′ ⎜ ⎟= ′′ ⎜ ⎟= ′′′ ⎜ ⎟=16⎝4⎠ ⎝4⎠ ⎝4⎠ ⎝4⎠2 3⎛ 8tan x 1 2 x π ⎞ ⎛2 x π ⎞ ⎛xπ ⎞≈ + ⎜ − ⎟+ ⎜ − ⎟ + ⎜ − ⎟⎝ 4⎠ ⎝ 4⎠ 3⎝ 4⎠23. f(1) = 3; f ′(1) = 2 + 3 = 5;f′′ (1) = 2 + 6 = 8; f′′′(1) = 62 3 2 31+ x + x = 3+ 5( x–1) + 4( x–1) + ( x–1)This is exact since24. f ( − 1) = 2+ 1+ 3+ 1=7;( n) f ( x ) = 0 for n ≥ 4.f ′( − 1) = –1− 6–3= − 10;f′′ ( − 1) = 6+ 6= 12; f′′′(1) = –62 3 2 32 – x+ 3 x – x = 7 − 10( x+ 1) + 6( x+ 1) − ( x+1)This is exact since( n) f ( x ) = 0 for n ≥ 4.25. The derivative of an even function is an oddfunction and the derivative of an odd function isan even function. (Problem 50 of Section 3.2).nSince f ( x)= ∑ anxis an even function, f′( x)is an odd function, so f′′ ( x)is an even function,hence f′′′ ( x)is an odd function, etc.( n) Thus f ( x ) is an even function when n is evenand an odd function when n is odd.nBy the Uniqueness Theorem, if f( x) = ∑ anx,( n) f (0)then an= . If g(x) is an odd function,n!g(0) = 0, thence a = 0 for all odd n sincen( n) f ( x ) is an odd function for odd n.n26. Let f ( x)= ∑ anxbe an odd function(f(–x) = –f(x)) for x in (–R, R). Then a n = 0 if nis even.The derivative of an even function is an oddfunction and the derivative of an odd function isan even function (Problem 50 of Section 3.2).nSince f( x)= ∑ anxis an odd function, f′( x)is an even function, so f′′ ( x)is an odd function,hence f′′′ ( x)is an even function, etc. Thus,27.( n) f ( x ) is an even function when n is odd and anodd function when n is even.nBy the Uniqueness Theorem, if f( x) = ∑ anx,( n) f (0)then an= . If g(x) is an odd function,n!g(0) = 0, hence a n = 0 for all even n since( n) f ( x ) is an odd function for all even n.12 –1/2= [1 + (– t )]21– t1 2 3 2 2 5 2 3= 1– (– t ) + (– t ) – (– t ) +2 8 162 4 6t 3t 5t= 1+ + + +2 8 16–1 x 1Thus, sin x = ∫ dt0 21– tx ⎛ 2 4 63 51t t t ⎞= ∫ + + + +dt0 ⎜ 2 8 16 ⎟⎝⎠3 5 7x⎡ t 3t 5t⎤= ⎢t+ + + + ⎥⎢⎣6 40 112 ⎥⎦03 5 7x 3x 5x= x + + + +6 40 112560 Section 9.8 Instructor’s Resource Manual

28.12 –1/2= (1 + t )21+t1 2 3 2 2 5 2 3= 1– t + ( t ) – ( t ) +2 8 162 4 6t 3t 5t= 1– + – +2 8 16–1 x 1Thus, sinh ( x)= ∫ dt0 21+tx ⎛ 2 4 63 51– t t –t ⎞= ∫ + +dt0 ⎜ 2 8 16 ⎟⎝⎠3 5 7x⎡ t 3t 5t⎤= ⎢t– + – + ⎥⎢⎣6 40 112 ⎥⎦03 5 7x 3x 5x= x – + – +6 40 11229.4 8 12 162 x x x xcos( x ) = 1– + – + – 2! 4! 6! 8!∫1 1⎛4 8 12 162x x x x ⎞cos( x ) dx = ∫ 1– + – + –dx0 0⎜2! 4! 6! 8! ⎟⎝⎠5 9 13 171⎡ x x x x ⎤= ⎢x– + – + – ⎥⎢⎣10 216 9360 685, 440 ⎥⎦01 1 1 1= 1– + – + – ≈ 0.9045210 216 9360 685,44030.31.3/2 5/2 7/2 9/2x x x xsin x = x – + – + –…3! 5! 7! 9!0.5 0.5⎛3/2 5/2 7/2 9/2x x x x ⎞∫ sin xdx = – – –0 ∫x + + …dx0 ⎜ 3! 5! 7! 9! ⎟⎝⎠5/2 7/2 9/2 11/20.5⎡2 3/2 2 x 2 x 2 x 2 x ⎤= ⎢ x – + – + –…⎥⎢⎣3 5 3! 7 5! 9 7! 11 9! ⎥⎦02 3/2 1 5/2 1 7/2 1 9/2 1 11/2(0.5) – (0.5) (0.5) – (0.5) (0.5) –3 15 420 22,680 1,995,840= + + … ≈ 0.224131 12 3 2 3= = 1 + (1– x) + (1– x) + (1– x) + = 1 −( x− 1) + ( x−1) −( x− 1) +x 1–(1– x)for –1 < 1 – x < 1, or 0 < x < 2.32.1/2 1 1 2 1 3 5 4 7 5(1 + x) = 1 + x– x + x – x + x – …2 8 16 128 2561/2 1 1 2 1 2 5 4 7 5(1 – x) = 1 – x– x – x – x – x – …2 8 16 128 256∞ ( n)1 2 5 4 f (0) nso f( x) = 2– x – x –…=∑x4 64 n=0 n!( n) Note that f (0) = 0 when n is odd.(4) (0) 5(51)ff (0)Thus, = – and = 0, so4! 64 51!(4) 5 15f (0) = – 4! = – and64 8f(51) (0) = 0.Instructor’s Resource Manual Section 9.8 561

33. a.b.c.d.2 2 2 3 2 42 ( x+ x ) ( x+ x ) ( x+x )f( x) = 1 + ( x+ x ) + + + +…2! 3! 4!2 1 2 3 4 1 3 4 5 6 1 4 5 6 7 8= 1 + ( x+ x ) + ( x + 2 x + x ) + ( x + 3x + 3 x + x ) + ( x + 4x + 6x + 4 x + x ) +…2 6 242 3 4 ∞ ( n)3x 7x 25 x f (0) n= 1+ x + + + +…=∑ x2 6 24 n!Thus(4)f (0) 25= so4! 24n=0(4) 25f (0) = 4! = 25.242 3 4sin x sin x sin xf( x) = 1+ sinx+ + + +…2! 3! 4!3 323334⎛ x ⎞ 1⎛ x ⎞ 1⎛ x ⎞ 1 ⎛ x ⎞= 1 + x– +… + x– +… + x– +… + x–+…⎜ 3! ⎟ 2 ⎜ 3! ⎟ 6 ⎜ 3! ⎟ 24 ⎜ 3! ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 3 4x ⎞ 1⎛ 2 x ⎞2 4 ∞ ( n)1 3 1 4x x f (0) n= 1 + x– +… + x –2 +… + ( x –…) + ( x –…)= 1 + x + – –⎜ 3! ⎟ 2 ⎜ 3! ⎟…=∑ x⎝ ⎠ ⎝ ⎠6 242 8 n=0 n!(4)f (0) 1 (4) 1Thus, = – so f (0) = – 4! = –3.4! 88t 2 ⎛ 4 6 82–1 –1 1t t t ⎞ 4 6 82 t t te = + + t + + + +…= t + + + +…⎜ 2! 3! 4! ⎟⎝⎠2! 3! 4!sot22 4 6e t t t= 1 + + + +…t2–12 6 24x⎛2 4 6t t t ⎞3 5x⎡ t t ⎤f ( x) = ∫1+ + + +…dt =0 ⎜ 2 6 24 ⎟⎢t+ + +… ⎥⎝⎠ ⎢⎣6 30 ⎥⎦0(4)f (0)= 0 so f(4) (0) = 0.4!Thus,2 3cos x –1 (cos –1) (cos 1)1 (cos –1)x x −e = + x + + +…2! 3!2 4 2 422 43⎛ x x ⎞ 1⎛ x x ⎞ 1⎛ x x ⎞= 1 + – + –… + – + –… + – + –… +…⎜ 2! 4! ⎟ 2 ⎜ 2! 4! ⎟ 6 ⎜ 2! 4! ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 2 4 4 6x x ⎞ 1⎛ x ⎞ 1⎛ x ⎞ 2 4x x= 1 + – + –… + –… + – +…= 1– + –…⎜ 2 24 ⎟ 2⎜ 4 ⎟ 6⎜ 8 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠2 6∞ ( n)e 2 e 4 f (0) nHence f( x) = e– x + x –…=∑x2 6 n=0 n!(4)f (0) e (4) eThus, = so f (0) = 4! = 4 e .4! 663 5 ∞ ( n)x x f (0) n= x + + +…=∑ x6 30 n=0 n!562 Section 9.8 Instructor's Resource Manual

2 2e. Observe that ln(cos x) = ln(1 − sin x).3 52⎛4 62 x x ⎞2 x 2xsin x = x− + − = x − + −⎜ 3! 5! ⎟⎝⎠3 454 62 2 sin x sin xln(1 − sin x) = −sinx− − −2 34 4243⎛2 x ⎞ 1⎛ 2 x ⎞ 1⎛ 2 x ⎞= − x − + − x − + − x − +⎜ 3 ⎟ 2⎜ 3 ⎟ 3⎜ 3 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 4 6 62 x 2x ⎞ 1⎛ 4 2x⎞4 61 (6=− x − + − − x − + − x − )⎜ 3 45 ⎟ 2⎜ 3 ⎟⎝ ⎠ ⎝ ⎠3 2 x 2x= −x− − −6 45∞ ( n)2 1 4 2 6 f (0) nHence f( x) =−x − x − x − =∑ x .6 45 n=0 n!(4)f (0) 1 (4) 1Thus, =− so f (0) = − 4! = − 4 .4! 6634.12sec x = = a0 + a1x+ a2x+… socos x⎛ 2 42 3 4 x x ⎞1 = ( a0 + a1x+ a2x + a3x + a4x+… ) 1– + –…⎜ 2 24 ⎟⎝⎠⎛ a0 ⎞ 2 ⎛ a1⎞3 ⎛ a2a0⎞ 4= a0 + a1x+ ⎜a2 – x a3 – x a4– x2⎟ + ⎜ ⎟ + + +…2⎜2 24⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠a0 a0Thus1 a2aa0 = 1, a1 = 0, a2 – = 0, a3 – = 0, a4– + = 0, so2 2 2 241 5a0 = 1, a1 = 0, a2 = , a3 = 0, a4=2 241 2 5 4and therefore sec x = 1+ x + x +…2 24.35.sinh x2tanh x = = a0 + a1x+ a2x+…cosh x20 1 2x 3 x 5 ⎛ 2 421x x ⎞x+ + +…= + + +… a0 + a1x+ a2x+…6 120 ⎜ 2 24 ⎟⎝⎠so sinh x = cosh x( a + a x+ a x +… )or ( )⎛ a0 ⎞ 2 ⎛ a1⎞3 ⎛ a2 a0 ⎞ 4 ⎛ a3a1⎞ 5= a0 + a1x+ ⎜a2 + x a3 x a4 x a5x2⎟ + ⎜ +2⎟ + ⎜ + + + + + +…2 24⎟ ⎜2 24⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠a0 a11Thus a0 = 0, a1 = 1, a2 + = 0, a3+ = ,2 2 6a2 a0 a3a11a4 + + = 0, a5+ + = , so2 24 2 24 1201 2a0 = 0, a1 = 1, a2 = 0, a3 = – , a4 = 0, a5= and therefore3 151 3 2 5tanh x = x– x + x –…3 15Instructor’s Resource Manual Section 9.8 563

36.12sech x = = a0 + a1x+ a2x+…cosh x20 1 2⎛ 2 4x x ⎞21= 1+ + +… a0 + a1x+ a2x+…⎜ 2 24 ⎟⎝⎠so 1= cosh xa ( + ax+ ax +… )or ( )⎛ a0 ⎞ 2 ⎛ a1⎞3 ⎛ a2 a0 ⎞ 4 ⎛ a3a1⎞ 5= a0 + a1x+ ⎜a2 + x a3 x a4 x a5x2⎟ + ⎜ +2⎟ + ⎜ + + + + + +2 24⎟ ⎜2 24⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ a0 ⎞ ⎛ a0 3 0Thus,1⎞⎛ a2a ⎞ ⎛ a a ⎞a0 = 1, a1 = 0, ⎜a2 + 0, a3 0, a4 0, a502⎟= ⎜ + ⎟= + + = + + =2⎜2 24⎟ ⎜2 24⎟ , so⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠1 5a0 = 1, a1 = 0, a2 = − , a3 = 0, a4 = , a5= 0and therefore2 241 2 5 4sech x = 1− x + x −2 2437. a. First define R 3 ( x)byf ''( a) f '''( a)R ( x) = f( x) − f( a) − f '( a)( x−a) − ( x−a) − ( x−a)2! 3!axwe define32 3For any t in the interval [ , ]4f ''( t) 2 f '''( t) 3 ( x−t)g() t = f( x) − f() t − f '()( t x−t) − ( x−t) − ( x−t) −R3 ( x)2! 3! 4( x − a)Next we differentiate with respect to t using the Product and Power Rules:12g '( t) = 0 − f '( t) −[ − f '( t) + f ''( t)( x−t) ] − ⎡−2 f ''( t)( x− t) + f '''( t)( x−t)⎤2! ⎣⎦31 2 (4) 3 4( x−t)− ⎡−3 f '''( t)( x− t) + f ( t)( x− t) ⎤+R3 ( x)3! ⎣⎦4( x−a)(4) 3 3f ( t)( x−t) ( x−t)=− + 4 R3 ( x)3! 4( x−a)Since gx= ( ) 0, g( a) = R3( x) − R3( x) = 0, and g()t is continuous on [ ax, , ] we can apply the Mean ValueTheorem for Derivatives. There exists, therefore, a number c between a and x such that g'( c ) = 0 . Thus,(4) 3 3f ( c)( x−c) ( x−c)0 = g'( c) =− + 4 R3 ( x)3! 4( x − a)which leads to:(4)f () c 4R3( x) = ( x−a)4!b. Like the previous part, first define Rn( x)by( n)f ''( a) 2 f ( a)nRn( x) = f( x) − f( a) − f '( a)( x−a) − ( x−a) −− ( x−a)2! n!For any t in the interval [ axwe , ] define( n) n+1f ''() t 2 f () t n ( x−t)g() t = f( x) − f() t − f '()( t x−t) − ( x−t) −− ( x−t) −Rn( x)2! n! n+1( x−a)Next we differentiate with respect to t using the Product and Power Rules:564 Section 9.8 Instructor's Resource Manual

38. a. For12g'( t) = 0 − f '( t) −[ − f '( t) + f ''( t)( x−t) ] − ⎡−2 f ''( t)( x− t) + f '''( t)( x−t)⎤−2! ⎣⎦n1 ( n) n− 1 ( n+1) n ( n+ 1)( x−t)− ⎡−nf ()( t x− t) + f ()( t x− t) ⎤+Rn( x)n! ⎣⎦n+1( x−a)( n+1) n nf ( t)( x−t) ( x−t)=− + ( n+1) Rn( x)n! n+1( x−a)Since gx= ( ) 0, g( a) = Rn( x) − Rn( x) = 0, and g()t is continuous on [ ax, , ] we can apply the Mean ValueTheorem for Derivatives. There exists, therefore, a number c between a and x such that g'( c ) = 0 . Thus,( n+1) n nf ( c)( x−c) ( x−c)0 = g'( c) =− + ( n+1) Rn( x)n! n+1( x−a)which leads to:( n+1)f () c n+1Rn( x) = ( x−a)( n + 1)!∞⎛ p ⎞ n ⎛ p ⎞ 1, limn px x+ ⎛ ⎞ρxn p( p– 1) …( p– n+ 1)( p– n) p( p– 1) …( p– n+1)∑ ⎜ ⎟ = ⎜ ⎟ ÷ ⎜ ⎟ = lim x÷1n 1 nn→∞n+n= ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ n→∞( n+1)! n!p–n= lim x = xn→∞n + 1∞⎛ p ⎞ nThus f ( x) = 1+ ∑ ⎜ ⎟xnconverges for x

39.⎧⎪ 0 if t < 0f′ () t = ⎨ 3⎪⎩ 4tif t ≥ 0,41.3 5 7x x xsin x = x– + – +6 120 5040⎧⎪ 0 if t < 0f′′ () t = ⎨ 2⎪⎩ 12tif t ≥ 0⎧0 if t < 0f′′′ () t = ⎨,⎩24tif t ≥ 0(4) ⎧0 if t < 0f () t = ⎨⎩24 if t ≥ 0(4)lim f ( t) 24+t→0,= while(4)lim f ( t) 0,–t→0= thusf(4) (0) does not exist, and f(t) cannot berepresented by a Maclaurin series.Suppose that g(t) as described in the text isrepresented by a Maclaurin series, so∞ ( n)2 g (0) ng()t = a0 + a1t+ a2t +…=∑ t for alln=0 n!t in (–R, R) for some R > 0. It is clear that, fort ≤ 0 , g(t) is represented by2gt () 0 0t 0 t .= + + +… However, this will notrepresent g(t) for any t > 0 since the car ismoving for t > 0. Similarly, any series thatrepresents g(t) for t > 0 cannot be 0 everywhere,so it will not represent g(t) for t < 0. Thus, g(t)cannot be represented by a Maclaurin series.42.43.44.2 3x xexp x = 1+ x+ + +2 632 5x3sin x– 2exp x = –2 + x– x – – 63 5 7x x x3sin x = 3 x– + – +2 40 168032 x–2exp x = –2 – 2 x– x – – 332 5xThus, 3sin x– 2exp x = –2 + x– x – – 64 6exp(2) 12 x xx = + x + + +2 62 2 2 32 2 ( x ) ( x )exp( x ) = 1+ x + + +2 64 62 x x= 1+ x + + +2 640. a.b.2–1/ h1ehf ′(0) = lim = lim2h→0 h h→01/ he2–1/ hhelim 0h→02= = (by l’Hôpital’s Rule)2⎧⎪–3 –1/ xf′ ( x)=2x e x ≠ 0⎨⎪⎩ 0 x = 0so2–1/ h2 42e4 2hhf ′′(0) = lim = lim = limh→0 4h h→0 1/ h h 0 1/ he → e2 24= lim = 0 (by using l’Hôpital’s Rule2h→01/ hetwice)( n) c. If f (0) = 0 for all n, then the Maclaurinseries for f(x) is 0.d. No, f(x) ≠ 0 for x ≠ 0. It only representsf(x) at x = 0.e. Note that for any n and x ≠ 0,1 xRx ( ) = e − .2566 Section 9.8 Instructor's Resource Manual

45.46.2 4 5x 5x 23xsin(exp x–1) = x+ – – – 2 24 1202 3 4x x xexp x–1= x+ + + +2 6 242 3 4 5 2 3325⎛ x x x x ⎞ 1⎛ x x ⎞ 1 ⎛ x ⎞sin(exp x–1) = x+ + + + +… –x+ + +… + x+ +…−⎜ 2 6 24 120 ⎟ 6 ⎜ 2 6 ⎟ 120 ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 2 3 4 5 4 5x x x x ⎞ 1⎛ 3 3x 5x⎞ 1 5= x+ + + + + – x + + + + ( x + ) −⎜ 2 6 24 120 ⎟ 6 ⎜ 2 4 ⎟⎝ ⎠ ⎝ ⎠120 ⎛ x 2 x 3 x 4 x 5 ⎞ ⎛ 3 4 5 55–x x x ⎞ ⎛ x ⎞2 4 55 23= x + + + + +… + + +… + +… −= x + x – x –x −⎜ 2 6 24 120 ⎟ ⎜ 6 4 24 ⎟ ⎜120⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠2 24 1202 4x xexp(sin x) = 1 + x+ – −2 83 5x xsin x = x– + −6 1203 5 323334⎛ x x ⎞ 1⎛ x ⎞ 1⎛ x ⎞ 1 ⎛ x ⎞exp(sin x) = 1 + x– + – + x– + + x– + + x–+ +⎜ 6 120 ⎟ 2 ⎜ 6 ⎟ 6 ⎜ 6 ⎟ 24 ⎜ 6 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 3 5 4 5 6x x ⎞ 1⎛ 2 x ⎞ 1⎛ 3 x ⎞ 1 ⎛4 2x⎞= 1 + x– + – + x – + + x – + + x – + +⎜ 6 120 ⎟ 2⎜ 3 ⎟ 6⎜ 2 ⎟ 24⎜ 3 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ 3 5 2 4 3 5 4 6x x ⎞ ⎛ x x ⎞ ⎛ x x ⎞ ⎛ x x ⎞2 4= 1 + x – + – + – + + – + + – + +⎜ 6 120 ⎟ ⎜ 2 6 ⎟ ⎜ 6 12 ⎟ ⎜ 24 36 ⎟ x x= 1 + x + – −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠2 847.3 52 x x(sin x)(exp x) = x+ x + – – 3 30⎛ 3 5 2 3 4x x ⎞⎛ x x x ⎞(sin x)(exp x) = x– + – ⎟⎜1+ x+ + + +⎜ 6 120 ⎟⎜ 2 6 24 ⎟⎝ ⎠⎝ ⎠⎛ 3 5 4 6 3 5 4 6 5 72– x x ⎞ ⎛– x x ⎞ ⎛– x – x ⎞ ⎛ x – x ⎞ ⎛ x –x ⎞= x + + + x + + + + + + + +⎜ 6 120 ⎟ ⎜ 6 120 ⎟ ⎜ 2 12 ⎟ ⎜ 6 36 ⎟ ⎜24 144 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠3 52 x x= x+ x + – – 3 3048.3 5sin x 2x– xx –xexp x = + 3 30+3 5 7– x x – xsin x x + +6 120 5040=2 3exp x 1+ x + x + x +2 63 52 x x= x– x + – +3 30Instructor’s Resource Manual Section 9.8 567

9.9 Concepts Review1. f(1); f′ (1); f′′(1)2.(6)f (0)6!3. error of the method; error of calculation4. increase; decreaseProblem Set 9.91.2.2xf( x)= e f (0) = 12f′ ( x) = 2ex f ′(0) = 22f′′ ( x) = 4ex f ′′(0) = 4(3) 2f ( x) = 8ex(4) 2f ( x) = 16exf(3) (0) = 8f(4) (0) = 164 2 8 3 16 42 4 3 2 4f( x) ≈ 1+ 2x+ x + x + x = 1+ 2x + 2x + x + x2! 3! 4!3 32 4 3 2 4f (0.12) ≈ 1+ 2(0.12) + 2(0.12) + (0.12) + (0.12) ≈ 1.27123 3–3xf( x)= e f (0) = 1–3xf′ ( x) = –3ef ′(0) = –3–3xf′′ ( x) = 9ef ′′(0) = 9(3) –3xf ( x) = –27e(4) –3xf ( x) = 81e(3) (0) –27f =f(4) (0) = 819 2 27 3 81 4 9 2 9 3 27 4f( x) ≈ 1–3 x+ x – x + x = 1–3 x + x – x + x2! 3! 4!2 2 89 2 9 3 27 4f (0.12) ≈ 1– 3(0.12) + (0.12) – (0.12) + (0.12) ≈ 0.69772 2 83. f( x) = sin2xf (0) = 0f′ ( x) = 2cos2xf ′(0) = 2f′′ ( x) = –4sin2xf ′′(0) = 0f(3) ( x) = –8cos2xf(3) (0) = –8f(4) ( x) = 16sin2xf(4) (0) = 08 3 4 3f( x) ≈ 2 x– x = 2 x–x3! 34 3f (0.12) ≈ 2(0.12) − (0.12) ≈ 0.237734. f( x) = tanxf (0) = 02f′ ( x) = sec x f ′(0) = 12f′′ ( x) = 2sec xtanxf ′′(0) = 0(3) 4 2 2f ( x) = 2sec x+4sec xtanxf(3) (0) = 2(4) 4 2 3f ( x) = 16sec xtanx+8sec xtanxf(4) (0) = 02 3 1 3f( x)≈ x+ x = x+x3! 31 3f (0.12) ≈ 0.12 + (0.12) ≈ 0.12063568 Section 9.9 Instructor Solutions Manual

5. f(x) = ln(1 + x) f(0) = 01f′ ( x)=1 + xf ′(0) = 11f′′ ( x) = –(1 )2+ xf ′′(0) = –1(3) 2f ( x)=3(1 + x)f(3) (0) = 2(4) 6f ( x) = –(1 )4+ xf(4) (0) = –61 2 2 3 6 4f( x) ≈ x– x + x – x2! 3! 4!1 2 1 3 1 4= x – x + x – x2 3 41 1 1f (0.12) ≈ 0.12 – (0.12) + (0.12) – (0.12)2 3 4≈ 0.11336. f ( x) = 1+ x f(0) = 17.2 3 41 –1/21f′ ( x) = (1 + x)f ′(0)=221 –3/21f′′ ( x) = – (1 + x)f ′′(0) = –44(3) 3 –5/ 2 (3) 3f ( x) = (1 + x)f (0) =88(4) 15 –7/ 2 (4) 15f ( x) = – (1 + x)f (0) = –16161 3 151 4 2 8 3 16 4f( x) ≈ 1 + x– x + x – x2 2! 3! 4!1 1 2 1 3 5 4= 1 + x – x + x – x2 8 16 1281 1 2f (0.12) ≈ 1 + (0.12) – (0.12)2 81 3 5 4+ (0.12) – (0.12) ≈1.058316 128–1f ( x) = tan x f(0) = 01f′ ( x)=21 + xf ′(0) = 12xf′′ ( x) = –(12 )2+ xf ′′(0) = 026 x –2f′′′ ( x)=2 3(1 + x )f ′′′(0) = –2f(4)3− 24x+ 24x( x)=2 4(1 + x )f (4) (0) = 02 3 1 3f ( x) ≈ x– x = x–x3! 31 3f (0.12) ≈0.12 – (0.12) ≈ 0.119438. f ( x) = sinhxf (0) = 0f ′( x) = coshxf ′(0) = 1f ′′( x) = sinhxf ′′(0) = 0f ′′′( x) = coshxf ′′′(0) = 1f (4) ( x) = sinhxf (4) (0) = 01 3 1 3f ( x)≈ x+ x = x+x3! 61 3f (0.12) ≈ 0.12 + (0.12) ≈ 0.12036x9. f ( x)= e f(1) = exf ′( x)= e f ′(1)= exf ′′( x)= e f ′′(1)= exf ′′′( x)= e f ′′′(1)= ee eP3 ( x) = e+ e( x–1) + ( x–1) + ( x–1)2 6⎛π⎞ 210. f(x) = sin x f ⎜ ⎟ =⎝4⎠2f ′( x) = cosxf ′′( x) = –sinxf ′′′( x) = –cosx⎛π⎞ 2f ′ ⎜ ⎟=⎝4⎠2π 2f ′′⎛ ⎞⎜ ⎟=–⎝4⎠2π 2f ′′′⎛ ⎞⎜ ⎟=–⎝4⎠22 322 2 ⎛ π⎞ 2⎛ π⎞P3( x) = + ⎜x– ⎟– ⎜x–⎟2 2 ⎝ 4⎠ 4 ⎝ 4⎠32 ⎛ π ⎞– ⎜x– ⎟12 ⎝ 4 ⎠⎛π⎞ 311. f(x) = tan x; f ⎜ ⎟ =⎝6⎠32 π 4f ′( x) = sec x;f ′⎛ ⎞⎜ ⎟=⎝6⎠32π 8 3f ′′( x) = 2sec xtanx;f ′′⎛ ⎞⎜ ⎟=⎝6⎠94 2 2 π 16f ′′′( x) = 2sec x+ 4sec xtanx;f ′′′⎛ ⎞⎜ ⎟=⎝6⎠32 33 4⎛ 4 3 8P3( x) x– π ⎞ ⎛x– π ⎞ ⎛x–π ⎞= + ⎜ ⎟+ ⎜ ⎟ + ⎜ ⎟3 3⎝ 6⎠ 9 ⎝ 6⎠ 9⎝ 6⎠Instructor's Resource Manual Section 9.9 569

⎛π⎞12. f(x) = sec x ; f ⎜ ⎟ = 2⎝4⎠πf ′( x) = secxtanx; f ′⎛ ⎞⎜ ⎟=2⎝4⎠3 2 πf ′′( x) = sec x+ secxtanx ; f ′′⎛ ⎞⎜ ⎟=3 2⎝4⎠13.3 3f ′′′( x) = 5sec xtan x+ sec xtanx ;πf ′′′⎛ ⎞⎜ ⎟=11 2⎝4⎠2⎛ π⎞ 3 2⎛ π⎞P3( x) = 2+ 2 ⎜x– ⎟+⎜x–⎟⎝ 4⎠ 2 ⎝ 4⎠311 2 ⎛ π ⎞+ ⎜x– ⎟6 ⎝ 4⎠–1 πf ( x) = cot x; f (1) =411f′ ( x) = – ; f ′(1) = –12+ x22x1f′′ ( x)= ; f ′′(1)=2 2(1 + x )22− 6x+ 21f′′′ ( x)= ; f ′′′(1) = –2 3(1 + x )2π 1 1 2 1 3P3( x) = – ( x–1) + ( x–1) – ( x–1)4 2 4 1214. f ( x)15.= x ; f (2) = 21 –1/2f′ ( x)= x ;21 –3/2f′′ ( x) = – x ; 4f ′(2)=242f ′′(2) = – 163 –5/2 3 2f′′′ ( x)= x ; f ′′′(2)=8642 2P3( x) = 2 + ( x–2)– ( x–2)4 322 ( –2)3+ x1283 2f( x) x –2x 3x5= + + ; f(1) = 72f ( x) 3 x –4x3 f ′ =f′′ ( x) = 6 x–4; f ′′(1) = 2′ = + ; (1) 2f(3) ( x ) = 6;f(3) (1) = 62 3P3( x) = 7+ 2( x− 1) + ( x− 1) + ( x−1)2 3= 5+ 3x − 2 x + x = f( x)216.17.f ( x)4x= ; f(2) = 163f ( x) 4x′ = ; f ′(2) = 322f ( x) 12x′′ = ; f ′′(2) = 48f(3) ( x) = 24x;f(3) (2) = 48f(4) ( x ) = 24;f(4) (2) = 242P4( x) = 16 + 32( x− 2) + 24( x−2)3 4+ 8( x− 2) + ( x−2)4= x = f( x)1f( x)= ; f(0) = 11– x1f′ ( x)= ; f ′(0) = 12(1 – x)2f′′ ( x)= ; f ′′(0) = 23(1 – x)(3) 6f ( x)= ; f(3) (0) = 64(1 – x)f(4)( n)f24( x)= ;5(1 – x)n!( x)=n 1(1 − x )+ ;(4) (0) 24 f =( n) f (0) = n !2 2 6 3 n!nf ( x) ≈ 1+ x+ x + x +…+ x2! 3! n!2 3 n= 1+ x + x + x +…+ xUsing n = 4, f ( x) ≈ 1+ x+ x + x + xa. f(0.1) ≈ 1.1111b. f(0.5) ≈ 1.9375c. f(0.9) ≈ 4.0951d. f(2) ≈ 3118. f(x) = sin x ; f(0) = 0f ′( x) = cosx; f ′(0) = 1f ′′( x) = –sinx; f ′′(0) = 0f(3) ( x) = –cosx;f(3) (0) = –1f(4) ( x) = sinx;f(4) (0) = 0When n is odd,2 3 43 5 7 ( n−1)/2nx x x ( −1)xsin x ≈ x− + − + +3! 5! 7! n!3 5x xUsing n = 5, sin x ≈ x− + .3! 5!a. sin(0.1) ≈ 0.0998570 Section 9.9 Instructor's Resource Manual

. sin(0.5) ≈ 0.479423.c. sin(1) ≈ 0.8417d. sin(10) ≈ 676.6719.3y24.3y−3 3 x−3 3 x20.−33y25.−3−3 3 x26.21.−33y27.−3 3 x22.3−3y28.3y−3 3 x−22x−3−3Instructor's Resource Manual Section 9.9 571

2 c –2c 2c1 6129. e + e ≤ e + ≤ e + 137. f( x) = ln(2 + x); f′( x) = ;2ce2 + x1f′′ (3) 2( x)=− ; f ( x) = ;30. tan c+ secc ≤ tan c + secc≤ 1+223(2 + x)(2 + x)(4) 6 (5) 244c4c2πf ( x) =− ; f ( x) = ;4531. = ≤ = 2 2π(2 + x)(2 + x)sin c sin c 12(6) 120 (7) 720f ( x) =− ; f ( x)=67(2 + x)(2 + x)4c4c432. = ≤ = 171 720 7 xc+ 4 c+4 4R6 ( x)= ⋅ x =7! 7 7(2 + c) 7(2 + c)c7c e 40.5−6eeR6 (0.5) ≤ ≈ 8.719×1033. = ≤77⋅2c+ 5 c+5 3cos c cos c38. f ( x)= e −x ; f ′( x) =− e −x ;134. = ≤−xc+ 2 c+2 2( n) ⎧⎪ e if n is evenf ( x)= ⎨−x⎪ ⎩ − e if n is odd2 22c + sin cc + sin c c + sin c−c7−e35.= ≤7 ( x−1)R6( x) = ( x− 1) = −10ln c 10ln c 10ln c7! c5040e16 + 1 177( −0.5)≤ =−7R6 (0.5) ≤ ≈ 9.402×1010ln 2 10ln 20.55040e22 –2c – cc c2 ⎛1⎞139. f(x) = sin x; f(7) ( x) =− cosx36. = ≤ 2 c – c ≤ 2 ⎜ ⎟ –cos c cos c⎝2 ⎠ 2722= (Note that x − x is maximum at 1 −cosc ⎛ π ⎞ −cosc( x−π) 7442 inR6( x)= ⎜x−⎟ =7! ⎝ 4 ⎠ 5040⎡⎣0, π / 4⎤⎦ .) cos 0.5( 0.5 − π) 74 −8R6(0.5) ≤ ≈ 2.685×10504040.1f( x) = 12; f′ ( x) =−x − 32 ; f′′ (3) 6( x) = ; f ( x) =− ;34( x − 3) ( x − 3)( x − 3)(5) 120 (6) 720 (7) 5040f ( x) =− ; f ( x) = ; f ( x)=−678( x − 3)( x − 3)( x − 3)71 5040 7 ( x −1)R6 ( x) = ⋅− ( x− 1) =−7! 8 8( c−3) ( c−3)7 7(0.5 −1) 0.5−5R6 (0.5) ≤ = ≈ 3.052×108 8(1 − 3) 2f(4)24( x) = ;5( x − 3)572 Section 9.9 Instructor's Resource Manual

141. If f( x)= , it is easily verified thatxn( n)( −1) n!f ( x)= . Thus for a = 142. If43.( n 1)x +7−( x −1)R6 ( x)8cThus,Therefore,= , where c is between x and 1 .7(0.5)R6 (0.5) = , where c∈ (0.5,1) .8c7(0.5)R6 (0.5) ≤ = 28(0.5)1f( x)= , it is easily verified that2xn( n)( − 1) ( n + 1)!f ( x)( n 2)x +a =7 7 7( −1) 8!( x−1) −8( x−1)R6 ( x)9 9(7!) cc= . Thus for 1= = , where c isbetween x and 1 . Thus,78(0.5)R6 (0.5) = , where c∈ (0.5,1) .9c78(0.5)R6 (0.5) ≤ = 329(0.5)Therefore,ce n+1Rn( x)= x( n + 1)!1Note that e < 3 .3R n (1) 399,999 .2n+ 1145. This is a Binomial Series ( p = ), so the third-order2Maclaurin polynomial is (see section 9.8, Thm D2 3and example 6) 3 ( ) 1 x x xP x = + − + ; further,2 8 1646.47.4−5xR3 ( x)=. Now if x ∈[ − 0.5,0.5] and c7128(1 + c)2is between 0 and x , then4 ⎛1⎞11+ c > 0.5 and x ≤ ⎜ ⎟ = so that, for all x,⎝2⎠16⎛ 1 ⎞⎜ ⎟⎝16⎠7 255 2R3( x) ≤ = ≈ 0.0276256128(0.5)3/2f( x) = (1 + x)f(0) = 13 1/23f′ ( x) = (1 + x)f ′(0)=223 –1/23f′′ ( x) = (1 + x)f ′′(0)=44(3) 3 –3/ 23f ( x) = – (1 + x)f ′′′(0) = –88(4) 9 –5/ 29f ( x) = (1 + x)f () c = (1 + c)16163/2 3 3 2 1 3(1 + x) ≈ 1 + x+x – x2 8 163 –5/2 4R3( x) = (1 + c)x1283 –5/2 4 –6R3( x) ≤ (0.9) (–0.1) ≈ 3.05×10128–1/2f( x) = (1 + x)f(0) = 14(4) –5/ 21 –3/21f′ ( x) = – (1 + x)f ′(0) = –223 –5/23f′′ ( x) = (1 + x)f ′′(0)=44(3) 15 –7 / 2 (3) 15f ( x) = – (1 + x)f (0) = –88(4) 105 –9/2105f ( x) = (1 + x)f () c = (1 + c)1616–1/2 1 3 2 5 3(1 + x) ≈ 1 – x+x – x2 8 1635 –9/2 4R3( x) = (1 + c)x12835 –9/2 4 –6R3( x) ≤ (0.95) (0.05) ≈ 2.15×10128(4) –9/ 2Instructor’s Resource Manual Section 9.9 573

48.49.50.⎛1+x ⎞f( x) = ln⎜ ⎟ f(0) = 0⎝ 1 − x ⎠2f′ ( x)= f ′(0) = 221– x4xf′′ ( x)= f ′′(0) = 02 2(1 – x )ffff(3)(4)(5)(5)24(1 + 3 x )( x)=2 3(1 – x )248 x(1 + x )( x)=2 4(1 – x )f (3) (0) = 42 448(1 + 10x+ 5 x )( x)=2 5(1 – x )2 448(1 + 10c+ 5 c )() c =2 5(1 – c )⎛1+x ⎞ 2 3ln ⎜ ⎟ ≈ 2x+ x⎝1−x ⎠ 32 42⎡1+ 10c+ 5c⎤R4 ( x)= ⎢ ⎥x5 2 5⎢⎣(1 – c ) ⎥⎦f (4) (0) = 02 42 ⎡1+ 10(0.5) + 5(0.5) ⎤R4 ( x) < ⎢ ⎥(0.5)5 2 5⎢⎣(1 – (0.5) ) ⎥⎦≈ 0.201cos c 5R4( x)= x5!5(0.5)R4( x) ≤ ≈0.00026042 ≤ 0.00026055!0.5 0.5⎛1 3 ⎞∫ sin x dx ≈ x – x dx0 ∫ 0⎜ ⎟⎝ 6 ⎠0.5⎡1 2 1 4⎤= ⎢ x – x ≈0.12242 24⎥⎣⎦0Error ≤ 0.0002605(0.5 – 0) = 0.00013025cos cR ( x) = – x 6!51R5( x) ≤ ≈ 0.0013896!1 1⎛2 4x x ⎞∫ cos x dx ≈ 1– dx0 ∫+0⎜2 24 ⎟⎝ ⎠61⎡ 3 5x x ⎤= ⎢x– + ⎥ ≈0.8417⎢⎣6 120⎥⎦0Error ≤ 0.001389(1 – 0) = 0.0013895551. Assume n is odd; that is n= 2m+ 1for m ≥ 0 .(2m+3)f () c 2m+3Then, Rn+ 1( x) = R2m+2( x)=x .(2m+ 3)!Note that, for all m,(4 m) (4m+1)f ( x) = sin x, f ( x) = cosx(4m+ 2) (4m+3)f ( x) =− sin x, f ( x) =−cos x;cos c 2m+3therefore, Rn+ 1( x)= x where c is(2m+ 3)!between 0 and x . For x ∈[0, π2], c∈(0, x)soπthat cos c < 1 and x≤ ; hence2Rπ2m+3⎛ ⎞⎜ ⎟⎝ 2 ⎠1 ( x ) . Now, for(2m+ 3)!n+ ≤π2πk = 2,3, …, 2m+ 3, ≤ so thatk 4Rn+x ∈ [0,π2m+3⎛ ⎞⎜ ⎟⎝ 2 ⎠1 ( ) π ⎛π⎞x ≤ ≤ ⎜ ⎟(2m+ 3)! 2 ⎝ 4 ⎠π]22m+2. Now2m+2π ⎛π⎞⎜ ⎟ ≤ 0.00005 ⇒2⎝4⎠⎛π⎞ ⎛2(0.00005)⎞(2m+ 2) ln ⎜ ⎟≤ln⎜ ⎟⇒⎝ 4 ⎠ ⎝ π ⎠2m+ 2 ≥ 42.8666 ⇒ n= 2m+ 1 > 42for all52. Assume n is even; that is n= 2mfor m ≥ 0 .(2m+2)f () c 2m+2Then, Rn+ 1( x) = R2m+1( x)=x .(2m+ 2)!Note that, for all m,(4 m) (4m+1)f ( x) = cos x, f ( x) =−sinx(4m+ 2) (4m+3)f ( x) =− cos x, f ( x) = sin x;cos c 2m+2therefore, Rn+ 1( x)= x where c is(2m+ 2)!between 0 and x . For x ∈[0, π2], c∈(0, x)soπthat cos c < 1 and x≤ ; hence2Rπ2m+2⎛ ⎞⎜ ⎟⎝ 2 ⎠1 ( x ) . Now, for(2m+ 2)!n+ ≤π2πk = 2,3, …, 2m+ 2, ≤ so thatk 4Rn+x ∈ [0,π2m+2⎛ ⎞⎜ ⎟⎝ 2 ⎠1 ( ) π ⎛π⎞x ≤ ≤ ⎜ ⎟(2m+ 2)! 2 ⎝ 4 ⎠π]2. Now2m+1for all574 Section 9.9 Instructor's Resource Manual

2m+1π ⎛π⎞⎜ ⎟ ≤0.00005⇒2⎝4⎠⎛π⎞ ⎛2(0.00005)⎞(2m+ 1)ln ⎜ ⎟≤ln⎜ ⎟⇒⎝ 4 ⎠ ⎝ π ⎠2m+ 1 ≥ 42.8666 ⇒ n= 2m>421 253. The area of the sector with angle t is tr . The2area of the triangle is1⎛ t ⎞⎛ t ⎞ 2 t t 1 2⎜rsin ⎟⎜2rcos ⎟ = r sin cos = r sin t2⎝ 2⎠⎝ 2⎠2 2 21 2 1 2A = tr – r sin t2 21 3Using n = 3, sin t ≈ t – t .61 2 1 2 1 3 1 2 3A tr – r ⎛ ⎞≈ ⎜t – t ⎟=r t2 2 ⎝ 6 ⎠ 1254.55. a.mv () =m0; m(0)= m21−v 02cmv 0m′ () v =; m′ (0) = 023/22⎛ v ⎞c1–⎜2c ⎟⎝ ⎠2 22mv0 + mc 0mm′′ 0() v =; m′′ (0) =25/224⎛ v ⎞cc1–⎜2c ⎟⎝ ⎠The Maclaurin polynomial of order 2 is:1 m0 2 m0⎛v⎞mv () ≈ m0 + v = m2 0 + ⎜ ⎟2 c2 ⎝c⎠ .12n⎛ r ⎞ln ⎜1+ ⎟ = ln 2⎝ 12 ⎠⎛ r ⎞12nln ⎜1+ ⎟=ln 2⎝ 12 ⎠ln 2n =⎛ r ⎞12ln ⎜1+⎟⎝ 12 ⎠256.57.b. f(x) = ln(1 + x); f(0) = 01f′ ( x)= ; f ′(0) = 11 + x1f′′ ( x) = – ; f ′′(0) = –1(1 )2+ xxln(1 + x)≈ x−2ln 2 ⎡ 24 ⎤n ≈ = ln 22 ⎢rr r(24 − r)⎥− ⎣ ⎦242ln 2 ln 2= +r 24 − r≈ ln 2 ln 2 0.693 0.029r+ 24≈ r+We let 24 – r ≈ 24 since the interest rate r isgoing to be close to 0.c. r n (exact) n (approx.) n (rule 72)0.05 13.8918 13.889 14.40.10 6.9603 6.959 7.20.15 4.6498 4.649 4.80.20 3.4945 3.494 3.6–(1 + k)xf( x) = 1– e ; f(0) = 0–(1 + k)xf′ ( x) = (1 + k)e ; f ′(0) = (1 + k)2 –(1 + k)xf′′ ( x) = –(1 + k)e ;2–(1 + k) x(1 + k)2f ′′(0) = –(1 + k)1– e ≈ (1 + k) x–x2For x = 2k, the polynomial is3 42k−4k −2k ≈ 2kwhen k is very small.(1 0.01)(0.02)1−e − + ≈0.019997 ≈ 0.024 3 2f( x) = x –3x + 2 x + x–2; f(1) = –13 2f ′( x) = 4 x –9x + 4x+ 1; f ′(1) = 02f′′ ( x) = 12 x –18x+ 4; f ′′(1) = –2f (3) ( x) = 24 x–18;f (4) ( x ) = 24;f (3) (1) = 6(4) (1) 24 f =f (5) ( x ) = 0Since f (5) ( x ) = 0, R5( x ) = 0.4 3 2x –3x + 2 x + x–2= –1–( x–1) + ( x–1) + ( x–1)2 3 42Instructor’s Resource Manual Section 9.9 575

58.f ''( a) 2 f '''( a) 3 f ( a)Pn( x) = f( a) + f '( a)( x− a) + ( x− a) + ( x− a) + + ( x−a)2! 3! n!( n)' f ''( a) f '''( a) 2 f ( a)n−1n ( ) = '( ) + 2( − ) + 3( − ) + + ( − )P x f a x a x a n x a2! 3! n!( n)2 n−1f '''( a) f ( a)= f '( a) + f ''( a)( x− a) + ( x− a) + + ( x−a)2! ( n −1)!P ' ( a) = f '( a) + 0 + 0 + + 0 = f '( a)n( n)( n)'' f '''( a) f ( a)Pn= 0 + f ''( a) + 2( x− a) + + ( n−1)( x−a)2! ( n −1)!f ( a)= f ''( a) + f '''( a)( x− a) + + ( x−a)( n − 2)!n−2n−2( n)nPn '' = f ''( a) + 0 + 0 + + 0 = f ''( a) ( n)( n) f ( a)0 ( n)n = − =P ( x) ( x a) f ( a)0!( n) ( n)Pn( a) = f ( a)59. f(x) = sin x ;f ′( x) = cosx;f ′′( x) = –sinx;f (3) ( x) = –cosx;f (4) ( x) = sinx;⎛π⎞ 2f ⎜ ⎟ =⎝4⎠2π 2f ′⎛ ⎞⎜ ⎟=⎝4⎠2⎛π⎞ 2f ′′ ⎜ ⎟=–⎝4⎠2f(3) π 2⎛ ⎞⎜ ⎟ = –⎝4⎠2f (4) () c = sincπ π43 °= – radians4 9022 2⎛ π⎞ 2⎛ π⎞sin x = + ⎜x– ⎟– ⎜x–⎟2 2 ⎝ 4⎠ 4 ⎝ 4⎠32 ⎛ π ⎞– ⎜x– ⎟ + R3( x)12 ⎝ 4 ⎠⎛π π ⎞ 2 2⎛ π ⎞ 2⎛ π ⎞sin ⎜ – ⎟= + ⎜– ⎟– ⎜–⎟⎝4 90⎠ 2 2 ⎝ 90⎠ 4 ⎝ 90⎠32 ⎛ π ⎞ ⎛π π ⎞– ⎜– ⎟ + R3⎜ – ⎟12 ⎝ 90 ⎠ ⎝4 90 ⎠≈ 0.681998 + R 3R34 4sin c ⎛ π ⎞ 1 ⎛ π ⎞= ⎜– ⎟ < ⎜ ⎟ ≈ 6.19×104! ⎝ 90 ⎠ 24 ⎝90⎠2–8π π60. 63°= + radians3 60( nSince f ) ( x ) is ±sin x or ±cos x,61.1 ⎛ π ⎞Rn( x) ≤ ⎜x–⎟( n + 1)! ⎝ 3⎠n+1⎛π π ⎞ 1 ⎛ π ⎞R n ⎜ + ⎟ ≤ ⎜ ⎟⎝3 60 ⎠ ( n + 1)! ⎝60⎠1 ⎛ π ⎞⎜ ⎟( n + 1)! ⎝60⎠f(x) = cos x ;n+1n+1≤ 0.0005 when n ≥ 2.⎛π⎞ 1f ⎜ ⎟ =⎝3⎠2f ′( x) = –sinx;π 3f ′⎛ ⎞⎜ ⎟=–⎝3⎠2f ′′ ( x) = –cosx;π 1f ′′⎛ ⎞⎜ ⎟=–⎝3⎠221 3⎛ π⎞ 1⎛ π⎞cos x = – ⎜x– ⎟– ⎜x– ⎟ + R3( x)2 2 ⎝ 3⎠ 4⎝ 3⎠1 3⎛ π ⎞ 1⎛ π ⎞cos 63 °≈ – ⎜ ⎟– ⎜ ⎟ ≈0.453972 2 ⎝60⎠ 4⎝60⎠1010 ⎛π⎞−51 1R9( x) ≤ x ≤ ⎜ ⎟ ≈ 2.5202×1010! 10! ⎝2⎠2576 Section 9.9 Instructor's Resource Manual

62. a.1 1 sincsin x = x– x + x – x6 120 7203 5 63x6x→0 5 x→0sin x– x+ ⎛ 1 sinc ⎞ 1lim = lim ⎜ – x ⎟=x⎝120 720 ⎠ 120( n+(i) Since f1) ( x)is continuous near c, then( n+f1) ( a) < 0 when a is near c.Thus Rn( x ) < 0 when x is near c, sof(x) < f(c) when x is near c. f(c) is a localmaximum.b.i=01 1 1 sinccos x = 1– x + x – x + x2 24 720 5040x→0x→02 4 6 72 4+ x x2 246cos x –1 –lim=x⎛ 1 sinc ⎞ 1lim ⎜– + x ⎟=–⎝ 720 5040 ⎠ 72063. The kth derivative of h(x)f(x) iskn 1∑ h ( x) f ( x).If hx ( ) = x + ,k () i ( k–)i( i )() i ( n + 1)! n+1– ih ( x) = x .( n+1– i)!Thus for i ≤ n + 1,() ih (0) = 0. Letn+1qx ( ) = x f( x).Thenk( k) k ( i) ( k– i)q (0) = ∑ ( i ) h (0) f (0) = 0i=0for k ≤ n + 1.( k) ( k) ( k)g ( x) = p ( x) + q ( x),so( k ) ( k ) ( k ) ( k )g (0) = p (0) + q (0) = p (0)for k ≤ n + 1.The Maclaurin polynomial of order n for g is( n)p′′(0) 2 p (0) np(0) + p′(0) x+ x +…+ x which2! n!is the Maclaurin polynomial of order n for p(x).Since p(x) is a polynomial of degree at most n,the remainder Rn( x ) of Maclaurin’s Formula forp(x) is 0, so the Maclaurin polynomial of order nfor g(x) is p(x).64. Using Taylor’s formula,f′′() c 2f() x = f() c + f′()( c x– c) + ( x– c)+…2!( n)f () c ( – )n+ x c + Rn( x )n!Since( n)f′ () c = f′′ () c = f′′′() c =…= f () c = 0,f ( x) = f( c) + R ( x)n( n+1)f ( a)Rn( x) = ( a)( x– c)( n + 1)!between x and c.n+1where a is( n+(ii) Since f1) ( x)is continuous near c, then( n+f1) ( a) > 0 when a is near c.Thus Rn( x ) > 0 when x is near c, sof(x) > f(c) when x is near c.f(c) is a local minimum.4Suppose f ( x) = x . f(x) > 0 when x > 0 andf(x) < 0 when x > 0. Thus x = 0 is a localminimum.(4)f′ (0) = f′′ (0) = f′′′(0) = 0, f (0) = 24 > 09.10 Chapter ReviewConcepts Test1. False: If b n = 100 and a n = 50 + (–1) n then⎧51 if n is evensince an= ⎨ ,⎩ 49 if n is odd≤ a ≤ b for all n and0 n nlim bn= 100 while lim a n does notn→∞n→∞exist.2. True: It is clear that n! ≤ n . The inequalitynn! ≤n ≤ (2 n–1)!is equivalent ton(2 –1)!1 ≤ n nn! ≤ n!.Expanding the terms givesnn n n n n n= ⋅ ⋅ ⋅⋅ ⋅n! 1 2 3 n–1n≤ ( n+ 1)( n+ 2) ⋅ ⋅ ( n+n–1)orn n nn⋅ ⋅ ⋅⋅ ≤ ( n+ 1)( n+ 2) ⋅2 3 n –1 ⋅ ( n+n–1)The left-hand side consists of n – 1terms, each of which is less than orequal to n, while the right-hand sideconsists of n – 1 terms, each of whichis greater than n. Thus, the inequalityis true so n! ≤n n ≤ (2 n–1)!nInstructor’s Resource Manual Section 9.10 577

3. True: If lim a n = L then for any ε > 0n→∞there is a number M > 0 such thatan– L < ε for all n ≥ M. Thus, forthe same ε, a3n+ 4 – L < ε forM – 43 n + 4 ≥ M or n ≥ . Since ε3was arbitrary, lim a3n+4 = L.n→∞4. False: Suppose a n = 1 if n = 2k or n = 3kwhere k is any positive integer anda n = 0 if n is not a multiple of 2 or 3.Then lim a2n= lim a3n= 1 butlim an→∞n→∞nn→∞does not exist.5. False: Let a n be given by⎧1 if n is primean= ⎨⎩ 0 if n is compositeThen a mn = 0 for all mn since m ≥ 2 ,hence lim a = 0 for m ≥ 2 .lim an→∞n→∞nmndoes not exist since for anyM > 0 there will be a n 's with a n = 1since there are infinitely many primenumbers.6. True: Given ε > 0 there are numbers M 1and M 2 such that a2 n – L < ε whenn≥ M 1 and a2n+ 1 – L < ε whenn≥ M 2 . Let M = max{ M1, M2},then when n ≥ M we havea2 n – L ε< and a2n+ 1 – L < ε soan– L < ε for alln ≥ 2M + 1 since every k ≥ 2M + 1is either even (k = 2n) or odd(k = 2n + 1).n n+8. False: {( 1) } and {( 1)1}− − both divergebut( 1) ( 1) + 1− + − = ( − 1) (1 − 1) = 0{ } { } {}converges.9. True: If { a n}converges, then for some N,10. False:there are numbers m and M withm≤ a ≤ M for all n ≥ N. Thusna nm M≤ ≤ for all n ≥ N. Sincen n n⎧m⎫⎨ ⎬⎩n⎭ and ⎧M⎫⎨ ⎬ both converge to 0,⎩ n ⎭⎧a n ⎫⎨ ⎬ must also converge to 0.⎩ n ⎭∞1∑ (–1) n converges.n=1 n1a (–1) n2 1n = so a n = andn ndiverges.∞∑n=111. True: The series converges by theAlternating Series Test.S = a , S = a – a , S = a – a + a ,1n1 1 2 1 2 3 1 2 3S = a – a + a – a , etc.4 1 2 3 40< a < a ⇒ 0 < a – a = S < a ;2 1 1 2 2 10 < a < a ⇒ – a < – a + a < 0 so3 2 2 2 30 < a – a < a − a + a = S < a ;1 2 1 2 3 3 10< a < a ⇒ 0 < a – a < a , so4 3 3 4 3– a2 < – a2 + a3 – a4 < – a2 + a3< 0,hence0 < a – a < a – a + a – a = S1 2 1 2 3 4 4< a1 – a2 + a3 < a1;etc.For each even n, 0 < Sn–1– anwhilefor each odd n, n > 1, S –1+ a < a1.nn1 17. False: Let a n = 1 + + + . Then2 n1an– an+ 1 = –n + 1solim ( an– a n + 1) = 0 but lim a nn→∞not finite sincewhich diverges.lim an→∞nn→∞∞1= ∑ ,kk = 1is12. True: For n ≥ 2, 1 1 n ≤ 2so∞ n ∞ n⎛1⎞ ⎛1⎞∑⎜ ⎟ ≤ ∑ ⎜ ⎟ whichn= 2⎝n⎠ n=2⎝2⎠converges since it is a geometric1series with r = .2∑∞ n⎛1⎞1 1⎜ ⎟ = 1+ + +…> 1n=1⎝n⎠ 4 27sinceall terms are positive.578 Section 9.10 Instructor’s Resource Manual

13. False:14. False:15. True:16. False:17. False:∞ n ∞ n ∞ n⎛1⎞ ⎛1⎞ ⎛1⎞⎜ ⎟ = 1+ ⎜ ⎟ ≤ 1+⎜ ⎟⎝n⎠ ⎝n⎠ ⎝2⎠∑ ∑ ∑n= 1 n= 2 n=21 1= 1+ + +… 4 8∞ n–11 ⎛1⎞1 1= – + ∑ ⎜ ⎟ = – +2n=⎝2⎠2 1–11 21 3= – + 2=2 2Thus, ∑ ∞⎛1 ⎞n⎜ ⎟ = Sn=1⎝n⎠with31< S ≤ < 2.2∞∑ (–1) n diverges but the partialn=1sums are bounded ( S n = –1 for odd nand S n = 0 for even n.)∞1 10 < 2n≤ nfor all n in N but 1∑ 2n=1 n∞1converges while ∑ diverges.n=1na= lim = 1, Ratio Test isn+1ρn→∞aninconclusive. (See the discussionbefore Example 5 in Section 9.4.)12> 0 for all n in N andn∞1∑ converges, but2n=1 n2an+1 nlim = lim = 1.n→∞a n→∞2( n + 1)nn⎛ 1⎞1lim ⎜1− ⎟ = ≠ 0 so the seriesn→∞⎝n⎠ecannot converge.n + 118. False: Since lim = 0, there is somen→∞nen 4number M such that e > n + 1 for4all n ≥ M, thus n> ln( n + 1) and19. True:41

21. True: If 0 ≤an+100 ≤ bnfor all n in N, then∞ ∞ ∞∑ an≤ ∑ bnso ∑ analson= 101 n=1n=1converges, since adding a finitenumber of terms does not affect theconvergence or divergence of a series.122. True: If ca n ≥ for all n in N with c > 0,n1then a n ≥ for all n in N socn∞ ∞ ∞1 1 1∑a≥ ∑ = ∑ whichcn c n23. True:nn= 1 n= 1 n=1∞diverges. Thus,Comparison Test.2 3∑ a diverges by thenn=1∞1 ⎛1⎞ ⎛1⎞ ⎛1⎞+ ⎜ ⎟ + ⎜ ⎟ +…= ∑ ⎜ ⎟3 ⎝3⎠ ⎝3⎠ ⎝3⎠n=1∞ n–11 1⎛ ⎞ 3 3⎜ ⎟ 1 2n=1 ⎝ ⎠ 1–3 31 1 1= ∑ = = = ,3 3 2so thesum of the first thousand terms is lessthan 1 .224. False: Consider the series with1( 1) n +−a n = . Thenn2n+1n ( −1) −1( − 1) a n = = son n∞n 1 1∑ ( − 1) a n = −1− − − whichn=12 3diverges.25. True: If b ≤ a ≤ 0 for all n in N thennn0 ≤ – an≤ – bnfor all n in N.∞∞– bn= (–1) bnn= 1 n=1∞since ∑ bnconverges.n=1Thus, by the Comparison Test,∞∑ – a n converges, hencen=1∞∞∑an= ( −1) ∑ (–an)alson= 1 n=1converges.∑ ∑ which convergesn26. True Since a ≥ 0 for all n,27. True:n∞( 1) n ∞− an = ann= 1 n=1∞( −1) n ann=1∑ ∑ so the series∑ converges absolutely.∞99n+ 1 1 n+1 1∑( −1) −∑( −1)n= 1nn=1n1 1 1 1=− + − − < = 0.01100 101 102 10028. True: Suppose ∑ anconverges. Thus,∑ 2 a n converges, so ∑ ( an+ an)converges since 0≤ an + an ≤ 2 an.But by the linearity of convergentseries ∑ an = ∑( an + an)−∑anconverges, which is a contradiction.29. True: 3 −( − 1.1) = 4.1, so the radius ofconvergence of the series is at least4.1.3− 7 = 4< 4.1 so x = 7 is within theinterval of convergence.30. False: If the radius of convergence is 2, thenthe convergence at x = 2 isindependent of the convergence atx = –2.31. True: The radius of convergence is at least1.5, so 1 is within the interval ofconvergence.111 ⎡ ∞ n+anx ⎤Thus ∫ f( x)dx = ⎢0∑ ⎥⎢⎣n= 0n + 1 ⎥⎦0∞an= ∑ .n +n=0132. False: The convergence set of a power seriesmay consist of a single point.33. False: Consider the function1⎧ −⎪ x2f( x)= ⎨e x ≠ 0 .⎪ ⎩ 0 x = 0The Maclaurin series for this functionrepresents the function only at x = 0.134. True: On (–1, 1), f( x) = .1 − x1f ′( x) = = [ f( x)]2 .2(1 − x)580 Section 9.10 Instructor’s Resource Manual

35. True:∞( −1)x −xd∑ = e , e−xe−x+ = 0n!dxn=0nf ′′ (0) 236. True: Px ( ) = f(0) + f′(0) x+x237. True: If p(x) and q(x) are polynomials ofdegree less than or equal to n,satisfying p(a) = q(a) = f(a) andn( k) ( k) ( k)p ( a) = q ( a) = f ( a)fork ≤ n, then p(x) = q(x).38. True: f(0) = f′ (0) = f′′(0) = 0, its secondorder Maclaurin polynomial is 0.39. True: After simplifying, P 3 ( x) = f( x).40. True: Any Maclaurin polynomial for cos xinvolves only even powers of x.41. True: The Maclaurin polynomial of an evenfunction involves only even powers ofx, so f ′(0) = 0 if f(x) is an evenfunction.42. True: Taylor’s Formula with Remainder forn = 0 is f ( x) = f( a) + f′( c)( x– a)which is equivalent to the Mean ValueTheorem.Sample Test Problems1.9n9lim = lim = 3→∞9n1n2 1+n→∞9 +2nThe sequence converges to 3.2. Using l’Hôpital’s Rule,3.1n nlim = lim = limn→∞ n n→∞ 1 n→∞2 nln 22= lim = 0 .n→∞nnnnn/4⎛ 4⎞ ⎛⎛ 4⎞⎞lim ⎜1+ ⎟ = lim ⎜1+ ⎟= en→∞⎝ n⎠ n→∞⎜⎝ n⎠⎟⎝ ⎠4The sequence converges to e .n + 1n + 14. an+1 = anthus for n > 3, since > 1,33a + > a and the sequence diverges.n 15. Letnn 1/n1y = n = n then ln y = ln n.n1n nnn→∞ n→∞ n→∞ n→∞1 ln 1lim ln = lim = lim = lim = 0 byn n 1 nusing l’Hôpital’s Rule. Thus,nln ylim n = lim e = 1. The sequencen→∞n→∞converges to 1.446.13lim = 0 whilen→∞n10n → so 11 ⎛1⎞= ⎜ ⎟3 ⎝3⎠n1/ n.As n →∞ ,1/ n1/ n 0⎛1⎞ ⎛1⎞ ⎛1⎞lim ⎜ ⎟ = lim ⎜ ⎟ = ⎜ ⎟ = 1.3 3 3n→∞⎝ ⎠ →0⎝ ⎠ ⎝ ⎠The sequence converges to 1.n7.2sin n 1an≥0; lim ≤ lim = 0n→∞n n→∞nThe sequence converges to 0.8. The sequence does not converge, since whenevern is an even multiple of 6, a n = 1 , whilewhenever n is an odd multiple of 6, a n = − 1.Instructor’s Resource Manual Section 9.10 581

9.S n⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ 1= ⎜ – ⎟+ ⎜ – ⎟+…+ ⎜ – ⎟+⎜ – ⎟ = 1– ,⎝ 1 2 ⎠ ⎝ 2 3 ⎠ ⎝ n–1 n ⎠ ⎝ n n+1⎠n + 1⎛ 1 ⎞lim Sn= lim ⎜1– ⎟=1 . The series converges to 1.→∞ n→∞⎝n + 1 ⎠nso10.⎛1 1⎞ ⎛1 1⎞ ⎛1 1⎞ ⎛ 1 1 ⎞ ⎛1 1 ⎞= ⎜ – ⎟+ ⎜ – ⎟+ ⎜ – ⎟+…+ ⎜ – ⎟+⎜ – ⎟⎝1 3⎠ ⎝2 4⎠ ⎝3 5 ⎠ ⎝n–1 n+ 1⎠ ⎝n n+2⎠⎛3 1 1 ⎞ 3lim Sn= lim ⎜ – – ⎟=.The series converges to 3 .→∞ n→∞⎝2 n+ 1 n+2⎠22S nn1 1 1= 1 + – – ,2 n+ 1 n+2so11.1 2 3nln + ln + ln +…= ∑ ln2 3 4 n + 1∞n=1∞∑= [ln n– ln( n+1)]n=1S = (ln1– ln 2) + (ln 2 – ln 3) +…+ (ln( n–1) – ln n) + (ln n– ln( n+1))n1As n →∞, →0son + 1The series diverges.1lim Sn= lim ln = – ∞.→∞ n→∞n + 1n1= ln1– ln( n + 1) = lnn + 112.13.14.15.16.⎧1 if k is evencos kπ=⎨ so⎩ –1 if k is odd17.∞∞∑coskπ=∑ (–1) k which diverges sincek= 0 k=0⎧1 if n is evenSn= ⎨ so { S n } does not⎩ 0 if n is oddconverge.18.∞ ∞ k2–2k⎛ 1 ⎞ 1 e∑e= ∑ ⎜1.15652 ⎟ = = ≈1 2k= 0 k=0⎝e⎠ 1–2 e –1e1since 1.2e .2 4 6x x xcos x = 1 − + − + , so2! 4! 6!2 4 62 2 21− + − + converges to2! 4! 6!cos 2 ≈ –0.41615.2 3x x x xe = 1+ + + + , so1! 2! 3!1 1 11− + − + = ≈0.36791! 2! 3!1 e − .n119. Let an= and b2 n = .1 + nn2ann1lim = lim = lim = 1;n→∞ b n 2 1n →∞1+n n→∞+ 12n0 < 1 < ∞ .By the Limit Comparison Test, since∞ ∞∞ ∞1n∑ bn= ∑ diverges, an=n= 1 n=1nn= 1 n=11+ndiverges.∑ ∑ alson + 5 120. Let an= and b3 n = .1 +2n n3 25a 1n n + 5n+nlim = lim = lim = 1;n→∞ b n 3 1n →∞ 1+n n→∞3+ 1n0 < 1 < ∞ .By the Limit Comparison Test, since∞ ∞∞ ∞1n + 5∑ bn= ∑ converges,2 ∑an= ∑ 2n= 1 n=1nn= 1 n=11+nalso converges.2582 Section 9.10 Instructor’s Resource Manual

1 121. Since the series alternates, > > 0 , and3n 3n+113lim = 0 , the series converges by then→∞nAlternating Series Test.22. The series diverges since1 −1/nlim = lim 3 = 1 .n→∞n3 n→∞23.∞n n ∞ n2 + 3 ⎛⎛1⎞ ⎛3⎞= +⎜⎜ ⎟ ⎜ ⎟⎝2⎠ ⎝4⎝ ⎠∑ ∑nn= 1 4 n=1⎛ 1 ⎞ ⎛ 1 ⎞= − 1 + − 1 = 1+ 3=4⎜1 1 ⎟ ⎜13 ⎟⎝− −2 ⎠ ⎝ 4 ⎠The series converges to 4. The 1’s must besubtracted since the index starts with n = 1.⎛ n+ 1 n ⎞ ⎛ n+1 ⎞24. ρ = lim÷ = lim2 2n→∞⎜( n+1) n ⎟ ⎜n 2n+1⎟⎝ee ⎠→∞⎝ne⎠⎛ 1+1 ⎞n= lim = 0 < 1, so the series converges.n→∞⎜2n+1e ⎟⎝ ⎠25.n + 1 1lim = ≠ 0 , so the series diverges.→∞10n+ 12 10n26. Let a27.28.=nnand+ 7n1 .3/2n1⎞⎟⎠nb2n =2annn→∞ b n 217n →∞ n + 7 n→∞+2nlim = lim = lim = 1;0 < 1 < ∞ .By the Limit Comparison Test, since∞ ∞1⎛3⎞∑bn= ∑ converges 1 ,3/2⎜ > ⎟n⎝2⎠n= 1 n=1∞ ∞n∑a= ∑ also converges.n 2n= 1 n=1n+ 72 2( n+ 1) n n+1ρ = lim ÷ = lim = 0 < 1 , son→∞( n+1)! n!n→∞2nthe series converges.3 n+1 3 n( n+1) 3 n 3ρ = lim ÷n→∞( n+ 2)! ( n+1)!( 1+)3 3 1n +n n3 n→∞23( 1)= lim = lim = 0 < 1n→∞n ( n+2) 1+The series converges.4n329.30.31.n+12 ( n+1)! 2 n!ρ = lim ÷n→∞( n+ 3)! ( n+2)!2( n + 1)= lim = 2 > 1n→∞n + 3The series diverges.n⎛ 1⎞1lim ⎜1− ⎟ = ≠ 0 , so the series does notn→∞⎝n⎠econverge.( 3 )n( )n2 n+12 2lim ( n + 1) lim 2 ( n + 1)ρ = =n→∞n→∞3 2nn2 232= < 1 , so the series converges.31 132. Since the series alternates, > ,1+ lnn1+ ln( n+1)1and lim = 0, the series converges by then→∞1+lnnAlternating Series Test.33.34.1 1 1a n = ; > so an> a n + 1;3 n–1 3 n–1 3n+21lim an= lim = 0, so the seriesn→∞n→∞3 n –1∞n 1∑ (–1) converges by the Alternatingn=13 n –1Series Test.1Let b n = , thennann 1 1lim = lim = lim = ;n→∞ b n→∞ 3 n–1 n→∞3–1 3n10 <

35.nn3 1 ⎛3⎞= ;n+ 8 8 ⎜ ⎟2 2 ⎝2⎠1 3 n1 3 n⎛ ⎞ ⎛ ⎞limlim8 ⎜ ⎟ =n→∞2 8 ⎜ ⎟ =∞2 ⎝ ⎠ 2 n→∞⎝2⎠The series is divergent.x xsince 3 1.2 >36. Let f( x) = , thenln x1/ 1/1 ⎡ xxxx ⎤f′ ( x) = ⎢ (1–ln x)ln x–⎥2 2(ln x)⎢⎣xx ⎥⎦1/ xx2= [ln x – (ln x) – x],for x ≥ 3, ln x > 12( xln x)2so (ln x)> ln x hence f(x) is decreasing onnn[3, ∞ ). Thus, if an = , an > an+1.ln nn 1/n 1Let y = n = n , so ln y = ln n.nUsing l’Hôpital’s Rule,1ln n n 1lim = lim = lim = 0, thusn→∞ n n→∞ 1 n→∞nln 0lim lim ynnnn = e = e = 1. Hence, limn→∞n→∞n→∞ln nis of the form 1 nnso lim = 0.∞ n→∞ln nThus, by the Alternating Series Test,∞ n n(–1) n∑ converges.n=2ln n∞ n ∞ n1 1n nln n < n, so > hence ∑ < ∑ .ln n nn= 2nn=2lnn∞ n∞ nnnThus if ∑ diverges, ∑n=2nn=2 ln also diverges.nnn 1Let an= and b n = . Thenn nannlim = lim n = 1 as shown above;n→∞bnn→∞∞ ∞10 < 1 < ∞ . Since ∑bn= ∑ diverges,n= 2 n=2n∞ nn∑ also diverges by the Limit Comparisonn=2n∞ nnTest, hence ∑n=2 ln also diverges.nThe series is conditionally convergent.37.38.n+1 nx xlimn→∞( 1)3 13n+ + n + 13ρ = ÷n + 1= lim x= xn→∞3( n + 1) + 1When x = 1, the series is∞ ∞ ∞1 1 1= 1+ ≤ 1+∑ ∑ ∑ , which3 3 3n= 0n + 1 n= 1n + 1 n=1nconverges.(–1)When x = –1, the series is ∑ which3n= 0 n + 1∞1converges absolutely since ∑ converges.3n= 0 n + 1The series converges on –1 ≤ x ≤ 1.∞n+ 2 n+ 1 n+1 n( −2) x ( −2)xρ = lim÷n→∞2n+ 5 2n+32n+ 3= lim 2x= 2xn→∞2n+ 5n; 2x < 1 when1 1– < x < . 2 21When x = , the series is21 nn∞ n+(–2)1 –2( 2) ∞ (–2)( 2 )∑ = ∑n= 02n+ 3n=02n+3∞ n+12(–1) 2 2 2= ∑ . an= ; > , son=02n+ 3 2n+3 2n+ 3 2n+5∞ n+122(–1)an> a n + 1; lim = 0 so ∑n→∞2n+ 3n= 02n+ 3converges by the Alternating Series Test.1When x = – , the series is21 nn∞ n+(–2)1 (– 2 ) ∞ (–2)( –2)∞2∑ – .n 02 n = ∑3n 02 n = ∑=+=+ 3n=02 n + 321a n = , let b n = then2n+ 3 nan2n2lim = lim = lim = 1;n→∞ b n→∞ 2n+ 3 n→∞2 +3n0< 1

39.40.41.n+1( x−4) ( x−4)ρ = lim ÷n→∞n+ 2 n+1n + 1= lim x− 4 = x−4; x – 4 < 1 whenn→∞n + 23 < x < 5.When x = 5, the series is∞ n n ∞ n(–1) (1) (–1)∑ = ∑ .n= 0n+ 1n=0n+11 1 1a n = ; > , so an> a n + 1;n+ 1 n+ 1 n+2∞ n1(–1)lim = 0 so ∑ converges by then→∞n + 1n= 0n + 1Alternating Series Test.When x = 3, the series is∞ n n ∞(–1) (–1) 1 1∑ = ∑ . a n ,n= 0n+ 1n=0n+1= n + 1let1 annb n = then lim = lim = 1;n n→∞bnn→∞n+1∞ ∞∞11hence since ∑bn= ∑ diverges, ∑n= 1 n=1nn= 0n + 1also diverges.The series converges on 3< x ≤ 5.n + 1 3 n + 3 n 3 n3 x 3 xρ = lim ÷n→∞(3 n+3)! (3 n )!3 1= lim 3x= 0n→∞(3n+ 3)(3n+ 2)(3n+1)The series converges for all x.n+1 nlim ( x−3) ( x−3)n→∞2 n+1 1 2 n+ + 1ρ = ÷1n21n21+ x − 3= lim x − 3 = ;n→∞2 + 2when 1 < x < 5.When x = 5, the series is∞n∞2 1∑ =n ∑n= 02 + 1 n=01+2so the series diverges.1( )n;nx –3< 12( )120< 1

1148. Let a k = and define f( x)= ; then f ( k) = a29 +2k and f is positive, continuous and non-increasingk9 + xA−2x∞ 1 ⎡1(since f′ −1x ⎤( x) = < 0) on [1, ∞ ) . Thus, by the Integral Test, Elim tan2 2n < ∫ dx =2(9 + x )n ⎢=9 x A→∞3 3⎥+⎣ ⎦n1⎡−1 A −1 n⎤π 1 −1n π 1lim tan tan tan3⎢− = −A→∞3 3⎥. Now tan−1n 0.00005 3 tan ( 3⎛π⎞− ≤ ⇒ n ≥ ⎡−0.00005⎤⎜ ⎟)≈ 20,000 .⎣⎦ 6 3 3 6 3 3⎢⎣⎝ 6 ⎠ ⎥⎦kx49. Let ak= and define f( x)= ; then f ( k) = a22k and f is positive, continuous and non-increasing (sincekxeeA21−2x∞ x ⎡ 1 ⎤f′ ( x) = < 0) on [1, ∞ ) . Thus, by the Integral Test, Elim2n < ∫ dx = ⎢− ⎥ =2 2xn x A→∞xee⎢⎣2e⎥⎦1⎡1 1 ⎤ 1⎢− lim + ⎥ =2 ⎢ A→∞⎣ 2e 2e ⎥⎦4eA 2 n 2 n 2. Now50. One million terms are needed to approximate the sum to within 0.001 since999,999 < n.51. a. From the Maclaurin series for14e2nb. In Example 6 of Section 9.8 it is shown thatc.2 1 2 1 41+ x = 1+ x − x + .2 82 3 4 52n≤ 0.000005 ⇒ e ≥50,000 ⇒ n ≥ln(50,000) ≈10.82⇒ n > 3 .11− x, we have 1 3 61 x x31− x = + + + .21< 0.001 is equivalent ton + 11 1 2 1 3 5 41+ x = 1+ x− x + x − x + so2 8 16 1282 3 4e− x = 1− x+ x − x + x − x + , so e−x − 1 + x = x − x + x − .2! 3! 4! 5!2! 3! 4!2 4 6x 5x 61xd. Using division with the Maclaurin series for cos x, we get sec x = 1+ + + + .2 4! 6!3 5x 5xThus, xsecx = x+ + +.2 4!e.f.⎛ 2 3 ⎞⎛ 3 5 7 ⎞−e x sin x = 1− x+ x − x + ⎟⎜x− x + x − x +⎜ 2! 3! ⎟⎜ 3! 5! 7! ⎟⎝ ⎠⎝ ⎠3 5 7x x x1+ sinx= 1+ x− + − + ; Using division, we get3! 5! 7!32 x= x− x + −3121 x x1+sinx = − + − .n52.f( x) = cos x f(0) = 1(1) (1)f ( x)=− sin x f (0) = 0(2) (2)f ( x)=− cos x f (0) =−12x∴ P2( x) = 1−22(0.1)Thus, cos(0.1) ≈1− = 1− 0.005 = 0.995253.f (0) = 02 2 2f '( x) = cos x−2x sin xf '(0) = 1px ( ) = xp ; (0.2) = 0.2; f(0.2) = 0.1998586 Section 9.10 Instructor’s Resource Manual

x54. a. f ( x)= xe f(0) = 055.56.x xf ′( x)= e + xe f ′(0) = 1f ′′( x) = 2e x + xex f ′′(0) = 2(3) ( ) 3x xf x = e + xe(4) ( ) 4x xf x = e + xe1 1f ( x)≈ x+ x + x + x2 6f(0.1) ≈ 0.110522 3 4f(3) (0) = 3f(4) (0) = 4b. f(x) = cosh x f(0) = 1f ′( x) = sinhxf ′(0) = 0f ′′( x) = coshxf ′′(0) = 1f(3) ( x) = sinhxf(3) (0) = 0f(4) ( x) = coshxf(4) (0) = 11 2 1 4f ( x) ≈ 1+ x + x 2 24f(0.1) ≈ 1.00500423 2gx ( ) x –2x 5 x–7= + g(2) = 32g′ ( x) = 3 x –4x+ 5 g′ (2) = 9g′′ ( x) = 6 x–4g′′ (2) = 8g(3) ( x ) = 6g(3) (2) = 6(4)Since g ( x) = 0, R3( x) = 0, so the Taylorpolynomial of order 3 based at 2 is an exactrepresentation.2 3gx ( ) = P4( x) = 3+ 9( x–2) + 4( x–2) + ( x–2)2 3g (2.1) = 3 + 9(0.1) + 4(0.1) + (0.1) = 3.94158.59.60.(5) 120f ( x) = – ,6( x + 1)so5( x –1)R4 ( x) = – .6( c + 1)5 5(0.2) (0.2)R4 (1.2) = ≤ = 0.0000056 6( c + 1) (2)1f ( x) = (1–cos2 x)2f(0) = 0f ′( x) = sin2xf ′(0) = 0f ′′( x) = 2cos2xf ′′(0) = 2f(3) ( x) = –4sin2xf(3) (0) = 0f(4) ( x) = –8cos2xf(4) (0) = –8f(5) ( x) = 16sin2xf(5) (0) = 0f(6) ( x) = 32cos2xf(6) () c = 32cos2c2 2 2 8 4 2 1 4sin x ≈ x – x = x – x2! 4! 332 6 2 6R4( x) = R5( x) = (cos2 c) x ≤ (0.2)6! 45–6< 2.85×10( n+1) (–1) n!f ( x)=n+1xn(–1)Rn( x) = ( x–1)n+1( n+1) cn+ 1 n+10.2 (0.25)≤ =n+1( n + 1)0.8 ( n + 1)n+1(0.25)< 0.00005( n + 1)nn+1when n ≥ 5 .57.11f( x)= f (1) =x + 1211f′ ( x) = – f ′(1) = –( 1)2x +42f′′ ( x)=3( x + 1)(3) 6f ( x) = –( x + 1)(4) 24f ( x)=5( x + 1)41f ′′(1)=4(3) 3f (1) = – 8(4) 3f (1) =41 1 1 2f( x) ≈ – ( x–1) + ( x–1)2 4 81 3 1 4– ( x–1) + ( x–1)16 3261. From Problem 60,1 1ln x ≈ ( x–1)– ( x–1) + ( x–1)2 31 4 1 5– ( x–1) + ( x–1) .4 52 361 6 0.2–5R5 ( x) = ( x–1) ≤ < 4.07×106 66c6⋅0.81.2 1.2 ⎡ 1 2 1 3∫ ln xdx≈ ( –1)– ( –1) ( –1)0.8 ∫0.8⎢ x x + x⎣ 2 31 4 1 5⎤– ( x –1) + ( x–1)dx4 5 ⎥ ⎦⎡1 2 1 3 1 4= ⎢ ( x–1) – ( x–1) + ( x–1)⎣2 6 121.21 5 1 6⎤– ( x–1) + ( x–1)20 30 ⎥ ⎦ 0.8≈ –0.00269867–5 –5Error ≤ (1.2 – 0.8)4.07× 10 < 1.63×10Instructor’s Resource Manual Section 9.10 587

Review and Preview Problems1.2.2xxf( x ) = so that f′ ( x)= and f ′(2) = 142a. The tangent line will be the line through thepoint (2,1) having slope m = 1. Using the pointslope formula: ( y− 1) = 1( x− 2) or y = x − 1.b. The normal line will be the line through the1point (2,1) having slope m =− =− 1. Using1the point slope formula: ( y− 1) = −1( x− 2) ory =− x + 3 or x+ y = 3 .2x xf( x) = y = , f′( x)=4 2a. The line y = xhas slope = 1, so we seek xsuch that f′ ( x) = 1 or x = 2. The point is (2,1) .xb. Since f′ ( x)= is the slope of the tangent22line at the point ( , x 2x4) , − will be the slopexof the normal line at the same point. Since y = xhas slope 1, we seek x such that2− = 1 or x =− 2. The point is ( − 2,1) .x3. Solving equation 1 for2 2 9 2y : y 9 x 16= − andputting this result into equation 2:2x 1 ⎛ 9 2 ⎞+ ⎜9− x ⎟=19 16⎝16 ⎠2 2175x = 1008 x = 5.76 x =± 2.4Putting these values into equation 1 we get2 9y = 9 − (5.76) = 9 − 3.24 = 5.76 so y =± 2.416also. Thus the points of intersection are(2.4,2.4), ( −2.4, 2.4), (2.4, −2.4), ( −2.4, − 2.4)4. Solving equation 2 forputting this result into equation 1:22 2 2y : y = 9− x and2 2x x− = 016 9Thus x = 0 and y = 9, y =± 3 . Thus the pointsof intersection are (0,3), (0, − 3)5. Since we are given a point, all we need is theslope to determine the equation of our tangentline.2d ⎛2 y ⎞ dx() 1dx ⎜+ 4 ⎟=⎝ ⎠ dxydy2x+ = 02 dxydy=−2x2 dxdy −4x=dx y⎛At the point⎜−⎝( − )3 ⎞ ,12 ⎟, we get⎠dy −4 3/2= = 2 3 = mtandx 1Therefore, the equation of the tangent line to the⎛ 3 ⎞curve at⎜− ,12 ⎟is given by⎝ ⎠⎛ ⎛ 3 ⎞⎞y− 1=2 3x− −⎜⎜2 ⎟⎟⎝ ⎝ ⎠⎠y− 1= 2 3x+3y = 2 3x+46. Since we are given a point, all we need is theslope to determine the equation of our tangentline.2 2d ⎛ x y ⎞ d() 1dx ⎜− 9 16 ⎟=⎝ ⎠ dx2xy dy− = 09 8 dxydy 2x− =−8 dx 9dy 16x=dx9yAt the point ( 9,8 2 ) , we get( )dy 16 9 22 mdx = = = =tan98 ( 2)2Therefore, the equation of the tangent line to thecurve at ( 9,8 2 ) is given by( x )y− 8 2 = 2 −9y− 8 2 = 2x−9 2y = 2x−2588 Review and Preview Instructor’s Resource Manual

7. Denote the curves asx 2 y 2 2 21 Cx y2C : + = 1 and : − = 1100 64 9 272 2a. From C 2 ,3 x − 27 = y and so, from C 1 ,2 216x+ 25 3x− 27 = 1600( )2 291x = 2275, x = 25, x = ± 52 2y = 3(25) − 27, y = 48, y = ± 4 3Thus the point of intersection in the first quadrantis P = (5, 4 3) .b. Slope m 1 of the line tangent to C1at P:x y 16xC′ 1 : + y′ = 0, y′=− so m 1 =−50 32 25yThe line T 1 tangent to C1at P is4 3T1: ( y− 4 3) =− ( x− 5) or154 3 x+ 15y= 80 3Slope m 2 of the line tangent to C2at P:2 2 3xC ′2 : x− yy′ = 0, y′= so9 27ym 15 5 32 = 4 3= 4.The line T 2 tangent to C2at P is4 3155 3T2:( y− 4 3) = ( x− 5) or 5 3x− 4y= 9 34c. To find the angles between the tangent lines, youcan use problem 40 of section 0.7 or consider thediagram below:Note thatα + β2 + (180 − β1) = 180 or α = β1− β2;further−1 −1⎛4 3⎞β1 = 180 − tan m1= 180 − tan ⎜155.215 ⎟=⎝ ⎠−1 −1⎛5 3⎞β 2= tan m2= tan ⎜= 65.24 ⎟⎝ ⎠Thus α = 155.2 − 65.2 = 90 so the tangent linesare perpendicular. This can be verified by noting1 15 15 3 5 3that − = = = = m2m 4 3 12 418. t x = 2cost y = 2sint0 2 0π 6 3 1π 4 2 2π 3 1 3π 2 0 2π -2 03π2 0 -22 π 20Note that2 2 2 2x + y = (2cos t) + (2sin t)=2 24(cos t+ sin t) = 4so all the points will lie on the circle of radius 2that is centered at the origin.9. By the Pythagorean Theorem,r = 9+ 16 = 5. Since3 3 −1sin θ = = , θ = sin (0.6) = 36.9r 510. By the Pythagorean Theorem,r = 4+ 25 = 29 . Since2 2 2r = 3 + 4 or2 2 2r = 2 + 5 or2 2 −12 29sin θ = = , θ = sin ( ) = 21.8r 292911. Since the triangle is an isosceles right triangle,2 2 22x = y and x + y = 8 . Thus 2x = 64 andx = y = 32 = 4 2π 1 y 112. Since sin = , = or y = 6 . Further6 2 12 22 2 2 2x + y = 12 or x = 144 − 36 = 108 .Hence x = 108 = 6 3 .Instructor’s Resource Manual Review and Preview 589

CHAPTER 10Conics and PolarCoordinates10.1 Concepts Review1. e < 1; e = 1; e > 12.2y = 4 px3. (0, 1); y = –14. parallel to the axis3.2x = –4(3) yFocus at (0, –3)Directrix: y = 3Problem Set 10.11.2y = 4(1) xFocus at (1, 0)Directrix: x = –14.2x = –4(4) yFocus at (0, –4)Directrix: y = 42.2y = –4(3) xFocus at (–3, 0)Directrix: x = 35.2 ⎛1⎞y = 4 ⎜ ⎟ x⎝ 4 ⎠⎛1 ⎞Focus at ⎜ , 0 ⎟⎝4⎠1Directrix: x = – 4590 Section 10.1 Instructor’s Resource Manual

6.2y = –3x2 ⎛3⎞y = –4⎜ ⎟x⎝ 4 ⎠⎛ 3 ⎞⎜– , 0⎟⎝ 4 ⎠3x =4Focus atDirectrix:8.23x= 9y2 ⎛3⎞x = 4 ⎜ ⎟ y⎝ 4 ⎠⎛ 3 ⎞⎜0, ⎟⎝ 4 ⎠3y = – 4Focus atDirectrix:7.22x= 6y2 ⎛3⎞x = 4 ⎜ ⎟ y⎝ 4 ⎠⎛ 3 ⎞⎜0, ⎟⎝ 4 ⎠3y = – 4Focus atDirectrix:9. The parabola opens to the right, and p = 2.2y = 8x10. The parabola opens to the left, and p = 3.2y = –12x11. The parabola opens downward, and p = 2.2x = –8y12. The parabola opens downward, andx2 4= – y 91p = .913. The parabola opens to the left, and p = 4.y2= –16x14. The parabola opens downward, and2x = –14y7p = .2Instructor’s Resource Manual Section 10.1 591

15. The equation has the form2(–1) = 3 c.1c =3y2 1= x32y = cx, so18. The equation has the form2(–3) = 5c .9c = ⇒5x2 9= y52x = cy, so16. The equation has the form2(4) = –2 c.c = –8y2= –8x2y = cx, so19.2y = 16x2yy′ = 1616y′ =2yAt (1, –4), y′ = –2.Tangent: y + 4 = –2(x – 1) or y = –2x – 211 9Normal: y+ 4 = ( x–1)or y = x–22 217. The equation has the form2(6) = –5 c.36c = – 52x = cy, so20.2x = –10y2 x = –10y′xy′ = – 52 36x = – y 52 52 5, – 2 , y′ = – .5At ( )2 5Tangent: y 2 – ( x–2 5)2 5y = – x+25+ = or55Normal: y 2 ( x–2 5)+ = or y =25x – 72592 Section 10.1 Instructor’s Resource Manual

21.2x = 2y2x = 2y′y′ = xAt (4, 8), y′ = 4 .Tangent: y – 8 = 4(x – 4) or y = 4x – 811Normal: y– 8 = – ( x–4)or y = – x+9442 5 21 5y = – x–5 522.2y = –9x2 yy′ = –99y′ = – 2 y3At (–1, –3), y′ =233 3Tangent: y+ 3 = ( x+ 1) or y = x–22 222 11Normal: y+ 3 = – ( x+ 1) or y = – x–33 324.2x = 4y2x = 4y′xy′ =2At (4, 4), y′ = 2.Tangent: y – 4 = 2(x – 4) or y = 2x – 411Normal: y– 4 = – ( x–4)or y = – x+62225.2x = –6y2 x = –6y′xy′ = – 323.2y = –15x2 yy′ = –1515y′ = – 2 yAt ( 3 2, –3), y′ = – 2.Tangent: y 3 – 2 ( x–3 2)Normal: y 32( x–3 2)+ = or y = – 2x+ 3+ = or y =22x – 62–3, – 3 5 , y′ =5.2Tangent: y+ 3 5 =5( x+ 3) or2y =5 3 5x–2 22 5Normal: y+ 3 5 = – ( x+ 3) or5At ( )Instructor’s Resource Manual Section 10.1 593

26.27.28.2y = 20x2yy′ = 2010y′ =y10At ( 2, – 2 10 ), y′ = – .210Tangent: y+ 2 10 = – ( x–2)or210y = – x– 10210Normal: y+ 2 10 = ( x–2)or5y =2y = 5x2yy′ = 55y′ =2y10 12 10x–5 55y′ = when y = 2 5, so x = 4.4The point is ( 4, 2 5 ).2x = –14y2 x = –14y′xy′ = – 72 7y′ = – when x = 2 7, so y = –2.7The point is ( 2 7, –2).29. The slope of the line is 3 2 .2y = − 18 x; 2yy′= − 18⎛3⎞2y⎜⎟ = − 18; y = − 6⎝2⎠2( − 6) =− 18 xx ; =− 2The equation of the tangent line is3 3y+ 6 = ( x+ 2) or y= x–3.2 230. Place the x-axis along the axis of the parabola2such that the equation y = 4 px describes the⎛ 2y ⎞0parabola. Let, y0be one of the⎜ 4 p ⎟⎝ ⎠⎛ 2y ⎞extremities and1, y1be the other.⎜ 4 p ⎟⎝ ⎠First solve for y 1 in terms of y 0 and p. Since thefocal chord passes through the focus (p, 0), wehave the following relation.y1y0=2 2y1 y0– p – p4 p 4 p2 2 2 2y1( y0 –4 p ) = y0( y1–4 p )2 2 2 2y0y1 – ( y0 –4 p ) y1 –4p y 0 = 02( y1– y0)( y0y1+ 4 p ) = 024y1 = y0or y1= – py0⎛ 3 24p4p⎞Thus, the other extremity is, – .⎜ 2y y ⎟⎝ 0 0 ⎠2Implicitly differentiate y = 4 px to get2yy′ = 4 p,so y′ =2 p .y⎛ 2y ⎞02 pAt, y0, y′ = . The equation of the⎜ 4 p ⎟⎝ ⎠y022 p ⎛ y ⎞0tangent line is y– y0= x– . Wheny ⎜0 4 p ⎟⎝ ⎠22 p y0x = –p, y = – + .y 20⎛ 3 24p4p⎞ y0At, – , y′ = – . The equation of⎜ 2y y ⎟⎝ 0 0 ⎠2 pthe tangent line is2 34py ⎛0 4p⎞y+ = – x– .y 20 2 p⎜y ⎟⎝ 0 ⎠594 Section 10.1 Instructor’s Resource Manual

When x = –p,2y02 py = – .2 y0Thus, the two tangent lines intersect on the⎛2y02 p ⎞directrix at– p, – .⎜ 2 y ⎟⎝0 ⎠31. From Problem 30, if the parabola is described by2the equation y = 4 px,the slopes of the tangent2 p y0lines are and – . Since they are negativey02 preciprocals, the tangent lines are perpendicular.32. Place the x-axis along the axis of the parabola2such that the equation y = 4 px describes the⎛ 1 ⎞parabola. The endpoints of the chord are ⎜1, ⎟⎝ 2 ⎠2⎛ 1 ⎞and ⎜1, – ⎟,⎝ 2 ⎠ so ⎛1⎞1⎜ ⎟ = 4(1) p or p = . The⎝2⎠16distance from the vertex to the focus is 1 .1633. Assume that the x- and y-axes are positioned suchthat the axis of the parabola is the y-axis with thevertex at the origin and opening upward. Then2the equation of the parabola is x = 4 py and(0, p) is the focus. Let D be the distance from apoint on the parabola to the focus.22⎛2 2 2 x ⎞D = ( x− 0) + ( y− p)= x + − p⎜4p ⎟⎝ ⎠4 2 2= x x 2 xp p216 p+ 2 + = 4p+x xD′ = ; = 0, x = 02p2pD′′ > 0 so at x = 0, D is minimum. y = 0Therefore, the vertex (0, 0) is the point closest tothe focus.34. Let the y-axis be the axis of the parabola, soEarth’s coordinates are (0, p) and the equation of2the path is x = 4 py , where the coordinates arein millions of miles. When the line from Earth tothe asteroid makes an angle of 90 with the axisof the parabola, the asteroid is at (40, p).2= p p p =(40) 4 ( ), 20The closest point to Earth is (0, 0), so the asteroidwill come within 20 million miles of Earth.35. Let the y-axis be the axis of the parabola, so theEarth’s coordinates are (0, p) and the equation of2the path is x = 4 py , where the coordinates arein millions of miles. When the line from Earth tothe asteroid makes an angle of 75 with the axisof the parabola, the asteroid is at( 40sin 75 , p + 40cos 75 ). (See figure.)2° = p p+ °(40sin75 ) 4 ( 40cos75 )2 2p + p °− °=40 cos 75 400sin 75 02 2− 40cos 75°± 1600cos 75°+ 1600sin 75°p == − 20cos 75°±20p = 20 − 20cos75°≈ 14.8 ( p > 0)The closest point to Earth is (0, 0), so the asteroidwill come within 14.8 million miles of Earth.236. Let the equation x = 4 py describe the cables.The cables are attached to the towers at(±400, 400).2(400) = 4 p(400), p = 100The vertical struts are at x =±300.2= y y =(300) 4(100) , 225The struts must be 225 m long.37. Let RL be the distance from R to the directrix.Observe that the distance from the latus rectum tothe directrix is 2p so RG = 2 p – RL . From thedefinition of a parabola, RL = FR . Thus,FR + RG = RL + 2 p– RL = 2 p.2Instructor’s Resource Manual Section 10.1 595

38. Let the coordinates of P be ( x0, y 0).2 p2yy′ = 4 p,so y′ = . Thus the slope of theyy0pnormal line at P is – .2The equation of the normal line isy0y– y0 = – ( x– x0).When y = 0,2 px = 2 p+ x0, so B is at (2 p+ x0, 0). A is at( x 0, 0). Thus, AB = 2 p+ x0 – x0= 2 p.40. Let C denote the center of the circle and r theradius. Observe that the distance from a point Pto the circle is PC – r . Let l be the line andPL the distance from the point to the line. Thus,PC – r = PL . Let l ' be the line parallel to l, runits away and on the side opposite from thecircle. Then PL′ , the distance from P to l ' , isPL + r; so PL = PL′– r.Therefore,PC – r = PL′– r or PC = PL′ . The set ofpoints is a parabola by definition.39. Let P 1 and P 2 denote ( x1, y 1)and ( x2, y 2),respectively. PP 1 2= PF 1 + P2Fsince the focalchord passes through the focus. By definition of aparabola, PF 1 = p+ x1and PF 2 = p+x2.Thus, the length of the chord isPP 1 2= x1+ x2+2 p.L = p+ p+ 2p = 4p41.dy δ x=dx Hy2δ x= +2HCδ xy(0) = 0 implies that C = 0. y =2HThis is an equation for a parabola.242. a. A( T 1)is the area of the trapezoid formed by2(, cc ), (c, 0) and by (c, 0),(a, 0), P, Q, (b, 0) minus the area the two trapezoids formed by (a, 0), P,a+ b b−c c−a b−aQ, (b, 0). Observe that since c = , = = .2 2 2 4b−a 2 2 c−a 2 2 b−c2 2 b−a 2 2 b−a2 2 2A( T 1)= [ a + b ] − [ a + c ] − [ c + b ] = [ a + b ] − [ a + 2 c + b ]2 2 22 4b 2– a 2 2 – 2 2[ ]– b a ⎡ ⎛2a +a b a b ⎞ ⎤= + ⎢ + ⎜ ⎟ + b ⎥b 2 2– a ⎡2 2– a – –b ⎤= ⎢a + b ab ⎥2 4 ⎢ ⎝ 2⎣⎠ ⎥ 4 2 2⎦ ⎢⎣⎥⎦b 2 2 3– a ⎛ a ( – )–b ⎞= ab + =b a4 ⎜ 2 2 ⎟⎝⎠82(, cc ),b.b– a b–a( ) ( )3 33 3( c– a) ( b– c)2 2AT ( 2)= + = +8 8 8 83( b– a)= =A( T1)32 4AT ( n–1)A( T1)c. Using reasoning similar to part b, AT ( n ) = , so AT ( n ) = .4n–14∞AT ( 1) ⎛ 1 ⎞ 4AS ( ) = AT ( 1) + AT ( 2) + AT ( 3)+… = ∑ (–1 1) (1 1)n 1 4 n = A T= A T⎜=1– ⎟ 3⎝ 4 ⎠d.2 2 3+b– a 2 2 ( b– a)( a b ) ( b– a)Area = [ a + b ] – A( S) =–2 2 6b– a ( b– a)1 3 3 b a[3a 3 b – ( b – 2 ab a )] [2a 2ab 2 b ] = ( b – a ) = –6 63 3 32 2 2 2 2 2= + + = + +3 3596 Section 10.1 Instructor’s Resource Manual

43.Using a CAS, we find that the largest verticaldistance between the catenary and theparabola is roughly 67 feet.44. a. Since the vertex is on the positive y-axis andthe parabola crosses the x-axis, its equation is2of the form: x = 4 py ( − k), where k is they-coordinate of the vertex; that is2x = 4 p( y− 630) . Since the point (315,0)is on the parabola, we have2 −315(315) = 4 p(0 − 630) or 4 p = = −157.52Thus the parabola has the equationx2=− −157.5( y 630)b. Solving for y, we get1 2y =− x + 630157.5The catenary for the Gateway Arch isxy = 758 − 128cosh .128650−350 0350xy = 758 −128cosh .1281 2y =− x + 630157.5c. Because of symmetry, we can focus on thelargest vertical distance between the graphsfor positive x.f x = y − y . That is,Let ( ) Arch parabolaf( x)⎛x ⎞= ⎜758 −128cosh ⎟⎝128 ⎠⎛ 1 2 ⎞−⎜− x + 630⎟⎝ 157.5 ⎠x 1 2=− 128cosh + x + 128128 157.510.2 Concepts Review1.2.xa3. foci2 2y+ = 1b2 22 2x y+ = 19 164. to the other focus; directly away from the otherfocusProblem Set 10.21. Horizontal ellipse2. Horizontal hyperbola3. Vertical hyperbola4. Horizontal hyperbola5. Vertical parabola6. Vertical parabola7. Vertical ellipse8. Horizontal hyperbola9.2 2x y+ = 1; horizontal ellipse16 4a = 4, b = 2, c = 2 3Vertices: (±4, 0)Foci: ( ± 2 3, 0)Instructor’s Resource Manual Section 10.2 597

10.2 2x y– 1;16 4= horizontal hyperbolaa = 4, b = 2, c = 2 5Vertices: (±4, 0)Foci: ( ± 2 5, 0)Asymptotes:1y =± x213.2 2x y+ = 1; vertical ellipse2 8a = 2 2, b= 2, c = 6Vertices: (0, ± 2 2)Foci: ( 0, ± 6 )11.2 24–91;y x = vertical hyperbolaa = 2, b = 3, c = 13Vertices: (0, ±2)Foci: ( 0, ± 13)14.2 2x y+ = 1; horizontal ellipse25 4a = 5, b = 2, c = 21Vertices: (±5, 0)Foci: ( ± 21, 0)Asymptotes:2y =± x312.2 2x y+ = 1; horizontal ellipse7 4a = 7, b= 2, c = 3Vertices: ( ± 7, 0)Foci: ( ± 3, 0)15.2 2x y– 1;10 4= horizontal hyperbolaa = 10, b= 2, c = 14Vertices: ( ± 10, 0)Foci: ( ± 14, 0)Asymptotes:y =±2x10598 Section 10.2 Instructor’s Resource Manual

16.2 2x y– 1;8 2= horizontal hyperbolaa = 2 2, b= 2, c = 10Vertices: ( ± 2 2, 0)Foci: ( ± 10, 0)Asymptotes:1y =± x222. This is a horizontal hyperbola with a = 4 andc = 5.b = 25 –16 = 32 2x y– = 116 923. This is a vertical hyperbola with a = 4 and c = 5.b = 25 –16 = 32 2y x– = 116 924. This is a vertical hyperbola with a = 3.⎛3⎞9 81 45c = ae= 3 ⎜ ⎟= , b = − 9 =⎝2⎠2 4 22 2y x− = 19 45417. This is a horizontal ellipse with a = 6 and c = 3.b = 36 − 9 = 272 2x y+ = 136 2718. This is a horizontal ellipse with c = 6.c 6a = = = 9, b= 81− 36 = 45e 232 2x y+ = 181 4519. This is a vertical ellipse with c = 5.c 5a = = = 15, b = 225 – 25 = 200e 132 2x y+ = 1200 22520. This is a vertical ellipse with b = 4 and c = 3.a = 16 + 9 = 52 2x y+ = 116 2521. This is a horizontal ellipse with a = 5.2 2x y+ = 125 2b4 9+ = 125 2b2 225b =212 2x y+ = 1252252125. This is a horizontal hyperbola with a = 8.1 b 1The asymptotes are y =± x , so = or b = 4.2 8 226.2 2x y– = 164 16c = ae= 6 2 2 2 3 2 2 1 2a, b = c – a = a – a = a2 2 22 2y x−2 1 2a 2a= 116 4− = 12 1 2a 2a2a = 82 2y x– = 18 427. This is a horizontal ellipse with c = 2.a a 28 = , 8 = , so a = 8c= 16.ecab = 16 − 4 = 122 2x y+ = 116 1228. This is a horizontal hyperbola with c = 4.a a 21 = , 1 = , so a = c = 4.ecab = 16 − 4 = 122 2x y− = 14 12Instructor’s Resource Manual Section 10.2 599

129. The asymptotes are y =± x . If the hyperbola is2b 1horizontala = 2or a = 2b. If the hyperbola isa 1vertical, = or b = 2a.b 2Suppose the hyperbola is horizontal.30.2 2x y– 12 24b b =16 9– 12 24b b =2b = –5This is not possible.Suppose the hyperbola is vertical.2 2y x–2 2a 4a = 19 16–2 2a 4a = 12a = 52 2y x– = 15 202 2x y+ = 12 2a b25 1 ⎫+ = 12 2 ⎪⎪a b 84⎬ – = –316 4 2+ = 1⎪a2 2a b ⎪⎭2a = 282 28b =32 2x y+ = 128 28331. This is an ellipse whose foci are (0, 9) and (0, - 9)and whose major diameter has length 2a = 26.Since the foci are on the y-axis, it is the major axisof the ellipse so the equation has the form2 2y x2+ = 1. Since 2a= 26, a = 169 and since2 2a b2 2 2 2 2a b c , b 169 (9) 882 2y xequation is + = 1169 88= + = − = . Thus the32. This is an ellipse whose foci are (4, 0) and (- 4, 0)and whose major diameter has length 2a = 14.Since the foci are on the x-axis, it is the major axisof the ellipse so the equation has the form2 2x y2+ = 1. Since 2a= 14, a = 49 and since2 2a b2 2 2 2 2a b c , b 49 (4) 332 2x yequation is + = 149 33= + = − = . Thus the33. This is an hyperbola whose foci are (7, 0) and(- 7, 0) and whose axis is the x-axis. So the2 2x yequation has the form − = 1. Since2 2a b2 2 2 2 22a = 12, a = 36 and b = c − a = (7 ) − 36 = 13Thus the equation is2 2x y− = 136 1334. This is an hyperbola whose foci are (0, 6) and(0, - 6) and whose axis is the y-axis. So the2 2y xequation has the form − = 1. Since2 2a b2 2 2 2 22a = 10, a = 25 and b = c − a = (6 ) − 25 = 11Thus the equation is2 2y x− = 125 1135. Use implicit differentiation to find the slope:2 2x+ y y′ = 0 . At the point27 92 2 6 6(3, 6), + y′ = 0, or y′= − so the9 9 6equation of the tangent line is6( y− 6) =− ( x− 3) or x+ 6y= 9.636. Use implicit differentiation to find the slope:1 yx+ y′ = 0 . At the point12 82 1(3 2, −2), − 0, or 24 4 y′ = y′= so theequation of the tangent line is( y+ 2) = 2( x− 3 2) or 2 x− y = 8.600 Section 10.2 Instructor’s Resource Manual

37. Use implicit differentiation to find the slope:2 2x+ y y′ = 0 . At the point27 92 2 6 6(3, − 6), − y′ = 0, or y′= so the9 9 6equation of the tangent line is6( y+ 6) =− ( x− 3) or x− 6y= 9 .638. Use implicit differentiation to find the slope:1x− y y′ = 0 . At the point ( 3, 2)223− 0, or 62 y′ = y′= so the equation of thetangent line is ( y− 2) = 6( x− 3) or6x− 6y= 4 3 .39. Use implicit differentiation to find the slope:2x+ 2y y′ = 0. At the point (5,12)510 + 24y′ = 0, or y′=− so the equation of the125tangent line is ( y− 12) =− ( x− 5) or125x+ 12y= 169.40. Use implicit differentiation to find the slope:2x− 2y y′ = 0. At the point ( 2, 3)62 2− 2 3y′ = 0, or y′= so the equation of36the tangent line is ( y− 3) = ( x− 2) or33y− 6 x = 3.41. Use implicit differentiation to find the slope:1 2x+ y y′ = 0 . At the point44 1692(0,13), 0, or 013 y′ = y′= . The tangent line ishorizontal and thus has equation y = 13 .42. Use implicit differentiation to find the slope:2 2x+ y y′ = 0 . At the point (7,0)49 332+ 0 y′ = 0, or y′is undefined. . The tangent line7is vertical and thus has equation x = 7 .43. Let the y-axis run through the center of the archand the x-axis lie on the floor. Thus a = 5 and2 2x yb = 4 and the equation of the arch is + = 1.25 162 2x (2) 5 3When y = 2, + = 1, so x =± .25 16 2The width of the box can at most be5 3 ≈ 8.66 ft.44. Let the y-axis run through the center of the archand the x-axis lie on the floor.The equation of the arch isWhen x = 2,2 2x y+ = 1.25 162 2(2) y4 21+ = 1, so y = ± .25 16 5The height at a distance of 2 feet to the right of thecenter is 4 21 ≈ 3.67 ft.545. The foci are at (±c, 0).2 2c = a – b2 2 2a – b y+ = 12 2a b4 22 b2 , by = y = ±a aThus, the length of the latus rectum is46. The foci are at (±c, 0)2 2c = a + b2 2 2a + b y– = 12 2a b4 22 b2 , by = y = ±a aThus, the length of the latus rectum is47. a = 18.09, b = 4.56,22b.a22b.a2 2c = (18.09) −(4.56) ≈ 17.51The comet’s minimum distance from the sun is18.09 – 17.51 ≈ 0.58 AU.48. a− c = 0.13, c = ae, a(1 − e) = 0.13,0.13a =≈17331−0.999925a+ c = a(1 + e) ≈ 1733(1 + 0.999925) ≈ 3466 AUInstructor’s Resource Manual Section 10.2 601

49. a – c = 4132; a + c = 45832a = 8715; a = 4357.5c = 4357.5 – 4132 = 225.5c 225.5e = = ≈ 0.05175a 4357.550. (See Example 5) Since a+ c = 49.31 anda− c = 29.65 , we conclude that2a= 78.96, 2c= 19.66 and soa = 39.48, c = 9.83 . Thus51.52.53.2 2b a c 1462.0415 38.24major diameter 2a78.96diameter = 2b= 76.48 .= − = ≈ . So the= = and the minor2 2x y+ = 14 9Equation of tangent line at ( x0, y 0):xx0 yy0+ = 14 93At (0, 6), y 0 = .223 x 1When y = , + = 1, x =± 3.2 4 4⎛ 3 ⎞The points of tangency are ⎜ 3, ⎟ 2⎛⎜−⎝3 ⎞3, ⎟ 2 ⎠2 2x y– = 14 36Equation of tangent line at ( x0, y 0):xx0 yy 0 =– 14 36At (0, 6), y 0 = –6 .When y = –6,⎝⎠ and2x 36– = 1, x =± 2 2.4 36The points of tangency are ( 2 2, –6)and( − 2 2, –6).2 22x 7y 35 0; 4x 14yy′0− − = − =2 2 2 7y′ = x ; − = x ; x =−y7y3 7y37ySubstitute x =− into the equation of the3hyperbola.98 2 27 35 0, 39 y − y − = y =±The coordinates of the points of tangency are7, 3 7, − 3 .( − ) and ( )154. The slope of the line is2 .2 2x + 2y − 2= 0; 2x+ 4yy′= 055.56.x 1 xy′ =− ; =− ; x =− 2 y2y2 2ySubstitute x =− 2yinto the equation of theellipse.2 2 12y + 2y − 2= 0; y =±2⎛ 1 ⎞The tangent lines are tangent at ⎜−1,⎟⎝ 2 ⎠ and⎛ 1 ⎞⎜1,− ⎟. The equations of the tangent lines are⎝ 2 ⎠1 1 1 1y− = ( x+ 1) and y+ = ( x− 1) or2 22 2x− 2y+ 2= 0 and x− 2 y− 2= 0.2y =± b 1−x2aa2xA = 4b∫1−dx0 2aLet x = a sin t then dx = a cos t d t. Then the limitsπare 0 and . 224 a1 x/2 20 24 πA = b∫ − dx = ab∫cos tdta0π /2π /2⎡ sin 2t⎤= 2 ab∫(1+ cos2 t) dt = 2ab t0⎢ +2⎥⎣ ⎦0= π ab2x =± a 1−y2b⎛ ⎞ ⎡ ⎤⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣⎢⎦⎥b2 32 y 2 yV = 2⋅π∫a 1− dy = 2πa y−0 ⎜ 2 2b ⎟3b=24πa b3b0602 Section 10.2 Instructor’s Resource Manual

57.58.2xy =± b − 1a2The vertical line at one focus is22 2a + b ⎛ 2x⎞V =π ⎜b 2 –1 ⎟∫dxa ⎜ a ⎟⎝ ⎠2 2x = a + b .2 22 2 22 a + b ⎛ xb2 –1⎞3a + b⎡2 x ⎤= π∫dx = b π xa ⎜a⎟⎢ − ⎥2⎝ ⎠ ⎢⎣3a ⎥⎦a⎡ 2 2 3/22 ( a + b ) 2 2 2 ⎤= b π ⎢– a + b + a⎥2⎢⎣3a3 ⎥⎦2πb2 2 3/2 2 2 2 3=⎡23( a + b ) –3 a a + b + 2 a⎤a⎢⎣⎥⎦y =± b21– x2a2a ⎛ 2x⎞V = 2⋅π ⎜b 1– ⎟∫dx0 ⎜ 2a ⎟⎝ ⎠2 3a2 a ⎛ x ⎞ ⎡2 x ⎤= 2π b ∫1– dx = 2πb x−0 ⎜ 2 2a ⎟⎢ ⎥⎢ 3a= 4 2⎝ ⎠ ⎣ ⎦⎥3 π ab059. If one corner of the rectangle is at (x, y) the sideshave length 2x and 2y.2x =± a 1−y2b2 4y2 yA= 4xy = 4ya 1− = 4a y −2 2bb34 y2a⎛2y⎞2dA⎜ − ⎟b dA=⎝ ⎠; = 0 whendy42 y dyy −2b32yy − = 02b⎛ 22y⎞y1− = 0⎜ 2b ⎟⎝ ⎠by = 0 or y =±2The Second Derivative Test shows thata maximum.b( ) 222x = a 1− =baTherefore, the rectangle is a 2 by b 2 .2by = is260. Position the x-axis on the axis of the hyperbola2 2x ysuch that the equation – = 1 describes the2 2a bhyperbola. The equation of the tangent line atxx 0 yy 0( x0, y 0)is – = 1. The equations of the2 2a bbasymptotes are y =± x.abSubstitute y = xaand y = – b x into theaequation of the tangent line.xx 0 02 – yx=a ab1 xx 0 yx 012a + ab=⎛bx0 – ay0⎞ ⎛bxx=0 + ay0⎞⎜12 ⎟ x⎜12 ⎟ =⎝ ab ⎠⎝ ab ⎠22ababx =x =bx0 – ay0bx0 + ay0Thus the tangent line intersects the asymptotes at⎛ 2 2ab ab ⎞,and⎜bx 0 – ay0 bx0 – ay ⎟⎝0 ⎠⎛ 2 2ab ab ⎞, – .⎜bx 0 ay0 bx0 ay ⎟⎝+ + 0 ⎠2 2 2 2Observe that b x0 – a y0= a b .2 21 ⎛ ab ab ⎞+2 ⎜bx 0 – ay0 bx0 ay ⎟⎝+ 0 ⎠2 2abx 0x2 2 2 2 00 – a y0= =b x2 21 ⎛ ab ab ⎞ aby–= = y2 ⎜bx 0 – ay0 bx0 + ay ⎟ 2 2 2 2⎝0 ⎠ b x0 – a y0Thus, the point of contact is midway between thetwo points of intersection.61. Add the two equations to gety =± 5 32 20029y = 675.Substitute y = 5 3 into either of the twoequations and solve for x ⇒ x = ±6The point in the first quadrant is ( 6, 5 3 ).Instructor’s Resource Manual Section 10.2 603

62. Substitute x = 6 – 2y into63.2 2(6 – 2 y) + 4y= 2028 y – 24y+ 16 = 0x2 2+ 4y= 20.y 2 – 3y+ 2=0(y – 1)(y – 2) = 0y = 1 or y = 2x = 4 or x = 2The points of intersection are (4, 1) and (2, 2).68.AP AB BP= +u v u2ucAP – BP =vThus the curve is the right branch of the horizontaluc ⎛ 2u ⎞hyperbola with a = , so b = 1– c.v ⎜ 2v ⎟⎝ ⎠The equation of the curve is2 2x y ⎛ uc⎞– = 1 x .2 2 2 ⎜ ≥ ⎟2c ⎝ v ⎠ucv( 1–u)2 2v64. If the original path is not along the major axis, theultimate path will approach the major axis.69. Let P(x, y) be the location of the explosion.3 AP = 3 BP + 12AP – BP = 4Thus, P lies on the right branch of the horizontalhyperbola with a = 2 and c = 8, so b = 2 15.2 2x y– = 14 60Since BP = CP , the y-coordinate of P is 5.2x 25 17– = 1, x = ±4 60 3⎛ 17 ⎞P is at⎜ , 5 .3 ⎟⎝ ⎠65. Written response. Possible answer: the ball willfollow a path that does not go between the foci.66. Consider the following figure.70.2 2lim⎛x a x⎞⎜ − − ⎟x→∞⎝⎠⎡⎛ 2 2 2 2x a x⎞ ⎛x a x⎞⎤⎢⎜ − − ⎟ ⎜ − + ⎟⎥= lim⎝ ⎠ ⎝ ⎠⎢⋅⎥x→∞⎢ 1 ⎛ 2 2x a x⎞⎜ − + ⎟⎥⎢⎣⎝⎠⎥⎦2−a= lim = 0x→∞2 2x − a + x71. 2a = p + q, 2 c = p–q2 22 2 2 ( p+q) ( p– q)b = a – c = – = pq4 4b = pqObserve that 2(α + β) = 180°, so α + β = 90°. Theellipse and hyperbola meet at right angles.67. Possible answer: Attach one end of a string to Fand attach one end of another string to F′ . Place aspool at a vertex. Tightly wrap both strings in thesame direction around the spool. Insert a pencilthrough the spool. Then trace out a branch of thehyperbola by unspooling the strings while keepingboth strings taut.72. x = acos t, y = asin t− bsin t = ( a−b)sintx ycos t = , sin t =a a − b2 2x y+ = 12 2a ( a−b)Thus the coordinates of R at time t lie on anellipse.604 Section 10.2 Instructor’s Resource Manual

73. Let (x, y) be the coordinates of P as the ladderslides. Using a property of similar triangles,2 2x b − y= .a b75. The equations of the hyperbolas areand2 2y x– 1.2 2b a =2 2c a + be = =a a2 2c a + bE = = b b2 2–2 –2 a be + E = + = 12 2 2 2a + b a + b2 2x y2 2a b =– 1Square both sides to get2 2 2x b − y 2 2 2 2 2 2= or b x + a y = a b or2 2a b2 2x y+ = 1.2 2a b76. Position the x-axis on the plane so that it makesthe angle φ with the axis of the cylinder and they-axis is perpendicular to the axis of the cylinder.(See the figure.)74. Place the x-axis on the axis of the hyperbola such2 2x ythat the equation is – = 1. One focus is at2 2a bb(c , 0) and the asymptotes are y =± x.Theaequations of the lines through the focus,perpendicular to the asymptotes, areay =± ( x– c).Then solve for x inbb ax = – ( x– c).a b2 2a + b acx =ab b2acx =2 2a + b22 2 2 aSince c = a + b , x = . The equation of thec2adirectrix nearest the focus is x = , so the linecthrough a focus and perpendicular to an asymptoteintersects that asymptote on the directrix nearestthe focus.77.2 2 2If P(x, y) is a point on C, ( xsin φ ) + y = rwhere r is the radius of the cylinder. Then2 2x y+ = 1.2r 2r2sin φWhen a < 0, the conic is an ellipse. When a > 0,the conic is a hyperbola. When a = 0, the graph istwo parallel lines.Instructor’s Resource Manual Section 10.2 605

10.3 Concepts Review1.2a42. 14; ellipse3.A − CB4. rotation, translationProblem Set 10.31.2.3.4.5.6.x2 + y2 – 2x+ 2y+1=02 2( x –2x+ 1) + ( y + 2y+ 1) = –1+ 1+12 2( x–1) + ( y+ 1) = 1This is a circle.2 2x + y + 6 x–2y+ 6=02 2( x + 6x+ 9) + ( y –2y+ 1) = –6+ 9+12 2( x+ 3) + ( y–1) = 4This is a circle.2 29x + 4y + 72 x–16y+ 124 = 02 29( x + 8x+ 16) + 4( y – 4y+ 4) = –124 + 144 + 162 29( x+ 4) + 4( y– 2) = 36This is an ellipse.2 216x − 9y + 192x+ 90y− 495 = 02 216( x + 12x+ 36) −9( y − 10y+25)= 495 + 576 −2252 216( x+ 6) −9( y− 5) = 846This is a hyperbola.2 29x + 4y + 72 x–16y+ 160 = 02 29( x + 8x+ 16) + 4( y – 4y+ 4) = –160 + 144 + 162 29( x+ 4) + 4( y– 2) = 0This is a point.2 216x + 9y + 192x+ 90y+ 1000 = 02 216( x + 12x+ 36) + 9( y + 10y+25)= –1000 + 576 + 2252 216( x+ 6) + 9( y+ 5) = –199This is the empty set.7.8.9.10.11.12.13.y 2 – 5 x–4 y –6=02( y –4y+ 4) = 5x+ 6+42( y– 2) = 5( x+2)This is a parabola.2 24x + 4y + 8 x–28 y–11=02 ⎛ 2 49 ⎞4( x + 2x+ 1) + 4 ⎜y – 7 y+ ⎟= 11+ 4 + 49⎝ 4 ⎠22 ⎛ 7 ⎞4( x+ 1) + 4 ⎜y– ⎟ = 64⎝ 2 ⎠This is a circle.2 23x + 3 y – 6x+ 12y+ 60 = 02 23( x – 2x+ 1) + 3( y + 4 y+ 4) = –60 + 3 + 122 23( x–1) + 3( y+ 2) = –45This is the empty set.2 24 x –4 y –2x+ 2y+ 1=0⎛ 2 1 1 ⎞ ⎛ 2 1 1 ⎞ 1 14 ⎜x – x+ ⎟–4 ⎜y – y+ ⎟= –1 + –⎝ 2 16⎠ ⎝ 2 16⎠4 42 21 14 ⎜ ⎛ x– ⎟ ⎞ –4 ⎜ ⎛ y– ⎟⎞ = −1⎝ 4⎠ ⎝ 4⎠2 21 14 ⎛ ⎜y– ⎞ ⎟ –4 ⎛ ⎜x– ⎞⎟ = 1⎝ 4⎠ ⎝ 4⎠This is a hyperbola.2 24 x –4y + 8x+ 12 y–5=02 ⎛ 2 9 ⎞4( x + 2x+ 1) – 4 ⎜y – 3y+ ⎟= 5 + 4 – 9⎝ 4 ⎠22 ⎛ 3 ⎞4( x+ 1) – 4 ⎜y– ⎟ = 0⎝ 2 ⎠This is two intersecting lines.2 24 x –4y + 8x+ 12 y–6=02 ⎛ 2 9 ⎞4( x + 2x+ 1) – 4 ⎜y – 3y+ ⎟= 6 + 4 – 9⎝ 4 ⎠22 ⎛ 3 ⎞4( x+ 1) – 4 ⎜y– ⎟ = 1⎝ 2 ⎠This is a hyperbola.24 x –24x+ 36=024( x – 6x+ 9) = –36 + 3624( – 3) 0 x =This is a line.606 Section 10.3 Instructor’s Resource Manual

14.24 x –24x+ 35=024( x – 6x+ 9) = –35 + 3624( x – 3) = 1This is two parallel lines.19.2( x+ 2) = 8( y−1)15.2 2( x+ 3) ( y+2)+ = 14 1620.2( x + 2) = 4x + 2 = ±2x = –4, x = 016.2 2( x+ 3) + ( y− 4) = 2521.2( y − 1) = 16y – 1 = ±4y = 5, y = –317.2 2( x+ 3) ( y+2)− = 14 1618.4( x+ 3) = ( y+2)222.2 2( x+ 3) ( y−2)+ = 04 8(–3, 2)Instructor’s Resource Manual Section 10.3 607

23.2 2x + 4y − 2x+ 16y+ 1=02 2( x − 2x+ 1) + 4( y + 4y+ 4) =− 1+ 1+162 2( x− 1) + 4( y+ 2) = 162 2( x− 1) ( y+2)+ = 116 426.2 2x − 4y −14x−32y− 11=02 2( x − 14x+ 49) − 4( y + 8y+ 16) = 11+ 49 − 642 24( y+ 4) −( x− 7) = 42 ( x − 7)( y + 4) − = 14224.2 225x + 9y + 150x− 18y+ 9 = 02 225( x + 6x+ 9) + 9( y − 2y+ 1) =− 9 + 225 + 92 225( x+ 3) + 9( y− 1) = 2252 2( x+ 3) ( y−1)+ = 19 2527.24x + 16x− 16y+ 32=024( x + 4x+ 4) = 16y− 32 + 1624( x+ 2) = 16( y−1)2( x+ 2) = 4( y−1)25.2 29x − 16y + 54x+ 64y− 127 = 02 29( x + 6x+ 9) −16( y − 4y+ 4) = 127 + 81−642 29( x+ 3) −16( y− 2) = 1442 2( x+ 3) ( y−2)− = 116 928.2x − 4x+ 8y= 02x − 4x+ 4= − 8y+4⎞( x− 2) = −8⎜y−⎟⎝ 2 ⎠2 ⎛ 1608 Section 10.3 Instructor’s Resource Manual

29.30.31.32.22y −4y− 10x= 022( y − 2y+ 1) = 10x+2⎞( y− 1) = 5⎜x+⎟⎝ 5 ⎠2 ⎛5⎞⎛ 1⎞( y− 1) = 4⎜ ⎟⎜x+⎟⎝ 4 ⎠⎝ 5 ⎠Horizontal parabola,2 ⎛ 15p =4⎛ 1 ⎞Vertex ⎜−, 1 ⎟⎝ 5 ⎠ ; Focus is at ⎛ 21 ⎞⎜ , 1 ⎟⎝20⎠ anddirectrix is at x =− 29 .202 2− x + x+ y + y =9 18 4 24 92 2− x − x+ + y + y+ = − +9( 2 1) 4( 6 9) 9 9 362 2y+ − x− =4( 3) 9( 1) 362 2y+ x−− =( 3) ( 1)9 42a = 9, a = 31The distance between the vertices is 2a = 6.2 216( x− 1) + 25( y+ 2) = 4002 2( x− 1) ( y+2)+ = 125 16Horizontal ellipse, center (1, –2), a = 5, b = 4,c = 25 − 16 = 3Foci are at (–2, –2) and (4, –2).2x − 6x+ 4y+ 3=02x − 6x+ 9= −4y− 3+92 ⎛ 3 ⎞( x− 3) = −4⎜y−⎟⎝ 2 ⎠Vertical parabola, opens downward, vertex⎛ 3 ⎞⎜3, ⎟, p = 1⎝ 2 ⎠⎛ 1 ⎞Focus is at ⎜3, ⎟⎝ 2 ⎠ and directrix is 5y = .233. a = 5, b = 42 2( x−5) ( y−1)+ = 125 1634. Horizontal hyperbola, a = 2, c = 3,b = 9− 4 = 52 2( x− 2) ( y+1)− = 14 535. Vertical parabola, opens upward, p = 5 – 3 = 22( x− 2) = 4(2)( y−3)2( x− 2) = 8( y−3)36. An equation for the ellipse can be written in the2 2( x−2) ( y−3)form + = 1.2 2a bSubstitute the points into the equation.16 4= 1, = 12 2a bTherefore, a = 4 and b = 2.2 2( x−2) ( y−3)+ = 116 437. Vertical hyperbola, center (0, 3), 2a = 6, a = 3,c = 5, b = 25 − 9 = 42 2( y−3) x− = 19 1638. Vertical ellipse; center (2, 6), a = 8, c = 6,b = 64 − 36 = 282 2( x−2) ( y−6)+ = 128 6439. Horizontal parabola, opens to the left10 − 2Vertex (6, 5), p = = 422( y− 5) =−4(4)( x−6)2( y− 5) =−16( x−6)40. Vertical parabola, opens downward, p = 12( x− 2) =−4( y−6)41. Horizontal ellipse, center (0, 2), c = 2Since it passes through the origin and center is at(0, 2), b = 2.a = 4+ 4 = 82 2x ( y− 2)+ = 18 442. Vertical hyperbola, center (0, 2), c = 2,2 2b = 4 − aAn equation for the hyperbola can be written in2 2( y−2) xthe form − = 1 .2 2a 4 − aSubstitute (12, 9) into the equation.49 144− = 12 2a 4 − a2 2 2 249(4 −a ) − 144 a = a (4 − a )4 2a − 197a+ 196 = 02 2( a −196)( a − 1) = 02 2a = 196, a = 1Since a < c, a = 1, b = 4− 1=322 x( y − 2) − = 13Instructor’s Resource Manual Section 10.3 609

43.44.2 2x + xy+ y = 645.cot 2θ = 0π π2 θ = , θ =2 42 2 2x = u – v = ( u– v)2 2 22 2 2y = u + v = ( u+v )2 2 21 2 1 1 2( u– v) + ( u– v)( u+ v) + ( u+ v) = 62 2 23 2 1 2u + v = 62 22 2u v+ = 14 122 23x + 10xy+ 3y+ 10=0π π46.cot 2θ = 0, 2 θ = , θ =2 42x = ( u – v )22y = ( u+v )23 2 3 2( u– v) + 5( u– v)( u+ v) + ( u+ v) + 10=02 22 28 u – 2 v = –10v 2 u 2– = 15 542 24x + xy+ 4y= 56π πcot 2θ = 0, 2 θ = , θ =2 42x = ( u – v )22y = ( u+v )22 122( u– v) + ( u– v)( u+ v) + 2( u+ v) = 5629 2 7 2u + v = 562 22 2u v+ = 1112 16 924 xy –3y = 643cot 2 θ = , r = 543cos 2θ =5351+2cosθ= =2 5351– 1sinθ = =2 51x = (2 u – v )51y = ( u+2 v )54 3 2(2 u– v)( u+ 2 v)– ( u+ 2 v) = 645 52 2u – 4v = 642 2u v– = 164 16610 Section 10.3 Instructor’s Resource Manual

47. 1 2 1 2– x + 7 xy– y –6 2 x–6 2y= 02 2π πcot 2θ = 0, 2 θ = , θ =2 42 2x = ( u – v ) ; y = ( u+v )221 2 7 1 2– ( u– v) + ( u– v)( u+ v)– ( u+ v) –6( u– v) − 6( u+ v) = 04 2 42 23 u –4 v –12u = 02 23( u –4u+ 4)–4v= 122 2( u–2)v– = 14 348. 3 2 3 2x + xy+ y + 2x+ 2y= 132 2π πcot 2θ = 0, 2 θ = , θ =2 42 2x = ( u – v ) ; y = ( u+v )223 2 1 3 2( u– v) + ( u– v)( u+ v) + ( u+ v) + ( u– v) + ( u+ v) = 134 2 42 22u + v + 2u= 13⎛ 2 1⎞2 12⎜u + u+ ⎟+ v = 13+⎝ 4⎠22⎛ 1⎞2 272⎜u+ ⎟ + v =⎝ 2⎠2( u + ) 21 22 v+ = 127 274 249. A= 4, B = − 3, C = D = E = 0, F = − 18yu4−0 4cot 2θ= =−−3 3Since 0≤2 θ ≤π, sin 2θ is positive, so cos 2θ is negative; using a 3-4-54right triangle, we conclude cos 2θ = − . Thus51−cos2θ1 −( −4 5) 3 10sinθ= = = and2 2 101+ cos2θ1 + ( −4 5) 10cosθ= = = . Rotating through the angle2 2 10v−55−55x1 cos−1( 0.8) 71.6θ = − = , we have2210 3 10 10 3 10 3 10 104 u− v −3 u− v u+ v = 18or10 10 10 10 10 10( ) ( )( )2 2v u2 2v u45 − 5 = 180 or − = 1. This is a hyperbola in standard position4 36in the uv-system; its axis is the v-axis, and a = 2, b = 6 .Instructor’s Resource Manual Section 10.3 611

50. A= 11, B = 96, C = 39, D = 240, E = 570, F = 87511−39 7cot 2θ= =−96 24Since 0≤2 θ ≤π, cos 2θ is negative; using a 7-24-25 right triangle, we7conclude cos 2θ =− .251−cos 2θ1 −( −7 25) 4Thus sinθ= = = and2 2 51+ cos 2θ1 + ( −7 25) 3cosθ= = = .2 2 51 1Rotating through the angle cos−θ = ( − 0.28) = 53.13 , we have23 423 4 4 3( u− v) + ( u− v5 5 5 5 )( u+ v5 5 ) +4 323 4 4 3( u v) ( u v) ( u v)11 9639 + + 240 − + 570 + = −8755 5 5 5 5 5or2 23u − v + 24u+ 6v= −352 23( u +8u+16) − ( v - 6v+ 9) = − 35 + 48 −92 23( u+ 4) −( v− 3) = 42 2( u+ 4) ( v−3)− = 144 3This is a hyperbola in standard position in the uv-system; its axis is the u-axis, its center is ( uv , ) = ( −4,3)and2 3a = , b = 2 .3vyux51. 2 234x + 24xy+ 41y + 250 y = –3257cot 2 θ = – , 2524 r =7cos 2 θ = – 257251– 3cosθ = = ;2 57251+4sinθ= =2 51 1x = (3 u – 4 v ) ; y = (4 u+3 v )5534 2 24 41 2(3 u– 4 v) + (3 u– 4 v)(4u+ 3 v) + (4u+ 3 v) + 50(4u+ 3 v) = –32525 25 252 250u + 25v + 200u+ 150 v = –3252 22u + v + 8u+ 6 v = –132 22( u + 4u+ 4) + ( v + 6v+ 9) = –13 + 8 + 92 22( u+ 2) + ( v+ 3) = 42 2( u+ 2) ( v+3)+ = 12 4This is an ellipse in standard position in the uv-system, with major axisparallel to the v-axis. Its center is ( uv , ) = ( −2, − 3)and a = 2 , b = 2 .612 Section 10.3 Instructor’s Resource Manual

52. 2 216x + 24xy+ 9 y – 20 x–15 y–150 = 07cot 2 θ = , r = 25247cos 2θ =251+725 41–725 3cosθ= = ; sinθ = =2 52 51 1x = (4 u – 3 v ) ; y = (3 u+4 v )5516 2 24 92 2(4u− 3 v) + (4u− 3 v)(3u+ 4 v) + (3u+4 v)25u− 25u= 15025 25 25−4(4u−3 v) − 3(3u+ 4 v) = 1502u − u = 62 1 1u – u+ = 6+4 42⎛ 1⎞25⎜u – ⎟ =⎝ 2 ⎠ 4The graph consists of the two parallel lines u = − 2 and u = 3 .53. a. If C is a vertical parabola, the equation for C2can be written in the form y = ax + bx+ c.Substitute the three points into the equation.2 = a – b + c0 = c6 = 9a + 3b + cSolve the system to get a = 1, b = –1, c = 0.2y = x − xb. If C is a horizontal parabola, an equation forC can be written in the form2x = ay + by + c . Substitute the three pointsinto the equation.–1 = 4a + 2b + c0 = c3 = 36a + 6b + c1Solve the system to get a = , b = –1, c = 0.41 2x = y − y4c. If C is a circle, an equation for C can be2 2 2written in the form ( x − h) + ( y− k)= r .Substitute the three points into the equation.2 2 2( −1 − h) + (2 − k)= r2 2 2h + k = r2 2 2(3 − h) + (6 − k)= r5 5Solve the system to get h = , k = , and2 22 25r = .2⎛ 2 25 5 25⎜x− ⎞ ⎟ + ⎛ ⎜y− ⎞⎟ =⎝ 2⎠ ⎝ 2⎠254. Let (p, q) be the coordinates of P. Byproperties of similar triangles and sinceKP ,x − p y−qα = α = and α = . SolveAP a − p b − qfor p and q to getx −αa andy −αp = q =b . Since P(p, q) is1−α1−αa point on a circle of radius r2 2 2centered at (0, 0), p + q = r2 2⎛ x−αa⎞ ⎛ y−αb⎞2Therefore, ⎜ ⎟ + ⎜ ⎟ = r or⎝ 1−α⎠ ⎝ 1−α⎠2 2 2 2( x αa) ( y αb) (1 α)r− + − = − is theequation for C.Instructor’s Resource Manual Section 10.3 613

55.2 2y Lx Kx = +⎛2 22 L L ⎞2 LK x + x+ – y =⎜ K 24K⎟⎝⎠4K2 2⎛ L ⎞ 2 LK⎜x+ ⎟ – y =⎝ 2K⎠ 4K2( x +L 22 K) y– = 12 2L L24K4KIf K < –1, the conic is a vertical ellipse. IfK = –1, the conic is a circle. If –1 < K < 0,the conic is a horizontal ellipse. If K = 0, the2original equation is y = Lx, so the conic isa horizontal parabola. If K > 0, the conic is ahorizontal hyperbola.If − 1< K < 0(a horizontal ellipse) the lengthof the latus rectum is (see problem 45,Section 10.2)2 22bL 1= 2 = La 4 K L2 KFrom general considerations, the result for avertical ellipse is the same as the one justobtained.For K =− 1 (a circle) we have2 2⎛ L⎞2 L L⎜x − ⎟ + y = ⇒ 2 = L⎝ 2⎠4 2If K = 0 (a horizontal parabola) we have2 2 L Ly = Lx; y = 4 x;p = , and the latus4 4rectum isL2 Lp = 2 L = L .4If K > 0 (a horizontal hyperbola) we can usethe result of Problem 46, Section 10.2. The22blength of the latus rectum is , which isaequal to L .56. Parabola: horizontal parabola, opens to the right,2p = c – a, y = 4( c−a)( x−a)2 2 2Hyperbola: horizontal hyperbola, b = c − a2 2x y− = 12 2a b2 2ybx= − 1a2 222 b 2 2y2 ( x aa)= − Now show that(hyperbola) is greater than2y2y (parabola).2 2 2b 2 2 c − a 2 2( x − a ) = ( x − a )2 2aa( c+ a)( c−a) = ( x + a)( x − a)2a( c+ a)( x+a) (2 a)(2 a)( c− a)( x− a) > ( c−a)( x−a)2 2aa(2 a)(2 a) ( c − a)( x − a) = 4( c − a)( x − a)2ac + a > 2a and x + a > 2a since c > a and x > aexcept at the vertex.57. x = u cos α – v sin αy = u sin α + v cos α(u cos α – v sin α) cos α + (u sin α + v cos α) sin α = d2 2u(cos + sin ) = dααu = dThus, the perpendicular distance from the origin is d.614 Section 10.3 Instructor’s Resource Manual

58.2 2x = ( u – v ) ; y = ( u+v )221/2 1/22 21/2⎢ ⎡ ( u– v) ⎤ ⎥ + ⎡ ⎢ ( u+ v)⎤⎥ = a⎣ 2 ⎦ ⎣ 2 ⎦1/22 1 2( u– v) 2 ⎡⎤+ ( u– v)( u v) ( u v)a2 ⎢ + + + =2 ⎥⎣⎦ 22 2 1/22u+ 2( u – v ) = a2 2 1/22( u – v ) = a– 2u2 2 2 22( u – v ) = a – 2 2au+2u2 1 2v = 2 au– a 2The corresponding curve is a parabola with x > 0 and y > 0.59. x = u cos θ – v sin θ; y = u sin θ + v cos θ2 2x(cos θ) + y(sin θ) = ( ucos θ – vcosθsin θ) + ( usin θ + vcosθsin θ)= u2 2x(– sin θ) + y(cos θ) = (– ucosθ sinθ + vsin θ) + ( ucosθsinθ + vcos θ)= vThus, u = x cos θ + y sin θ and v = –x sin θ + y cos θ.60.5 3 3u = 5cos60°− 3sin 60°= − ;2 2⎛5 3 3 5 3 3⎞( u, v) = ⎜− ,–−2 2 2 2⎟⎝⎠5 3 3v = –5sin 60°− 3cos60 °= – − 2 261. Rotate to eliminate the xy-term.2 2x + 14xy+ 49y= 10024cot 2 θ = – 724cos 2 θ = – 2524251– 1cosθ = = ;2 5 224251+7sinθ= =2 5 21 1x = ( u –7 v ) ; y = (7 u+v )5 25 21 2 14 49 2( u–7) v + ( u–7)(7 v u+ v) + (7 u+ v) = 10050 50 50250u = 1002u = 2u =±2Thus the points closest to the origin in uv-coordinates are ( 2, 0 ) and (– 2, 0 ).x = 1( 2 ) = 1 or x = 1 ( ) =1– 2 –5 2 5 5 2 5y = 1( 7 2)75 2= 5or y = 1 ( –7 2 ) = – 75 25⎛1 7⎞The points closest to the origin in xy-coordinates are ⎜ , ⎟⎝5 5⎠ and ⎛ 1 7⎞⎜– , – ⎟.⎝ 5 5⎠Instructor’s Resource Manual Section 10.3 615

62. x = u cos θ – v sin θy = u sin θ + v cos θ2 2 2 2 2 2Ax = A( u cos θ – vsin θ) = A( u cos θ – 2uv cosθsinθ + v sin θ)2 2 2 2Bxy = B( u cos θ – vsin θ)( u sinθ + vcos θ) = B( u cosθsin θ + uv(cos θ – sin θ) – v cosθ sin θ)2 2 2 2 2 2Cy = C( u sinθ + v cos θ) = C( u sin θ + 2uv cosθsinθ + v cos θ)2 2 2 2 2 2 2Ax + Bxy+ Cy = ( Acos θ + Bcosθ sinθ + Csin θ) u + (–2Acosθ sin θ + B(cos θ – sin θ) + 2Ccosθ sin θ)uv2 2 2+ ( Asin θ – Bcosθsinθ + Ccos θ)v2 22 2Thus, a = Acos θ + Bcosθsinθ + Csinθ and c = Asin θ – Bcosθsinθ + Ccos θ.2 2 2 2a+ c = A(cos θ + sin θ) + B(cosθsin θ – cosθ sin θ) + C(sin θ + cos θ)= A+C2 263. From Problem 62, a = Acos θ + Bcosθsinθ + Csin θ,2 2b= –2Acosθ sin θ + B(cos θ – sin θ) + 2Ccosθsin θ,and2 2c = Asin θ – Bcosθsinθ + Ccos θ.2 2 4 3 2 2 2 2 2b = B cos + 4(– AB + BC)cos sin + 2(2 A – B –4AC + 2 C )cos sin3 2 4+ 4( AB – BC)cos sin + B sinθ θ θ θ θθ θ θ4 3 2 2 2 2 2 3 44ac = 4AC cos θ + 4(– AB + BC)cos θ sinθ + 4( A – B + C )cos θsin θ + 4( AB – BC)cosθ sin θ + 4ACsin θ2 2 4 2 2 2 2 4b – 4 ac = ( B – 4 AC)cos θ + 2( B – 4 AC)cos θsin θ + ( B – 4 AC)sinθ2 2 2 2 2 2 2 2= ( B – 4 AC)(cos θ )(cos θ + sin θ) + ( B – 4 AC)(sin θ)(cos θ + sin θ)2 2 2 2= ( B –4 AC)(cos θ + sin θ) = B –4AC64. By choosing an appropriate angle of rotation, thesecond-degree equation can be written in the form2 2au + cv + du + ev + f = 0. From Problem 63,2− 4 ac = B –4AC.a. If B 2 – 4AC = 0, then 4ac = 0, so the graphis a parabola or limiting form.b. If B 2 – 4AC < 0, then 4ac > 0, so the graphis an ellipse or limiting form.2c. If B – 4AC > 0, then 4ac < 0, so the graphis a hyperbola or limiting form.65. a. From Problem 63, –4 ac = B –4 AC = – Δ or1 4 .ac = Δb. From Problem 62, a + c = A + C.1 1 a+ c 4( A+C)+ = =a c ac Δ2c.2 ⎛2 2A + C± ( A– C)+ B⎞⎜⎟Δ ⎝ ⎠2 ⎡ Δ⎛1 1⎞2 2 2 ⎤= ⎜ + ⎟± A + 2AC+ C + B −4ACΔ⎢4 a c⎥⎣ ⎝ ⎠⎦1⎛1 1⎞2 ( )2= ⎜ + ⎟± A+ C −Δ2 ⎝ac⎠Δ1⎛1 1⎞ 2 Δ ⎛1 1⎞= ⎜ + ⎟± ⎜ + ⎟ −Δ2⎝a c⎠ Δ 16 ⎝a c⎠1⎛1 1⎞ 1 ⎛1 1⎞ ⎛ 4⎞= ⎜ + ⎟± ⎜ + ⎟ −4⎜ ⎟2⎝a c⎠ 2 ⎝a c⎠ ⎝Δ⎠1⎛1 1⎞ 1 2 1 ⎛ 1 ⎞= ⎜ + ⎟± + + −42 a c 2 ac 2 ⎜ ⎟⎝ ⎠ a c ⎝ac⎠1⎛1 1⎞ 1 ⎛1 1⎞= ⎜ + ⎟± ⎜ − ⎟2⎝a c⎠ 2 ⎝a c⎠1⎛1 1 1 1 ⎞= ⎜ + ± − ⎟2 ⎝a c a c ⎠The two values are 1 a and 1 .c2222616 Section 10.4 Instructor’s Resource Manual

68.66.2 2Ax + Bxy + Cy = 1 can be transformed to2Δ = 4(25)(1) – 8 = 362– –1 –3 –2= 0, 20 0 02 2au + cv = 1. Since 4ac =Δ> 0, the graph is an1 22 2Since c < a, =⎛A C ( A– C)B⎞ellipse or a limiting form. 1 + 1 = 4 ⎜ + + + ⎟c Δ ⎝ ⎠( A+ C) > 0,a c Δ1 1 2 2 1=⎛25 1 24 8⎞⎜ + + + ⎟= ( 13 + 4 10 )so a > 0 and c > 0. Thus, the graph is an ellipse (orc 18 ⎝⎠ 9circle).⎛ 9 ⎞⎛13–4 10 ⎞2 2c =The area of au + cv = 1 is⎜ ⎜= 13 – 4 10⎝13 + 4 10⎟⎜13 – 4 10 ⎟⎠⎝ ⎠π1 4 2 π=π = .2π 2π πac Δ= =ΔΔ 6 3Thus, the distance between the foci is 26 – 8 10π67. cot 2θ = 0, θ =π4and the area is .32x = ( u – v )269. From Figure 6 it is clear thaty =2 v = rsin φ and u = rcosφ.( u+v )2Also noting that y = rsin( θ + φ)leads us to1 2 B1 2y = rsin( θ + φ) = rsinθ cosφ + rcosθ sinφ( u– v) + ( u– v)( u+ v) + ( u+ v) = 12 2 2= ( rcosφ)( sinθ) + ( rsinφ)( cosθ)2+B 2 2– B 2u + v = 1= usinθ+ vcosθ2 2a. The graph is an ellipse if 2 + B> 0 and22– 10.4 Concepts Review> 0, 21. simple; closed; simpleb. The graph is a circle if 2 + B 2– B= , so2 2B = 0.2. parametric; parameter3. cycloidc. The graph is a hyperbola if 2 + B> 0 and24.g′( dy dt) ( dx dt)=() tf ′()t2– 0+ B or if < 0 and 2– B > 0, so222B < –2 or B > 2.Problem Set 10.4d. The graph is two parallel lines if 2 + B1. a. t x y= 0 or2–2 –6 –41 3 22 6 4b. Simple; not closedInstructor’s Resource Manual Section 10.4 617

c.xt = ⇒32y = x32. a. t x y–2 –4 –6–1 –2 –30 0 01 2 32 4 6b. Simple; not closedc.1t = ( x+ 2) ⇒41y = ( x+2)2b. Simple; not closedc.xt = ⇒2y =32x5. a. t x y0 4 01 3 12 2 23 1 34 0 23. a. t x y0 –1 01 2 12 5 23 8 34 11 4b. Simple; not closedb. Simple; not closedc. t = 4 – xy = 4– x6. a. t x y0 –3 02 –1 24 1 2 26 3 2 38 5 4c.1t = ( x+ 1) ⇒31y = ( x+1)34. a. t x y0 –2 01 2 22 6 43 10 6b. Simple; not closedc. t = x + 3y = 2x+6618 Section 10.4 Instructor’s Resource Manual

7. a. s x y1 1 113 335155717791999. a. t x y–3 –15 5–2 0 0–1 3 –30 0 –41 –3 –32 0 03 15 5b. Simple; not closedc.1s =x1y =x8. a. s x y1 1 13 3135 51517 779 919b. Not simple; not closedc.2 6 4 2x = t – 8t + 16t2t = y+42 3 2x = y+ y+ + y+( 4) – 8( 4) 16( 4)2 3 2x = y + y410. a. t x y–3 –21 15–2 –4 8–1 1 30 0 01 –1 –12 4 03 21 3b. Simple; not closedc. s = x1y =xb. Simple; not closedc.t 2 – 2 t – y = 0t = 1± 1+y( 1 1 3) –2 ( 1 1 )x = ± + y ± + yx = 2+ 3 y± ( y+ 2) 1+y2 2x y = y+ y+( – 3 – 2) ( 1)( 2)Instructor’s Resource Manual Section 10.4 619

11. a. t x y2 0 3 23 2 34 2 2 013. a. t x y0 0 3π22 0π 0 –33π2–2 02π 0 3b. Simple; not closedc.1 2t = x + 241 2t = 4– y 91 1x + 2=4–4 92 22 2x y+ =8 1812. a. t x y1y3 0 27 322 24 3 0b. Simple; closedc.2sin2cos2xt =42yt =92 2x y+ =4 914. a. r x y10 0 –2π23 0π 0 23π2–3 02π 0 –2b. Simple; not closedc.1 2t = x + 391 2t = 4– y 41 1x + 3=4–9 42 22 2x y+ =9 41yb. Simple; closedc.2sin2cos2xr =92yr =42 2x y+ =9 41620 Section 10.4 Instructor’s Resource Manual

15. a. r x y0 0 –3π2–2 0π 0 33π22 02π 0 –35π2–2 03π 0 37π22 04π 0 –3c.2cos2sinr =r =x2y3x y+ = 12 317. a. θ x y0 0 9π49292π20 93π49292π 0 9b. Not simple; closedc.2sin2cos2xr =42yr =92 2x y+ =4 916. a. r x y10 2 0π20 3π 2 03π20 32π 2 0b. Not simple; closedc.2sin2cosθ =θ =x9y9x y+ = 19 9x + y = 918. a. θ x y0 9 0π49292π20 93π49292π 9 0b. Not simple; closedb. Not simple; closedInstructor’s Resource Manual Section 10.4 621

c.2 xcos θ =92 ysin θ =9x y+ = 19 9x + y = 919. a. θ x y0 + 2π n 1 0π3 + 2πn 12– 3 2π2 + 2πn 0 02π3 + 2πn – 1 2– 3 2π + 2πn –1 04π3 + 2πn – 1 2– 3 23π2 + 2πn 0 05π3 + 2πn 1– 3 2 2b. Not simple; not closedc. sin θ = xcosθ =dx21. 6τdτ =dy 212τdτ =dy= 2τdxdy′ = 2dτ21– x22 2 2y = 2(cos θ – sin θ )2 2y = 2(2 x –1)d y 1=2dx 3τdx22. 12sb. Not simple; not closedc. cos θ = x2sinθ = 1– x2 2y = –8sin θ cos θ2 2y = –8 x (1– x )20. a. θ x y0+ 2πn 0 2π6 + 2πn 1 12 2π2 + 2πn 1 25π6 + 2πn 1 12 2π + 2πn 0 27π6 + 2πn – 1 12 23π+ 2πn –1 2211π6 + 2πn − 1 12 2ds =dy 2–6sds =dy 1– sdx = 2dy′ 1 = –ds 22d y 1= –2dx 24 sdx23. 4θdθ =dy 23 5θdθ =dy 3 5= θdx 4dy′ 3 5 =dθ 42d y 3 5=2dx 16θ622 Section 10.4 Instructor’s Resource Manual

dx24. 2 3θdθ =dy 2–3 3θdθ =dy 3= – θdx 2dy′ 3 = –dθ 22d y 3= –2dx 4 θdx25. sin tdt =dycostdt =dycot tdx =dy′ =2– csc tdt2d y 3= – csc t2dxdx26. 2sintdt =27.dy5costdt =dy 5 cot tdx = 2dy′ 5 =2– csc tdt 22d y 5 3= – csc t2dx 4dx 23sec tdt =dy5secttantdt =dy 5 sin tdx = 3dy′ 5 = cos tdt 32d y 5 cos3= t2dx 928.29.30.31.dx 2– csc tdt =dy2csctcottdt =dy–2costdx =dy′ = 2sintdt2d y 3= –2sin t2dxdx 2t= –dt (12 )2+ tdy 2 t –1=dt 2 2t (1 – t)2 2dy (1 – 2 t)(1 + t )=dx 3 22 t (1– t)5 4 3 2dy′ 3t + 7 t –6t + 10 t –9t+ 3=−dt 4 32 t (1– t)d y (3t + 7 t – 6t + 10 t – 9t+ 3)(1 + t )=2 5 3dx 4 t (1– t)2 5 4 3 2 2 2dx 4t= –dt (12 )2+ t2dy 2(3t+ 1)= –dt 2 2 2t (1 + t )2dy 3t+ 1=dx 32t2dy′ 3( t + 1)= –dt 42t2 2 3d y 3( t + 1)=2 5dx 8tdx dy 2= 2, t = 3tdt dtdy 3tdx = 2dyAt t = 2, x = 4, y = 8, and 3.dx =Tangent line: y – 8 = 3(x – 4) or 3x – y – 4 = 0Instructor’s Resource Manual Section 10.4 623

32.dx dy 2= 3, = 24tdt dtdy 28tdx =1 3dyAt t =− , x =− , y =− 1, and 22 2dx = .⎛ 3 ⎞Tangent line: y+ 1= 2⎜x+⎟ or 2x – y + 2 = 0⎝ 2 ⎠34.dx t dy 1 −t= 2 e , = − edt dt 3dy 1 −2t=− edx 61 dy 1At t = 0, x = 2, y = , and – .3 dx = 61 1Tangent line: y− =− ( x− 2) or3 6x + 6y – 4 = 033.dxdy 2= 2secttan t,= 2sec tdtdtdycsctdx =π 4 2 dyAt t =− , x = , y =− , and =− 2.6 3 3 dx2 ⎛ 4 ⎞Tangent line: y+ =−2⎜x−⎟3 ⎝ 3⎠ or2 3x+ 3y− 6=0dx dy35. = 2, = 3dt dt3 3 30 00∫ ∫L = 4+ 9dt = 13 dt = 13[ t] = 3 13dx dy36. = –1, = 2dt dt37.3 3 3–3 –3–3∫ ∫L = 1+ 4dt = 5 dt = 5[ t] = 6 5dx dy 3= 1, = tdt dt 21/23 9L = ∫ 1+tdt0 41 3= 4 9tdt2∫ +01 ⎡ 2= (4 9 t )18⎢ +⎣ 33/21 3/2 1= (31 − 8) = (31 31 − 8)27 27dx dy38. = 2cos t, = –2sintdt dt⎤⎥⎦30π 2 2 π π0 00∫ ∫L = 4cos t+ 4sin tdt = 2 dt = 2[ t] = 2π39.dx dy= 6, t = 3tdt dt22 2 4 2 236 9 3 40 022 3/2∫ ∫L = t + t dt = t + t dt⎡⎣⎤⎦13 (4 + t ) = 16 2−830624 Section 10.4 Instructor’s Resource Manual

40.41.42.43.44.dx 1 dy 2= 1– , =dt 2t dt t4 ⎛ 2 1 ⎞ 4L = ∫ 1– + +1⎜⎝ ⎠2 4 ⎟ dt2t t t4 2 1 4 ⎛ 1 ⎞= ∫ 1+ + dt = 1 dt1 2 4 ∫ 1⎜ +2 ⎟t t ⎝ t ⎠44⎛ 1 ⎞ ⎡ 1⎤15= ∫ 1 dt t –1 ⎜ +2 ⎟ = ⎢ =t t⎥⎝ ⎠ ⎣ ⎦14dx t dy 9 3/2 t= 2 e , = edt dt 22ln3 2t 81 3t 2ln3 t 81 tL = ∫ 4e + e dt = e 4 e dtln 3 4∫ +ln 3 43/22ln3⎡ 8 ⎛ 81 t ⎞ ⎤=⎢ ⎜4+ e ⎟ ⎥⎢243 4⎣ ⎝ ⎠ ⎥⎦ln 3=745 745 − 259 259243dx t dy– , –1dt= 21– t dt=1/42t1/4 1L = ∫ + 1dt =dt0 2 ∫1– t0 21– t–1 1/4 –11= [sin t] 0 = sin4dx 2 dy 1= , = 2 t –dt t dt 22t1 4 ⎛ 2 2 1 ⎞L = ∫ + 4 t – dt1/4 ⎜ +t t 4 ⎟⎝ 4t⎠1 2 2 1= ∫ 4t+ + dt1/4 t 44t21 ⎛ 1 ⎞= ∫ 2tdt1/4 ⎜ +2 ⎟⎝ 2t⎠11 ⎛ 1 ⎞ ⎡ 2 1 ⎤ 39= ∫ 2 t dt t –1/4⎜ +2 ⎟ = ⎢ =2t2 t⎥⎝ ⎠ ⎣ ⎦1/416dx 2 dy= sech t, = 2 tanh tdt dtL = 3 4 2∫ sech t+4 tanh tdt–3= 3 2 4∫ 4 – 4sech t+sech tdt–33 2 2 3 2= ∫ (2 – sech t) dt = (2 – sech t)dt–3 ∫ –33= [2 t – tanh t] –3 = 12 – 2 tanh 32dx45. sin t,dt =− 2dy sect tan t + sec t= − cost = sect−costdt sect + tan tπ /4 2 2 2L = ∫ sin t+ (sec t− 2 + cos t)dt0π /4= ∫ tan tdt0π /4 1 1=⎡− ⎣ ln cost⎤ ⎦ =− ln = ln 20 2 2dx dy46. = tsin t, = tcostdt dt/2 2 2 2 2L = ∫πt sin t+t cos tdtπ /4π /2π /22⎡1 2 ⎤ 3π= ∫ tdt = tπ /4 ⎢2⎥ =⎣ ⎦π/4 32dx dy47. a. = cos θ , = – sinθdθdθ2π2 2 2πL = ∫ cos + sin d = d0 ∫ 02= [ θ ] 0π = 2πθ θ θ θdx dyb. = 3cos3 θ , = –3sin3θdθdθ2π2 2L = ∫ 9cos 3θ+ 9sin 3θdθ02π2π= 3∫dθ = 3[ θ] 00 = 6πc. The curve in part a goes around the unit circleonce, while the curve in part b goes aroundthe unit circle three times.48. Δ S = 2πxΔs2 2⎛dx⎞ ⎛dy⎞ds = ⎜ ⎟ + ⎜ ⎟ dt⎝ dt ⎠ ⎝ dt ⎠2 2bb ⎛dx⎞ ⎛dy⎞S = ∫ 2π xds = 2 x dta ∫ π +a ⎜ ⎟ ⎜ ⎟⎝ dt ⎠ ⎝ dt ⎠See Section 5.4 of the textdx dy49. = – sin t, = costdt dtS 2π2 2= ∫ 2 π (1 + cos t ) sin t + cos tdt02π2π2= 2 π ∫ (1 + cos tdt ) = 2 π [ t+ sin t] 00 = 4πInstructor’s Resource Manual Section 10.4 625

dx dy55. dx = dt; when x = 0, t = –1; when x = 1, t = 0.50. = – sin t, = cost1dt dt2 0 2 3∫ ( x – 4 ydx ) = S 2π0 ∫ [( t+ 1) – 4( t + 4)] dt–12 2= ∫ 2 π (3 + sin t ) sin t + cos tdt0 = 0 3 2∫2π–1 ( − 4t + t + 2t−15)dt2π2= 2 π ∫ (3 + sin tdt ) = 2 π [3 t– cos t] 00 = 12π0⎡ 4 1 3 2 ⎤ 44= ⎢− t + t + t − 15t3⎥ =−⎣⎦−13dx dy51. = – sin t, = costdt dt2π56. dy = sec t dt; when y = 1, t = ;S 2π2 2= ∫ 2 π (1 + sin t ) sin t + cos tdt40 π2πwhen y = 3, t = .2π2= 2 π ∫ (1+ sin tdt ) = 2 π [ t–cos t] 00 = 4π33 /3 2∫ xydy=1 ∫π(sec t)(tan t)sectdtπ /4dx dy 152. = t , =π /3dt dt t⎡1 3 ⎤ 8 2 2= ⎢ sec t3⎥ = −2 3 ⎛2 3/2 ⎞ 1⎣ ⎦π/4 3 3S = ∫ 2π t t+dt0 ⎜ ⎟⎝3⎠ t257. dx = 2e t dt4π2 3 2= t t + 1dt3∫25 ln 50t t ln 5A= ∫ ydx =1 ∫ 2 e dt = [2 e ] 0 = 802 34π⎡1 2 3/2⎤4π= ( t + 1) = (13 13−1)3 ⎢3 ⎥⎣ ⎦09dx dy53. = 1, = t + 7dt dtS = ∫7 2π ( t+ 7 ) 1+ ( t+7 ) 2 dt– 73/27⎡1⎛2 ⎞ ⎤= 2π ⎢ ⎜1+ ( t + 7)⎟ ⎥⎢⎣3 ⎝⎠ ⎥x⎦– 758. a. t =v0 cosα2π= ( 29 29 –1)23xxy =− 16 + ( v0sin α)v0cosαv0cosdx dy54. = t+ a, = 1⎛ 16 ⎞dt dt2y = – x + (tan α)x⎜ 2 2av20 cos α ⎟⎝ ⎠S = ∫ 2 π ( t+ a) ( t+ a) + 1dt– aThis is an equation for a parabola.a⎡12 3/2⎤= 2 π ⎢ (( t+ a) + 1)b. Solve for t when y = 0.3⎥⎣⎦–a2− 16 t + ( v0sin α) t = 02π ⎡ 23/2t( − 16t+ v= ( a + 2a a + a+10 sin α) = 0)3⎢⎣v0 sin αt = 0,23/2⎤16+ + ⎥ ⎦( a a a a )– –2 1⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ α ⎠The time of flight isv0 sin α seconds.16v0sinαv0sinαc. At t = , x = ( v0cos α)⎛ ⎞16⎜16⎟⎝ ⎠2 20 v0v sinα cosα sin 2α= = .16 32626 Section 10.4 Instructor’s Resource Manual

59.60.d. Let R be the range as a function of α .2v0 sin 2αR =322dR v0 cos 2 αdα = 162v0 cos 2 α= 0, cos 2α = 0, α =π16 42 22d R v0sin 2 α d R=− ; < 0 at α = π .2 8 2dαdα4The range is the largest possible when α = π .4Let the wheel roll along the x-axis with P initiallyat (0, a – b).ON = arc NQ = atx = OM = ON − MN = at − bsinty = MP = RN = NC + CR = a − bcostLet the wheel roll along the x-axis with P initiallyat (0, a – b).ON = arc NQ = atx = OM = ON − MN = at − bsinty = MP = RN = NC + CR = a − bcost61. The x- and y-coordinates of the center of the circleof radius b are (a – b)cos t and (a – b)sin t,respectively. The angle measure (in a clockwisedirection) of arc BP is a t . The horizontal changebfrom the center of the circle of radius b to P is⎛ ⎛a ⎞⎞⎛a–b ⎞bcos ⎜– ⎜ t – t⎟⎟=bcos⎜ t⎟and the vertical⎝ ⎝b⎠⎠⎝ b ⎠a a bchange is bsin ⎛ ⎛ ⎞t t ⎞ ⎛bsin− ⎞⎜−⎜ − ⎟⎟=−⎜ t⎟.⎝ ⎝b⎠⎠⎝ b ⎠⎛Therefore, x ( a b)cost bcos a − b ⎞= − + ⎜ t⎟⎝ b ⎠ and⎛y ( a b)sint bsin a − b ⎞= − − ⎜ t⎟⎝ b ⎠ .62. From Problem 61,⎛x ( a b)cost bcos a − b ⎞= − + ⎜ t⎟⎝ b ⎠ and⎛y ( a b)sint bsin a − b ⎞= − − ⎜ t⎟⎝ b ⎠ .aSubstitute b = .4⎛3a⎞ ⎛a⎞x = ⎜ ⎟cost+⎜ ⎟cos(3 t)⎝ 4 ⎠ ⎝4⎠⎛3a⎞ ⎛a⎞= ⎜ ⎟cost+ ⎜ ⎟cos(2 t+t)⎝ 4 ⎠ ⎝4⎠⎛3a⎞ ⎛a⎞= ⎜ ⎟cos t+ ⎜ ⎟(cos 2tcost−sin 2tsin t)⎝ 4 ⎠ ⎝4⎠⎛3a⎞ ⎛a⎞3 2 2= ⎜ ⎟cos t+⎜ ⎟(cos t – sin tcos t – 2sin tcos t)⎝ 4 ⎠ ⎝4⎠⎛3a⎞ ⎛a⎞ 3 ⎛3a⎞2= ⎜ ⎟cost+ ⎜ ⎟cos t−⎜ ⎟costsint⎝ 4 ⎠ ⎝4⎠ ⎝ 4 ⎠⎛3a⎞ 2 ⎛a⎞3 3= ⎜ ⎟(cos t)(1 − sin t) + ⎜ ⎟cos t = acost⎝ 4 ⎠ ⎝4⎠⎛3a⎞ ⎛a⎞y = ⎜ ⎟sin t−⎜ ⎟sin(3 t)⎝ 4 ⎠ ⎝4⎠Instructor’s Resource Manual Section 10.4 627

⎛3a⎞ ⎛a⎞2= sin t− sin(2 t+t)64. x = 2acost− acos2t = 2acost− 2acost+a⎜ ⎟ ⎜ ⎟⎝ 4 ⎠ ⎝4⎠= 2acos t(1− cos t)+ a⎛3a⎞ ⎛a⎞= ⎜ ⎟sin t− ⎜ ⎟(sin 2tcos t+cos 2tsin t)y = 2asint− asin2t = 2asint−2asintcost⎝ 4 ⎠ ⎝4⎠= 2asin t(1−cos t)⎛3a⎞ ⎛a⎞2 2 3= ⎜ ⎟sin t− ⎜ ⎟(2sin tcos t+ cos tsin t−sin t)x − a = 2acos t(1−cos t)⎝ 4 ⎠ ⎝4⎠2 2 2 2( x − a) + y = 4 a (1−cos t)⎛3a⎞ ⎛3a⎞ 2 ⎛a⎞3= ⎜ ⎟sin t− ⎜ ⎟sin tcos t+⎜ ⎟sint2 2( x− a) + y + 2 a( x− a)=⎝ 4 ⎠ ⎝ 4 ⎠ ⎝4⎠2 2 2⎛3a⎞ 2 ⎛a⎞3 34 a (1− cos t) + 4 a (1−cos t)cost= ⎜ ⎟(sin t)(1 − cos t) + ⎜ ⎟sin t = asint⎝ 4 ⎠ ⎝42 2 2⎠( x − a) + y + 2 a( x− a) = 4 a (1−cos t)2 2 2 2 2 2[( x − a) + y + 2 a( x− a)] = 4a ⋅4 a (1 −cos t)2 2 2 2 2 2[( x − a) + y + 2 a( x− a)] = 4 a [( x− a) + y ]dx ⎛a⎞65. = ⎜ ⎟( −2sin t−2sin 2 t),dt ⎝3⎠dy ⎛a⎞= ⎜ ⎟(2cost−2cos 2 t)dt ⎝3⎠2/3 2/3 2/3 2 2/3 2 2/3x + y = a cos t+ a sin t = a2 2⎛dx⎞ ⎛a⎞ 2 2⎜ ⎟ = ⎜ ⎟ (4sin t+ 8sin tsin 2t+4sin 2 t)63. Consider the following figure similar to the one in⎝ dt ⎠ ⎝3⎠2 2the text for Problem 61.⎛dy⎞ ⎛a⎞ 2 2⎜ ⎟ = ⎜ ⎟ (4cos t− 8costcos 2t+4cos 2 t)⎝ dt ⎠ ⎝3⎠2 2 2⎛dx ⎞ ⎛dy ⎞ ⎛a⎞⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ (8 + 8sin tsin 2t−8cos tcos 2 t)⎝ dt ⎠ ⎝ dt ⎠ ⎝3⎠2⎛a⎞2 3 2= ⎜ ⎟ (8 + 16sin tcost− 8cos t+8sin tcos t)⎝3⎠2⎛a⎞2 3= ⎜ ⎟ (8 + 24cos tsin t−8cos t)⎝3⎠2⎛a⎞3= ⎜ ⎟ (8 + 24cos t−32cos t)⎝3⎠The x- and y-coordinates of the center of the circle2 22 π /3 ⎛dx⎞ ⎛dy⎞of radius b are (a + b)cos t and (a + b)sin tL = 3∫0 ⎜ ⎟ + ⎜ ⎟ dtrespectively. The angle measure (in a counterclockwisedirection) of arc PB is a 2 π /3⎝ dt ⎠ ⎝ dt ⎠tb . The3= a∫8 + 24cost−32cost dt016ahorizontal change from the center of the circle ofUsing a CAS to evaluate the length, L = .radius b to P is3⎛a ⎞ ⎛a+b ⎞bcos⎜ t+ t+π ⎟= −bcos⎜ t⎟and the⎝b⎠ ⎝ b ⎠vertical change is⎛a ⎞ ⎛a+b ⎞bsin⎜ t+ t+π ⎟=−bsin⎜ t⎟⎝b⎠ ⎝ b ⎠ . Therefore,⎛x ( a b)cost bcos a + b ⎞= + − ⎜ t⎟⎝ b ⎠ and⎛y ( a b)sint bsin a + b ⎞= + − ⎜ t⎟⎝ b ⎠ .628 Section 10.4 Instructor’s Resource Manual

66. a. Let x = a cos t and y = b sin t.dx dy=− asin t,= bcostdt dtb.2 2⎛dx⎞ ⎛dy⎞ 2 2 2 2⎜ ⎟ + ⎜ ⎟ = a sin t+b cos t⎝ dt ⎠ ⎝ dt ⎠2 2 2 2= a + ( b − a )cos t⎛ 22 2 2 2 c ⎞2= a − c cos t = a 1−cos t⎜ 2a ⎟⎝⎠⎛ 22 ⎛c⎞ ⎞2 2 2 2= a 1− cos t = a (1−e cos t)⎜ ⎜ ⎟⎝a⎠ ⎟⎝⎠2 2π /2 ⎛dx⎞ ⎛dy⎞P = 4∫dt0 ⎜ ⎟ + ⎜ ⎟⎝ dt ⎠ ⎝ dt ⎠π /2 2 2= 4a∫1−e cos t dt0π /22cos tP = 4∫1−dt0 16π /2 2= ∫ 16 – cos tdt≈6.18380(The answer is near 2π because it is slightlysmaller than a circle of radius 1 whoseperimeter is 2π ).c.d.The curve touches a horizontal border sixtimes and touches a vertical border twice.Note that the curve is traced out five times.The curve touches a horizontal border 18times and touches a vertical border four times.68. This is a closed curve even thought the graph doesnot look closed because the graph retraces itself.1yc.π /2 2P = ∫ 16 – cos tdt ≈6.18380−11x67. a.−169. a. x = cost; y = sin 2ty1b.The curve touches a horizontal border twiceand touches a vertical border twice.−11x−1b. x = cos 4t; y = sin8tyThe curve touches a horizontal border fivetimes and touches a vertical border threetimes.1−11x−1Instructor’s Resource Manual Section 10.4 629

a. The graph touches a horizontal side ifc. x = cos5t; y = sin10tterms is u v . yksin bt = ± 1 or bt π ( k odd)12kt = ,2b where 1, 3, , 4 1 Hence, H = 2b.−11 xb. The graph touches a vertical side ifcos at = ± 1 or at nπ( n an integer)−1nt = ,a π where 0,1, , 2 1 d. x = cos 2t; y = sin 3tHence, V = 2a.y1kat0 = nπ, bt0= π so that2−11 xu a n 2nv bk2 k−1is the case, write u 2re.kx = cos 6t; y = sin 9ta corner at t 0 ; then t 0 y2b 1akat0 n ,( n an integer).2b ak 2rwk rk2b 2vw v−11 xf.−1b π wherex = cos12t; y = sin18tym = v,3 v,5 v, ,(4w−1)v1given by−11 x⎧ 2a a uV- 1 C = = if C=02 ⎪ 2b b v=1⎨H- C 2 2 12 ⎪ a−w u−= if C≠0−1⎪⎩2b−w 2v−1x = cos at, y = sin bt, t∈ [0, 2 π ) ; we assume aand b are integers. This graph will be contained inthe box with sides x =± 1, y =± 1. Let H be thenumber of times the graph touches a horizontalside, V the number of times it touches a verticalside, and C the number of times it touches a corner(right now, C is included in H and V). Finally, letw = the greatest common divisor of a and b;awrite a = u⋅ w , b= v⋅ w. Note: in lowestb70. Consider the curve defined parametrically by= ; that is,= ; that is,c. If t 0 yields a corner, then (see a. and b.) then= = = . Thus corners can onlyoccur if u is even and v is odd. Assume that= , and assume we have= = is an integer; hence v is afactor of rk. But v and r have no factors incommon, so v must be a factor of k.Conclusion: k is an odd multiple of v. Thuscorners occur at 2m… ; therefore C = 2w.Now if we count corner contacts as halfhorizontal and half vertical, the ratio ofvertical contacts to horizontal contacts is630 Section 10.4 Instructor’s Resource Manual

71. a.1d.0.5−1 −0.5 0.5 1−0.5b.−110.5Given a parameterization of the form x = cosf(t) and y = sin f(t), the point moves aroundthe curve (which is a circle of radius 1) at aspeed of f ′().t The point travels clockwisearound the circle when f(t) is decreasing andcounterclockwise when f(t) is increasing.Note that in part d, only part of the circle willbe traced out since the range of f(t) = sin t is[–1, 1].−1 −0.5 0.5 172. a.−0.5c.−11b.The curve traced out is the graph offor 0≤ x ≤ 2y = x20.5−1 −0.5 0.5 1The curve traced out is the graph offor 0≤ x ≤ 8.y = x2−0.5c.−1The curve traced out is the graph offor −16 ≤ x ≤ 0 .y = − x2Instructor’s Resource Manual Section 10.4 631

d.Some possible graphs for different a and b areshown below.a = 3, b = 1The curve traced out is the graph offor 0≤ x ≤ 32y = x2a = 5, b = 2All of the curves lie on the graph of y =± x 2 ,but trace out different parts because of theparameterization73. a. 0≤t≤ 2πa = 5, b = 4b. 0≤t≤ 2πa = 2, b = 3c. 0≤t≤ 4πLet p = a where p q b qais the reduced fraction of .bThe length of the t-interval is 2qπ . The numberof times the graph would touch the circle of radiusa during the t-interval is p. If a bis irrational, thecurve is not periodic.d. 0≤t≤ 8π75.3t3tx = , y =3 3t + 1 t + 12Let p = a where p is the reduced fractionq b qaof . The length of the t-interval is 2qπ .bThe number of times the graph would touchthe circle of radius a during the t-interval is p.If a bis irrational, the curve is not periodic.74.When x > 0, t > 0 or t < –1.When x < 0, –1 < t < 0.When y > 0, t > –1. When y < 0, t < –1.Therefore the graph is in quadrant I for t > 0,quadrant II for –1 < t < 0,quadrant III for no t, and quadrant IV for t < –1.632 Section 10.4 Instructor’s Resource Manual

10.5 Concepts Review5.1. infinitely many2. r cos θ ; r sin θ ;3. circle; line4. conicProblem Set 10.51.2.2ra.b.c.d.⎛ 3 5 1 3⎜1, – π ⎞ ⎟, ⎛ ⎜1, π ⎞ ⎟, ⎛ ⎜–1, – π ⎞ ⎟, ⎛ ⎜–1,π⎞⎟⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠⎛ 3 5 7 9⎜1, – π ⎞ ⎟, ⎛ ⎜1, π ⎞ ⎟, ⎛ ⎜–1, – π ⎞ ⎟, ⎛ ⎜–1,π⎞⎟⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠⎛ 7 ⎞ ⎛ 5 ⎞ ⎛ 4 ⎞⎜ 2, – π⎟, ⎜ 2, π⎟, ⎜– 2, – π⎟,⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠⎛ 2 ⎞⎜– 2, π⎟⎝ 3 ⎠⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞⎜ 2, – π⎟, ⎜ 2, π⎟, ⎜– 2, – π⎟,⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠⎛ 1 ⎞⎜– 2, π⎟⎝ 2 ⎠6.3.4.a.b.c.d.⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞⎜3 2, – π⎟, ⎜3 2, π⎟, ⎜–3 2, – π⎟,⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠⎛ 1 ⎞⎜–3 2, π⎟⎝ 2 ⎠⎛ 5 3 1 7⎜1, – π ⎞ ⎟, ⎛ ⎜1, π ⎞ ⎟, ⎛ ⎜–1, – π ⎞ ⎟, ⎛ ⎜–1,π⎞⎟⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠⎛ 5 ⎞ ⎛ 1 ⎞ ⎛ 8 ⎞⎜ 2, – π⎟, ⎜ 2, π⎟, ⎜– 2, – π⎟,⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠⎛ 4 ⎞⎜– 2, π⎟⎝ 3 ⎠⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞⎜2 2, – π⎟, ⎜2 2, π⎟, ⎜–2 2, – π⎟,⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠⎛ 1 ⎞⎜–2 2, π⎟⎝ 2 ⎠Instructor’s Resource Manual Section 10.5 633

7. a.b.c.d.1x = 1cos π= 021y = 1sin π= 12(0, 1)1 2x = –1cos π= –4 21 2y = –1sin π= –4 2⎛ 2 2 ⎞⎜– ,–2 2 ⎟⎝⎠⎛ 1 ⎞ 2x = 2cos ⎜– π ⎟=⎝ 3 ⎠ 2⎛ 1 ⎞ 6y = 2sin ⎜– π ⎟=–⎝ 3 ⎠ 2⎛ 2 6 ⎞⎜, –2 2 ⎟⎝ ⎠5x = – 2cos π= 025y = – 2sin π= – 22( 0, – 2 )d.29x = –2 2cos π= 0229y = –2 2sin π= –2 22( 0, – 2 2 )2 29. a. ( ) 2r = 3 3 + 3 = 36, r = 61 πtan θ = , θ =3 6⎛ π ⎞⎜6, ⎟⎝ 6 ⎠2 2b. ( ) 2r = –2 3 + 2 = 16, r = 42 5πtan θ = , θ =–2 3 6⎛ 5π⎞⎜4,⎟⎝ 6 ⎠2 22c. ( ) ( )r = – 2 + – 2 = 4, r = 2– 2 5πtan θ = , θ =– 2 4⎛ 5π⎞⎜2, ⎟⎝ 4 ⎠8. a.b.c.7x = 3 2cos π= 027y = 3 2sin π= –3 22( 0, – 3 2 )15 2x = –1cos π= –4 215 2y = –1sin π= 4 2⎛ 2 2 ⎞⎜– ,2 2 ⎟⎝ ⎠⎛ 2 ⎞ 2x = – 2cos ⎜– π ⎟=⎝ 3 ⎠ 2⎛ 2 ⎞ 6y = – 2sin ⎜– π ⎟=⎝ 3 ⎠ 2⎛ 2 6 ⎞⎜,2 2 ⎟⎝ ⎠d.10. a.b.2 2 2r = 0 + 0 = 0, r = 0tan θ = 0, θ = 0(0, 0)2 22 ⎛ 3 ⎞ ⎛ 1 ⎞ 10 10r = ⎜– ⎟ + ⎜ ⎟ = , r =⎝ 3⎠ ⎝ 3⎠3 31tan 3 , tan –1 ⎛– 1 ⎞θ = θ = π+ ⎜ ⎟⎝ 3 ⎠⎛⎜⎝r–3310 –1 ⎛ 1 ⎞⎞, π+ tan ⎜– ⎟3 ⎝ 3 ⎠⎟⎠2 22 ⎛ 3⎞ ⎛ 3⎞6 6= ⎜– + = , r =2 ⎟ ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠ 4 232 3πtan θ = , θ =3– 42⎛ 6 3π⎞⎜ ,2 4 ⎟⎝ ⎠634 Section 10.5 Instructor’s Resource Manual

c.d.2 2 2r = 0 + (–2) = 4, r = 22 3πtan θ = – , θ =0 2⎛ 3π⎞⎜2, ⎟⎝ 2 ⎠2 2 2r = 3 + (–4) = 25, r = 54 –1 ⎛ 4⎞tan θ = – , θ = tan ⎜–⎟3 ⎝ 3⎠⎛ –1 ⎛ 4 ⎞⎞⎜2, tan ⎜– ⎟⎟⎝ ⎝ 3 ⎠⎠14. x – y = 0r cos θ – r sin θ = 0tan θ = 1πθ =411. x – 3y + 2 = 015.x2 2+ = 4yr cos θ – 3r sin θ + 2 = 02r = – cos θ – 3sin θ2r =3sin θ – cosθ12. x = 016.2 2( cos ) + ( sin ) = 4r2r =r = 2xθ42= 4pyrθπθ =213. y = –217.2( rcos θ ) = 4 p( rsin θ )4psinθr =2cosθr = 4p sec θ tan θπθ =2cot θ = 0x0y =x = 0r sin θ = –22r = – sin θr = –2 csc θ18. r = 32r = 92 2x + y = 9Instructor’s Resource Manual Section 10.5 635

19. r cos θ + 3 = 0x + 3 = 0x = –320. r – 5 cos θ = 0r2 – 5rcosθ = 0x 2 + y 2 – 5 x = 0⎛ 2 25 ⎞ 2 25⎜x –5x+ ⎟+ y =⎝ 4 ⎠ 4⎛ 25 2 25⎜x– ⎞ ⎟ + y =⎝ 2 ⎠ 426.4r =−cosθ4r =cos θ − π , line( )21. r sin θ – 1 = 0y – 1 = 0y = 127. r = 4sinθ⎛ π ⎞r = 2(2)cos⎜θ− ⎟⎝ 2 ⎠ , circle22.r2 – 6rcos θ –4rsinθ+ 9=0x2 + y2 – 6 x–4y+9=02 2( x –6x+ 9) + ( y –4y+ 4) = –9+ 9+42 2( x–3) + ( y–2) = 423. r = 6, circle28. r = − 4cosθr = 2(2)cos( θ −π ) , circle24.2πθ = , line329.4r = 1 + cosθ(1)(4)25.3r =sinθ3r =cos θ −π( )2, liner = 1 + (1)cosθ, parabolae =1636 Section 10.5 Instructor’s Resource Manual

30.4r1 2sinθ(2)(2)r = , hyperbola1 + 2cos θ − π= + ( )e = 2233.4r = 2 + 2cosθr = (1)(2)1 + (1)cosθ, parabolae = 131.r62 sinθ= + ( 1 ) 261 ( ) cos( θπ)r = , ellipse+ −1e =212 234.4r =2 + 2cos( θ −π3 )2(1)r = , parabola1 + (1)cos( θ − π3 )e = 132.6r = 4 − cosθ6 ( 1 )r = 4, ellipse1 + 1( 4 ) cos( θ −π )1e =435.r4=12+ cos( −π )θ4(2)r = 1 + 2cos( θ −π ),hyperbolae = 236.4r =3cos( θ − π)3, lineInstructor’s Resource Manual Section 10.5 637

37. By the Law of Cosines,2 2 2a = r + c −2rccos( θ − α)(see figure below).38. r = asinθ+ bcosθ2r = arsinθ+ brcosθ2 2x + y = ay+bx2 2x − bx+ y − ay = 0⎛ 2 2 2 22 b ⎞ ⎛2 a ⎞ a + bx − bx+ + y − ay+ =⎜ 4 ⎟ ⎜ 4 ⎟⎝ ⎠ ⎝ ⎠42 2 2 2⎛ b ⎞ ⎛ a ⎞x ya +⎜ − ⎟ + ⎜ − ⎟ =b⎝ 2⎠ ⎝ 2⎠4This is an equation of a circle with radius2 2a + b⎛b a⎞and center ⎜ , ⎟2⎝2 2⎠ .39. Recall that the latus rectum is perpendicular to theaxis of the conic through a focus.⎛ π ⎞ edr⎜θ0+ ⎟= = ed⎝ 2 ⎠ 1+ecosπ2Thus the length of the latus rectum is 2ed.40. a. The point closest to the pole is at θ 0.rrθeded1 = 1( 0)= =1+ ecos(0) 1+eThe point furthest from the pole is at θ 0 +π .red ed= r( θ +π ) = =1+ ecosπ1– e2 0b. The length of the major diameter is2 2– +r1+ r2 = + = +1+1– 2 21– 1–2ed= .21–eeda =21– e2edc = ea =1– eed ed ed e d ed e de e e e22 222 2 2 ⎛ ed ⎞ ⎛ e d ⎞b = a – c = ⎜ –2 ⎟ 2⎝1– e ⎠⎜1–e ⎟⎝ ⎠2 2 2 2 2ed (1 – e)ed= =2 2 2(1 – e ) 1 – eb =ed21– eThe length of the minor diameter is41. a + c = 183, a – c = 172a = 200, a = 1002c = 166, c = 83ce = = 0.83a42.185.8a = = 92.9,2c = ea = (0.0167)92.9 = 1.55143Perihelion = a – c ≈ 91.3 million miles2ed.21– e43. Let sun lie at the pole and the axis of the parabolalie on the pole so that the parabola opens to theleft. Then the path is described by the equationdr = . Substitute (100, 120 ) into the1 + cosθequation and solve for d.d100 =1 + cos120 °d = 50The closest distance occurs when θ = 0°.50r = = 25 million miles1+ cos0°638 Section 10.5 Instructor’s Resource Manual

44. a.4d= =1 + cosπ( − θ0)+π( − θ0)2⎛ π π ⎞4+ 4⎜cos cosθ0 + sin sinθ0⎟=d⎝ 2 2 ⎠31 cosd4⎛ π π ⎞3+ 3⎜cos cosθ0 + sin sinθ0⎟=d⎝ 4 4 ⎠d = 4+ 4sinθ0d = 3+ 3 2 cosθ 3 20 + sinθ02 23 2 3 24+ 4sinθ0 = 3+ cosθ0 + sinθ02 23 2 ⎛ 3 2 ⎞cosθ0 + −4 sinθ0− 1 = 02 ⎜2 ⎟⎝ ⎠θ( )3 2cos 0 + 3 2−8 sin 0 − 2=0θθθ4.24cos 0 −3.76sin 0 − 2 = 0b. A graph shows that a root lies near 0.5. Using Newton’s Method, θ0 ≈ 0.485 .d = 4+ 4sinθ0≈ 5.86d dc. The closest the comet gets to the sun is r = 2.93 AU1+ cos( θ −θ) = 2≈0 045.4e4ex = cos t,y = sin t1+ ecost 1+ecoste = 1e = 0.10.40.2-0.4-0.20.2e = 1.1-0.2-0.4e = 0.521e = 1.3-4-3 -2-11-1-2e = 0.9642-35-25-15-5-2-4-6Instructor’s Resource Manual Section 10.5 639

10.6 Concepts Review1. limaçon2. cardioid3. rose; odd; even3. r sin θ + 4 = 04r = – sin θSince sin( − θ ) =− sinθ, test 2 is passed. Theother two tests fail so the graph has only y-axissymmetry.4. spiralProblem Set 10.61.22 π – 016θ =πθ = ± 4Changing θ →−θor r →−ryields anequivalent set of equations. Therefore all 3 testsare passed.4. r = –4 sec θ4r = – cos θSince cos( − θ ) = cosθ, the graph is symmetricabout the x-axis. The other symmetry tests fail.π2. ( r –3) ⎜⎛ θ – ⎟⎞ = 0⎝ 4 ⎠πr = 3 or θ =4θ = θ 0 defines a line through the pole. Since aline forms an angle of π radians, changingθ → π + θ results in an equivalent set ofequations, thus passing test 3. The other twotests fail so the graph has only origin symmetry.5. r = 2 cos θSince cos( − θ ) = cosθ, the graph is symmetricabout the x-axis. The other symmetry tests fail.6. r = 4 sin θSince sin( − θ ) =− sinθ, the graph is symmetricabout the y-axis. The other symmetry tests fail.640 Section 10.6 Instructor’s Resource Manual

7.2r =1–cosθSince cos( − θ ) = cosθ, the graph is symmetricabout the x-axis. The other symmetry tests fail.11. r = 1 – 1 sin θ (cardioid)Since sin( π − θ) = sinθ, the graph is symmetricabout the y-axis. The other symmetry tests fail.8.4r =1 + sinθSince sin( π − θ) = sinθ, the graph is symmetricabout the y-axis. The other symmetry tests fail.12. r = 2– 2sinθ(cardioid)Since sin( π − θ) = sinθ, the graph is symmetricabout the y-axis. The other symmetry tests fail.9. r = 3 – 3 cos θ (cardioid)Since cos( − θ ) = cosθ, the graph is symmetricabout the x-axis. The other symmetry tests fail.13. r = 1 – 2 sin θ (limaçon)Since sin( π − θ) = sinθ, the graph is symmetricabout the y-axis. The other symmetry tests fail.10. r = 5 – 5 sin θ (cardioid)Since sin( π − θ) = sinθ, the graph is symmetricabout the y-axis. The other symmetry tests fail.14. r = 4 – 3 cos θ (limaçon)Since cos( − θ ) = cosθ, the graph is symmetricabout the x-axis. The other symmetry tests fail.Instructor’s Resource Manual Section 10.6 641

15. r = 2 – 3 sin θ (limaçon)Since sin( π − θ) = sinθ, the graph is symmetricabout the y-axis. The other symmetry tests fail.19.r2= –9cos2θ(lemniscate)r =± 3 –cos2θSince cos( − 2 θ ) = cos 2θandcos(2( π − θ)) = cos(2π − 2 θ) = cos( − 2 θ) = cos 2θthe graph is symmetric about both axes and theorigin.16. r = 5 – 3 cos θ (limaçon)Since cos( − θ ) = cosθ, the graph is symmetricabout the x-axis. The other symmetry tests fail.20.2r=− 16cos 2θ(lemniscate)r =± 4 − cos2θSince cos( − 2 θ ) = cos 2θandcos(2( π − θ)) = cos(2π − 2 θ) = cos( − 2 θ) = cos 2θthe graph is symmetric about both axes and theorigin.17.r2= 4cos2θ(lemniscate)r =± 2 cos2θSince cos( − 2 θ ) = cos 2θandcos(2( π − θ)) = cos(2π − 2 θ) = cos( − 2 θ) = cos 2θthe graph is symmetric about both axes and theorigin.21. r = 5cos3θ(three-leaved rose)Since cos( − 3 θ ) = cos(3 θ ) , the graph issymmetric about the x-axis. The other symmetrytests fail.18.r2= 9sin2θ(lemniscate)( θ )r =± 3 sin 2Since sin(2( π + θ)) = sin(2π + 2 θ) = sin 2θ, thegraph is symmetric about the origin. The othersymmetry tests fail.22. r = 3sin3θ(three-leaved rose)Since sin( − 3 θ ) =− sin(3 θ ) , the graph issymmetric about the y-axis. The other symmetrytests fail.642 Section 10.6 Instructor’s Resource Manual

23. r = 6sin2θ(four-leaved rose)Sincesin(2( π − θ)) = sin(2π −2 θ)= sin( − 2 θ ) = −sin(2 θ )and sin( − 2 θ ) = − sin(2 θ ), the graph is symmetricabout both axes and the origin.27.1r = θθ , ≥ 0 (spiral of Archimedes)2No symmetry. All three tests fail.28. r = 2 θ , θ ≥ 0 (spiral of Archimedes)No symmetry. All three tests fail.24. r = 4cos2θ(four-leaved rose)Since cos( − 2 θ ) = cos 2θandcos(2( π − θ)) = cos(2π − 2 θ) = cos( − 2 θ) = cos 2θthe graph is symmetric about both axes and theorigin.29. r = e θ , θ ≥ 0 (logarithmic spiral)No symmetry. All three tests fail.25. r = 7cos5θ(five-leaved rose)Since cos( − 5 θ ) = cos5θ, the graph is symmetricabout the x-axis. The other symmetry tests fail.30.r = e θ /2 , θ ≥ 0 (logarithmic spiral)No symmetry. All three tests fail.26. r = 3sin5θ(five-leaved rose)Since sin( − 5 θ ) =− sin 5θ, the graph issymmetric about the y-axis. The other symmetrytests fail.31.2 r = , θ > 0 (reciprocal spiral)θNo symmetry. All three tests fail.Instructor’s Resource Manual Section 10.6 643

32.1 r =− , θ > 0 (reciprocal spiral)θNo symmetry. All three tests fail.35. r = 3 3cos θ , r = 3sinθ33. r = 6, r = 4+4cosθ6= 4+4cosθ1cosθ =2π 5πθ = , θ =3 3⎛ π⎞ ⎛ 5π⎞⎜6, ⎟, ⎜6,⎟⎝ 3⎠ ⎝ 3 ⎠34. r = 1− cos θ , r = 1+cosθ1− cosθ= 1+cosθcosθ = 0π 3πθ = , θ =2 2⎛ π⎞ ⎛ 3π⎞⎜1, ⎟, ⎜1,⎟⎝ 2⎠ ⎝ 2 ⎠(0, 0) is also a solution since both graphs includethe pole.36.37.3 3cosθ= 3sinθtanθ = 3π 4πθ = , θ =3 3⎛3 3 π ⎞ ⎛ 3 3 4π⎞⎜, = − ,2 3⎟ ⎜ 2 3 ⎟⎝ ⎠ ⎝ ⎠(0, 0) is also a solution since both graphs includethe pole.5r = 5, r =1 − 2cosθ55 = 1−2cosθcosθ = 0π 3π⎛ π ⎞ ⎛ 3π⎞θ = , θ = ; ⎜5, ⎟, ⎜5,⎟2 2 ⎝ 2⎠ ⎝ 2 ⎠Note that r = –5 is equivalent to r = 5.5− 5 = 1−2cosθcosθ = 1θ = 0 ; (–5, 0)6r = 6sin θ , r =1 + 2sinθ644 Section 10.6 Instructor’s Resource Manual

66sinθ=1 + 2sinθ212sin θ + 6sinθ− 6 = 06(2sin − 1)(sin + 1) = 0θθ1sin θ = , sinθ= − 125 3θ = π , θ = π , θ =π6 6 2⎛ π⎞ ⎛ 5π⎞ ⎛ 3π⎞⎜3, ⎟, ⎜3, ⎟, ⎜–6,⎟⎝ 6⎠ ⎝ 6 ⎠ ⎝ 2 ⎠ or ⎛6, π ⎞⎜ ⎟⎝ 2 ⎠1 1 1 1 1r = cos ( π – θ ) = cos π cos θ + sin πsinθ2 2 2 2 21= sin θ 21r = sin θ 240. Consider the following figure.38.2r = 4cos2 θ , r = 2 2sinθ( 2 2sinθ) 24cos2θ=2 24 − 8sin θ = 8sin θ2 1 1sin θ = ⇒ sinθ= ±4 25 7 11θ = π , θ = π , θ = π , θ =π6 6 6 6⎛ π⎞ ⎛ 7π⎞⎜ 2, ⎟= ⎜−2, ⎟,⎝ 6⎠ ⎝ 6 ⎠⎛ 5π⎞ ⎛ 11π⎞⎜ 2, ⎟= ⎜−2, ⎟⎝ 6 ⎠ ⎝ 6 ⎠(0, 0) is also a solution since both graphs includesthe pole.ar = − bcosθrcosθ= a−bcosθx = a−bcosθxr = ar − br cosθ( x − ar ) =− bx2 2 2 2( − ) =x a r b x2 2 2 2 2( − ) ( + ) =x a x y b x2 22 b x 2= − x2( − )yxif b>aa39. Consider1r = cos θ.2if b

41.43. a. y = 45r sin θ = 4545r =sinθb.x2 2+ = 36y2r = 36r = 62 2= ( − cos ) + ( sin )PF a r θ r2 2= + − 2 cosr a ar θθ′ 2 2= ( + cos ) + ( sin )PF a r θ r2 2= + + 2 cosr a ar θ′ 2 2 2 2 2 2 2= ( + ) − 4 cos =PF PF r a a r θ a4 2 2 4 2 2 2 4( + 2 + ) − 4 cos =r a r a a r θ a4 2 2 2 2 2− 4 cos θ + 2 = 02 2 2 2( −2 (2cos − 1)) = 0r a r a rr r a θ2 2 2−2 (2cos − 1) = 0r a θ2 2= 2 (1+ cos2 − 1)r a θ2 2r = 2a cos2θThis is the equation of a lemniscate.42. Consider the following figure.AP r cosθThen tanθ= =BA 2acosθ− rsinθsinθr cosθ=cosθ 2acosθ − rsinθθ2asin cos2− rsin 2= rcos2 2θ θ θ θrcos θ + rsin θ = 2asinθ cosθr = asin 2θThis is a polar equation for a four-leaved rose.c.2 2x – y = 12 2 2 2r cos θ – r sin θ = 12 1r =cos 2θ1r =±cos 2θd. 4xy = 124r cosθsinθ = 12 1r =2sin2θ1r =±2sin2θe. y = 3x + 2r sin θ = 3r cos θ + 2r(sin θ – 3 cos θ) = 22r =sin θ – 3cosθf.g.23x+ 4y= 22 23rcos θ + 4rsinθ= 22 2(3cos ) r + (4sin ) r – 2 = 0θθ2 2–4sinθ ± 16sin θ + 24cos θr =26cos θ2 2–2sinθ ± 4sin θ + 6cos θr =23cos θ2 2x + 2 x+ y –4 y–25=02r + 2rcos θ –4rsin θ –25=0r2+ (2cos θ – 4sin θ) r – 25 = 02–2cosθ + 4sin θ ± (2cos θ – 4sin θ) + 100r =22= – cos + 2sin ± (cos – 2sin ) + 25r θ θ θ θ646 Section 10.6 Instructor’s Resource Manual

44.⎛θ⎞49. r = 1+ 3cos⎜ ⎟⎝ 3 ⎠The curve repeats itself after period p iff θ + p = f θ .( ) ( )( θ + p)⎛8 ⎞ ⎛8θp ⎞cos⎜ ⎟= cos⎜ + ⎟⎝ 5 ⎠ ⎝ 5 5 ⎠pWe need = 2π.545. a. VIIb. Ic. VIIId. IIIe. Vf. II⎛ π ⎞50. a. The graph of r = 1+ sin ⎜θ– ⎟ is the⎝ 3 ⎠πrotation of the graph of r = 1 + sin θ by 3counter-clockwise about the pole. The graph⎛ π ⎞of r = 1+ sin⎜θ+ ⎟ is the rotation of the⎝ 3 ⎠πgraph of r = 1 + sin θ by clockwise about3the pole.b. r = 1 – sin θ = 1 + sin(θ – π )The graph of r = 1 + sin θ is the rotation ofthe graph of r = 1 – sin θ by π about thepole.46.47.48.g. VIh. IVr =21–0.5sin⎛13θ⎞r = cos⎜ ⎟⎝ 5 ⎠⎛5θ⎞r = sin ⎜ ⎟⎝ 7 ⎠θ⎛ π ⎞c. r = 1+ cosθ = 1+ sin⎜θ+ ⎟⎝ 2 ⎠The graph of r = 1 + sin θ is the rotation ofthe graph of r = 1 + cos θ by 2π counterclockwiseabout the pole.d. The graph of r = f(θ) is the rotation of thegraph of r = f(θ – α) by a clockwise aboutthe pole.51. a. The graph for φ = 0 is the graph for φ ≠ 0rotated by φ counterclockwise about thepole.b. As n increases, the number of “leaves”increases.c. If a > b , the graph will not pass throughthe pole and will not “loop.” If b < a , thegraph will pass through the pole and willhave 2n “loops” (n small “loops” and n large“loops”). If a = b , the graph passesthrough the pole and will have n “loops.” Ifab ≠ 0, n > 1, andφ = 0 , the graph will beπsymmetric about θ = k,where k = 0, n−1.nInstructor’s Resource Manual Section 10.7 647

52. The number of loops is 2n.2. r = 2a cos θ, a > 053. The spiral will unwind clockwise for c < 0. Thespiral will unwind counter-clockwise for c > 0.54. This is for c = 4π .The spiral will wind in the counter-clockwisedirection.55. a. IIIb. IV1 π2 2 π 2A = (2acos θ ) dθ 2a cos θ dθ2∫ =0 ∫ 0π2 π 2 ⎡ 1 ⎤ 2= a ∫ (1 + cos 2 θ) dθ = a θ sin 2θa0⎢ + = π2⎥⎣ ⎦03. r = 2 + cos θc. Id. IIe. VIf. V10.7 Concepts Review1.2.3.1 22 r θ1 β 2[ f ( )] d2∫ θ θα1 2π2(2 2cos θ ) dθ2∫ +04. f(θ) = 01 2π2A = (2 cos θ ) dθ2∫ +01 2π2= (4 4cosθ cos θ)dθ2∫ + +01 2π ⎡1 ⎤= 4 4cos (1 cos2 ) d2∫+ + +0 ⎢2 ⎥⎣⎦1 2π⎛9 1 ⎞= 4cosθ cos2θ dθ2∫0 ⎜ + + ⎟⎝2 2 ⎠2π1⎡9 1 ⎤ 9= θ 4sinθ sin2θ2⎢ + +2 4⎥ = π⎣⎦024. r = 5 + 4 cos θθ θ θProblem Set 10.71. r = a, a > 01 2π2 2A = a dθa2∫ = π01 2π2A = (5 4cos θ ) dθ2∫ +01 2π2= (25 40cosθ 16cos θ)dθ2∫ + +01 2π= [25 40cos 8(1 cos 2 )] d2∫ + + +01 2π= (33 40cosθ 8cos 2 θ)dθ2∫ + +02= 1 [33 θ + 40sin θ + 4sin2 θ ] 0π = 33 π2θ θ θ648 Section 10.7 Instructor’s Resource Manual

5. r = 3 – 3 sin θ7. r = a(1 + cos θ)6.1 2π2A = (3 – 3sin θ ) dθ2∫ 01 2π2= (9 –18sinθ 9sin θ)dθ2∫+01 2π ⎡9 ⎤= 9–18sin θ (1–cos2 θ)dθ2∫+0 ⎢2 ⎥⎣⎦1 2π⎛27 9 ⎞= –18sin θ – cos2θ dθ2∫0 ⎜ ⎟⎝ 2 2 ⎠2π1⎡27 9 ⎤ 27= θ 18cos θ – sin 2θ2⎢ +2 4⎥ = π⎣⎦028.1 2π2A = [ a(1 cos θ )] dθ2∫ +02a 2π2= (1 2 cos cos ) d2∫ + +02a 2π ⎡ 1=2∫ + + +0 ⎢ 22θ θ θ⎤1 2cos θ (1 cos2 θ)⎥ dθ⎣⎦a 2π⎛3 1 ⎞= 2cosθ cos2θ dθ2∫ 0⎜ + + ⎟⎝2 2 ⎠2 2π2a ⎡3 1 ⎤ 3πa= θ 2sinθ sin2θ2⎢ + +2 4⎥ =⎣⎦022r = 6cos2θ1 2π2A = (3 3sin θ ) dθ2∫ +01 2π2= (9 18sinθ 9sin θ)dθ2∫ + +01 2π ⎡9 ⎤= 9 18sin (1 cos2 ) d2∫+ + −0 ⎢2 ⎥⎣⎦1 2π⎛27 9 ⎞= 18sinθ cos 2θ dθ2∫0⎜ + − ⎟⎝ 2 2 ⎠2π1⎡27 9 ⎤= θ 18cosθ sin 2θ2⎢ − −2 4⎥⎣⎦027= π2θ θ θ9.1 π/4 π/4A = 2⋅ 6cos2 62∫ d =– π/4 ∫ – π/4cos2 d/4= 3[sin 2 θ ] – π /4 = 62r = 9sin2θθ θ θ θ1 π/2 π/2A = 2⋅ 9sin2 d 9 sin2 d2∫ =0 ∫ 0θ θ θ θ/2= 9 [– cos 2 θ ] 0π = 92Instructor’s Resource Manual Section 10.7 649

10.2r = acos 2θ12. r = 2−4cosθ1 π/4 π/4A = 2⋅ cos 22∫ a d = a– π/4 ∫ −π/4cos 2 da /4= [sin 2 θ ] – π /4 = a211. r = 3 – 4 sin θ–1 341 π /2 2A = 2 ⋅–1(3–4sin θ ) dθ32∫ sin4π /2 2= ∫ (9 – 24sin + 16sin )dsin3 – 4 sin θ = 0, θ = sin–1 34π /2sin∫–1 34θ θ θ θθ θ θ= [9 – 24sinθ + 8(1– cos 2 θ)]dθ= ∫π /2sin–1 34(17 – 24sin θ – 8cos 2 θ)dθ= [17θ + 24 cos θ – 4 sin 2 θ]π/2–1 3 sin4= [17θ + 24cos θ – 8sinθ cos θ]π/2–1 3 sin417π ⎡–1 3 ⎛ 7 ⎞ ⎛3 ⎞⎛ 7 ⎞⎤= – ⎢17sin + 24 – 82 4 ⎜ ⎥4 ⎟ ⎜ ⎟⎝4⎠⎜ 4 ⎟⎢⎣⎝ ⎠ ⎝ ⎠⎥⎦17π–1 3 9 7= –17sin –2 4 2π2 – 4 cos θ = 0, θ =31 π /3 2A = 2 ⋅ (2–4cos θ ) dθ2∫ 0π /3 2= ∫ (4 –16cosθ + 16cos θ)dθ0π /3= ∫ [4 –16cos + 8(1 + cos2 )]d0π /3= ∫ (12 − 16cosθ + 8cos 2 θ)dθ0θ θ θ[ 12θ 16sinθ 4sin 2θ] π /3= − += 4π−6 313. r = 2 – 3 cos θ–1 22 – 3 cos θ = 0, θ = cos31 π2A = 2 ⋅ (2–3cos θ ) dθ2∫∫π–1cos 23–1cos 23= (4 –12cosθ + 9cos θ)dθπ ⎡ 9 ⎤= ∫ –14 –12cos θ (1 cos 2 θ)dθcos 2 ⎢ + +32 ⎥⎣⎦π ⎛17 9 ⎞= ∫ –1–12cosθ cos2θ dθcos 2 ⎜+ ⎟3 ⎝ 2 2 ⎠π⎡17 9 ⎤= ⎢ θ –12sinθ + sin2θ2 4⎥⎣⎦cos⎡17θ9 ⎤= ⎢ –12sinθ + sinθ cosθ2 2⎥⎣⎦02–1 23πcos–1 2317π ⎡17 –1 2 ⎛ 5 ⎞ 9 ⎛ 5 ⎞⎛2⎞⎤= – ⎢ cos –122 2 3 ⎜ +⎥3 ⎟ ⎟⎜ ⎟2⎜ 3 ⎟⎢⎣⎝ ⎠ ⎝ ⎠ ⎝ 3 ⎠ ⎥⎦17π17 –1 2= – cos + 3 52 2 3650 Section 10.7 Instructor’s Resource Manual

14. r = 3 cos 2θ18. r = 3sin θ , r = 1+sinθ1 π/4 2 π/42= 2 ⋅ (3cos2 ) 9 cos 22∫ =0 ∫ 0π /41= 9 ∫ (1+cos4 θ ) dθ0 2π /49⎡1 ⎤ 9π= θ sin 4θ2⎢ +4⎥ =⎣ ⎦08A θ dθ θ dθ15. r = 4cos3θSolve for the θ -coordinate of the firstintersection point.3sinθ= 1+sinθ1sinθ =2πθ =61 π /2 2 2A = 2 ⋅ [(3sin θ ) –(1 sin θ) ] dθ2∫+π /6π /2 2= ∫ (8sin θ – 2sin θ –1)d θ= ∫π /6π /2π /6(3 – 4cos 2 θ – 2sin θ )d θ/2/6= [3 θ – 2sin 2θ + 2cos θ]π π =π1 π/6 2 π/62A = 6 ⋅ (4cos3 θ ) dθ 48 cos 3θ dθ2∫ =0 ∫ 0π /6π /6⎡ 1 ⎤= 24 ∫ (1 + cos6 θ) dθ = 24 θ sin 6θ40⎢ + = π6⎥⎣ ⎦019.2r = 2, r = 8cos 2θ16. r = 2sin3θ17.1 π/3 2 π/32= 3 ⋅ (2sin 3 ) 6 sin 32∫ =0 ∫ 0π /3A θ dθ θ dθ∫= 3 (1−cos6 θ )dθ0π /3⎡ 1 ⎤= 3⎢θ− sin6θ=π6⎥⎣ ⎦01 2π1 2πA= 100 dθ– 49dθ512∫ = π0 2∫0Solve for the θ -coordinate of the firstintersection point.4=8cos2θ1cos 2θ =2π2θ=3πθ =61 π /6A = 4 ⋅ (8cos 2 θ – 4) dθ2∫ 0=/6θ π2[4sin 2 θ – 4 ] 04π= 4 3−3Instructor’s Resource Manual Section 10.7 651

20. r = 3 − 6sinθ22. r = 2+ 2sin θ , r = 2+2cosθLet A 1 be the area inside the large loop and letA 2 be the area inside the small loop.1 π /6 2A1= 2 ⋅ (3–6sin θ ) dθ2∫ – π /2∫π /6 2– π /2π /6– π /2= (9 – 36sin θ + 36sin θ )d θ= ∫(27 – 36sin θ –18cos 2 θ )d θ[ ] /6 27 3= 27θ + 36cosθ − 9sin 2θ π = 18π+−π /221 π /2 2A2= 2 ⋅ (3–6sin θ ) dθ2∫ π /6/2 27 3= [27θ + 36cos θ – 9sin 2 θ] π π /6 = 9 π –2A= A1− A2 = 9π+27 321. r = 3+ 3cos θ , r = 3+3sinθSolve for the θ -coordinate of the intersectionpoint.3+ 3cosθ= 3+3sinθtanθ = 1πθ =41 π /4 2 2A = [(3 3cos θ ) – (3 3sin θ) ] dθ2∫ + +01 π /4 2 2= (18 cosθ 9 cos θ – 18sin θ – 9 sin θ)dθ2∫+01 π /4= (18cos θ –18sinθ 9cos 2 θ)dθ2∫+0π /41⎡9 ⎤= 18sinθ 18cosθ sin 2θ2⎢ + +2⎥⎣⎦27= 9 2−401 π2 2A = [(2 2sin θ ) (2 2cos θ) ] dθ2∫ + − +π /21 π2 2= (8sinθ 4sin θ – 8cos θ – 4cos θ)dθ2∫ +π /21 π= (8sin θ – 8cos θ – 4cos 2 θ)dθ2∫ π /21= [ −8cosθ −8sinθ − 2sin2θ] π /2= 82π23. a. f( θ ) = 2cos θ, f′( θ) =− 2sinθ(2cos θ )cos θ + ( −2sin θ)sinθm = − (2cos θ )sin θ + ( − 2sin θ)cosθ2 22cos θ − 2sin θ cos2θ= =−4cosθ sinθ −sin2θ1232π − 1At θ = , m = = .3 − 3b. f( θ ) = 1+ sin θ, f′( θ) = cosθ(1 + sin θ )cos θ + (cos θ)sinθm = − (1 + sin θ )sin θ + (cos θ)cosθcosθ + 2sinθ cosθ cosθ + sin 2θ= =2 2cos θ −sin θ −sinθcos 2θ− sinθAt θ π, 13 m+= = = − .− −312 21 32 2c. f( θ ) = sin2 θ, f′( θ) = 2cos2θ(sin 2 θ )cos θ + (2cos 2 θ)sinθm = − (sin 2 θ )sin θ + (2cos 2 θ) cosθπAt θ = , .33 3( )( ) ( 1) ( )3 3 1( 2 )( 2 ) + ( −1)( 2)1+ − 32 2 2 −4−54m = = =−35652 Section 10.7 Instructor’s Resource Manual

d. f( θ ) = 4− 3cos θ, f′( θ) = 3sinθ(4 − 3cos θ )cos θ + (3sin θ)sinθm = − (4 − 3cos θ )sin θ + (3sin θ)cosθ2 24cosθ − 3cos θ + 3sin θ=− 4sinθ + 6sinθ cosθ= 4cosθ− 3cos2θ− 4sinθ+ 3sin 2θπAt θ = ,3( ) − ( − )4 1 3 1 72 2 2 7m = = =−3− 4 + 3 − 323 3( 2 ) ( 2 )24. f( θ ) = a(1+ cos θ), f′( θ) = − asinθa(1 + cos θ )cos θ + ( −asin θ)sinθm = − a(1 + cos θ )sin θ + ( − asin θ)cosθ2 2 2cosθ + cos θ − sin θ 2cos θ + cosθ−1= =−sinθ −2sinθ cosθ − sin θ(1 + 2cos θ)a. m = 0 when 2cos θ + cosθ− 1=0(2cosθ− 1)(cosθ+ 1) = 0.21cos θ = , cosθ= − 12π πθ = , − , θ = π; when3 3θ =π , f ( θ) = 0, so θ =π is the tangent line.⎛3aπ⎞ ⎛3aπ⎞⎜ , ⎟, ⎜ , − ⎟, (0, π)⎝ 2 3⎠ ⎝ 2 3⎠.b. m is undefined when sin θ (1 + 2cos θ ) = 02and 2cos θ + cosθ−1≠ 0.2π4πθ = 0, ,3 3⎛a2π⎞ ⎛a4π⎞( 2 a, 0 ), ⎜ , ⎟, ⎜ , ⎟⎝2 3 ⎠ ⎝2 3 ⎠There is no vertical tangent at θ = π sincelim m θ = 0 (see part (a)).θ →0( )25. f( θ ) = 1− 2sin θ, f′( θ) = − 2cosθ(1−2 sin θ ) cos θ + ( −2 cos θ) sinθm = − (1 − 2sin θ )sin θ + ( − 2cos θ)cosθcosθ − 4sinθ cosθ=−2 2sin + 2sin − 2coscos θ(1 − 4sin θ)=−2 2sin + 2sin − 2cosθ θ θθ θ θm = 0 when cos θ (1 – 4 sin θ) = 0cosθ= 0, or 1− 4sinθ= 0π 3π −1⎛1⎞θ = , θ = , θ = sin ⎜ ⎟≈0.25,2 2 ⎝4⎠−1 ⎛1⎞θ =π−sin ⎜ ⎟≈2.89⎝4⎠⎛π⎞ ⎛3π⎞ ⎛ −1⎛1⎞⎞1⎜ ⎟=− 1, ⎜ ⎟= 3, ⎜sin ⎜ ⎟⎟=,⎝2⎠ ⎝ 2 ⎠ ⎝ ⎝4⎠⎠2f f f⎛ −1 ⎛1⎞⎞1f ⎜π − sin ⎜ ⎟⎟=⎝ ⎝4⎠⎠2⎛ π⎞ ⎛ 3π⎞ ⎛1 ⎞ ⎛1⎞⎜−1, ⎟, ⎜3, ⎟, ⎜ , 0.25 ⎟, ⎜ , 2.89⎟⎝ 2⎠ ⎝ 2 ⎠ ⎝2 ⎠ ⎝2⎠2 2b ⎛dx⎞ ⎛dy⎞26. Recall from Chapter 5 that L = ∫ dta ⎜ ⎟ + ⎜ ⎟ for x and y functions of t and a ≤ t ≤ b.⎝ dt ⎠ ⎝ dt ⎠x = rcos θ = f( θ)cos θ, y = rsin θ = f( θ)sinθdxdy= f′ ( )cos f( )sin , f ( )sin f( )cosdθ θ − θ θ = ′θdθθ θ + θ θβ2 2L = ∫ ( f′ ( θ )cos θ − f( θ)sin θ) + ( f′( θ)sin θ + f( θ)cos θ)dθαβ= + + +α2 2 2 2 2 2∫ [ f ( θ )] (sin θ cos θ) [ f′( θ) ] (sin θ cos θ)dθ= ∫ [ ( )] + [ ′( )]βα2 2f θ f θ dθ27. f( θ ) = a(1+ cos θ), f′( θ) = − asinθ2π2 2 2π2π1+cosθ2πθL = ∫ [ a(1 + cos θ )] + [– asin θ]dθ= a 2+2cosθ dθ0∫= 2a dθ2a cos dθ0∫ =0 2∫ 0 2⎡ π θ 2πθ ⎤= 2a⎢∫cos dθ– cos dθ0 2∫π2 ⎥⎣⎦⎛ π2π⎡ θ ⎤ ⎡ θ ⎤ ⎞= 2a2sin − 2sin = 8a⎜⎢ 2⎥ ⎢0 2⎥ ⎟⎝⎣ ⎦ ⎣ ⎦π⎠Instructor’s Resource Manual Section 10.7 653

28.1f ( θ) = e , f′( θ)= e2θ /2 θ /222 π /2 2 1 /2L [ e θ ⎡ θ ⎤ 2π5 θ 2 π 5 θ /2θ /22ππ= ∫ ] + e dθ0⎢2 ⎥ = ∫ e dθ= e dθ⎣ ⎦ 0 4∫= ⎡ 5e⎤ = 5( e −1) ≈ 49.510 2 ⎣ ⎦029. If n is even, there are 2n leaves.1 π /2n2A = 2 n ( cos )2∫ a nθdθ= na– π /2n∫π /2n2 ⎡1 sin2nθ⎤ 1 2= na ⎢ θ + a2 4n⎥ = π⎣⎦−π/2n2If n is odd, there are n leaves.1 π /2n22 naA = n⋅ ( acos nθ) dθ2∫=– π /2n2∫2 π /2n2– π /2ncosπ /2ncos 2– π /2nnθdnθd2θ = na ∫2π /2n– π /2n1+cos2nθd θ2π /2nna ⎡1 sin2nθ⎤ 1 2θ = a2⎢ θ + = π2 4n⎥⎣⎦ 4– π /2n30. r = secθ − 2cosθ31. a. Sketch the graph.Solve for the θ -coordinate when r = 0.secθ− 2cosθ= 02 1cos θ =21cosθ =±2π 3πθ =± , ±4 4π πNotice that the loop is produced for − ≤θ≤ .4 41 π /4 2A = (secθ 2cos θ)dθ2∫ −– π /41 π /4 2 2= (sec θ 4 4cos θ)dθ2∫ − +– π /41 π /4 2= (sec θ – 2 2cos 2 θ)dθ2∫+– π /4= 1 [ tan θ − 2 θ + sin 2 θ] π /42−π /41 ⎡ ⎛1 π ⎞ ⎛1 1 π ⎞= 1 ⎤ 2π2⎢⎜ − + ⎟−⎜− + − ⎟ = −2 2⎥⎣⎝ ⎠ ⎝ ⎠⎦2Solve for the θ -coordinate of the intersection.2a sinθ = 2b cosθb−1⎛b⎞tanθ = ; θ = tan ⎜ ⎟a⎝a⎠−1⎛b⎞Let θ0 = tan ⎜ ⎟⎝a⎠ .A 1 θ02 1 /2 2(2 sin ) (2 cos )2 0a d π= ∫ θ θ +2∫θb θ d θ0θπ /2= 2a sin θ dθ + 2b cos θ dθ∫ ∫2 0 2 2 20θ02 θ02 π /2∫0 ∫θ0θ0sin 2θsin 2= a (1 – cos 2 θ ) d θ + b (1+cos 2 θ)d θ⎡ ⎤ ⎡ θ ⎤= ⎢ − θ2 ⎥ + ⎢ +2 ⎥⎣ ⎦ ⎣ ⎦θ2 2a θb0π /22 22 2⎛π ⎞ a + ba θ0 b θ0 sin 2θ0= + ⎜ − ⎟−⎝2 ⎠ 22 2⎛π⎞ 2 2= a θ0 + b ⎜ −θ0⎟− ( a + b )sinθ0cosθ0⎝2⎠⎛π⎞= + − −2 −1⎛b⎞ 2 −1⎛b⎞a tan ⎜ ⎟ b ⎜ tan ⎜ ⎟⎟ab.⎝a⎠ ⎝2⎝a⎠⎠batan θ = , cos θ =a 2 2a + bbθ =2 2a bNote that sinceand sin .+0654 Section 10.7 Instructor’s Resource Manual

. Let m 1 be the slope of r = 2asinθ.2asinθ cosθ + 2acosθsinθm1= − 2asinθ sinθ + 2acosθ cosθ2sinθcosθ=2 2cos θ − sin θ−1⎛b⎞2abAt θ = tan ⎜ ⎟,m1 = .⎝a ⎠2 2a − bAt θ = 0 (the pole), m 1 = 0 .Let m 2 be the slope of r = 2bcosθ.2bcosθ cosθ − 2bsinθsinθm2= − 2bcosθ sinθ − 2bsinθ cosθ2 2cos θ − sin θ= − 2sinθcosθ2 2−1⎛b⎞a − bAt θ = tan ⎜ ⎟,m2=− .⎝a⎠2abπAt θ = (the pole), m 2 is undefined.2Therefore the two circles intersect at rightangles.32. The area swept from time t 0 to t 1 isθ ( t1) 1 2A = ∫ r dθ.θ ( t0) 2By the Fundamental Theorem of Calculus,dA 1 2 dθ= r .dt 2 dtdA kSo = where k is the constant angulardt 2mmomentum.Therefore, dAkis a constant so A = ( t1− t0).dt2mEqual areas will be swept out in equal time.33. The edge of the pond is described by the equation r = 2acosθ.Solve for intersection points of the circles r = ak and r = 2acosθ.ak = 2a cosθk −1⎛k⎞cos θ = , θ = cos ⎜ ⎟2 ⎝2⎠Let A be the grazing area.1 2 1 π /2 2 2 1 2 2 2 π /2 2 2A = π ( ka) + 2 ⋅ [( ) (2 cos ) ]2 2 coskka − a θ dθ= k a π+ a–1 ( –4cos )2∫ coskk θ dθ1= π+ − −2∫ –1( )( )2 2 2 π /2 2k a a k d1= π+ − −212 2 2 2∫ –1 (( 2) 2cos 2 )coskθ θ = k a π+ a ⎡( k −2) θ −sin2θ⎤−1( 2 )2 ⎣⎦cosk( )2 2 2 2π /2k a a ⎡k θ 2θ 2sinθ cosθ⎤− k2 ⎣⎦cos21( )1⎡4= π+ −π− cos ⎜ ⎟+ 2cos ⎜ ⎟+2 ⎢ 2 ⎝2⎠ ⎝2⎠2⎣⎡22 2 2 1 k k 4 k⎤− ⎛ ⎞ −= a ⎢( k −1) π+ (2 − k )cos ⎜ ⎟+⎥⎢⎝2⎠2 ⎥⎣⎦2 22 2 2 k π 2 −1 k −1k a a kk k −⎢⎛ ⎞ ⎛ ⎞ k⎤⎥⎥⎦2π /22Instructor’s Resource Manual Section 10.7 655

34. PT = ka − φa; φ goes from 0 to k.k1 k2 1⎡13⎤A= ( ) ( )2∫ka− φa dφ = ka φa02⎢− −3a⎥⎣⎦0= 12 36 a kThe grazing area is2 32 2 k kka A a ⎛ π ⎞1π ( ) + 2 = +.2 ⎜ 2 3 ⎟⎝ ⎠35. The untethered goat has a grazing area of π a .From Problem 34, the tethered goat has a grazing2 32 k karea of a ⎛ π+⎞.⎜ 2 3 ⎟⎝ ⎠2 32 2 k ka a ⎛ π ⎞π = +⎜ 2 3 ⎟⎝ ⎠2 3πk kπ= +2 33 2k k2 + 3π −6π=0Using a numerical method or graphing calculator,k ≈ 1.26. The length of the rope is approximately1.26a.36. f( θ ) = 2+ cos θ, f′( θ) = − sinθ37.π∫0π02 2dL = 2 [2+ cos θ ] + [ −sin θ]θ= 2∫ 5+4cosθdθ≈ 13.36f( θ ) = 2+ 4cos θ, f′( θ) = − 4sinθπ∫ 0π0[ ] [ ]2 2dL = 2 2+ 4cosθ + −4sinθ θ= 4∫ 5+4cosθdθ≈ 26.731 π/3 2 π/32A = 3 ⋅ (4sin3 θ ) dθ 24 sin 3θ dθ2∫ =0 ∫ 0π /3= 12 ∫ (1 −cos 6 θ ) dθ0π /3⎡ sin 6θ⎤= 12 ⎢θ− = 4π6⎥⎣ ⎦0f( θ ) = 4sin3 θ, f′( θ) = 12cos3θπ /3 2 2L = 3 ∫ (4sin3 θ ) + (12cos3 θ)dθ0π /3 2 2= 3∫16sin 3θ + 144cos 3θ dθ0π /3 2= 12∫1+ 8cos 3θdθ≈ 26.730238.39.1 π /4 π /4A= 4⋅ 8cos2θ dθ = 2[4sin2 θ] 00 = 82∫−8sin2θf( θ) = 8cos2 θ, f′( θ)=8cos2θπ /4 8sin 2θL = 4∫8cos2θ+ dθ0cos 2θπ /4 8= 4∫0 cos 2θdθπ /4 1= 8 2∫14.830 cos 2θdθ ≈⎛3θ⎞r = 4sin ⎜ ⎟, 0 ≤θ≤ 4π⎝ 2 ⎠⎛3θ⎞ ⎛3θ⎞f( θ) = 4sin ⎜ ⎟, f′( θ) = 6cos⎜ ⎟⎝ 2 ⎠ ⎝ 2 ⎠22 2θd4π ⎡ ⎛3θ⎞⎤ ⎡ ⎛3⎞⎤L = ∫ 4sin 6cos0 ⎢ ⎜ ⎟ + ⎜ ⎟2⎥ ⎢2⎥⎣ ⎝ ⎠⎦ ⎣ ⎝ ⎠⎦4π 2⎛3θ⎞ 2⎛3θ⎞= ∫ 16sin 36cos dθ0⎜ ⎟+⎜ ⎟⎝ 2 ⎠ ⎝ 2 ⎠4π 2 ⎛3θ⎞= ∫ 16 + 20cos dθ≈63.460⎜ ⎟⎝ 2 ⎠10.8 Chapter ReviewConcepts Test1. False: If a = 0, the graph is a line.2. True: The defining condition of a parabola isPF = PL . Since the axis of a parabolais perpendicular to the directrix and thedistance from the vertex to the directrixis equal to the distance to the focus, thevertex is midway between the focus andthe directrix.3. False: The defining condition of an ellipse isPF = e PL where 0 < e < 1. Hence thedistance from the vertex to a directrix isa greater than the distance to a focus.θ4. True: See Problem 33 in Section 10.1.656 Section 10.8 Instructor’s Resource Manual

5. True: The asymptotes for both hyperbolas areby =± x.a6. True:2π2 2 2 2C = ∫ a sin t+b cos t dt;02π2π b = ∫ 02 2 2 2b sin t+b cos t dt2π< C < ∫ 02 2 2 2a sin t+ a cos t dt = 2πa7. True: As e approaches 0, the ellipse becomesmore circular.8. False: The equation can be rewritten as2 2x y+ = 1 which is a vertical ellipse4 6with foci on the y-axis.9. False: The equation x 2 – y 2 = 0 represents thetwo lines y =± x .10. True:2y x22y x( − 4 + 1) = 0 implies− 4 + 1= 0 which is an equation fora parabola.11. True: If k > 0, the equation is a horizontalhyperbola; if k < 0, the equation is avertical hyperbola.12. False: If k < 0, there is no graph.13. False: If b > a, the distance is14. True: If y = 0,22 22 b – a .x = –2 which is not possible.915. True: Since light from one focus reflects to theother focus, light emanating from apoint between a focus and the nearestvertex will reflect beyond the otherfocus.16. True:8 2a = = 4, c = = 1,2 2b = 16 − 1 = 15 . The length of theminor diameter is 2b = 60 .17. True: The equation is equivalent to2 2 2 2C D C Dx y F⎛ ⎞ ⎛ ⎞⎜ + ⎟ + ⎜ + ⎟ = – + + .⎝ 2 ⎠ ⎝ 2 ⎠ 4 4Thus, the graph is a circle if2 2C D– F + + > 0, a point if4 42 2C D– F + + = 0, or the empty set if4 42 22 C D– F + + < 0.4 418. False: The equation is equivalent to2 2 2 2C D C Dx y F⎛ ⎞ ⎛ ⎞2 ⎜ + ⎟ + ⎜ + ⎟ = – + + .⎝ 4 ⎠ ⎝ 2 ⎠ 8 4Thus, the graph can be a point if2 2C D– F + + = 0.8 419. False: The limiting forms of two parallel linesand the empty set cannot be formed insuch a manner.20. True: By definition, these curves are conicsections, which can be expressed by anequation of the form2 2Ax + Cy + Dx + Ex + F = 0.21. False: For example, xy = 1 is a hyperbola withcoordinates only in the first and thirdquadrants.22. False: For example, the graph of2 2x + 3xy+ y = 1 is a hyperbola thatpasses through the four points.23. False: For example, x = 0, y = t, and x = 0,y = –t both represent the line x = 0.24. True: Eliminating the parameter givesx = 2y.25. False: For example, the graph ofx = t 2 , y = t does not represent y as afunction of x. y =± x , buthx ( )=± xis not a function.26. True: When t = 1, x = 0, and y = 0.27. False: For example, ify = x so2d y2dx3 3x t y t= , = then= 0 , butg′′() t= 1 .f′′() tInstructor’s Resource Manual Section 10.8 657

28. True: For example, the graph of the fourleavedrose has two tangent lines atthe origin.29. True: The graph of r = 4 cos θ is a circle ofradius 2 centered at (2, 0). The graph⎛ π ⎞r = 4cos ⎜θ– ⎟ is the graph of⎝ 3 ⎠πr = 4 cos θ rotated 3counter-clockwise about the pole.30. True: (r, θ) can be expressed as (r, θ + 2π n)for any integer n.f.g.1 1+ 4 = ; – + + 4 = ;4 42 2 2 2x y x x x y( x – ) 21 22 y+ = 11 14 16(8) An ellipse2 2 2 1 2 1x + 4 y = − x; x + x+ + 4 y = ;4 4( x + ) 21 22 y+ = 11 14 16(8) An ellipse31. False: For example, if f ( θ ) = cosθandg( θ ) = sinθ, solving the two equationssimultaneously does not give the pole⎛ π ⎞( which is ⎜0, ⎟ for f ( θ ) and (0, 0)⎝ 2 ⎠for g( θ ) ).32. True: Since f is odd f(–θ) = –f(θ). Thus, if wereplace (r, θ) by (–r, –θ), the equation –r= f(–θ) is –r = –f(θ) or r = f(θ).Therefore, the graph is symmetric aboutthe y-axis.33. True: Since f is even f(–θ) = f(θ). Thus, if wereplace (r, θ) by (r, –θ), the equationr = f(–θ) is r = f(θ). Therefore, the graphis symmetric about the x-axis.34. True: The graph has 3 leaves and the area isexactly one quarter of the circle r = 4.(See Problem 15 of Section 12.8.)2.h.i.j.2 2x + 4y=− 1(1) No graph2 2 2( x + 4y− 1) = 0; x + 4y− 1=0(7) A parabola2 2 2 2 2 13x + 4y =− x + 1; x + y =4(6) A circle2 2 2 ⎛3⎞y − 6x = 0; y = 6 x; y = 4 ⎜ ⎟ x⎝ 2 ⎠3Horizontal parabola; opens to the right; p =2⎛3 ⎞Focus is at ⎜ , 0 ⎟ and vertex is at (0, 0).⎝2⎠Sample Test Problems1. a.b.2 2 xx − 4y = 0; y =±2(5) Two intersecting lines2 22 2x yx – 4y = 0.01; – = 10.01 0.0025(9) A hyperbola3.2 22 2 x y9x+ 4y− 36= 0; + = 14 9Vertical ellipse; a = 3, b = 2, c = 5Foci are at ( 0, ± 5 ) and vertices are at (0, ±3).c.2x− 4= 0; x = ± 2(4) Two parallel linesd.2x x x− 4 + 4= 0; = 2(3) A single linee.2 2x y+ 4 = 0; (0, 0)(2) A single point658 Section 10.8 Instructor’s Resource Manual

4.2 22 2 y x25x− 36y+ 900 = 0; − = 125 36Vertical hyperbola; a = 5, b = 6, c = 61Foci are at ( 0, ± 61)and vertices are at (0, ±5).7.2 22 2 x y9x+ 25y− 225= 0; + = 125 9Horizontal ellipse, a = 5, b = 3, c = 4Foci are at (±4, 0) and vertices are at (±5, 0).5.2 2 2 ⎛9⎞x + 9 y = 0; x = –9 y; x = –4 ⎜ ⎟y⎝ 4 ⎠9Vertical parabola; opens downward; p =4⎛ 9 ⎞Focus at ⎜0,− ⎟ and vertex at (0, 0).⎝ 4 ⎠8.2 2 2 29x + 9y − 225= 0; x + y = 25Circle; r = 56.2 22 2 x yx −4y− 16= 0; − = 116 4Horizontal hyperbola; a = 4, b = 2, c = 2 5Foci are at ( ± 2 5, 0)and vertices are at (±4, 0).9.5( )(1)2( )5r = =2+ 2sinθ 1 + (1)cos θ −π2e = 1; parabola⎛ 5 ⎞Focus is at (0, 0) and vertex is at ⎜0, ⎟⎝ 4 ⎠ (inCartesian coordinates).10.( )123r(2 + cos θ ) = 3; r =1 + 12cosθ1e = , ellipse2At θ = 0, r = 1. At θ =π , r = 3. .1+3a = = 2, c = ea = 12Center is at (–1, 0) (in Cartesian coordinates).Foci are (0, 0) and (–2, 0) and vertices are at (1,0) and (–3, 0) (all in Cartesian coordinates).Instructor’s Resource Manual Section 10.8 659

22 ⎛ 9 ⎞( x− 3) + ⎜y+ ⎟ = 9;⎝ 2 ⎠circle11. Horizontal ellipse; center at (0, 0), a = 4,c 1e = = , c = 2, b = 16 − 4 = 2 3a 22 2x y+ = 116 1212. Vertical parabola; opens downward; p = 32x =− 12y13. Horizontal parabola;2 2y = ax, (3) = a( − 1), a = − 92y =− 9x14. Vertical hyperbola; a = 3, c = ae = 5,b = 25 − 9 = 4 , center at (0, 0)2 2y x− = 19 1615. Horizontal hyperbola, a = 2,ax =± 2 y, = 2, b=1b2 2x y− = 14 116. Vertical parabola; opens downward; p = 12( x− 3) = −4( y−3)17. Horizontal ellipse; 2a = 10, a = 5, c = 4 – 1 = 3,b = 25 − 9 = 42 2( x−1) ( y−2)+ = 125 1618. Vertical hyperbola; 2a = 6, a = 3, c = ae = 10,b = 100 − 9 = 91 , center at (2, 3)2 2( y−3) ( x−2)− = 19 9120.21.22.2 24x + 9 y – 24 x– 36y+ 36 = 02 24( x – 6x+ 9) + 9( y – 4y+ 4) = –36 + 36 + 362 24( x– 3) + 9( y– 2) = 362 2( x–3) ( y–2)+ = 1; ellipse9 42x + 8x+ 6y+ 28=02( x + 8x+ 16) = –6 y– 28 + 162( x 4) –6( y 2);+ = + parabola2 23x − 10y + 36x− 20y+ 68=02 23( x + 12x+ 36) − 10( y + 2y+ 1) = − 68 + 108 −102 23( x+ 6) − 10( y+ 1) = 302 2( x+ 6) ( y+1)− = 1 ; hyperbola10 319.2 24x + 4y − 24x+ 36y+ 81 = 02 ⎛ 2 81⎞4( x − 6x+ 9) + 4⎜y + 9y+ ⎟= − 81+ 36 + 81⎝ 4 ⎠22 ⎛ 9 ⎞4( x− 3) + 4⎜y+ ⎟ = 36⎝ 2 ⎠660 Section 10.8 Instructor’s Resource Manual

23.24.2x = ( u – v )22y = ( u+v )21 2 3 1 2( u– v) + ( u– v)( u+ v) + ( u+ v) = 102 2 25 2 1 2u – v = 102 25 1r = , s = –2 22 2u v– = 1; hyperbola4 20a = 2, b = 2 5, c = 4+ 20 = 2 6The distance between foci is 4 6.2 27x + 8xy+ y = 93cot 2θ =43cos 2θ =526.27.yt =42y 2x = or y = 4x4x + 2 y −1sin t = , cost=4 32 2( x+ 2) ( y−1)+ = 116 91+35 2cosθ= =2 51– 35 1sinθ = =2 5−1 1θ = sin ≈ 0.463651x = (2 u – v )528.2x 2 2 2= sec t, y = tan t42x 2− y = 1425.1y = ( u+2 v )57 2 8 1 2(2 u– v) + (2 u– v)( u+ 2 v) + ( u+ 2 v) = 95 5 52 29 u – v = 922 vu – = 1; hyperbola91t = ( x –2)61y = ( x –2)329.dx 2 1 26 4, dy 1t += t − = + =dt dt t + 1 t + 1t+2dy t+ 1 t + 2= =dx 2 26t − 4 ( t+ 1)(6t−4)At t = 0, x = 7, y = 0, anddy 1dx = − 2.1Tangent line: y =− ( x− 7) or x+ 2 y− 7 = 02Normal line: y = 2(x – 7) or 2x – y – 14 = 0.Instructor’s Resource Manual Section 10.8 661

30.dx −tdy 1=− 3 e , = edt dt 21 tdy e2 1 2t= edx t3e − = −− 61 ,2At t = 0, x = 3,ty = anddy 1dx =− 6.33 r = 6cosθ1 1Tangent line: y− =− ( x− 3) or x + 6y – 6 = 02 61Normal line: y– = 6( x– 3) or212x – 2y – 35 = 0.34.5r =sinθ31. One approach is to use the arc lengthformulab 2 2 9⎛dx ⎞ ⎛dy ⎞ 9t 9tL = ∫ ⎜ ⎟ + ⎜ ⎟ dt = dtdt dt∫ + =4 4a ⎝ ⎠ ⎝ ⎠ 0993 2⎡ 3tdt 2 t 2 ⎤= = 27 22∫ ⎢ ⎥0 ⎣ ⎦0Another way is to note that whent = 0, ( x, y) = (1, 2) , when t = 9 ,( xy , ) = (28, 29) , and y = x + 1, which is astraight line. Thus the curve length is simply thedistance between the points (1, 2) and (28,29)2 2or (28 − 1) + (29 − 2) = 27 235. r = cos 2θdx32. sin t sin t tcost tcostdt =− + + =dycost cost tsin t tsintdt = − + =2 2 2 2L ππ= ∫ (cos) t t + (sin) t t dt = tdt0 ∫ 02π⎡12 222 t ⎤= ⎢ ⎥ = π⎣ ⎦036.3r =cosθ37. r = 4662 Section 10.8 Instructor’s Resource Manual

38. r = 5−5cosθ43.2r = 16sin 2θr =± 4 sin2θ39. r = 4−3cosθ44. r = −θ , θ ≥ 040. r = 2−3cosθ45.r2 – 6(cos r θ + sin θ)+ 9=0x2 + y2 – 6 x–6y+9=02 2( x – 6x+ 9) + ( y – 6y+ 9) = –9 + 9 + 92 2( x– 3) + ( y– 3) = 941.2θ = π342. r = 4sin3θ46.r2 cos 2θ = 92 2 2 2r cos θ − r sin θ = 92 2x − y = 92 2x y− = 19 9Instructor’s Resource Manual Section 10.8 663

47. f( θ ) = 3+ 3cos θ, f′( θ) = − 3sinθ(3 + 3cos θ )cos θ + ( −3sin θ)sinθm = − (3 + 3cos θ )sin θ + ( − 3sin θ)cosθ2 2cos + cos − sin cos + cos 2= =−sin −2cos sin −sin −sin 2cosπcosππ +6 3At θ = , m = = −1.6 −sin π−sinπ6 3θ θ θ θ θθ θ θ θ θ48. r = 5sin θ , r = 2 + sinθ51.2 2x y x yy′+ = 1; + = 0400 100 200 50x 2y′ ; y′4y3= − = − at (16, 6)2Tangent line: y− 6 = − ( x−16)32 22When x = 14, y =− (14 − 16) + 6 = .3 322k =352 a. III b. IVc. I d. II53. a. I b. IVc. III d. II5sinθ= 2 + sinθ1 π 5πsinθ = ⇒ θ = ,2 6 6⎛5 π⎞ ⎛5 5π⎞⎜ , ⎟, ⎜ , ⎟⎝2 6⎠ ⎝2 6 ⎠Review and Preview Problems1. x = 2t, y = t − 3 ; 1≤ t ≤ 43y49.50.1 π2A = 2 ⋅ (5 5cos θ ) dθ2∫ −0π2= 25 ∫ (1 − 2cosθ + cos θ)dθ0π ⎛3 1 ⎞= 25∫2cosθ cos 2θ dθ0⎜ − + ⎟⎝2 2 ⎠π⎡3 1 ⎤ 75π= 25 ⎢ θ − 2sinθ + sin 2θ2 4⎥ =⎣⎦021 π /2 2 2A = 2 ⋅ ⎡(5sin ) (2 sin ) ⎤d2∫− +π /6 ⎣⎦π /2 2= ∫ (24sin θ −4sin θ −4)dπθ/6θ θ θπ /2= ∫ (8 −12cos 2 θ −4sin θ )dπθ/6[ 8θ 6sin2θ 4cosθ] π /2 8= − + = π+ 3π /6 32.−3−3tx = ,22y = t ; −1≤t≤ 242y9x−3−3x664 Review and Preview Instructor’s Resource Manual

3. x = 2cost, y = 2sint; 0≤t≤ 2π3y8. x = h⋅cosθy = h⋅sinθ−3−3 x−34. x = 2sint, y =− 2cost; 0≤t≤ 2π3yFor problems 9-12, L = ∫9.dx dy 9= 1, = tdt dt 24 81 4 1∫0 4 ∫021 328 2 1162 ∫ udu43 1623 2b dx2dy2⎛ ⎞ ⎛ ⎞dta ⎜ ⎟ + ⎜ ⎟⎝ dt ⎠ ⎝ dt ⎠L = 1+ t dt = 4+ 81t dt =u= 4+81tdu = 81dt328⎡32 ⎤= ⋅⎢u =⎣⎥⎦41 ⎡ ⎤(328) −8 ≈ 24.4129243 ⎢⎣⎥⎦−3−3x10.dx dy= 1, = 2dt dt55⎡ ⎤1 ⎣ ⎦1∫L = 1+ 4dt = 5t= 4 5 ≈8.94−35. x = t , y = tan 2t;y6π π− < t 0 so that the absolute minimum ofthe (square of the) distance occurs at the point(0.8,2.6) where the distance is(0,3) is [ ] 22d (0.8) = 5(0.8) − 8(0.8) + 4 = 0.8 ≈ 0.894Instructor’s Resource Manual Review and Preview 665

14. Results will vary. The only limitation ona , a , b , b is that the t value that makes{ }1 2 1 2x = 1 must also make y =− 1 and the t value thatmakes x = 3 must also make y = 3 . Lett = 0 yield (1, −1)and let t = 1 yield (3,3) ; then1 = a1(0) + b1 − 1 = a2(0)+ b23 = a1(1) + b1 3 = a2(1)+ b2From this we get b1 = 1, b2 = − 1, a1 = 2, a2= 4 ;thus one parametric representation isx = 2t+ 1, y = 4t− 1. Using other t values toyield (1, -1) and (3, 3) will give otherrepresentations.18.2 2x y+ = 14 96−6 6−6yx15.2s() t = t − 6t+8a. vt () = s′() t = 2t−6at () = v′() t = 219.x2 2− 4y= 06yb. The object is moving forward (in positive x-direction) when vt () > 0 or t> 3.16. at () = 2−6 6xa. vt () = ∫ at () dt= 2 t+v(0); since the object isinitially at rest, v (0) = 0 so vt () = 2t.2s() t = ∫ v() t dt = t + s(0); since s (0) = 20 ,2st () = t + 2020.x2 2− y = 4−6y2b. st ( ) = 100 ⇒ t + 20 = 100 ⇒ t= 80 ≈ 8.944The object will reach position 100 after about8.944 time units.617.8x = y2y−6 6x12−6−6 6 x21. r = 23y−12−33x−3666 Review and Preview Instructor’s Resource Manual

22.πθ =6y3−33x−323. r = 4sinθy5−3−13x24.1r =11+cosθ23y−33x−3Instructor’s Resource Manual Review and Preview 667

CHAPTER11.1 Concepts Review1. coordinates2.2 2 2( x+ 1) + ( y–3) + ( z–5)3. (–1, 3, 5); 44. plane; 4; –6; 3Problem Set 11.111Geometry in Spaceand Vectors1. A(1, 2, 3), B(2, 0, 1), C(–2, 4, 5), D(0, 3, 0),E(–1, –2, –3)1A 3, –3,3 , B(0, π, –3), C ⎛ ⎜–2, , 2 ⎞⎟,D(0, 0, e)⎝ 3 ⎠2. ( )7. P(2, 1, 6), Q(4, 7, 9), R(8, 5, –6)2 2 2PQ = (2 – 4) + (1– 7) + (6 – 9) = 72 2 2PR = (2 – 8) + (1 − 5) + (6 + 6) = 142 2 2QR = (4 – 8) + (7 – 5) + (9 + 6) = 2452 2 2PQ + PR = 49 + 196 = 245 = QR , so thetriangle formed by joining P, Q, and R is a righttriangle, since it satisfies the PythagoreanTheorem.8. a. The distance to the xy-plane is 1 since thepoint is 1 unit below the plane.b. The distance isc.2 2 2(2–0) + (3–3) + (–1–0) = 5 sincethe distance from a point to a line is thelength of the shortest segment joining thepoint and the line. Using the point (0, 3, 0)on the y-axis clearly minimizes the length.2 2 2(2 – 0) + (3− 0) + (–1– 0) = 149. Since the faces are parallel to the coordinateplanes, the sides of the box are in the planesx = 2, y = 3, z = 4, x = 6, y = –1, and z = 0 and thevertices are at the points where 3 of these planesintersect. Thus, the vertices are (2, 3, 4), (2, 3, 0),(2, –1, 4), (2, –1, 0), (6, 3, 4), (6, 3, 0), (6, –1, 4),and (6, –1, 0)3. x = 0 in the yz-plane. x = 0 and y = 0 on the z-axis.4. y = 0 in the xz-plane. x = 0 and z = 0 on the y-axis.5. a.2 2 2(6 –1) + (–1– 2) + (0 – 3) = 43b.2 2 2(–2 – 2) + (–2 + 2) + (0 + 3) = 52 2c. ( ) ( 4) ( 0– 3) 2e +π + π+ + ≈ 9.3996. P(4, 5, 3), Q(1, 7, 4), R(2, 4, 6)2 2 2PQ = (4 − 1) + (5 – 7) + (3 – 4) = 142 2 2PR = (4 – 2) + (5 – 4) + (3 – 6) = 142 2 2QR = (1–2) + (7–4) + (4–6) = 14Since the distances are equal, the triangle formedby joining P, Q, and R is equilateral.10. It is parallel to the y-axis; x = 2 and z = 3. (If itwere parallel to the x-axis, the y-coordinate couldnot change, similarly for the z-axis.)11. a.b.2 2 2x y z( –1) + ( –2) + ( –3) = 252 2 2x y z( + 2) + ( + 3) + ( + 6) = 5c. ( x– ) 2 ( y– e) 2 ( z– 2) 2π + + =π668 Section 11.1 Instructor’s Resource Manual

12. Since the sphere is tangent to the xy-plane, the point (2, 4, 0) is on the surface of the sphere. Hence, the radius of2 2 2the sphere is 5 so the equation is ( x– 2) + ( y– 4) + ( z– 5) = 25.13.14.15.16.2 2 2x x y y z z( –12 + 36) + ( + 14 + 49) + ( – 8 + 16) = –1+ 36 + 49 + 162 2 2x y z( – 6) + ( + 7) + ( – 4) = 100Center: (6, –7, 4); radius 102 2 2x x y y z z( + 2 + 1) + ( – 6 + 9) + ( –10 + 25) = –34 + 1+ 9 + 252 2 2x y z( + 1) + ( –3) + ( –5) = 1Center: (–1, 3, 5); radius 12 2 2 13x y z x y z+ + – + 2 + 4 =4⎛ 2 1⎞2 2 13 1⎜x – x+ ⎟+ ( y + 2y+ 1) + ( z + 4z+ 4) = + + 1+4⎝ 4⎠4 42⎛ 1⎞ 2 2 17⎜x– ⎟ + ( y + 1) + ( z + 2) =⎝ 2⎠2⎛1 ⎞Center: ⎜ ,–1,–2 ⎟ ; radius⎝2⎠172.922 ≈2 2 2( x + 8x+ 16) + ( y – 4y+ 4) + ( z – 22z+ 121) = –77 + 16 + 4 + 1212 2 2( x+ 4) + ( y–2) + ( z–11) = 64Center: (–4, 2, 11); radius 817. x-intercept: y = z = 0 ⇒ 2x = 12, x = 6y-intercept: x = z = 0 ⇒ 6y = 12, y = 2z-intercept: x = y = 0 ⇒ 3z = 12, z = 419. x-intercept: y = z = 0 ⇒ x = 6y-intercept: x = z = 0 ⇒ 3y = 6, y = 2z-intercept: x = y = 0 ⇒ –z = 6, z = –618. x-intercept: y = z = 0 ⇒ 3x = 24, x = 8y-intercept: y = z = 0 ⇒ –4y = 24, y = –6z-intercept: x = y = 0 ⇒ 2z = 24, z = 1220. x-intercept: y = z = 0 ⇒ –3x = 6, x = –2y-intercept: x = z = 0 ⇒ 2y = 6, y = 3z-intercept: x = y = 0 ⇒ z = 6Instructor’s Resource Manual Section 11.1 669

21. x and y cannot both be zero, so the plane isparallel to the z-axis.x-intercept: y = z = 0 ⇒ x = 88y-intercept: x = z = 0 ⇒ 3y = 8, y =3For problems 25-36, L = ∫dx dy dz25. = 1, = 1, = 2dt dt dt2 2 2 2 20 ∫020b dx2dy2dz2⎛ ⎞ ⎛ ⎞ ⎛ ⎞a ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠L = ∫ 1 + 1 + 2 dt = 6dt=⎡ 6t⎤ ⎣ ⎦= 2 6 ≈ 4.899dt22. x and z cannot both be zero, so the plane isparallel to the y-axis.x-intercept: y = z = 0 ⇒ 3x = 12, x = 4z-intercept: x = y = 0 ⇒ 4z = 12, z = 323. This is a sphere with center (0, 0, 0) and radius 3.24. This is a sphere with center (2, 0, 0) and radius 2.26.27.28.29.30.31.dx 1 dy 1 dz 1= , = , =dt 4 dt 3 dt 23 121212⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 3 611 ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟4 3 2∫⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 144∫L = dt = dt =361 61t144 ⎦1144⎡ ⎤ = 2 ≈ 1.302⎣dx 3 dy dz= t , = 3, = 4dt 2 dt dt1 13618 ∫1094 ⎛9⎞4 11 ⎜ ⎟4∫⎝ ⎠ 1 2u= 9t+100du = 9dt∫L = t + 916 + dt = 9t+ 100dt=1 7636 ∫401361 ⎡32 ⎤udu= u ≈16.5927 ⎢⎣⎥⎦109dx 3 dy 3 dz= t, = t, = 12dt 2 dt dt4 ⎛9 ⎞ ⎛9⎞ 4 12 ⎜ ⎟ ⎜ ⎟4 4∫⎝ ⎠ ⎝ ⎠ 2 2u= 18t+4du = 18dt∫L = t + t + 1dt = 18t+ 4dt=761 ⎡32 ⎤udu= u ≈7.58554 ⎢⎣⎥⎦40dx dy dz= 2, t = 2 t, = 1dt dt dtL = t2+ t+ dt = t+ dt =8 8 24 4 10 ∫ (2 1)0∫8 28t dt ⎡t t⎤0 ⎣ ⎦0∫(2 + 1) = + = 72dx dy dz= 2, t = 2 3, t = 3dt dt dtL = t2+ t+ dt = t+ dt =4 4 24 12 91 ∫ (2 3)1∫4 24t dt ⎡t t⎤1 ⎣ ⎦1∫(2 + 3) = + 3 = 28− 4 = 24dx dy dz=− 2sin t, = 2cos t, = 3dt dt dtπ−π2 2L = ∫ 4sin t+ 4cos t+ 9 dt = ∫ 13dt=π⎡ 13 t⎤ ⎣ ⎦= 2π13 ≈ 22.654−ππ−π670 Section 11.1 Instructor’s Resource Manual

32.33.34.dx dy dz 1=− 2sin t, = 2cos t,=dt dt dt 20L = t+ t+ dt =8π2 2 1∫ 4sin 4cos04008π1 18π1601 dt ⎡ 1601t⎤0 40 ⎣40⎦0∫π51601 ≈ 25.14= =dx 1 dy dz= , = 1, = 1dt 2 t dt dtL = + + dt = + dt6 ⎛ 1 ⎞6 ⎛ 1 ⎞11 21 ⎜ ⎟4t∫ ⎜ ⎟⎝ ⎠1 ⎝4t⎠∫By the Parabolic Rule (n = 10):i xif ( x i ) cici⋅ f ( xi)0 1 1.5000 1 1.50001 1.5 1.4720 4 5.88782 2 1.4577 2 2.91553 2.5 1.4491 4 5.79664 3 1.4434 2 2.88685 3.5 1.4392 4 5.75706 4 1.4361 2 2.87237 4.5 1.4337 4 5.73498 5 1.4318 2 2.86369 5.5 1.4302 4 5.720810 6 1.4289 1 1.4289approximation 7.2273dx dy dz= 1, = 2 t, = 3tdt dt dt2L = 1 4t2 9t4∫ + + dt1By the Parabolic Rule (n = 10):i xif ( x i ) cici⋅ f ( xi)0 1 3.7417 1 3.74171 1.1 4.3608 4 17.44332 1.2 5.0421 2 10.08413 1.3 5.7849 4 23.13954 1.4 6.5890 2 13.17795 1.5 7.4540 4 29.81616 1.6 8.3799 2 16.75987 1.7 9.3664 4 37.46558 1.8 10.4134 2 20.82689 1.9 11.5208 4 46.083210 2 12.6886 1 12.6886approximation 7.7075235.36.dx dy dz= − 2sin t, = cos t, = 1dt dt dtL = 4sin t+ cos t+ 1 dt =6π2 2∫06π23sin t+2 dt0∫By the Parabolic Rule (n = 10):i xif ( x i ) cici⋅ f ( xi)0 0 1.4142 1 1.41421 1.88 2.1711 4 8.68432 3.77 1.7425 2 3.48513 5.65 1.7425 4 6.97024 7.54 2.1711 2 4.34215 9.42 1.4142 4 5.65696 11.3 2.1711 2 4.34217 13.2 1.7425 4 6.97028 15.1 1.7425 2 3.48519 17 2.1711 4 8.684310 18.8 1.4142 1 1.4142approximation 34.8394dx dy dz= cos t, =− sin t, = costdt dt dtL = t+ t+ t dt =2π2 2 2∫ cos sin cos02π2cos t+1 dt0∫By the Parabolic Rule (n = 10):i xif ( x i ) cici⋅ f ( xi)0 0 1.4142 1 1.41421 0.63 1.2863 4 5.14512 1.26 1.0467 2 2.09333 1.88 1.0467 4 4.18664 2.51 1.2863 2 2.57265 3.14 1.4142 4 5.65696 3.77 1.2863 2 2.57267 4.4 1.0467 4 4.18668 5.03 1.0467 2 2.09339 5.65 1.2863 4 5.145110 6.28 1.4142 1 1.4142approximation 7.640537. The center of the sphere is the midpoint of thediameter, so it is⎛–2+ 4 3–1 6+5⎞ ⎛ 11⎞⎜ , , ⎟=⎜1, 1, ⎟.The radius is⎝ 2 2 2 ⎠ ⎝ 2 ⎠1 2 2 2 53(–2 – 4) + (3 + 1) + (6 – 5) = . The2 222 2 ⎛ 11⎞53equation is ( x− 1) + ( y–1) + ⎜z– ⎟ = .⎝ 2 ⎠ 4Instructor’s Resource Manual Section 11.1 671

38. Since the spheres are tangent and have equalradii, the radius of each sphere is 1 of the2distance between the centers.1 (–3 – 5)2 (1 3)2 (2 – 6)2r = + + + = 2 6. The22 2 2spheres are ( x+ 3) + ( y–1) + ( z–2) = 24 and2 2 2( x– 5) + ( y+ 3) + ( z– 6) = 24.39. The center must be 6 units from each coordinateplane. Since it is in the first octant, the center is(6, 6, 6). The equation is2 2 2( x− 6) + ( y– 6) + ( z– 6) = 36.40. x + y = 12 is parallel to the z-axis. The distancefrom (1, 1, 4) to the plane x + y = 12 is the sameas the distance in the xy-plane of (1, 1, 0) to theline x + y – 12 = 0. That distance is1+1–12= 5 2. The equation of the sphere is1/2(1 + 1)2 2 2( x– 1) + ( y– 1) + ( z– 4) = 50.41. a. Plane parallel to and two units above thexy-planeb. Plane perpendicular to the xy-plane whosetrace in the xy-plane is the line x = y.c. Union of the yz-plane (x = 0) and thexz-plane (y = 0)d. Union of the three coordinate planese. Cylinder of radius 2, parallel to the z-axisf. Top half of the sphere with center (0, 0, 0)and radius 342. The points of the intersection satisfy both2 2 2( x– 1) + ( y+ 2) + ( z+ 1) = 10 and z = 2, so2 2 2( x–1) + ( y+ 2) + (2+ 1) = 10. This simplifies2 2to ( x–1) + ( y+ 2) = 1, the equation of a circleof radius 1. The center is (1, –2, 2).2 2 2 2 2 243. If P(x, y, z) denotes the moving point, ( x–1) + ( y–2) + ( z+ 3) = 2 ( x–1) + ( y–2) + ( z–3) ,2 2 2which simplifies to ( x–1) + ( y–2) + ( z–5) = 16, is a sphere with radius 4 and center (1, 2, 5).2 2 2 2 2 244. If P(x, y, z) denotes the moving point, ( x–1) + ( y–2) + ( z+ 3) = ( x–2) + ( y–3) + ( z–2) ,which simplifies to x + y + 5z = 3/2, a plane.2 ⎡ ⎛h⎞⎤45. Note that the volume of a segment of height h in a hemisphere of radius r is πh⎢ r− ⎜ ⎟ .3 ⎥⎣ ⎝ ⎠⎦The resulting solid is the union of two segments, one for each sphere. Since the two spheres have the same radius,each segment will have the same value for h. h is the radius minus half the distance between the centers of the twospheres.1 2 2 2 3 1h = 2 − (2− 1) + (4− 2) + (3− 1) = 2− =2 2 2⎡ 2⎛1⎞ ⎛ 1⎞⎤ 11πV = 2⎢π⎜ ⎟ ⎜2− ⎟⎥ =⎢ ⎝2⎠ ⎝ 6⎠⎥12⎣⎦46. As in Problem 45, the resulting solid is the union of two segments. Since the radii are not the same, the segmentswill have different heights. Let h 1 be the height of the segment from the first sphere and let h 2 be the height fromthe second sphere. r 1 = 2 is the radius of the first sphere and r 2 = 3 is the radius of the second sphere .2 2 2Solving for the equation of the plane containing the intersection of the spheres ( x− 1) + ( y− 2) + ( z−1) − 4=02 2 2and ( x− 2) + ( y− 4) + ( z−3) − 9= 0, we get x + 2y + 2z – 9 = 0.The distance from (1, 2, 1) to the plane is 2 ,3 and the distance from (2, 4, 3) to the plane is 7 .32 4 7 2h1 = 2 − = ; h2= 3− =3 3 3 32 2⎛4⎞ ⎛ 4⎞ ⎛2⎞ ⎛ 2⎞V =π⎜ ⎟ ⎜2− ⎟+π⎜ ⎟ ⎜3− ⎟= 4π⎝3⎠ ⎝ 9⎠ ⎝3⎠ ⎝ 9⎠672 Section 11.1 Instructor’s Resource Manual

47. Plots will vary. We first note that the sign of c will influence the vertical direction an object moves (along thehelix) with increasing time; if c is negative the object will spiral downward, whereas if c is positive it will spiralupward. The smaller c is the “tighter” the spiral will be; that is the space between successive “coils” of the helixdecreases as c decreases.48. Plots will vary. We first note that the sign of a will influence the rotational direction that an object moves (alongthe helix) with increasing time; if a is negative the object will rotate in a clockwise direction, whereas if a ispositive rotation will be counterclockwise. The smaller a is the narrower the spiral will be; that is the circlestraced out will be of smaller radius as a decreases11.2 Concepts Review1. magnitude; direction2. they have the same magnitude and direction.3. the tail of u; the head of v4. 3Problem Set 11.21.7.8.9.10.1 1w = u cos60°+ v cos 60°= + = 12 22 2w = u cos 45°+ v cos 45°= + = 22 2u+ v = − 1+ 3,0+ 4 = 2,4u− v = −1−3,0− 4 = −4, −42 2u = ( − 1) + (0) = 1=12 2v = (3) + (4) = 25 = 5u+ v = 0 + ( − 3),0+ 4 = −3,4u− v = 0 −( −3),0− 4 = 3, −4uv2 2= (0) + (0) = 0 = 02 2= ( − 3) + (4) = 25 = 52.11.u+ v = 12 + ( − 2),12 + 2 = 10,14u− v = 12 −( −2),12 − 2 = 14,10uv2 2= (12) + (12) = 288 = 12 22 2= ( − 2) + (2) = 8 = 2 23.12.u+ v = ( − 0.2) + ( − 2.1),0.8 + 1.3 = −2.3, 2.1u− v = ( −0.2) −( −2.1),0.8 − 1.3 = 1.9, −0.54. 0uv2 2= ( − 0.2) + (0.8) = 0.68 ≈0.8252 2= ( − 2.1) + (1.3) = 6.10 ≈ 2.475.6.1 1 1w = ( u+ v)= u+v2 2 21 1 1n = ( v− u)= v−u2 2 21 1 1m = v– n = v– ( v− u)= v+u2 2 213.u+ v = − 1+ 3,0 + 4,0 + 0 = 2, 4,0u− v = −1−3,0 −4,0 − 0 = −4, −4,02 2 2u = ( − 1) + (0) + (0) = 1=12 2 2v = (3) + (4) + (0) = 25 = 5Instructor’s Resource Manual Section 11.2 673

14.15.16.u+ v = 0 + ( − 3), 0 + 3, 0 + 1 = −3,3,1u− v = 0 −( −3), 0 −3, 0 − 1 = 3, −3, −12 2 2u = (0) + (0) + (0) = 0 = 02 2 2v = ( − 3) + (3) + (1) = 19 ≈ 4.359u+ v = 1 + ( − 5),0+ 0,1+ 0 = −4,0,1u− v = 1 −( −5),0 −0,1− 0 = 6,0,12 2 2u = (1) + (0) + (1) = 2 ≈1.4142 2 2v = ( − 5) + (0) + (0) = 25 = 5u+ v = 0.3 + 2.2,0.3 + 1.3,0.5 + ( − 0.9) =2.5,1.6, −0.4u− v = 0.3 −2.2,0.3 −1.3,0.5 −( − 0.9) =−1.9, −1.0,1.42 2 2u = (0.3) + (0.3) + (0.5) = 0.43 ≈0.6562 2 2v = (2.2) + (1.3) + ( − 0.9) = 7.34 ≈ 2.70917. Let θ be the angle of w measured clockwise fromsouth.w cosθ = u cos30°+ v cos 45°= 25 3 + 25 2= 25( 3 + 2 )w sinθ = v sin 45 ° – u sin 30°=25 2 – 25= 25( 2 –1)2 2 2 2 2w = w cos θ + w sinθ( ) ( )2 2= 625 3 + 2 + 625 2 –1= 625( 8 − 2 2 + 2 6 )w = 625( 8 − 2 2 + 2 6 ) = 25 8 − 2 2 + 2 6≈ 79.34w sinθ2–1tanθ= =w cosθ3+2–1 ⎛ 2–1 ⎞θ = tan ⎜= 7.5°3+2 ⎟⎝ ⎠w has magnitude 79.34 lb in the directionS 7.5° W.18. Let v be the resulting force. Let θ be the angle ofv measured clockwise from south.vcosθ = 60cos30°+ 80cos 60°= 30 3 + 40= 10( 3 3 + 4)v sinθ = 80sin 60 ° – 60sin 30°=40 3 – 30= 10( 4 3 – 3)2 2 2 2 2v = v cos θ + v sinθ( ) ( )2 2= 100 3 3 + 4 + 100 4 3 – 3= 100(100) = 10,000v = 10,000 = 100v sinθ4 3–3tanθ= =v cosθ3 3+4–1 ⎛4 3–3⎞θ = tan ⎜≈ 23.13°3 3+4⎟⎝ ⎠The resultant force has magnitude 100 lb in thedirection S 23.13° W.19. The force of 300 N parallel to the plane hasmagnitude 300 sin 30° = 150 N. Thus, a force of150 N parallel to the plane will just keep theweight from sliding.20. Let a be the magnitude of the rope that makes anangle of 27.34°. Let b be the magnitude of therope that makes an angle of 39.22°.1. a sin 27.34° = b sin 39.22°2. a cos 27.34° + b cos 39.22° = 258.5Solve 1 for b and substitute in 2.sin 27.34°acos 27.34°+ a cos39.22°=258.5sin 39.22°258.5a =≈178.15cos 27.34°+ sin 27.34° cot 39.22°a sin 27.34°b = ≈129.40sin 39.22°The magnitudes of the forces exerted by theropes making angles of 27.34° and 39.22° are178.15 lb and 129.40 lb, respectively.21. Let θ be the angle the plane makes from north,measured clockwise.425 sin θ = 45 sin 20°9sinθ = sin 20°85–1 ⎛ 9 ⎞θ = sin ⎜ sin 20° ⎟≈ 2.08°⎝85⎠Let x be the speed of airplane with respect to theground.x = 45 cos 20° + 425 cos θ ≈ 467The plane flies in the direction N 2.08° E, flying467 mi/h with respect to the ground.674 Section 11.2 Instructor’s Resource Manual

22. Let v be his velocity relative to the surface. Let θbe the angle that his velocity relative to thesurface makes with south, measured clockwise.v cosθ= 20, v sinθ= 32 2 2 2 2v = v cos θ + v sin θ = 400 + 9 = 409v = 409 ≈20.22sinθ3tanθ= vv cosθ= 20–1 3θ = tan ≈ 8.53°20His velocity has magnitude 20.22 mi/h in thedirection S 8.53° W.23. Let x be the air speed.x cos 60° = 4040x = = 80cos 60°The air speed of the plane is 80 mi/hr24. Let x be the air speed. Let θ be the angle that theplane makes with north measured counterclockwise.x cos θ = 837 + 63 cos 11.5°x sin θ = 63 sin 11.5°2 2 2 2 2x = x cos θ + x sin θ2 2= (837 + 63cos11.5 ° ) + (63sin11.5 ° )= 704,538 + 105,462 cos 11.5°x = 704,538 + 105,462cos11.5° ≈ 898.82xsinθ63sin11.5°tanθ= =x cosθ837 + 63cos11.5°–1 ⎛ 63sin11.5°⎞θ = tan ⎜⎟≈ 0.80°⎝837 + 63cos11.5°⎠The plane should fly in the direction N 0.80° Wat an air speed of 898.82 mi/h.25. Let u = u1, u2 , v = v1, v2, and w = w1,w2a. u+ v = u1+ v1,u2 + v2=v + u , v + u = v+u1 1 2 2b. ( u v) w u1 v1, u2 v2 w1,w2+ + = + + + =( u1+ v1) + w1,( u2 + v2)+ w2u1+ ( v1+ w1), u2 + ( v2 + w2)==u , u v + w , v + w = u+ ( v+w)1 2 1 1 2 2c. u1 0, u2 0 u1,u2u+ 0= + + = = uu+ 0= 0+u by part a.u + (-u) = u , u + −u , − u =u + ( − u ), u + ( − u ) = 0,0 = 0d. 1 2 1 21 1 2 2e. u ( )ab ( ) = a bu, bu = abu ( ), abu ( ) =( ab) u ,( ab) u = ( ab)u1 21 2 1 2f. a( u v) a u1 v1,u2 v2+ = + + =au ( + v), au ( + v)=1 1 2 2au + av , au + av =1 1 2 2au , au + av , av = au+av1 2 1 2g. ( a b) u ( a b) u1,( a b)u2+ = + + =au + bu , au + bu =1 1 2 2au , au + bu , bu = au+bu1 2 1 2h. 1u1 u1,1 u2 u1,u2i.= ⋅ ⋅ = = u2 21, 2 ( 1) ( 2)au= au au = au + au =2 2 2 2 2 2 21 2 1 2au + au = a( u + u ) =2 2 21 2a u + u = a ⋅ u26. Let u = u1, u2, u3 , v = v1, v2, v3, andw = w , w , w1 2 3u+ v = u + v , u + v , u + v =a. 1 1 2 2 3 3v + u , v + u , v + u = v+u1 1 2 2 3 3b. ( u+ v) + w = u1+ v1, u2 + v2,u3 + v3+ w , w , w =1 2 3( u + v ) + w ,( u + v ) + w ,( u + v ) + w =1 1 1 2 2 2 3 3 3u + ( v + w ), u + ( v + w ), u + ( v + w ) =1 1 1 2 2 2 3 3 3u , u , u v + w , v + w , v + w =1 2 3 1 1 2 2 3 3u+ ( v+w)c. u1 0, u2 0, u3 0 u1, u2,u3u+ 0= + + + = = uu+ 0= 0+u by part a.d. u (-u) u1, u2, u3 u1, u2,u3+ = + − − − =u + ( − u ), u + ( − u ), u + ( − u ) =1 1 2 2 3 30,0,0 = 0e. ab ( u) a( bu1, bu2,bu3)= =abu ( ), abu ( ), abu ( ) =1 2 3( ab) u ,( ab) u ,( ab) u = ( ab)u1 2 3g. ( a b) u ( a b) u1,( a b) u2,( a b)u3+ = + + + =au + bu , au + bu , au + bu =1 1 2 2 3 3au , au , au + bu , bu , bu = au+bu1 2 3 1 2 3h. 1u1 u1,1 u2,1 u3 u1, u2,u3= ⋅ ⋅ ⋅ = = uInstructor’s Resource Manual Section 11.2 675

27. Given triangle ABC, let D be the midpoint of ABand E be the midpoint of BC. u = AB , v = BC ,w = AC , z = DEu + v = w1 1 1 1z = u+ v = ( u+ v)= w2 2 2 2Thus, DE is parallel to AC.28. Given quadrilateral ABCD, let E be the midpointof AB, F the midpoint of BC, G the midpoint ofCD, and H the midpoint of AD.ABC and ACD are triangles. From Problem 17,EF and HG are parallel to AC. Thus, EF isparallel to HG. By similar reasoning usingtriangles ABD and BCD, EH is parallel to FG.Therefore, EFGH is parallelogram.29. Let P i be the tail of v i.Thenv + v +…+ v1 2 n⎯⎯→ ⎯⎯→ ⎯⎯→= PP 1 2+ P2P3+ +PnP1⎯⎯→= PP 11 =0 .30. Consider the following figure of the circle.2πα =nThe vectors have the same length. Consider thefollowing figure for adding vectors v i and v i+ 1.2πThen β =π – . Note that the interior angle ofn2πa regular n-gon is π – . Thus the vectorsn(placed head to tail from v1 to v n ) form a regularn-gon. From Problem 19, the sum of the vectorsis 0.31. The components of the forces along the linescontaining AP, BP, and CP are in equilibrium;that is,W = W cos α + W cos βW = W cos β + W cos γW = W cos α + W cos γThus, cos α + cos β = 1, cos β + cos γ = 1, andcos α + cos γ = 1. Solving this system of1equations results in cosα = cos β = cos γ = .Hence α = β = γ = 60°.Therefore, α + β = α + γ = β + γ = 120°.32. Let A′ , B′ , C′ be the points where the weights areattached. The center of gravity is locatedAA′ + BB′ + CC′units below the plane of the3triangle. Then, using the hint, the system is inequilibrium when AA′ + BB′ + CC′ismaximum. Hence, it is in equilibrium whenAP + BP + CP is minimum, because the totallength of the string isAP + AA′ + BP + BB′ + CP + CC′.33. The components of the forces along the linescontaining AP, BP, and CP are in equilibrium;that is,5w cos α + 4w cos β = 3w3w cos β + 5w cos γ = 4w3w cos α + 4w cos γ = 5wThus,5 cos α + 4 cos β = 33 cos β + 5 cos γ = 43 cos α + 4 cos γ = 5.Solving this system of equations results in3 4cos α = , cos β = 0, cos γ = , from which it5 54 3follows that sin α = , sin β = 1, sin γ = .5 54Therefore, cos( α + β) = – , cos( α + γ) = 0,53cos( β + γ) = – , so5⎞α + β = cos ⎜– ⎟≈ 143.13 ° ,⎝ 5 ⎠–1 ⎛ 3 ⎞β + γ = cos ⎜– ⎟≈ 126.87 ° .⎝ 5 ⎠–1 ⎛ 42α + γ = 90 ° ,676 Section 11.2 Instructor’s Resource Manual

This problem can be modeled with three stringsgoing through A, four strings through B, and fivestrings through C, with equal weights attached tothe twelve strings. Then the quantity to beminimized is 3 AP + 4 BP + 5 CP .34. Written response.35. By symmetry, the tension on each wire will bethe same; denote it by T where T can be thetension vector along any of the wires. Thechandelier exerts a force of 100 lbs. verticallydownward. Each wire exerts a vertical tension ofT sin 45 upward. Since a state of equilibriumexists,1004⋅ T sin 45 = 100 or T = ≈35.362 2The tension in each wire is approximately 35.36lbs.36. By symmetry, the tension on each wire will bethe same; denote it by T where T can be thetension vector along any of the wires. Thechandelier exerts a force of 100 lbs. verticallydownward. Each wire exerts a vertical tension ofT sin 45 upward. Since a state of equilibrium200exists, 3⋅ T sin45 = 100 or T = ≈ 47.143 2The tension in each wire is approximately 47.14lbs.e. aca ⋅ = 4 + 9[(0)( − 2) + ( − 5)(3)] = −15 13f. bb ⋅ − b = (2)(2) + ( −3)( −3) − 4 + 9= 13 − 132. a. –4a + 3b = − 12,4 + 3, − 3 = − 9, 13. a.b. bc ⋅ = (1)(0) + (–1)(5) = –5c. ( a+ b) ⋅ c = 4, −2 ⋅ 0,5= (4)(0) + (–2)(5) = –10d. 2 c⋅ (3a+ 4 b ) = 2 0,5 ⋅ ( 9,–3 + 4,–4)= 2 0, 5 ⋅ 13, – 7 = 2[(0)(13) + (5)(–7)]= –70e. bba ⋅ = 1+ 1[(1)(3) + ( −1)( − 1)] = 4 22f. c c c ( ) 2= 0− ⋅ = 0 + 25 − [(0)(0) + (5)(5)]ab ⋅ (1)( − 1) + ( −3)(2) 7cosθ= = = −a b507= – ≈ –0.98995 2( 10 )( 5)11.3 Concepts Review1. uv 1 1 + u2v2 + uv 3 3 ; u v cos θ2. 03. F•D4. A, BC ,Problem Set 11.31. a. 2a – 4b = (–4i + 6j) + (–8i + 12j)= –12i + 18jb. ab ⋅ = (–2)(2) + (3)(–3) = –13c. a⋅ ( b+ c) = (–2i+ 3 j) ⋅(2 i–8 j )= (–2)(2) + (3)(–8) = –28d. (–2a+ 3 b) ⋅ 5c= 5[(10 i–15 j) ⋅(–5 j )]= 5[(10)(0) + (–15)(–5)] = 375b.ab ⋅ ( − 1)(6) + ( −2)(0) 6cosθ= = =−a b 5 (6) 6 51= – ≈− 0.44725( )Instructor’s Resource Manual Section 11.3 677

c.ab ⋅ (2)( − 2) + ( −1)( −4) 0cosθ= = = = 0a b10( 5)( 2 5)c.ab ⋅ ( − 1)(2) + (3)( −6) −20cosθ= = = =−1a b20−1θ = cos − 1 = 180( 10 )( 2 10 )d.ab ⋅ (4)( − 8) + ( −7)(10)cosθ= =a b( 65)( 2 41)–102 51= = – ≈ –0.98792 2665 2665d.( ) + ( )(2)( 2 3)ab ⋅ 3 (3) (1) 3 4 3cosθ= = = = 1a b4 3θ−1= cos 1 = 04. a.ab ⋅ (12)( − 5) + (0)(0)cosθ= =a b (12)(5)–60= = − 160−1θ = cos − 1 = 1805. a. ab= i (1)(0) + (2)(1) + ( − 1)(1) = 1b. ( a+ c) ib = (3 j+k ) i (j+k)= (0)(0) + (3)(1) + (1)(1) = 4c.a 1= ( i+ 2j−k)a 2 2 21 + 2 + ( −1)=6i+ 6j−6k6 3 6d. ( b− c) ia = ( i− k ) i (i+ 2j−k)= (1)(1) + (0)(2) + ( −1)( − 1) = 2b.ab ⋅ (4)( − 8) + (3)( −6) −50cosθ= = = =−1a b (5)(10) 50−1θ = cos − 1 = 180e.ab ia b=(1)(0) + (2)(1) + ( −1)(1)2 2 2 2 2 21 + 2 + ( − 1) 0 + 1 + 11 3= =6 2 6f. By Theorem A (5), bb i − b =06. a. ac i = ( 2)( − 2) + ( 2)(2) + (0)(1) = 0b. ( a− c) ib= 2 −( −2), 2−2,0−1 i 1, − 1,1 =(2 + 2)(1) + ( 2 −2)( − 1) + ( − 1)(1) = 32678 Section 11.3 Instructor’s Resource Manual

c.aa1= 2, 2,0 =2 2 2( 2) + ( 2) + 01 2 22, 2,0 = , ,02 2 2d. ( b− c) ia= 3, − 3, 0 i 2, 2,0 =e.(3)( 2) + ( − 3)( 2) + (0)(0) = 0bc i (1)( − 2) + ( − 1)(2) + (1)(1)= =b c 2 2 2 2 2 21 + ( − 1) + 1 ( − 2) + 2 + 1−3 3=−3 9 3f. By Theorem A (5), aa i − a =07. The basic formula is cosθ = uv iu v−1⎛ 2− 2+0⎞θab, = cos⎜4 3 ⎟=⎝ ⋅ ⎠−1cos 0 = 90−1⎛− 2 2+ 2 2+0⎞θac, = cos⎜4 9 ⎟=⎝ ⋅ ⎠−1cos 0 = 90−1⎛−2− 2+1⎞θbc, = cos ⎜ ⎟=⎝ 3⋅9 ⎠−1⎛ 3 ⎞cos ⎜−≈125.263 ⎟⎝ ⎠8. The basic formula is cosθ uv , =⎛ 3 3 ⎞⎜ − + 0 ⎟−1θ3 3, cos ⎜⎟ab = =⎜ 1⋅2 ⎟⎜⎟⎝⎠−1cos 0 = 902uv iu v⎛ −2 3 −2 3 3 ⎞⎜ + + ⎟−1θ3 3 3, cos ⎜⎟ac = =⎜ 1⋅9 ⎟⎜⎟⎝⎠−13cos − ≈125.263−1⎛− 2+ 2+0⎞θbc, = cos ⎜ ⎟=⎝ 2⋅9 ⎠−1cos 0 = 909. The basic formulae areu1 u2u3cosαu = cos βu = cosγu=u u ua. a = 2, 2,0 a = 2a1 2cos αa= = , αa= 45a 2a2 2cos βa= = , βa= 45a 2a3 0cosγa= = = 0, γa= 90a 2b. b = 1, − 1,1 b = 3b1 1cosαb= = ≈0.577, αb≈54.74b 3b2 −1cos βb= = ≈−0.577, βb≈125.26b 3b3 1cosγb= = ≈0.577, γb≈54.74b 3c. c= − 2, 2,1 c = 3c1 2cos αc= =− , αc≈131.81c 3c2 2cos βc= = , βc≈ 48.19c 3c3 1cos γc= = , γc≈70.53c 310. The basic formulae areu1 u2u3cosαu = cos βu = cosγu=u u ua.3 3 3a = , , a = 13 3 3a1 3cosαa= = ≈0.577, αa≈54.74a 3a2 3cos βa= = ≈0.577, βa≈54.74a 3a3 3cosγa= = ≈0.577, γa≈54.74a 3Instructor’s Resource Manual Section 11.3 679

. b = 1, − 1, 0 b = 2b1 2cos αb= = , αb= 45b 2b2 2cos βb= =− , βb≈135b 2b3 0cosγb= = = 0, γb= 90b 2c. c= −2, − 2,1 c = 3c1 2cos αc= =− , αc≈131.81c 3c2 2cos βc= = − , βc≈131.81c 3c3 1cos γc= = , γc≈70.53c 311. 6,3 i − 1, 2 = (6)( − 1) + (3)(2) = 0Therefore the vectors are orthogonal12. ab i = (1)(1) + (1)( − 1) + (1)(0) = 0ac i = (1)( − 1) + (1)( − 1) + (1)(2) = 0bc i = (1)( − 1) + ( −1)( − 1) + (0)(2) = 0Therefore the vectors are mutually orthogonal.13. ab i = (1)(1) + ( − 1)(1) + (0)(0) = 0ac i = (1)(0) + ( − 1)(0) + (0)(2) = 0bc i = (1)(0) + (1)(0) + (0)(2) = 0Therefore the vectors are mutually orthogonal.14. ( u+ v) ⋅ ( u– v) = u⋅u– u⋅ v+ v⋅u–v⋅v= u2− v2=0Thus, u2= v2or u =15. If xi + yj + zk is perpendicular to –4i + 5j + k and4i + j, then –4x + 5y + z = 0 and 4x + y = 0 sincethe dot product of perpendicular vectors is 0.Solving these equations yields y = –4x andz = 24x. Hence, for any x, xi – 4xj + 24xk isperpendicular to the given vectors.2 2 2xi xj xkx x x− 4 + 24 = + 16 + 576= x 593This length is 10 when x =±10. The vectors593are10 40 240i–j+k and593 593 593–10 40 240i+j– k .593 593 593v16. If x, yz , is perpendicular to both 1, -2, -3 and-3, 2, 0 , then x – 2y – 3z = 0 and –3x + 2y = 0.Solving these equations for y and z in terms of x3 2yields y = x, z = – x.Thus, all the vectors2 33 2have the form x, x, – x where x is a real2 3number.17. A vector equivalent to BA isu = 1+ 4,2−5,3− 6 = 5, −3, − 3 .A vector equivalent to BC isv = 1 + 4, 0 - 5, 1 - 6 = 5, -5, -5 .u⋅v5⋅ 5 + (–3)(–5) + (–3)(–5)cosθ= =u v 25 + 9 + 9 25 + 25 + 2555 11 ,–1 11= = so θ = cos ≈ 14.4 ° .43 75 12912918. A vector equivalent to BA isu = 6 - 3, 3 - 1, 3 + 1 = 3, 2, 4 .A vector equivalent to BC isv = -1 - 3, 10 - 1, -2.5 + 1 = -4, 9, -1.5 .u⋅ v = 3(–4) + 2⋅ 9 + 4 ⋅ (–1.5) = 0 so u isperpendicular to v. Thus the angle at B is a rightangle.19.c,6 i c, − 4 = 0⇒c− 24 = 0⇒c2= 24 ⇒ c =± 2 620. (2ci− 8 j) i (3 i+ cj) = 0 ⇒6c− 8c= 0 ⇒− 2c= 0⇒ c = 0221. ( ci+ j+ k) i (0i+ 2 j+ dk) = 0⇒ 0c+ 2+ d = 0⇒c is any number, d =−222. a,0,1 i 0,2, b = 0⇒ b = 0a,0,1 i 1, c,1 = 0 ⇒ a+ 1 = 00, 2, b i 1, c,1 = 0 ⇒ 2c+ b=0Thus: a = − 1 and c = b=0For problems 23-34, the formula to use is⎛ ba i ⎞proj = ⎜ ⎟aba ⎜2 ⎟⎝ a ⎠23. u = 1, 2 , v = 2, − 1 , w = 1, 5(1)(2) + (2)( −1)projv u = 2, − 1 =2 22 + ( −1)0680 Section 11.3 Instructor’s Resource Manual

24. u = 1, 2 , v = 2, − 1 , w = 1, 534. u = 3, 2,1 , v = 2,0, − 1 , w = 1,5, −3(2)(1) + ( −1)(2)u v = =2 21 + 20(3)(1) + (2)(0) + (1)(0)i u 2 2 21 + 0 + 0⎛ ⎞(1)(1) + (5)(2) 11 22u w = =35. a. proj ⎜uu ⋅⎟⎛uu⋅ ⎞uu = u =2 22 ⎜ ⎟u = u1 + 25 5⎜ ⎟ ⋅⎝ u ⎝uu⎠⎠⎛( )⎞b. proj −⎜u⋅−u ⎟⎛u⋅uu( )( − 1)(1) + (6)(2) 11 22u w − v⎜ 2 ⎟⎜⋅= =⎝ −u⎝uu⎠2 21 + 25 5⎛( −u)⋅u⎞36. a. proj ( ) ⎜ ⎟u − u = ( −u)⎜ 2 ⎟(1)(0) + (2)(1)projj u⎝ u ⎠= 0,1 = 0,22 20 + (1)⎛−uu⋅ ⎞= ⎜ ⎟( − u)= u⎝ uu ⋅ ⎠⎛( −u) ⋅( −u)⎞(1)(1) + (2)(0)proji ub. proj= 1,0 = 1,0− ( ) ⎜ ⎟u − u = ( −u)2 2⎜ 2 ⎟1 + (0)⎝ −u⎠⎛uu⋅ ⎞= ⎜ ⎟( − u)=−u⎝uu⋅ ⎠(3)(2) + (2)(0) + (1)( −1)v u = − =2 2 22 + 0 + ( −1)37. u⋅ v = (–1)(–1) + 5⋅ 1+ 3(–1) = 3−v = 1+ 1+ 1=3u⋅v 3= = 3v 3(2)(3) + (0)(2) + ( −1)(1)u v = =2 2 23 + 2 + 138. u⋅ v = 5 (– 5)+ 5( 5) + 2(1) = 2v = 5+ 5+ 1=11u⋅v 2=v 11(1)(3) + (5)(2) + ( −3)(1)u w = =21/22 2 239.3 + 2 + 1u = (4 + 9 + z ) = 5 and z > 0, soz = 2 3 ≈ 3.4641.2 2 240. cos (46 °+ ) cos (108 °+ ) cos γ = 13, 2,1 , 2,0, 1 , 1,5, 3⇒cosγ≈ ± 0.6496v 3,5, 4⇒ γ ≈ 49.49° or γ ≈ 130.51°(3)(3) + (5)(2) + ( −4)(1)u w + v = =2 2 23 + 2 + 1v = − 2, 1, c is a candidate.–4, 2, 5 ⋅ –2, 1, c = 8 + 2 + 5c(3)(0) + (2)(0) + (1)(1)k u = =2 2 2proj 1,225. u = 1, 2 , v = 2, − 1 , w = 1, 5proj 1,2 ,26. u = 1, 2 , v = 2, − 1 , w = 1, 5 , w− v = −1, 6proj ( ) 1, 2 ,27. u = 1, 2 , v = 2, − 1 , w = 1, 528. u = 1, 2 , v = 2, − 1 , w = 1, 529. u = 3, 2,1 , v = 2,0, − 1 , w = 1,5, −3proj 2,0, 12,0, 130. u = 3, 2,1 , v = 2,0, − 1 , w = 1,5, −3proj 3, 2,115 10 5, ,14 14 1431. u = 3, 2,1 , v = 2,0, − 1 , w = 1,5, −332.proj 3,2,115 10 5, ,7 7 7u = v = − w = −w+ = −proj ( ) 3,2,145 30 15, ,14 14 1433. u = 3, 2,1 , v = 2,0, − 1 , w = 1,5, −3proj 0,0,1 0,0,10 + 0 + (1)proj = 1,0,0 = 3,0,0⎞u = − u = ⎟u = u⎠41. There are infinitely many such pairs. Note that–4, 2, 5 ⋅ 1, 2, 0 = –4 + 4 + 0 = 0, sou = 1, 2, 0 is perpendicular to − 4, 2,5. Forany c, –2, 1, c ⋅ 1, 2, 0 = –2 + 2 + 0 = 0 so8 + 2 + 5c = 0 ⇒ c = –2, so one pair isu = 1, 2, 0 , v = 2,1, 2 − − .Instructor’s Resource Manual Section 11.3 681

42. The midpoint is⎛3+ 5 2–7 –1+2⎞ ⎛ 5 1⎞⎜ , , ⎟=⎜4, – , ⎟,⎝ 2 2 2 ⎠ ⎝ 2 2⎠5 1vector is 4, – , .2 2so the52. r = ka + mb ⇒ 6 = k(–4) + m(2) and–7 = k(3) + m(–1)–4k + 2m = 63k – m = –7Solve the system of equations to getk = –4, m = –5.43. The following do not make sense.a. v⋅w is not a vector.b. u⋅w is not a vector.44. The following do not make sense.c. u is not a vector.d. u+v is not a scalar.45. au+ bu= a u1, u2 + b u1,u2= au1, au2 + bu1, bu2 = au1+ bu1,au2 + bu2( + ) ,( + ) = ( + ) ,a b u1 a b u2 a b u1 u2a b= ( + ) u46. uv uv 11 u2v2 vu 11 vu 2 2i = + = + = vu i47. c( u⋅ v ) = c( u , u ⋅ v , v )1 2 1 2= cuv ( 1 1+ uv 2 2) = cuv ( 1 1) + cuv ( 2 2)= ( cu1) v1+ ( cu2) v2 = cu1, cu2 ⋅ v1,v2( )= cu, u ⋅ v, v = ( cu)⋅v1 2 1 248. u⋅ ( v+ w ) = u , u ⋅ ( v , v + w , w )1 2 1 2 1 2= u1, u2 ⋅ v1+ w1,v2 + w2= u1( v1+ w1) + u2( v2 + w2)= ( uv 11+ u2v2) + ( uw 1 1+ u2w2)= u⋅ v+ ⋅49. 0u= i 0u1+ 0u2= 050. ( ) 2 2 2 2 2 2uu i = u + u = u + u = u1 2 1 251. r = ka + mb ⇒ 7 = k(3) + m(–3) and–8 = k(–2) + m(4)3k – 3m = 7–2k + 4m = –8Solve the system of equations to get2 5k = , m = – .3 3u w53. a and b cannot both be zero. If a = 0, then the lineax + by = c is horizontal and n = bj is vertical, son is perpendicular to the line. Use a similarargument if b = 0. If a ≠ 0 and b ≠ 0 , then⎛c⎞P1 ⎜ , 0⎟⎝ a ⎠ and ⎛P2 0, c ⎞⎜ ⎟ are points on the line.⎝ b ⎠54.55.⎯⎯→⎛ c c ⎞n⋅ PP 1 2 = ( ai+ bj) ⋅ ⎜– i+ j ⎟= – c+ c=0⎝ a b ⎠2 2u+ v + u− v = ( u+ v) ⋅ ( u+v)+ ( u−v) ⋅( u− v) = [ u⋅ u+ 2( u⋅ v) + v⋅v]+ [ uu ⋅ −2( uv ⋅ ) + vv ⋅ ] = 2( uu ⋅ ) + 2( vv ⋅ )2 2= 2 u + 2 v2 2u+ v − u− v = ( u+ v) ⋅ ( u+v)u v u v u u u v v vuu uv vv uv−( − ) ⋅( − ) = [ ⋅ + 2( ⋅ ) + ⋅ ]−[ ⋅ −2( ⋅ ) + ⋅ ] = 4( ⋅ )1 2 1 2so u⋅ v = u+ v − u−v4 456. Place the cube so that its corners are at the points(0, 0, 0), (1, 0, 0), (0,1, 0), (1, 1, 0), (0, 0, 1),(1, 0, 1), (0,1, 1), and (1, 1, 1). The maindiagonals are (0, 0, 0) to (1, 1, 1), (1, 0, 0) to(0, 1, 1), (0, 1, 0) to (1, 0, 1), and (0, 0, 1) to(1, 1, 0). The corresponding vectors are1, 1, 1 , − 1,1,1 , 1, − 1,1 , and 1,1, − 1 .Because of symmetry, we need only address onesituation; let’s choose the diagonal from(0,0,0) to (1,1,1) and the face that lies in thexy − plane. A vector in the direction of thediagonal is d = 1,1,1 and a vector normal to thechosen face is n = 0,0,1 . The angle betweenthe diagonal and the face is the complement ofthe angle between d and n; that is −1⎛ dn i ⎞90 − θ = 90 − cos=⎜⎟⎝ d n ⎠ −1⎛1 ⎞ −1⎛3⎞90 − cos ⎜ = 90 −cos3 1⎟ ⎜ 3 ⎟⎝ ⎠ ⎝ ⎠ ≈90 − 54.74 = 35.26682 Section 11.3 Instructor’s Resource Manual

57. Place the box so that its corners are at the points(0, 0, 0), (4, 0, 0), (0, 6, 0), (4, 6, 0), (0, 0, 10),(4, 0, 10), (0, 6, 10), and (4, 6, 10). The maindiagonals are (0, 0, 0) to (4, 6, 10), (4, 0, 0) to(0, 6, 10), (0, 6, 0) to (4, 0, 10), and (0, 0, 10) to(4, 6, 0). The corresponding vectors are4, 6, 10 , − 4,6,10 , 4, − 6,10 , and4,6, − 10 .All of these vectors have length16 + 36 + 100 = 152. Thus, the smallest angleθ between any pair, u and v of the diagonals isfound from the largest value of u⋅v, sinceu⋅vu⋅vcos θ = = .u v 152There are six ways of pairing the four vectors.The largest value of u⋅v is 120 which occurswithu = 4, 6, 10 and v = − 4,6,10 . Thus,120 15–1 15cosθ = = so θ = cos ≈ 37.86 ° .152 191958. Place the box so that its corners are at the points(0, 0, 0), (1, 0, 0), (0,1, 0), (1, 1, 0), (0, 0, 1),(1, 0, 1), (0,1, 1), and (1, 1, 1). The maindiagonals are (0, 0, 0) to (1, 1, 1), (1, 0, 0) to(0, 1, 1), (0, 1, 0) to (1, 0, 1), and (0, 0, 1) to(1, 1, 0). The corresponding vectors are1, 1, 1 , − 1,1,1 , 1, − 1,1 , and 1,1, − 1 .All of these vectors have length 1+ 1+ 1=3.Thus, the angle θ between any pair, u and v ofthe diagonals is found fromu⋅vu⋅vcos θ = = .u v 3There are six ways of pairing the four vectors,but due to symmetry, there are only two cases weneed to consider. In these cases, u⋅ v = 1or1u⋅ v = −1. Thus we get that cosθ = or31cosθ =− . Solving for θ gives3θ ≈70.53 or θ ≈109.47 .59. Work = F⋅ D= (3i+ 10 j) ⋅(10 j )= 0 + 100 = 100 joules60. F = 100 sin 70°i – 100 cos 70°jD = 30iWork = F⋅D= (100sin 70 ° )(30) + ( − 100cos70 ° )(0)= 3000 sin 70° ≈ 2819 joules61. D = 5i + 8jWork = F · D = (6)(5) + (8)(8) = 94 ft-lb62. D = 12jWork = F· D = (–5)(0) + (8)(12) = 96 joules63. D = (4 – 0)i + (4 – 0)j + (0– 8)k = 4 +4j – 8 kThus, W = F⋅ D= 0(4) + 0(4) −4( − 8) = 32 joules.64. D = (9 – 2)i + (4 – 1)j + (6 – 3)k = 7i + 3j + 3kThus, W = F⋅ D = 3(7) – 6(3) + 7(3) = 24 ft-lbs.65. 2( x−1) −4( y− 2) + 3( z+ 3) = 02x− 4y+ 3z=−1566. 3( x+ 2) − 2( y+ 3) −1( z− 4) = 03x−2y− z =−467. ( x− 1) + 4( y− 2) + 4( z− 1) = 0x+ 4y+ 4z= 1368. z + 3=0z = −369. The planes are 2x – 4y + 3z = –15 and3x – 2y – z = –4. The normals to the planes areu = 2, − 4,3 and v = 3, −2, − 1 . If θ is the anglebetween the planes,u⋅ v 6+ 8−3 11cos θ = = = , sou v 29 14 406–1 11θ = cos ≈ 56.91 ° .40670. An equation of the plane has the form2x + 4y – z = D. And this equation must satisfy2(–1) + 4(2) – (–3) = D, so D = 9. Thus anequation of the plane is 2x + 4y – z = 9.71. a. Planes parallel to the xy-plane may beexpressed as z = D, so z = 2 is an equation ofthe plane.b. An equation of the plane is2(x + 4) – 3(y + 1) – 4(z – 2) = 0 or2x – 3y – 4z = –13.72. a. Planes parallel to the xy-plane may beexpressed as z = D, so z = 0 is an equation ofthe plane.73.b. An equation of the plane isx + y+ z = D; since the origin is in theplane, 0+ 0+ 0= D . Thus an equation isx+ y+ z = 0 .(1) + 3(–1) + (2) – 7 7Distance = = ≈ 2.11061+ 9+1 11Instructor’s Resource Manual Section 11.3 683

74. The distance is 0 since the point is in the plane.( (–3)(2) + 2(6) + (3) – 9 = 0)75. (0, 0, 9) is on –3x + 2y + z = 9. The distance from(0, 0, 9) to 6x – 4y – 2z = 19 is6(0) – 4(0) – 2(9) –19 37= ≈4.9443is the36 + 16 + 4 56distance between the planes.76. (1, 0, 0) is on 5x – 3y – 2z = 5. The distance from(1, 0, 0) to –5x + 3y + 2z = 7 is–5(1) + 3(0) + 2(0) – 7 12= ≈1.9467.25 + 9 + 4 3877. The equation of the sphere in standard form is2 2 2x y z( + 1) + ( + 3) + ( –4) = 26, so its center is(–1, –3, 4) and radius is 26. The distance fromthe sphere to the plane is the distance from thecenter to the plane minus the radius of the sphereor3(–1) + 4(–3) + 1(4) –15 – 26 = 26 – 26 = 0,9+ 16+1so the sphere is tangent to the plane.78. The line segment between the points isperpendicular to the plane and its midpoint,(2, 1, 1), is in the plane. Then6 −( −2),1−1, −2− 4 = 8, 0, − 6 isperpendicular to the plane. The equation of theplane is8(x – 2) + 0(y – 1) – 6(z – 1) = 0 or 8x – 6z = 10.79. u⋅ v = cosθ u v ≤ u v since cosθ ≤ 1.80.2 2 2u+ v = u + 2u⋅ v+v2 2≤ u + 2 u⋅ v + v2 2( ) 2≤ u + 2 u v + v = u + vTherefore, u+ v ≤ u + v .81. The 3 wires must offset the weight of the object,thus(3i + 4j + 15k) + (–8i – 2j + 10k) + (ai + bj + ck)= 0i + 0j + 30kThus, 3 – 8 + a = 0, so a = 5;4 – 2 + b = 0, so b = –2;15 + 10 + c = 30, so c = 5.82. Let v 1 , v 2 ,…, v n represent the sides of thepolygon connected tail to head in successionaround the polygon.Then v1+ v2 +…+ vn= 0 since the polygon isclosed, soF⋅ v1+ F⋅ v2 +…+ F⋅ vn= F⋅ ( v1+ v2+…+ vn)= F · 0 = 0.83. Let x = x, y,z , so( x− a) ⋅( x− b)=x −a1, y−a2, z−a3 ⋅ x−b1, y−b2,z−b32 2= x − ( a1+ b1) x+ a1b1+ y − ( a2 + b2)y+a2b22+ z − ( a3 + b3)z+a3b3Setting this equal to 0 and completing the squaresyields2 22⎛ a1+ b1⎞ ⎛ a2 + b2⎞ ⎛ a3 + b3⎞⎜x− 2⎟ + ⎜y− 2⎟ + ⎜z−2⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ .1 (2 2 2= ⎡ a1− b1 ) + ( a2 − b2 ) + ( a3 −b3) ⎤4 ⎣⎦⎛a 3 3A sphere with center1+ b1 a, 2 + b 2 a + b ⎞⎜,2 2 2⎟⎝⎠and radius1 ⎡ 2 2 22( a1 – b 11) + ( a2 – b2) + ( a3 – b3) ⎤ = a–b4⎣⎦ 484. Let Px ( 0 , y 0 , z 0 ) be any point onAx + By + Cz = D, so Ax0 + By0 + Cz0= D. Thedistance between the planes is the distance fromPx ( 0, y0, z 0)to Ax + By + Cz = E, which isAx0 + By0 + Cz0– E D – E=.2 2 2 2 2 2A + B + C A + B + C85. If a, b, and c are the position vectors of thevertices labeled A, B, and C, respectively, thenthe side BC is represented by the vector c – b.The position vector of the midpoint of BC is1 1b+ ( c– b) = ( b+c ). The segment from A to2 2the midpoint of BC is 1 ( b+c)– a . Thus, the2position vector of P is2⎡1 ⎤ a+ b+ca+ ( + )– =3⎢b c a2 ⎥⎣ ⎦ 3If the vertices are (2, 6, 5), (4, –1, 2), and(6, 1, 2), the corresponding position vectors are2, 6, 5 , 4, − 1, 2 , and 6, 1, 2 . The positionvector of P is1 12+ 4+ 6, 6–1+ 1, 5+ 2+ 2 = 12, 6, 9 =3 34, 2, 3 . Thus P is (4, 2, 3).684 Section 11.3 Instructor’s Resource Manual

86. Let A, B, C, and D be the vertices of thetetrahedron with corresponding position vectorsa, b, c, and d. The vector representing thesegment from A to the centroid of the oppositeface, triangle BCD, is 1 ( b+ c+d)–a by3Problem 85. Similarly, the segments from B, C,and D to the opposite faces are1 1( a+ c+ d)– b, ( a+ b+d)– c , and3 31 ( a+ b+c )– d , respectively. If these segments3meet in one point which has a nice formulation assome fraction of the way from a vertex to thecentroid of the opposite face, then there is somenumber k, such thata ⎡1 ⎤ ⎡1⎤+ k⎢ ( + + )– = + k ( + + )–3 ⎥ ⎢3⎥⎣ b c d a ⎦ b ⎣ a c d b⎦⎡1 ⎤ ⎡1⎤= c + k⎢ ( + + )– = + k ( + + )– .3 ⎥ ⎢3⎥⎣ a b d c ⎦ d ⎣ a b c d⎦k 3Thus, a(1 – k ) = a , so k = . Hence the3 4segments joining the vertices to the centroids ofthe opposite faces meet in a common point whichis 3 of the way from a vertex to the43corresponding centroid. With k = , all of the4above formulas simplify to 1 ( a+ b+ c+d ), the4position vector of the point.87. After reflecting from the xy-plane, the ray hasdirection ai + bj – ck. After reflecting from thexz-plane, the ray now has direction ai – bj – ck.After reflecting from the yz-plane, the ray nowhas direction –ai – bj – ck, the opposite of itsoriginal direction. ( a < 0 , b < 0 , c < 0 )11.4 Concepts Reviewi j k2 1 –1 1 –1 21. –1 2 1 = i–j+k1 –1 3 –1 3 13 1 –1= (–2 – 1)i – (1 – 3)j + (–1 – 6)k= –3i + 2j – 7k = -3, 2, -7Problem Set 11.4i j k1. a. a× b = –3 2 –2–1 2 –42 –2 –3 –2 –3 2= i – j + k2 –4 –1 –4 –1 2=(–8 + 4)i – (12 – 2)j + (–6 + 2)k= –4i – 10j – 4kb. b + c = 6i + 5j – 8k, soi j ka× ( b+ c ) = –3 2 –26 5 –82 –2 –3 –2 –3 2= i–j+k5 –8 6 –8 6 5= (–16 + 10)i – (24 + 12)j + (–15 – 12)k= –6i – 36j – 27kc. a⋅ ( b+ c ) = –3(6) + 2(5) – 2(–8) = 8i j kd. b× c = –1 2 –47 3 –42 –4 –1 –4 –1 2= i–j+k3 –4 7 –4 7 3= (–8 + 12)i – (4 + 28)j + (–3 – 14)k= 4i – 32j – 17ki j ka× ( b× c ) = –3 2 –24 –32 –172 –2 –3 –2 –3 2= i–j+k–32 –17 4 –17 4 –32= (–34 – 64)i – (51 + 8)j + (96 – 8)k= –98i – 59j + 88ki j k2. a. a× b = 3 3 1–2 –1 03 1 3 1 3 3= i–j+k–1 0 –2 0 –2 –1= (0 + 1)i – (0 + 2)j + (–3 + 6)k = i – 2j + 3k= 1, − 2, 32. u v sinθ3. –(v × u)4. parallelInstructor’s Resource Manual Section 11.4 685

. b + c = -2 - 2, -1 - 3, 0 - 1= -4, -4, -1i j ka× ( b+ c ) = 3 3 1–4 –4 –13 1 3 1 3 3= i – j + k–4 –1 –4 –1 –4 –4= (–3 + 4)i – (–3 + 4)j + (–12 + 12)k= i – j = 1, - 1, 0i j kc. b× c = –2 –1 0–2 –3 –1–1 0 –2 0 –2 –1= i–j+k–3 –1 –2 –1 –2 –3= (1 – 0)i – (2 – 0)j + (6 – 2)k = 1, -2, 4a⋅ ( b× c ) = 3(1) + 3( − 2) + 1(4) = 1i j kd. a× ( b× c ) = 3 3 11 –2 43 1 3 1 3 3= i–j+k–2 4 1 4 1 –2= (12 + 2)i – (12 – 1)j + (–6 – 3)k= 14, −11, − 9i j k3. a× b = 1 2 3–2 2 –42 3 1 3 1 2= i–j+k2 –4 –2 –4 –2 2= (–8 – 6)i – (–4 + 6)j + (2 + 4)k = –14i – 2j + 6kis perpendicular to both a and b. Allperpendicular vectors will have the formc(–14i – 2j + 6k) where c is a real number.i j k4. a× b = –2 5 –23 –2 45 –2 –2 –2 –2 5= i–j+k–2 4 3 4 3 –2= (20 – 4)i – (–8 + 6)j + (4 – 15)k= 16i + 2j – 11kAll vectors perpendicular to both a and b willhave the form c(16i + 2j – 11k) where c is a realnumber.5. u = 3−1, −1−3,2− 5 = 2, −4, − 3 andv = 4−1,0−3,1− 5 = 3, −3, − 4 are in theplane.i j ku× v = 2 –4 –33 –3 –4–4 –3 2 –3 2 –4= i–j+k–3 –4 3 –4 3 –3= (16 – 9)i – (–8 + 9)j + (–6 + 12)k= 7, − 1, 6 is perpendicular to the plane.1 7 1 6± 7, –1, 6 = ± , – ,49 + 1+36 86 86 86are the unit vectors perpendicular to the plane.6. u = 5+ 1,1−3,2− 0 = 6, − 2, 2 andv = 4+ 1, −3−3, −1− 0 = 5, −6, − 1 are in theplane.i j ku× v = 6 –2 25 –6 –1–2 2 6 2 6 –2= i – j+k–6 –1 5 –1 5 –6= (2 + 12)i – (–6 – 10)j + (–36 + 10)k= 14,16, − 26is perpendicular to the plane.±1196 + 256 + 67614, 16, – 267 8 13=± , ,–282 282 282are the unit vectors perpendicular to the plane.i j k7. a× b = –1 1 –34 2 –41 –3 –1 –3 –1 1= i – j + k2 –4 4 –4 4 2= (–4 + 6)i – (4 + 12)j + (–2 – 4)k= 2i – 16j – 6kArea of parallelogram = a×b= 4 + 256 + 36 = 2 74686 Section 11.4 Instructor’s Resource Manual

i j k8. a× b = 2 2 –1–1 1 –42 –1 2 –1 2 2= i–j+k1 –4 –1 –4 –1 1= (–8 + 1)i – (–8 – 1)j + (2 + 2)k= –7i + 9j + 4kArea of parallelogram = a× b = 49 + 81+16=1469. a = 2–3,4–2,6–1 = –1,2,5 andb = −1−3,2−2,5− 1 = − 4,0, 4 are adjacentsides of the triangle. The area of the triangle ishalf the area of the parallelogram with a and b asadjacent sides.i j ka× b = –1 2 5–4 0 42 5 –1 5 –1 2= i – j + k0 4 –4 4 –4 0= (8 – 0)i – (–4 + 20)j + (0 + 8)k = 8, -16, 8( )1 1Area of triangle = 64 + 256 + 64 = 8 62 2= 4 610. a = 3−1,1−2,5− 3 = 2, − 1, 2 andb = 4−1,5−2,6− 3 = 3, 3, 3 are adjacentsides of the triangle.i j ka× b = 2 –1 23 3 3–1 2 2 2 2 –1= i–j+k3 3 3 3 3 3= (–3 – 6)i – (6 – 6)j + (6 + 3)k = -9, 0, 91 1 9 2Area of triangle = a× b = 81+ 81 =2 2 211. u = 0−1,3−3,0− 2 = −1, 0, − 2 andv = 2−1,4−3,3− 2 = 1, 1, 1 are in the plane.i j ku× v = –1 0 –21 1 10 –2 –1 –2 –1 0= i – j+k1 1 1 1 1 1= (0 + 2)i – (–1 + 2)j + (–1 – 0)k = 2, −1, − 1The plane through (1, 3, 2) with normal2, −1, − 1 has equation2(x – 1) – 1(y – 3) – 1(z – 2) = 0 or2x – y – z = –3.12. u = 0−1,0−1,1− 2 = −1, −1, − 1 andv = −2−1, −3−1,0− 2 = −3, −4, − 2 are in theplane.i j ku× v = –1 –1 –1–3 –4 –2–1 –1 –1 –1 –1 –1= i–j+k–4 –2 –3 –2 –3 –4= (2 – 4)i – (2 – 3)j + (4 – 3)k = − 2,1,1The plane through (0, 0, 1) with normal − 2,1,1has equation –2(x – 0) + 1(y – 0) + 1(z – 1) = 0 or–2x + y + z = 1.13. u = 0− 7,3−0,0− 0 = − 7,3,0 andv = 0− 7,0−0,5− 0 = − 7,0,5 are in theplane.i j ku× v =− 7 3 0 =−7 0 53 0 7 0 7 30 5 − −− +−7 5 −7 0= (15 – 0)i – (–35 – 0)j + (0 + 21)k = 15,35, 21The plane through (7, 0, 0) with normal15,35, 21 has equation 15(x – 7) +35(y – 0) +21(z – 0) = 0 or 15x + 35y +21 z = 105.Instructor’s Resource Manual Section 11.4 687

14. u = 0 −ab, −0,0− 0 = − ab , ,0 andv = 0 −a,0−0, c− 0 = − a,0,c are in theplane.i j ku× v = − a b 0 =−a0 cb 0 a 0 a b0 c − −− +−a c −a0= bci – (–ac)j + abk = bc, ac,abThe plane through (a, 0, 0) with normalbc, ac,ab has equation bc(x – a) + ac(y – 0) +ab(z – 1) = 0 or bcx +ac y +abz = abc.15. An equation of the plane is1(x –2) – 1(y – 5) +2(z – 1) = 0 orx –y +2z = –1.16. An equation of the plane is1(x –0) +1(y – 0) +1(z – 2) = 0 orx + y + z = 2.17. The plane’s normals will be perpendicular to thenormals of the other two planes. Then a normal is1, − 3, 2 × 2, −2, − 1 = 7, 5, 4 . An equationof the plane is 7(x + 1) + 5(y + 2) + 4(z – 3) = 0or 7x + 5y + 4z = –5.18. u = 1,1,1 is normal to the plane x+ y+ z = 2and v = 1, −1, −1is normal to the planex− y− z = 4 . A normal vector to the requiredplane must be perpendicular to both the othernormals; thus one possibility isi j ku× v = 1 1 1 = 0,2, −2The plane has an1 −1 −1equation of the form: 2y − 2 z = D.Since thepoint (2, −1,4)is in the plane, 2( −1) − 2(4) = D;thus D = - 10 and an equation for the plane is2y− 2z= − 10 or y− z = − 5.19. (4i + 3j – k) × (2i – 5j + 6k) = 13i – 26j – 26k= 13(i – 2j – 2k)is normal to the plane. An equation of the plane is1(x – 2) – 2(y + 3) – 2(z – 2) = 0 orx – 2y – 2z = 4.20. k = 0,0,1 is normal to the xy-plane andv = 3, −2,1is normal to the plane3x− 2y+ z = 4. A normal vector to the requiredplane must be perpendicular to both the othernormals; thus one possibility is:i j kk× v = 0 0 1 = 2,3,0 The plane has an3 −2 1equation of the form: 2x + 3 y = D.Since thepoint (0,0,0)is in the plane, 2(0) + 3(0) = D;thus D = 0 and an equation for the plane is2x+ 3y= 0 .21. Each vector normal to the plane we seek isparallel to the line of intersection of the givenplanes. Also, the cross product of vectors normalto the given planes will be parallel to both planes,hence parallel to the line of intersection. Thus,4, − 3, 2 × 3, 2, − 1 = − 1,10,17 is normal tothe plane we seek. An equation of the plane is–1(x – 6) + 10(y – 2) + 17(z + 1) = 0 or–x + 10y + 17z = –3.22. a × b is perpendicular to the plane containing aand b, hence a × b is perpendicular to both a andb. (a × b) × c is perpendicular to a × b hence itis parallel to the plane containing a and b.23. Volume = 2, 3, 4 ⋅ ( 0, 4, –1 × 5, 1, 3 ) = 2, 3, 4 ⋅ 13, – 5, – 20 = –69 = 6924. Volume = (3 i– 4j+ 2 k) ⋅ [(– i+ 2 j+ k) × (3 i– 2j+5 k ) = (3 i–4j+ 2 k) ⋅ (12i+ 8 j–4 k ) = –4 = 425. a. Volume = u⋅ ( v×w ) = 3, 2, 1 ⋅ –3, –1, 2 = –9 = 9b. Area = u× v = 3, −5, 1 = 9+ 25+ 1=35688 Section 11.4 Instructor’s Resource Manual

c. Let θ be the angle. Then θ is the complement of the smaller angle between u and v × w.⎛π( )cos sin2 θ ⎞θ u⋅ v×w9 9⎜ − ⎟= == = , θ ≈ 40.01 °⎝ ⎠ u v×w 9+ 4+ 1 9+ 1+4 1426. From Theorem C, a⋅ ( b×c) = ( a×b)⋅c , which leads to a⋅ ( b×c) = c⋅( a×b ) . Again from Theorem C,c⋅ ( a× b) = ( c× a) ⋅ b = −( a×c)⋅b , which leads to c⋅ ( a×b) = b⋅( a×c ) . Therefore, we have thata⋅ ( b×c) = b⋅ ( a×c) = c⋅( a×b ).27. Choice (c) does not make sense because ( abis ⋅ ) a scalar and can't be crossed with a vector. Choice (d) does notmake sense because ( a×b)is a vector and can't be added to a constant.28. a × b and c × d will both be perpendicular to the common plane. Hence a × b and c × d are parallel so(a × b) × (c × d) = 0.29. Let b and c determine the (triangular) base of the tetrahedron. Then the area of the base is 1 2the area of the parallelogram determined by b and c. Thus,1 1⎡1⎤(area of base)(height) = (area of corresponding parallelogram)(height)3 3 ⎢ 2⎥⎣⎦1 1= (area of corresponding parallelpiped) = a⋅ ( b×c )66b×c which is half of30. a = 4+ 1, −1−2,2− 3 = 5, −3, − 1 ,b = 5+ 1,6−2,3− 3 = 6, 4, 0 ,c = 1+ 1,1−2, −2− 3 = 2, −1, − 51Volume = a⋅ ( b×c )61= 5, – 3, –1 ⋅ –20, 30, –1461 88= –176 =6 331. Let u = u1, u2,u3and v = v1, v2,v3thenu× v =u23 v – uv 32, uv 31– uv 13, uv 12–u21v2 2 2 2 2 2 2 2 2 2 2 2 2u× v = u2v3 − 2u2u3v2v3+ u3v2 + u3v1 − 2u1u3v1v3+ u1v3 + u1v2 − 2u1u2v1v2+u2v12 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2= u1 ( v1 + v2 + v3 )– u1 v1 + u2 ( v1 + v2 + v3 )– u2 v2 + u3 ( v1 + v2 + v3)2 2– u3v3 –2 u2uv 3 2v3–2 uuvv 1 3 1 3–2uu 1 2vv1 22 2 2 2 2 2 2 2 2 2 2 2= ( u1 + u2 + u3)( v1 + v2 + v3)–( u1v1 + u2v2 + u3v3 + 2u2uv 3 2v3+ 2uuvv 1 3 1 3+2 uu 1 2vv1 2)2 2 2 2 2 2 2 2 2 2= ( u1 + u2 + u3 )( v1 + v2 + v3 )–( u1v1+ u2v2 + u3v3)= u v −( u⋅v)32. u = u1, u2, u3 , v = v1, v2, v3, w = w1, w2,w3u× ( v× w ) = ( uv 2 3 – uv 3 2) + ( uw 2 3 – uw 3 2), ( uv 3 1 – uv 1 3) + ( uw 3 1 – uw 1 3), ( uv 1 2 – uv 2 1) + ( uw 1 2 – uw 2 1)= (u × v) + (u × w)33. (v + w) × u = –[u × (v + w)]= –[(u × v) + (u × w)] = –(u × v) – (u × w)= (v × u) + (w × u)34. u × v = 0 ⇒ u and v are parallel. u⋅ v = 0 ⇒ u and v are perpendicular. Thus, either u or v is 0.Instructor’s Resource Manual Section 11.4 689

35. PQ = – a, b, 0 , PR = – a,0, c ,The area of the triangle is half the area of theparallelogram with PQ and PR as adjacentsides, so area1( Δ PQR) = − a, b, 0 × − a, 0, c21 1 2 2 2 2 2 2= bc, ac,ab = b c + a c + a b .2 236. The area of the triangle is1x2 −x1, y2 − y1, 0 × x3 −x1, y3 − y1, 0 =21 0, 0, ( x2 −x1 )( y3 − y1 ) − ( y2 − y1 )( x3 −x1)2= 1 ( 2 3 – 3 2 )–( 1 3 – 3 1 ) ( 1 2 – 2 1 )2 x y x y x y x y + x y x ywhich is half of the absolute value of thedeterminant given. (Expand the determinantalong the third column to see the equality.)2 1 (2 2 2 2 2 2D = b c + a c + a b )42 2 2⎛1 ⎞ ⎛1 ⎞ ⎛1⎞ 2 2 2= ⎜ bc ⎟ + ⎜ ac⎟ + ⎜ ab⎟= A + B + C .⎝2 ⎠ ⎝2 ⎠ ⎝2⎠37. From Problem 35,38. Note that the area of the face determined by aand b is 1 a×b .21Label the tetrahedron so that m = ( a×b ),21 1n = ( b×c ), and p = ( c×a ) point outward.22The fourth face is determined by a – c and b – c,soq = 1 [( b – c ) × ( a – c )]21= [( b× a ) – ( b× c ) – ( c× a ) + ( c×c )]21= [–( a× b ) – ( b× c ) − ( c×a )].21m+ n+ p+ q = [( a× b ) + ( b× c ) + ( c×a )2− ( a× b) − ( b× c) − ( c× a)]= 0139. The area of the triangle is A = a×b . Thus,22 1 2 1 2 2 2A = a× b = ( a b −( a⋅b) )4 41⎡2 2 1 2 2 2( ) 2 ⎤= ⎢ a b − a + b − a−b⎥4⎣4⎦1 4 2 2 –( 2 2 – 2 ) 2= ⎡ a b a + b c ⎤16 ⎣⎦1 (22 2 –4 22 2 –4 22 2 –4= ab a + ac b + bc c ).161Note that s – a = ( b+c– a),211s – b = ( a+ c– b),and s – c = ( a+b– c).22ss ( − a)( s−b)( s−c)1= ( a+ b+ c )( b+ c− a )( a+ c− b )( a+ b−c)161 (2 2 2 – 4 2 2 2 – 4 2 2 2 – 4= ab a + ac b + bc c )16which is the same as was obtained above.40. u× v = ( u1i+ u2j+ u3k) × ( v1i+ v2j+v3k)= ( uv 11)( i× i) + ( uv 12)( i× j) + ( uv 13)( i× k)+( uv 21)( j× i) + ( uv 22)( j× j) + ( uv 23)( j× k)+( uv 31)( k× i) + ( uv 32)( k× j) + ( uv 33)( k×k)= ( uv 11)(0) + ( uv 12)( k) + ( uv 13)( − j)+( uv 21)( − k) + ( uv 22)(0) + ( uv 23)( i)+( uv 31)( j) + ( uv 32)( − i) + ( uv 33)(0)= ( u23 v – uv 32) i+ ( uv 31– uv 13) j+( uv 12– u21v)k690 Section 11.4 Instructor’s Resource Manual

11.5 Concepts Review1. a vector-valued function of a real variable7.3 2lim ln( t ), t ln t,t+t→03lim ln( t ) = – ∞ .+t→0does not exist because2. f and g are continuous at c; f ′() t i+g′() t j3. position8.2−1/t tlim e , , t−t→0t4. r′ (); t r ′′();t tangent; concaveProblem Set 11.51.2.3.4.5.6.2 2lim[2 ti– t j] = lim(2 t) i– lim( t ) j=2 i–jt→1 t→1 t→12 3lim[2( t−3) i−7 t j]t→32 3= lim[2( t−3) ] i− lim(7 t ) j=−189jt→3 t→3⎡ 2t− 1 t + 2t−3⎤lim ⎢ i−j⎥t→12⎢⎣t−1t −1⎥⎦( t 1)( t 3)⎛ t −1⎞ ⎛ − + ⎞= limlimt→1⎜( t 1)( t 1)⎟i−⎜ ⎟j⎝ − + ⎠ t→1⎝t−1⎠⎛ 1 ⎞1= lim⎜⎟i− lim( t + 3) j= i−4jt→1⎝t+ 1⎠t→12⎡ 2 32t −10t−28 7t⎤lim ⎢i−j⎥t→−2⎢⎣t+ 2 t−3⎥⎦⎛ 2 32t −10t−28⎞ ⎛ 7t⎞= limi−limjt→−2⎜ t 2 ⎟ t→−2⎜t3⎟⎝+⎠ ⎝−⎠56 56= lim (2t−14) i− j=−18i−jt→−25 5⎡3sin tcost 7t t ⎤lim ⎢ i− j+k⎥t t⎢⎣e t+1 ⎥⎦t→03⎛sin tcost⎞⎛7t⎞= lim ⎜ ⎟i−limjt→0⎝ t ⎠ t→0⎜ te ⎟⎝ ⎠⎛ t ⎞+ lim ⎜ ⎟k= it→0⎝t+ 1⎠⎡3tsin t 7t sin t ⎤lim ⎢ i−j−k⎥t→∞2 3⎢⎣t t − 3tt ⎥⎦3⎛tsin t⎞ ⎛ 7t ⎞ ⎛sint⎞= lim ⎜ lim limt→∞ 2 ⎟i− j −t 3⎜ ⎟k⎝ t ⎠ →∞⎜t − 3t⎟ t→∞⎝ ⎠ ⎝ t ⎠⎛sint⎞ ⎛sint⎞= lim ⎜ ⎟i−7j− lim ⎜ ⎟k =−7jt→∞⎝ t ⎠ t→∞⎝ t ⎠t= lim , lim , lim t = 0, −1,02−1/te− −t 0 t 0 t−→ → t→029. a. The domain of f()t = t – 4is( −∞,4) ∪(4, ∞ ) . The domain ofg() t = 3– t is (– ∞ , 3]. The domain ofht () = ln4− t is ( −∞,4) ∪ (4, ∞ ) . Thus, thedomain of r is (– ∞ , 3] or { t∈ : t ≤ 3}.2= t b. The domain of f () t is ( −∞ , ∞ ) . Thedomain of g() t = 20– t is (– ∞ , 20]. Thedomain of ht () = 3is ( −∞, ∞ ). Thus, thedomain of r is (– ∞ , 20] or { t∈ : t ≤ 20}.c. The domain of f() t = costis ( −∞, ∞ ). Thedomain of g() t = sintis also ( −∞, ∞ ). The2domain of ht () 9 t − 3,3 . Thus,= − is [ ]the domain of r is [ − 3,3], or{ t∈ : −3≤t≤3}.10. a. The domain of f(t) = ln(t – 1) is (1, ∞ ). Thedomain of g() t = 20– t is (– ∞ , 20]. Thus,the domain of r is (1, 20] or{ t∈ : 1< t ≤ 20}.b. The domain of f () t = ln( t ) is (0, ∞ ) ).–1The domain of g() t = tan t is (– ∞ , ∞ ).The domain of ht () = t is ( −∞, ∞ ). Thus,the domain of r is (0, ∞ ) or { t∈ : t > 0}.2g() t = 1/ 1− t is ( − 1,1) .2The domain of ht () = 1/ 9− t is ( − 3,3).(The function f is f( x ) = 0 which hasdomain ( −∞, ∞ ).) Thus, domain of r isc. The domain of( − 1,1).–1Instructor’s Resource Manual Section 11.5 691

11. a.b.2f()t = t – 4is continuous on( −∞,4) ∪ (4, ∞ ) . g() t = 3– t iscontinuous on (– ∞ , 3]. ht () = ln4−tis continuous on ( −∞ ,4) and on(4, ∞ ) . Thus, r is continuous on(– ∞ , 3] or { t∈: t ≤3}.f () t2= t is continuous on(– n 1, – n ) ( n , n 1)+ ∪ + wheren is a non-negative integer.g() t = 20– t is continuous on( ,20)−∞ or { t∈ : t < 20}. ht () = 3is continuous on ( −∞, ∞ ). Thus, r iscontinuous on(– n + 1, – n ) ∪ ( k , k + 1)wheren and k are non-negative integers andk < 400 or2{ t∈ : t < 20, t not an integer}.c.13. a.2g() t = 1/ 1− t is continuous on ( − 1,1) .2ht () = 1/ 9− t is continuous on ( − 3,3)function f is f( x ) = 0 which is continuous on( , )−∞ ∞ .) Thus, r is continuous on ( 1,1)22 tDtr( t) = 9(3t+ 4) i+ 2tej+0k22 2Dtr( t) = 54(3t+ 4) i+ 2(2t + 1) et jb. Dtr() t = sin2ti− 3sin3tj+2tk2Dtr ( t ) = 2cos 2 ti− 9cos3 tj+2 k214. a. ( t) ( e 2 te )− (ln 2)22 2t ( e t e e )b.′ t t tr = − i+ j+k− .t 2 – t – t 2 tr′′ ( ) = + 4 – 2 i+(ln 2) 2 j2 1r′ () t = 2sec 2ti+ j21 + t2 2tr′′ () t = 8tan2sec t 2 ti– j(12 )2+ t. (Thec. f () t = costand f () t = sintarecontinuous on ( −∞,∞ ).2ht () = 9− t is continuous on[ − 3,3]. Thus, r is continuous on[ − 3,3].12. a. f(t) = ln(t – 1) is continuous on ( 1, ∞ ).b.g() t = 20– t is continuous on (– ∞ ,20). Thus, r is continuous on (1, 20) or{ t∈ : 1< t < 20}.–1f() t ln( t )( 0,∞ ).= is continuous on(– ∞ , ∞ ). ht ()–1g() t tan t= is continuous on= t is continuous on( −∞, ∞ ).Thus, r is continuous on (0,∞ ) or { t∈ : t > 0}.15.′−t2r () t = – e i–j ; r''( t)= e i+jt2t−2t2 2r() t ⋅ r ''() t = e − ln2 ( t )t− t ⎡ 2 1 4Dt[() r t ⋅ r''()] t = −2e −⎢⋅ ⋅2t−lnt2 2 3⎣t t t– t 22( t )( )2 24ln−2t4=−2e− +t t16. r′ () t = 3cos3ti+3sin3tjr() t ⋅ r '() t = 017.D [() r t ⋅ r '()] t = 0t3 3–3 t–3t2ht () r() t = e t–1i+e ln(2 t ) j–3te ⎛6 t –7⎞Dt[ h( t) r( t)]= – ⎜ ⎟i2 ⎝ t –1 ⎠–3t⎛2 – 3ln(22 ⎞+ e ⎜ t ) ⎟j⎝ t ⎠⎤⎥⎦18. h(t)r(t) = ln(3t – 2)sin 2ti + ln(3t – 2) coshtj⎡3sin2t⎤Dt[ h( t) r( t)] = ⎢2ln(3 t –2)cos2t+3–2 t⎥ i⎣⎦⎡3cosht⎤+ ⎢ln(3 t – 2)sinh t+3–2 t ⎥ j⎣⎦692 Section 11.5 Instructor’s Resource Manual

19. v() t = r′() t = 4i+ 10tj+2ka() t = r′′() t = 10jv(1) = 4i + 10j + 2k; a(1) = 10j;s (1) = 16 + 100 + 4 = 2 30 ≈ 10.95420.21.22.23.24.2v() t = i+ 2( t –1) j+3( t –3) ka() t = 2j+6( t –3) kv(0) = i – 2j + 27k; a(0) = 2j – 18k;s (0) = 1+ 4 + 729 = 734 ≈ 27.0921 2t4v() t = – i– j+5tk2 2 2t ( t –1)22 2+6t3a() t = i+ j+20tk3 2 3t ( t –1)1 4v(2) = – i– j+80 k ;4 91 26a(2) = i+ j+160 k ;4 271 16 8,294,737s (2) = + + 6400 =16 81 36≈ 80.0025 2 5() t = 6t + 72(6 t t –5) +v i j k4 2 2 4( t) = 30t + 72(66 t – 5)(6 t – 5)a i jv(1) = 6i+ 72 j+ k; a(1) = 30i+4392 j ;s (1) = 36 + 5184 + 1 = 5221 ≈ 72.2562 –4/3v j k ; a() t = 2 tj– t k 92 2 –1/3() t = t + t32/32(2) = 4 +3−1/32v j k ; a(2) = 4 j–k94/32s (2) = 16 + ≈ 4.03592/32 1v(2) = 4 j+ k; a(2) = 4 j– k ;3 39 24/32s (2) = 16 + ≈ 4.03592 3v() t = t i+ 5( t –1) j+ sinπtk2a() t = 2ti+ 15( t –1) j+πcosπtkv(2) = 4i+ 5 j; a(2) = 4i+ 15 j+πk;s (2) = 16 + 25 = 41 ≈ 6.40325. v() t = − sinti+ costj+ka() t = −costi−sintjv(π ) = –j + k; a(π ) = i; s( π ) = 1+ 1= 2 ≈ 1.41426. v(t) = 2 cos 2t i – 3 sin 3t j – 4 sin 4t ka(t) = –4 sin 2t i – 9 cos 3t j – 16 cos 4t kv ⎛π⎞ ⎛π⎞⎜ ⎟= –2 + 3 ; ⎜ ⎟=–16 ;⎝2⎠ i j a ⎝2⎠ks ⎛⎜π ⎞ ⎟= 4 + 9 = 13 ≈3.606⎝2⎠27.28.2 tv( t) = sec ti+3 e j– 4sin 4tk2ta() t = 2sec ttanti+3 e j–16cos4tkv ⎛π⎞ π/4 ⎛π⎞π/4⎜ ⎟= 2 + 3 e ; ⎜ ⎟= 4 + 3e+ 16 ;⎝4⎠ i j a ⎝4⎠i j k⎛π⎞ /2s⎜⎟ = 4 + 9e π ≈ 6.877⎝4⎠t 2 –1/3v() t = – e i–sinπ tj + t k32 –4/3a() t = – e t i– πcos πtj– t k 92/32 2v(2) = – e i+ 2 k; a(2) = – e i– πj– 1 k ;3 39 24/34 2s(2) = e + ≈ 7.408929. v() t = ( πtcosπ t+ sin πt)it+ (cos πt−πtsin πt)j−e − k2a() t = (2πcosπt−π tsin πt)i2t+ ( −2πsinπt−π tcos π t)j+e − k30.–2 2 –2v(2) = 2 π i+ j– e k; a(2) = 2 πi–2 π j+e k ;2 –4s(2) = 4π + 1+ e ≈ 6.3641 2 3v()t = i+ j+kt t t1 2 3a() t = – i– j–k2 2 2t t t1 3 1 1 3v(2) = i+ j+ k; a(2) = – i– j– k ;2 2 4 2 41 9 14s (2) = + 1+ = ≈ 1.8714 4 2231. If v = C,then v = v⋅ v =C Differentiateimplicitly to get D t ( v⋅ v) = 2v⋅ v ′ = 0. Thus,v⋅ v′= v⋅ a = 0, so a is perpendicular to v.Instructor’s Resource Manual Section 11.5 693

32. If r () t = C,similar to Problem 31,33.34.35.36.37.r() t ⋅ r ′() t = 0. Conversely, ifr() t ⋅ r ′() t = 0, then 2() r t ⋅ r ′() t = 0. Butsince 2 r() t ⋅ r′() t = D [ r() t ⋅r ()], t thistt t tmeans that r() ⋅ r() = r () is a constant,so r () t is constant.2 2 2 2s = ∫ 1 + cos t+(–sin t)dt02 2 2 2= ∫ 1+ cos t+ sin tdt = 1+1dt0 ∫02= 2∫dt = 2(2–0) = 2 202 2 2s = ∫ (cos t – tsin t) + (sin t+ tcos t) + 2 dt02 2= ∫ t + 3dt02⎡t2 3 2 ⎤= ⎢ t + 3+ ln t+ t + 32 2⎥⎣⎦0( )3 3= 7 + ln 2+ 7 – ln 3 ≈ 4.1262 2Use Formula 44 with u = t and a = 3 for∫t2+ 3 dt.6 2 436 4 23∫∫s = 24t + 4t + 36 dt= 2 t + 6t + 9dt6 ⎡ 32 t ⎤= ∫ 2( t + 3) dt = 2 ⎢ + 3t⎥3⎢⎣3 ⎥⎦= 2[72 + 18 – (9 + 9)] = 1441 2 4 4∫ 4 36 324011 2 ⎡ 1 2 3/2⎤∫ 2t 1 90 t dt (1 90 t )0⎢135⎥⎣⎦01 (913/2 –1) 6.423s = t + t + t dt= + = += ≈1351 4 4 4∫ 9 36 32401 2 31t dt ⎡ t ⎤0 ⎣ ⎦0s = t + t + t dt∫= 3 41 = 41 = 41≈6.40326339.40.3uf i jF′ () t = f′ ( g()) t g′() t′( u) = –sinu + 3 e ; g′( t) = 6t22 9 t –12= –6tsin(3 t – 4) i+18te2f′ ( u) = 2ui+ sin2 uj; g′( t) = sec tF′ () t = f′ ( g()) t g′() t2 2= 2 tan tsec ti+sec tsin(2 tan t)j41. 1 t −t t t1( e e ) dt ⎡ −e e ⎤∫ i+ j = i−j0 ⎣ ⎦0−1= ( e− 1) i+ (1 −e) j42. 1 3/2 3/2∫ [(1 + t) i+ (1 −t) j]dt−11⎡2 5/2 2 5/2 ⎤= ⎢ (1 + t) i− (1 −t)j5 5 ⎥⎣⎦−18 2 8 2= i+j5 543. r(t) = 5 cos 6ti + 5 sin 6tjv(t) = –30 sin 6ti + 30 cos 6tj2 2v ( t) = 900sin 6t+ 900cos 6t= 30a(t) = –180 cos 6ti – 180 sin 6tj44. a. r′ () t = cos ti–sin tj+(2 t –3) kb.c.32t – 3 < 0 for t < , so the particle moves23downward for 0 ≤ t < .22 2 2r ′() t = cos t+ sin t+(2 t –3)2= 4 t –12t+1024 t –12t+ 10= 0 has no real-number solutions, sothe particle never has speed 0, i.e., it never stopsmoving.t2 – 3t+ 2= 12 whent 2 – 3–10 t = ( t+ 2)(–5) t = 0, t = –2, 5. Since t ≥0, the particle is 12 meters above the ground whent = 5.d. v(5) = cos 5i – sin 5j + 7kj38.∫1 12 12 12s = 343 t + 98 t + 1764 t dt∫01 6 7= 21 5t dt = ⎡3 5t⎤ = 3 5 ≈6.708⎣ ⎦0 01694 Section 11.5 Instructor’s Resource Manual

45. a. The motion of the planet with respectto the sun can be given byx = Rpcos t,y = Rpsin t.Assume that when t = 0, both theplanet and the moon are on the x-axis.Since the moon orbits the planet 10times for every time the planet orbitsthe sun, the motion of the moon withrespect to the planet can be given byx = Rm cos10 t,y = Rmsin10 t.Combining these equations, the motionof the moon with respect to the sun isgiven by x = R cost+R cos10 t,y = R sin t+R sin10 t.ppmmπ10t = π + t or t = .9Thus, when the radius of the planet’s orbit aroundthe sun is ten times the radius of the moon’s orbitπaround the planet and t = , the moon is9motionless with respect to the sun.46. a. Yearsb.b.The moon’s orbit is almost indistinguishable froma circle.c. The sun orbits the earth once each year while themoon orbits the earth roughly 13 times each year.c. x′ () t = – R sint−10R sin10tpy′ () t = Rpcost+10Rmcos10tThe moon is motionless with respect tothe sun when x′ () t and y′ () t are both0.Solve x′ () t = 0 for sin t and y′ () t = 010Rmfor cos t to get sin t = – sin10 t,Rp10Rmcos t = – cos10t.Rp2 2Since sin t+ cos t = 12 2100Rm2 100Rm21= sin 10t+cos 10t2 2RRp2Rm.2p10 Rm.1002 2= Thus, Rp= 100RmorRRp= Substitute this intox′ () t = 0 and y′ () t = 0to get– Rp(sin t+ sin10 t) = 0 andRp (cost+ cos10 t) = 0.πIf 0 ≤t≤ , then to have sin t + sin210t = 0 and cos t + cos 10t = 0 it mustbe thatmpd. r(0) = 93.24i = 93.24 million mi, the sum ofthe orbital radii.e. 93 – 0.24 = 92.76 million mif. No; since the moon orbits the earth 13 times foreach time the earth orbits the sun, the moon couldnot be stationary with respect to the sun unless theradius of its orbit around the earth were 1 th the13radius of the earth’s orbit around the sun.g. v(t) = [–186π sin (2π t) – 6.24π sin(26π t)]i +[186π cos (2π t) + 6.24π cos (26π t)]j2 2a( t) = [ −372π cos(2 πt) −162.24π cos(26 πt)]i2 2+− [ 372π sin(2 πt) −162.24π sin(26 πt)]j1v⎜⎛ ⎟⎞ = 0 i + (–186 π – 6.24 π ) j = –192.24 π j⎝2⎠s 1⎜⎛ ⎟⎞ 192.24⎝2⎠= π million mi/yr12 2 2a⎜⎛ ⎟⎞ = (372 π + 162.24 π ) i + 0j = 534.24 π i⎝2⎠Instructor’s Resource Manual Section 11.5 695

47. a. Winding upward around the right circular cylinder x = sin t, y = cos t as t increases.b. Same as part a, but winding faster/slower by a factor ofc. With standard orientation of the axes, the motion is winding to the right around the right circular cylinder x =sin t, z = cos t.23t .d. Spiraling upward, with increasing radius, along the spiral x = t sin t, y = t cos t.48. For this problem, keep in mind that r, θ , u 1 , and u 2 are all functions of t and that prime indicates differentiationwith respect to t .a. u1 = cosθi+sinθj and u2 =− sinθi+cosθj. Applying the Chain Rule to u 1 gives'u1= ( − sin θ) θ' i+( cos θ)θ'j= θ' ( − sinθi+ cos θ j)= θ'u2Similarly, applying the Chain Rule to u 2 givesu = ( − cos θ) θ' i+ ( −sin θ)θ'j=− θ'cosθi + sin θ j =−θ'u'2( ) 1b. ( )'v() t = r'() t = Dtru = ru + r'u= r'u + rθ ' ua() t = v'() t = D r' u + r ' u1 1 11 2t ( 1 θ 2)''1 1 θ 2 θ 2 θ22( θ ) u1 θ θ u2= r' u + r'' u + r ' u + r '' u + ' r'u( r'' r ' ) ( 2 r' ' r '')= − + +c. The only force acting on the planet is the gravitational attraction of the sun, which is a force directed along theline from the sun to the planet. Thus, by Newton’s Second Law,ma= F = − cu + 0u1 2From Newton’s Law of GravitationGMmF =2rso from part (b)GMmma=−u2 1r2GM( r'' − r( θ') ) u + ( 2 r' θ' + rθ'') u = a= − u2r1 2 1Equating the coefficients of the vectors u 1 and u 2 gives( ) 2 GMr''− r θ ' = −2r2 r' θ' rθ'' = 0d. r×r ' is a constant vector by Example 8. Call it D= r×r ' . Thus,D= r× r'= r× ( r1u1+rθ' u2)= r1r× u1+ rθ' r×u2= 0+ rθ' ru × u( )( )'( u u )1 22= r θ 1×22= r 'θ k696 Section 11.5 Instructor’s Resource Manual

e. The speed at t = 0 is the distance from the sun times the angular velocity, that is, v0 = r0 θ '(0) . Thus,θ = . Substituting these into the expression from part (d) gives'(0) v0 / r02( r )(0) θ '(0) k = D( r(0))v2 0r0k = DD=rv 0 0kSince D is a constant vector (i.e., constant for all t), we conclude that2r θ ' k = D=r v k2r θ ' = r vfor all t.0 00 0f. Let q = r'. From (c)drq = r'=dtd dr dq dq dr dqr''= = = = qdt dt dt dr dt dr( ) 2 GMr''− r θ ' =−2r20 0⎞2 ⎟ 2dq ⎛rvGMq − r⎜=−dr ⎝ r ⎠ r20 0⎞2 ⎟ 2dq ⎛rvGMq = r⎜−dr ⎝ r ⎠ r2 20 03 2dq r v GMq = − dr r rg. Integrating the result from (f) gives2 2dq ⎛r 0v0GM ⎞∫ q dr = drdr∫−⎜ 3 2r r ⎟⎝⎠2 22 r0v021GMq = + + C2 ( −2)r r2 22 r0v022GMq =− + + C1r rWhen t = 0 , r'(0) = q(0) = 0 since the rate of change of distance from the origin is 0 at the perihelion. Also,when t = 0 , r(0)= r0.ThusC2 20 02r0r v 2GM 0 =− + + Cr2GM21 =− + v0r001Thus,2 2 2 2 22 r0v0 2GM r0v0 2GM 2GM2 ⎛1 1 ⎞ ⎛2 r ⎞0q =− + + C2 1 =− + − + v2 0 = 2GM ⎜ − ⎟+ v01−r r r 2r r 0 ⎝r r ⎜0 ⎠ r ⎟⎝ ⎠Instructor’s Resource Manual Section 11.5 697

1h. Let p = 1/ r. Then r = 1/ p and r'=− p'. Thus,2p22 ⎛ ( )22p ' ⎞ p 'q = ( r') = − =⎜ 2 4p ⎟⎝ ⎠ p2Dividing both sides of the equation from (g) by ( ) 2'q2 2 2( r θ' 0) ( r θ' ) ( r θ')q2( r θ ') ( rv 0 0) 0 ( rv 0 0)q( r θ ')22 22GM⎛1 1 ⎞ v0 r0= 12 2⎜ − ⎟+ ⎜ −2 222 22GM⎛1 1 ⎞ v0 r0= 12 2⎜ − ⎟+ ⎜ −2 22⎛ ⎞⎝r r ⎠⎜ r ⎟⎝ ⎠⎛ ⎞⎝r r ⎠⎜ r ⎟⎝ ⎠2GM⎛1 1 ⎞ ⎛ 1 1 ⎞= ⎜ − ⎟+ −⎝ ⎠⎜ ⎟⎝ ⎠2 2 2 2 22 r0v r r0 0 r0rNow substitute the result from above to get1 2( p ')4p 2GM2 2= ( p− p0) +2 2 ( p0− p1)2( θ ')v0r04p2 2 2v0r0⎛dp⎞2 ⎜ 2GM p p v r p pdt⎟ = − + −⎝ ⎠2 2 22v0r0⎛dp ⎞⎛2 p ⎞2GM ( p p0)v2 ⎜ 0 1dt⎟ = − + −2⎝ ⎠ ⎜ p ⎟⎝ 0 ⎠( θ ')( θ ')2 2 2 2( 0) 0 0 ( 0 )i. Continuing with the equation from (h), and using the Chain Rule gives2 22 2⎛dp / dt ⎞⎛2 p ⎞v0r0 ⎜ 2GM ( p p0)v0 1dθ/ dt⎟ = − + −2⎝ ⎠ ⎜ p ⎟⎝ 0 ⎠⎛ dp ⎞⎜dθ⎟⎝ ⎠2 2( p p0) ( p0p )r θ and using the result from (e) that2r θ ' = r v gives2 22GMp0= − + −2v022 22 2 2 22 2GMp ⎛0 GMp ⎞⎛0 2 2GMp ⎛0 GMp ⎞0= p0 − p2 0 + − p − p+⎜ 2 ⎟ 2 ⎜ 2 ⎟0 0 0 0⎞⎛ dp ⎞ ⎜ ⎟⎜dθ⎟⎝ ⎠ v v v v⎝ ⎠ ⎜ ⎟⎝⎝ ⎠⎠22 22⎛ dp ⎞ ⎛ GMp ⎞ ⎛ 2p 0 GMp ⎞⎜ = 0 −0dθ⎟ −p −⎜ 2⎝ ⎠ v ⎟ ⎜ 2⎝ 0 ⎠ v ⎟⎝ 0 ⎠j. Taking the square root of both sides from part (i) gives2 22 2dp ⎛ GMp ⎞ ⎛0 GMp ⎞0=± p0 − − p−dθ⎜ 2 2v ⎟ ⎜0 v ⎟⎝ ⎠ ⎝ 0 ⎠22From (e) we have r θ ' = r0v0, so θ ' = rv 0 0/ r > 0. Recall that the planet is at its perihelion at time t = 0this is as close as it gets to the sun. Thus, for t near 0, the distance from the sun r must increase with t .2dr dr / dt r '1dp dp / dt − p r 'Thus, = = > 0 and from the beginning of (h) r'=− p'. Thus, = =

k. Separating variables in this differential equation gives−dp= dθ2 2⎛ 2 2GMp ⎞ ⎛0 GMp ⎞p00 − − p−⎜ 2 2v ⎟ ⎜0 v ⎟⎝ ⎠ ⎝ 0 ⎠2 2 2 2( 0 0 0) ( 0 0)2 21/ p −GMp / v − p− GMp / v dp = d∫ ∫⎛2 2−1 p−GMp0 / v ⎞0cos⎟= θ + C⎜2 2 2p0 GMp0 / v ⎟⎝ − 0 ⎠p pThe initial condition for this differential equation is = 0 when t = 0 . Thus,⎛2 2−1 p0 − GMp0 / v ⎞0cos ⎟= θ (0) + C⎜2 2p0 GMp0 / v ⎟⎝ − 0 ⎠−1cos (1) = 0 + C0 = CThe solution is therefore θ = cosl. Finally (!)−20 /200 −20 /20p GMp vcosθ=p GMp v2222 2−1 ⎛ p−GMp0 / v0⎜2 2p0 − GMp0 / v0Solving this for p gives ( )( θ )Recall that p = 1/ r, so that( )( θ )2rv 0 00⎝⎞⎟⎠0 0 2 / 0 2 cos 0 2 / 02p = p − GMp v + GMp v12 2 2 2= p0 − GMp0 / v0 cos + GMp0 / v0r1r0r = =2 2 2 2GMp0 / v0 + ( p0 −GMp0 / v0)( cosθ) GM ⎛ GM ⎞+ 1−cosθ2 2rv ⎜0 0 rv ⎟⎝ 0 0⎠rGM r0( 1+e)= =⎛ 2rv ⎞ 1 + ecosθ1+ 0 0−1 cosθ⎜GM⎟⎝ ⎠2rv 0 0where e = − 1 is the eccentricity. This is the polar equation of an ellipse.GMθInstructor’s Resource Manual Section 11.5 699

11.6 Concepts Review1. 1 + 4t; –3 – 2t; 2 – t2.3.x –1 y + 3 –2= =z4 –2 –122 ti–3j+3tk4. 2, − 3,3 ;Problem Set 11.6x –1 y + 3 –1= =z2 –3 31. A parallel vector isv = 4−1, 5 + 2, 6− 3 = 3, 7, 3 .x = 1 + 3t, y = –2 + 7t, z = 3 + 3t2. A parallel vector isv = 7 −2, − 2 + 1, 3 + 5 = 5, − 1, 8x = 2 + 5t, y = –1 – t, z = –5 + 8t3. A parallel vector isv = 6−4, 2−2, −1− 3 = 2, 0, − 4 or1, 0, − 2 .x = 4 + t, y = 2, z = 3 – 2t4. A parallel vector isv = 5 − 5, 4 + 3, 2 + 3 = 0, 7, 5x = 5, y = –3 + 7t, z = –3 + 5t5. x = 4 + 3t, y = 5 + 2t, z = 6 + tx– 4 y–5 z–6= =3 2 16. x = –1 – 2t, y = 3, z = –6 + 5tSince the second direction number is 0, the linedoes not have symmetric equations.7. Another parallel vector is 1, 10, 100 .x = 1 + t, y = 1 + 10t, z = 1 + 100tx –1 y–1 z–1= =1 10 1008. x = –2 + 7t, y = 2 – 6t, z = –2 + 3tx + 2 y –2 z + 2= =7 –6 39. Set z = 0 . Solving 4x+ 3y= 1and10x+ 6y= 10 yields x = 4, y =− 5 . ThusP1(4, −5,0)is on the line. Set y = 0 . Solving4x− 7z= 1and 10x− 5z= 10 yields13 3 13 3x = , z = . Thus 2 ,0,10 5P ⎛⎜⎞⎟ is on the line.⎝10 5 ⎠ 13 3 27 3PP 1 2= −4,0 −( −5), − 0 = − ,5, is a10 5 10 5direction vector. Thus,27, −50, − 6 = −10PP 1 2is also a directionvector. The symmetric equations are thusx − 4 y + 5= =z27 −50 −610. With x = 0, y – z = 2 and –2y + z = 3 yield(0, –5, –7).With y = 0, x – z = 2 and 3x + z = 3 yield⎛5 3⎞⎜ ,0,– ⎟.⎝4 4⎠A vector parallel to the line is5 3 5 25,5,– + 7 = ,5, or 1, 4, 5 .4 4 4 4x y + 5 z + 7= =1 4 511. u = 1, 4, − 2 and v = 2, −1, − 2 are bothperpendicular to the line, so u × v =−10, −2, − 9 , and hence 10, 2,9 is parallel tothe line.With y = 0, x – 2z = 13 and 2x – 2z = 5 yield⎛ 21⎞⎜−8, 0, − ⎟.The symmetric equations are⎝ 2 ⎠x+ 8 y z += =10 2 921212. u = 1, -3, 1 and v = 6, -5, 4 are bothperpendicular to the line, so u × v = -7, 2, 13is parallel to the line.With x = 0, –3y + z = –1 and –5y + 4z = 9 yield⎛ 13 32 ⎞⎜0, , ⎟.⎝ 7 7 ⎠13 32x y– z–=7 =7–7 2 1313. 1, -5, 2 is a vector in the direction of the line.x– 4 y z–6= =1 –5 2700 Section 11.6 Instructor’s Resource Manual

14. 2,1, − 3 × 5, 4, − 1 = 11, − 13,3 is in thedirection of the line.x + 5 y –7 z + 2= =11 –13 315. The point of intersection on the z-axis is (0, 0, 4).A vector in the direction of the line is5−0, −3−0,4− 4 = 5, − 3, 0 . Parametricequations are x = 5t, y = –3t, z = 4.16. 3,1, − 2 × 2,3, − 1 = 5, − 1,7 is in thedirection of the line since the line isperpendicular to 3,1, − 2 and 2,3, − 1 .x – 2 y + 4 –5= =z5 –1 717. Using t = 0 and t = 1, two points on the first lineare (–2, 1, 2) and (0, 5, 1). Using t = 0, a point onthe second line is (2, 3, 1). Thus, two nonparallelvectors in the plane are0+ 2,5−1,1− 2 = 2, 4,1 and2+ 2,3−1,1− 2 = 4, 2, − 1Hence, 2, 4, − 1 × 4, 2, − 1 = −2, −2, − 12 is anormal to the plane, and so is 1, 1, 6 . Anequation of the plane is1(x + 2) + 1(y – 1) + 6(z – 2) = 0 orx + y + 6z = 11.x–1 y–2x – 2 y –118. Solve = and =–4 3 –1 1simultaneously to get x = 1, y = 2. From the firstline 1–1 z= –4 , so z = 4 and (1, 2, 4) is on the–4 –2first line. This point also satisfies the equations ofthe second line, so the lines intersect.−4,3, − 2 and − 1,1, 6 are parallel to the planedetermined by the lines, so−4,3, − 2 × − 1,1, 6 = 20,26, − 1 is a normalto the plane. An equation of the plane is20(x – 1) + 26(y – 2) – 1(z – 4) = 0 or20x + 26y – z = 68.19. Using t = 0, another point in the plane is(1, –1, 4) and 2, 3, 1 is parallel to the plane.Another parallel vector is1−1, − 1+ 1,5− 4 = 0, 0, 1 . Thus,2, 3, 1 × 0, 0, 1 = 3, − 2, 0 is a normal tothe plane. An equation of the plane is3(x – 1) – 2(y + 1) + 0(z – 5) = 0 or 3x – 2y = 5.20. Using t = 0, one point of the plane is (0, 1, 0).2, − 1,1 × 0, 1, 1 = −2, − 2, 2 = –2 1,1, − 1 isperpendicular to the normals of both planes,hence parallel to their line of intersection.3, 1, 2 is parallel to the line in the plane weseek, thus 3, 1, 2 × 1,1, − 1 = − 3,5, 2 is anormal to the plane. An equation of the plane is–3(x – 0) + 5(y – 1) + 2(z – 0) = 0 or–3x + 5y + 2z = 5.21. a. With t = 0 in the first line, x = 2 – 0 = 2,y = 3+ 4⋅ 0= 3, z = 2⋅ 0= 0, so (2, 3, 0) ison the first line.b. − 1, 4, 2 is parallel to the first line, while1, 0, 2 is parallel to the second line, so− 1, 4, 2 × 1, 0, 2 = 8, 4, − 4 = 4 2,1, − 1is normal to both. Thus, π has equation2(x – 2) + 1(y – 3) – 1(z – 0) = 0 or2x + y – z = 7, and contains the first line.c. With t = 0 in the second line,x = –1 + 0 = –1, y = 2, z = –1+ 2⋅ 0 = –1, soQ(–1, 2, –1) is on the second line.d. From Example 10 of Section 11.3, thedistance from Q to π is2(–1) + (2) – (–1) – 7 6= = 6 ≈ 2.449.4+ 1+1 622. With t = 0, (1, –3, –1) is on the first line.2, 4, − 1 × − 2,3, 2 = 11, − 2,14 isperpendicular to both lines, so11(x – 1) – 2(y + 3) + 14(z + 1) = 0 or11x – 2y + 14z = 3 is parallel to both lines andcontains the first line.With t = 0, (4, 1, 0) is on the second line. Thedistance from (4, 1, 0) to 11x – 2y + 14z = 3 is11(4) – 2(1) + 14(0) – 3 39= ≈2.1768.121+ 4 + 196 321⎛π⎞ π ⎛ π ⎞23. r⎜⎟ = i + 3 3 j + k , so ⎜1, 3 3, ⎟ is on the⎝3⎠3 ⎝ 3 ⎠tangent line.r′ () t = –2sinti+ 6cos tj+k , so⎛π⎞r′ ⎜ ⎟= – 3i+ 3j+k is parallel to the tangent⎝3⎠πline at t = . The symmetric equations of the3–1 –3 3 z – πx y3line are = = .– 3 3 1Instructor’s Resource Manual Section 11.6 701

2 324. The curve is given by r() t = 2t i+ 4 tj+t k .r(1) = 2i + 4j + k, so (2, 4, 1) is on the tangentline.2r′ () t = 4ti+ 4j+3 t k , so r′ (1) = 4i+ 4j+3k isparallel to the tangent line. The parametricequations of the line are x = 2 + 4t, y = 4 + 4t,z = 1 + 3t.2 525. The curve is given by r() t = 3ti+ 2 t j+t k .r(–1) = –3i + 2j – k, so (–3, 2, –1) is on theplane.4r′ () t = 3i+ 4tj+5 t k , so r′ (–1) = 3 i– 4j+5k isin the direction of the curve at t = –1, hencenormal to the plane. An equation of the plane is3(x + 3) – 4(y – 2) + 5(z + 1) = 0 or3x – 4y + 5z = –22.26.27. a.⎛π⎞ π 3π⎛π3π⎞r⎜⎟ = i + j , so ⎜ , , 0⎟is on the⎝2⎠2 2 ⎝2 2 ⎠plane.r′ ( t) = ( tcost+ sin t) i+ 3 j+(2cos t – 2tsin t)k so⎛π⎞r′ ⎜ ⎟= i+ 3 j–πk is in the direction of the⎝2⎠πcurve at , hence normal to the plane. An2equation of the plane is⎛ π⎞ ⎛ 3π⎞1 ⎜x– ⎟+ 3 ⎜y– ⎟– π ( z–0) = 0 or⎝ 2⎠ ⎝ 2 ⎠x + 3y – π z = 5π .b.2 2 2[ xt ( )] + [ yt ( )] + [ zt ( )]( )222= (2 t) + 7t +⎛9 7t 4t⎞⎜ − − ⎟⎝⎠2 2= 4t + 7t+ 9−7t− 4t= 9Thus, the curve lies on the sphere2 2 2x + y + z = 9 whose center is at theorigin.r() t = 2ti+ 7t j+ 29−7t−4tk1r(1/4) = i+ 27/4j+ 9−7/4−1/4k1 7= i+ j+2 27 k7 −7−8tr'( t) = 2i+ j+2 t 22 9−7t−4tkr'(1/ 4) = 2i+ 97 j−k2 7The tangent line is therefore228. a.b.1x = + 2t27y = + 7 t29z = 7 − t2 7This line intersects the xz-plane when y = 0 ,7which occurs when 0= + 7t, that is,211when t = − . For this value of t, x = − ,229 37y = 0 , and z = + 7 = . The4 7 4 7point of intersection is therefore⎛ 1 37 ⎞⎜−,0, ⎟.⎝ 2 4 7 ⎠2 2 2[ xt ( )] + [ yt ( )] + [ zt ( )]2 2 2 2= (sin tcos t) + (sin t) + cos t2 2 4 2= sin tcos t+ sin t+cos t2 2 2 2= sin t(cos t+ sin t) + cos t2 2= sin t+ cos t = 1Thus, the curve lies on the sphere2 2 2x + y + z = 1whose center is at the origin.r ⎛π⎞ ⎛1⎞⎛3⎞1 3⎜ ⎟= ⎜ ⎜+ +⎝6⎠ ⎝2 ⎟⎜ 2 ⎟⎠⎝ i ⎠ 4 j 2k3 1 3 ⎛ 3 1 3⎞= i+ j+k , so, ,4 4 2 ⎜4 4 2 ⎟is on⎝ ⎠the tangent line.2 2r′ ( t) = (cos t – sin t) i+2costsin tj– sin tk⎛π⎞ 1 3 1so r′ ⎜ ⎟= i+j–k is parallel to the⎝6⎠2 2 2line. The line has equations3 1 3x = + t, y = + 3 t, z = – t.4 4 2The line intersects the xy-plane when z = 0,3so t = , hence23 3 3 3 1 3 7x = + = , y = + = .4 2 4 4 2 4⎛3 3 7 ⎞The point is⎜ , , 0 .4 4 ⎟⎝ ⎠702 Section 11.6 Instructor’s Resource Manual

29. a.2 2r() t = 2 ti+ t j+ (1 −t) k. Notice that2 2x() t + y() t + z() t = 2t+ t + 1− t = 2t+ 1.Since x = 2t, we have x + y+ z = x+ 1. , sothis curve lies on the plane with equationy + z = 1.=2 2⎯⎯→ ⎛⎯⎯→⎞2PQ ⎜ PQ ⎟nn−⎜⎝2⋅n⎟⎠by Lagrange’s Identity. Thus,=2⎯⎯→PQ × nn2b. Since r(2) = 4i+ 4j−3k ,r'( t) = 2 i+ 2tj−2tk andr'(2) = 2i+ 4j−4k , the equation of thetangent line is x = 4+ 2t, y = 4+ 4t,z =−3− 4t. The line intersects the xy-planewhen z = 0 , that is, when3t =− , which45gives x = , y = 1, and z = 0 . The point of2⎛5 ⎞intersection is therefore ⎜ ,1,02⎟⎝ ⎠ .30. In Figure 7, d is the magnitude of the scalar⎯⎯→projection of ⎯⎯→ ⎯⎯→PQ ⋅nPQ on n. prnPQ =2 n,nso⎯⎯→⎯⎯→ PQ ⋅n⎯⎯→PQ ⋅nprnPQ = n = n2 2n n=⎯⎯→PQ ⋅ nThe point (0, 0, 1) is on the plane4x – 4y + 2z = 2. With P(0, 0, 1), Q(4, –2, 3), and1n = 2, – 2, 1 = 4, – 4, 2 ,2⎯⎯→PQ = 4 – 0, – 2 – 0, 3 –1 = 4, – 2, 2 and4, – 2, 2 ⋅ 2, – 2, 1 14d = = .4+ 4+1 3From Example 10 of Section 11.3,4(4) – 4(–2) + 2(3) – 2 28 14d = = = .16 + 16 + 4 6 3⎯⎯→ ⎯⎯→31. Let PR be the scalar projection of PQ on n.Then2 2⎯⎯→ ⎯⎯→2PQ PR d= + sond =⎯⎯→PQ × n.na. P(3, –2, 1) is on the line, so⎯⎯→PQ = 1–3, 0 + 2, –4–1 = –2, 2, –5while n = 2, − 2,1 , so−2, 2, − 5 × 2, −2, 1d =4+ 4+1−8, −8, 0 8 2= = ≈3.7713 3b. P(1, –1, 0) is on the line, so⎯⎯→PQ = 2 – 1, – 1 + 1, 3 – 0 = 1, 0, 3 whilen = 2,3, − 6 .1, 0, 3 × 2, 3, −6 −9, 12, 3d = =4+ 9+3673 26= ≈ 2.185732. d is the distance between the parallel planescontaining the lines. Since n is perpendicular toboth n 1 and n 2, it is normal to the planes. Thus,d is the magnitude of the scalar projection of PQPQ ⋅non n, which is .na. P(3, –2, 1) is on the first line, Q(–4, –5, 0) ison the second line, n 1 = 1, 1, 2 , andn 2 = 3, 4, 5 .PQ = –4–3, –5+ 2,0–1 = –7, –3, –1n = 1, 1, 2 × 3, 4, 5 = –3, 1, 1–7, –3, –1 ⋅ –3,1,1 17d = = ≈5.1269+ 1+1 112 2 2⎯⎯→ ⎯⎯→ ⎯⎯→2d PQ PR PQ= − = −2⎯⎯→PQ ⋅nn2Instructor’s Resource Manual Section 11.6 703

. P(1, –2, 0) is on the first line, Q(0, 1, 0) is onthe second line, n 1 = 2, 3, – 4 , andn 2 = 3, 1, – 5 .PQ = 0–1,1+ 2,0–0 = –1,3,0n = 2, 3, – 4 × 3, 1, – 5 = –11, – 2, – 7−1, 3, 0 ⋅ −11, −2, −7d =121+ 4 + 495= ≈0.37917411.7 Concepts Review1.dTds2. 1/ a ; 03.4. 022 ;d sdt⎛⎜⎝dsdt⎞⎟⎠2κProblem Set 11.71.2r () t = t,tv() t = r′() t = 1,2tv(1) = 1, 2a() t = v′() t = 0,2a(1) = 0, 2T() t =1,2t1+ 4t1+4tT(1) =1 2,5 52 24t2T′ () t = −,2 2( 1+ 4t) ( 1+4t)T′ (1) =4 2− ,3/2 3/25 5T′ (1)2=5v(1) = 5κ =T′(1) 2 2= =v(1) 5 5 53/2 3/23/22.2r () t = t ,1+2tv() t = r′() t = 2,2 tv(1) = 2, 2a() t = v′() t = 2,0a(1) = 2, 0T() t =2t,24+ 4t4+4t=2 2t,11+ t 1+t1 1T(1) = ,2 22 21T′ () t = , −2 2( 1+ t ) ( 1+t )3/2 3/2T′ (1) =1 1, −2 2 2 2T′ (1)1=2v() t = 2 2κ =T′(1) 1/2 1= =v(1)2 2 4 2t704 Section 11.7 Instructor’s Resource Manual

3. r () t = t,2cos,2sint tv() t = r′() t = 1, −2sin t,2costv( π ) = 1,0, −2a() t = v′() t = 0, −2cos, t −2sinta( π ) = 0,2,0T() t =1 2sint2cost, − ,5 5 5T( π ) =1 2,0, −5 5T′ () t =2cost2sint0, − , −5 5T′ ( π ) = 0,2,05T′ ( π ) =25v() t = 5κ =T′( π ) 2/ 5 2= =v( π ) 5 54. r () t = 5cos,2,5sin t t tv() t = r′() t = −5sin t,2,5costv( π ) = 0,2, −5a() t = v′() t = −5cos,0, t −5sinta( π ) = 5,0,0T() t =5sint2 5cost− , ,29 29 29T( π ) = 0,2 5, −29 29T′ () t =5cost5sint− ,0, −29 29T′ ( π ) =5,0,0295.T′ ( π ) =529v() t = 29κ =T′( π ) 5/ 29 5= =v( π ) 29 292tr () t = ,5cos,5sin t t8tv() t = r′() t = , −5sin t,5cost4πv( π ) = ,0, −541a() t = v′() t = , −5cos, t −5sint41a( π ) = ,5,04T() t =t, −20sin t,20cost400 + t 400 + t 400 + t2 2 2π20T( π ) = ,0, −2 2400 + π 400 + π400T′() t =,3/22( 400 + t )2− 20 ((400 + t )cost−tsint)3/22( 400 + t )2− 20 ((400 + t )sin t+tcost)3/22( 400 + t ),Instructor’s Resource Manual Section 11.7 705

400T ′( π ) =,=3/22( 400 + π )2− 20 ((400 + π )cosπ −π sinπ)3/22( 400 + π )2− 20 ((400 + π )sinπ + π cosπ)3/22( 400 + π )400 −20, ,2 2( 400 + π ) ( 400 + π )20π2( 400 + π )3/2 1/23/22 2400 400 400πT′ ( π ) = + +2 2( ) ( )2400 + π 400 + π( 400 + π )≈ 0.9890912πv() t = + 2516T′( π )κ = ≈0.195422v( π )3 3,T () t =t,−4sint ,4cost16 + t 16 + t 16 + t2 2 2π −4T( π ) = ,0,2 216 + π 16 + π16 (2)2( 16 + t )2( 16 + t )2− 4cos t t+ 416 ( + t ) sint3/22( 16 + t )16− (2+ π )2( 16 + π )2( 16 + π )− 416+ t cost+4sin t tT′ () t =, ,2( 16 + π )3/2 3/2416 cosπT′ ( π ) =, ,−4πcosπ3/2 3/23/2( π )2( 16 + π )22 2 2 2 216 + 4 16 + + 4 πT′ ( π ) =322 24 16+ π + ( 16+π )=≈ 0.801495( 16 + π )23/2πv() t = 4+4T′( π )κ = ≈0.315164v(π )6.2tr () t = ,2cos,2sin t t4tv() t = r′() t = , −2sin t,2cost2πv( π ) = ,0, −221a() t = v′() t = , −2cos, t −2sint21a( π ) = ,2,02706 Section 11.7 Instructor’s Resource Manual

7. u′ () t = 8ti+4j8.2u ′() t = 4 4t+ 1T()t =u′() t 2t1= i+ju′() t 2 24t + 1 4t+ 1⎛1⎞ 1 1T⎜⎟ = i + j⎝2 ⎠ 2 2x′ () t = 8ty′ () t = 4x′′ () t = 8y′′ () t = 0xy ′′′– yx ′′′ 32κ () t = =⎡ 2 2x′ + y′⎤ 64(4t+ 1)⎣ ⎦1=2 3/22(4t+ 1)κ ⎛ 1 1 1⎜⎞ ⎟⎝2 ⎠= 3/22(2)=4 22r′ () t = t i+tj2r ′() t = t t + 13/2 2 3/2T()t =r′() t t 1= i+jr′() t 2 2t + 1 t + 1T(1)=1 1i+j2 22x′ () t = ty′ () t = tx′′ () t = 2ty′′ () t = 1κxy ′′′ − yx ′′′2t1() t = = =2 23/2 3 2 3/2 2 3/2⎡x′ + y′⎤ t ( t + 1) t( t + 1)⎣ ⎦1 1κ (1) = =3/21(2) 2 29. z′ () t = –3sinti+4costjz′ 2 2() t = 9sin t+16cos tT()t =z′() tz′() t=−3sint4costi+2 2 2 29sin t+ 16cos t 9sin t+16cos t⎛π⎞ 3 4T⎜⎟ = – i + j⎝4⎠5 5x′ () t = –3sinty′ () t = 4costx′′ () t = –3costy′′ () t = –4sintxy ′′′ − yx ′′′12κ()t = =⎡x′ + y′⎤t+t⎣ ⎦12 24 2κ⎛ ⎜π ⎞ ⎟= =⎝4⎠25 125( 9sin 16cos )2 23/22 23/2( ) 3/22j10. r′ () t = e i+e j11.r ′() t = e 2tttT()t =r′() t 1 1= i+jr′() t 2 2T(ln 2) =1 1i+j2 2tx′ () t = ety′ () t = etx′′ () t = ety′′ () t = exy ′ ′′ – yx ′ ′′κ () t = = 02 23/2⎡x′ + y′⎤⎣ ⎦κ(ln 2) = 02 3x = 1 − t , y = 1−tx′ () t = − 2, t y′() t =− 3tx′′ () t = − 2, y′′() t =− 6t2 3( ) ( )r() t = 1− t i+ 1−tj2r'( t) =−2ti−3tjr′ () t −2ti−3tjT()t = =r′() t 2 44t + 9t2() i−()() + ()−21 31 j 2 3T()1 = = − i−j2 441 91 13 13κ()t ==xy ′′′ − xy ′′′2 2(( x′ ) + ( y′) )2( −2t)( −6t) −( −2)( −3t)⎛⎜⎝23/222 2( − 2t) + ( −3t)3/22 ⎞⎟⎠2 212t− 6t26t= =3/2 3/22 4 2 4( 4t + 9t ) ( 4t + 9t)When t = 1, the curvature is6 6κ = 3/2 = 3/2 ≈0.1280084+9 13( )Instructor’s Resource Manual Section 11.7 707

12. r′ ( t) = cosh ti+sinh tjr ′() t =2 2sinh t+cosh tT()t =r′() t cosht=ir′() t 2 2sinh t+cosh t+sinh tj2 2sinh t+cosh tT( ln 3)=5 4i+j41 41x′ () t = coshty′ () t = sinhtx′′ () t = sinhty′′ () t = coshtxy ′′′– yx ′′′1κ () t = =2 23/2 2 2 3/2⎡x′ + y′⎤ (sinh t+cosh t)⎣ ⎦1 27κ (ln 3) =41( ) 3/2 =41 419;15.y′ = 4 x, y′′= 442 3/2κ =(1+16 x )4At (1, 2), κ = and17 1717 17R = .413.– ttr′ ( ) = –(cos + sin ) i+(cos – sin ) jt t t e t t e–() t 2etr ′ =r′ ( t) cost+ sin t cost−sintT()t = = − i+jr′() t 2 21 1T(0) = – i+j2 216.x′ () t = –(cos t –sin t)e −t y′ () t = (cos t –sin t)e −tx′′ () t = (2sin t)e −t y′′ () t = (–2cos) t e −txy ′′′– yx ′′′ –2tt2e eκ () t = = =2 23/2 –3t⎡x′ + y′⎤ 2 2e2⎣ ⎦1κ (0) =214. r′ ( t) = (cos t – tsin t) i+ (sin t+tcos t)j2r ′() t = t + 1r′ () t cost− tsint sint+tcostT()t = = i+jr′() t 2 2t + 1 t + 1cos1– sin1 sin1+cos1T(1)= i+j2 2x′ () t = cos t – tsinty′ () t = sint+tcostx′′ () t = –2sin t – tcosty′′ () t = 2cos t – tsintxy ′′′– yx ′′′ t + 2κ () t = =⎡ 2 2x′ + y′⎤ ( t + 1)⎣ ⎦3κ (1) =2 223/2 2 3/22y′ = 3 x –16x+ 16, y′′= 6 x–166 x –16κ =2 23/2⎡1 + (3 x –16x+16) ⎤⎣⎦81At (4, 0), κ = = 8 and R = .18708 Section 11.7 Instructor’s Resource Manual

17.19.y′ = cos x,y′′= – sin xsin xκ =(1+cos x)2 3/2⎛π2 ⎞ 2 2At⎜, , κ4 2 ⎟= =33/2⎝ ⎠3 3( )21andR =3 3 .22 yy′ –8x= 02 24 x 4( – ) 4( –4 ),y xy ′y′ = y′′= =y xy 2 3y y2 2x34 y –162 2y 4( y – 4 x )κ = =3/2 2 2 3/2⎛ 216 ( 16 )1 x ⎞ y + x⎜ +2 ⎟⎝ y ⎠At (2, 6),80 2κ = = and1000 25R =25 .218.20.2yy′ = 11 y′1y′ = , y′′= – = –2y 2y 4y14 y2 332κ = =⎛ 1 (4y1)1⎞ +⎜ +2 ⎟⎝ 4y⎠2At (1, 0), κ = = 2 and13/2 2 3/21R = .22 yy′ –8x= 02 24 x 4( – ) 4( –4 ),y xy ′y′ = y′′= =y xy 2 3y y2 24( y –4 x )32 2y 4( y – 4 x )κ = =3/2 2 2 3/2⎛ 216 ( 16 )1 x ⎞ y + x⎜ +2 ⎟⎝ y ⎠At (2, –6),80 2κ = = and1000 2525R = .2Instructor’s Resource Manual Section 11.7 709

21.23y′ = –2sin2 x,y′′= –4cos2xκ =4cos2x2 3/2(1+4 sin 2 x)⎛π1 ⎞At ⎜ , ⎟⎝6 2⎠ , 2 1κ = = and R = 4.8 42 2sec , 2sec tan22sec x tan x4 3/2(1 sec )y′ = x y′′= x xκ =+x⎛π⎞ 4At ⎜ , 1 ⎟,κ = and R =⎝4 ⎠ 5 55 5 .422242 2– x2 – xy′ = –2 xe , y′′= (4 x – 2) e22 – x22x2(4 x – 2) e e 4 x –2κ = =(1+ 4 xe ) ( e + 4 x)2 22 –2x3/2 2x2 3/2⎛ 1⎞2eAt ⎜1, ⎟,κ =⎝ e ⎠ ( e + 4)22 3/22 3/2( e + 4)and R =22e.y′ = 1 , y′′= –12 x3/24x13/24x2κ = =3/2 3/2⎛ 1 ⎞ (4x+ 1)1 4⎜⎝+⎟x ⎠2 5 5At (1, 1), κ = and R = .5 5 2710 Section 11.7 Instructor’s Resource Manual

2526y′ = 1 , y′′= –22/3 5/33x9x25/31/39x6xκ = =3/2 4/3 3/21 (9 1)( 1+x +4/39x)6 3At (1, 1), κ = = and R =10 10 5 105 10 .32 2sech , –2sech tanh22sech x tanh x4 3/2(1 sech )y′ = x y′′= x xκ =+x96⎛ 3 ⎞ 125 12,000At ⎜ln 2, ⎟,κ = =⎝ 5 ⎠ 8813/2 881 881R =881 881 .12,000( )625and27.2r′ () t = ti+ j + t kr′′ () t = i+2tkr′ (2) 2i+ j + 4k2 1 4T(2)= == i + j+kr′ (2) 4+ 1+16 21 21 21r′ (2) ⋅ r′′(2) (2i+ j + 4 k) ⋅ ( i+4 k)18a T (2) = ==r′(2) 2121r′ (2) × r′′(2) (2i+ j + 4 k) × ( i+4 k)1 33 11a N (2) = == 4i−4j− k = =r′(2) 2121 21 718 ⎛ 2 1 4 ⎞′′(2) – a (2) (2)N(2)= r T T ( i+ 4 k)–21⎜ i+ j+k21 21⎟⎝21=⎠ 7 ⎛ 15 18 12 ⎞= ⎜– i – j + k ⎟aN(2)1111 ⎝ 21 21 21 ⎠77 ⎛ 5 6 4 ⎞ 5 6 4= ⎜– i – j + k ⎟ = – i – j + k11 ⎝ 7 7 7 ⎠ 77 77 77r′ (2) × r′′(2) 33 33 11 11κ(2)= = = = =3 3r′(2) 219261 3087 21 7B(2) = T(2) × N (2)( )⎛ 2 1 4 ⎞ ⎛ 5 6 4 ⎞= ⎜ + + ⎟× ⎜– – + ⎟⎝ 21 i 21 j 21 k ⎠ ⎝ 77 i 77 j 77k 4 4 1= i– j–k⎠ 33 33 33Instructor’s Resource Manual Section 11.7 711

28. r(t) = sin 3t, cos 3t, tr ′() t = 3cos3,–3sin3,1t tr ′′() t = –9sin3,–9cos3,0t tπ( 9 )π( ), ,1⎛π⎞ r′T⎜⎟ = =⎝9 ⎠ r′ + + 1a Ta Nπ( ) ⋅rπ9 ( 9)r π( )⎛π⎞ r′ ′′⎜ ⎟ =⎝9⎠ ′⎛π⎞⎜ ⎟ =⎝9⎠ ′3 3 3−2 2=9 279 4 43 3 3 9 3⋅92 2 2 293 3 3 1,– ,2 10 2 10 10,– ,1 – ,– ,0= = 010π π 3 3 3 9 3r′ ( 9) × r′′( 9),– ,1 × – ,–9,02 2 2 2 1 9 9 3 1= ( )rπ( 9 )1010 2 210π( ) – aπT ( ) T π( ) 1 9 3 9 3 1⎛π⎞ r′′9 9 9N⎜⎟ == – ,– ,0 = – ,– ,0⎝9⎠ aN9 2 2 2 2′ π( ) ′′ π⎛π⎞r × r9 ( 9)9 10 9κ ⎜ ⎟ = = =⎝9⎠ 3 3π10r′( 9 ) ( 10 )B ⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞⎜ ⎟= ⎜ ⎟×⎜ ⎟⎝9⎠ T ⎝9⎠ N 3 3 3 1 3 1= ,– , × – ,– ,0 =⎝9⎠2 10 2 10 10 2 229. r () t = 7 sin 3t, 7 cos 3t, 14tr ′( t) = 21cos3 t, −21sin 3 t, 14r ′′() t = –63sin3,–63cos3,0t tπ⎛π⎞ r′ ( 3 ) −21, 0, 14 1T⎜⎟ = = = − 21, 0, 14 = −⎝3 ⎠ r′π( 3 ) 441+196 7 13π π⎛π⎞ r′ ( 3) ⋅r′′( 3)−21, 0, 14 ⋅ 0, – 63, 0a T ⎜ ⎟ == = 0⎝3⎠ r′π( 3 )7 13′ π( ) ′′ π⎛π⎞r × r3 ( 3)− 21, 0, 14 × 0, −63, 0a N ⎜ ⎟ ==⎝3⎠ r′π( 3 )7 13π( ) – aπ π⎛π⎞ r′′3 T ( 3) T( 3)1N ⎜ ⎟ = = 0, 63, 0 = 0, 1, 0⎝3⎠aπ( 3 ) 63Nπ( )⎛π⎞r × r′′3 ⎜ ⎟⎛π⎞ ⎝3 ⎠ 441 13 9κ ⎜ ⎟= = =⎝3⎠ 3 3′91rπ( ) ( 7 13)3B ⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞⎜ ⎟= ⎜ ⎟×⎜ ⎟⎝3⎠ T ⎝3⎠ N 3 2= − ,0, × 0,1,0⎝3⎠13 13= ,– ,–27 = 9 10 = 93 2,0,13 131 3 3,– ,–2 10 2 10 10882, 0, –1323 441 13= = = 637 13 7 132 3= − ,0, −13 13712 Section 11.7 Instructor’s Resource Manual

30.2 2r t t ti t tk′( ) = –3cos sin + 3sin cos2 3 2 3r t t t t i t t t k′′( ) = (6cos sin – 3cos ) + (6cos sin – 3sin )⎛π⎞πr′ ⎜ ⎟=0 so the object is motionless at t 1 = .⎝2⎠2κ, T, N, and B do not exist.t31. r′ () t = sinh i+j 31 tr′′ () t = cosh i3 31i+ j sinh1i+j3 32 11+33r′(1) sinhT(1)= = =r′(1) sinh 1 cosha Ta N( sinh1i+ j 1 1) ⋅( cosh i)r′ (1) ⋅r′′(1)(1) = =r′(1) cosh(1) =r′ (1) × r′′(1)r′(1)=′′(1) – a (1) (1)N(1)= r T Ta (1)N1 1= sech i – tanh j3 31 1r cosh3 33 3cosh133 3 31313+ ×1cosh13 3cosh13( sinh ) ( )1 1= tanh i+sech j3 31 1= sinh3 3i j i −1cos 1k3 3 1= =cosh 313⎡1 1 1 1 1 1 ⎤⎛2 1⎛⎞ 1 sinh ⎞= 3⎢cosh i– sinh ⎜tanh i+sech j3 1⎟3 3 3 3 3 3⎥ = ⎜cosh – ⎟i– tanh j⎣⎝⎠1⎦ ⎜ 3 cosh ⎟ 3⎝3 ⎠r′ (1) × ′′(1) 1 2 1κ (1) = = = sechr′(1)3 31 1 1 1B(1) = T(1) × N(1)⎛ tanh sech ⎞ ⎛ sech – tanh⎞ ⎛ 2 1 2 1⎞= ⎜ i+ j⎟×⎜ i j⎟= ⎜ –sech – tanh ⎟⎝ 3 3 ⎠ ⎝ 3 3 ⎠ ⎝ 3 3⎠ k = –k32.7t 7t 7tr t e t t i e t t j e k′( ) = (7 cos 2 – 2sin 2 ) + (7sin 2 + 2cos 2 ) + 77t 7t 7tr t e t t i e t t j e k′′( ) = (45cos2 − 28sin 2 ) + (45sin 2 + 28cos2 ) + 49π( 3 )π( )77 3( 2 ) ( 2 )7 π /3 277 3( 2 ) ( 2 )7 π/3 7 π/3 7 π/3e − − 3 + e − 1 + 7e⎛π⎞ r′i j kT⎜⎟ = =⎝3⎠ r′23 e − − 3 + − 1 + 49aaTN( ) ⋅r( )rπ( )π π 14 π / 33 3 714e7 π /37 102e7 π /3e3⎛π⎞ r′ ′′⎜ ⎟ = = =⎝3 ⎠ ′10214 π / 3( 49 + 14 3)iπe ( )e⎛π⎞ ⎛π⎞r′ ⎜ ⎟×r′′⎜ ⎟ 14 / 3 14 π/ 3⎛π⎞ ⎝3⎠ ⎝3⎠+ 14 − 49 3 j+106ek⎜ ⎟= = = 2 53e⎝3 ⎠ ⎛π⎞7 π /3r′ e 102⎜ ⎟⎝3⎠π( ) aπT ( ) Tπ( )⎛π ⎞ r′′–3 3 3 2–7 3 7+2 3N⎜⎟ = = i–j ;⎝3 ⎠ a2 53 2 53NB ⎛π⎞ ⎛π⎞ ⎛π ⎞ 42 + 49 3 14 3 −147 53⎜ ⎟= ⎜ ⎟× ⎜ ⎟= + +⎝3 ⎠ T ⎝3 ⎠ N ⎝3 ⎠ 6 1802 i 6 1802j 102k⎛ 7+2 3⎞ ⎛7 3–2⎞7= ⎜– + +2 102 ⎟ i ⎜ 2 102 ⎟j k⎝ ⎠ ⎝ ⎠ 102( ) × r3 ( 3)rπ( )7 π /3π π 14 π / 3′ ′′⎛π⎞r2 5406eκ ⎜ ⎟ = =⎝3 ⎠ ′102 e33 3 7π53 –7 /351 e π= ;Instructor’s Resource Manual Section 11.7 713

−2 233. '()2 t tr t =− e i+ 2ej+2 2k−2t2tr''( t) = 4e i+4ejr′ (0) − 2i+ 2j+2 2kT(0)= =r′ (0) 4+ 4+8r′ (0) ⋅r′′(0)a T (0) ==r′(0)( )1 1 2= – i + j+k2 2 2–2i + 2j + 2 2 k ⋅ (4i+4 j)= 04r′ (0) × r′′(0) − 8 2i + 8 2j −16k16 2a N (0) = == = 4 2r′(0) 44r′′ (0) – a (0) (0) 4 4(0)T T i + j 1 1N = = = i + jaN(0) 4 2 2 2r′ (0) × r′′(0) 16 2 2κ (0) = = =3r′(0) 64 4B(0) = T(0) × N(0)⎛ 1 1 2 ⎞ ⎛ 1 1 ⎞ 1 1 1= ⎜– i + j+ k × ⎜ + ⎟2 2 2 ⎟i j = – i + j – k⎝⎠ ⎝ 2 2 ⎠ 2 2 234.2r () t = ln t,3,t t1r ′() t = ,3,2tt1r ′′() t = – ,0,22t1 1r′(2) ,3,4 ,3,42 21 6 8T (2) = = = = , ,r′ (2) 1 + 9+16 101 101 101 1014 2r′ (2) ⋅r′′(2) 2 ⎛ 1 1 ⎞ 2 ⎛63⎞63a T (2) == ,3,4 – ,0,2r′⎜ ⋅ ⎟=⎜ ⎟ =(2) 101 ⎝ 2 4 ⎠ 101 ⎝ 8 ⎠ 4 101r′ (2) × r′′(2) 2 3 2 ⎛ 649 ⎞ 649a N (2) = = 6, −2,= r′(2) 101 4 101 ⎜=4 ⎟⎝ ⎠ 2 101′′(2) – a (2) (2)(2) = r T T 2 101⎛1 63 1 6 8 ⎞N= ⎜ – , 0, 2 – , , ⎟aN(2) 649 ⎝ 4 4 101 101 101 101 ⎠κ2 101 41 189 76= – , – ,649 101 202 1013r′ (2) × r′′ (2) ⎛ 649 ⎞⎛ 2 ⎞ 2 649(2) =3=⎜ ⎜ ⎟⎜ ⎟′ (2) 4 ⎟=r ⎝ ⎠ ⎝ 101 ⎠ 101 101=82 189 152– , – ,65,549 65,549 65,5491 6 8 82 189 152B(2) = T(2)× N(2) = , , × – , – ,101 101 101 65,549 65,549 65,549=24 8 3,– ,649 649 649714 Section 11.7 Instructor’s Resource Manual

35.y′ = 1 , y′′= –1x 2x12xxκ = =3/2 2 3/21 ( 1)( 1+x +2x)xSince 0 < x < ∞ , κ =2 3/2( x + 1).2 3/2 2 2 1/2 2( x + 1) –3 x ( x + 1) –2x+ 1κ ′ = =2 3 2 5/2( x + 1) ( x + 1)κ ′ = 0 when1x = . Since κ ′ > 0 on2⎛ 1 ⎞⎜0,⎟⎝ 2 ⎠ and κ ′ < 0 on ⎛ 1 ⎞⎜ , ∞⎟, so κ is maximum when⎝ 2 ⎠1 1 ln2⎛ 1 ln2⎞x = , y = ln = – . The point of maximum curvature is , – .2 2 2⎜ ⎟⎝ 2 2 ⎠36 y′ = cos x, y′′= – sin xsin xκ =2 3/2(1+cos x)κ ′ =sin x2 3/2 2 1/2cos x(1 + cos x) + 3 sin x cos xsin x(1 + cos x)sin x2 3(1+cos x)22sinx cot x(2+cos x)=2 5/2(1+cos x)π πk′ = 0 when x =− , . κ ′ is not defined when x = – π , 0. Since κ ′⎛ π⎞ ⎛ π⎞> 0 on ⎜– π− , ⎟∪⎜0,⎟2 2⎝ 2⎠ ⎝ 2⎠ and κ ′ < 0 on⎛ π ⎞ ⎛π⎞ππ⎜− , 0 ⎟∪⎜ , π⎟,so κ has local maxima when x = – , y = –1 and x = , y = 1.⎝ 2 ⎠ ⎝2⎠22⎛ π⎞ ⎛π⎞κ ⎜– ⎟= κ ⎜ ⎟=1⎝ 2⎠ ⎝2⎠⎛π⎞The points of maximum curvature are ( 0, 1 ) and ⎜ , 1 ⎟.⎝2⎠37. y′ = sinh x, y′′= cosh xcosh x2κ = = sech x2 3/2(1+sinh x)2κ ′ = –2sech x tanh xκ ′ = 0 when x = 0. Since κ ′ > 0 on (– ∞ , 0) and κ ′ < 0 on (0, ∞ ), so κ is maximum when x = 0, y = 1. The pointof maximum curvature is ( 0, 1 ).Instructor’s Resource Manual Section 11.7 715

38. y′ = cosh x, y′′= sinh xsinh xκ =2 3/2(1+cosh x)κ ′ =sinh x2 3/2 2 1/2cosh x(1 + cosh x) – 3 sinh x cosh xsinh x(1 + cosh x)sinh x2 3(1 + cosh x)κ ′ is not defined when x = 0 and κ ′ = 0 when cosh x = 2 or ln ( 2 1 ).(– ∞ , – ln ( 2 + 1)) ∪ ( 0, ln ( 2 + 1)) and κ ′ < 0 on ( ( ) ) ( ( ) )x = – ln ( 2 + 1 ), y = –1 and x = ln ( 2 + 1 ), y = 1.1κ (–ln( 2+ 1)) = κ ( ln( 2+ 1)) =3 3The points of maximum curvature are (– ln ( 2 + 1 ), –1)and ( ln ( 2 + 1 ), 1 ).22sinhx coth x(2–cosh x)=2 5/2(1 + cosh x)x = ± + Since κ ′ > 0 on– ln 2 + 1 , 0 ∪ ln 2 + 1 , ∞ , κ has local maxima whenxx39. y′ = e , y′′= exeκ =2x3/2(1 + e )x 2x 3/2 3x 2x1/2e (1 + e ) – 3 e (1 + e )κ ′ =2x3(1 + e )x 2xe (1 – 2 e )=2x5/2(1 + e )40.κ ′ = 0 when1x =− ln 2 . Since κ ′ > 0 on2⎛ ⎞⎛ 1 ⎞⎜−∞, − ln2⎟⎝ 2 ⎠ and κ ′ < 0 on 1⎜−ln 2, ∞⎟⎝ 2 ⎠ , soκ is maximum when x = – 1 2 ln 2, 1y = . The2⎛ 1 1 ⎞point of maximum curvature is ⎜– ln 2, ⎟.⎝ 2 2 ⎠2y′ =− tan x,y′′=− sec x2sec xκ = = cos x2 3/2(1+tan x)π πSince – < x < , κ = cos x.2 2κ ′ = – sin xκ ′ = 0 when x = 0. Since κ ′⎛ π ⎞> 0 on ⎜– , 0⎟⎝ 2 ⎠and κ ′⎛ π ⎞< 0 on ⎜0, ⎟,κ is maximum when⎝ 2 ⎠x = 0, y = 0. The point of maximum curvature is(0, 0).41. r′ () t = 3i+6tjr′′ () t = 6jds= r ′() t = 3 1+4tdt2d s 12taT= =2dt21+4t22 2 2 144t36aN= r ′′() t – aT= 36– =1 42 1 42+ t + t6a N =21+4t1 1218At t 1 = , a T = and a N = .3 131342. r′ () t = 2ti+jr′′ () t = 2ids2= r ′() t = 4t+ 1dt2d s 4taT= =2 2dt4t+ 122 () 2 24 16t4aN= r ′′ t − aT= − =4 2 1 4 2t + t + 12a N =24t+ 142At t 1 = 1, a T = and a N = .552716 Section 11.7 Instructor’s Resource Manual

43. r′ () t = 2i+2tjr′′ () t = 2jds= r ′() t = 2 1+tdt2d s 2taT= =2dt21+t22 2 2 4t4aN= r ′′() t − aT= 4− =12 12+ t + t2a N =21+tAt t 1 = –1, a T = – 2 and a N = 2.44. r′ () t = – asinti+acostjr′′ () t = – acos ti– asintjds=dtr ′()t = a2d saT= = 02dt2aN2 2 2= r ′′() t – aT= aaN= aπAt t 1 = , a T6= 0 and aN= a.45. r′ ( t) = asinh ti+acoshtjr′′ ( t) = acosh ti+asinhtj2ds2 2= r ′( t) = a sinh t+cosh tdt2d s 2acoshtsinhtaT= =2dt 2 2sinh t+cosh t2 2 2aN= r ′′()t aT2 2 22 (cosh 2 sinh 2) – 4a cosh tsinht= a t+tsinh 2 cosh 2t+t2 2 2 2aNa (cosh t – sinh t )=2 2sinh t+cosh t2 2a(cosh t – sinh t)=2 2sinh t+cosh t5 4At t 1 = ln 3, cosh t = , sinh t = , and3 32 2 41sinh t+ cosh t = , so33aa N = .4140aa T = and3 4146. x′ () t = 3y′ () t = –6x′′ () t = 0y′′ () t = 0ds 2 2= x′ () t + y′() t = 3 5dt2d saT= = 02dt2 2 2 2aN= [ x′′ ( t) + y′′( t) ]– aT= 0; a N = 0At t 1 = 0, aT= 0 and aN= 0.47. r′ () t = i+ 3j + 2tkr′′ () t = 2k( i+ 3j+ 2 tk) ⋅(2 k) 4taT() t = =2 21+ 9+ 4t10+4t4aT(1) =14( i+ 3j+ 2 tk) × (2 k) 6i−2jaN() t = =2 210 + 4t10 + 4t36 + 4 10 5= = 2 = 2210 + 4t10 + 4t5 + 2t5aN(1) = 2 748.2 3r () t = t, t , t2r ′() t = 1,2,3 t tr ′′() t = 0,2,6t2 2aT() t =21, 2 t, 3t ⋅ 0, 2, 6t34t+18t=2 4 2 41+ 4t + 9t 1+ 4t + 9taT(2) =15216121, 2 t, 3t × 0, 2, 6t 26 t , −6 t, 2aN() t = =2 4 2 41+ 4t + 9t 1+ 4t + 9t=4 2 4 236t + 36t + 4 9t + 9t+ 1= 22 4 2 41+ 4t + 9t 1+ 4t + 9t181a N (2) = 2 161Instructor’s Resource Manual Section 11.7 717

49.– t tr () t = e ,2, t e ;−tr ′′() t = e ,0, e ;−2t2tr ′() t = e + 4+et −tr′ () t × r ′′() t = 2 e ,2, −2et– t tr ′() t = – e ,2, e== 22t−2t4e+ 4+4e2t−2te + 1+eaT() t =2t−2te − e2t−2te + 4 + eaT(0) = 0aN () t = 22t−2te + 1+e2t −2te + 4 + eaN(0) = 23=6250. r′ ( t) = 2( t−2) i− 2tj+kr′′ () t = 2i−2j51.–2t2tr′ () t ⋅ r ′′() t = – e + e[2( t−2) i− 2 tj+ k] ⋅(2i−2 j)aT() t =2 24( t− 2) + 4t+ 14( t− 2) + 4t 8t−8= =2 28t − 16t+ 17 8t − 16t+178aT(2) =17[2( t−2) i− 2 tj+ k] × (2i−2 j)aN() t =28t+− 16t+172i+ 2j+8k6 2= =2 28t − 16t+ 17 8t − 16t+176 2aN(2) =172 2r′ () t = (1– t ) i–(1 + t ) j+kr′′ () t = –2 ti–2tj2 2r′ () t ⋅ r′′() t = –2(1– t t ) + 2(1 t + t )3= 4t2 2 2 2 4r ′() t = (1 − t ) + (1 + t ) + 1= 2t+ 3r′ () t × r′′() t = 2ti−2tj−4tk2 2 2= 4t + 4t + 16t= 2 6 t34t3T( ) = ; T(3) = 36455a t a2t+ 32 6 t2aN( t) = ; aN(3) = 642t+ 35552.53.2 1r′ () t = i+ t j– k , t > 02t2r''( t) = 2tj+k3t3 2 2 8r′ () t ⋅ r ′′() t = 2 t – = ( t –1)5 5t tr ′() t =4 1 1 4 81+ t + = t + t + 14 2t tr′ () t × r′′() t =4 2i− j+2tkt 3t=4 16 2 2 4 8+ + 4t = 1+ 4t + t6 2 3t t t2 8( t −1)85t2( t −1)aT() t = =1 4 8 3 8 4t + t + 1 t t + t + 12taT(1) = 02 4 81+ 4t+ t 8 43t2 t + 4t+ 1aN() t = =1 4 88 4t + t + 1 t t + t + 12taN(1) = 26= 2 23v(t) = cos t, 2cos 2t , a(t) = −sin t, − 4sin 2ta(t) = 0 if and only if –sin t = 0 and –4 sin 2t = 0,which occurs if and only if t = 0, π , 2π , so itoccurs only at the origin.a(t) points to the origin if and only if a(t) = –kr(t)for some k and r(t) is not 0. This occurs if andπ 3πonly if t = , , so it occurs only at (1, 0) and2 2(–1, 0).718 Section 11.7 Instructor’s Resource Manual

54. v(t) = − sin t+ tcost+ sin t,cost+ tsin t−cost= t cos ti + t sin tja(t) = − tsin t+ cos t, tcost+sin ta.b.ds2 2 1/2( t) t(cos t sin t)tdt = v = + =(since t ≥ 0 )aT2d s ⎛ d ⎞= = () t 12 ⎜ ⎟ =dt ⎝dt⎠2 2 2N = – Ta a a2 2 2 2 2 2= [ t (sin t+ cos t) + (cos t+ sin t)]–1=tTherefore, a = t.N55. s′′ ( t) = aT= 0 ⇒ speed = s′( t)= c (a constant)2⎛ds⎞dsκ ⎜ ⎟ = aN= 0⇒ κ = 0 or 0⎝ dt ⎠dt = ⇒ κ = 056. r(t) = a cos ωt i + b sin ωt j;v(t) = –aω sin ωt i + bω cos ωt j2 2 2() t = – aω cos ωt, – bω sin ωt =−ω() ta rvT =1/2( v⋅v);dT ( v⋅v) a–( v⋅a)v=dt3/2( v⋅v)– abω= ( bcosωti+asin ωtj)2 2 2 3/2( a sin ωt+b cos ωt)2 2 2 2 1/2dTabω( b cos ωt+a sin ωt)=dt 2 2 2 3/2( a sin ωt+b cos ωt)abω=2 2 2 2a sin ωt+b cos ωtThendTdt−1=dT2 2 2 2 1/2( a sin ωt+b cos ωt)dt× ( bcosωti+asin ωtj)Note that this was done assuming ab > 0; ifab < 0, drop the negative sign in the numerator.57. v(5) is tangent to the helix at the point where theparticle is 12 meters above the ground. Its path isdescribed by v(5) = cos 5i – sin 5j + 7k.ds58. a N = 0 wherever κ = 0 or 0.dt = κ , thecurvature, is 0 at the inflection points, whichπ dsoccur at multiples of . However, 02dt ≠ onπthis curve. Therefore, a N = 0 at multiples of .259. It is given that at (–12, 16), s′ () t = 10ft/s ands′′ () t = 5ft/s 2 1. From Example 2, κ = .20⎛ 1 ⎞ 2Therefore, a T = 5 and a N = ⎜ ⎟ (10) = 5, so⎝20⎠a = 5T + 5N.Let r(t) = 20cos t,20sint describe the circle.34r(t) = − 12, 16 ⇒ cos t = – and sin t = .55v(t) = − 20sin t,20cost, so v () t = 20.() tThen () t = vTsin t, cos t .v()t= −4 3Thus, at (–12, 16), T = – ,–and5 53 4N = , – since N is a unit vector5 5perpendicular to T and pointing to the concaveside of the curve.Therefore,4 3 3 4a = 5 – , – + 5 , – = – i– 7 j .5 5 5 560. s′ () t = 4and s′′ () t = 0.y′′2κ = =2 3/2 2 3/2[1 + ( y′) ] (1 + 4 x )Therefore,2 2 32a = (0) T+ (4) N =N .2 3/2 2 3/2(1+ 4 x ) (1+4 x )2mv61. Let μ mg = R . Then vR= μgR.At theRvalues given,v R = (0.4)(32)(400) = 5120 ≈ 71.55 ft/s(about 49.79 mi/h).62. a.R F sinθF cosθ= (from the given2v gRequations, equating m in each.)Therefore, vR= Rgtan θ .b. For the values given,v = (400)(32)(tan10 ° ) ≈ 47.51 ft/s.R63. tanφ = y′2 3/2 2 3/2 3(1 + ′ ) = (1 + tan ) = secy′′ y′′3κ = = = y′′cos23/2 3⎡1+ y′⎤ sec φ⎣ ⎦y φ φφInstructor’s Resource Manual Section 11.7 719

dTdφ−sin φ, cosφ( ds )ds64. N = =dT2 2 1/2 dφds (sin φ + cos φ)dsdφds= – sin φ, cosφdφdsdφIf > 0, N = – sin φ, cosφand ifdsdφ< 0, N = sin φ, −cos φ , so N points to thedsconcave side of the curve in either case.65. B = T×N. Left-multiply by T and useTheorem 11.4CT× B = T× ( T×N)= ( TN i ) T−( TT i ) N= 0T−1N=−NThus, N =− T× B = B×T.To derive a result for T in terms of N and B,begin with N = B×T and left multiply by B:( )( BT i ) B ( i )B× N = B× B×T= − BB T= 0B−1T=−TThus, T=− B× N = N×B.66. Sincelim y = lim y = 0 = y (0), y is–x→0 x→0continuous.⎧⎪ 0 if x ≤ 0y′ ( x)= ⎨ 2⎪⎩ 3xif x > 0is continuous sincelim y′ = lim y′ = 0 = y′(0).–+0 0x→x→⎧0 if x ≤ 0y′′ ( x)= ⎨⎩6xif x > 0is continuous sincelim y′′ = lim y′′ = 0 = y′′(0). Thus,–+0 0x→x→y′′κ =2 3/2(1 + y′)+is continuous also. If x ≠ 0 theny’ and κ are continuous as elementary functions.67. Let2 3 4 55 = 0 + 1 + 2 + 3 + 4 + 5P ( x) a a x a x a x a x a x .P (0) = 0 ⇒ a = 05 02 3 45 1 2 3 4 5P′ ( x) = a + 2a x+ 3a x + 4a x + 5 a x , soP′ (0) = 0 ⇒ a = 0.5 12 32 3 4 5P′′ ( x) = 2a + 6a x+ 12a x + 20 a x , soP′′ (0) = 0 ⇒ a = 0.5 23 4 5P5 x = a3x + a4x + a5x2 3 45( ) 3 3 4 4 5 5 ,2 35′′ ( ) = 6 3 + 12 4 + 20 5 .Thus,68. Let( ) ,P′ x = a x + a x + a x andP x a x a x a xP (1) = 1, P ′ (1) = 0, and P ″ 5 (1) = 0 ⇒5 53 + 4 + 5 = 1a a a3a + 4a + 5a= 03 4 56a3 + 12a4 + 20a5= 0The simultaneous solution to these equations isa = 10, a = –15, a = 6, so3 4 53 4 5P5( x) = 10 x –15x + 6 x .2 3 4 55 = 0 + 1 + 2 + 3 + 4 + 5P ( x) a a x a x a x a x a x .Then P5( x ) must satisfyP ( ) = ; ( )5 0 0 P 5 1 = 1;P′ 5 (0) = 0 ; P′ 5 (1) = 1 ;P′′ 5 (0) = 0 ; P′′ 5 (1) = 0As in Problem 67, the three conditions at 0 implya0 = a1 = a2 = 0 . The three conditions at 0 leadto the system of equationsa + a + a = 13 4 53a + 4a + 5a= 13 4 56a3 + 12a4 + 20a5= 0The solution to this system isa = 6, a =− 8, a = 3 . Thus, the required3 4 5polynomial is3 4 5P5( x) = 6x − 8x + 3x.720 Section 11.7 Instructor’s Resource Manual

69. Let the polar coordinate equation of the curve ber = f(θ). Then the curve is parameterized by x = r cos θ and y = r sin θ.x′ = – rsinθ+ r′cosθy′ = rcosθ+ r′sinθx′′ = – rcos θ – 2r′ sinθ + r′′cosθy′′ = – rsinθ + 2r′ cosθ + r′′sinθBy Theorem A, the curvature isx′′′ y – yx ′′′κ =2 23/2x′ + y′(– rsinθ + r′ cos θ)(– rsinθ + 2r′ cosθ + r′′ sin θ) – ( rcosθ + r′ sin θ)(– rcos θ – 2r′ sinθ + r′′cos θ)=2 2 3/2[(– rsinθ + r′ cos θ) + ( rcosθ + r′sin θ) ]2 2r + 2 r′ – rr′′=.2 2 3/2( r + r′)70. r′ = –4sin θ , r′′= –4cosθ2 2 216cos θ + 32sin θ + 16cos θ 32 1κ = = =2 2 3/2(16cos θ + 16sin θ)64 271. r′ = – sin θ , r′′= – cosθAt θ = 0, r = 2, r′ = 0, and r′′ = –1.4+ 0+2 6 3κ = = =3/2(4 + 0) 8 472. r′ = 1, r′′= 0At θ = 1, r = 1, r′ = 1, and r′′ = 0.1+2–0 3κ = =3/2(1 + 1) 2 273. r′ = –4sin θ , r′′= –4cosθAt θ π= , 4, –4,2 r = r ′ = and r′′ = 0.16 + 32 – 0 48 3κ = = =3/2(16 + 16) 128 2 8 274.3 3r′ 3 e θ θ= , r′′= 9e3 3At θ = 1, r = e , r′= 3 e , and r′′ = 9 e .6 6 6e + 18 e – 9e610e1κ = = =( e + 9 e ) 10 10e 10e6 6 3/2 9 975. r′ = 4cos θ , r′′=− 4sinθAt θ π= , 8, 0, 4.2 r = r ′ = r ′′ = −64 + 0 + 32 96 3κ = = =32(64 + 0) 512 1676.6 θ 6θr′ = 6 e , r′′= 36e12θ 12θ 12θe + 72 e – 36e37eκ = =( e + 36 e ) 37 37e312θ12θ 12θ 3/ 2 18θ1 ⎛ 1 ⎞ 1 ⎛1⎞= ⎜376e θ ⎟=⎜ ⎟⎝ ⎠ 37 ⎝r⎠77. r = cos 2θ;sin 2 θ cos 2 θ +r′ = – , r′′= –1cos 2 θ3/2(cos 2 θ )2 22sin 2θcos 2θ+1 3cos 2θ cos 2θ cos 2θ3/2 3/2cos 2θ+ +κ = =21( cos 2θ+sin 2θ) ( cos 2 )cos 2θθ( θ )= 3 cos2 = 3r278. r(t) = f(t)i + g(t)j, where x = f(t) and y = g(t); v() t = r ′()tv()t x′ y′T()t = = i+jv()t 2 2 2 2x′ + y′ x′ + y′T i j i j2 2 3/2 2 2 3/2 2 2 3/22 2 2 2i2 2 3/2 2 2 3/2x′′ ( x′ + y′ )– x′ ( xx ′ ′′ + yy ′ ′′) y′′ ( x′ + y′ )– y′ ( xx ′ ′′ + yy ′ ′′)= +j( x′ + y′ ) ( x′ + y′)( x′′ y′ – xy ′ ′′) y′ ( y′′ x′ – yx ′ ′′ ) x′ ( xy ′ ′′ – yx ′ ′′ )'( t) = + = ( − y′ + x′)( x′ + y′ ) ( x′ + y′ ) ( x′ + y′)x′ y′′ − yx ′ ′′2 2 xy ′ ′′ − yx ′ ′′T′ () t xy ′′′ − yx ′′′T ′()t = x′ + y′=⇒ κ = =2 2 3/2 2 2( x′ + y′ )x′ + y′r′() t ( x′ + y′)2 2 3/2Instructor’s Resource Manual Section 11.7 721

79.Maximum curvature ≈ 0.7606,minimum curvature ≈ 0.124880. B⋅ B = 1ddB( B⋅ B) = 2B⋅ = 0dsdsThus, d B is perpendicular to B.ds81. d B d ( )d T d N= T× N = × N+ T ×ds ds ds dsdTSince = ds dTN , 0,dTds× N = dBdNso = T × .ds dsdsdB ⎛ dN⎞T⋅ = T⋅ T× =dNT× T ⋅ = soThus ⎜ ⎟ ( ) 0,ds ⎝ ds ⎠ dsdB is perpendicular to T.ds82. N is perpendicular to T, and B = T × N isperpendicular to both T and N. Thus, since d Bdsis perpendicular to both T and B, it is parallel toN, and hence there is some number τ(s) such thatdB= – τ () s N .ds83. Let ax + by + cz + d = 0 be the equation of theplane containing the curve. Since T and N lie inai+ bj+ckthe plane B =±. Thus, B is a2 2 2a + b + cconstant vector and d Bds = 0 , so τ(s) = 0, since N′ = + +84. r () t a0i b0j c0kr 0′′()t =Thus, r′ () t × r′′() t = 0 and sincer′ () t × r′′() tκ = , κ = 0.3r′() tTo show that τ = 0 , note that the curve isconfined to a plane. This means that the curve istwo-dimensional and thus τ = 0 .85. r(t) = 6 cos π ti + 6 sin π tj, + 2tk, t > 02 2 2Let (6cos π t) + (6sin π t) + (2 t) = 100.2 2 2Then 36(cos π+ t sin π t) + 4t= 100;86.24 64;t = t = 4.r(4) = 6i + 8k, so the fly will hit the sphere at thepoint (6, 0, 8).r′ () t = –6πsinπ ti+ 6πcosπ tj+2 k , so the fly willhave traveled4 2 2 2( −6πsin π ) + (6πcos π ) + (2)04 2∫ 36 4 dt0∫= π +t t dt22= 36π + 4(4 – 0) = 8 9π + 1 ≈ 75.8214⎛34⎞r ( t) = 10cos t, 10sin t,⎜ ⎟t⎝2π⎠Using the result of Example 1 with a = 10 and34c = , the length of one complete turn is2 π2 ⎛34⎞2 π (10) + ⎜ ⎟⎝2π⎠2–8 2 2angstroms= 10 400π + 34 cm. Therefore, the totallength of the helix is8 –8 2 2(2.9)(10 )(10 ) 400π + 34 ≈ 207.1794 cm.will not necessarily be 0 everywhere.722 Section 11.7 Instructor’s Resource Manual

11.8 Concepts Review1. traces; cross sections2. cylinders; z-axis3. ellipsoid4. elliptic paraboloidProblem Set 11.85.2 2( x–4) + ( y+ 2) = 7Circular cylinder1.2 2x y+ = 136 4Elliptic cylinder6. Two planes2. Circular cylinder7.2 2 2x y z+ + = 1441 196 36Ellipsoid3. Plane8.2 2 2x y z– + = 11 9 1Hyperboloid of one sheet4. Parabolic cylinderInstructor’s Resource Manual Section 11.8 723

9.2 2x yz = +8 2Elliptic paraboloid14.2 2 2x y z– – + = 14 4 1Hyperboloid of two sheets10. Circular cone15.2 2x zy = +4 9Elliptic Paraboloid11. Cylinder16.2 2 2x y z+ + = 125 9 25Ellipsoid12. Plane17. Plane13. Hyperbolic paraboloid724 Section 11.8 Instructor’s Resource Manual

18. Cylinder25. All central hyperboloids of two sheets aresymmetric with respect to (a) the origin, (b)the z-axis, and (c) the yz-plane.26. a. 1, 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15, 16, 18b. 1, 2, 6, 7, 8, 9, 10, 14, 16, 19, 2019. Hemisphere20. One sheet of a hyperboloid of two sheets.21. a. Replacing x by − x results in an equivalentequation.b. Replacing x by − x and y by − y results inan equivalent equation.c. Replacing x by − x , y by − y , and z with− z , results in an equivalent equation.22. a. Replacing y by − y results in an equivalentequation.b. Replacing x with − x and z with − zresults in an equivalent equation.c. Replacing y by − y and z with − z resultsin an equivalent equation.23. All central ellipsoids are symmetric withrespect to (a) the origin, (b) the x-axis, andthe (c) xy-plane.24. All central hyperboloids of one sheet aresymmetric with respect to (a) the origin, (b)the y-axis, and (c) the xy-plane.27. At y = k, the revolution generates a circle ofy kradius x = = . Thus, the cross section in2 22 2 kthe plane y = k is the circle x + z = or22 22x + 2 z = k.The equation of the surface is2 2y = 2x + 2 z .28. At z = k, the revolution generates a circle ofz kradius y = = . Thus, the cross section in the2 2planez = k is the circle2 2 24x 4 y k .22 2 kx + y = or4+ = The equation of the surface is2 2 2z = 4x + 4 y .29. At y = k, the revolution generates a circle of3 2 3 2radius x = 3– y = 3– k . Thus, the4 4cross section in the plane y = k is the circle2 2 3 22 2 2x + z = 3– k or 12 – 4 x – 4z = 3 k . The42 2 2equation of the surface is 4x + 3y + 4z= 12.30. At x = k, the revolution generates a circle of4 2 4 2radius y = x – 4 = k –4. Thus, the3 3cross section in the plane x = k is the circle2 2 4 22 2 2y + z = k – 4 or 12 + 3y + 3z = 4 k . The32 2 2equation of the surface is 4x = 12+ 3y + 3 z .2 2x y31. When z = 4 the equation is 4 = + or4 92 2x y 2 21 = + , so a 36, b 16,16 362 2 2c a – b 20,= = and= = hence c =± 2 5. The majoraxis of the ellipse is on the y-axis so the foci areat ( 0, ± 2 5, 4 ).Instructor’s Resource Manual Section 11.8 725

216 yz = + or4 92 99y = 9( z– 4) = 4 ⋅ ( z– 4), hence p = . The4432. When x = 4, the equation isvertex is at (4, 0, 4) so the focus is⎛ 9 25⎜4, 0, 4 + ⎞ ⎟=⎛ ⎜4, 0,⎞⎟⎝ 4⎠ ⎝ 4 ⎠ .2 2 2x y h33. When z = h, the equation is + + = 1 or2 2 2a b c2 2 2 2x y c – h+ = which is equivalent to2 2 2a b c2 22 2x yx y+ = 1, which is + = 12 2 2 2 2 2a ( c – h ) b ( c – h )2 2A B2 2cca 2 2 b 2 2with A = c – h and B = c – h . Thus,ccthe area is2 2⎛a 2 2 ⎞⎛b 2 2 ⎞ πab( c – h )π ⎜ c – h ⎟⎜ c – h ⎟=⎝c ⎠⎝c ⎠2c34. The equation of the elliptical cross section is2 2x y+ = 1, for each z in [0, h).2 2a ( h– z) b ( h– z)Therefore, ΔV ≈π( a h– z)( b h–z)Δ z=πab( h – z)Δ z , using the area formulamentioned in Problem 33.2hh⎡ z ⎤Therefore, V = ∫ π ab( h– z) dz = πab⎢hz– ⎥0⎢⎣2 ⎥⎦0⎡⎛ 2 2h ⎞ ⎤ πabh=π ab ⎢h – –0 ⎥ = , which is the⎢⎜2 ⎟⎣⎝⎠ ⎥⎦2height times one half the area of the base (z = 0),π ( a h)( b h) =π abh.2 2 235. Equating the expressions for y, 4–x = x + z2 2x zor 1 = + which is the equation of an2 4ellipse in the xz-plane with major diameter of2 4 = 4 and minor diameter 2 2.36.37.y = x intersects the cylinder when x = y = 2. Thus,the vertices of the triangle are (0, 0, 0), (2, 2, 1),and (2, 2, 4). The area of the triangle with sidesrepresented by 2, 2, 1 and 2, 2, 4 is1 12, 2, 1 × 2, 2, 4 = 6, − 6, 02 21= ( 6 2 ) = 3 2 .22 2 2(cos) t t + (sin) t t −t2 2 2 2 2 2= t (cos t+ sin t) − t = t − t = 0,hence every point on the spiral is on the cone.For r = 3t cos ti + t sin tj + tk, every point2 2 2satisfies x + 9 y –9z= 0 so the spiral lies onthe elliptical cone.38. It is clear that x = y at each point on the curve.Thus, the curve lies in the plane x = y. Since2z = t = y2 , the curve is the intersection of theplane x = y with the parabolic cylinder z = y 2 .Let the line y = x in the xy-plane be the u-axis,then the curve determined by r is in the uz-plane.The u-coordinate of a point on the curve, (y, y, z)is the signed distance of the point (y, y, 0) from2 2the origin, i.e., u = 2 y.Thus, u = 2y = 2z⎞or u = 4 ⎜ ⎟ z.This is a parabola in the⎝2⎠uz-plane with vertex at u = 0, z = 0 and focus at1 ⎛ 1 ⎞u = 0, z = . The focus is at ⎜0, 0, ⎟.2⎝ 2 ⎠2 ⎛1726 Section 11.8 Instructor’s Resource Manual

11.9 Concepts Review1. circular cylinder; sphere2. plane; cone3.2 2 2ρ = r + z⎛π⎞ ⎛π⎞4. a. x = 8sin⎜ ⎟cos⎜ ⎟=2 2⎝6⎠ ⎝4⎠⎛π⎞ ⎛π⎞y = 8sin⎜ ⎟sin⎜ ⎟=2 2⎝6⎠ ⎝4⎠⎛π⎞z = 8cos⎜⎟=4 3⎝6⎠4.2 2 2x + y + z = 4, z so2 2x + y + z 2 – 4z+ 4= 4 or2 2 2x + y + ( z– 2) = 4.Problem Set 11.91. Cylindrical to Spherical:2 2ρ = r + zzcosφ=2 2r + zθ = θSpherical to Cylindrical:r = ρ sinφz = ρ cosφθ = θ2. a. ( ρθφ)⎛ π π ⎞, , = ⎜ 2, , ⎟⎝ 2 4⎠b.5. a.⎛3π⎞ ⎛π⎞x = 4sin⎜ ⎟cos⎜ ⎟=⎝ 4 ⎠ ⎝3⎠2⎛3π⎞ ⎛π⎞y = 4sin⎜ ⎟sin⎜ ⎟=⎝ 4 ⎠ ⎝3⎠6⎛3π⎞z = 4cos ⎜ ⎟=–2 2⎝ 4 ⎠2 2 2ρ = x + y + z= 4+ 12+ 16 = 4 2y –2 3tan θ = = = – 3 and (x, y) is in thex 25 π4th quadrant so θ = .3z 4 2 πcosφ= = = so φ = .ρ 4 2 2 4⎛ 5ππ⎞Spherical: ⎜4 2, , ⎟⎝ 3 4⎠⎛ π ⎞ ⎛ 5π⎞b. Note: ⎜− 2, , 2⎟=⎜2, , 2⎟⎝ 4 ⎠ ⎝ 4 ⎠⎛ 5π π ⎞( ρθφ , , ) = ⎜ 2 2, , ⎟⎝ 4 4⎠⎛π⎞3. a. x = 6cos⎜⎟=3 3⎝6⎠⎛π⎞y = 6sin⎜⎟=3⎝6⎠z = –2b. ρ = 2+ 2+ 12 = 42tan θ = = –1 and (x, y) is in the 2nd– 23 πquadrant so θ = .42 3 3 πcosφ = = so .4 2 6⎛ 3ππ⎞Spherical: ⎜4, , ⎟⎝ 4 6⎠b.⎛4π⎞x = 4cos ⎜ ⎟=–2⎝ 3 ⎠⎛4π⎞y = 4sin ⎜ ⎟=–2 3⎝ 3 ⎠z = –86. a. r = 4+ 4 = 2 22πtanθ = = 1, x > 0, y > 0, so θ = . z = 324b. r = 48 + 16 = 84 1tan θ = – = – , x > 0, y < 0, so4 3 311πθ = . z = 66Instructor’s Resource Manual Section 11.9 727

7. r = 5Cylinder11. r = 3 cos θCircular cylinder8. ρ = 5Sphere12. r = 2 sin 2θ4-leaved cylinder9.πφ =6Cone13. ρ = 3 cos φ22 2 ⎛ 3⎞9x + y + ⎜z– ⎟ =⎝ 2 ⎠ 4Sphere10.πθ =6Plane14. ρ = sec φρ cos φ = 1z = 1Plane728 Section 11.9 Instructor’s Resource Manual

15.2 2r + z = 92 2 2x + y + z = 9Sphere23.2 2 2 2 2 2( + ) + = 4; + cos = 4;r z z ρ ρ φ2 4ρ =21 + cos φ24.2 2 2 2 2= 2 cos ; r + z = 2 z; r = 2 z– z ;ρ ρ φ2r = 2 z–z25. r cos θ + r sin θ = 4;4r =sinθ+ cosθ16.2 2 2r cos θ + z = 42 2x + z = 4Circular cylinder26. ρ sin φ cos θ + ρ sin φ sin θ + ρ cos φ = 1;1ρ =sin φ(sinθ + cos θ) + cosφ27.2 2 2 2( x + y + z )– z = 9;2 2 2 2(1 – cos ) = 9; sin = 9ρ φ ρ φρ sinφ = 32 2 2ρ – ρ cos φ = 9;28.2r = 2rsinθ;2 2x + y = 2y;2 2x + ( y− 1) = 117.18.19.20.21.22.2 2 2x y 9; r 9;+ = = r = 32 2 2 2 2r cos θ – r sin θ = 25; r cos 2θ= 25;2 2r = 25sec θ ; r = 5sec2 2r + 4z= 10θ2 2 2 2 2 2 2( + + ) + 3 = 10; + 3 cos = 10;x y z z ρ ρ φ2 10ρ =21 + 3cos φ2 2 2 2 2 2 2( + + )–3 = 0; –3 cos = 0;x y z z ρ ρ φ2 1cos32 2sin2 13φ = (pole is not lost); cos φ = (orφ = or32tan 2) φ =2 2 2 2 2 2ρ [sin φcos θ – sin φsin θ – cos φ ] = 1;2 2 2 2 2 2[sin cos – sin sin –1+ sin ] = 1;ρ φ θ φ θ φ2 2 2 2 2[sin cos –1+ sin (1– sin )] = 1;ρ φ θ φ θ2 2 2 2 2[sin cos –1+ sin cos ] = 1;ρ φ θ φ θ2 2 2ρ [2sin φcos θ –1] = 1;ρ2 1=2 22sin cos –1φθ29.2 2 2 2r cos 2 = z; r (cos – sin ) = z;θ θ θ2 2( rcos ) –( rsin ) z;θ θ =2 2x – y = z30. ρ sin φ = 1 (spherical); r = 1 (cylindrical);2 2x + y = 1 (Cartesian)31.2 2 2 2z 2x 2y 2( x y )= + = + (Cartesian);(cylindrical)2 2 232. 2x + 2 y – z = 2 (Cartesian);(cylindrical)2z = 2r2 22 r – z = 2Instructor’s Resource Manual Section 11.9 729

33. For St. Paul:ρ = 3960,θ = 360° – 93.1° = 266.9° ≈ 4.6583 radφ = 90° – 45° = 45° = 4π radπx = 3960sin cos 4.6583 ≈ –151.44πy = 3960sin sin 4.6583 ≈ –2796.04πz = 3960cos ≈ 2800.14For Oslo:ρ = 3960, θ = 10.5° ≈ 0.1833 rad,φ = 90° – 59.6° = 30.4° ≈ 0.5306 radx = 3960 sin 0.5306 cos 0.1833 ≈ 1970.4y = 3960 sin 0.5306 sin 0.1833 ≈ 365.3z = 3960 cos 0.5306 ≈ 3415.5As in Example 7,(–151.4)(1970.4) + (–2796.0)(365.3) + (2800.1)(3415.5)cosγ≈ ≈ 0.525723960so γ ≈ 1.0173 and the great-circle distance is d ≈ 3960(1.0173) ≈ 4029 mi34. For New York:ρ = 3960, θ = 360° – 74° = 286° ≈ 4.9916 radφ = 90° – 40.4° = 49.6° ≈ 0.8657 radx = 3960 sin 0.8657 cos 4.9916 ≈ 831.1y = 3960 sin 0.8657 sin 4.9916 ≈ –2898.9z = 3960 cos 0.8657 ≈ 2566.5For Greenwich:ρ = 3960, θ = 0, φ = 90° – 51.3° = 38.7° ≈ 0.6754 radx = 3960 sin 0.6754 cos 0 ≈ 2475.8y = 0z = 3960 cos 0.6754 ≈ 3090.6(831.1)(2475.8) + (–2898.9)(0) + (2566.5)(3090.6)cosγ= ≈ 0.637023960so γ ≈ 0.8802 and the great-circle distance is d ≈ 3960(0.8802) ≈ 3485 mi35. From Problem 33, the coordinates of St. Paul are P(–151.4, –2796.0, 2800.1).For Turin:πρ = 3960, θ = 7.4° ≈ 0.1292 rad, φ = rad4πx = 3960sin cos 0.1292 ≈ 2776.84πy = 3960sin sin 0.1292 ≈ 360.84πz = 3960cos ≈ 2800.14(–151.4)(2776.8) + (–2796.0)(360.8) + (2800.1)(2800.1)cosγ≈ ≈ 0.408823960so γ ≈ 1.1497 and the great-circle distance is d ≈ 3960(1.1497) ≈ 4553 mi730 Section 11.9 Instructor’s Resource Manual

π36. The circle inscribed on the earth at 45° parallel (φ = 45°) has radius 3960cos .4St. Paul and Turin is 93.1° + 7.4° = 100.5° ≈ 1.7541 rad⎛ π ⎞Thus, the distance along the 45° parallel is ⎜3960cos ⎟(1.7541) ≈ 4912 mi⎝ 4 ⎠The longitudinal angle between37. Let St. Paul be at P 1(–151.4, – 2796.0, 2800.1) and Turin be at P 2(2776.8, 360.8, 2800.1) and O be the center ofthe earth. Let β be the angle between the z-axis and the plane determined by O, P 1, and 2.⎯⎯→⎯⎯→to the plane. The angle between the z-axis and OP1× OP2is complementary to β. Henceβ⎛⎛⎯⎯→⎯⎯→ ⎞ ⎞⎜OP1× OP2⋅k⎟⎜⎜⎟ ⎟6π −1cos⎝ ⎠ π ⎛⎜⎟–1 7.709×10 ⎞= −≈ – cos ≈0.56892 ⎜ ⎯⎯→ ⎯⎯→ ⎟ 2 ⎜71.431 10 ⎟⎝ × ⎠OP1×OP2k⎜⎟⎝⎠.⎯⎯→ ⎯⎯→P OP1 OP2The distance between the North Pole and the St. Paul-Turin great-circle is 3960(0.5689) ≈ 2253 mi38. xi = ρisinφi cos θi, yi = ρisinφisin θi,zi = ρi cosφifor i = 1, 2.2 2 2d = ( ρ2sinφ2cosθ2 − ρ1sinφ1cos θ1) + ( ρ2sinφ2sinθ2 −ρ1sinφ1sin θ1)2+ ( ρ2cosφ2 −ρ1cos φ1)2 2 2 2 2 2 2 2 2 2 2 2= ρ2 sin φ2(cos θ2 + sin θ2) + ρ1 sin φ1(cos θ1+ sin θ1) + ρ2 cos φ2 + ρ1 cos φ1–2ρ1ρ2sinφ1sin φ2(cosθ1cosθ2 + sinθ1sin θ2) – 2ρ1ρ2cosφ1cosφ22 2 2 2 2 2 2 2= ρ2 sin φ2 + ρ1 sin φ1+ ρ2 cos φ2 + ρ1 cos φ1−2ρ1ρ2sinφ1sin φ2[cos( θ1−θ2)]−2ρρ 1 2cosφ1cosφ22 2 2 2 2 2= ρ2(sin φ2 + cos φ2) + ρ1 (sin φ1+ cos φ1) + 2 ρ1ρ2[– cos( θ1 – θ2)sinφ1sin φ2 – cosφ1cos φ2]2 2= ρ1 + ρ2 + 2 ρ1ρ2[– cos( θ1 – θ2)sinφ1sin φ2 – cosφ1cos φ2]2 2= ρ1 – 2ρρ 1 2 + ρ2 + 2 ρρ 1 2[1– cos( θ1 – θ2)sinφ1sin φ2 – cosφ1cos φ2]2= ( ρ1– ρ2) + 2 ρ1ρ2[1– cos( θ1 – θ2)sinφ1sin φ2 – cosφ1cos φ2]2 1/2Hence, d = {( ρ1 – ρ2) + 2 ρ1ρ2[1– cos( θ1 – θ2)sinφ1sin φ2 – cosφ1cos φ2]}× is normal⎯⎯→ ⎯⎯→39. Let P 1 be ( a1, θ1, φ 1)and P 2 be ( a2, θ2, φ 2).If γ is the angle between OP1and OP2then the great-circledistance between P 1 and P 2 is aγ. OP1 = OP2= a while the straight-line distance between P 1 and P 2 is (from2 2 2Problem 38) d = ( a– a) + 2 a [1– cos( θ – θ )sinφ sin φ – cosφ cos φ ]1 2 1 2 1 22= 2 a {1– [cos( θ1 – θ2)sinφ1sinφ2 + cosφ1cos φ2]}.Using the Law Of Cosines on the triangle OP1P2,2 2 2 2= + – 2 cos2d a a a γ = 2 a (1–cos γ ).Thus, γ is the central angle and cosγ = cos( θ1 – θ2)sinφ1sinφ2 + cosφ1cos φ2.Instructor’s Resource Manual Section 11.9 731

40. The longitude/latitude system (α, β) is related to a spherical coordinate system (ρ, θ, φ) by the following relations:ρ = 3960; the trigonometric function values of α and θ are identical butπ–π ≤ α ≤ π rather than 0 ≤ θ ≤ 2π , and β = – φ so sin β = cos φ and cos β = sin φ.2From Problem 39, the great-circle distance between (3960, θ1, φ 1)and (3960, θ2, φ 2)is 3960γ where 0 ≤γ ≤ πand cosγ = cos( θ1 – θ2)sinφ1sin φ2 + cosφ1cosφ2= cos( α1 – α2)cos β1cos β2 + sin β1sin β2.41. a. New York (–74°, 40.4°); Greenwich (0°, 51.3°)cosγ = cos(–74 ° – 0 ° )cos(40.4 ° )cos(51.3 ° ) + sin(40.4 ° )sin(51.3 ° ) ≈ 0.637Then γ ≈ 0.880 rad, so d ≈ 3960(0.8801) ≈ 3485 mi.b. St. Paul (–93.1°, 45°); Turin (7.4°, 45°)cos γ = cos(–93.1° – 7.4°) cos (45°) cos (45°) + sin (45°) sin (45°) ≈ 0.4089Then γ ≈ 1.495 rad, so d ≈ 3960(1.1495) ≈ 4552 mi.c. South Pole (7.4°, –90°); Turin (7.4°, 45°)Note that any value of α can be used for the poles.cos γ = cos 0° cos (–90°) cos (45°) + sin (–90°)sin (45°) = − 123π⎛3π⎞thus γ = 135°= rad, so d = 3960⎜⎟≈9331mi.4⎝ 4 ⎠d. New York (–74°, 40.4°); CapeTown (18.4°, –33.9°)cosγ = cos (–74° – 18.4°) cos (40.4°) cos (–33.9°) + sin (40.4°) sin (–33.9°) ≈ –0.3880Then γ ≈ 1.9693 rad, so d ≈ 3960(1.9693) ≈ 7798 mi.e. For these points α 1 = 100° and α 2 = –80° while β1 = β2 = 0, hencecos γ = cos 180° and γ = π rad, so d = 3960π ≈ 12,441 mi.42. ρ = 2a sin φ is independent of θ so the cross section in each half-plane, θ = k, is a circle tangent to the origin andwith radius 2a. Thus, the graph of ρ = 2a sin φ is the surface of revolution generated by revolving about the z-axisa circle of radius 2a and tangent to the z-axis at the origin.11.10 Chapter ReviewConcepts Test1. True: The coordinates are defined in terms ofdistances from the coordinate planes insuch a way that they are unique.2 2 22. False: The equation is ( x– 2) + y + z = –5,so the solution set is the empty set.3. True: See Section 11.2.4. False: See previous problem. It represents aplane if A and B are not both zero.5. False: The distance between (0, 0, 3) and(0, 0, –3) (a point from each plane) is 6,so the distance between the planes isless than or equal to 6 units.6. False: It is normal to the plane.7. True: Let t =1 .2a b c8. True: Direction cosines are , ,u u u9. True: (2 i– 3 j) ⋅ (6i+ 4 j ) = 0 if and only if thevectors are perpendicular.10. True: Since u and v are unit vectors,u⋅vcos θ = = u⋅v.u v11. False: The dot product for three vectors( ab ⋅ ) ⋅c is not defined.732 Section 11.10 Instructor’s Resource Manual

12. True: u⋅ v = u v cosθ ≤ u v sincecosθ ≤ 1.13. True: If u⋅ v = u v , then cosθ = 1 since14. False: If u = –i andu⋅ v = u v cos θ . Thus, u is a scalarmultiple of v. If u is a scalar multiple of2v, u⋅ v = u⋅ ku = k u= u ku = u v .1 3v = i+j , then2 21 3u+ v = – i+j and2 2u = v = u+ v =1.26. True: The vectors are both parallel andperpendicular, so one or both mustbe 0.27. True: (2i× 2 j) ⋅ ( j× i) = 4 ( k) ⋅(– k )= 4( k⋅ k ) = 428. False: Let u = v = i, w = j.Then (u × v) × w = 0 × w = 0; butu × (v × w) = i × k = –j.29. True: Since b 1 , b 2 , b 3 is normal to theplane.30. False: Each line can be represented byparametric equations, but lines with anyzero direction number cannot berepresented by symmetric equations.15. True:16. True:2 2( u+ v) ⋅( u− v) = u − v = 02 2u = v or u = v .so2u+ v = ( u+ v) ⋅ ( u+v )2 2= u + v + 2u⋅v31. True:r′ × r′′κ = = 0 ⇒ r′ × r′′3r′= r′ r ′′ sinθ= 0. Thus, either r ′ andr ′′ are parallel or either r ′ or r ′′ is 0,which implies that the path is a straightline.17. True: Theorem 11.5A18. True: Dt[ F( t) ⋅F( t)]= F() t ⋅ F′ () t + F() t ⋅F′()t= 2 F( t) ⋅F′( t)19. True:uu = u u = u20. False: The dot product of a scalar and a vectoris not defined.21. True: u× v = − v× u = − 1 v×u= v×u22. True: ( kv) × v = k( v× v) = k( 0)= 023. False: Obviously not true if u = v. (Moregenerally, it is only true when u and vare also perpendicular.)232. True: An ellipse bends the sharpest at pointson the major axis.33. False: κ depends only on the shape of thecurve.34. True: x′ = 3y′ = 2x′′ = 0y′′ = 0xy ′′′–yx ′′′Thus, κ = =2 23/2⎡x′ + y′⎤⎣ ⎦35. False: x′ = –2sinty′ = 2costx′′ = –2costy′′ = –2sintxy ′′′– yx ′′′ 4 1Thus, κ = = = .2 2 3/2[ x′ + y′] 8 20.24. False: It multiplies v by a; it multiplies thelength of v by a .25. True:u×v u v sinθ= = tanθ( u⋅v) u v cosθInstructor’s Resource Manual Section 11.10 733

36. True:r′() tT() t = ;r ′ () tr′ () t ⋅r′′() tT′ () t =− r′() t3r′() t1+ r′′(); tr ′ () tT() t ⋅T′()t1⎡r′ ( t) ⋅r′′( t)= r′ () t ⋅⎢−r′() tr′ () t ⎢ 3⎣ r′() t1 ⎤+ r′′() t ⎥r′ () t ⎥⎦r′ () t ⋅r′′ () t r′ () t ⋅r′′() t=− + = 02 2r′ () t r′() t37. False: Consider uniform circular motion:dv / dt = 0 but v = aω .38. True:κy′′= = 023/2[1 + y′]39. False: If y′′ = k then y′ = kx+ C andy′′kκ = =[1 + y′] [1 + ( kx+C) ]not constant.2 3/2 2 3/240. False: For example, if u = i and v = j, thenu⋅ v =0.2 241. False: For example, if r() t = cost i+sin t j ,2 2then r′ () t = –2sin t t i+2cos t t j , sor () t = 1but r ′() t = 2. t42. True: If v⋅ v = constant, differentiate bothsides to get v⋅ v′ + v⋅ v′ = 2v⋅ v ′ = 0, sov⋅ v ′ = 0.43. True If r() t = acosωti+ asinωtj+ctk , thenr′ () t =− aω sinωti+ aωcosωtj+ck soT() t = r′ ()/ t r′() t− aωsinωti + aωcosωtj+ck=2 2 2a ω + c2 2−aω cosωti−aω sinωtjT ′()t =which2 2aω+ cpoints directly to the z-axis. ThereforeN() t = T′ ()/ t T ′()t points directly to thez-axis.is44. False: Suppose v(t) = cos ti + sin tj, thenv () t = 1but a(t) = –sin ti + cos tjwhich is non-zero.45. True: T depends only upon the shape of thecurve, hence N and B also.46. True: If v is perpendicular to a, then T is alsoperpendicular to a, sod ⎛ds⎞ ⎜ ⎟ = aT= Ta ⋅ = 0. Thusdt ⎝ dt ⎠dsspeed = is a constant.dt47. False: If r(t) = a cos ti + a sin tj + ctk then v isperpendicular to a, but the path ofmotion is a circular helix, not a circle.48. False: The circular helix (see Problem 27) hasconstant curvature.49. True: The curves are identical, although themotion of an object moving along thecurves would be different.50. False: At any time 0< t < 1, r 1 () t ≠ r 2 () t51. True: The parameterization affects only therate at which the curve is traced out.52. True: If a curve lies in a plane, then T and Nwill lie in the plane, so T × N = B willbe a unit vector normal to the plane.53. False: For r(t) = cos ti + sin t j, r () t = 1, butr′ () t ≠ 0 .54. True: The plane passes through the origin soits intersection with the sphere is a greatcircle. The radius of the circle is 1, so iscurvature is 1 1.1 =55. False: The graph of ρ = 0 is the origin.56. False: It is a parabolic cylinder.57. False: The origin, ρ = 0, has infinitely manyspherical coordinates, since any value ofθ and φ can be used.734 Section 11.10 Instructor’s Resource Manual

Sample Test Problems1. The center of the sphere is the midpoint⎛–2 + 4 3 + 1 3 + 5 ⎞⎜ , , ⎟ = (1, 2, 4) of the diameter.⎝ 2 2 2 ⎠2 2 2The radius is r = (1 + 2) + (2 – 3) + (4 – 3)= 9+ 1+ 1= 11. The equation of the sphere is2 2 2x y z( –1) + ( – 2) + ( – 4) = 11b.(–5)(2) + (–3)(–1) 7cos θ = = –25 + 9 4 + 1 170≈ –0.53692.2 2 2x x y y z z( –6 + 9) + ( + 2 + 1) + ( –8 + 16)2 2 2= 9+ 1+16; ( x–3) + ( y+ 1) + ( z–4) = 26Center: (3, –1, 4); radius: 263. a. 3 2, −5 − 2 1,1 = 6, −15 − 2, 2 = 4, − 17b. 2, – 5 ⋅ 1, 1 = 2 + (–5) = –3c. 2, −5 ⋅ ( 1, 1 + − 6, 0 ) = 2, −5 ⋅ − 5, 1= –10 + (–5) = –15d. ( 42,–5+ 51,1 ) ⋅ 3–6,0= 13, –15 ⋅ –18, 0 = –234 + 0 = –234e. 36 + 0 −6, 0 ⋅ 1, − 1 = 6( − 6 + 0) =− 36c.(7)(5) + (0)(1) 5cosθ= = ≈0.980649 + 0 25 + 1 26f. –6, 0 ⋅ –6, 0 – 36 + 0= (36 + 0) – 6 = 305. a. a+ b+ c= 2i+ j+4k4. a.(3)( − 1) + (2)(4) 5cosθ= = ≈0.33639 + 4 1+16 221b. bic= (0)(3) + (1)( − 1) + ( − 2)(4) = −9c.ib× c= 0j1k− 2 = 2i−6j−3k3 −1 4( ) ( )a⋅ ( b× c) = − i+ j+ 2k ⋅ 2i−6j −3k=−2−6− 6=−14d. bc ⋅ is a scalar, and a crossed with a scalardoesn’t exist.e.a− b = − i+ 0j+4k2 2 2= 1 + 0 + 4 = 17f. From part (c), b× c= 2i−6j−3k so2 2 2b× c = 2 + 6 + 3 = 49 = 7.Instructor’s Resource Manual Section 11.10 735

6. a.7.b.ab ⋅ 0+ 5−3cosθ= =a b( 3 3)( 10)≈ 0.121716−1θ = cos 0.121716 ≈83.009ab ⋅ −1− 0+6cosθ= =a b( 5)( 11)≈ 0.67200−1θ = cos 0.67200 ≈ 47.6088. a. −5, − 5,5 = –5 1,1, − 1b. 2, − 1,1 × 0, 5, 1 = 6, − 2,10c. 2, −1,1⋅ −7,1, − 5 = –20d. 2, − 1,1 × −7,1, − 5 = 4,3, − 59. c 3,3, − 1 × −1, − 2, 4 = c 10, −11, − 3 for any cin R .10. Two vectors determined by the points are−1, 7, − 3 and 3, −1, − 3 . Then−1, 7, − 3 × 3, −1, − 3 = –4 6, 3, 5 and6, 3, 5 are normal to the plane.± 6, 3, 5 ± 6, 3,5= are the unit vectors36 + 9 + 25 70normal to the plane.11. a. y = 7, since y must be a constant.b. x = –5, since it is parallel to the yz-plane.a. a = 4+ 1+ 4 = 3;b.b = 25 + 1+ 9 = 35a 2 1 2= i− j+k , direction cosinesa 3 3 32 1,– ,3 3 and 2 .3b 5 1 3= i+ j−kb 35 35 35cosines5 1 3, , –35 35 35, directionc. z = –2, since it is parallel to the xy-plane.d. 3x – 4y + z = –45, since it can be expressedas 3x – 4y + z = D and D must satisfy3(–5) – 4(7) + (–2) = D, so D = –45.12. a. 4+ 1,1− 5,1+ 7 = 5, − 4,8 is along theline, hence normal to the plane, which hasequation x– 2, y+ 4, z+ 5 ⋅ 5, – 4, 8 = 0 .b. 5(x – 2) – 4(y + 4) + 8(z + 5) = 0 or5x – 4y + 8z = –14c.c.d.2 1 2i–j+k3 3 3ab ⋅ 10 −1−6cosθ= =a b 3 353 1= =3 35 35–1 1θ = cos ≈1.4010 ≈ 80.27°35736 Section 11.10 Instructor’s Resource Manual

13. If the planes are perpendicular, their normals willalso be perpendicular. Thus0= 1, 5, C ⋅ 4, –1,1 = 4–5 + C,so C = 1.14. Two vectors in the same plane are 3, −2, − 3 and3, 7, 0 . Their cross product, 3 7, − 3,9 , isnormal to the plane. An equation of the plane is7(x – 2) – 3(y – 3) + 9(z + 1) = 0 or7x – 3y + 9z = –4.15. A vector in the direction of the line is 8,1, − 8 .Parametric equations arex = –2 + 8t, y = 1 + t, z = 5 – 8t.16. In the yz-plane, x = 0. Solve –2y + 4z = 14 and 2y– 5z = –30, obtaining y = 25 and z = 16.In the xz-plane, y = 0. Solve x + 4z = 14 and–x – 5z = –30, obtaining x = –50 andz = 16. Therefore, the points are(0, 25, 16) and (–50, 0, 16).19. 5, −4, − 3 is a vector in the direction of the line,and 2, − 2,1 is a position vector to the line.Then a vector equation of the line is r(t)= 2, − 2,1 + t 5, −4, − 3 .20. t x y z–2 –2 2 − 8/3–1 –1 1/2 − 1/30 0 0 01 1 1/2 1/32 2 2 8/33 3 9/2 917. (0, 25, 16) and (–50, 0, 16) are on the line, so50, 25, 0 = 25 2, 1, 0 is in the direction ofthe line. Parametric equations are x = 0 + 2t,y = 25 + 1t, z = 16 + 0t or x = 2t,y = 25 + t, z = 16.18. 3, 5, 2 is normal to the plane, so is in thedirection of the line. Symmetric equations of thex– 4 y–5 z–8 line are = = .3 5 221.′ 2() t = 1, t, t , ′(2) = 1,2,4r r and8r (2) = 2, 2, . Symmetric equations for the3tangent line areplane is8x–2 y–2z –3.1 2 4⎛ 8 ⎞1( x– 2) + 2( y– 2) + 4 ⎜z– ⎟=0⎝ 3 ⎠3x + 6y + 12z = 50.= = NormalorInstructor’s Resource Manual Section 11.10 737

22. r(t) = t cos t, t sin t, 2t ;r ′( t) = – tsin t+ cos t, tcost+sin t, 2 ;r′′ ( t) = −tcost−2sin t, − tsin t+2cos t, 0⎛π,⎞ π ⎛π⎞πr′ ⎜ ⎟= – ,1,2 ; r′′⎜ ⎟ = –2, – ,0⎝ 2 ⎠ 2 ⎝2⎠22⎛π⎞π + 20r ′ ⎜ ⎟ = ;⎝2⎠2π⎛π⎞ r′ ( 2 ) −π,2,4T⎜⎟ = =⎝2 ⎠ r′ π 2( 2 ) π + 2023. r ′() t = e t cost+ sin t,–sint+cos,1 tr ′() t = 3et5 t t5Length is 3edt⎡ 3e⎤∫ =1 ⎣ ⎦ 15= 3( − ) ≈ 252.3509.24. –( F1+ F2) = –5 i–9j25. Let the wind vector bew = 100cos30 ° , 100sin 30°26.= 50 3, 50 .Let p = p1,p2be the plane’s air velocityvector.We want w + p = 450 j = 0, 450 .p50 3, 50 + 1, 2 = 0, 450⇒ 50 3 + p1 = 0, 50 + p2= 450p⇒ 1 = –50 3, 2 = 400ppTherefore, p = –50 3, 400 . The angle βformed with the vertical satisfiesp⋅j 400cos β = = ; β ≈ 12.22°. Thus,p j 167,500the heading is N12.22°W. The air speed isp = 167,500 ≈ 409.27 mi/h.2 – 2 –r() t = e t , e t ; r ′() t = 2 e t ,–eta.b.2 – 2 –lim e t , e t = lim e t , lim et = 1,1t→0 t→0 t→0r(0 + h) – r(0)lim = r ′(0) = 2, –1h→0heec.ln 2 2t−t∫ e , e dt0ln 2⎡ ⎛1⎞ 2t−t⎤= ⎢ ⎜ ⎟e, −e⎥⎣ ⎝2⎠ ⎦01 1 3 1= 2, – – , –1 = ,2 2 2 2d. Dt[ tr()] t = tr′() t + r () t2 t – t 2 t – t= t 2 e ,– e + e , e2 t – t= e (2t+ 1), e (1 − t)e. Dt[ r(3t+ 10)] = [ r ′(3t+10)](3)f.27. a.28.6t+20 –3 t–10= 3 2 e ,–e6t+ 20 −3t−10= 6 e , − 3e4 t –2tDt[() r t ⋅ r ′()] t = Dt[2 e – e ]4 t –2t= 8e+ 2e1–2r′ () t = ,–6 t ; r ′′() t = – t ,–6tb. r ′() t = cos, t –2sin2 t ;c.r ′′() t = –sin t, –4cos2t2 3r ′() t = sec t, –4 t ;2 2r ′′() t = 2sec ttan t, –12t–() t tv t = e ,– e ,2–() t ta t = e , e ,0 ;1a (ln 2) = 2, , 02κ(ln 2) =v(ln 2) × a(ln 2)=v(ln2)1v (ln 2) = 2, – , 22−1, 4, 233( 4 )3 38 21 8 7= 21 = 8 =333/2 35937 11979≈ 0.1934738 Section 11.10 Instructor’s Resource Manual

29.2t t t t tv() = 1,2,3 ; a () = 0,2,6v(1) = 1, 2, 3v(1) = 14a(1) = 0, 2, 6a Ta Nva ⋅ 0 + 4 + 18 22= = = ≈5.880;v 14 14v×a 6, − 6, 2= =v 142 19= ≈ 2.3301433. Circular paraboloid34. Plane30. Circular cylinder35. Plane31. Sphere36. Hyperboloid of one sheet32. Parabolic cylinderInstructor’s Resource Manual Section 11.10 739

37.2 2 2x y z+ + = 112 9 4Ellipsoidc.2 2 2 22 x –( x + y + z ) = 1;2 2 2 22 sin cos – 1;ρ φ θ ρ =212 2ρ =2sin φcos θ –1d.2 2x + y = z;2 2 2 2 2 2sin cos + sin sin = cosρ φ θ ρ φ θ ρ φ2 2 2 2sin (cos + sin ) = cos ;ρ φ θ θ ρ φ2sin cos ;ρ φ = φ ρ = cot φ csc φ2 2 238. The graph of 3x + 4y + 9 z = –36 is the emptyset.39. a.b.c.d.2r = 9; r = 32 2 2( x + y ) + 3y= 162 2 2 2 16r + 3r sin θ = 16, r =21 + 3sin θr2= 9z2 2r + 4z= 10(Note that when we divided through by ρ in partc and d we did not lose the pole since it is also asolution of the resulting equations.)42. Cartesian coordinates are ( 2 2, 2 2, 4 3 ) and( 2, 6, – 2 2 ) . Distance1/2⎡2 2⎤⎢2 + ( 2 2 – 6 ) + ( 4 3 + 2 2 ) ⎥≈9.8659.⎣⎦43. (2, 0, 0) is a point of the first plane. The distancebetween the planes is2(2) – 3(0) + 3(0) – 9 5= = 1.254+ 9+3 1640. a.b.c.41. a.b.2 2 2x + y + z = 92 2x + z = 42 2 2 2r (cos θ – sin θ ) + z = 1;2 2 2x – y + z = 1ρ2= 4; = 2ρ2 2 2 2x + y + z – 2z= 0;2 2 2ρ – 2ρ cos φ = 0;2 2ρ (1 – 2 cos φ ) = 0;22 1 π1–2cos φ = 0; cos φ = ; φ = or2 43 πφ = .4Any of the following (as well as others)would be acceptable:⎛ π⎞⎛ 3π⎞ ⎜φ– ⎟⎜φ– ⎟ = 0⎝ 4⎠⎝ 4 ⎠2 1cos φ =22sec φ = 22tan 1 φ =44. 2, − 4,1 and 3, 2, − 5 are normal to therespective planes. The acute angle between thetwo planes is the same as the acute angle θbetween the normal vectors.6–8–5 7cos θ = = ,21 38 798so θ ≈ 1.3204 rad ≈ 75.65°ds45. If speed = = c,a constant, thendt222d s ⎛ds⎞2 d sa = T+ κ = c κ2 ⎜ ⎟ N N since = 0.dt ⎝ dt ⎠2dtT is in the direction of v, while N isperpendicular to T and hence to v also. Thus,2a= c κ N is perpendicular to v.740 Section 11.10 Instructor’s Resource Manual

Review and Preview1.2 2 2x + y + z = 64840-4-8-8-4048 -8 -40482.x2 2+ z = 4210-1-2-2-1012 -2 -1012Instructor’s Resource Manual Review & Preview 741

3.2 2z = x + 4y86420-2 -1 01102 -14.2 2z x y = −1680-8-16-4-2024 -4 -20245. a.b.c.d.d 3 22x= 6xdxd 3 25x= 15xdxd kxdx3 2= 3 kxd ax3 2= 3 axdx6. a.dsin 2x= 2cos 2xdxb.dsin17t= 17cos17tdtc.dsin at = a cos atdtd.dsin bt = b cosbtdt742 Review & Preview Instructor’s Resource Manual

d7. a. sin 2a= 2cos 2ada8. a.9.db. sin17a= 17cos17adadc. sin ta = t costadadd. sin sa = scossadab.c.d.d e4t+ 1 4t14 e+=dtd edxd edx− 7x+ 4 − 7x+4=− 7 eax+ b ax+b= aed etx+ s tetx+s=dx12f( x)= is both continuous andx −1differentiable at x = 2 since rational functionsare continuous and differentiable at every realnumber in their domain.10. f ( x) tan xf ( π 2)is undefined.= is not continuous at x = π 2 since11. f( x) = x− 4 is continuous at x = 4 sincex− = f = f is not differentiable atlim 4 (4) 0.x→4x = 4 .12. As x approaches 0 from the right, 1/ x approaches+∞,so sin ( 1/ x ) oscillates between − 1 and 1and lim f ( x)does not exist. Since the limit atx→00x = does not exist, f is not continuous at x = 0 .Consequently, f is not differentiable at x = 0 .14.15.′ 3 2( ) = 4 − 54 + 226 −288f x x x x2( x )( x x )= 2 2 −9 − 9 + 169 9±17f '( x ) = 0 when x = or x = (using2 2the quadratic formula).9−17In [2,6], f′ ( x) = 0 when x = ≈ 2.438292or x = = 4.5. f′′( x) = 12x − 108x+226. Since2f ′′(2.438) > 0, a local minimum occurs at9−17x = . Since f′′(4.5) < 0, a local2maximum occurs at x = 4.5.⎛9−17 ⎞f(2) = f(6) = 0, f⎜4, and2 ⎟=−⎝ ⎠f (4.5) = 14.0625. Thus, the minimum valueof f on [2,6] is − 4 and the maximum value off on [2,6] is 14.0625.2S = 2πr + 2πrhSince the volume is to be 8 cubic feet, we haveV = 82π r h = 88h =2π rSubstituting for h in our surface area equationgives us2 ⎛ 8 ⎞ 2 16S = 2πr + 2πr⎜2πr2 ⎟= +⎝πr ⎠ rThus, we can write S as a function of r:2 16( ) = 2πr +S rr16. The area of two of the sides of the box will bel⋅ h. Two other sides will have area wh ⋅ . Thearea of the base is l⋅ w.Thus, the total cost, C,of the box will be C = 2lh+ 2wh+ 3 lw,where C27 54 54is in dollars. Since h = , C = + + 3 lw.lw w l13.′ 2( ) 3 3( 1) ; ′( ) 0f x = − x− f x = when x = 0, 2 ;f′′ ( x) =−6( x− 1); Since f ′′(2) =−6,a localmaximum occurs at x = 2 . Since f (0) = − 1,f (2) = 5, and f (4) =− 15, the maximum valueof f on [0, 4] is 5 while the minimum value is− 15 .Instructor’s Resource Manual Review & Preview 743

CHAPTER12Derivatives for Functions ofTwo or More Variables12.1 Concepts Review1. real-valued function of two real variables2. level curve; contour map4. a. 6b. 12c. 23. concentric circles4. parallel lines5.d.1/2(3cos 6) + 1.44 ≈ 3.13722 2 2 2 2F( tcos t, sec t) = t cos tsec t = t , cos t ≠ 0Problem Set 12.11. a. 5b. 0c. 66.2 t /2F( f( t), g( t)) = F(ln t , e )2 t/2 2 2 t= exp(ln t ) + ( e ) = t + e , t ≠ 07. z = 6 is a plane.d.a6 2+ ae.22 x , x ≠ 0f. UndefinedThe natural domain is the set of all (x, y) suchthat y is nonnegative.2. a. 4b. 17c.17168. x + z = 6 is a plane.d.2+ ≠1 a , a 0e.3x + x, x ≠ 0f. UndefinedThe natural domain is the set of all (x, y) suchthat x is nonzero.9. x + 2y + z = 6 is a plane.3. a. sin(2 π ) = 0⎛π⎞b. 4sin⎜⎟ = 2⎝6⎠⎛π⎞c. 16sin ⎜ ⎟ = 16⎝2⎠d.2 2π sin( π ) ≈ –4.2469744 Section 12.1 Instructor’s Resource Manual

10.2z = 6– x is a parabolic cylinder.15.2 2z = exp[–( x + y )]11.2 2 2x + y + z = 16 , z ≥ 0 is a hemisphere.16.2xz = , y > 0y12.2 2 2x y z+ + = 1, z ≥ 0 is a hemi-ellipsoid.4 16 1617.2 2 2 2x + y = z x + y = k2; 213.2 2z = 3– x – y is a paraboloid.18. x = zy, y ≠ 0 ; x = ky, y ≠ 014.z = 2– x–y219.2 2x = zy, y ≠ 0; x = ky, y ≠ 0Instructor’s Resource Manual Section 12.1 745

20.2 2x = –( y– z); x = –( y– k)24.2 2 16( x–2) + ( y+ 3) =2V21.2x +z = 1 , k = 1,2,42 2x + y2k = 1: y = 1 or y =± 1 ;two parallel lines2 2 2k = 2: 2x + 2y = x + 12 2x y11 12+ = ; ellipse2 2 2k = 4: 4x + 4y = x + 12 2x y+ = 1; ellipse1 13 422. y = sin x + z; y = sin x + k25. a. San Francisco and St. Louis had atemperature between 70 and 80 degreesFahrenheit.b. Drive northwest to get to coolertemperatures, and drive southeast to getwarmer temperatures.c. Since the level curve for 70 runs southwestto northeast, you could drive southwest ornortheast and stay at about the sametemperature.26. a. The lowest barometric pressure, 1000millibars and under, occurred in the regionof the Great Lakes, specifically nearWisconsin. The highest barometricpressure, 1025 millibars and over, occurredon the east coast, from Massachusetts toSouth Carolina.b. Driving northwest would take you to lowerbarometric pressure, and driving southeastwould take you to higher barometricpressure.23. x = 0, if T = 0:2 ⎛ 1 ⎞y –1 x2= ⎜ ⎟ ,⎝T⎠if y ≠ 0 .27.c. Since near St. Louis the level curves runsouthwest to northeast, you could drivesouthwest or northeast and stay at about thesame barometric pressure.2 2 2x + y + z ≥ 16 ; the set of all points on andoutside the sphere of radius 4 that is centered atthe origin28. The set of all points inside (the part containingthe z-axis) and on the hyperboloid of one sheet;2 2 2x y z+ – = 1.9 9 929.2 2 2x y z+ + ≤ 1 ; points inside and on the9 16 1ellipsoid746 Section 12.1 Instructor’s Resource Manual

30. Points inside (the part containing the z-axis) or onthe hyperboloid of one sheet,2 2 2x y z+ – = 1,9 9 16excluding points on the coordinate planes31. Since the argument to the natural logarithmfunction must be positive, we must have2 2 2x y z 0x, yz ,+ + > . This is true for all ( )except ( xyz , , ) = ( 0,0,0). The domain consistsall points in3 except the origin.32. Since the argument to the natural logarithmfunction must be positive, we must have xy > 0 .This occurs when the ordered pair ( x,y ) is in thefirst quadrant or the third quadrant of thexy-plane. There is no restriction on z. Thus, thedomain consists of all points ( x, yz , ) such that xand y are both positive or both negative.37.38.2 24 x –9 y k,= k in R;2 2x y– 1,k k4 9= if k ≠ 0;2xplanes y =± (for k = 0) and all hyperbolic3cylinders parallel to the z-axis such that the ratio1 1a:b is⎛ ⎜ ⎞ ⎟:⎛ ⎜ ⎞⎟ or 3:2 (where a is associated⎝2⎠ ⎝3⎠ with the x-term)x2+ y2+ z2e = k, k > 02 2 2x + y + z = ln kconcentric circles centered at the origin.39. a. All ( wxyz , , , ) except ( )0,0,0,0 , whichwould cause division by 0.b. All ( x x x ), ,…, n in n-space.1 233.34.35.2 2 2x + y + z = k, k > 0; set of all spherescentered at the origin2 2 2100x 16y 25 z k,2 2 2x y z1;k k k100 16 25+ + = k > 0;+ + = set of all ellipsoids centeredat origin such that their axes have ratio⎛ 1 1 1⎜ ⎞ ⎟: ⎛ ⎜ ⎞ ⎟:⎛ ⎜ ⎞⎟ or 2:5:4.⎝10 ⎠ ⎝4 ⎠ ⎝5⎠2 2 2x y z–1 1 116 4k;2x2y2z9 9 16+ = the elliptic cone+ = and all hyperboloids (one and twosheets) with z-axis for axis such that a:b:c is⎛ 1 1 1⎜ ⎞ ⎟: ⎛ ⎜ ⎞ ⎟:⎛ ⎜ ⎞⎟ or 3:3:4.⎝4⎠ ⎝4⎠ ⎝3⎠ c. All ( x x x ), ,…, n that satisfy1 22 2 21 + 2 + + ≤1x1, x2, , xnx x x n ; other values of( … ) would lead to the squareroot of a negative number.40. If z = 0, then x = 0 or x =± 3y.41. a. AC is the least steep path and BC is the moststeep path between A and C since the levelcurves are farthest apart along AC andclosest together along BC.36.2 2 2x1 y1z19 42y2z2x9 36 4= k ; the elliptical cone+ = and all hyperboloids (one and twosheets) with x-axis for axis such that a:b:c is⎛1⎞ ⎛1⎞⎜ ⎟: ⎜ ⎟:1or 2:3:6⎝3⎠ ⎝2⎠b.2 2AC ≈ (5750) + (3000) ≈ 6490 ft2 2BC ≈ (580) + (3000) ≈ 3060 ft42. Completing the squares on x and y yields theequivalent equation2 2f( x, y) + 25.25 = ( x– 0.5) + 3( y+ 2) , anelliptic paraboloid.Instructor’s Resource Manual Section 12.1 747

43.2 2 2(2 x – y )exp(– x – y )2 2sin 2x + y46.44.sin xsiny2 2(1 + x + y )2 2sin( x + y )2 2x + y12.2 Concepts Review1.[( f ( x0 + h, y0) – f( x0, y0)]lim; partialh→0hderivative of f with respect to x2. 5; 145.3.2 fy x∂∂ ∂4. 0Problem Set 12.21.2.3 3fx( x, y) = 8(2 x– y) ; fy( x, y) = –4(2 x– y)21/2f x y x yx ( , ) = 6(4 – ) ;21/2fy ( x, y) = –3 y(4 x– y )748 Section 12.2 Instructor’s Resource Manual

2 2 2 2( xy)(2 x)–( x – y )( y)x + y3. fx( x, y)= =2 2( xy)x y2 2( x + y )=−2xy2 2( xy)( −2 y) −( x − y )( x)fy( x, y)=2( xy)4.xxfx( xy , ) = e cos y; fy( x, y) = – e sin y5.yyfx( x, y) = e cos x; fy( x, y) = e sinx14.t2– s2t2– s2fs(,) s t = –2 se ; fs(, s t) = 2te15. Fx( x, y) = 2cosxcos y; Fy( x, y) = –2sinxsiny16.17.2 3fr(, r ) = 9r cos2; fθ(, r ) = –6rsin2θ θ θ θ3 2 5fx ( x, y) = 4 xy –3 x y ;2 2 4fxy ( x, y) = 12 xy –15x y2 2 3 4f y ( xy , ) = 6 x y –5 xy;2 2 4f yx ( x, y) = 12 xy –15x y6.7.1 2 2 –4/3fx( x, y) ⎛ ⎞= ⎜– ⎟(3 x + y ) (6 x)⎝ 3 ⎠2 2 –4/3= –2 x(3 x + y ) ;1 2 2 –4/3fy( x, y) ⎛ ⎞= ⎜– ⎟(3 x + y ) (2 y)⎝ 3 ⎠⎛ 2y⎞ 2 2 –4/3= ⎜– ⎟(3 x + y )⎝ 3 ⎠2 2 −1/2fx ( x, y) = x( x − y ) ;2 2 –1/2fy ( x , y ) = – y ( x – y )uvuv8. fu( u, v) = ve ; fv( u, v)= ue9.10.– xy– xygx( x, y) = – ye ; gy( x, y) = – xe2 2 –1fs (, s t) = 2( s s – t ) ;2 2 −1ft(, s t) =−2( t s − t )18.19.20.3 2 4 2fx ( x, y) = 5( x + y ) (3 x );2 3 2 3fxy ( x, y) = 60 x ( x + y ) (2 y)2 3 2 3= 120 x yx ( + y )3 2 4f y ( x, y) = 5( x + y ) (2 y);3 2 3 2f yx ( x, y) = 40 y( x + y ) (3 x )2 3 2 3= 120 x yx ( + y )2x2xxyf ( x, y) = 6e cos y; f ( x, y) = –6e sin yx2x2xf y( x, y) = –3e sin y; fyx( x, y) = –6e sin y2 2 –1fx ( x, y) = y(1 + x y ) ;2 2 2 2 −2fxy ( x, y) = (1 − x y )(1 + x y )2 2 –1fx ( x, y) = x(1 + x y ) ;2 2 2 2 −2fxy ( x, y) = (1 − x y )(1 + x y )11.12.13.2 –1fx ( x, y) = 4[1 + (4 x–7 y) ] ;2 –1fy ( x, y) = –7[1 + (4 x–7 y) ]1 ⎛1⎞ –1 ⎛w⎞Fw( w, z) = w ⎜ ⎟+sin ⎜ ⎟21– w ⎝ z⎠ ⎝ z ⎠( z )wz–1 ⎛w⎞= + sin ⎜ ⎟;21– w ⎝ z ⎠( z )w2( z )w( )1 ⎛ w ⎞ –Fz= ( w, z) = w ⎜–2 2 ⎟=21– w ⎝ z ⎠( )1–zz2 2fx ( x, y) = –2xysin( x + y );2 2 2 2 2f y ( x, y) = –2y sin( x + y ) + cos( x + y )21.22.2( xy)(2) – (2 x – y)( y) y 1Fx( x, y) = = = ;2 2 2 2( xy)x y x1Fx(3, − 2) =92( xy)(–1)–(2 x – y)( x) –2x2Fy( x, y) = = = – ;2 2 2 2( xy)x y x1Fy(3, − 2) =−22 2 –1Fx ( x, y) = (2 x+ y)( x + xy+y ) ;2F x (–1, 4) = ≈ 0.1538132 2 –1Fy ( x , y ) = ( x+ 2 y )( x + xy+y ) ;7F y (–1, 4) = ≈ 0.538513Instructor’s Resource Manual Section 12.2 749

23.2 2 4 –1fx ( x, y) = – y ( x + y ) ;4f x ( 5, – 2 ) = – ≈ –0.1905212 4 –1fy ( x , y ) = 2 xy ( x + y ) ;( )4 5f y 5, – 2 = – ≈ –0.42592124. fx( x, y) = e y sinh x;fx (–1, 1) = esinh(–1) ≈ –3.1945f ( x, y) = e cosh x;y25. Let26. Let27.28.yfy (–1, 1) = ecosh(–1) ≈ 4.19452 2x yz = f( x, y) = + .9 4yfy( x, y ) =2The slope is f y (3, 2) = 1.2 2 1/2z = f( x, y) = (1/3)(36–9 x –4 y ) .⎛ 4 ⎞ 2 2 −1/2fy( x, y) = ⎜− ⎟y(36−9 x − y )⎝ 3 ⎠8f y (1, – 2) = ≈ 0.8040.3 11The slope is⎛1⎞ 2 2 1/2z = f( x, y) = ⎜ ⎟(9x + 9y−36)⎝2⎠9xfx( x, y)=2 2 1/22(9x+ 9 y – 36)f x (2, 1) = 3⎛5⎞ 21/2z = f( x, y) = ⎜ ⎟ (16– x ) .⎝4⎠⎛ 5 ⎞ 2 –1/2fx( x, y) = ⎜ – ⎟ x(16– x )⎝ 4 ⎠5f x (2, 3) = – ≈ –0.72174 329. Vr(, r h) = 2 π rh;V r (6, 10) = 120π≈ 376.99 in. 230.2= y (3, 2) 12Ty ( x, y) 3 y ; T = degrees per ftkT31. PV ( , T)=VkPT( V, T) = ;VkP T (100, 300) = lb/in. 2 per degree10032. V[ PV( V, T)] + T[ PT( V, T)]–2 –1= V(– kTV ) + T( kV ) = 033.34.35.36.37. a.PVTV T P⎛ kT ⎞⎛ k ⎞⎛V ⎞ kT PV= ⎜– – – –12 ⎟⎜ ⎟⎜ ⎟= = =⎝ V ⎠⎝P⎠⎝ k ⎠ PV PV2 3fx ( x, y) = 3 x y– y ; fxx( x, y) = 6 xy;3 2f y ( x , y ) = x –3 xy ; f yy ( x, y) = –6xyTherefore, fxx( x, y) + fyy( x, y) = 0.2 2 –1fx ( x, y) = 2 x( x + y ) ;2 2 2 2 –1fxx ( x, y) = –2( x – y )( x + y )2 2 –1fy ( x , y ) = 2 y ( x + y ) ;2 2 2 2 −1fyy ( x, y) = 2( x − y )( x + y )4 4 2 2Fy ( x, y) = 15 x y –6 x y ;4 3 2Fyy ( x, y) = 60x y − 12 x y;4 2 2Fyyy ( x, y) = 180 x y –12x2 2fx ( x, y) = [–sin(2 x – y )](4 x)2 2= –4xsin(2 x – y )2 2fxx( x, y) = (–4 x)[cos(2 x – y )](4 x)2 2+ [sin(2 x – y )](–4)2 2 2fxxy( xy , ) = –16 x[– sin(2 x – y)](–2 y)2 2– 4[cos(2 x – y )](–2 y)2 2 2 2 2= –32x ysin(2 x – y ) + 8ycos(2 x – y )b.3∂ f3∂ y3∂ y∂ y∂x2c.4∂ y3∂ y ∂ x750 Section 12.2 Instructor’s Resource Manual

38. a. f yxxb. f yyxxc. f yyxxx39. a. f ( x, y, z) = 6 xy–yzb.x2 2f y ( x , y , z ) = 3 x – xz + 2 yz ;f y (0, 1, 2) = 845. Domain: (Case x < y)The lengths of the sides are then x, y – x, and1 – y. The sum of the lengths of any two sidesmust be greater than the length of the remainingside, leading to three inequalities:1x + (y – x) > 1 – y ⇒ y >21(y – x) + (1 – y) > x ⇒ x y – x ⇒ y < x+2c. Using the result in a, f ( x, y, z) = 6 x– z.xy40. a.b.2 3 2 312 x ( x + y + z)3 2 3f y ( x , y , z ) = 8 y ( x + y + z ) ;f y (0,1,1) = 64c.3 2 3f ( x, y, z) = 4( x + y + z) ;z2 2 2f ( x, y, z) = 12( x + y + z)zz41. fx( xyx , , ) =42.f43. If– xyz2 –1– yze – y( xy – z )–1/ 2⎛1⎞⎛ xy⎞ ⎛ y⎞x ( x, y, z) = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ;2⎝ ⎠⎝ z ⎠ ⎝ z ⎠–1/ 2⎛1⎞⎛1⎞ ⎛ 1⎞1f x (–2, –1, 8) = ⎜ ⎟⎜ ⎟ ⎜– ⎟=–⎝2⎠⎝4⎠ ⎝ 8⎠84 3 2( , ) = + + 12, y ( , ) = 3 ;f x y x xy f x y xyf y (1, – 2) = 12. Therefore, along the tangent lineΔ y = 1⇒ Δ z = 12, so 0, 1, 12 is a tangentvector (since Δx = 0). Then parametric equations⎧ x = 1 ⎫⎪ ⎪of the tangent line are ⎨y= –2 + t ⎬. Then the⎪z= 5+12t⎪⎩ ⎭point of xy-plane at which the bee hits is(1, 0, 29) [since y = 0⇒ t = 2⇒ x = 1, z = 29].44. The largest rectangle that can be contained in thecircle is a square of diameter length 20. The edgeof such a square has length 10 2, so its area is200. Therefore, the domain of A is2 2{( x, y) : 0 x y 400},(0, 200].≤ + < and the range isThe case for y < x yields similar inequalities(x and y interchanged). The graph of D A,thedomain of A is given above. In set notation it is⎧ 1 1 1⎫DA= ⎨( x, y): x< , y > , y < x+⎬⎩ 2 2 2⎭⎧ 1 1 1⎫∪ ⎨( x, y): x > , y < , x< y+⎬.⎩ 2 2 2⎭Range: The area is greater than zero but can bearbitrarily close to zero since one side can bearbitrarily small and the other two sides arebounded above. It seems that the area would belargest when the triangle is equilateral. Anequilateral triangle with sides equal to 1 3 has3area .36 Hence the range of A is ⎛ 3 ⎤⎜0, ⎥.(In⎝ 36 ⎦Sections 8 and 9 of this chapter methods will bepresented which will make it easy to prove thatthe largest value of A will occur when the triangleis equilateral.)46. a. u = cos (x) cos (ct):u = – sin( x)cos( ct); u = – ccos( x)sin( ct)xtuxx= – cos( x)cos( ct)2utt= – c cos( x)cos( ct)2Therefore, cuxx= utt.xu = e cosh( ct) :xxux= e cosh( ct), ut= ce sinh( ct)x2 xuxx= e cosh( ct), utt= c e cosh( ct)2Therefore, cuxx= utt.Instructor’s Resource Manual Section 12.2 751

.– ctu = e sin( x) :– ctux= e cos x– ctuxx= – e sin xctut= – ce sin xTherefore, cuxx= ut.–1/ 2 – x2/ 4ctu = t e :–1/ 2 – x2/ 4ct⎛ x ⎞ux= t e ⎜– ⎟⎝ 2 ct ⎠2uxx( x –2 ct)=25/2 x2/4ct(4 ct e )2( x –2 ct)ut=5/2 x2/4ct(4 ct e )Therefore, cu = u .xx47. a. Moving parallel to the y-axis from the point(1, 1) to the nearest level curve andΔzapproximating , we obtainΔy4–5f y (1, 1) = = –4.1.25 –1b. Moving parallel to the x-axis from the point(–4, 2) to the nearest level curve andΔzapproximating , we obtainΔx1–0 2fx(–4, 2) ≈ = .–2.5 – (–4) 3t48. a.b.c.d.2sin( x + y )2Dx (sin( x+y ))2Dy (sin( x+y ))D ( D (sin( x+y) ))xy2c. Moving parallel to the x-axis from the point(–5, –2) to the nearest level curve andΔzapproximately , we obtainΔx1–0 2fx(–4, – 5) ≈ = .–2.5 – (–5) 5d. Moving parallel to the y-axis from the point(0, –2) to the nearest level curve andΔzapproximating , we obtainΔy0–1 8f y (0, 2) ≈ = .–(–2) 3–19849. a. f ( x, y,z)yf ( xy , +Δyz , ) − f( xyz , , )= limΔ→ y 0 Δyb. f ( x, y,z)zf ( xyz , , +Δz) − f( xyz , , )= limΔ→ z 0 Δzc. G ( w, x, y,z)xGwx ( , +Δxyz , , ) −Gwxyz( , , , )= limΔ→ x 0Δx752 Section 12.2 Instructor’s Resource Manual

d.∂λ( xyzt , , , )∂zλ( x, yz , +Δzt , ) −λ( xyzt , , , )= limΔ→ z 0Δz∂Sb ( , b, b, …, b)=e. 0 1 2∂b2=⎛Sb ( , b, b +Δb, …, b)⎞⎟Sb ( , b, b, …, bn) ⎟⎟⎜⎟⎝⎠0 1 2 2 n⎜0 1 2lim ⎜−Δb2 →0⎜Δb2∂∂w50. a. ( sin w sin x cos y cos z)= cos w sin x cos y cos z∂∂x⎣b. ⎡x ln ( wxyz) ⎤ = x ⋅ + 1⋅ln( wxyz)= 1+ln( wxyz)⎦nwyzwxyzc. λ t ( x, yzt , , )( 1+ xyzt ) cosx −t ( cosx)xyz=( 1+xyzt ) 2cos x=1+xyzt( ) 212.3 Concepts Review1. 3; (x, y) approaches (1, 2).2.lim f( x, y) = f(1, 2)( x, y) →(1, 2)3. contained in S4. an interior point of S; boundary pointsProblem Set 12.31. –182. 34. The limit does not exist because of Theorem A.The function is a rational function, but the limitof the denominator is 0, while the limit of thenumerator is -1.5.5– 26. − 17. 18.2 2tan( x + y )lim( x, y) →(0, 0) 2 2( x + y )2 2sin( x + y ) 1= lim( x, y) →(0, 0) 2 2 2 2( x + y ) cos( x + y )= (1)(1) = 19. The limit does not exist since the function is notdefined anywhere along the line y = x. That is,there is no neighborhood of the origin in whichthe function is defined everywhere exceptpossibly at the origin.10.2 2 2 2( x + y )( x – y )lim( x, y) →(0, 0) 2 2x + y= lim2 2( x – y ) = 0( x, y) →(0, 0)11. Changing to polar coordinates,xy r cosθ⋅ r sinθlim= lim( xy , ) →(0,0) 2 2x + yr→0r= lim r cosθ ⋅ sinθ= 0r→012. If ( x,y ) approaches (0, 0) along the line y2x1lim= lim =+∞( xx , ) →(0,0) 2 2 2 ( , ) (0,0) 2( x + x ) xx→4xThus, the limit does not exist.13. Use polar coordinates.7/3 7/3 7/3x r ( cosθ)= = r2 2 2x + y r( ) 7/31/3( cosθ)7/3= x,r 1/3 cosθ → 0 as r → 0 , so the limit is 0.3.⎡ 2 ⎛ xy ⎞⎤lim ⎢xcos xy– sin ⎜ ⎟ 3⎥⎣⎝ ⎠⎦( x , y ) →(2, π)2 ⎛2π⎞ 3= 2cos 2 π–sin⎜ ⎟ = 2 – ≈ 1.1340⎝ 3 ⎠ 214. Changing to polar coordinates,2 2 2 22 r cos θ − r sinlim r cosθsinθ⋅r→02r2= lim r cosθsinθcos 2θ= 0r→0θInstructor’s Resource Manual Section 12.3 753

15.4 2 2r cos θ sin θf( x, y)=2 2 4 4r cos θ + r sin θ⎛ 2 22 cos θsinθ ⎞= r⎜ 2 2 4cos θ + r sin θ ⎟⎝⎠If cos 0 , 0θ = , then ( )f x y = . If cosθ ≠ 0 ,hen this converges to 0 as r → 0 . Thus thelimit is 0.26.2 2 2Require 4 − x − y − z > 0;2 2 2x + y + z < 4. S is the space in the interiorof the sphere centered at the origin with radius 2.27. The boundary consists of the points that form theouter edge of the rectangle. The set is closed.16. As ( x,y ) approaches (0,0) along x = y 2 ,4y 1lim = .( xx , ) →(0,0)4 4y + y 2however,not exist.0lim 0.( x,0) →(0,0)2xAlong the x-axis,= Thus, the limit does17. f( x, y) is continuous for all ( x, y ) since2 2for all ( xy , ), x + y + 1 ≠ 0.28. The boundary consists of the points of the circleshown. The set is open.18. f( x, y) is continuous for all ( x, y ) since2 2for all ( xy , ), x + y + 1 > 0.2 2 2 219. Require 1– x – y > 0; x + y < 1. S is theinterior of the unit circle centered at the origin.20. Require 1+ x+ y > 0; y >−x−1. S is the setof all ( xy , ) above the line y=−x−1.21. Requirethe parabola2y– x ≠ 0. S is the entire plane excepty = x2 .29. The boundary consists of the circle and theorigin. The set is neither open (since, forexample, (1, 0) is not an interior point), norclosed (since (0, 0) is not in the set).22. The only points at which f might bediscontinuous occur when xy = 0.sin( xy)lim = 1 = f( a, 0) for all nonzero( x, y) →( a, 0) xya in , and thensin( xy)lim = 1 = f (0, b)for all b in .( x, y) →(0, b)xyTherefore, f is continuous on the entire plane.23. Require x – y + 1 ≥ 0; y ≤ x + 1. S is the regionbelow and on the line y = x + 1.2 2 2 224. Require 4– x – y > 0; x + y < 4. S is theinterior of the circle of radius 2 centered at theorigin.25. f( x, y, z) is continuous for all ( x, y, z) ≠ ( 0,0,0)2 2 2since for all ( xyz , , ) ≠ ( 0,0,0), x + y + z >0.754 Section 12.3 Instructor’s Resource Manual

30. The boundary consists of the points on the linex = 1 along with the points on the line x = 4. Theset is neither closed nor open.31. The boundary consists of the graph of⎛1⎞y = sin ⎜ ⎟ along with the part of the y-axis for⎝ x ⎠which y ≤ 1. The set is open.35. Along the x-axis (y = 0):Along y = x:0lim = 0( x, y) →(0, 0) 2x + 02x1 1lim = lim = .( x, y) →(0, 0) 22x( x, y) →(0, 0) 2 2Hence, the limit does not exist because for somepoints near the origin f(x, y) is getting closer to 0,but for others it is getting closer to 1 .20lim = 0.x→02x + 02 3x + x 1+x 1lim = lim = .x→0 2 2x + x x→02 236. Along y = 0:37. a.Along y = x:2 3x ( mx)mxlim= limx→0 4 2 0 4 2 2x + ( mx)x→x + m xlim mx= =0 2 20x→x + m.b.2 2 4x ( x ) x 1 1lim = lim = lim =x→0 4 2 2 0 4x + ( x ) x→ 2xx→02 232. The boundary is the set itself along with theorigin. The set is neither open (since none of itspoints are interior points) nor closed (since theorigin is not in the set).33.2 2x –4 y ( x+2 y)( x–2 y)= = x + 2y(if x ≠ 2y)x–2 y x–2yIf x = 2y, x + 2y = 2x. Take g ( x) = 2x.34. Let L and M be the latter two limits.[ f( x, y) + g( x, y)]–[ L+M]≤ f( x, y)– L + f( x, y)–M ≤ ε +ε2 2for (x, y) in some δ-neighborhood of (a, b).Therefore,lim [ f ( x , y ) + g ( x , y )] = L + M .( x, y) →( a, b)c.2x ylim( x, y) → (0, 0) 4x + ydoes not exist.38. f is discontinuous at each overhang. Moreinteresting, f is discontinuous along theContinental Divide.39. a.b.2 2{( x, y, z) : x y 1, z in [1, 2]}+ = [For2 2x + y < 1, the particle hits the hemisphereand then slides to the origin (or bounds2 2toward the origin); for x + y = 1, it2 2bounces up; for x + y > 1, it falls straightdown.]2 2{( x, y, z) : x y 1, z 1}+ = = (As one movesat a level of z = 1 from the rim of the bowltoward any position away from the bowlthere is a change from seeing all of theinterior of the bowl to seeing none of it.)c. {(x, y, z): z = 1} [f(x, y, z) is undefined(infinite) at (x, y, 1).]d. φ (Small changes in points of the domainresult in small changes in the shortest pathfrom the points to the origin.)Instructor’s Resource Manual Section 12.3 755

40. f is continuous on an open set D and P 0 is in D implies that there is neighborhood of P 0 with radius r on which fis continuous. f is continuous at P0 ⇒ lim f( P) = f( P0).Now let ε = f ( P0) which is positive. Then there is a δ41. a.P→P0f ( p)– f( P ) f( P )such that 0 < δ < r and 0 < 0 if P is in the δ-neighborhood of P 0. Therefore,– f ( P ) < f( p)– f( P ) < f( P ), so 0 < f(p) (using the left-hand inequality) in that δ-neighborhood of P 0.0 0 0⎧ 2 2 1/2⎪( x + y ) + 1 if y ≠ 0⎫⎪f( x, y) = ⎨ ⎬.⎪⎩x–1 if y = 0⎪⎭2 2 1/2As y = 0, ( x y ) 1 x 1,yields x – x,Check discontinuities where y = 0.+ + = + so f is continuous if x + 1 = x–1.Squaring each side and simplifying= so f is continuous for x ≤ 0. That is, f is discontinuous along the positive x-axis.b. Let P = (u, v) and Q = (x, y).⎧ ⎪ OP + OQ if P and Q are not on same ray from the origin and neither is the origin⎫⎪f( u, v, x, y) = ⎨ ⎬.⎪⎩PQ otherwise⎪⎭This means that in the first case one travels from P to the origin and then to Q; in the second case one travelsdirectly from P to Q without passing through the origin, so f is discontinuous on the set{( u, v, x, y) : u, v = k x, y for some k > 0, u, v ≠ 0, x, y ≠ 0 }.42. a.b.c.d.⎛ hy( h2– y2) ⎞⎜ – 02 2 ⎟2 2h + yfx(0, y) lim ⎜⎟yh ( − y )== lim= −yh→0⎜h ⎟ h→02 2h + y⎜⎟⎝⎠⎛ xh( x2– h2) ⎞⎜ – 02 2 ⎟2 2x + hyx ( – h)fy( x, 0) = lim ⎜⎟ = lim = xh→0h2 2h→0x + y⎜⎟⎝⎠fyxf y(0 + h, y) – fy(0, y)h –0(0, 0) = lim = lim = 1h→0hh→0hf ( , 0 )– ( , 0)xy (0, 0) lim x x + h f x xf =h→0hTherefore, fxy(0, 0) ≠ fyx(0, 0).– h –0= lim = –1h→0h43.b.44. a.45.756 Section 12.3 Instructor’s Resource Manual

46. A function f of three variables is continuous at apoint ( a , b,c)if f ( a,b,c)is defined and equal tothe limit of f ( x,y,z)as ( x,y,z)approaches( a , b,c). In other words,lim f ( x,y,z)= f ( a,b,c).( x , y , z ) → ( a , b , c )A function of three variables is continuous on anopen set S if it is continuous at every point in theinterior of the set. The function is continuous ata boundary point P of S if f (Q)approachesf (P) as Q approaches P along any path throughpoints in S in the neighborhood of P.47. If we approach the point ( 0,0,0)along a straightpath from the point ( x, xx , ) , we have3xx ( )( x) x 1lim= lim =3 3 3 3( xxx , , ) →(0,0,0) x + x + x ( xxx , , ) →(0,0,0)3x3Since the limit does not equal to f (0,0,0) , thefunction is not continuous at the point (0,0,0) .48. If we approach the point (0,0,0) along thex-axis, we get2 2 2( x − 0 )xlim (0 + 1) = lim = 1( x,0,0) →(0,0,0) 2 2 ( ,0,0) (0,0,0) 2( x + 0 ) x → xSince the limit does not equal f (0,0,0) , thefunction is not continuous at the point (0,0,0).12.4 Concepts Review1. gradient2. locally linear3.∂∂∂pi pjy i j∂f f 2( ) + ( ) ; + 2xyx y4. tangent planeProblem Set 12.41.2.22xy+ 3 y, x + 3x2 3 23 x y, x –3y3. ∇ f( x, y) = ( x)( e xy y) + ( e xy )(1), xe xy xxy 2= e xy+1, x4.22xycos y, x (cos y– ysin y )5.–2 2x( x+ y) y( x+2), x6.2 2 2 2 2 2 2∇ f ( x, y) = 3[sin ( x y)][cos( x y)](2 xy), 3[sin ( x y)][cos( x y)]( x )2 2 2= 3xsin ( x y)cos( x y) 2 y,xInstructor’s Resource Manual Section 12.4 757

7.8.2 2 2 –1/2( x + y + z ) x, y,z2 2 22 xy+ z , x + 2 yz, y + 2xz9.2 x– z x– z 2 x– z 2 x–z∇ f( x, y) = ( x y)( e ) + ( e )(2 xy), x e , x ye (–1)x–z= xe y( x+2), x, – xy10.11.–1 –1 –1xz( x+ y+ z) + zln( x+ y+ z), xz( x+ y+ z) , xz( x+ y+ z) + xln( x+ y+z)2 2∇ f ( x, y) = 2 xy – y , x –2 xy ; ∇ f (–2, 3) = –21, 16z = f(–2, 3) + –21, 16 ⋅ x+ 2, y– 3 = 30 + (–21 x– 42 + 16 y– 48)z = –21x + 16y – 6012.2 2 3f ( x, y) 3x y 3 y , x 6 xy ,∇ = + + so ∇ f (2, – 2) = (–12, –16).Tangent plane:z = f(2,–2) +∇(2,–2) ⋅ x–2, y+ 2 = 8 + –12, –16 ⋅ x– 2, y+ 2 = 8 + (–12x + 24 – 16y – 32)z = –12x – 16y13. ∇ f ( x, y) = – πsin( πx)sin( πy), πcos( πx)cos( π y) + 2πcos(2 π y)1∇ f ⎛⎜–1, ⎞⎟= 0, – 2π⎝ 2 ⎠⎛ 1⎞1z = f ⎜–1, ⎟+ 0, – 2π ⋅ x+ 1, y– = –1 + (0 – 2 π y+π);⎝ 2⎠2z = –2π y + (π – 1)14.15.2∇ 2xxf( x, y) = , ; f(2, 1) 4, 4y− 2y∇ − = − −z = f(2, –1) + −4, – 4 ⋅ x– 2, y+1= –4 + (–4x + 8 –4y – 4)z = –4x – 4y2f( x, y, z) 6 x z , –4 y, 2 xz ,∇ = + so ∇ f (1,2,–1) = 7,–8,–2Tangent hyperplane:w= f(1,2,–1) +∇f(1,2,–1) ⋅ x–1, y–2, z+ 1 = –4+ 7, –8, –2 ⋅ x–1, y–2, z+1= –4 + (7x – 7 – 8y + 16 – 2z – 2)w = 7x – 8y – 2z + 316. ∇ f ( x, y, z) = yz + 2 x, xz, xy ; ∇ f (2, 0, – 3) = 4, – 6, 0w= f(2, 0, – 3) + 4, – 6, 0 ⋅ x– 2, y, z+ 3 = 4 + (4x – 8 – 6y + 0)w = 4x – 6y – 417.⎛ f ⎞∇ ⎜ ⎟=⎝ g ⎠gfx − fgx, gfy − fgy,gfz − fgz2gg fx, fy, fz – f gx, gy,gz=2gg∇f – f∇g=2g18.r r–1 r–1 r–1∇ ( f ) = rf fx, rf fy,rf fzr–1 r–1= rf fx, f y,fz= rf ∇ f758 Section 12.4 Instructor’s Resource Manual

19. Let20. Let2 2F( x, y, z) = x − 6x+ 2y − 10y+ 2xy− z = 0∇ F( x, y, z) = 2x− 6+ 2 y,4y− 10+ 2 x, − 1The tangent plane will be horizontal if∇ F( xyz , , ) = 0,0, k , where k ≠ 0 . Therefore,we have the following system of equations:2x+ 2y− 6=02x+ 4y− 10=0Solving this system yields x = 1 and y = 2 .Thus, there is a horizontal tangent plane atxy , = 1,2 .( ) ( )3F( x, y, z) = x − z = 02∇ F( x, y, z) = 3 x ,0, − 1The tangent plane will be horizontal if∇ F( xyz , , ) = 0,0, k , where k ≠ 0 . Therefore,we need only solve the equation 3x = 0. Thereis a horizontal tangent plane at ( x, y) = (0, y).(Note: there are infinitely many points since ycan take on any value).21. a. The point (2,1,9) projects to (2,1,0) on the xyplane. The equation of a plane containing thispoint and parallel to the x-axis is given byy = 1. The tangent plane to the surface at thepoint (2,1,9) is given byz = f(2,1) +∇f(2,1) ⋅ x−2, y−1= 9 + 12,10 x−2, y−1= 12x+ 10y−25The line of intersection of the two planes isthe tangent line to the surface, passingthrough the point (2,1,9) , whose projection inthe xy plane is parallel to the x-axis. This lineof intersection is parallel to the cross productof the normal vectors for the planes. Thenormal vectors are 12,10, −1and 0,1,0 forthe tangent plane and vertical planerespectively. The cross product is given by12,10, − 1 × 0,1,0 = 1,0,12Thus, parametric equations for the desiredtangent line are x = 2 + ty = 1z = 9+12tb. Using the equation for the tangent plane fromthe previous part, we now want the verticalplane to be parallel to the y-axis, but still passthrough the projected point (2,1,0) . Thevertical plane now has equation x = 2 . Thenormal equations are given by 12,10, −1and1, 0, 0 for the tangent and vertical planes2respectively. Again we find the cross productof the normal vectors:12,10, − 1 × 1,0,0 = 0,10,10Thus, parametric equations for the desiredtangent line are x = 2y = 1+10tz = 9+10tc. Using the equation for the tangent plane fromthe first part, we now want the vertical planeto be parallel to the line y = x, but still passthrough the projected point (2,1,0) . Thevertical plane now has equation y− x+ 1= 0.The normal equations are given by12,10, −1and 1, − 1, 0 for the tangent andvertical planes respectively. Again we findthe cross product of the normal vectors:12,10, − 1 × 1, − 1,0 = −1, −1, − 22Thus, parametric equations for the desiredtangent line are x = 2 −ty = 1−tz = 9−22t22. a. The point (3, 2,72) on the surface is the point(3, 2,0) when projected into the xy plane. Theequation of a plane containing this point andparallel to the x-axis is given by y = 2 . Thetangent plane to the surface at the point(3,2,72) is given byz = f(3, 2) +∇f(3, 2) ⋅ x−3, y−2= 72 + 48,108 x−3, y−2= 48x+ 108y−288The line of intersection of the two planes isthe tangent line to the surface, passingthrough the point (3,2,72) , whose projectionin the xy plane is parallel to the x-axis. Thisline of intersection is parallel to the crossproduct of the normal vectors for the planes.The normal vectors are48,108, −1and 0, 2,0 for the tangent planeand vertical plane respectively. The crossproduct is given by48,108, − 1 × 0,2,0 = 2,0,96Thus, parametric equations for the desiredtangent line arex = 3+2ty = 2z = 72 + 96tInstructor's Resource Manual Section 12.4 759

. Using the equation for the tangent plane fromthe previous part, we now want the verticalplane to be parallel to the y-axis, but still passthrough the projected point (3,2,72) . Thevertical plane now has equation x = 3 . Thenormal equations are given by48,108, −1and 3, 0, 0 for the tangent andvertical planes respectively. Again we findthe cross product of the normal vectors:48,108, − 1 × 3,0,0 = 0, −3, − 324Thus, parametric equations for the desiredtangent line arex = 3y = 2−3tz = 72 −324tc. Using the equation for the tangent plane fromthe first part, we now want the vertical planeto be parallel to the line x =− y , but still passthrough the projected point (3,2,72) . Thevertical plane now has equation y+ x− 5= 0.The normal equations are given by48,108, −1and 1,1, 0 for the tangent andvertical planes respectively. Again we findthe cross product of the normal vectors:48,108, − 1 × 1,1,0 = 1, −1, − 60Thus, parametric equations for the desiredtangent line arex = 3 + ty = 2 −tz = 72 −60t23.⎛1 xy ⎞ ⎛1 xy ⎞∇ f ( x, y) = –10 ⎜ y⎟, –10⎜ x⎟⎜2 xy xy ⎟ ⎜2xy xy ⎟⎝ ⎠ ⎝ ⎠–5xy⎡ a a ⎤= y,x Note that .3/2=xy⎢⎣a a ⎥⎦∇ f (1, – 1) = –5, 5Tangent plane:z = f(1, – 1) +∇f(1, – 1) ⋅ x– 1, y+ 1 = –10 + –5, 5 ⋅ x–1, y+ 1 = –10 + (–5x+ 5 + 5y+5)z = –5x + 5y24. Let a be any point of S and let b be any otherpoint of S. Then for some c on the line segmentbetween a and b:f( b) − f( a) =∇f( c) ⋅( b− a) = 0 ⋅( b− a ) = 0, sof(b) = f(a) (for all b in S).25. f(a) – f(b) = f 2,1 − f 0,0 = 4 − 9 =− 5∇ f ( xy , ) = −2 x, − 2 y;b – a = 2,1The value c = c , c will be a solution to− 5= −2 cx, − 2cy2,1c ∈ { cx, cy :4cx + 2cy= 5}xyIn order for c to be between a and b, c must lieon the line y = 1 x. Consequently, c will be the2solution to the following system of equations:4cx+ 2cy= 5 and c 1y = cx. The solution isc = 1, 1.2226. f(b) – f(a) = f 2,6 − f 0,0 = 0 − 2 =− 2−x∇ f( x, y) = ,0 ; b – a = 2,624 − xThe value c = cx, cywill be the solution to−cx− 2 = ,0 2,624 − cx−2cx− 2= ⇒ cx= 224 − cxSince c must be between a and b, c must lie onthe line y = 3x. Since c x = 2, c y = 3 2.Thus, c = 2,3 2 .27. ∇ f( p) =∇g( p) ⇒∇ [ f( p)– g( p)]= 0⇒ f( p)– g( p ) is a constant.760 Section 12.4 Instructor’s Resource Manual

28. ∇ f ( p) = p ⇒∇ f( x, y) = x,y⇒ fx( x, y) = x, fy( x, y)= y1 2⇒ f ( x, y) = x + α( y)for any function of y,21 2and f ( x, y) = y + β ( x)for any function of x.21 2 2⇒ f ( x, y) = ( x + y ) + C for any C in .231. a. (i)∂ ( f + g) ∂ ( f + g) ∂ ( f + g)∇ [ f + g]= i+ j+k∂x ∂y ∂z∂f ∂g ∂f ∂g ∂f ∂g= i+ i+ j+ j+ k+k∂x ∂x ∂y ∂y ∂z ∂z∂f ∂f ∂f ∂g ∂g ∂g= i+ j+ k+ i+ j+k∂x ∂y ∂z ∂x ∂y ∂z=∇ f +∇g29.(ii)∂[ α f] ∂[ α f] ∂[ α f]∇ [ α f ] = i+ j+k∂x ∂y ∂z∂[ f ] ∂[ f] ∂[ f]= α i+ α j+α k∂x ∂y ∂z= α∇f30.− xya. The gradient points in the direction ofgreatest increase of the function.b. No. If it were, 0 + h – 0 = 0 + h δ ( h)whereδ ( h) →0 as h→ 0, which is possible.sin(x) + sin(y) – sin(x + y)(iii)∂( fg) ∂( fg) ∂( fg)∇ [ fg]= i+ j+k∂x ∂y ∂z⎛ ∂g ∂f ⎞ ⎛ ∂g ∂f⎞= ⎜ f + g ⎟i+ ⎜ f + g ⎟j⎝ ∂x ∂x⎠ ⎝ ∂y ∂y⎠⎛ ∂g∂f⎞+ ⎜ f + g ⎟k⎝ ∂z∂z⎠⎛∂g ∂g ∂g⎞= f ⎜ i+ j+k⎟⎝ ∂x ∂y ∂z⎠⎛∂f ∂f ∂f⎞+ g ⎜ i+ j+k⎟⎝∂x ∂y ∂z⎠= f∇ g+ g∇fb. (i)∂ ( f + g) ∂ ( f + g)∇ [ f + g]= i1+i2∂x1 ∂x2∂ ( f + g)+ +in∂xn∂f ∂g ∂f ∂g= i1+ i1+ i2 + i2∂x1 ∂x1 ∂x2 ∂x2∂f∂g+ + in+ in∂xn∂xn∂f ∂f ∂f= i1+ i2+ +in∂x1 ∂x2∂xn∂g ∂g ∂g+ i1+ i2+ +in∂x1 ∂x2∂xn=∇ f +∇gInstructor's Resource Manual Section 12.4 761

(ii)∂[ α f] ∂[ α f] ∂[ α f]∇ [ α f ] = i1+ i2+ +in∂x1 ∂x2∂xn∂[ f] ∂[ f] ∂[ f]= α i1+ α i2+ +α in∂x1 ∂x2∂xn= ∇fα(iii)∂( fg) ∂( fg) ∂( fg)∇ [ fg] = i1+ i2+ +in∂x1 ∂x2∂xn⎛ ∂g ∂f ⎞ ⎛ ∂g ∂f⎞= ⎜ f + g ⎟i+ ⎜ f + g ⎟i⎝ ∂x ∂x ⎠ ⎝ ∂x ∂x⎠⎛ ∂g∂f⎞+ + ⎜ f + g ⎟in⎝ ∂xn∂xn⎠⎛ ∂g ∂g ∂g⎞= f ⎜ i1+ i2+ +in⎟⎝∂x1 ∂x2∂xn⎠⎛ ∂f ∂f ∂f⎞+ g ⎜ i1+ i2+ +in⎟⎝∂x1 ∂x2∂xn⎠= f∇ g+ g∇f12.5 Concepts Review1.[ f( p+hu)– f( p)]h2. u1fx( x, y) + u2fy( x, y)3. greatest increase4. level curveProblem Set 12.51.2.1 21 1 2 22 3 4 8Du f( x, y) = 2 xy, x ⋅ , − ; Duf(1, 2) =5 5 5–1 2 ⎡⎛1 ⎞ ⎤Duf( x, y) = x y , 2ylnx⋅ ⎢⎜⎟ 1, –1 ⎥ ;⎣⎝2 ⎠ ⎦Duf (1, 4) = 8 2 ≈11.3137⎛ a ⎞3. Duf( x, y) = f( x, y)⋅u where u =⎜⎟⎝ a ⎠1, – 1= 4 x+ y, x–2 y ⋅ ;21, – 1 3Duf (3, – 2) = 10, 7 ⋅ = ≈ 2.12132 24. Duf( x, y)5.6.⎡⎛1 ⎞ ⎤= 2 x–3 y, –3x+ 4y⋅⎢⎜⎟ 2, –1 ⎥;⎣⎝5 ⎠ ⎦27Duf (–1, 2) = – ≈ –12.07485x⎡⎛1⎞ ⎤Duf( x, y) = e sin y,cosy⋅⎢⎜ ⎟ 1, 3 ;2⎥⎣⎝⎠ ⎦( 2+6)⎛ π ⎞Duf ⎜0, ⎟ = ≈ 0.9659⎝ 4⎠4– xy – xyD f ( x, y) = – ye – xe ⋅u–1, 3–1, 3 – e– e 3Duf(1,–1) = e,–e ⋅ =2 2≈ –3.71327. Duf( x, y, z ) =2 3 2 2 ⎡⎛18.⎞ ⎤= 3 x y, x –2 yz ,–2y z ⋅⎢⎜ ⎟ 1,–2,2 ;3⎥⎣⎝⎠ ⎦52Duf (–2, 1, 3) =3⎡⎛1⎞⎤Duf( x, y, z) = 2 x, 2 y, 2z⋅⎢⎜ ⎟ 2, –1, –1 ;2⎥⎣⎝⎠⎦Duf (1, –1, 2) = 2 –1 ≈0.41429. f increases most rapidly in the direction of the2 4gradient. ∇ f( x, y) = 3 x , –5 y ;∇ f (2,–1) = 12,–512, – 5is the unit vector in that direction. The13rate of change of f(x, y) in that direction at thatpoint is the magnitude of the gradient.12, – 5 = 1310. ∇ f ( x, y) = e cos x, e sin x ;yy⎛5π⎞ 3 1∇ f ⎜ ,0 ⎟=– , , which is a unit vector.⎝ 6 ⎠ 2 2The rate of change in that direction is 1.2762 Section 12.5 Instructor’s Resource Manual

11.2 2xyz x z x y∇ f ( x, y, z) = 2 , , ;f (1, –1, 2) = –4, 2, –1A unit vector in that direction is⎛ 1 ⎞⎜ ⎟ − 4, 2, − 1 . The rate of change in that⎝ 21 ⎠direction is 21 ≈ 4.5826.16. At (2, 1),2 2x 4y8∇ f ( x, y) = 2 x, 8y+ = is the level curve.∇ f (2, 1) = 4 1, 2 , which is perpendicular to thelevel curve at (2, 1).12. f increases most rapidly in the direction of theyz yz yzgradient. ∇ f ( x, y, z) = e , xze , xye ;∇ f (2,0,–4) = 1,–8,01, – 8, 0is a unit vector in that direction.651, – 8, 0 = 65 ≈ 8.0623 is the rate of changeof f(x, y, z) in that direction at that point.13. – ∇ f ( x, y) = 2 x, y ; – ∇ f (–1, 2) = 2 –1, 2 isthe direction of most rapid decrease. A unit⎛ 1 ⎞vector in that direction is u = ⎜ ⎟ –1, 2 .⎝ 5 ⎠14. – ∇ f ( x, y) = –3cos(3 x– y), cos(3 x– y) ;⎛π π⎞⎛ 1 ⎞– ∇ f ⎜ , ⎟=⎜ ⎟ –3,1 is the direction of⎝6 4⎠ ⎝ 2 ⎠most rapid decrease. A unit vector in that⎛ 1 ⎞direction is ⎜ ⎟ –3, 1 .⎝ 10 ⎠15. The level curves arey2x= k.For p = (1, 2),k = 2, so the level curve through (1, 2) isor2y = 2x(x ≠ 0).–3 –2∇ f( x, y) = –2 yx , xy22x =∇ f (1, 2) = –4, 1 , which is perpendicular to theparabola at (1, 2).17.u =2 2 1,– ,3 3 32 2 1Duf( x, y, z) = y, x, 2 z ⋅ , – ,3 3 32Duf (1,1,1) =3⎛ π ⎞18. ⎜0, ⎟ is on the y-axis, so the unit vector toward⎝ 3 ⎠the origin is –j.– x– xDu( x, y) = – e cos y, – e sin y ⋅ 0, –1– x= e sin y ;⎛ π ⎞ 3Du⎜0, ⎟ =⎝ 3 ⎠ 219. a. Hottest if denominator is smallest; i.e., at theorigin.b.–200 2 x, 2 y, 2z∇ T( x, y, z) =;2 2 2 2(5 + x + y + z )⎛ 25 ⎞∇ T (1,–1,1) = ⎜ – ⎟ 1,–1,1⎝ 4 ⎠−1,1, − 1 is one vector in the direction ofgreatest increase.c. Yes20. – ∇ V( x, y, z)–( x2+ y2+ z2)= –100 e –2 x, – 2 y, – 2z–( x2+ y2+ z2)200 ex, y,z= is the direction ofgreatest decrease at (x, y, z), and it points awayfrom the origin.Instructor’s Resource Manual Section 12.5 763

21.∇f( x, y, z)12 2 2−2 2 2 2( )= x x + y + z cos x + y + z ,12 2 2−2 2 2 2( )y x + y + z cos x + y + z ,12 2 2−2 2 2 2( )z x + y + z cos x + y + z⎛( ) 12 2 2−⎞2 2 2 2= ⎜ x + y + z cos x + y + z ⎟ x, y,z⎜⎟⎝⎠which either points towards or away from theorigin.22. Let D =2 2 2x + y + z be the distance. Then wehave∇ T =∂T ∂T ∂T dT ∂D dT ∂D dT ∂D, , = , ,∂x ∂y ∂z dD ∂x dD ∂y dD ∂z1 1dT 2 2 2− dT 2 2 2−= x 2 2( x + y + z ) y( x + y + z )dDdD1dT 2 2 2−z 2( x + y + z )dD, ,⎛dT( ) 12 2 2− ⎞= ⎜ x + y + z 2 ⎟ x , y , z⎜dD⎟⎝⎠which either points towards or away from theorigin.23. He should move in the direction of1 1– ∇ f( p) = – fx( p), fy( p ) = – – , –2 4⎛1⎞= ⎜ ⎟ 2, 1 . Or use 2, 1 . The angle α formed⎝4⎠–1 ⎛1⎞with the East is tan ⎜ ⎟ ≈ 26.57 ° (N63.43°E).⎝2⎠24. The unit vector from (2, 4) toward (5, 0) is3 4,– . Then5 53 4Duf (2, 4) = –3, 8 ⋅ , – = –8.2.5 525. The climber is moving in the direction of⎛ 1 ⎞u = ⎜ ⎟ −1, 1 . Let⎝ 2 ⎠( x22 y2) /100f( x, y) = 3000 e − + .–( x2+ 2 y2) /100 x y∇ f( x, y) = 3000 e– , – ;50 25–3f(10, 10) = –600e1, 2She will move at a slope of–3 ⎛ 1 ⎞Du(10, 10) = –600e1, 2 ⋅ ⎜ ⎟ –1, 1⎝ 2 ⎠( ) e–3= –300 2 ≈ –21.1229 .She will descend. Slope is about –21.dx dydt dt dx dy26. = ; = ; ln x = – ln y + C2 x –2 y x – yAt t = 0: ln –2 = – ln 1 + C ⇒ C = ln 2.2 2ln x = – ln y + ln 2 = ln ; x ; xy 2y= y=Since the particle starts at (–2, 1) and neither xnor y can equal 0, the equation simplifiesto xy = − 2. ∇ T (–2, 1) = –4, – 2 , so the particlemoves downward along the curve.27. ∇ T( x, y) = –4 x, –2ydx dy= –4 x, = –2ydt dtdx dydt dt2= has solution x = 2 y . Since the–4 x –2yparticle starts at (–2, 1), this simplifies to2x = –2 y .764 Section 12.5 Instructor’s Resource Manual

28. f(1, –1) = 5 -1, 1D f (1, – 1) = u , u ⋅ –5, 5 = –5 u + 5 uu1,u21 2 1 2⎛ 1 ⎞a. -1, 1 (in the direction of the gradient); u = ⎜ ⎟⎝ 2 ⎠⎛b. ± 1, 1 (direction perpendicular to gradient); u = ⎜ ±⎝–1, 1 .1 ⎞⎟ 1, 12 ⎠c. Want Du f (1, – 1) = 1 where u = 1. That is, want –5u1+ 5u2= 1 andand4 3− , − .5 52 2u1 u2 1.+ = Solutions are u =3 4,5 529. a.10(2 x) 10(2 y) 10(2 z)∇ T( x, y, z) = − , − , −2 2 2 2 2 2 2 2 2 2 2 2( x + y + z ) ( x + y + z ) ( x + y + z )=20– x, y,z2 2 2 2( x + y + z )rt () = tcos πt,t sin πt, t , so r(1) = -1, 0, 1 . Therefore, when t = 1, the bee is at (–1, 0, 1), and∇ T (–1, 0, 1) = –5 –1, 0, 1 .r ′( t) = cos πt – πtsin πt, sin π t+πtcos πt, 1 , so r ′(1) = –1, – π, 1 .r′(1) –1, – π, 1U = = r ′(1) 22 +πDT u (–1, 0, 1) = u ⋅∇T(–1, 0, 1)–1, – π, 1 ⋅ 5, 0, – 5 10= = – ≈ –2.90262 22+π 2+πis the unit tangent vector at (–1, 0, 1).Therefore, the temperature is decreasing at about 2.9°C per meter traveled when the bee is at (–1, 0, 1); i.e.,when t = 1 s.b. Method 1: (First express T in terms of t.)10 10 10 5T = = = =2 2 2 2 2 2 2 2x + y + z (cos t π t) + (sin t π t) + () t 2t t−2 −3Tt () = 5 t ; T′ () t = − 10 t ; t′(1) = − 10Method 2: (Use Chain Rule.)dT ds102DT t () t = = ( DuT) ( r '() t ),so DT t () t = [ D T( 1,0,1)] ( '(1)⎛) 2⎞u − r = − ⎜ +π ⎟= −10ds dt22 +π ⎝ ⎠Therefore, the temperature is decreasing at about 10°C per second when the bee is at (–1, 0, 1); i.e., whent = 1 s.30. a.3 4Du f = ,– ⋅ fx⋅ fy= –6, so5 53 fx–4 f y = –30.4 3Dv f = , ⋅ fx, fy= 17, so5 54fx+ 3fy= 85.The simultaneous solution isf = 10, f = 15, so ∇ f = 10, 15 .xyInstructor's Resource Manual Section 12.5 765

31.b. Without loss of generality, letu = i and v = j. If θ and φ are the anglesbetween u and ∇ f , and between v and ∇ f ,then:π1. θ + φ = (if ∇ f is in the 1st quadrant).2π2. θ = + φ (if ∇ f is in the 2nd quadrant).23π3. φ + θ = (if ∇ f is in the 3rd quadrant).2π4. φ = + θ (if ∇ f is in the 4th quadrant).2In each case cos φ = sin θ or cos φ = –sin θ,2 2so cos φ = sin θ.Thus,2 2 2 2( Du f) + ( Dv f) = ( u⋅∇ f) + ( v ⋅∇f)2 2 2 2=∇ f cos θ +∇ f cos φ2 2 2=∇ +f (cos θ cos φ)2 2 2 2=∇ f + =∇ fcos θ sin θ .Therefore, f is not differentiable at the origin. ButDu f(0, 0) exists for all u sincef(0 + h, 0) – f(0, 0) 0 – 0fx (0, 0) = lim = limh→0 hh→0h= lim (0) = 0, andh→0f(0, 0 + h) – f(0, 0) 0 – 0f y (0, 0) = lim = limh→0 hh→0h= lim (0) = 0, so ∇ f (0, 0) = 0, 0 =0 . Thenh→0Du f(0, 0) = ∇f(0, 0) ⋅ u = 0⋅ u = 0.33. Leave: (–0.1, –5)2 2x – ya. A′ (100, 120)b. B′ (190, 25)34. Leave (–2, –5)c.20 – 30 1fx( C) ≈ = – ; fy( D) = 0;230 – 200 340 – 30 2Duf( E)≈ =25 532. Graph of domain of f⎧0, in shaded region⎫f( x, y)= ⎨ ⎬⎩1, elsewhere ⎭3x – x29– ylim f ( x, y)does not exist since( x, y) →(0, 0)( x, y) → (0, 0):along the y-axis, f(x, y) = 0, but along thecurve, f(x, y) = 1.4y = x35. Leave: (3, 5)36. (4.2, 4.2)766 Section 12.5 Instructor’s Resource Manual

12.6 Concepts Review1.∂zdx∂ zdy+∂ x dt ∂ y dt2. 2y cost+2 xy(– sin t)3 2= cos t – 2sin tcost3.∂ z ∂ x z y+∂ ∂∂ x ∂t ∂ y ∂t4. 12Problem Set 12.61.2.dw 3 2 2 2(2 xy )(3 t ) (3 x y )(2 t)dt = +9 2 10 11= (2 t )(3 t ) + (3 t )(2 t) = 12tdw2 2(2 xy– y )(– sin t) ( x – 2 xy)(cos t)dt = += (sin t+ cos t)(1 − 3sin tcos t)dw3. ( e x sin y e y cos x)(3) ( e x cos y e y sin x)(2)dt = + + + 3t 2t 3t 2t= 3e sin2t+ 3e cos3t+ 2e cos2t+2e sin3t4.dw ⎛1⎞2 ⎛ 1⎞2= ⎜ ⎟sec t+⎜– ⎟(2sec ttan t)dt ⎝ x ⎠ ⎝ y ⎠2 2 2 2sec sec – 2 tan 1– tan= t –2tant= t t =ttan t tan t tan t5.dw 2 2 2 2 2 2[ yz (cos( xyz )](3 t ) [ xz cos( xyz )](2 t) [2xyz cos( xyz )](1)dt = + +22 2 2 6 6 6 76 7= (3yz t + 2xz t + 2 xyz)cos( xyz ) = (3t + 2t + 2 t )cos( t ) = 7tcos( t )dw6. ( y z)(2 t) ( x z)( 2 t) ( y x)( 1)dt = + + + − + + − 2 2= 2 t(2– t – t )–2 t(1– t+t )–(1)3= –4t+ 2 t –17.8.9.10.∂ w= (2 xy)( s) + ( x )(–1) = 2 sts ( – ts ) – st∂ t2 2 2= s 2 t(2 s– 3 t)∂ w –1 –2 2 2 ⎡ –3 ⎛= (2 x – x y)(– st ) + (– ln x)( s ) s 1–2 t –ln s ⎞⎤=∂ t⎢ ⎜ ⎟t⎥⎣ ⎝ ⎠⎦w x2y2x2y2e+ x s t e+= + y s∂2 2(2 )( cos ) (2 )(sin ) 2 x + y= e ( xscost+ysin s)∂ t2 2 2 2 2 2= 2( s sin tcost+ tsin s)exp( s sin t+t sin s)1 1 1 1[( ) ( ) ]( sst ( + 1)∂ w− − − − st 2 e ( st –1)= x + y − x− y e ) + [( x+ y) + ( x−y) ]( se ) =∂ t2 2s2stt e – eInstructor's Resource Manual Section 12.6 767

11.∂ w x(– ssin st) y( scos st) z( s )= + +∂ t ( x + y + z ) ( x + y + z ) ( x + y + z )2 2 2 1/2 2 2 2 1/2 2 2 2 1/224 4 2 –1/2= st(1 + st)∂ w12. ( xy + z )(1) ( xy +e y e z x)( 1) ( e xy += + − +z )(2 t)∂ txy+zs2= e ( y− x+ 2 t) = e (0) = 013.∂ z2(2 xy)(2) ( x )(–2 st)∂ t= + 2 2 ⎛∂z ⎞= 4(2 t + s)(1 −st ) − 2 st(2 t + s) ; ⎜ ⎟ = 72⎝ ∂ t ⎠(1, −2)∂ z14. ( y 1)(1) ( x 1)( rt)∂ s= + + + ⎛∂z ⎞= 1 + rt(1+ 2 s+ r+ t); ⎜ ⎟ = 5⎝∂s ⎠(1, –1, 2)15.16.dw2(2 u– tan v)(1) (– usec v)( )dx = + π 2= 2 x –tan πx– πxsecπ xdw ⎛1⎞ ⎛π ⎞ 1+π= –1– – –2.0708dx 1 ⎜ ⎟ ⎜ ⎟= ≈x = ⎝2⎠ ⎝2⎠24∂w2(2 xy)(– ρsinθsin φ) ( x )( ρcosθsin φ) (2 z)(0)∂θ = + + 3 3 2 2= ρ cosθsin φ(–2sin θ + cos θ);⎛∂ w ⎞⎜ ⎟⎝ ∂θ ⎠(2, π, π)2= –817.2 drV(, r h) r h, 0.5 dtdh8=π = in./yr,19. The stream carries the boat along at 2 ft/s withrespect to the boy.dt = in./yrdV 2= (2 π rh) ⎛ ⎜ dr ⎞ ⎟+ ( πr) ⎛ ⎜ dh ⎞⎟;dt ⎝ dt ⎠ ⎝ dt ⎠⎛dV⎞⎜ ⎟ = 11200πin. 3 /yr⎝ dt ⎠(20, 400)311200 π in. 1 board ft= × ≈ 244.35 board ft/yr1 yr 3144 in.– x–3y18. Let T = e .dT – x –3 y dx – –3(–1) x y dy= e+ e (–3)dt dt dt– x –3 y – –3(–1)(2) x y– –3= e+ e (–3)(2) = –8ex ydT= –8, so the temperature is decreasingdt (0, 0)at 8°/min.20.dx dy 2 2 2= 2, = 4, s = x + ydt dt⎛ds ⎞ ⎛dx ⎞ ⎛dy⎞2s⎜ ⎟= 2x⎜ ⎟+2y⎜ ⎟⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠ds (2x + 4 y)=dt sWhen t = 3, x = 6, y = 12, s = 6 5. Thus,⎛ds⎞⎜ ⎟ = 20 ≈ 4.47 ft/s⎝ dt ⎠t=3⎛1⎞ 2 dhV(, r h) = ⎜ ⎟π r h, = 3⎝3⎠ dtdr2in./min,dt = in./mindV 2 1 2= ⎛ ⎜ ⎞ ⎟π rh⎛ ⎜ dr ⎞ ⎟+ ⎛ ⎜ ⎞ ⎟πr⎛ ⎜ dh ⎞⎟;dt ⎝3⎠ ⎝ dt ⎠ ⎝3⎠ ⎝ dt ⎠⎛⎜⎝dVdt⎞ 20,800π⎟ = ≈ 21,782⎠(40, 100) 3in. 3 /min768 Section 12.6 Instructor’s Resource Manual

21. LetThen22. Let3 2 3F( x, y) = x + 2 x y– y = 0.F 2 2dy − ∂∂x(3x + 4 xy) 3x + 4xy= = − = .dx ∂F2 2 2 2∂y2x −3y 3y −2x–F( x, y) = ye x + 5 x–17=0.–dy (– yex + 5) x= – = y–5edx – xe23. Let F(x, y) = x sin y + y cos x = 0.dy (sin – sin ) sin – sin= –y y x =y x ydx x cos y + cos x x cos y + cos x24. LetThen25. Let2 2F( x, y) = x cos y– y sin x = 0.∂ F2dy x –(2x cos y – y cos x)= – ∂=dx ∂ F 2∂ y – x sin y– 2ysinx22x cos y– y cosx=.2x sin y+2ysinx∂∂26. Let2 3 3F( x, y, z) = 3x z + y − xyz = 0.3 3z (6 xz– yz ) yz – 6xz= – =x 32 –32 32 –32x xyz x xyz–f( x, y, z) = ye x + zsinx= 0.∂ x –sinx sinx= =∂ z – x– x– ye + z cos x ye – z cos xT T x T y T z T w27.∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ +∂ ∂∂ s ∂x ∂s ∂ y ∂s ∂ z ∂s ∂w ∂s∂ z28. We use the z r notation for .∂ rz = z x + z y = z cosθ+ z sinθr x r y r x yzθ = zxxθ + zyyθ = zx(– rsin θ ) + zy( rcos θ ), so–1r zθ – zxsinθ zycos θ.= + Thus,2 –2 2 2( zr) + ( r )( z ) = ( zx cos + zysin )2 2( zx) ( zy)θ θ θ2+ (– zxsinθ+ zycos θ )= + (expanding and using2 2sin + cos = 1).θθ129. y ⎛ ⎞= ⎜ ⎟ [ f ( u ) + f ( v )],⎝2⎠where u = x – ct, v = x + ct.1 1yx= ⎜ ⎛ ⎟ ⎞ [ f′ ( u)(1) + f′ ( v)(1)] = ⎜ ⎛ ⎟⎞ [ f′ ( u) + f′( v)]⎝2⎠ ⎝2⎠⎛1 ⎞yxx= ⎜ ⎟ [ f′′ ( u )(1) + f′′( v )(1)]⎝2⎠⎛1 ⎞= ⎜ ⎟ [ f ′′ ( u ) + f′′( v )]⎝2⎠⎛1 ⎞yt= ⎜ ⎟ [ f′ ()(–) u c + f′()()] v c⎝2⎠⎛ c ⎞= ⎜ – ⎟ [ f ′( u)– f′( u)]⎝ 2 ⎠⎛ c ⎞ytt= ⎜− ⎟[ f′′ ( u)( −c) − f′′( v)( c)]⎝ 2 ⎠⎛ 2c ⎞2= [ f ′′( u) + f′′( v)]= c y⎜xx2 ⎟⎝ ⎠30. Let w = f(x, y, z) where x = r – s, y = s – t,z = t – r. Thenw + w + w = ( w x + w x ) + ( w y + w y )r s t x r x s y s y t+ ( wz+wz)z tz r= [ w (1) + w ( − 1)] + [ w (1) + w ( −1)]= 0x x y y+ [ w (1) + w ( −1)]zz31. Let w yx= f ( udu ) = – f ( udu ) ,xyy = h(t).Thendw ∂w dx ∂w dy= + =− f ( xg ) ′() t + f( yh ) ′()tdt ∂x dt ∂y dt= f ( ht ()) h′ ()– t f( gt ()) g′() t .∫ ∫ where x = g(t),Thus, for the particular function given2 4F′ () t = 9 + ( t ) (2) t4+ ( πt) ( π πt)( 2 ) (5)( 2 2 ) –(3)( 2 )– 9 sin 2 2 cos 2 ;F′ = π= 10 2 – 3 2π≈0.8135.32. If f ( tx, ty) = tf ( x, y),thendd[ f ( tx, ty)] = [ tf ( x, y)].dtdtThat is,[ f ( tx, ty)][ x] + [ f ( tx, ty)][ y] = f ( x, y).txLetting t = 1 yields the desired result.tyInstructor's Resource Manual Section 12.6 769

33.2 2c = a + b 2 – 2abcos40° (Law of Cosines) where a, b, and c are functions of t.aa′ + bb′ – ( a′ b + ab′)cos40 °2cc′ = 2aa′ + 2 bb′ – 2( a′ b + ab′)cos 40° so c′ =.c2 2 2When a = 200 and b = 150, c = (200) + (150) – 2(200)(150)cos 40°= 62,500 – 60,000 cos 40°.It is given that a′ = 450 and b′ = 400, so at that instant,(200)(450) + (150)(400) − [(450)(150) + (200)(400)]cos 40°c′ = ≈ 288 .62,500 – 60,000cos 40°Thus, the distance between the airplanes is increasing at about 288 mph.234. r = x, y,z , so r 2 = r = x 2 + y 2 + z2 .GMmF =2 2 2x + y + z, soF′ () t = Fmm′ () t + Fxx′ () t + Fyy′ () t + Fzz′() tGMm′ () t 2 GMmxx′() t=–2 2 2 2 2 2 2x + y + z ( x + y + z )2 GMmyy′ ( t) 2 GMmzz′( t)– +(2 2 2 )2 (2 2 2 )2x + y + z x + y + z2 2 2GM[( x + y + z ) m′ ()–2 t m( xx′ () t + yy′ () t + zz′()] t=.2 2 2 2( x + y + z )12.7 Concepts Review1. perpendicular2. 3,1, − 13. x−+ 4( y− 1) + 6( z− 1) = 04.∂ f ∂ fdx + dy∂x∂ yProblem Set 12.71. ∇ F( x, y, z) = 2 x, y, z ;( )∇ F 2, 3, 3 = 2 2, 3, 3Tangent Plane:2( x– 2) + 3( y– 3) + 3 z– 3 = 0, or2x+ 3y+ 3z= 16( )2. ∇ F( x, y, z) = 2 8 x, y, 8 z ;2∇ F ⎛1, 2, ⎞⎜= 4 4, 1, 2 22 ⎟⎝ ⎠Tangent Plane:⎛ 2 ⎞4( x–1) + 1( y–2) + 2 2 ⎜z– ,2 ⎟or⎝ ⎠3. Let4x+ y+ 2 2z= 8.2 2 2F( x, y, z) = x – y + z + 1=0.∇ F( x, y, z) = 2 x, –2 y, 2z = 2 x, – y,z( )∇ F 1, 3, 7 = 2 1, – 3, 7 , so 1, – 3, 7 isnormal to the surface at the point. Then thetangent plane is1( x– 1) – 3( y– 3) + 7 z– 7 = 0, or moresimply, x− 3y+ 7 z = –1.( )770 Section 12.7 Instructor’s Resource Manual

4. ∇ f ( x, y, z) = 2 x, y, – z ;5.∇ f (2, 1, 1) = 2 2, 1, − 1Tangent plane:2( x− 2) + 1( y−1) −1( z− 1) = 0, or 2x+ y− z = 4 .⎛1⎞∇ f( x, y) = ⎜ ⎟ x, y ; ∇ f(2, 2) = 1,1⎝2⎠Tangent plane: z− 2= 1( x− 2) + 1( y− 2) , orx+ y− z = 2 .6. Let7.8.–2 yf( x, y) = xe .−2y−2y∇ f( x, y) = e , − 2xe∇ f (1, 0) = 1, − 2Then 1, – 2, – 1 is normal to the surface at(1, 0, 1), and the tangent plane is1( x−1) −2( y−0) −1( z− 1) = 0 , or x− 2y− z = 0.3y3y∇ f ( x, y) = –4e sin2 x, 6e cos2 x ;⎛π⎞∇ f ⎜ ,0 ⎟=–2 3,–3⎝3⎠⎛ π ⎞Tangent plane: z+ 1 = –2 3 ⎜x– ⎟–3( y–0),⎝ 3 ⎠( )2 3 π –3or 2 3x+ 3 y+ z = .3⎛1⎞1 1∇ f( x, y) = ⎜ ⎟ , ;⎝2⎠ x y1 1∇ f (1, 4) = ,2 4⎛1⎞ ⎛1⎞Tangent plane: z− 3 = ⎜ ⎟( x− 1) + ⎜ ⎟( y−4),⎝2⎠ ⎝4⎠or 1 x+ 1 y− z =− 3 .2 4 29. Let2 3z = f( x, y) = 2 x y ;3 2 2dz 4xy dx 6 x y dy.= + For the points given,dx =− 0.01, dy = 0.02 ,dz = 4( − 0.01) + 6(0.02) = 0.08.Δ z = f(0.99, 1.02) – f(1, 1)2 3 2 3= 2(0.99) (1.02) −2(1) (1) ≈ 0.0801799210. dz = (2x − 5 y) dx + ( − 5x + 1) dy= ( − 11)(0.03) + ( −9)( − 0.02) = − 0.15Δ z = f(2.03,2.98) − f(2,3) =− 0.146111.12. Let−1 −1 ⎛1⎞dz = 2 x dx + y dy = ( − 1)(0.02) + ⎜ ⎟( −0.04)⎝4⎠= –0.03Δ z = f(–1.98, 3.96) – f(–2.4)2= ln[(–1.98) (3.96)] – ln16 ≈ –0.030151–1z = f( x, y) = tan xy;yxdz = dx + dy;2 2 2 21+ x y 1+x y( −0.5)( − 0.03) + ( −2)( −0.01)= = 0.0175.1 + (4)(0.25)Δ z = f( −2.03, −0.51) − f( −2, −0.5) ≈ 0.01734213. Let2 2F( x, y, z) = x –2 xy– y –8x+ 4 y– z = 0;∇ F( x, y, z) = 2 x–2 y–8, –2 x–2y+4, –1Tangent plane is horizontal if ∇ F = 0, 0, k forany k ≠ 0 .2x – 2y – 8 = 0 and –2x – 2y + 4 = 0 if x = 3 andy = –1. Then z = –14. There is a horizontaltangent plane at (3, –1, –14).14. 8, −3, − 1 is normal to 8x – 3y – z = 0.F xyz , , = 2x + 3y − z.15. ForLet ( )2 2∇ F( x, y, z) = 4 x, 6 y, –1 is normal to2 2z 2x 3y= + at (x, y, z). 4x = 8 and16y = –3, if x = 2 and y = – ; then2⎛ 1 ⎞z = 8.75 at ⎜2, – , 8.75 ⎟.⎝ 2 ⎠2 2F( x, y, z) = x + 4y+ z = 0,∇ F( x, y, z) = 2 x, 4, 2z = 2 x, 2, z .F(0, –1, 2) = 0, and∇ F(0,–1,2) = 20,2,2 = 40,1,1.2 2 2For Gx ( , y, z) = x + y + z –6z+ 7 = 0,∇ Gx ( , y, z) = 2 x, 2 y, 2 z–6 = 2 x, y, z–3 .G(0, –1, 2) = 0, and∇ G(0, – 1, 2) = 2 0, – 1, – 1 = –2 0, 1, 1 .0, 1, 1 is normal to both surfaces at(0, –1, 2) so the surfaces have the same tangentplane; hence, they are tangent to each other at(0, –1, 2).Instructor’s Resource Manual Section 12.7 771

16. (1, 1, 1) satisfies each equation, so the surfacesintersect at (1, 1, 1). For2 2z = f( x, y) = x y: ∇ f( x, y) = 2 xy, x ;∇ f (1,1) = 2,1 , so 2, 1, –1 is normal at(1, 1, 1).2For f( x, y, z) = x − 4y+ 3=0;∇ f( x, y, z) = 2, –4, 0 ;∇ f (1, 1, 1) = 2, – 4, 0 so 2, – 4, 0 is normal at(1, 1, 1).2, 1, –1 ⋅ 2, – 4, 0 = 0, so the normals, hencetangent planes, and hence the surfaces, areperpendicular at (1, 1, 1).2 2 2F( x, y, z) = x + 2y + 3 z –12=0;17. Let∇ F( x, y, z) = 2 x, 2 y, 3zis normal to theplane.A vector in the direction of the line,2, 8, – 6 = 2 1, 4, – 3 , is normal to the plane.x,2 y,3 z = k 1,4,-3 and (x, y, z) is on thesurface for points (1, 2, –1) [when k =1] and(–1, –2, 1) [when k = –1].2 2 218. Let F( x, y, z) = x + y + z = 1.2 2 2a b c2x 2y 2z∇ F( x, y, z) = , ,2 2 2a b cx0 y0 z0∇ F( x0, y0, z0) = 2 , ,2 2 2a b cThe tangent plane at ( x0, y0, z 0)isx0( x– x0) y0( y– y0) z0( z– z0)+ + = 0.2 2 2a b c2 2 2xx 0 yy 0 zz ⎛0 x0 y0 z ⎞0+ + – + + = 02 2 2 2 2 2a b c ⎜a b c ⎟⎝⎠xx 0 yy 0 zz 0Therefore, + + = 1, since2 2 2a b c2 2 2x0 y0 z0+ + = 1.2 2 2a b c19. ∇ f ( x, y, z) = 2 9 x, 4 y, 4 z ;∇ f (1, 2, 2) = 2 9, 8, 8∇ g( x, y, z) = 2 2 x, – y, 3 z ;∇ f (1,2,2) = 41,–1,39, 8, 8 × 1, –1, 3 = 32, –19, –17Line: x = 1+ 32 t, y = 2-19 t, z = 2-17t20. Letf ( x, y, z) = x– z , and∇ f ( x, y, z) = 1, 0, − 2zand∇ g( x, y, z) = 0,1, –3z∇ f (1,1,1) = 1,0, –2 and∇ g(1,1,1) = 0,1, –31,0, − 2 × 0,1, − 3 = 2,3,1Line: x = 1+ 2 t, y = 1+ 3 t, z = 1+t22g( x, y, z) = y– z .21. dS = SAdA + SWdWW A − WdA+AdW=− dA + dW =2 2 2( A−W) ( A−W) ( A−W)At W = 20, A = 36:− 20dA + 36dW − 5dA + 9dWdS = =.256 645 dA + 9 dW 5(0.02) + 9(0.02)Thus, dS ≤ ≤= 0.00437564 641 122. V = lwh, dl = dw = , dh = , l = 72, w = 48,2 4h = 36dV = whdl + lhdw + lwdh = 3024 in. 3 (1.75 ft 3 )23.2 2V =π r h, dV = 2π rhdr+πr dh2 2dV ≤2π rh dr +πr dh ≤2 π rh(0.02 r) +πr (0.03 h)2 2= 0.04π r h+ 0.03π r h=0.07VMaximum error in V is 7%.24. T = f( L, g) = 2πdT = f dL + f dgLgLg⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟1 ⎛ 1⎞1 ⎛ L ⎞= 2π ⎜ ⎟ dL + 2π⎜ ⎟−dg⎜ L ⎟⎜ ⎟gL2g2⎝ ⎠⎜ ⎟⎜ ⎟2 ⎝ ⎠⎜ g ⎟ ⎜ g ⎟⎝ ⎠ ⎝ ⎠π( gdL – Ldg) = , so2 LggdT π( gdL – Ldg) gdL – Ldg= =T ⎛ L ⎞⎛ 2 L ⎞ 2gL⎜2πgg ⎟⎜ g ⎟⎝ ⎠⎝ ⎠1 ⎛dLdg ⎞= ⎜ – ⎟.2 ⎝ L g ⎠Therefore,dT 1⎛dL dg ⎞ 1≤ ⎜ + ⎟= (0.5% + 0.3%) = 0.4%.T 2⎝L g ⎠ 23772 Section 12.7 Instructor’s Resource Manual

25. Solving for R, R RR= 1 2R + R∂∂2R R= 2R 21 ( R1+R2)andso,1 22∂ R R= 1∂ R 22 ( R1+R2)2 22 1 1 2R dR + R dRTherefore, dR =;2( R + R )26. Let1 22 22 1 + 1 22( R1+R2)R dR R dRdR ≤.Then at R 1 = 25,(25)(100)R2 = 100, dR1 = dR2= 0.5, R = = 2025 + 1002 2(100) (0.5) + (25) (0.5)and dR ≤ = 0.34.2(125)2 2 2F( x, y, z) = x + y + 2 z .∇ F( x, y, z) = 2 x, 2 y, 4 z ;∇ F(1, 2, 1) = 2 1, 2, 2 ;∇ F =∇F1 2 2, ,3 3 31 2 2Thus, u = , , is the unit vector in the3 3 3direction of flight, and1 2 2x, y, z = 1, 2,1 + 4 t , , is the location3 3 3of the bee along its line of flight t seconds aftertakeoff. Using the parametric form of the line offlight to substitute into the equation of the planeyields t = 3 as the time of intersection with theplane. Then substituting this value of t into theequation of the line yields x = 5, y = 10, z = 9 sothe point of intersection is (5, 10, 9)..27. Let F(x, y, z) = xyz = k; let (a, b, c) be any pointon the surface of F.k k k∇ F( x, y, z) = yz, xz, xy = , ,x y z1 1 1= k , , x y z1 1 1∇ Fabc ( , , ) = k , , a b cAn equation of the tangent plane at the point is⎛1⎞ ⎛1⎞ ⎛1⎞⎜ ⎟( x– a) + ⎜ ⎟( x– b) + ⎜ ⎟( x– c) = 0, or⎝a⎠ ⎝b⎠ ⎝c⎠x y z+ + = 3.a b cPoints of intersection of the tangent plane on thecoordinate axes are (3a, 0, 0), (0, 3b, 0), and(0, 0, 3c).The volume of the tetrahedron is⎛ 1 1 1⎜ ⎞ ⎟(area of base)(altitude)= ⎛ ⎜ 3 3 ( 3 )3 3 2 a b ⎞⎟⎝ ⎠ ⎝ ⎠c9 abc 9 k= = (a constant).2 228. If F( x, y, z) = x + y + z,then1 1 10.5 , , – , – , – 0,x y z0 0 0Intercepts are 0, 0, 0;1 1 1∇ F( x, y, z) = 0.5 , , . The equation of the tangent isx y zx y z+ + = x + y + z = a.⋅ x x0 y y0 z z0= or 0 0 0x0 y0 z02a x a y a z so the sum is a( x + y + z ) = a0 0 0 .Instructor’s Resource Manual Section 12.7 773

29.2 2 1/2f( x, y) = ( x + y ) ; f(3, 4) = 52 2 –1/2 3x2 2 –1/2 4fx( x, y) = x( x + y ) ; f (3, 4) = = 0.6;fy= ( x, y) = y( x + y ) ; fx(3, 4) = = 0.8552 2 2 –3/2162 2 –3/2fxx ( x, y) = y ( x + y ) ; f x (3, 4) = = 0.128 ; fxy ( x, y) = – xy( x + y ) ;12512f xy (3, 4) = – = –0.0961252 2 2 –3/29fyy= x ( x + y ) ; f xx (3, 4) = = 0.072125Therefore, the second order Taylor approximation is2 2f( x, y) = 5 + 0.6( x– 3) + 0.8( y– 4) + 0.5[0.128( x– 3) + 2(–0.096)( x– 3)( y– 4) + 0.072( y– 4) ]a. First order Taylor approximation: f( x, y) = 5 + 0.6( x- 3) + 0.8( y- 4) .Thus, f (3.1,3.9) ≈ 5 + 0.6(0.1) + 0.8(-0.1) = 4.98 .b.2 2f (3.1,3.9) ≈ 5 + 0.6(–0.1) + 0.8(0.1) + 0.5[0.128(0.1) + 2(–0.096)(0.1)(–0.1) + 0.072(–0.1) ] = 4.98196c. f (3.1, 3.9) ≈ 4.981967530.⎛ 2 2x + y ⎞f( x, y) = tan ;⎜ 64 ⎟⎝ ⎠f(0,0) = 02 2x ⎛2 x + y ⎞fx( x, y) = ⋅ sec ;32 ⎜ 64 ⎟⎝ ⎠fx(0,0) = 02 2y ⎛2 x + y ⎞fy( x, y) = ⋅ sec ;32 ⎜ 64 ⎟⎝ ⎠fy(0,0) = 02 2 2 2 2 2 22x ⎛2 x + y ⎞ ⎛ x + y ⎞ 1 ⎛2 x + y ⎞1fxx( x, y) = sec tan + sec ; f (0,0)2xx =32 ⎜ 64 ⎟ ⎜ 64 ⎟ 32 ⎜ 64 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠322 2 2 2 2 2 22y ⎛2 x + y ⎞ ⎛ x + y ⎞ 1 ⎛2 x + y ⎞1fyy( x, y) = sec tan + sec ; f (0,0)2yy =32 ⎜ 64 ⎟ ⎜ 64 ⎟ 32 ⎜ 64 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠32fxyxy , will contain either x or y, resulting in xy ( 0,0)0When computed, each term of ( )second-order Taylor approximation is1 1 1 1 1f ( xy , ) 0 0 x 0 y ⎡⎤= + ⋅ + ⋅ + x 2 0 x y y x y2⎢+ ⋅ ⋅ ⋅ +32 32 ⎥ = +⎣⎦ 64 642 2 2 2f = . Therefore, thea. The first-order Taylor approximation is f( x, y) = 0 + 0⋅ x+ 0⋅ y = 0; Thus, f(0.2, −0.3) ≈ 0.b.1 2 1 3f (0.2, −0.3) ≈ (0.2) + ( − 0.3) = 0.0020312564 64c. f (0.2, −0.3) ≈ 0.0020312528774 Section 12.7 Instructor’s Resource Manual

12.8 Concepts Review1. closed bounded2. boundary; stationary; singular3. f( x0, y0)∇ =04.2fxx ( x0, y0) fyy ( x0, y0)– fxy( x0, y 0)Problem Set 12.81. ∇ f( x, y) = 2 x–4, 8y= 0, 0 at (2, 0), a stationary point.2 2D fxx fyy – fxy(2)(8) – (0) 16 0= = = > and f = 2> 0. Local minimum at (2, 0).2. ∇ f( x, y) = 2x− 2, 8y+ 8 = 0, 0 at (1, –1), a stationary point.f xx = 2> 0. Local minimum at (1, –1).xx2D = fxx fyy – fxy= (2)(8) – (0)2 = 16 > 0 and3.3 2∇ f ( x, y) = 8 x –2 x, 6y = 2 x(4 x –1), 6y= 0,0 , at (0,0),(0.5,0),( − 0.5,0) all stationary points.fxx2= 24 x – 2;2 2 2 2D = fxx fyy – fxy= (24 x – 2)(6) – (0) = 12(12 x –1).At (0,0) : D =− 12 , so (0, 0) is a saddle point.− 0.5,0 : D = 24 and f = 6 , so local minima occur at these points.At ( 0.5,0 ) and ( )xx4.2f( x, y) y –12 x, 2 xy–6y0, 0∇ = = at stationary points (0,0),(3, − 6) and (3,6).2 2 2D = fxx fyy – fxy= (–12)(2 x– 6) – (2 y)= –4( y + 6 x–18), f xx = –12At (0, 0): D = 72, and f xx = –12, so local maximum at (0, 0).At (3, ± 6) : D = − 144, so (3, ± y)are saddle points.5. ∇ f( x, y) = y, x = 0, 0 at (0, 0), a stationary point.2 2xx yy xyD = f f – f = (0)(0) – (1) = –1, so (0, 0) is a saddle point.6. Let2 2∇ f( x, y) = 3 x –6 y, 3 y –6x= 0, 0 . Then23 x –6y = 0 and23y− 6x= 0.2 2 2 4 2 1 4 23x − 6y = 0→ 3x = 6y → x = 2y → x = 4y → x = y42 ⎛1 4⎞3 4 3 3 323y − 6x = 0→3⎜x ⎟− 6x = 0→ x − 6x = 0→ x( x − 8) = 0→ x( x− 2)( x + 2x+ 4)= 0→ x= 0, x=2⎝4 ⎠ 4 4 42x = 0: 3x − 6y = 0→3( 0)− 6y = 0→− 6y = 0→ y = 02x = 2 : 3x − 6y = 0 →3( 2) 2−6y →12 − 6y = 0 → 12 = 6y → 2 = ySolving simultaneously, we obtain the solutions (0, 0) and (2, 2).2 2fxx= 6 x;D = fxx fyy – fxy= (6 x)(6 y) – (–6) = 36 ( xy–1); At (0, 0): D = − 36 < 0 , so (0, 0) is a saddle point.At (2, 2): D = 108 > 0 , f xx > 0, so a local minimum occurs at ( 2, 2 ) .Instructor’s Resource Manual Section 12.8 775

7.8.2 2x y−2 xy −4f( x, y) , 0, 02 2x y∇ = = at (1, 2).2 2xx yy – xy (4 – 3)(8 – 3) – (1)D = f f f = x y−3 −3 −3= 32x y − 1, fxx= 4xAt (1, 2): D = 3> 0, and f xx > 0, so a local minimum at (1, 2).2 2f( x, y) –2exp(– x – y 4 y) x, y–2∇ = + = 0, 0 at (0,2).2D = fxx fyy – fxy2 2 2fxx= (4 x – 2)exp(– x – y + 4 y)At (0,2) : D > 0 , and xx 0,2 2 2 2 2= exp 2( −x − y + 4 y)[(4x −2)(4y − 16y+ 14) −(4xy−8 x) ],f < so local maximum at (0, 2).9. Let ∇ f ( x, y)= – sin x –sin( x+ y), –sin y–sin( x+ y)= 0, 0⎛– sin x– sin( x+ y) = 0 ⎞Then ⎜⎟ . Therefore,⎝ sin y+ sin( x+ y) = 0 ⎠πsin x = sin y , so x = y = . However, these values4satisfy neither equation. Therefore, the gradient isdefined but never zero in its domain, and theboundary of the domain is outside the domain, sothere are no critical points.10. ∇ f( x, y) = 2 x–2acos y, 2axsin y = 0, 0 at⎛ π ⎞⎜0, ± ⎟, ( a, 0)⎝ 2 ⎠2 2D = fxx fyy – fxy= (2)(2axcos y) – (2asin y) ,f xx = 2⎛ π ⎞2 ⎛ π ⎞At ⎜0, ± ⎟:D = –4a< 0, so ⎜0,± ⎟⎝ 2 ⎠⎝ 2 ⎠ are2saddle points. At (a, 0): D = 4a> 0 andf xx > 0, so local minimum at ( a ,0).11. We do not need to use calculus for this one. 3x isminimum at 0 and 4y is minimum at –1. (0, –1) isin S, so 3x + 4y is minimum at (0, –1); theminimum value is –4. Similarly, 3x and 4y areeach maximum at 1. (1, 1) is in S, so 3x + 4y ismaximum at (1, 1); the maximum value is 7. (Usecalculus techniques and compare.)12. We do not need to use calculus for this one. Each2 2of x and y is minimum at 0 and (0, 0) is in S,2 2so x + y is minimum at (0, 0); the minimum2 2value is 0. Similarly, x and y are maximum atx = 3 and y = 4, respectively, and (3, 4) is in S, so2 2x y+ is maximum at (3, 4); the maximumvalue is 25. (Use calculus techniques andcompare.)13. ∇ f( x, y) = 2 x, –2y= 0, 0 at (0, 0).2xx yy xyD = f f – f = (2)(–2) – (0)2 < 0, so (0, 0) isa saddle point. A parametric representation of theboundary of S is x = cos t, y = sin t, t in[0,2 π ].2 2f ( x, y) = f( x( t), y( t)) = cos t –sin t+1= cos 2 t –1cos 2 t –1is maximum if cos 2t = 1, which occursfor t = 0, π , 2π . The points of the curve are( ± 1,0). f ( ± 1,0) = 2f(x, y) = cos 2 t –1 is minimum if cos 2t = –1,π 3πwhich occurs for t = , . The points of the2 2curve are (0, ± 1). f (0, ± 1) = 0 . Global minimumof 0 at (0, ± 1); global maximum of 2 at (±1, 0).14. ∇ f( x, y) = 2x−6, 2 y− 8 = 0,0 at (3, 4),which is outside S, so there are no stationarypoints. There are also no singular points.x = cos t,y = sin t , t in [0,2 π ] is a parametricrepresentation of the boundary of S.f ( x, y) = f( x( t), y( t))2 2= cos t – 6cost+ sin t – 8sin t+7= 8−6cost− 8sin t = F( t)4F′ () t = 6sin t –8cost= 0iftan t = . t can be3in the 1st or 3rd quadrants. The corresponding⎛ 3 4⎞points of the curve are ⎜± , ± ⎟.⎝ 5 5⎠⎛ 3 4⎞ ⎛3 4⎞f ⎜– , – ⎟= 18; f ⎜ , ⎟ = –2⎝ 5 5⎠ ⎝5 5⎠⎛3 4⎞Global minimum of –2 at ⎜ , ⎟;global⎝5 5⎠⎛ 3 4⎞maximum of 18 at ⎜– , – ⎟.⎝ 5 5⎠776 Section 12.8 Instructor’s Resource Manual

15. Let x, y, z denote the numbers, so x + y + z = N.Maximize2 2P = xyz = xy( N – x – y) = Nxy – x y – xy .2 2Let ∇ Px ( , y) = Ny–2 xy– y, Nx– x –2xy= 0, 0 .⎛2Ny – 2 xy – y = 0 ⎞Then.⎜ 2Nx – x –2xy= 0⎟⎝⎠2 2N( x, − y) = x − y = ( x+ y)( x− y).x yN = x+y.= orTherefore, x = y (since N = x + y would mean thatP = 0, certainly not a maximum value).2Then, substituting into Nx – x –2xy = 0, we2 2obtain Nx – x –2x = 0, from which we obtainx(N – 3x) = 0, soPxx= –2 y;2D = PxxPyy – Pxy2= ( −2 y)( −2 x) −( N −2x−2 y)2= 4 xy −( N −2x −2 y)AtNx = (since x = 0 ⇒ P = 0).32N N 2Nx = y = : D = > 0, Pxx= – < 0 (so3 3 3local maximum)N NIf x = y = , then z = .3 3NConclusion: Each number is . (If the intent is3to find three distinct numbers, then there is nomaximum value of P that satisfies thatcondition.)16. Let s be the distance from the origin to (x, y, z) on2 2 2 2the plane. s = x + y + z andx+ 2y+ 3z= 12. Minimize2 2 2 2s = f( y, z) = (12–2 y–3 z) + y + z .∇ f ( y, z) = –48 + 12x+ 10 y, – 72 + 12y+20z⎛12 18 ⎞= 0, 0 at ⎜ , ⎟.⎝ 7 7 ⎠2= yy zz – yz = 56 > 0 and f yy = 10 > 0;D f f f⎛local maximum at ⎜⎝12 18,7 7⎞⎟⎠2 504s = , so the shortest distance is496 14s = ≈ 3.2071.717. Let S denote the surface area of the box withdimensions x, y, z.S = 2xy + 2xz + 2yz and V0 = xyz,so–1 –1S = 2( xy+ V0y + V0x).–1 –1Minimize f ( x, y)= xy+ V0y + V0xsubjectto x > 0, y > 0.–2 –2∇ f( x, y) = y– V0x , x– V0y= 0, 0 at1/3 1/3( V0 , V 0 ).D = fxx fyy 2fxy=2 –3 –3V0x yfxx–3= 2 Vx 0 .1/3 1/30 , 0 :At ( )– 4 –1,V V D = 3 > 0, f xx = 2> 0, solocal minimum.Conclusion: The box is a cube with edge1/3V 0 .18. Let L denote the sum of edge lengths for a box ofdimensions x, y, z. Minimize L = 4x + 4y + 4z,subject to V0 = xyz.4 0Lx ( , y) = 4x+ 4 y+ V , x> 0, y>0xyLet−1 −1 −1 2 −1 20 0∇ Lx ( , y) = 4 x y x ( x y−V), y ( xy −V)= 0,0 .2Then x y = V 0 andfollows that x = y. Therefore8V0Lxx= ;3x y2xy = V 0 , from which it1/3x = y = z = V 0 .22 ⎛8V ⎞⎛0 8V ⎞ ⎛0 4V⎞0D = LxxLyy − Lxy= −⎜⎟⎜3 3 2 2x y⎟⎜ xy⎟ ⎜x y⎟⎝ ⎠⎝ ⎠ ⎝ ⎠1/3 1/3V0 , V 0 : D > 0, L xx > 0 (so localAt ( )minimum).There are no other critical points, and as(x, y) → boundary, L → ∞ . Hence, the optimal1/3box is a cube with edge V 0 .19. Let S denote the area of the sides and bottom ofthe tank with base l by w and depth h.S = lw + 2lh + 2wh and lwh = 256.⎛256 ⎞ ⎛256⎞Sl (, w) = lw+ 2l⎜ ⎟+2 w⎜ ⎟,w > 0, l > 0.⎝ lw ⎠ ⎝ lw ⎠–2 –2Slw ( ) = w– 5121 , l– 512w= 0, 0 at(8, 8). h = 4 there. At (8, 8) D > 0 and 11 0, S >so local minimum. Dimensions are 8’ × 8’ × 4’.Instructor’s Resource Manual Section 12.8 777

20. Let V denote the volume of the box and (x, y, z) denote its 1st octant vertex.2 2 2V = (2x)(2y)(2z) = 8xyz and 24x + y + z = 9.2 1 2 2 2 2V 64 ⎡ ⎛ ⎞⎤= ⎢⎜ ⎟(9 – y – z ) y z24⎥⎣⎝⎠⎦2 2 2 2Maximize f ( y, z) = (9– y – z ) y z , y > 0,z > 0.2 2 2 2 2 2∇ = 0, 0f ( y, z) 2 yz (9–2 y – z ), y z(9– y –2zAt ( 3, 3 ),⎛ 2 ⎞⎜ ⎟( )( )8⎜3 3 = 6 2.4 ⎟⎝ ⎠2yy zz yz= at ( 3, 3).D = f f – f > 0 and f < 0, so local maximum. The greatest possible volume is21. Let x, y, z denote the vector; let S be the sum of its components.2 2 22 2 1/2x + y + z = 81, so z = (81 – x – y ) .2 2 1/2 2 2S( x, y) = x+ y+(81– x – y ) , 0≤ x + y ≤ 9.2 2 –1/2 2 2 –1/2∇ S( x, y) = 1– x(81– x – y ) , 1– y(81– x – y ) = 0, 0 .MaximizeLet2 2 1/2yy2 2 1/2Therefore, x = (81 – x – y ) and y = (81– x – y ) . We then obtain x = y = 3 3 as the only stationarypoint. For these values of x and y, z = 3 3 and S = 9 3 ≈ 15.59.The boundary needs to be checked. It is fairly easy to check each edge of the boundary separately. The largestvalue of S at a boundary point occurs at three places and turns out to be 18 12.73.2 ≈Conclusion: the vector is 3 3 1,1,1 .22. Let Pxxz ( , , ) be any point in the plane 2x+ 4y+ 3z= 12. The square of the distance between the origin and P isd2 = x2 + y2 + z2 . Consequently,2 2 2 2d = f( x, y) = x + y + (12−2x− 4 y) 9. To find the critical points, setf 2x ( x, y) = 2 x+ (12−2x−4 y)( − 2) = 0 and f 29y ( x, y) = 2 y+ (12−2x−4 y)( − 4) = 0The resulting system of9equations is 13x+ 8y= 24 and 8x+ 25y= 48, which leads to a critical point of24 48( ,29 29 ). Since f26xx ( x , y ) =9,f50yy ( x , y ) =9, and f16xy ( x , y ) =9, D 24 48 116( ,29 29 ) = Since D 24 489 ( ,29 29 ) > 0 and f 24 48xx ( ,29 29 ) > 0, 24 48( ,29 29 )yields a minimum distance. The point on the plane 2x+ 4y+ 3z= 12 that is closest to the origin is24 48 36( , ,29 29 29 )and this minimum distance is approximately 2.2283.23. Let Pxyz ( , , ) be any point onz = x2 + y2 . The square of the distance between the point (1, 2, 0) and P can be2 2 2 2expressed as d = f( x, y) = ( x− 1) + ( y− 2) + z . To find the critical points, set f ( xy , )3 23 2= 4x + 2x+ 4xy− 2= 0 and fy ( x , y ) 4 y 2 y 4 x y 4 0.second equation by x and summing the results leads to the equation 2y4x0.310x+ x− 1 = 0, whose solution is 0.393.into the first equation yieldsx == + + − = Multiplying the first equation by y and the2 2fxx ( xy , ) = 2 + 12 x + 4 y ,2 2f yy ( xy , ) = 2 + 12 y + 4 x , and fxy( xy , ) 8 xy.(0.393, 0.786) is approximately 57 and since xx (0.393,0.786) 0,The point on the surface24x− + = Thus, y = 2x. Substitutingx ≈ Consequently, y ≈ 0.786.= The value of D for the critical pointf > (0.393, 0.786) yields a minimum distance.2 2z = x + y is (0.393, 0.786, 0.772) and this minimum distance is approximately 1.56.778 Section 12.8 Instructor’s Resource Manual

24. Let (x, y, z) denote a point on the cone, and s denote the distance between (x, y, z) and (1, 2, 0).2 2 2 2 2 2 2s = ( x− 1) + ( y− 2) + z and z = x + y . Minimize2 2 2 2 2s = f( x, y) = ( x–1) + ( y–2) + ( x + y ), x, y in R.∇ f( x, y) = 2 2x−1, 2y− 2 = 0, 0 at⎛1 ⎞⎜ , 1 ⎟ .⎝2⎠ At ⎛1 ⎞⎜ , 1 ⎟ ,⎝2⎠5Conclusion: Minimum distance is s = ≈ 1.5811.2D > 0 and f xx > 0, so local minimum.25.⎛1 ⎞A= ⎜ ⎟ [ y+ ( y+2 x sin α )]( x cos α ) and⎝2⎠2 ⎛1⎞ 2⎛ π ⎞2x + y = 12. Maximize Ax ( , α) = 12xcos α –2x cosα + ⎜ ⎟ x sin2 α,x in (0, 6], a in ⎜0, ⎟.⎝2⎠⎝ 2 ⎠2 2⎛ π ⎞Ax ( , α) = 12cos α – 4xcosα + 2xsinαcos α, –12xsinα + 2x sinα + x cos 2α= 0, 0 at ⎜4, ⎟.⎝ 6 ⎠⎛ π ⎞At ⎜4, ⎟,D > 0 and A xx < 0, so local maximum, and A = 12 3 ≈ 20.78. At the boundary point of x = 6, we get⎝ 6 ⎠α = π , 18.4 Aπ π 2 π= Thus, the maximum occurs for width of turned-up sides = 4”, and base angle = + = .2 6 326. The lines are skew since there are no values of s and t that simultaneously satisfy t – 1 = 3s, 2t = s + 2, andt + 3 = 2s – 1. Minimize f, the square of the distance between points on the two lines.2 2 2f(, s t) = (3 s– t+ 1) + ( s+ 2–2) t + (2 s–1– t –3)Let∇ f( s, t) = 2(3 s– t+ 1)(3) + 2( s– 2t+ 2)(1) + 2(2 s– t – 4)(2), 2(3 s– t+ 1)(–1) + 2( s– 2t+ 2)(–2) + 2(2 s– t – 4)(–1)= 28s−14t−6, − 14s+ 12t− 28 = 0,0 .5Solve 28s – 14t – 6 = 0, –14s + 12t – 2 = 0, obtaining s = , t = 1.72 2D = fss ftt– fst= (28)(12) – (–14) > 0; ss 28 0.f = > (local minimum)The nature of the problem indicates the global minimum occurs here.2 2 2f ⎛ 5 15 5 25 875⎜ , 1 ⎞ ⎟= ⎛ ⎜ ⎞ ⎟ + ⎛ ⎜ ⎞ ⎟ + ⎛ ⎜–⎞⎟ =⎝7 ⎠ ⎝ 7 ⎠ ⎝7⎠ ⎝ 7 ⎠ 49Conclusion: The minimum distance between the lines is 875 / 7 ≈ 4.2258.27. Let M be the maximum value of f(x, y) on thepolygonal region, P. Then ax + by + (c – M) = 0is a line that either contains a vertex of P ordivides P into two subregions. In the latter caseax + by + (c – M) is positive in one of the regionsand negative in the other. ax + by + (c – M) > 0contradicts that M is the maximum value ofax + by + c on P. (Similar argument forminimum.)a. x y 2x + 3y + 4–1 2 80 1 71 0 6–3 0 –20 –4 –8Maximum at ( − 1, 2)b. x y –3x + 2y + 1–3 0 100 5 112 3 14 0 –111 –4 –10Minimum at (4, 0)28. x y 2x + y0 0 02 0 41 4 60 14 / 3 14 / 3Maximum of 6 occurs at ( 1, 4 )Instructor’s Resource Manual Section 12.8 779

29.The edges of P are segments of the lines:1. y = 02. 4x + y = 83. 2x + 3y = 14, and4. x = 02 2zx ( , y) = y – xzx ( , y) = –2 x, 2y= 0, 0 at (0, 0).There are no stationary points and no singularpoints, so consider boundary points.On side 1:2 2 2y = 2x, so z = 4 x – x = 3xz′ ( x) = 6x= 0 if x = 0.Therefore, (0, 0) is a candidate.On side 2:y = –4x + 6, so2 2 2z = (–4x+ 6) – x = 15 x – 48x+36.z′ ( x) = 30 x–48= 0 if x = 1.6.Therefore, (1.6, –0.4) is a candidate.On side 3:2 2y = –x, so z = (– x) – x = 0.Also, all vertices are candidates.x y z0 0 01.6 –0.4 –2.42 –2 01 2 3Minimum value of –2.4; maximum value of 3∂fn∂2∂mi=1 ∂mn= 2 ∑( yi −mxi −b)( −xi)i=1n=−2∑− −i=130. a. = ∑ ( yi−mxi−b)b.2( xiyi mxi bxi)Setting this result equal to zero yieldsi=1ni=1n∑2( xiyi mxi bxi)0=−2− −∑2( xy i i mxi bxi)0 = − −or equivalently,n n n2∑ xiyi = m∑xi + b∑xii= 1 i= 1 i=1n∂f∂= ∑ i − i −∂b∂bi=1n( y mx b)∑( yimxib)= 2 − − ( −1)i=1n∑( yimxib)=−2− −i=1Setting this result equal to zero yields∑( yimxib)0= −2− −i=1n∑( yimxib)0 = − −i=1nor equivalently,nn∑ ∑m x + nb=yiii= 1 i=1nTherefore,nnb = y −m xb =∑i ii= 1 i=1n∑ny − m xi ii= 1 i=1n∑∑n n n2∑xy i i = m∑xi + b∑xii= 1 i= 1 i=1⎛⎜= +⎝nn2∑ xy i i m∑xii= 1 i=12y − m⎞x ⎟ x⎠nn n n∑ ∑ ∑i i ⎟ ii= 1 i= 1 i=1780 Section 12.8 Instructor’s Resource Manual

c.This simplifies inton n n1∑xiyi − ∑xi∑yini= 1 i= 1 i=1m =nn22 1 ⎛ ⎞∑xi− xin ⎜∑⎟i= 1 ⎝i=1 ⎠2n2∑ xii=1∂ f2∂m= 22∂ f2∂b= 2n2 n∂ f= 2∑xi∂mb∂i=1Then, by Theorem C, we have⎛ n22 1 ⎛ n ⎞ ⎞D = 4n⎜xi− xi⎟∑i= 1 n ⎜∑ .⎜⎟⎝i=1 ⎠ ⎟⎝⎠Assuming that all the xiare not the same, wefind that D > 0 and2∂ f2∂m> 0 .Thus, f ( mb , ) is minimized.31. x i32.y i2x i xiyi3 2 9 64 3 16 125 4 25 206 4 36 247 5 49 35∑5i=125 18 135 97m(135) + b(25) = (97) and m(25) + (5)b = (18).Solve simultaneously and obtain m = 0.7, b = 0.1.The least-squares line is y = 0.7x + 0.1.2 2z = 2 x + y –4 x–2y+ 5, so∇ z = 4x−4, 2y− 2 =0 .∇ z =0 at (1, 1) which is outside the region.Therefore, extreme values occur on the boundary.Three critical points are the vertices of thetriangle, (0, 0), (0, 1), and (4, 0). Others mayoccur on the interior of a side of the triangle.On vertical side: x = 02zy ( ) y –2y5,= + y = [0, 1]. z′ ( y) = 2 y–2,soz′ ( y) = 0 if y = 1. Hence, no additional criticalpoint.On horizontal side: y = 02zx ( ) = 2 x –4x+ 5, x in [0, 4]. z′ ( x) = 4 x–4,so z′ ( x) = 0 if x = 1. Hence, an additional criticalpoint is (1, 0).On hypotenuse: x = 4 – 4y2 2zy ( ) = 2(4–4 y) + y –4(4–4 y)–2y+5233 y – 50y21,= + y in [0, 1].25z′ ( y) = 66 y–50,so z′ ( y) = 0 if y = .33⎛32 25 ⎞Hence, an additional critical point is ⎜ , ⎟.⎝33 33 ⎠x y z0 0 54 0 210 1 41 0 332 / 33 25/ 33 2.06Maximum value of z is 21; it occurs at (4, 0).Minimum value of z is about 2.06; it occurs at⎛32 25 ⎞⎜ , ⎟.⎝33 33 ⎠33. Let x and y be defined as shown in Figure 4 fromSection 12.8. The total cost is given by2 2Cxy ( , ) = 400 x + 50 + 200(200 −x−y)2 2+ 300 y + 100Taking partial derivatives and setting them equalto 0 gives2 2 −1 2C ( x, y) = 200( x + 50 ) (2 x) − 200 = 0x2 2 −1 2Cy ( x, y) = 150( y + 100 ) (2 y) − 200 = 0The solution of these equations is50100x = ≈ 28.8675 and y = ≈ 89.442731.25We now apply the second derivative test:2 2 2 2 2400 x + 50 − 400x x + 50Cxx( x, y)=2 2x + 50300 y + 100 − 300y y + 100Cyy( x, y)=2 2y + 100Cxy ( x , y ) = 0Evaluated at x = 50 3 and y = 100 1.25 ,2 2 2 2 22D ≈ (5.196)(1.24) − 0 > 0Thus, a local minimum occurs withC ( 50 3 ,100 1.25 ) ≈ $79,681We must also check the boundary. When x = 0,2 2C 1 ( ) (0, ) 200(200 ) 300 y + 100and when y = 0,Instructor’s Resource Manual Section 12.8 781

C 2 22 ( x ) = C ( x ,0) = 400 x + 50 + 200(200 − x )Using the methods from Chapter 3, we find thatC reaches a minimum of about $82,361 when1y = 8000 and C 2 reaches a minimum of about$87,321 when x = 2500 3 . Addressing theboundary x + y = 200, we find thatC 2 23 ( x ) = C ( x ,200 − x ) = 400 x + 502 2+ 300 (200 − x) + 100 This function reaches aminimum of about $82,214 when x ≈ 41.08 .Thus, the minimum cost path is whenx = 50 3 ≈ 28.8675 ft andy = 100 1.25 ≈89.4427ft, which produces acost of about $79,681.34. Let x and y be defined as shown in Figure 4 fromSection 12.8. The total cost is given by2 2Cxy ( , ) = 500 x + 50 + 200(200 −x−y)2 2+ 100 y + 100Taking partial derivatives and setting them equalto 0 gives2 2 −1 2C ( x, y) = 500( x + 50 ) (2 x) − 200 = 0x2 2 −1 2Cy ( x, y) = 100( y + 100 ) (2 y) − 200 = 0There is, however, no solution to Cy( x, y ) = 0Now we check the boundary. When x = 0,C 2 21 ( y ) = C (0, y ) = 200(200 − y ) + 100 y + 100There is, however, no solution to C ′1 ( y)= 0 .When y = 0 ,C 2 22 ( x ) = C ( x ,0) = 500 x + 50 + 200(200 − x )C ′2 ( x)= 0 yields x = 100 21 andC ( 100 21,0 ) ≈ $72,913On the boundary x+ y = 200 , we find thatC 2 23 ( x ) = C ( x ,200 − x ) = 500 x + 502 2+ 100 (200 − x) + 100 This function reaches aminimum of about $46,961 when x ≈ 9.0016 .Thus, the minimum cost path is whenx ≈ 9.0016 ft and y ≈ 190.9984 ft, whichproduces a cost of about $46,961.35. ( )f xy , = 10+ x+y∇ f = 1,1 ≠0; thus no interior criticalpoints exist. Lettingx = 3cos t, y = 3sin t, 0 ≤t ≤ 2 π ,g() t = f(3cos,3sin t t)andg′ () t = 3cost− 3sin. t Setting g′ () t = 036.yields t = π 4 or 5π4.The critical points are ( 3 2,3 2)and( −3 2, − 3 2).Since f ( 3 2,3 2)= 10+ 6 2 andf ( −3 2, − 3 2)= 10− 6 2, theminimum value of f on2 2x + y ≤3is10 − 6 2 and the maximum value of f is 10 + 6 2 .2 2f ( xy , ) = x + y;∇ f = 2 x,2 y .∇ f = 0 at (0,0).2D (0,0) = 2⋅2 − 0 = 4 > 0 and f xx (0,0) = 2 > 0,Thus, f (0,0) = 0 is a minimum.In order to optimize g() t = f( acos, t bsin t)where 0 ≤ t ≤ 2 π , we findg′ () t = 2 x( − asin t) + 2 y( bcos)t2 2= 2b sintcost−2a sintcost= ( b 2 a 2 )sin2 t.g t− Setting ′()= 0, wehave t = 0, π 2, π , or 3π 2. The resulting criticalpoints are ( a ,0), (0, b ) , ( − a,0), and (0, − b).2f ( a,0) = f( − a,0) = a ;2f (0, b) = f(0, − b) = b .Since a > b, the maximum value of f on the givenregion is a 2 and the minimum value of f is 0.37. The volume of the box can be expressed asVlwh (, , ) = lwh= 2 and the surface area as( )S l, w, h = 2lh + 2wh + lw + lw . Since h =2 ,lwSlw (, ) = 4 + 4+ w llw + lw When cost is factored,1 1Clw (, ) = 0.65lww+ l+ with w> 0, l > 01Cl(, l w) = − + 0.65w=02l1Cw (, l w) = − + 0.65l= 02wSolving this system of equations leads to0.65w = 3 ≈1.1544and l = w≈ 1.1544 .0.4225Consequently, h ≈ 1.501. Applying the second2derivative test with Cll(, l w) = ,3l2Cww (, l w ) = and C (, ) 0.65,3 lw l w =wD ≈ 1.268 > 0. Thus, the minimum cost occurswhen the length is approximately 1.1544 feet, thewidth is approximately 1.1544 feet and the heightis approximately 1.501 feet.782 Section 12.8 Instructor’s Resource Manual

38. The cost function in three variables isC(, l w, h) = 4lw + 2lh + 2wh + 6l + 6w + 4 h,39.60where lwh = 60. Substituting h = yieldslw120 240Clw (, ) = 4lw+ + 6l+ 6w+ withwlwl > 0 and w > 0.120 240Cl(, l w) = 4w− + 6− = 0l2 wl2120 240Cw (, l w ) = 4 l− +2 6 − =2w lw0Multiplying both sides of the first equation bywl 2 , multiplying both sides of the second2equation by lw , and subtracting the resultingequations produces − 120w+ 120l= 0 or l = w.120 240Consequently, 4w − + 6− = 0 orw2 w34 32w + 3w −60w− 120 = 0 Using a CAS, thisequation yields w ≈ 3.2134240 480240 480Cll(, l w) = + ; C3 3 ww 3 3l wl(, l w ) = +w lw;240Clw (, l w ) = 4 +2 2l w; Using the critical point(3.2134, 3.2134), D ≈ 131.44 > 0Thus, w= l ≈3.2yields a minimum. Theminimum cost involved with making this box isapproximately $177.79. This minimum costoccurs when the length and width areapproximately 3.2 feet and the height isapproximately 5.8 feet.2 2T( x, y) = 2 x + y – y∇ T = 4 x, 2 y–1=01 ⎛ 1 ⎞If x = 0 and y = , so ⎜0, ⎟ is the only interior2 ⎝ 2 ⎠critical point.1T′ ( y) = –1–2y= 0 if y = – , so on the2boundary, critical points occur where y is1–1, – , 1.2⎛ 1 ⎞Thus, points to consider are ⎜0, ⎟,(0, –1),⎝ 2 ⎠⎛ 3 1⎞ ⎛ 3 1⎞⎜, – , – , –2 2⎟ ⎜ 2 2⎟and (0, 1). Substituting⎝ ⎠ ⎝ ⎠these into T(x, y) yields that the coldest spot is⎛ 1 ⎞⎜0, ⎟⎝ 2 ⎠ where the temperature is 1– , and there4⎛ 3 1⎞is a tie for the hottest spot at⎜± , –2 2⎟where⎝ ⎠the temperature is 9 .440. Let x 2 , y 2 , z 2 denote the areas enclosed by thecircle, and the two squares, respectively. Thenxthe radius of the circle is , and the edges ofπthe two squares are y and z, respectively. We2 2wish to optimize A(x, y, z) = x + y + z2 ,⎛ x ⎞subject to 2π ⎜ ⎟+ 4y+ 4 z = k,or⎝ π ⎠equivalently 2 π x + 4y+ 4 z = k,with each of x,y, and z nonnegative. Geometrically: we seek thesmallest and largest of all spheres with center atthe origin and some point in common with thetriangular region indicated.On the boundary xof y there.2 2= 1– y , so T is a function2 2 2T( y) = 2(1– y ) + y – y = 2– y– y ,y = [–1, 1]k kSince > , the largest sphere will intersect2 π 4⎛ k ⎞the region only at point ⎜ , 0, 0⎟and will⎝2π ⎠kthus have radius . Thus A will be maximum2 πkif x = , y = z = 0 (all of the wire goes into2 πthe circle). The smallest sphere will be tangent tothe triangle. The point of tangency is on thenormal line through the origin,Instructor’s Resource Manual Section 12.8 783

x, y, z = t π , 2, 2 . Substituting x = π ,y = 2, z = 2 into the equation of the plane yieldskthe value t = , so the minimum value of2( π+ 8)k πA is obtained for the values of x = ,2( π+ 8)ky = z = . Thus the circle will have radiusπ+ 8⎡ k π ⎤⎢⎣2( π+ 8) ⎥⎦ k = , and the squares will eachπ 2( π+ 8)khave sides . Therefore, the circle will use( π+ 8)πkunits and the squares will each use( π+ 8)4kunits.( π+ 8)[Note: sum of the three lengths is k.]41. Without loss of generality we will assume that α ≤ β ≤ γ.We will consider it intuitively clear that for a triangle ofmaximum area the center of the circle will be inside or on the boundary of the triangle; i.e., α, β , and γ are in theinterval[0, π ] . Along with α + β + γ = 2 π , this implies that α + β ≥π .1 2The area of an isosceles triangle with congruent sides of length r and included angle θ is sin .2 r θ1 2 1 2 1 2Area( Δ ABC) = r sinα + r sin β + r sinγ2 2 21 2= r (sinα + sin β + sin[2 π− ( α + β)]21 2= r [sinα + sin β − sin( α + β)]2Area( ΔABC)will be maximum if (*) A( α, β) = sinα + sin β − sin( α + β)is maximum.Restrictions are 0 ≤α≤ β ≤π , and α + β ≥π .Three critical points are the vertices of the triangular domain of A : ⎜⎛π , π⎟⎞ , (0, π), and ( ππ , ). We will now search⎝2 2⎠for others.Δ A( α, β ) = cosα − cos( α + β), cos β − cos( α + β) = 0 ifcosα = cos( α + β) = cos β.Therefore, cosα = cos β, so α = β [due to the restrictions stated]. Thencosα = cos( α + α) = cos 2α = 2cos α − 1, so cosα= 2cos α − 1.2Solve for α : 2cos α −cosα− 1= 0; (2cosα+ 1)(cosα− 1) = 0;1 2πcos α =− or cosα = 1; α = or α = 0.2 322784 Section 12.8 Instructor’s Resource Manual

⎛2π2π⎞(We are still in the case where α = β.)⎜ , ⎟ is a new critical point, but (0, 0) is out of the domain of A.⎝ 3 3 ⎠There are no critical points in the interior of the domain of A.On the β =π edge of the domain of A; A( α) = sinα −sin( α −π ) = 2sin α so A′( α) = 2cos α.A′ ( α) 0 if α π ⎛π⎞= = . ⎜ , π⎟is a new critical point.2 ⎝2⎠On the β =π− α edge of the domain of A:A( α) = sinα + sin( π−α) −sin(2 α −π ) = 2sinα + sin 2 α,soA′ ( α) = 2cosα + 2cos 2α = 2[cos α + (2cos α − 1)] = 2(2cosα − 1)(cos α + 1) .1πA′ ( α) = 0 if cos α = or cosα = − 1, so α = or α = π.2 3⎛π2π⎞⎜ , ⎟ and ( π ,0) are outside the domain of A.⎝3 3 ⎠(The critical points are indicated on the graph of the domain of A.)2α β Aπ2π220 π 0π π 02π 2π3 33 32ππ 22Maximum value of A. The triangle is equilateral.42. If the plane through (a, b, c) is expressed asAx + By + Cz = 1, then the intercepts are 1 , 1 , 1 ;A B Cvolume⎛1⎞⎡⎛1⎞⎛ 1 ⎞⎛ 1 ⎞⎤⎛ 1 ⎞ 1of tetrahedron is V = ⎜ ⎟ .3⎢⎜ ⎟⎜ ⎟⎜ ⎟ =2 A B⎥⎜ ⎟⎝ ⎠⎣⎝ ⎠⎝ ⎠⎝ ⎠⎦⎝C⎠6ABCTo maximize V subject to Aa + Bb + Cc = 1 is equivalent tomaximizing z = ABC subject to Aa + Bb + Cc = 1.1 − aA − bB (1 ), so AB −CzaA −= =bB .cc⎛1⎞ 2 21 1 ⎡ 1 ⎤∇ z = ⎜ ⎟ B−2 aAB−bB , A−2 bAB− aA = 0 if A= , B = C = .c 3a 3b ⎢ 3c⎥⎝ ⎠ ⎣ ⎦⎛ 1 1 ⎞⎜ , ⎟ is the only critical point in the first quadrant. The second partials test yields that z is maximum at this⎝3a3b⎠ point. The plane is 1 1 1 x y zx+ y+ z = 1, or + + = 3.3a 3b 3c a b c1 9abcThe volume of the first quadrant tetrahedron formed by the plane is= .⎡6 1 1 1( 3 )( 3 )( 3 ) ⎤ 2⎣ a b c ⎦43. Local max: f(1.75, 0) = 1.15Global max: f(–3.8, 0) = 2.3044. Global max: f(0, 1) = 0.5Global min: f(0, –1) = –0.545. Global min: f(0, 1) = f(0, –1) = –0.1246. Global max: f (0,0) = 1Global min: f (2, − 2) = f( − 2,2) = e −≈ 0.0001234147. Global max: f(1.13, 0.79) = f(1.13, –0.79) = 0.53Global min: f(–1.13, 0.79) = f(–1.13, –0.79)= –0.539Instructor’s Resource Manual Section 12.8 785

48. No global maximum or global minimum49. Global max: f(3,3) = f( −3,3) ≈ 74.9225Global min: f(1.5708,0) = f( − 1.5708,0) =− 850. Global max: f(1, 43, 0) = 0.13Global min: f(–1.82, 0) = –0.2352. Global max: f(–5.12, –4.92) = 1071Global min: f(5.24, –4.96) = –65853. Global max: f(2.1, 2.1) = 3.5Global min: f(4.2, 4.2) = –3.551. Global max: f(0.67, 0) = 5.06Global min: f(–0.75, 0) = –3.5454. a.1k( α, β) = [80sinα + 60sin β + 48sin(2 π −α −β)]2= 40sinα + 30sin β − 24sin( α + β)1/2 1/2L( α, β) = (164 − 160cos α) + (136 −120cos β)1/2+ (100 − 96cos( α + β))b. (1.95, 2.04)c. (2.26, 2.07)12.9 Concepts Review1. free; constrained2. parallel3. g(x, y) = 04. (2, 2)Problem Set 12.91. 2 x, 2y = λ y,x2x = λy, 2y = λx, xy = 3Critical points are( ± 3, ± 3 ), f ( ± 3, ± 3)= 6.It is not clear whether 6 is the minimum ormaximum, so take any other point on xy = 3, forexample (1, 3). f(1, 3) = 10, so 6 is the minimumvalue.2. y,x = λ 8 x, 18y2 2y = 8λx, x = 18λy, 4x+ 9y= 36⎛ 3 2 ⎞ ⎛ 3 2 ⎞Critical points are ⎜ , ± ⎟, ⎜– , ± ⎟.⎝ 2 2 ⎠ ⎝ 2 2 ⎠⎛ 3 2 ⎞Maximum value of 3 occurs at ⎜± , ± ⎟.⎝ 2 2 ⎠3. Let ∇ f ( x, y) = λ∇ g( x, y),where2 2gx ( , y) = x + y –1=0.8x -4 y,-4x+ 2y= λ 2 x, 2y1. 4x – 2y = λx2. –2x + y = λy2 23. x + y = 14. 0 = λx + 2λy (From equations 1 and 2)5. λ = 0 or x + 2y = 0 (4)λ = 0: 6. y = 2x (1)7. x =±15(6, 3)8. y =±25(7, 6)x + 2y = 0: 9. x = –2y10. y =±15(9, 3)11. x =25(10, 9)⎛ 1 2 ⎞ ⎛ 1 2 ⎞Critical points: ⎜ , ⎟,⎜– ,– ⎟,⎝ 5 5⎠⎝ 5 5⎠⎛ 2 1 ⎞ ⎛ 2 1 ⎞⎜ ,– ⎟,⎜– , ⎟⎝ 5 5⎠⎝ 5 5⎠f(x, y) is 0 at the first two critical points and 5 atthe last two. Therefore, the maximum value off(x, y) is 5.786 Section 12.9 Instructor’s Resource Manual

4. 2x + 4 y,4x+ 2y= λ 1, -12x + 4y = λ, 4x + 2y = –λ, x – y = 6Critical point is (3, –3).5. 2 x,2 y,2z = λ 1, 3, -22x = λ, 2y = 3λ, 2z = –2λ, x + 3y – 2z = 12⎛6 18 12⎞Critical point is ⎜ , , – ⎟.⎝7 7 7 ⎠f ⎛ 6 18 12 72⎜ , , –⎞ ⎟7 7 7 = is the minimum.⎝⎠ 76. Let ∇ f( x, y, z) = λ∇ g( x, y, z),where2 2gx y z x y z( , , ) = 2 + –3 = 0.4, − 2,3 = λ 4 x,2 y, − 31. 4 = 4λx2. –2 = 2λy3. 3 = –3λ2 24. 2 x + y –3z= 05. λ = –1 (3)6. x = –1, y = 1 (5, 1, 2)7. z = 1 (6, 4)Therefore, (–1, 1, 1) is a critical point, andf(–1, 1, 1) = –3. (–3 is the minimum rather thanmaximum since other points satisfying g = 0 havelarger values of f. For example, g(1, 1, 1) = 0, andf(1, 1, 1) = 5.)7. Let l and w denote the dimensions of the base, hdenote the depth. Maximize V(l, w, h) = lwhsubject to g(l, w, h) = lw + 2lh + 2wh = 48.wh, lh,lw = λ w+ 2 h, l+ 2 h,2l+2wwh = λ(w + 2h), lh = λ(l + 2h), lw = λ(2l + 2w),lw + 2lh + 2wh = 48Critical point is (4, 4, 2).V(4, 4, 2) = 32 is the maximum. (V(11, 2, 1) = 22,for example.)8. Minimize the square of the distance to the plane,2 2 2f( x, y, z) = x + y + z , subject tox + 3y – 2z – 4 = 0.2 x,2 y,2z = λ 1, 3, -22x = λ, 2y = 3λ, 2z = –2λ, x + 3y – 2z = 4⎛2 6 4⎞Critical point is ⎜ , , – ⎟.The nature of the⎝7 7 7⎠problem indicates that this will give a minimumrather than a maximum. The least distance to the1/2 1/2⎡ ⎛2 6 4⎞⎤⎛8⎞plane is ⎢f⎜ , , – ⎟ = ⎜ ⎟ ≈1.0690.7 7 7⎥⎣ ⎝ ⎠⎦⎝7⎠9. Let l and w denote the dimensions of the base, hthe depth. Maximize V(l, w, h) = lwh subject to0.601w + 0.20(lw + 2lh + 2wh) = 12, whichsimplifies to 21w + lh + wh = 30, org(l, w, h) = 2lw + lh + wh – 30.Let ∇ V(, l w, h) = λ∇g(, l w, h);wh, lh,lw = λ 2 w+ h,2 l+ h,l+ w .1. wh = λ(2w + h)2. lh = λ(2l + h)3. lw = λ(l + w)4. 2lw + lh + wh = 305. (w – l)h = 2λ(w – l) (1, 2)6. w = l or h = 2λw = 1:7. l = 2λ = w (3) Note: w ≠ 0 , for thenV = 0 .8. h = 4λ (7, 2)9. λ =52(7, 8, 4)10. l = w= 5, h= 2 5(9, 7, 8)h = 2λ:11. λ = 0 (2)12. l = w = h = 0 (11, 1 – 3)(Not possible since this does not satisfy 4.)( 5, 5, 2 5 ) is a critical point andV ( 5, 5, 2 5)= 10 5 ≈ 22.36 ft 3 is themaximum volume (rather than the minimumvolume since, for example, g(1, 1, 14) = 30 andV(1, 1, 14) = 14 which is less than 22.36).10. Minimize the square of the distance,2 2 2f( x, y, z) = x + y + z , subject to2 2gx ( , y, z) = xy– z + 9=0.22 x, 2 y, 2z = λ 2 xy, x , –2z2x = 2λxy, 2 y = λx, 2z = –2λz,22 2x y– z + 9=0Critical points are (0, 0, ±3) [case x = 0];( 2, –1, 7)( 3 6 3)± ± [case x ≠ 0 , λ = –1]; and± − [case x ≠ 0, λ ≠− 1].Evaluating f at each of these eight points yields 9(case x = 0), 10 (case x ≠ 0 , λ = –1), and3 3 3( ) 22 9 (case x ≠ 0, λ ≠ − 1). The latter is2the smallest, so the least distance between theorigin and the surface is 6 3 3 2.8596.4 ≈Instructor’s Resource Manual Section 12.9 787

11. Maximize f(x, y, z) = xyz, subject to2 2 2 2 2 2 2 2 2 2 2 2g( x, y, z) = bcx + acy + abz – abc= 02 2 2 2 2 2yz, xz, xy = λ 2 b c x, 2 a c y, 2a b z2 2 2 2 2 2yz = 2 b c x, xz = 2 a c y, xy = 2 a b z,2 2 2 2 2 2 2 2 2 2 2 2bcx + acy + abz = abc⎛ a b c ⎞Critical point is ⎜ , , ⎟.⎝ 3 3 3⎠a b c 8abcV ⎛ ⎜ , , ⎞ ⎟=, which is the⎝ 3 3 3⎠3 3maximum.12. Maximize V(x, y, z) = xyz, subject tox y zgx ( , y, z) = + + –1= 0. Leta b c∇ V( x, y, z) = λ∇ g( x, y, z),so1 1 1yz, xz, xy = λ , , . Thena b cλx λy λz= = (each equals xyz).a b cλ ≠ 0 since λ = 0 would imply x = y = z = 0which would not satisfy the constraint.x y zThus, = = . These along with thea b ca b cconstraints yield x = , y = , z = .3 3 3abcThe maximum value of V = .2713. Maximize f(x,y,z) = x + y + z with the constraint2 2 2gxyz ( , , ) = x + y + z − 81 = 0. Let∇ f ( xyz , , ) = λ∇ gxyz ( , , ), so1,1,1 = λ 2 x, 2 y, 2 z ; Thus, x = y = zand 3x 2 = 81 or x = y = z = ± 3 3.The maximum value of f is 9 3 whenxyz , , = 3 3,3 3,3 32 2 2 214. Minimize d = f( x, y, z)= x + y + z withthe constraint gxyz ( , , ) = 2x+ 4y+ 3z− 12=0∇ f ( xyz , , ) = λ∇gxyz ( , , )2 x,2 y,2z = λ 2,3,4 ;2x = 2 λ;2y = 4 λ;2z= 3λleads to a criticalpoint of24 48 36( ), , The nature of the problem29 29 29indicates this will give a minimum rather than amaximum value. The minimum distance is24 2 48 2 36 2+ + ≈ 2.228329 29 2915. Minimize d2= f( x, y, z)2 2 2( x 1) ( y 2) z= − + − + with the constraint2 2gxyz ( , , ) = x + y − z=0;2x− 2,2y− 4,2z = λ 2 x,2 y, − 1Setting up, solving each equation for λ , and2 2substituting into equation x + y − z = 0produces λ ≈ − 1.5445; The resulting criticalpoint is approximately (0.393, 0.786, 0.772). Thenature of this problem indicates this will give aminimum value rather than a maximum. Theminimum distance is approximately 1.5616.2= f( x, y, z)2 2 2( x 1) ( y 2) z2 2 2gxyz ( , , ) = x + y − z = 02x −2,2y− 4,2z = λ 2 x,2 y, − 2z16. Minimize d= − + − + with the constraint151, x , y 1, z2 21 5,1, and1 52 2 ( 2 2 )10 .2λ =− = = =± ; The critical pointsare ( )a minimum distance of,1,− which both lead to17. (See problem 37, section 12.8). Let thedimensions of the box be l, w, and h . Then thecost of the box is.25(2hl + 2 hw + lw) + .4( lw)orClwh ( , , ) = .5 hl+ .5 hw+ .65lw.We want to minimize C subject to the constraintlhw = 2 ; set Vlhw (, , ) = lwh− 2.Now:∇ Clwh ( , , ) = (.5 h+ .65 w) i+ (.5 h+ .65 l) j+ .5( l+w)kand∇ V (, l w, h)= whi+ lh j+ lwkThus the Lagrange equations are.5 h+ .65w= λwh(1).5 h+ .65l = λlh(2).5( l+ w)= λlw(3)lwh = 2(4)Solving (4) for h and putting the result in (1) and(2), we get1 2λ+ .65w = (5)lw l1 2λ+ .65l = (6)lw wMultiply (5) by l and (6) by w to get1+ .65lw= 2λ(7)w1+ .65lw= 2 λ(8)l788 Section 12.9 Instructor’s Resource Manual

from which we conclude that l = w. Putting thisresult into (3) we have2l = λl(9)1Since V ≠ 0, l ≠ 0 and (9) tells us that l = ;λ1 1 2 2thus l = , w l , h 2λλ= = λ= lw= .Putting these results into equation (1), weconclude2 ⎛ 1⎞ ⎛ 1⎞2.5(2 λ ) + .65 ⎜ ⎟=λ⎜ ⎟(2 λ ) or⎝λ⎠ ⎝λ⎠⎛ 1 ⎞ 2.65⎜⎟ = λ .⎝λ⎠Hence: λ = 3 .65 ≈ .866 , so the minimum cost isobtained when:12l = w= ≈ 1.154 and h = 2λ≈1.5λ18. (See problem 38, section 12.8). Let thedimensions of the box be l, w, and h . Then thecost of the box is1(2hl + 2 hw) + 4( lw) + 3(2l + 2 w) + 4horC(, l w, h) = 2hl + 2hw + 4lw + 6l + 6w + 4h.We want to minimize C subject to the constraintlhw = 60 ; set Vlhw (, , ) = lwh− 60.Now:∇ C( l, w, h) = ( 2h+ 4w+ 6) i+ ( 2h+ 4l+6)j and+ ( 2l+ 2w+4)k∇ V (, l w, h)= whi+ lh j+ lwkThus the Lagrange equations are2h+ 4w+ 6= λwh(1)2h+ 4l+ 6= λlh(2)2l+ 2w+ 4= λlw(3)lwh = 60(4)60Solving (4) for h = and putting the result inlw(1) and (2), we get120 60λ+ 4w+ 6= lwl(5)120 60λ+ 4l+ 6= lww(6)Multiply (5) by l and (6) by w to get120 + 4 lw + 6 l = 60 λw(7)120+ 4lw+ 6w= 60 λl(8)from which we conclude that120 120+ 6w= + 6 ll wor ( l− w)( lw+ 20) = 0 .Since lw cannot be negative (= − 20 ), weconclude that l = w; putting this result intoequation (3), we have2 ⎛ 12w+ 2w+ 4= λwor 4 w + ⎞λ = ⎜ 2 ⎟⎝ w ⎠ .Therefore, from equation (1), we have120 ⎛4 6 4 w + 1 ⎞ ⎛ww60 ⎞+ + =2 ⎜ 2 ⎟ ⎜ 2 ⎟ orw ⎝ w ⎠ ⎝w⎠3(multiplying through by w and simplifying)4 32w + 3w −60w− 120 = 0Using one of several techniques available tosolve, we conclude that w= l = 3.213 and60h = ≈ 5.812 .2(3.213)19. (See problem 40, section 12.8)Letc = circumference of circlep = perimeter of first squareq = perimeter of second squareThen the sum of the areas is2 2 2 2 2 2c p q 1 ⎡c p qAc (, pq , ) = + + = ⎢ + +4π16 16 4⎢⎣ π 4 4so we wish to maximize and minimize2 2 2c p qAc (, pq , ) = + + subject to theπ 4 4constraint Lc (, pq , ) = c+ p+ q− k= 0.Now2c p q∇ Ac (, pq , ) = i+ j+kπ 2 2∇ Lc (, pq , ) = i+ j+kso the Lagrange equations are2cλπ = (1)p= λ (2)2q= λ (3)2c+ p+ q = k (4)Putting (1), (2) and (3) into (4) we getπ 2k(4 + ) λ = k or λ =2 8+ πTherefore:π kc0= ≈0.282k8 + π4kp0= ≈0.359k8 + π4kq0= ≈0.359k8 + π2Now A( c0, p0, q0) ≈ 0.0224kwhile2A( k,0,0) ≈ .079k, so we conclude that⎤⎥⎥⎦Instructor’s Resource Manual Section 12.9 789

A( c0, p0, q 0)is a minimum value. There is alsoa maximum value (see problem 40, section 12.8)but our Lagrange approach does not capture this.The reason is that the maximum exists because c,p, and q must all be ≥ 0 . Our constraint,however, does not require this and allowsnegative values for any or all of the variables.Under these conditions, there is no globalmaximum.20. (See problem 42, section 12.8). Let P be thex y zplane + + = 1 . This plane will cross theA B Cfirst octant, forming a triangle, T, in P; thevertices of this triangle occur where P intersectsthe coordinate axes. They are:V = ( A,0,0), V = (0, B,0), V = (0,0, C).x y za. Define the vectors g = −AB , ,0 andh = −A,0,C . From example 3 in 11.4, weknow the area of T is1 1 ( ) 2 ( ) 2 ( ) 2g× h = BC + AC + AB .2 2b. The height of the tetrahedron in question isthe distance is the distance between (0,0,0)and P . By example 10 in 11.3, this distanceis21 ( ABC)h = =1 1 12 2 2+ + ( BC) + ( AC) + ( AB)A2B2C2c. Finally, the volume of the tetrahedron is1 (area of )3 h T , or 1 2 2 2V ( A, B, C) = ( BC) + ( AC) + ( AB)6⎡2( ABC)⎤⋅ ⎢⎥⎢ 2 2 2( BC) + ( AC) + ( AB)⎥⎣⎦1 2 1That is, V( A, B, C) = ( ABC)= ABC .6 6Hence we want to minimizeV ( A, B, C)= ABC subject to the constrainta b c+ + = 1 ; defineA B Ca b cg( A, B, C) = + + − 1.A B CNowA B C∇ V( A, B, C)= i+ j+kABC ABC ABC−a −b −cand ∇ g( A, B, C)= i+ j+k . Thus2 2 2A B Cthe Lagrange equations areA −λa= (1)ABC 2ABABCCABC−λb= (2)2B−λ= (3)c2Ca b c+ + = 1 (4)A B CFrom (1) – (3) we have3 3 3−A −B −Cλ ABC = = = (5)a b cSolving in pairs we get⎛ b ⎞ ⎛ c ⎞B = 3 A,C 3⎜ =Aa ⎟ ⎜ a ⎟(6)⎝ ⎠ ⎝ ⎠and putting these results into (4) we obtain3 2 3 2A = a + ab + ac3 3 2 3 2 3 2= a⎛a + b + c⎞⎜⎟⎝⎠Similarly, we have3 2 3 2B = a b + b+bc3= b⎛⎜ a + b + c⎝3 2 3 2 3 23 2 3 2C = a c + b c + c3 3 2 3 2 3 2= c⎛a + b + c⎞⎜⎟⎝⎠Finally, the volume of the tetrahedron is33 3 2 3 2 3 2abc⎛a b c⎞ABC ⎜ + + ⎟=⎝⎠.6 621. Finding critical points on the interior first:∂ f ∂ f = 1 ≠ 0 = 1 ≠ 0; There are no critical∂x∂ypoints on the interior. Finding critical points onthe boundary: ∇ f ( xy , ) = λ∇gxy ( , );1,1 = λ 2 x, 2 y ; The solution to the system⎞⎟⎠2 21 2 x,1 2 y, x y 1= λ⋅ = λ⋅ + = is λ =± 12 ,1, y12 2x =± =± The four critical points are1( , ± 12 2)and ( − 1 ± 1) 2 21( , 12 2)10 21( , 1) 10 2, .f = + is the maximum value.f − − = − is the minimum value.2 2790 Section 12.9 Instructor’s Resource Manual

22. Finding critical points on the interior:∂ f = 1 − y = 0 ⇒ y = 1;∂x∂ f = 1 − x = 0 ⇒ x = 1; The only critical point on∂ythe interior is c 1 = (1,1). Finding critical pointson the boundary: Solve the system of equations2 21− y = λ ⋅ 2 x;1− x = λ ⋅2 y;x + y = 9Using substitution, it can be found that thecritical points on the boundary are⎛ 3 3 ⎞c2= ⎜ , ⎟⎝ 2 2 ⎠ , ⎛ 3 3 ⎞c3= ⎜− , − ⎟⎝ 2 2 ⎠ ,c = (2.56155, -1,56155),4c 5 = (-1.56155, 2.56155)The maximum value of 5 is obtained substitutingeither c 4 or c 5 into f. The minimum value ofabout -8.7426 is obtained by substituting c 3 intof.23. Finding critical points on the interior:∂ f2x3 y 0;∂ f= + − = = 2y− x = 0∂x∂yThe solution to this system is the only criticalpoint on the interior, c 1 = (-2,-1).Critical points on the boundary will come fromthe solutions to the following system ofequations:2x + 3− y = λ ⋅ 2 x,2y − x = λ ⋅ 2 y,2 2x y 9the critical points are c 2 = (0,3),⎛3 3 3⎞⎛ 3 3 3⎞c3= ⎜, −,2 2⎟c4= ⎜− , −⎝ ⎠ 2 2⎟⎝ ⎠f( c 1) =− 3, f( c 2) = 9, f( c3) ≈ 20.6913,f( c4) ≈− 2.6913 The max value of f is+ = . From the solutions to this system,≈ 20.6913 and the min value is -3.2 224.2{ 4 9 }( , ) xon the set ( , ): x yf x y = S = x y + ≤ 11+yWe first find the max and min for f on the setx2y2S = ( x, y ): + < 1 using the methods of{ 4 9 }section 12.8:1 −2xy∇ f( x, y)= i +1 + y (1 + y )2 2 21∇ f( x, y)=0 we have = 021+yThus f has no max or min on S .j so setting(impossible).We now look for the max and min of f on thex2y2boundary S {( x, y ): + 14 9 }= = ; this is doneusing Lagrange multipliers. Let2 2x y( , ) 14 9gxy = + − ; then1 −2xy∇ f( x, y)= +1 + y (1 + y )i2 2 2x 2y∇ gxy ( , ) = i+j2 9The Lagrange equations are1 λx=21+y 2−2xy2λy=2 2(1 + y ) 92 29x4y36j and(1)(2)+ = (3)Putting (1) into (2) yields2 3−λ x y 2λy= (4)2 9One solution to (4) is y = 0 which yields, from(3), x = ± 2 . Thus (2,0) and ( −2,0)arecandidates for optimization points.If y ≠ 0 , (4) can be reduced to2 3x 22 9−4λ39x1 =−22 21+y 9x−λ λ= (5)so thatyields= . Putting this result into (1), which has no solutions(left side always +, right side always -).Therefore the only two candidates for max/minare (2,0) and ( − 2,0) . Since f (2,0) = 2 andf ( − 2,0) =− 2 we conclude that the max value off on S is 2 and the min value is − 2 .∂f∂f25. = = 2(1 + x+ y) = 0⇒ x+ y = −1∂x∂yThere is no minimum or maximum value on theinterior since there are an infinite number ofcritical points. The critical points on theboundary will come from the solutions to thefollowing system of equations:12(1 + x + y)= λ ⋅ x212(1 + x + y)= λ ⋅ y8Solving these two equations for λ leads toy = −x− 1 or y = 4x. Together with the2 2x yconstraint + − 1= 0 leads to the critical4 16Instructor’s Resource Manual Section 12.9 791

points on the boundary:⎛−1−2 19 − 4 + 2 19 ⎞,,⎜ 5 5 ⎟⎝⎠⎛− 1+2 19 −4 −2 19 ⎞ ⎛ 2 8 ⎞,, ,⎜ 5 5 ⎟ ⎜−− ⎟⎝⎠ ⎝ 5 5⎠ and⎛ 2 8 ⎞⎜ , ⎟. Respectively, the maximum value is⎝ 5 5⎠ ≈ 29.9443 and the minimum value is 0.26. It is clear that the maximum will occur for atriangle which contains the center of the circle.(With this observation in mind, there areadditional constraints: 0 < α < π , 0 < β < π ,0 > γ < π .)Note that in an isosceles triangle, the sideopposite the angle θ which is between thecongruent sides of length r has length⎛θ⎞2rsin ⎜ ⎟.Then we wish to maximize⎝2⎠⎡ ⎛α ⎞ ⎛β ⎞ ⎛γ⎞⎤P( α, β, γ) = 2r⎢sin ⎜ ⎟+ sin ⎜ ⎟+sin ⎜ ⎟2 2 2⎥⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦subject to g(α, β, γ) = α + β + γ − 2π= 0 = 0.⎛α ⎞ ⎛β ⎞ ⎛γ⎞Let r cos ⎜ ⎟, cos ⎜ ⎟, cos⎜ ⎟ = λ 1, 1, 1 .⎝ 2⎠ ⎝ 2 ⎠ ⎝2⎠⎛α ⎞ ⎛ β ⎞ ⎛γ⎞Then λ = rcos⎜ ⎟= rcos⎜ ⎟=rcos ⎜ ⎟,so⎝ 2 ⎠ ⎝ 2 ⎠ ⎝2⎠⎛ α β γ ⎞α = β = γ ⎜since + + = π⎟.⎝ 2 2 2 ⎠2 π2 π3αa = 2π , so α = ; then β = γ = .3327. Let α + β + γ = 1, α > 0, β > 0, and γ > 0.α β γMaximize Px ( , y, z) = kx y z , subject to g(x, y, z) = ax + by + cz – d = 0.Let ∇ Px ( , y, z) = λ∇ gx ( , y, z).Thenα–1 β γ α β–1 γ α β γ –1kαx y z , kβx y z , kγx y z = λ a, b, c .λax λby λczα β γTherefore, = = (since each equals kx y z ).α β γλ ≠ 0 since λ = 0 would imply x = y = z = 0 which would imply P = 0.Therefore, ax = by =cz (*).α β γ⎛ax ⎞ ⎛by ⎞ ⎛cz⎞The constraints ax + by + cz = d in the form α⎜⎟ + β⎜ ⎟ + γ ⎜ ⎟ = d becomes⎝ α ⎠ ⎝ β ⎠ ⎝ γ ⎠⎛ax ⎞ ⎛ax ⎞ ⎛ax⎞α⎜ ⎟+ β⎜ ⎟+ γ ⎜ ⎟ = d,using (*).⎝ α ⎠ ⎝ α ⎠ ⎝ α ⎠⎛ax⎞Then ( α + β + γ) ⎜ ⎟=d,or ax = d (since α + β + γ = 1).⎝ α ⎠ ααdβdγ dx = (**); y = and z = then following using (*) and (**).a b cSince there is only one interior critical point, and since P is 0 on the boundary, P is maximum whenαd βd γdx = , y = , z = .a b c28. Let (x, y, z) denote a point of intersection. Letf(x, y, z) be the square of the distance to the2 2 2origin. Minimize f ( x, y, z)= x + y + zsubject to gxyz ( , , ) = x+ y+ z-8= 0andh(x, y, z) = 2x – y + 3z – 28 = 0.Let ∇ f ( x, y, z) = λ∇ g( x, y, z) + μ∇h( x, y, z).2 x,2 y,2z = λ 1,1,1 + μ 2,-1,31. 2x = λ + 2μ2. 2y = λ – μ3. 2z = λ + 3μ4. x + y + z = 85. 2x – y + 3z = 286. 3λ + 4μ = 16 (1, 2, 3, 4)7. 2λ + 7μ = 28 (1, 2, 3, 5)8. λ = 0, μ = 4 (6, 7)9. x = 4, y = –2, z = 6 (8, 1-3)f(4, –2, 6) = 56, and the nature of the problemindicates this is the minimum rather than themaximum.Conclusion: The least distance is 56 ≈ 7.4833.792 Section 12.9 Instructor’s Resource Manual

29. -1, 2, 2 = λ 2 x,2 y ,0 + μ 0,1, 22 2–1 = 2λx, 2 = 2λy + μ, 2 = 2μ, x + y = 2,y + 2z = 1Critical points are (–1, 1, 0) and (1, –1, 1).f(–1, 1, 0) = 3, the maximum value;f(1, –1, 1) = –1, the minimum value.30. a. Maximizewx ( 1, x2, …, xn) = xx 1 2, …, xn, ( xi> 0)subject to the constraintgx ( 1, x2, …, xn) = x1+ x2+…+ xn–1= 0. Let ∇ w( x1, x2, …, xn) = λ∇ g( x1, x2, …, xn).x2… xn, x1x3… xn, x1 … xn–1= λ 1, 1, …, 1 .Therefore, λx1 = λx2 =…= λxn(since eachequals x1x2 … x n ). Then x1 = x2 =…= x n .(If λ = 0, some x i = 0, so w = 0.)1Therefore, nxi= 1; xi= .nn⎛1⎞The maximum value of w is ⎜ ⎟ , and⎝n⎠1occurs when each x i = .nn⎛1⎞b. From part a we have that xx 1 2… x n ≤ ⎜ ⎟ .⎝n⎠Therefore, n1xx 1 2… x n ≤ .naiaiIf xi= = for each i, thena1+…+ anAa1 a2 an1n … ≤ , so nAaa 1 2 … an≤ , orA A A nnna1+ a2+… anaa 1 2 … an≤.n31. Let a 1 , a 2 ,… an= λ 2 x 1 , 2 x 2 ,…, 2 xn.Therefore, ai= 2 λxi,for each i = 1, 2, …, n(since λ = 0 implies a i = 0, contrary to thehypothesis).x xi j ⎛1 ⎞= for all i, j ⎜since each equals ⎟.aiaj⎝2λ ⎠The constraint equation can be expressed2 222⎛ x1⎞ 2⎛ x2⎞2⎛xn⎞a1 ⎜ ⎟ + a2⎜ ⎟ +…+ an⎜ ⎟ = 1.⎝a1⎠ ⎝a2⎠ ⎝an⎠22 2 2 ⎛ x1⎞+ +…+ ⎜ ⎟ = 1.⎝a1⎠Therefore, ( a1 a2an)22 a1x1 =2 2a1+…+ an;similar for each otherx i2 .The function to be maximized in a hyperplanewith positive coefficients and constant (sointercepts on all axes are positive), and theconstraint is a hypersphere of radius 1, so themaximum will occur where each x i is positive.There is only one such critical point, the oneobtained from the above by taking the principalsquare root to solve for x i.Then the maximum value of w is⎛ a1 ⎞ ⎛ a2⎞ ⎛ an⎞ Aa1⎜ ⎟+ a2⎜ ⎟+…+ an⎜ ⎟= = A⎝ A⎠ ⎝ A⎠ ⎝ A⎠A2 2 2where A = a1 + a2 +…+ a n .32. Max: f(–0.71, 0.71) = f(–0.71, – 0.71) = 0.7133. Min: f(4, 0) = –434. Max: f(1.41, 1.41) = f(–1.41, –1.44) = 0.03735. Min: f(0, 3) = f(0, –3) = –0.9912.10 Chapter ReviewConcepts Test1. True: Except for the trivial case of z = 0,which gives a point.xy2. False: Use f(0, 0) = 0; f( x, y)=x + yelsewhere for counterexample.3. True: Since g′ (0) = f x (0, 0)2 24. True: It is the limit along the path, y = x.5. True: Use “Continuity of a Product”Theorem.6. True: Straight forward calculation of partialderivatives7. False: See Problem 25, Section 12.4.8. False: It is perpendicular to the level curvesof f. The gradient ofF(x, y, z) = f(x, y) – z is perpendicularto the graph of z = f(x, y).9. True: Since 0,0, − 1 is normal to thetangent planeex10. False: C : For the cylindrical surface3f( x, y) = y , f(p) = 0 for every p onthe x-axis, but f(p) is not an extremevalue.Instructor’s Resource Manual Section 12.10 793

11. True: It will point in the direction of greatestincrease of heat, and at the origin,∇ T (0,0) = 1,0 is that direction.b. 2x− y−1≥012. True: It is nonnegative for all x, y, and it hasa value of 0 at (0, 0).13. True: Along the x-axis, f( x, 0)x →±∞ .→±∞ as14. False: Du f( x, y ) = 4, 4 ⋅u≤4 2⎛ 1 ⎞ ⎞(equality if u = ⎜ ⎟ 1,1 ⎟⎝ 2 ⎠ ⎠15. True: – D f( x, y) = –[ ∇f( x, y) ⋅u]u=∇ ⋅ u = – uf ( x, y) (– ) D f( x, y)16. True: The set (call it S, a line segment)contains all of its boundary pointsbecause for every point P not in S(i.e., not on the line segment), there isan open neighborhood of P (i.e., acircle with P as center) that containsno point of S.17. True: By the Min-Max Existence Theorem18. False: ( x0, y 0)could be a singular point.2.3.2x + y = kx – k = – y23 2 7fx ( x, y) = 12x y + 14xy2 2 7fxx ( x, y) = 36x y + 14y3 6fxy ( x, y) = 24x y+98xy⎛π⎞ ⎛π⎞19. False: f ⎜ ,1⎟= sin⎜ ⎟=1, the maximum⎝2 ⎠ ⎝2⎠π /2,1 is in the set.value of f, and ( )4. fx( x, y) = –2cos xsin x = – sin 2xfxx ( x, y) = –2xcos 2xfxy ( x, y ) = 020. False: The same function used in Problem 2provides a counterexample.Sample Test Problems1. a.x2 2+ 4y−100≥02 2x y+ ≥ 1100 255.6.7.8.– y 2fx( x, y) = e sec x– y 2fxx( x, y) = 2e sec xtanx– y 2fxy( x, y) = – e sec x– xfx( x, y) = – e sin y– xfxx( x, y) = e sin y– xfxy( x, y) = – e cosy3 5 6Fy ( x, y) = 30 x y –7xy3 4 5Fyy ( x, y) = 150 x y –42xy2 4 5Fyyx ( x, y) = 450 x y –42y3 4fx ( x, y, z) = y –10xyz2 2 4f y ( x, y, z) = 3 xy –5x z2 3fz( x, y, z) =− 20x yzTherefore, f x (2, –1, 1) = 19 ;y (2, –1, 1) –14 f z =f = ; (2, –1, 1) 80794 Section 12.10 Instructor's Resource Manual

y29. zy( x, y ) = ; z y (2, 2) = = 12210. Everywhere in the plane except on the parabola2x = y.11. No. On the path y = x,12. a.path y = 0,x –0lim = 1.x + 0x→0x–xlim = 0.x→0x+x2x –2y4–4lim = = 0( x, y) →(2, 2) 2x + 2y4+4On theb. Does not exist since the numerator lends to 4and the denominator to 0.c.13. a.14.15.2 2 2 2( x + 2 y )( x –2 y )lim( x, y) →(0, 0) 2 2x + 2y= lim2 2( x – y ) = 0( x, y) →(0, 0)3 2 3 2 2∇ f ( x, y, z) = 2 xyz , x z , 3x yzf (1, 2, − 1) = −4, − 1, 6b. ∇ f ( x, y, z)=2 2y zcos xz, 2ysin xz, xy cos xz∇ f (1, 2, – 1) = –4 cos(1), sin(1), – cos(1)≈ −2.1612, − 3.3659,2.16122 2 –1 2 2 –1Duf( x, y) = 3 y(1 + 9 x y ) , 3 x(1 + 9 x y ) ⋅u6 12 3 1Duf (4, 2) = , ⋅ , −577 577 2 2≈− 0.0013932 2z = f( x, y)= x + y=( 3 3−6)5771, – 3, 0 is horizontal and is normal to thevertical plane that is given. By inspection,3, 1, 0 is also a horizontal vector and is16. In the direction of ∇ f (1, 2) = 4 9, 422 2x 2 x y+ y = 9, or + = 1.218 9b. ∇ f( x, y) = x, 2 y , so f(4,1) = 4,2 .17. a. f(4, 1) = 9, soc.18. Fx = Fu u x + Fv v xv y u 1= +2 222 21+ u v xy 1+u v 2 x19.v y + u=2 22(1 + u v ) xFy = Fu u y + Fv v yv x u −1= +2 222 21+ u v xy 1+u v 2 yv x – u=2 22(1 + u v )y⎛1 ⎞ ⎛ u ⎞fx = fu u x + fu v y = ⎜ ⎟(2x ) + ⎜– ( yz )v2 ⎟⎝ ⎠ ⎝ v ⎠–2 –1 –1 2= x y z ( x + 3 y–4 z)⎛1 ⎞ ⎛ u ⎞f y = fu u y + fv v y = ⎜ ⎟(–3)+ ⎜–xzv2 ⎟⎝ ⎠ ⎝ v ⎠−1 −2 −1 2=− x y z ( x + 4 z)⎛1 ⎞ ⎛ u ⎞fz = fu u z + fv v z = ⎜ ⎟(4)+ ⎜– ( xy )v2 ⎟⎝ ⎠ ⎝ v ⎠−1 −1 −2 2= x y z (3 y−x )( )perpendicular to 1, –3, 0 and therefore isparallel to the vertical plane. Then u =3 1,2 2is the corresponding 2-dimensional unit vector.D f( x, y) =∇f( x, y)⋅uu3 1= 2 x, 2 y ⋅ , = 3x+y2 2Du f (1, 2) = 3 + 2 ≈3.7321is the slope of thetangent to the curve.Instructor’s Resource Manual Section 12.10 795

20.dF dF dx dF dy= +dt dx dt dy dt2 2 3= (3 x − y )( − 6sin 3 t) + ( −2xy−4 y )(3cos t)⎛dF⎞t = 0 ⇒ x = 2 and y = 0, so ⎜dt ⎟ = 0.⎝ ⎠t= 021. Ft = Fx x t + Fy y t + Fz z t1/2 2 2⎛10xy ⎞⎛3t ⎞ ⎛5x ⎞⎛1⎞⎛ 15x y ⎞3t= ⎜ – (3 e )3 ⎟ + +2 3 ⎜ ⎟t4⎝ z ⎜ ⎟ ⎜ z ⎟ ⎝ ⎠ ⎜ z ⎟⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠2 2 3t15xy t 5x 45x ye= + –3 3 4z z t zdc db dα22. = 3, = –2, = 0.1dt dt dt⎛1⎞Area = Ab ( , c, α) = ⎜ ⎟ cb ( sin α)⎝2⎠dA ⎡⎛ b ⎞ ⎛(sin ) dc ⎞ ⎛ c ⎞ ⎛(sin ) db ⎞ ⎛ b ⎞ ⎛( bc cos )d α ⎞⎤= α α αdt⎢⎜ ⎟ ⎜ ⎟+ ⎜ ⎟ ⎜ ⎟+⎜ ⎟ ⎜ ⎟2 dt 2 dt 2dt⎥⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦⎛⎜⎝dAdt23. Let( + )⎞7 4 3⎟ = ≈6.9641in. 2 /s⎠⎛π ⎞8, 10, 2⎜ ⎟⎝ 6 ⎠2 2 2F( x, y, z) = 9x + 4y + 9 z –34=0∇ F( x, y, z) = 18 x, 8 y,18 z , so ∇ f (1,2,–1) = 29,8,–9.Tangent plane is 9(x – 1) + 8(y – 2) – 9(z + 1) = 0, or 9x + 8y – 9z = 34.24.25.26.V =π r 2 h;2dV = Vrdr + Vhdh = 2π rh dr +π r dhIf r = 10, dr ≤ 0.02, h = 6, dh = 0.01, then2dV ≤2π rh dr +πr dh ≤ 2π (10)(6)(0.02) + π (100)(0.01) = 3.4πV(10, 6) = π (100)(6) = 600πVolume is 600π ± 3.4π ≈ 1884.96 ± 10.682 2 –1 2 –1 2 2 –2df = y (1 + z ) dx + 2 xy(1 + z ) dy – 2 xy z(1 + z ) dzIf x = 1, y = 2, z = 2, dx = 0.01, dy = –0.02, dz = 0.03, then df = –0.0272.Therefore, f(1.01, 1.98, 2.03) ≈ f(1, 2, 2) + df = 0.8 – 0.0272 = 0.7728∇ f( x, y) = 2 xy–6 x, x –12y= 0, 0at (0, 0) and (±6, 3).D = f f – f = (2 y– 6)(–12) – (2 x)2 2xx yy xy2= 4(18 −6 y− x ); fxx= 2( y– 3)2At (0, 0): D = 72 > 0 and f xx < 0, so local maximum at (0, 0).At (±6, 3): D < 0, so (±6, 3) are saddle points.796 Section 12.10 Instructor’s Resource Manual

27. Let (x, y, z) denote the coordinates of the 1stoctant vertex of the box. Maximizef(x, y, z) = xyz subject to2 2 2gx ( , y, z) = 36x + 4y + 9 z –36=0(where x, y, z > 0 and the box’s volume isV(x, y, z) = f(x, y, z).Let ∇ f( x, y, z) = λ∇g( x, y, z).yz, xzxy , 8 = λ 72 x,8 y,18z1. 8yz = 72λx2. 8xz = 8λy3. 8xy = 18λz2 2 24. 36x + 4y + 9z= 36yz 72λx2 25. = , so y = 9 x . (1, 2)xz 8λyyz 72λx2 26. = , so z = 4 x . (1, 3)xz 18λy2 2 217. 36x + 36x + 36x= 36, so x = .3(5, 6, 4)3 28. y = , z = (7, 5, 6)3 31 3 2 1 3 2V ⎛ ⎜ , , ⎞ ⎟=8⎛ ⎜ ⎞⎛ ⎟⎜ ⎞⎛ ⎟⎜ ⎞⎟⎝ 3 3 3⎠ ⎝ 3⎠⎝ 3⎠⎝ 3⎠16= ≈ 9.23763The nature of the problem indicates that thecritical point yields a maximum value rather thana minimum value.28. y,x = λ 2 x,2y2 2y = 2λx, x = 2λy, x + y = 1⎛ 1 1 ⎞Critical points are ⎜ , ± ⎟⎝ 2 2 ⎠ and⎛ 1 1 ⎞⎜– , ± ⎟.Maximum of 1⎝ 2 2 ⎠2 at⎛ 1 1 ⎞1⎜± , ± ⎟;minimum of – at⎝ 2 2 ⎠2⎛⎜±⎝1 1 ⎞, ⎟.2 2 ⎠Review and Preview Problems1.2.3.4.29. Maximize V(, r h) =π r h,subject to2Sr (, h) = 2π r + 2 πrh–24π=0.22 rh, r = 4 r + 2 h,2rπ π λ π π π22rh = λ(2r + h), r = 2λ, r + rh = 12Critical point is (2, 4). The nature of the problemindicates that the critical point yields a maximumvalue rather than a minimum value. Conclusion:The dimensions are radius of 2 and height of 4.Instructor’s Resource Manual Review and Preview 797

5.9.6.10.7.11.8.798 Review and Preview Instructor’s Resource Manual

12.2 2 ⎡ 1 2 1 2 3⎤a+ bx+ c x dx =0⎢ax+ bx + c x2 3 ⎥⎣⎦8 2= 2a+ 2b+c3218. ∫ ( )2019.⎡1 1 ⎤ πxdx= x− x2 4 ⎥ =⎣⎦ 2π2sin sin 20 ⎢∫π020.∫3/41/ 43/41 1 x + 1 1 ⎛21⎞dx = ln = ln ⎜ ⎟1 2 1 2 52− x x −1/ 4 ⎝ ⎠13.21.1 1 1ln 1 ln 2+ u3/423/4du = + x =x=1/41 21/42∫22.∫11+u4−1x4du ⎡ ex=0 2 ⎣⎤⎦0= tan ≈0.767123.2u = 4r + 1; du = 8rdr3 2 1 3∫ r 4r + 1dr =0 8∫ u dur=033/23/2⎡1 2 2 ⎤ − 1+37= ⎢ i ( 4r+ 1)=8 3⎥⎣⎦01214.24.2 2u = a − r ; du =−2rdra/2 ar a a/21∫dr =− du0 2 2 2∫r= 0a − rua /2 2a ⎡12 2 ⎤a=− a r2⎢ −2⎥ =⎣ ⎦08( 2−3)25.π /2⎛1 1 ⎞∫ cos 20 ⎜ + θ2 2⎟dθ⎝⎠π /2⎡1 1 ⎤= ⎢ θ + sin 2 θ =2 4⎥⎣⎦04π15.16.17.∫12−2x−2xe dx =− e + C1 12 21 −2x1 −2x=− xe − e + C2 4∫ ∫∫−2x −2x −2xxe dx =− xe + e dx+Ca /2−a/2a /2⎛ xπ⎞ a ⎛ xπ⎞ 2acos⎜ ⎟dx= sin ⎜ ⎟ =⎝ a ⎠ π ⎝ a ⎠ π−a/226.27.2π /2⎛1 1 ⎞∫0⎜ + cos 2 θ2 2⎟ dθ⎝⎠2π /2⎛1 1 1 2 ⎞= ∫ cos 2 cos 20 ⎜ + +4 2 4⎟ d⎝⎠π /2⎛3 1 1 ⎞= ∫ cos 2θ cos 4θ dθ0 ⎜ + +8 2 8⎟⎝⎠π /2⎡3 1 1 ⎤ 3= ⎢ θ + sin2θ + sin4θ=8 4 32⎥⎣⎦0162 2 2 22π ⎛ ⎜ a −b − a −c⎞ ⎟⎝⎠θ θ θθ is not part of the integrand.πInstructor’s Resource Manual Review and Preview 799

28. The area is an equilateral triangle of length 2 .1 6 3A = 2 =2 2 229. The solid is half of a right circular cylinder ofradius 3 and height 8.1 2 πV = π r h= ( 9)( 8)= 36π2 230. The solid is a sphere of radius 7.4 3 4π3 1372πV = π r = 7 = ≈ 1436.8 31.3 3 3The solid looks similar to a football.2π2 1 1 πV = π∫sin xdx = π ⎡ x sin 2x⎤0⎢ − =2 4 ⎥⎣⎦ 232. The solid is a right circular cylinder of radius 7and height 100.2V = π r h = 4900π33. The solid is half an elliptic paraboloid.In the xz-plane, we can consider rotating the2graph of z = 9 − x around the z-axis for0≤ x ≤ 3. Using the Shell Method, we would get3 2V = 2πx 9−x dx∫( )0⎡ 2 43x x ⎤ ⎡ ⎤9 81 81 81π= 2π⎢ − ⎥ = 2π2 4 ⎢ − =2 4 ⎥⎢⎣⎥⎦⎣ ⎦ 2034. The solid is half of a hollow sphere of radius 1inside half of a solid sphere of radius 4.1 4 3 4 3V ⎛⎞= π 4 π1 42π2⎜ − =3 3⎟⎝⎠π0800 Review and Preview Instructor’s Resource Manual

CHAPTER 13Multiple Integrals13.1 Concepts Review1.n∑k=1f ( x , y ) ΔAk k k2. the volume of the solid under z = f(x, y) andabove R3. continuous4. 123.∫∫ ∫∫ ∫∫ ∫∫f ( x, y) dA = 2dA + 1dA + 3dAR R1 R2 R3= 2 A( R ) + 1 A( R ) + 3 A( R )1 2 3= 2(2) + 1(2) + 3(2) = 12Problem Set 13.11.∫∫ ∫∫2dA + 3dA = 2 A( R ) + 3 A( R )R1 R2= 2(4) + 3(2) = 141 24.∫∫ ∫∫ ∫∫2dA + 3dA + 1dAR1 R2 R3= 2 A( R ) + 3 A( R ) + 1 A( R )1 2 3= 2(3) + 3(2) + 1(1) = 132.∫∫ ∫∫(–1) dA+ 2 dA= (–1) AR ( ) + 2 AR ( )R1 R2= (–1)(3) + 2(3) = 31 2∫∫ ∫∫5. 3 f( x, y) dA– g( x, y) dA= 3(3)–(5) = 4RR∫∫ ∫∫6. 2 f ( xydA , ) + 5 gxydA ( , )R= 2(3) + 5(5) = 31R7.8.∫∫ ∫∫RgxydA ( , ) – gxydA= ( , ) (5)–(2) = 3R1∫∫ ∫∫2 g( xydA , ) + 3dA= 2(2) + 3 AR ( )R1 R1= 4 + 3(2) = 101801 Section 13.1 Instructor’s Resource Manual

9.17.[f(1, 1) + f(3, 1) + f(5, 1) + f(1, 3) + f(3, 3)+ f(5, 3)](4) = [(10) + (8) + (6) + (8) + (6)+ (4)](4) = 16810. 4(9 + 9 + 9 + 1 + 1 + 1) = 12018.11. 4(3 + 11 + 27 + 19 + 27 + 43) = 52012.⎡⎛41⎞ ⎛33 ⎞ ⎛25 ⎞ ⎛35 ⎞ ⎛27 ⎞ ⎛19⎞⎤⎢⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟+⎜ ⎟ (4)6 6 6 6 6 6⎥⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦= 12013. 4( 2 + 4 + 6 + 4 + 6 + 8)≈ 52.566514.3 5 3 9 154( e+ e + e + e + e + e ) ≈ 1310924715.19.16.20.802 Section 13.1 Instructor’s Resource Manual

21.Then c = 2 5 + 2 2 + 2(2) + 4(1) ≈ 15.300622.z = 6 – y is a plane parallel to the x-axis. Let T bethe area of the front trapezoidal face; let D be thedistance between the front and back faces.(6 – ydA ) = volume of solid = ( T)( D)∫∫R⎡⎛1 ⎞ ⎤= ⎢⎜⎟ (6 + 5) (1) = 5.52⎥⎣⎝⎠ ⎦For C, take the sample point in each square to bethe point of the square that is farthest from theorigin. Then,C = 2 13+ 2 10+ 2 8+ 4 5+ 2 2 ≈ 30.9652.26. The integrand is symmetric with respect to they-axis (i.e. an odd function), so the value of theintegral is 0.27. The values of x yand x y+ are indicatedin the various square subregions of R. In eachcase the value of the integral on R is the sum ofthe values in the squares since the area of eachsquare is 1.a. The integral equals –6.z = 1 + x is a plane parallel to the y-axis.∫∫ (1 + x)dA is the product of the area of aRtrapezoidal side face and the distance between theside faces.⎡⎛1 ⎞ ⎤= ⎢⎜⎟ (1 + 3)(2) (1) = 42⎥⎣⎝⎠ ⎦b. The integral equals 6.∫∫23. 0 dA = 0 A ( R ) = 0RThe conclusion follows.∫∫ ∫∫ ∫∫24. m dA < f ( x, y)dA < M dAR R R(Comparison property)Therefore, ma( R) < f ( x, y) dA < MA( R)25.∫∫RFor c, take the sample point in each square to bethe point of the square that is closest to the origin.28. Mass of the plate in grams29. Total rainfall in Colorado in 2005; averagerainfall in Colorado in 2005.30. For each partition of R, each subrectanglecontains some points at which f(x, y) = 0 andsome points at whichf(x, y) = 1. Therefore, for each partition there aresample points for which the Riemann sum is 0and others for which the Riemann sum is(1)[Area (R)] = 12.Instructor’s Resource Manual Section 13.1 803

31. To begin, we divide the region R (we will use the outline of the contour plot) into 16 equal squares. Then we canapproximate the volume by16V = f( x, y) dA≈ f( x , y ) ΔA∑∫∫ k k k .Rk = 1Each square will have Δ A = (1⋅ 1) = 1 and we will use the height at the center of each square as f ( xk, y k).Therefore, we get16V ≈ f( x , y ) = 20+ 21+ 24+ 29+ 22+ 23+ 26+ 32+ 26+ 27+ 30+ 35+ 32+ 33+ 36+42∑k=1= 458 cubic units13.2 Concepts Reviewkk7.1π⎡2 ⎤π0 ⎢ ⎥0⎣ ⎦x=0⎛1⎞ ⎛1⎞⎜ ⎟x sin y dy = ⎜ ⎟sin ydy = 1⎝2⎠ ⎝2⎠∫ ∫1. iterated2.⎡⎢⎣2 2−1 0f ( x, y)dy⎤dx⎥⎦∫ ∫ ;3. signed; plus; minus⎡⎢⎣2 2∫ ∫0 −1f ( x, y)dx⎤dy⎥⎦8.ln 3 ln 2 x y ln 3 x y ln 2= [ ]0 0 0 y = 0ln 3 x x ln 3 x∫ [ e (2)– e (1)] dx e dx0 ∫ 0x ln 3[ e ] 0 3–1 2∫ ∫ ∫e e dy dx e e dx= == = =4. is below the xy-planeProblem Set 13.29.π/2 1 π/20y= 0 =0∫ ∫[– cos xy] dx (1– cos x)dxπ= –1 ≈ 0.570822 231. ∫ [ 9 − ] =0 ∫ [ 27−3]2.3.y xy dx x dx0 02⎡ 3 2 ⎤= ⎢27x− x = 482 ⎥⎣ ⎦02 21 2 29 − = 9−−2 0 −22⎡ 1 3 ⎤ 92∫ ∫⎡ y yx ⎤ dx ⎡ x ⎤dx⎣ ⎦ ⎣ ⎦= ⎢9x− x =3 ⎥⎣ ⎦ 3−232⎡⎛⎞ 2 2⎤2 240 ⎢⎜⎟ ⎥0⎣⎝⎠ ⎦y=11 32x y dx = x dx =2 3∫ ∫10.11.1 xy 1 10y 00∫ ∫∫x[ e ] = dx = ( e –1) dx = e– 2 ≈0.718330= ∫⎡2( x + y)⎢⎢⎣3302 3/21⎤⎥⎥⎦dyx=03/2 3/22[(1 + y) – y ]dy330( )⎡ 5/2 5/24[(1 + y) – y ] ⎤ 432–9 3 –4= ⎢⎥ =⎢⎣15 ⎥⎦15( )431–9 3=15≈ 4.10974.5.24 ⎡3 ⎤4–1 ⎢ ⎥–1⎣ ⎦y=1⎛1⎞ ⎛ 7⎞115xy + ⎜ ⎟y dx = ⎜x + ⎟dx=⎝3⎠ ⎝ 3⎠6∫ ∫232 x y 2 2 9y2+ = + 31 2 1⎢⎣⎥⎦⎝ 2x=0⎡ ⎤ ⎛ ⎞⎢ xy ⎥ dx ⎜ y ⎟dy⎠∫ ∫2⎡ 29y3⎤ 13 55= ⎢ + y ⎥ = 17 – = = 13.75⎢⎣4 ⎥⎦4 4112.13.1 –1 1 1⎛1 ⎞∫ [–( xy+ 1) ]0x=0 dy = ∫ 1– dy0⎜ ⎟⎝ y + 1⎠= 1 – ln 2 ≈ 0.30691ln 3 ⎡2 ⎤ln 30 ⎢⎥ =0⎣⎦y=0⎛1⎞ ⎛1⎞x⎜ ⎟exp( xy ) dx ⎜ ⎟( e –1) dx⎝2⎠ ⎝2⎠∫ ∫⎛1⎞= 1– ⎜ ⎟ln 3 ≈0.4507⎝2⎠6.21 ⎡ 3 2⎤1 2–1 ⎢⎥–1⎣⎦x=1⎛1⎞ ⎛7 ⎞ 16⎜ ⎟x + xy dy = ⎜ + y ⎟dy=⎝3⎠ ⎝3 ⎠ 3∫ ∫14.221⎡⎤1∫ y⎢ ⎥ dx = ∫0 2 0 2⎢⎣2(1 + x ) ⎥⎦1+xy=0–1 1 ⎛π⎞ π= [2tan x] 0 = 2 ⎜ ⎟– 0 =⎝4⎠22dx804 Section 13.2 Instructor’s Resource Manual

15.3π ⎡1 2 2 ⎤ π 9 2coscos0 ⎢2 ⎥ =⎣ ⎦00 2π∫ ∫y x dx xdx⎡9 9 ⎤ 9π= ⎢ x+ cos 2x=4 8 ⎥⎣⎦ 41 ⎡12 ⎤1−1⎣⎦0−11⎡⎣y( e 1)⎤⎦−1e 11 1e dy e dy2 21= − = −2x16. ∫ ⎢ ⎥ = ∫ ( −1)17.18.19.100dx = 03∫ (since xy defines an odd0function in y).21 ⎡ 2 3⎤1 22–1 ⎢⎥–1⎣⎦y=0x y ⎛1⎞ 83 y dx ⎛ x ⎞+ ⎜ ⎟ = ⎜ + ⎟⎝ ⎠ ⎝ 3⎠dx∫ ∫⎡⎛2⎞ 3 ⎛8⎞⎤ 20= ⎢⎜ ⎟x+ ⎜ ⎟x=3 3⎥⎣⎝ ⎠ ⎝ ⎠ ⎦ 31–1π/2 π/2sin( x + )0 0π /2 π /2∫ [– cos( + )]0x=0∫ ∫=ydxdyx y dyπ /2 ⎡ ⎛–cosπ ⎞= ∫+ y + cosy ⎤ dy0 ⎢ ⎜ ⎟2⎥⎣ ⎝ ⎠ ⎦π /2 π /2(sin y cos y) dy [– cos y sin y]00∫= + = += (0 + 1) – (–1 + 0) = 22 3 2 222. = ( 25 − − )∫∫V x y dydx0 032⎛2 1 3⎞∫0⎜25⎟⎝3 ⎠02 2∫ ( x ) dx0= y−x y−y dx= 75 −3 − 9 = 12423. = 3 4 ( 1+ 2 +2)24.25.∫∫V x y dxdy0 033⎛1 3 2⎞∫0⎜⎟⎝ 3 ⎠03⎛64 2 ⎞∫y0⎜⎟= x+ x + xy dy= 4 + + 4 dy = 112⎝ 3 ⎠3 2 2−x∫∫ 50 023 ⎛ 1 −2x1 −2x⎞∫ 50⎜⎟⎝ 2 4 ⎠0V = xye dx dy= y − xe − e dy4( e − )3 ⎡ 5 45 1−41⎤= ∫ 5y e dy11.04390 ⎢− −4 4⎥= ≈4⎣ ⎦ 4exz = is a plane.2x – 2z = 020.32⎡2 3/2⎤2(1 )1 ⎢ + ⎥ =1⎣⎦x=02⎡⎛7⎞ 2 ⎤⎛1⎞ ⎛7⎞⎜ ⎟y x dy ⎜ ⎟ydy⎝3⎠ ⎝3⎠∫ ∫= ⎢⎜⎟ 3.56 y ⎥ =⎣⎝⎠ ⎦126. z = 2 – x – y is a plane.x + y + z = 22 321. = ∫∫ ( 20 − − )V x y dydx0 02⎛1 2 ⎞= ∫ 20y−xy−y dx0⎜⎟⎝ 2 ⎠2⎛9 ⎞= ∫ 60 −3x− dy = 1050⎜ ⎟⎝ 2 ⎠30Instructor’s Resource Manual Section 13.2 805

27.2 2z = x + y is a paraboloid opening upward withz-axis.b d b da c a cd bhydy ( ) gxdx ( )ca∫∫ ∫ ∫= ∫ ∫33. g( xhydydx ) ( ) = gx ( ) hydydx ( )(First step used linearity of integration withrespect to y; second step used linearity ofintegration with respect to x; now commute.)28.2z = 4– y is a parabolic cylinder parallel to thex-axis.34.ln 2 2x 1 2 –1∫ ∫xe dx y (1 + y ) dy0 0ln 2 11 2x 1 2eln(1 y )2020⎡ ⎛ ⎞ ⎤ ⎡ ⎛ ⎞= ⎢⎜ ⎟ ⎥ ⎢⎜ ⎟ +⎤⎥⎣⎝ ⎠ ⎦ ⎣⎝ ⎠ ⎦⎡1⎤⎡⎛1⎞ ⎤ ⎛1⎞= ⎢ ln 2 ln 2 0.17332⎥⎢⎜ ⎟ = ⎜ ⎟ ≈2⎥⎣ ⎦⎣⎝ ⎠ ⎦ ⎝4⎠29.30.31.32.13 1 3⎡⎛1⎞ 2 ⎤( + + 1) =1 0 1 ⎢⎜⎟ + +2⎥⎣⎝⎠ ⎦x=0∫∫ ∫x y dxdy x yx x dy3⎛3 ⎞= ∫ y+ dy = 71⎜ ⎟⎝ 2 ⎠42 4 2⎡⎛3⎞ 2 ⎤(2 + 3 ) = 21 0 1 ⎢ + ⎜ ⎟2 ⎥⎣ ⎝ ⎠ ⎦y=0∫∫ ∫∫12x ydydx xy y dx= (8 x+24) dx = 36x2 2∫ ∫+ y + 2>11 1 2 2–1 011 ⎡ 2 ⎛1⎞ 3 ⎤∫ –1⎢⎜ ⎟3⎥⎣ ⎝ ⎠ ⎦y=0[( x + y + 2) –1] dydx= x y+ y + y dx1 ⎛ 2 4⎞10= ∫ x dx–1⎜+ ⎟ =⎝ 3⎠3⎡ 3x ⎤(4 – x ) dx dy = ⎢4 x – ⎥ dy⎢⎣3 ⎥⎦2 2 220 0 0∫∫ ∫2⎛16 ⎞ ⎡16y⎤ 32= ∫ dy0⎜ ⎟ = =3⎢3⎥⎝ ⎠ ⎣ ⎦ 3202035.36.37.38.39.1 1 2 2 1 2 1 2x y x yxye e dy dx⎛xe dx⎞⎛ye dy⎞∫∫ = ⎜ ⎟⎜ ⎟0 0 ⎝∫0 ⎠⎝∫0 ⎠1 2 2x=⎛xe dx⎞⎜ ⎟⎝∫ (Changed the dummy variable y0 ⎠to the dummy variable x.)2⎛ 2 1⎡ x2e⎤ ⎞⎜ ⎟ ⎛e–1⎞= ⎢ ⎥0.73812 = ⎜ ⎟ ≈⎜⎢ ⎥ ⎟ ⎝ 2 ⎠⎝⎣⎦0⎠ππV cos xcosy dxdy= ∫ ∫– π – ππ π ⎛ π– π – π – π= cos x dx cos y dy cos x dx⎞∫ ∫ =⎜ ⎟⎝∫⎠π( x0)π /22/22=⎛4 cosxdx⎞⎜⎟ = 4[sin ] = 16⎝∫ 0 ⎠2 2 1 3 2 2 1 3x dx y dy⎛2 x dx⎞⎛2 y dy⎞∫−2 ∫ = ⎜ ⎟⎜ ⎟−1 ⎝∫0 ⎠⎝∫ 0 ⎠⎛8⎞ ⎛1⎞8= 2⎜ ⎟2⎜ ⎟=⎝3⎠ ⎝4⎠32 2 1 3x∫ dx y dy = 0−2∫ (since the second integral−1equals 0). x dx y dy x dx y dy2 2 1 3 2 2 1 3 2 2−2 = −1 0 0∫ ∫ ∫ ∫1 2 3 2 1= 2 ⎡ ⎢∫ 0 dx + 1 dx 2 dx 3 dx ⎤⎡ 2 ⎛ ⎞⎤0 ∫ + +1 ∫ 2 ∫ 3 ⎥⎢ ⎜ ⎟ 4 ⎥⎣ ⎦⎣ ⎝ ⎠⎦1= 2⎡0+ ( 2 –1) + 2( 3– 2) + 3( 2– 3)⎤ ⎡ ⎤⎣⎦⎢ 2⎥⎣ ⎦= 5 – 3 – 2 ≈ 1.85372806 Section 13.2 Instructor’s Resource Manual

40.41.1 3 2 2 –2 18 ( + + 1) = [–4( 2 + 2 + 1) –1 ] 3 0 0 0x = 0∫∫ ∫x x y dx dy x y dy⎡ 1 ⎛ y ⎞ ⎤= 4 ⎢– arctan ⎜ ⎟+arctan( y)2 2⎥⎣ ⎝ ⎠ ⎦0 [ f ( xgy ) ( )– f( ygx ) ( )] dxdy101⎡–1 1= 4∫⎢ +0⎢⎣4+ y 1+y2 2⎤⎥ dy⎥⎦⎡⎛1 ⎛1⎞π⎞ ⎤⎛1⎞= 4 ⎢⎜– arctan ⎜ ⎟+⎟– 0⎥=π– 2arctan ⎜ ⎟≈2.2143⎣⎝2 ⎝2⎠4⎠⎦⎝2⎠b b2= b b 2 2 2 2a a ∫∫+a a b 2 2 2 2f ( xdx ) b g ( ydy ) –2 b f( xgxdx ) ( ) b f( yg ) ( ydy ) b f ( ydy ) bg ( xdx )a a a a a ab 22 b 2f xdx g xdx⎡ bf xgxdx⎤∫a ∫a ⎢∫a⎥≤ ∫∫∫ ∫ ∫ ∫ ∫ ∫= += 2 ( ) ( ) –2 ( ) ( )⎣Therefore,⎡⎢⎣b 2⎤ b 2 b 2≤a ⎥ a a∫ ∫ ∫f ( xgxdx ) ( ) f ( xdx ) g ( xdx ) .⎦⎦[ f ( xg ) ( y)–2 f( xgx ) ( ) f( ygy ) ( ) f ( yg ) ( x)]dxdy42. Since f is increasing, [y – x][f(y) – f(x)] > 0. Therefore,0 < ∫∫ b b[ y – x ][ f ( y )– f ( x )] dxdyb b ( ) – b b ( ) – b b ( ) b b= ( )a a ∫∫ yf y dx dya a ∫∫ yf x dx dya a ∫∫ xf y dx dy + xf x dx dya a ∫∫ a ab2 2 2 2b – a b b – a b b= ( b– a) ∫ yf( ydy ) – ( ) – ( ) ( – ) ( )a 2∫ f xdx f ydy+b a xf xdxa 2∫a ∫ a2 22( b – a b) xf ( xdx b) –( b – a ) f ( xdxbb= ∫ )a∫ = ( b – a )⎡2 xf ( xdx ) –( b + a ) f ( xdx )⎤a⎢⎣∫a∫ a ⎥⎦Therefore, ( b a ) bf ( x b) dx 2 xf ( x ) dxb+ ∫ < .a ∫ Now divide each side by the positive number 2 ( )a∫ f xdx to obtain theadesired result.Interpretation:If f is increasing on [a, b] and f( x) ≥ 0, then the x-coordinate of the centroid (of the region between the graph of fand the x-axis for x in [a, b]) is to the right of the midpoint between a and b.Another interpretation:If f(x) is the density at x of a wire and the density is increasing as x increases for x in [a, b], then the center of massof the wire is to the right of the midpoint of [a, b].13.3 Concepts Review1. A rectangle containing S; 02. φ1 ≤ ≤ φ23.4.( x) y ( x)b∫∫aφ2xφ ( x)11 1– x∫∫0 0( )f ( x, y)dy dx12 x dy dx; 3Problem Set 13.31.3x y = dx = x dx =41 2 3x1 3[ ]0y 0 30∫ ∫3.4.5.33y3 ⎡x2⎤3 3 3⎢ + ⎥ = +–1 3–1⎢⎣⎥⎦x= 0∫ ∫4 3–1y x dy (9y 3 y ) dy= [3 y ] = 243 – 3 = 240x1 2 ⎛1⎞ 4 1 3 ⎛1⎞4– ––3⎜ ⎟ =4 –3⎜ ⎟⎝ ⎠ y=0⎝4⎠⎡ ⎤ ⎡ ⎤⎢x y y ⎥ dx ⎢x x ⎥dx⎣ ⎦ ⎣ ⎦∫ ∫= –32.22 y3⎡⎛1⎞ 2 3 ⎤3⎛3⎞2 31 ⎢⎜ ⎟2⎥ =1⎜ ⎟⎣⎝ ⎠ ⎦x=– y ⎝2⎠∫ ∫x exp( y ) dy y exp( y ) dy⎛ 1 ⎞= ⎜ ⎟ ( e – e ) ≈ 2.660 × 10⎝2⎠27 112.x–12⎡⎛ ⎞ 2⎤2⎛ ⎞ 2( –1)1 ⎢⎜ ⎟ ⎥1⎜ ⎟⎣⎝ ⎠ ⎦y=0 ⎝ ⎠1 1 1y dx = x dx =2 2 6∫ ∫Instructor’s Resource Manual Section 13.3 807

6.x5 –1 ⎛ y ⎞5tan1 ⎢ ⎜ ⎟⎥dx =1⎣ ⎝ ⎠⎦y=0⎡3 ⎤ 3 πdxx x x 4∫ ∫5⎡3πlnx⎤ 3πln5= ⎢ = ≈3.79214⎥⎣ ⎦ 411 115.2∫ ∫–1xxy dy dx = 07.1 2 2x12π1/2y=0 = π1/2∫ ∫[ y cos( x )] dx 2x cos( x ) dx2= – ≈ –0.22512π8.2cosθπ/4 ⎡⎛1⎞ 2⎤π/420 ⎢⎜⎟2⎥ =0⎣⎝⎠ ⎦r=2∫ ∫r dθ (cos θ –1) dθ(2 – π)= ≈ –0.1427816.1 4x x∫∫( + y) dydx = 60 49.π/9 3rπ/9θ0θ =π /4 =0π /9∫ ∫[tan ] dr (tan 3 r –1) dr⎡ ln cos3r⎤= ⎢– – r⎥⎣ 3 ⎦0⎛−ln1( 2 ) π⎞⎛ ln(1) ⎞= ⎜ – ⎟– – –0⎜⎜ ⎟3 9⎟ ⎝ 3 ⎠⎝ ⎠3ln2– π= ≈ –0.1180917.10.11.12.13.2x2 2 2 2 2⎡ −x x xye⎤ −dx 2xe dx⎡ −e⎤0 ⎢= = −⎣ ⎥⎦0⎢−x⎣ ⎥⎦0−4∫ ∫= 1−eπ/2 x sin y π/20 x= 0=0∫ ∫sin y[ e cos y ] dy ( e cos y – cos y ) dy= e – 2≈0.718323x52∫ 2⎡y⎤ 2 x ⎡ 1 6 ⎤⎢ ⎥ dx dx x1 3x∫ 1 3 ⎢18⎥⎢⎣⎥⎦⎣ ⎦012 2+0 ⎢ ⎜ ⎟∫= = = 3.524– x⎡ ⎛1⎞ ⎤xy y dx2 ⎥⎣ ⎝ ⎠ ⎦y=02 ⎡ 2 1/2 ⎛1⎞ 2⎤= ∫ x(4 – x ) + 2 – x dx0 ⎢⎜ ⎟ 2⎥⎣⎝ ⎠ ⎦16=318.1 x 2 1 2 2 x2( + 2 ) = [ + ]20 x0y=x∫∫ ∫∫x ydydx x y y dx1 5/2 4 4= [( x + x ) – ( x + x )] dx0⎡ 7/2 2 52x x 2x⎤ 2 1 2= ⎢ + – ⎥ = + –⎢⎣7 2 5 ⎥⎦7 2 527= ≈ 0.38577022 3x−x2∫∫0x( )3 210( x − xy)dydx2−x x − 4x+4 8= ∫dx =−0 2 1514.π/2 2 sinθπ/22/6r=0 =ππ/63 /2 7[sin θ ] π π /6 0.875∫ ∫[3r cos θ ] dθ 3sin θ cosθ dθ= = =819.2 2 2 –1 –1∫∫ 2(1 + x ) dydx = 4tan 2–ln5 ≈ 2.81920x808 Section 13.3 Instructor’s Resource Manual

20.24.21.Since S is symmetric with respect to the originand the integrand is an odd function in x, thevalue of the integral is 0.25.2 2x4∫∫0 0− + ⎡ ⎛1⎞20⎢2x+ ⎜ ⎟y ⎤ dydx =4⎥⎣ ⎝ ⎠ ⎦ 322.3 (–2/3) x+2∫∫0 0(6 – 2 x– 3 y) dydx = 626.22 (1/2) 36–9x∫∫0 0⎛1 ⎞⎜ ⎟ (9 x + 4 y ) dy dx = 10⎝6⎠23.4 (–3/4) x+3∫∫0 0(12 – 3 x – 4 y) dy dx = 244– y4– ydy dx⎛1dx ⎞⎛ ⎞= ⎜ ⎟⎜ dy ⎟2 ⎝ ⎠⎝ 2 ⎠5 4 5 4∫∫ ∫ ∫0 0 0 04⎡ 2y ⎤= 5⎢2 y – ⎥ = 5(8–4) = 20⎢⎣4 ⎥⎦023 9– x 2 2∫∫0 0(9 – x – y ) dydx239−x3⎡ 32 3/22 y ⎤ 2(9 − x )⎢(9 ) ⎥0 3 0⎢⎣⎥⎦3y=0∫ ∫= −x y− dx =dxπ /2 3π /2 4= ∫ 18cos t3costdt= 54cos tdt0∫ 0/2 81 27 cos 4t= ∫π ⎛ 27 cos 2t⎞ dt0⎜ + + ⎟⎝ 4 4 ⎠⎡81t 27sin 2t 27sin 4t⎤= ⎢ + +4 2 16⎥⎣⎦π /281π= ≈ 31.80868(At the third step, the substitution x = 3 sin t wasused. At the 5th step the identity2 ⎛1⎞cos A = ⎜ ⎟(1 + cos 2 A)was used a few times.)⎝2⎠0Instructor’s Resource Manual Section 13.3 809

27.31.28.1∫∫0 09 2∫∫0 0y4(1 – ydxdy= )15y⎡ ⎛5⎞ ⎤⎢5– ⎜ ⎟y dxdy = 729⎥⎣ ⎝ ⎠ ⎦32.⎡ 4 ⎤4– x – y dy dx = 3π9⎥⎣ ⎝ ⎠ ⎦22 (3/2) 4– x 2 ⎛ ⎞ 20 0 ⎢ ⎜ ⎟∫∫≈ 9.424829.Making use of symmetry, the volume is2 (16 – 2 ) 1/2 2 4 x(16 – 2 ) 1/2x dA =x dy dx∫∫ ∫ ∫R14 2 1/200 0= 2 ∫ [(16– x ) y] x y=0 dx2 3/24 2 1/2⎡ –2(16 – x ) ⎤= 2 ∫ (16– x ) xdx = ⎢ ⎥0⎢⎣3 ⎥⎦2(64) 128= 0 + = ≈ 42.66673 3401 x 2 1 2 x∫∫ =0 0 ∫ 0y = 01⎡21 x ⎤2tan x dy dx [ y tan x ] dxln cos⎛ 1 ⎞= xtan x dx ⎢– ⎥∫ = = – ln(cos1)0⎜ ⎟2 ⎝ 2⎠⎢⎣⎥⎦0≈ 0.307833.1 10 y∫∫f ( x, y)dx dy30.1 1– x x – y–1∫∫0 0⎛1 ⎞e dydx = ⎜ ⎟ ( e+ e – 2) ≈ 0.5431⎝2⎠810 Section 13.3 Instructor’s Resource Manual

34.4∫∫x0 x /2f ( x, y)dy dx1 038.2∫ ∫–1 x –1f ( x, y)dy dx1∫∫35.40yyf ( x, y)dxdy39.2 2– y 2 4 y–22xy dx dy0 0 2 0∫∫ ∫∫≈ 17.0667+ xy dx dy =2561536.37.1/3 1/31/2 y1 y∫ ∫ ∫ ∫f ( x, y) dx dy + f ( x, y)dx dy1/8 1/2 1/20 1 1 1∫ ∫ ∫ ∫f ( x , y ) dy dx + f ( x , y ) dy dx–1 – x0 xy1 – x+2 1/ 2 1/2–2 x–1/ 2 x∫ ∫ ∫ ∫40.2xydydx–245= – = –5.6258xydydx441. The integral over S of x y is 0 since this is anodd function of y. Therefore,∫∫ ∫∫S2 4 2+ =S( x x y)dA x dA2 2= 4⎛x dA x dA⎞⎜ + ⎟⎝∫∫S∫∫1 S2⎠⎛2 21 4– x 2 2 4– x ⎞2= 42x dy dx + x dy dx⎜∫∫ 0 1– x ∫∫ 1 0⎟⎝⎠2 2 2 1 2 2= 4⎛x 4– x dx– x 1– x dx⎞⎜ ⎟⎝∫0 ∫ 0⎠π/2 2 2 π/22 2= 4⎛16 sin θ cos θ dθ – sin φcosφdφ⎞⎜ ⎟⎝∫0 ∫0⎠(using x = 2 sin θ in 1st integral; x = sin φ in 2nd)π /2 2 2 15π= 60∫sin θ cos θ dθ=04(See work in Problem 42.)≈ 11.7810Instructor’s Resource Manual Section 13.3 811

42.43.2z = f( x, y) = sin( xy ) is symmetric with respectto the x-axis, as is the annulus. Therefore, theintegral equals 0.Then dx = 2 cos θ dθπx = 2 ⇒ θ =2x = 0 ⇒ θ = 0( 2– 2) π /2 2= (2sin θ ) (2cos θ)2cosθ dθ2∫ 0π /2 2 2= 82– ( 2)∫ sinθ cosθ dθ0( 2– 2)⎛ π ⎞π* = 8( 2– 2)⎜ ⎟=⎝16 ⎠ 2Therefore,⎡( )xdA π 2– 2 ⎤24 ⎢ ⎥∫∫ = = 2π( 2 – 2 ) ≈3.6806S ⎢ 2 ⎥⎣ ⎦22 y 3 2 3=0 0 0322 2 3⎡ cos( y ) ⎤∫ y sin( y ) dy –0⎢⎣3 ⎥⎦0∫∫ ∫2yx=0sin( y ) dxdy [ xsin( y )] dy= = ⎢ ⎥1–cos8= ≈ 0.3818344. Let S ' be the part of S in the first quadrant.∫∫22 2 4– x 2x dA =2S′∫0∫4– x / 222⎛ 2 2 4– x∫⎜ 4– –0x dy dx⎞= x x ⎟dx⎜2 ⎟⎝⎠2 2 2 ⎛ 1 ⎞= ∫ x 4– x 1– dx0⎜ ⎟⎝ 2 ⎠( 2– 2) 2= x 24– x 2dx2∫ 0⎛ π π⎞Let x = 2 sin θ, θ in ⎜– , ⎟.⎝ 2 2⎠π /2 2 20* = ∫ sin θ cos θ dθπ /2⎡1 ⎤⎡1⎤= ∫ (1 – cos 2 θ ) (1 cos 2 θ)dθ0 ⎢ +2 ⎥⎢2⎥⎣ ⎦⎣ ⎦1 π /2 2= (1 – cos 2 θ ) dθ4∫ 01 π 1 π /21= – (1 cos 4 θ ) dθ42 4∫+0 2/2θ ππ 1 π 1⎡sin4⎤= – +8 8 2 8 ⎢4 ⎥⎣ ⎦= π – π + 0=π8 16 1645. We first slice the river into eleven 100’ sectionsparallel to the bridge. We will assume that thecross-section of the river is roughly the shape ofan isosceles triangle and that the cross-sectionalarea is uniform across a slice. We can thenapproximate the volume of the water by11 111V ≈ A ( y ) Δ y = ( w )( d )100∑kk k k k= 1 k=1211∑= 50 ( w )( d )k=1kk∑0where wkis the width across the river at the leftside of the kth slice, and dkis the center depth ofthe river at the left side of the kth slice. ThisgivesV ≈ 50[300⋅ 40 + 300⋅ 39 + 300⋅ 35 + 300⋅31+ 290⋅ 28 + 275⋅ 26 + 250⋅ 25 + 225⋅24+ 205⋅ 23 + 200⋅ 21+ 175⋅19]3= 4,133,000 ft812 Section 13.3 Instructor’s Resource Manual

46. Since f is continuous on the closed and boundedset R, it achieves a minimum m and a maximumM on R. Suppose ( x1,y 1)and ( x2,y 2)are suchthat f ( x1,y1)= m and f ( x2, y2)= M.Then,m≤f ( x,y)≤ M∫∫ mdA≤∫∫ f ( x,y)dA≤∫∫M dAR R RmA( R) ≤∫∫f ( x,y) dA ≤ MA( R)R1m≤f ( x,y)dA≤MA( R)∫∫RLet C be a continuous curve in the plane from, x , y that is parameterized by( x1 y 1)to ( 2 2)x = xt (), y y()tht () f xt () yt ( )= , c≤t ≤ d.Let( )= , . Since f is continuous, so ish. By the Intermediate Value Theorem, thereexists a t 0 in ( cd , ) such that1ht = ∫∫ f xydA. But( 0 )( , )A( R)R( 0) ( 0) ( 0)a = x( t 0 ) and ( )( ) ( )ht = f xt , yt = f ab , , where1b = y t 0 . Thus,f ( ab , ) = f( xy , ) dAA( R)∫∫Ror, ∫∫ f ( x, y) dA= f ( a,b) ⋅A( R).R13.4 Concepts Review1. a ≤ r ≤b;α ≤θ ≤ β2. r dr dθ3.π∫∫4. 4π2 30 0 rdrdθProblem Set 13.42.3.4.5.6.7.sinθπ/2 ⎡⎛1⎞ 2⎤π/2⎛1⎞sin20 ⎢⎜ ⎟2⎥ =0⎜ ⎟⎣⎝ ⎠ ⎦r=0 ⎝2⎠∫ ∫π= ≈ 0.39278r dθ θ dθ3sinθ3π⎡r⎤πsin⎢ ⎥ dθ=0 3 0⎢⎣⎥⎦3r=02π (1 – cos ) sind0 3⎡3θ θ ⎤π∫ ∫= ∫–cos cos= ⎢ +⎢⎣3 9θ θ θ⎥⎥⎦1 1 1 1 4= ⎜ ⎛ – ⎟ ⎞ – ⎜ ⎛ – + ⎟⎞ =⎝3 9⎠ ⎝ 3 9⎠901–cosθ1 ⎞ 2 ⎤⎟rsinθ2⎥⎠ ⎦r=0θdθπ ⎡⎛∫dθ0 ⎢⎜⎣⎝π⎛1 ⎞ (1 – cos )2= ∫θ sin θ dθ0⎜ ⎟⎝2⎠4=32π ⎡12 θ ⎤ πcos θ0 ⎢ ⎥ = ∫⎣ ⎦00π∫θr d 2cos dθ2 4 4⎡ θ ⎤= ⎢8sin = 4 24⎥⎣ ⎦0θ2π2π⎡ 1 2⎤ 2π1 2 ⎡13⎤r d d0 ⎢ θ θ θ θ2⎥ = ∫ =00 2⎢6⎥⎣ ⎦ ⎣ ⎦03∫=4π31.cosθπ /2⎡⎛1⎞ 3 ⎤r sinθ0 ⎢⎜⎟3⎥⎣⎝⎠ ⎦r=0∫dθπ /2⎛1 ⎞ cos3= ∫θ sin θ dθ0⎜ ⎟⎝3⎠1= ≈ 0.083312π /3 4cosθ⎡2π⎤2∫ rdrdθ2 3 7.65290 ∫= + ≈2⎢ 3 ⎥⎣ ⎦Instructor’s Resource Manual Section 13.4 813

8.23 π/2 2–4sinθ3 π/2⎡r⎤2∫ rdrdθ25 π/6∫ =0 ∫ ⎢ ⎥5 π/6⎢⎣2 ⎥⎦3 π /2= 2 ∫ (6–8sin θ –4cos2 θ)dθ5 π /63 /2= 2[6θ + 8 cos θ – 2 sin 2 θ] π 5 π /6= 2( 4π+ 3 3)≈ 35.5252–4sinθ0dθ24sinθπ/6 4sin θπ/6⎡ ⎤rrdrdθ= ⎢ ⎥⎢⎣2 ⎥⎦∫ ∫ ∫0 0 0π/6 2 π/6d0 0∫ ∫0dθ= 8sin θ θ = 4(1–cos2 θ)dθ/6 2[4 θ – 2sin 2 θ] 0π π= = – 3 ≈ 0.3623312.9.4−1(1/ 2) cos (4 / 9) 3 cos 2θ∫ ∫0 2−1⎛4⎞= 65 −4cos⎜ ⎟⎝9⎠≈ 3.6213rdrdθπ /2 a sin2θ∫ ∫0 0 82a πrdrdθ=13.π 2/4 ⎡12 ⎤/4r d π 2dπ0 ⎢ θ θ2⎥ = =⎣ ⎦002∫ ∫10.2π6–6sinθ∫ ∫0 0rdrdθ= 54π≈169.646014.32π⎡12 24 80 2 r ⎤d π⎢ ⎥ =0d =⎣ ⎦1∫ ∫θ θ π11.814 Section 13.4 Instructor’s Resource Manual

15.θπ /2 ⎡1 2⎤π /2120 ⎢ r2⎥ dθ = θ dθ⎣ ⎦ 00 2π /2 3⎡13 ⎤ π= ⎢ θ6⎥ =⎣ ⎦048∫ ∫18.θ23 π /2⎡1 2⎤3 π /214r d d0 ⎢2⎥ =⎣ ⎦ 002∫ ∫3 π /2 5⎡ 1 5 ⎤ 243π= ⎢ θ10⎥ =⎣ ⎦0320θ θ θ19.16.cosθπ /2 ⎡1 2⎤π /21 2∫cos0 ⎢ r2⎥ d = ∫d⎣ ⎦002π /21 ⎡1 1 ⎤= ∫cos 2 θ dθ0 2⎢ +2 2⎥⎣⎦π /2⎡1 1 ⎤ π= ⎢ θ + cos 2 θ =4 8⎥⎣⎦08θ θ θπ 2 2r4e rdrdθe0 0∫∫2 =π( −1) ≈168.383620.17.sinθπ ⎡1 2⎤π 1 sin2∫0 ⎢ r2⎥ d = ∫ d⎣ ⎦00 2π 1 ⎡1 1 ⎤= ∫ sin 2 θ dθ0 2⎢ −2 2⎥⎣ ⎦π⎡1 1 ⎤ π= ⎢ θ + cos 2 θ =4 8⎥⎣⎦04θ θ θπ /4 2 2 1/2∫ ∫(4 – r )0 0rdrdθ2 3/22π /4⎡ ⎤(4 – r )= ∫ ⎢ ⎥ dθ0⎢⎣–3 ⎥⎦0π /4π /4⎛8⎞ ⎡8θ⎤ 2π= ∫ dθ2.09440⎜ ⎟ = = ≈3⎢3⎥⎝ ⎠ ⎣ ⎦ 3021.π /4 2 2 –1∫ ∫0 0⎛π⎞(4 + r ) rdrdθ = ⎜ ⎟ln 2 ≈0.2722⎝8⎠Instructor’s Resource Manual Section 13.4 815

22.π /2 2∫ ∫0 17rsinθrdrdθ=326.23.π/4 2cos θ –1 π/42cosθθ = []0 secθ0secθπ /4(2cos θ – sec θ)dθ0/4⎣2sin θ – ln secθ tanθ ⎦π0∫ ∫ ∫= ∫r rdrd r dθ= ⎡ + ⎤( )= 2 – ln ( 2 + 1)≈ 0.5328= ⎡ 2 – ln 2 + 1 ⎤– [0 – ln(1 + 0)]⎣⎦27.24.π /2 1 2 –1/2∫ ∫(4 – r ) rdrdθ0 0π= ∫/2 [–(4 – r2 ) 1/2 ]1 dθ00π /2⎛π⎞= ∫ (– 3 + 2 ) dθ = (– 3 + 2)≈0.42090⎜ ⎟⎝2⎠⎛π⎞[sin( )] rdrdθ = ⎜ ⎟ (1– cos1)⎝4⎠π /2 1 2r0 0∫ ∫≈ 0.3610R2 2 π /2 3 2+ =0 0∫∫ ∫ ∫( x y ) dA r rdrdθ81π= ≈ 31.8086828.25.π /2 cscθ2 2∫ ∫π /4 0r1cos θ rdrdθ= ≈0.083312∫∫2 2 1/2Rπ /2 2 2 1/24 (18–2 x –2 y ) dA= ∫ ∫4 (18–2 r )0 0rdrdθ⎛π⎞ 3/2 3/2= ⎜ ⎟(18 –10 ) ≈ 46.8566⎝3⎠816 Section 13.4 Instructor’s Resource Manual

29.⎡ 3y ⎤y dydx = ⎢ ⎥ dx⎢⎣3 ⎥⎦0 −x20( )−5 3x−5∫ ∫ ∫−x3x(–1– 3 3)⎡40 –1– 3 3x ⎤= 3xdx ⎢⎥∫ –5 3 = ⎢ 12 ⎥⎣⎦( + )1 3 3 625= ≈ 322.71631230. a. The solid bounded by the xy-plane and2 2z = sin x + y for x + y ≤ 4π is thesolid of revolution obtained by revolvingabout the z-axis the region in the xz-planethat is bounded by the x-axis and the graphof z = sin x for 0≤ x ≤ 2π.0–52 2 2∫∫ ∫∫2 2 2 2sin– sinS1 S2W = x + y dA + x + y dA2π π 2(sin r ) r drπ=⎡(sin r ) r dr ⎤∫ −d θ0 ⎢⎣∫0 ∫ π ⎥⎦2= 2 π[( π) – (–3 π )] = 8π ≈ 78.956831. This can be done by the methods of this section,but an easier way to do it is to realize that theintersection is the union of two congruentsegments (of one base) of the spheres, so (seeProblem 20, Section 5.2, with d = h and a = r) the⎡⎛1 ⎞ 2 ⎤ 2 (3 a– d)volume is 2 ⎢⎜⎟ π d (3 a– d) = 2 π d .3⎥⎣⎝⎠⎦ 332.2π10 – r/10 10 – r/10θ0 0 0– r /10∫ ∫ ∫100 = ke rdrd = 2πke rdrLet u = r and dv = e dr .– r /10Then du = dr and v = –10 e .2 ⎛–10 – r/10 10 1010 – r/10⎞= π k⎜⎡ re ⎤ + e dr⎣ ⎦0∫ 0⎟⎝⎠⎛ –1 – /10102 k –100 e – 100er ⎞= π ⎜⎡ ⎤⎣ ⎦0⎟⎝⎠–1 –1= 2 π k(–100 e –100e+ 100)–1e= 200 π k(1– 2 e ), so k = ≈0.6023.2 π( e –2)33.b.c.Regions A and B are congruent but region Bis farther from the origin, so it generates alarger solid than region A generates.Therefore, the integral is negative.2π 2π 2πθ0 0 0∫ ∫ ∫V = (sin r) rdrd = 2 π (sin r)rdrNow use integration by parts.2= 2 π(–2 π ) = –4 π ≈ –39.4784π /2 a sinθ2 2Volume 4 a – r rdrdθ= ∫ ∫0 0π /2⎡⎛1 ⎞ 3 3 3 ⎤= ∫ – ( a cos θ – a ) dθ0 ⎢⎜ ⎟3⎥⎣⎝⎠⎦⎛ 4⎞ 3⎡2 π⎤ ⎛2⎞3= ⎜– ⎟a– a (3 –4)3 ⎢ = ⎜ ⎟ π3 2⎥⎝ ⎠ ⎣ ⎦ ⎝9⎠Instructor’s Resource Manual Section 13.4 817

34. Normal vector to plane is 0,-sin a,cosa . Therefore, an equation of the plane is(–sin α)y + (cos α)z = 0, or z = (tan α)y, or z = (tan α)(r sin θ).π /2aVolume 2 (tan α) rsinθ rdrdθ= ∫ ∫0 0π /2 a 2d0 0= 2(tan α) ∫ sinθ θ∫r dr⎡ 3a ⎤ ⎛2⎞= 2(tan α)[1] ⎢ ⎥ = ⎜ ⎟atanα⎢⎣3 ⎥⎦⎝3⎠335. Choose a coordinate system so the center of the sphere is the origin and the axis of the part removed is the z-axis.Volume (Ring) = Volume (Sphere of radius a) – Volume (Part removed)2 22 24 3 2 π a – b 2 2 4 3 a – b 2 2 1/2= πa –2 a – r rdrdθ3∫0 ∫= πa –2(2 ) ( a – r ) rdr03π∫ 04 ⎡1= π a + 4 π ( a r )3⎢ −⎣33 2 2 3/2⎤⎥⎦02 2a −b4 1 4= π a + 4 π ( b – a ) = π b3 3 33 3 3 336.2 2 2EF = a – b2 2CD = a – ( h – b)2 2 2AB = b – ( h– b)37.38.2 2 2 2Area of left cross-sectional region =π[ a – ( h– b) ] – π [ a – b ]2 2=π [ b – ( h– b) ] = area of right cross-sectional region⎛4⎞ 3 ⎛1⎞2⎛1⎞ 2Volume = ⎜ ⎟πb – ⎜ ⎟π(2 b– h) [3 b– (2 b– h)]= ⎜ ⎟πh (3 b– h)⎝3⎠ ⎝3⎠⎝3⎠hAlternative: 2 2 1V = π ⎡b ( t b) ⎤∫ − − dt = πh 2 ( 3b−h)0 ⎢⎣⎥⎦3⎡1 1lim (1 r ) rdr⎤ ⎛ ⎡⎛ ⎞ ⎛ ⎞⎤⎞d lim ⎜– ⎟(1 b ) – ⎜–⎟ ⎟dθ⎣ ⎦ 2 2⎥⎝ ⎣⎝ ⎠ ⎝ ⎠⎦⎠π/2 b 2 –2 π/22 –1+ θ = +0 ⎢b→∞0 ⎥ ⎜0 ⎢b→∞∫ ∫ ∫1 2 1 2A=r2 ( θ2 – θ1)– r1 ( θ2 – θ1)2 22 2=1 ( θ – 1 θ )( – )2 r2 r12= 1 ( θ – 2 θ )( 1 r – )( )2 r1 r2 + r12r1+r2= ( r2 – r1)( θ2 – θ1)2π /2⎛1⎞ π= ∫ dθ 0.78540⎜ ⎟ = ≈⎝2⎠4818 Section 13.4 Instructor’s Resource Manual

x − μ39. Using the substitution u = we getσ 2dxdu = . Our integral then becomesσ 2∞ 1 2 2 2−( x−μ) /2σ∞ 1 −u∫ e dx = e du−∞∫σ 2π −∞ π2 ∞ 2u= e −∫ duπ 0Using the result from Example 4, we see that∞ 2ue − π∫ du = . Thus we have0 2∞ 12 2−( x−μ) /2σ∫ e dx−∞ σ 2π2 π= ⋅ = 1.π 213.5 Concepts Review2.3.22 4– x–2 0∫ ∫m = y dy dx =M = 0 (symmetry)yMx22 4– x–2 0∫ ∫163= yydxdy = 2π⎛ 3π⎞( x, y) = ⎜ 0, ⎟⎝ 8 ⎠1.2.∫∫S∫∫S2 4x ydA2 5x ydAm3.∫∫S4 4x ydA4. greaterProblem Set 13.51.3 4∫∫m = ( y + 1) dx dy = 300 03 4M = x( y+ 1) dxdy = 60y∫∫0 03 4M = y( y+ 1) dxdy = 54x∫∫0 0( x, y ) = (2, 1.8)2sin xπ sin xπ⎡ ⎤ym = ∫∫ y dy dx = dx0 0 ∫ ⎢ ⎥0⎢⎣2 ⎥⎦2πsin x π1– cos 2x= ∫ dx =dx0 2∫ 0 4⎡ x sin 2–x ⎤π = π⎢=4 8⎥⎣ ⎦ 4003sin xπ sin xπ⎡y⎤M x = ∫∫ yy dy dx = dx0 0 ∫ ⎢ ⎥0⎢⎣3 ⎥⎦03πsin x 1 dx π 2(1 – cos x ) sin x dx0 0∫ ∫= =3 331⎡cos x ⎤ 4= ⎢–cosx+ ⎥ =3⎢⎣3 ⎥⎦9M x49π416y = = = ≈0.5659;m 9ππx = (by symmetry)2Thus , My= x⋅ m= π ⋅ π =π4 2 8π02Instructor’s Resource Manual Section 13.5 819

4.1 x2 1/ x0 0 1 0∫∫ ∫∫m = x dy dx + x dy dx =1 x 2 2 1/ x 2 7M y = ∫∫ x dy dx + x dy dx0 0 ∫∫ =1 041 x2 1/ x ⎛1 ⎞M x = ∫∫ xy dy dx + xy dy dx0 0 ∫∫= (1 4 ln 2)1 0⎜ ⎟ +⎝8⎠⎛21 ⎛ 3 ⎞ ⎞( x, y) = ⎜ , ⎜ ⎟(1 + 4ln 2) ⎟≈(1.3125, 0.3537)⎝16 ⎝32⎠ ⎠435.– x1 e 2 ⎛1 ⎞ (1 ––3m = ∫∫ y dy dx = e )0 0⎜ ⎟⎝9⎠− x1 e 3 ⎛ 1 ⎞ (1−4Mx= ∫∫ y dydx = −e) ≈ 0.06140 0⎜ ⎟⎝16⎠– x1 e 2 ⎛ 1 ⎞ (1 – 4–3M y = ∫∫ xy dy dx = e )0 0⎜ ⎟ ≈⎝27⎠0.0297( x, y ) =⎛⎛1⎞ ⎛ 9 ⎞⎜⎜ ⎟( e –4)( e –1) , ⎜ ⎟e ( e –1)( e –1)⎝⎝3⎠ ⎝16⎠3 3 –1 –1 4 3 –1⎞⎟⎠≈ (0.2809, 0.5811)6.820 Section 13.5 Instructor’s Resource Manual

x21 e(2 – )1⎡y ⎤m = ∫∫ x+y dydx = (2 – x)y0 0 ∫ ⎢ + ⎥0⎢⎣2 ⎥⎦2x12⎡x x x e ⎤ e + 8 e–13= ⎢2 e –( xe – e ) + ⎥ =⎢⎣4 ⎥⎦40xey=0xx2 3e1 eM x = ∫∫ (2 – x+y )1⎡2 xy y ⎤ydydx = y –0 0 ∫ ⎢ + ⎥0⎢⎣2 3 ⎥⎦⎡ 2x 2x 2x 3xe ⎛ xe e ⎞ e ⎤= ⎢ – –+ ⎥⎢ 2 ⎜ 4 8 ⎟⎣ ⎝ ⎠ 9 ⎥⎦101⎛2xdx 2 x – x e ⎞= ∫e xe +dx0 ⎜2 ⎟⎝⎠y=01⎡2x32x xe edx = ∫ ⎢e– +0⎢⎣2 3⎛ e 2 e 2 e 2 e 3 ⎞ ⎛1 1 1⎞x⎤⎥ dx⎥⎦3 28e+ 27 e –53= – + + – ⎜ –0+ + ⎟ =⎜ 2 4 8 9 ⎟⎝⎠ ⎝2 8 9⎠72x21 e(2 – )1⎡2 xy ⎤1⎛2x2M y = ∫∫ x+y xdydx = ⎢2 xy–x y+⎥ dx0 0 ∫0⎢⎣20⎥2 x – x xe ⎞= ∫xe x e + dx⎜⎦2 ⎟⎝⎠⎡⎛= ⎢(2 xe – 2 e ) – ( x e – 2xe + 2 e ) +–⎢⎜⎣⎝ 4 82x2xx x 2 x x x xe exey=002 3 2+ M x +y2 21⎞⎤⎟⎥⎠⎥⎦e 2 – 8e+33=8M y e –8e 33 8e 27 e –53x = = ≈ 0.5777; = = ≈1.0577m 2( e + 8 e–13) m 18( e + 8 e–13)7.8.π 2sinθ0 0∫∫m= rrdrdθ=329π 2sinθ64Mx= ∫∫ ( rsin θ)rrdrdθ=0 015M = 0 (symmetry)y( x, y ) = (0,1.2)5m= 2∫∫rrdrdθ=0 03π 1+ cosθππ 1 + cosθ7 π∫∫My= 2 ( rcos θ)rrdrdθ=0 04M = 0 (symmetry)x( x, y ) = (1.05, 0)Instructor’s Resource Manual Section 13.5 821

9.π 21π 2m = ∫0∫rdrdθ = drd1 r∫0∫θ = π1π 2 1 π 3Mx= ∫ rcosθ rdrdθ cosθdθ30∫= =1 r∫02M y = 0bysymmetry⎛ 3 ⎞, = ⎜ 0,π ⎟⎝ ⎠( x y)11.10.2π2+2cosθm = ∫0 ∫ rrdrdθ02+2cosθ2π⎡13 ⎤= ∫0⎢ r3⎥ dθ⎣ ⎦02π1 8 24cos 24cos2 8cos3= ⎡ θ θ θ⎤∫ + + + dθ0 3 ⎣⎦1 40π= [ 0+ 24π+ 16π]=3 3M x = 0bysymmetry2π2+2cosθM y = ∫( cos )0 ∫ r r θ rdrdθ02+2cosθ2π⎡14 ⎤= ∫ r cosθdθ0 ⎢4⎥⎣ ⎦0M y = 28π⎛ 28π⎞ ⎛21⎞, = ⎜ ,0 = ,040 π / 3⎟ ⎜10⎟⎝ ⎠ ⎝ ⎠( x y)12.3 9 202yI = y ( x+y)dxdyx∫∫3⎛2 681y3 y 5⎞7533= ∫+ 9 y – − y dy = ≈ 2690 ⎜2 2 ⎟⎝⎠ 289 x932⎛7/2 x ⎞I y = ∫∫ x ( x+ y)dydx = x dx0 0 ∫+0⎜2 ⎟⎝ ⎠41553= ≈ 51948305937Iz = Ix + Iy= ≈ 5463564 y 3 2048Ix= ∫∫ y dxdy = ≈227.560 – y94 y 2 512Iy= ∫∫ x ydxdy = ≈ 24.380 – y2115872Iz = Ix + Iy= ≈ 251.946313.a a⎛ 5 ⎞Ix= ∫∫ ( x+ y)y dxdy = a0 0⎜ ⎟⎝12⎠⎛ 5 ⎞ 5 ⎛5⎞ 5I y = ⎜ ⎟ a ; Iz= ⎜ ⎟ a⎝12⎠ ⎝6⎠2 5822 Section 13.5 Instructor’s Resource Manual

14.xa a– y 2 2 2( )0 0∫∫I = x + y y dxdy1 = + −3∫Iya 3 2 2 2 5( a y –3 a y 6 ay 4 y ) dy06 667a=1807a7a= ; Iz= (Same result for a < 0)180 9015. The density is constant , ( x,y)( x y)δ = k .22 ⎡k2 ⎤m = ∫ kxdx =0 ⎢ x 22⎥ = k⎣ ⎦0x2 x2⎡k2 ⎤M x = ∫∫ ky dy dx =0 0 ∫0⎢ y2⎥ dx⎣ ⎦022 k 2 ⎡k 3⎤4k= ∫ x dx =0 2⎢ x6⎥ =⎣ ⎦032 x2 xM y = ∫∫ kx dy dx = [ kxy]dx0 0 ∫0022 2 ⎡k3⎤8k= ∫ kx dx = x0 ⎢3⎥ =⎣ ⎦03⎛8 k/3 4 k/3⎞ ⎛4 2⎞, = ⎜ , = ,2k2k⎟ ⎜3 3⎟⎝ ⎠ ⎝ ⎠16. The density is proportional to the distance fromδ x,y = ky .the x-axis, ( )( 1 )11⎡k 2⎤1k 2 km = ∫ 0⎢y dx x2⎥ = ∫ − =⎣ ⎦ 0x 2 311 x 2 1⎡k3⎤x = ∫∫ =0 0 ∫0⎢ 3⎥⎣ ⎦xk 1 3 k=0( 1 )3∫− x dx =411 1 1⎡kxy2⎤y = ∫∫ =0 x ∫ ⎢0 2⎥⎢⎣⎥⎦xk 1 3 k= ∫0( x− x ) dx =M ky dy dx y dxM kxy dy dx dx2 8⎛k/8 k/4⎞ ⎛3 3⎞, = ⎜ , ,k/3 k/3 ⎟=⎜8 4⎟⎝ ⎠ ⎝ ⎠( x y)17. The density is proportional to the squaredδ x,y = k x 2 + y2 .distance from the origin, ( ) ( )3 9−x22 2m= k( x + y ) dydx∫ ∫−3 09−x23 ⎡ 2 1 3⎤∫ k x y y dx−3⎢3⎥⎣ ⎦03 ⎡ 2 4 1 6⎤∫ k 246 72x 8x x dx−3⎢3⎥⎣⎦3⎡3 8 5 1 7⎤25596kk⎢246x 24x x x5 21⎥⎣⎦−3353 9−x22 2x = ∫−3∫0( + )9−x23 ⎡1 2 2 1 4⎤∫ k x y y dx−3⎢2 4⎥⎣⎦03 ⎡2 6 86561 1377x 225 4 17x x∫ ⎢−3= += − + −= − + − =M ky x y dydx= +⎤= k − + x − + ⎥ dx⎢⎣4 2 2 2 4 ⎥⎦29160k=7Instructor’s Resource Manual Section 13.5 823

M y( x y)= 0bysymmetry⎛ 29160 k / 7 ⎞ ⎛ 450 ⎞, = ⎜0, = 0,25596 k / 35⎟ ⎜79⎟⎝ ⎠ ⎝ ⎠20. The density is constant, ( r,)δ θ = k .18. The density is constant, ( x,y)( x y)δ = k .π /2 π /2m= ∫ kcos xdx = [ ksin x]= 2k−π/2−π/2π /2 cosxMx= ∫ kydydx−π/2∫0cos xπ /2 ⎡k y2 ⎤= ∫dx−π/2⎢ 2⎥⎣ ⎦0k π /2 2 kπ= cos xdx2∫=−π/24M y = 0bysymmetry⎛ kπ/4⎞ ⎛ π ⎞, = ⎜0, = 0,2k⎟ ⎜8⎟⎝ ⎠ ⎝ ⎠19. The density is proportional to the distance fromδ r,θ = k⋅ r .the origin, ( )3π 3 2 π ⎡k3⎤m= ∫0∫kr drdθ=1 ∫0⎢ r3⎥ dθ⎣ ⎦1π 26 26kπ= ∫ kdθ=0 3 3π 3 2Mx= ∫ kr rsindrd0∫θ θ13π ⎡k r4 ⎤ π= ∫ sin d 20 k sin d0 ⎢4⎥ = ∫⎣ ⎦ 01π= [ − 20kcosθ] = 40k0M y = 0 bysymmetry⎛ 40k⎞ ⎛ 60 ⎞, = ⎜0, 0,26 kπ/ 3⎟=⎜13π⎟⎝ ⎠ ⎝ ⎠( x y)θ θ θ θθπ /2 θ π /2⎡k2 ⎤m= ∫0 ∫ krdrdθ=0 ∫0⎢ r2⎥ dθ⎣ ⎦0π /2π /23k 2 ⎡k 3⎤kπ= ∫ θ dθ =0 2⎢ θ6⎥ =⎣ ⎦048π /2 θ 2Mx= ∫ kr sinθdrdθ0 ∫0θπ /2 ⎡k3 ⎤ π /2k3= ∫ r sin d = sin d0 ⎢3⎥ ∫⎣ ⎦ 00 32( π − 8)θ θ θ θ θk=4π /2 θ 2My= ∫ cos0 ∫ kr θ drdθ0θπ /2 ⎡k3 ⎤ π /2k3= ∫ r cos d cos d0 ⎢=3⎥ ∫⎣ ⎦ 00 3( x y)3( π − 24π+ 48)θ θ θ θ θk=24⎛ 3 22( π − 24π + 48) 12( π −8⎞), =⎜,⎟3 3⎜ ππ ⎟⎝⎠824 Section 13.5 Instructor’s Resource Manual

21.a∫∫am = ( x+ y ) dxdy = a0 0I x1/2 1/2⎛ ⎞ ⎛ 5 ⎞r = ⎜ = ⎜ ⎟ a ≈0.6455am⎟⎝ ⎠ ⎝12⎠324.22.23.⎛1⎞m = ∫∫ ( x + y ) dxdy = a0 0⎜ ⎟⎝6⎠a a– y 2 2 4I y1/2 1/2⎛ ⎞ ⎛ 7 ⎞r = ⎜ ⎟ = ⎜ ⎟ a ≈0.4830a⎝ m ⎠ ⎝30⎠b/2 a/2 2 2∫ ∫I = I z = ( – b/2 – a/2 + y ) kdxdy⎛ k ⎞ 3 3= ⎜ ⎟( ab+ab)⎝12⎠25.2m = δπaThe moment of inertia about diameter AB is2πa 2 2xsin0 04 2 42π2πI I δ r θ rdrdθ= =∫ ∫δasin θ δa= ∫ dθ = (1 – cos 2 θ)dθ0 4 8∫ 04 2π4δa⎡ sin 2θ ⎤ δaπ= θ –8⎢=2⎥⎣ ⎦ 40Ix= ∫∫= ∫ ∫2δ ydASπ /2 2asinθ22 δ ( rsin θ)rdrdθ0 0π /2 4 6= 2δ ∫ 4asin θ dθ04 (1)(3)(5) π 5aδπ= 8aδ=(2)(4)(6) 2 44r1/2I⎛⎛ ⎞= ⎜ ⎟ = ⎜ ⎟ =⎝m⎠ ⎜δπa⎝ ⎠4 1/2δ a π ⎞4 a2 ⎟ 226. x = 0 (by symmetry)π /2 a(1+sin θ )3a(1+sin θ )π /2 ⎡r⎤M x = ∫∫ 1ydA=2 k ( rsin θ ) rdrdθS ∫– π /2∫= 2ksinθdθ0∫ ⎢ ⎥– π /2⎢⎣3 ⎥⎦r= 032kaπ /2 3= (1 sin θ ) sinθ dθ3∫ +– π /232kaπ /2 2 3 4= (sinθ 3sin θ 3sin θ sin θ)dθ3∫ + + +– π /234kaπ /2 2 4= (3sin θ sin θ)dθ3∫ +(using the symmetry property for odd and even functions.)03 34ka⎛ 1 π 1⋅3 π⎞5πka= ⎜3+ ⎟=3 ⎝ 2 2 2⋅4 2⎠4M x 5 aTherefore, y = = .m 6(using Formula 113)Instructor’s Resource Manual Section 13.5 825

2 π /2 a(1+sin θ )42π /2 ⎡r2⎤Ix= ∫∫ ky dA=2 k ( rsin θ ) rdrdθS ∫– π /2∫= 2ksin θ0∫ ⎢ ⎥– π /2⎢⎣4 ⎥⎦441(1+sin θ )ka π /2 4 2 ka π /2 2 3 4 5 6= (1 + sin θ ) sin θ dθ2∫= (sin θ 4sin θ 6sin θ 4sin θ sin θ)dθ– π /2 2∫ + + + +– π /2π /2 2 4 6= ka∫ (sin θ + 6sin θ + sin θ)dθ(symmetry property for odd and even functions)04 ⎡1 π 1⋅3 π 1⋅3⋅5 π⎤49πka= ka ⎢ + 6 + =22 2 42 2 4 62⎥⎣ ⋅ ⋅ ⋅ ⎦ 324(using Formula 113)r=0dθ27.x = 0 (by symmetry)π /2 a(1+sin θ )π /2⎛ 3a ⎞2 3 4M x = ∫∫ ky dA = 2 k ( r sin θ ) r dr dθS ∫– π /2∫= 2 k (3sin θ 3sin θ sin θ)dθ0∫+ +0 ⎜ 3 ⎟⎝ ⎠⎛2 ⎞ 3⎡(15π+32) ⎤ ⎛ 1 ⎞ 3= ⎜ ⎟ka= ka (15π+32)3 ⎢⎜ ⎟16 ⎥⎝ ⎠ ⎣ ⎦ ⎝24⎠π /2 a(1+sin θ )π /2⎛1⎞ 2 22 ⎡(8 +π) ⎤ ⎛1⎞ 2m = ∫∫ k dA = 2k r dr dθS ∫– π /2∫= 2 k a (2sinθ + sin θ)dθ0∫0⎜ ⎟= kaka ( 8)⎝2⎢ = ⎜ ⎟ π+⎠4 ⎥⎣ ⎦ ⎝4⎠1 3M ( 24 ) ka (15π+ 32)xa(15π+32)Therefore, y = = =≈ 1.1836am 1 2ka ( π+ 8) 6( π+ 8)( )428.∫∫∫∫∫∫ ∫∫ ∫∫22 2SS2 2S S S2 2 2y0I′ = ( x+d) δ ( x, y)dA = ( x + 2 xd + d ) δ ( x, y)dA= x δ( x, y) dA + 2 xdδ( x, y) dA + d δ( x, y)dA= I + M + d m = I + + d m= I + d m826 Section 13.5 Instructor’s Resource Manual

29. a.32.b.∫∫ ∫ ∫m = ( x+ y) dA= ( x+y)dxdyS0 02a ⎞a2⎛a ⎡x⎤ ⎛a⎞= ⎜ xy ⎟∫ ⎢ + ⎥ dy = + ay dy0 ⎜ 2 ⎟ ∫ 0⎢ ⎥⎜ 2 ⎟⎝⎣ ⎦x=0 ⎠ ⎝ ⎠⎡ 2 2a y ay ⎤= ⎢ + ⎥ = a⎢⎣2 2 ⎥⎦a03aa a 2y ∫∫S∫0∫0⎛2a ⎞a22 a⎜ xy ⎟20 2 ⎜⎜⎟ ∫0⎢ ⎥2y=0M = x( x + y) dA = ( x + xy)dy dx⎡ ⎤ ⎛ a x ⎞= ∫ ⎢x y+ ⎥ dx = ax +dx⎜ ⎟⎝⎣ ⎦ ⎠ ⎝ ⎠3 2 2a4⎡ax a x ⎤ 7a= ⎢ + ⎥ =⎢⎣3 4 ⎥⎦120M y 7 aTherefore, x = = .m 12a33.The square of the distance of the corner from thecenter of mass is dI = I(Prob. 16) + md2a=22 2+ b43 3 2 2 3 3kab ( + ab ) ( )( )a + b kab += + kab =ab12 4 3M x δ ( x , y ) dAy= ∫∫S1∪S2∫∫ ∫∫∫∫ δ= xδ( x , y ) dA+xδ( x , y ) dAS1 S2m x ( x , y ) dA m x δ( x , y ) dA= +∫∫1S21 S2m1 m2= mx 1 1+m2x2M y mx 1 1+m2x2Thus, x = =which is equal tom m1+m2what we are to obtain and which is what wewould obtain using the center of mass formulafor two point masses. (Similar result can beobtained for y .).c.34.30.31.2 5a⎛7a⎞3I y = IL+ d m,so = IL+ ⎜ ⎟ ( a );12 ⎝12⎠511aI L =14452 4 2 2I25 = I23 + md = 0.25 δa π+ ( δπa ) a4= 1.25δa πIx4ka π= 2[ I23]= ;2I = 2[ I + md ]y234 2 2 4= 2[0.25 a π+ ( kπ a )(2 a) ] = 8.5kaπI = I + I = 9kaπz x y4222 2 3(– a)( δπ a ) + ( ta)[ δπ( ta) ] a( t –1)x = =2 2 2δπ a + δπ ( ta) t + 12( a)( δπ a ) + (0) ay = =2 2 2δπ a + δπ ( ta) t + 1Instructor’s Resource Manual Section 13.5 827

35.3.⎛2 2a a – x⎜2 2– a – a – x 2 2 2∫ ∫⎜⎝aa – x – y2π a ⎛ ar ⎞= ∫ dr d ;0 ∫θ0 ⎜ 2 2 ⎟⎝ a – r ⎠⎞⎟dy dx⎟⎠22πa36.ab , is perpendicular to the line ax + by = 0.Therefore, the (signed) distance of (x, y) to L isthe scalar projection of x,y onto ab , , whichx, y ⋅ a,b ax + byis d( x, y) = = .a, b a,bM d( x, y) δ ( x, y)dAL= ∫∫= ∫∫SSax + by( x, ydA )a,bδab= x ( x, y) dA y ( x, y)dAa, b∫∫ δ +δSa,b∫∫ Sa b= (0) + (0) = 0a, b a,b[since ( x, y ) = (0, 0)]4. 2πahProblem Set 13.61.2.1 2z = 2 − x−y 2 31 2fx( x, y) = − ; fy( x,y)= −2 32 1 1 4A( G) = ∫∫ + + 1dydx0 0 4 961= ≈ 2.60343The equation has the form x + y = b.3a3a3 a+ =b so b = .2π2ππ3 aTherefore, the equation is x+ y = , or ππ x + π y = 3a.13.6 Concepts Review1. u×v2.∫∫S2 2x yf + f + 1 dA1 1z = 2 − x+y 2 31 1fx( x, y) = − ; fy( x, y)=2 334 − x+620 01 1A( G) = ∫∫ + + 1dydx4 97 4 3 76∫x dx0 2 6⎛ ⎞ ⎛ ⎞= ⎜− + 6 ⎟ = ⎜ ⎟(12) = 14⎝ ⎠ ⎝ ⎠828 Section 13.6 Instructor’s Resource Manual

3.6.4.2 1/2z = f( x, y) = (4– y ) ; f ( x, y) = 0;fy ( x , y ) = – y (4– y )2 –1/21 2 2 2 –1= ∫∫+0 11 2 21 2dx dy =0 1 2∫ 0 2A( G) y (4– y ) 1dxdy= ∫∫4– y10x4 − ydy⎡ −1⎛ y ⎞⎤ ⎛π⎞π= ⎢2sin ⎜ ⎟ = 2⎜ ⎟−2(0)2⎥= ≈ 1.0472⎣ ⎝ ⎠⎦⎝6⎠32 2Let z = f( x, y) = x + y ; fx( x, y) = 2 x;f y ( x, y) = 2 y.∫∫2 2AG ( ) = 4 4x + 4y + 1dydxS22 4−y∫∫ 0 02x2y dy dxπ /2 2 2 1/2∫0 ∫ r rdrdθ02 3/22π /2⎡3/2r ⎤∫ ⎢ ⎥ dr0⎢⎣⎥⎦0= 4 4 + 4 + 1= 4 (4 + 1)(4 + 1) (17 −1)π= 4=12 3 2≈ 36.17695.22 4– y 2 –1/2∫∫AG ( ) = 2(4– y ) dxdy=40 0(See problem 3 for the integrand.)2 1/2Let z = f( x, y) = (9– x ) .2 –1/2f ( x, y) = – x(9– x ) , f ( x, y) = 0xy2 3 2 2 –1( ) = ∫∫ [ (9– ) + 1]0 02 3 2 1/2 1∫∫ 3(9 x ) dydx 9sin0 0A G x x dydx= −− − ⎛2⎞= ⎜ ⎟⎝3⎠≈ 6.5675Instructor’s Resource Manual Section 13.6 829

7.2 2 1/2z = f( x, y) = ( x + y )2 2 –1/ 2 2 2 –1/ 2yf ( x, y) = x( x + y ) , f ( x, y) = y( x + y )x4 4– x 2 2 2 –1 2 2 2 –1 1/2∫∫A( G) = [ x ( x + y ) + y ( x + y ) + 1] dydx0 04 4– x∫∫= 2dy dx = 8 20 08.π /2 b a= 8∫0 ∫ rdrdθ0 2 2a – r⎛π⎞ b 2 2 –1/2= 8 a⎜ ⎟ ( a – r ) rdr2∫ ⎝ ⎠0( )2 2 1/2b2 2= –4 aπ ⎡( a – r ) ⎤ = 4 πa a– a – b⎣ ⎦010.9.⎛1⎞ 2z = f( x, y) = ⎜ ⎟x+ 4⎝4⎠xfx= ( x, y) = ; fy( x, y) = 021/21 2 1 2A( G) ⎡ ⎛ ⎞ ⎤= ∫∫ x + 1 dy dx0 0 ⎢⎜⎟4⎥⎣⎝⎠ ⎦( 5+1)⎡ ⎤5= + 2ln⎢⎥ ≈ 2.08052 ⎢ 2 ⎥⎣ ⎦2 2xyfx( x, y) = ; f ( x, y)=a −x − y a −x − y(See Example 3.)A( G) 8= ∫∫2 2 2 y2 2 22 2a ( b/ a) a – x a0 0 2 2 2a – x – ya –1 ⎛b⎞ 2 –1 ⎛b⎞= 8a∫sin dx 8asin0⎜ ⎟ = ⎜ ⎟⎝a⎠ ⎝a⎠dydx11.2 2xyfx( x, y) = ; f ( x, y)=a −x − y a −x − y(See Example 3.)aA( G) = 8∫∫dAS 2 2 2a – x – y2 2 2 y2 2 2830 Section 13.6 Instructor’s Resource Manual

12.2x2y2 2 2 y2 2 2fx( x, y) = ; f ( x, y)=a −x − y a −x − y(See Example 3.)π /2 a sinθaA( G) = 4∫ rdrd0 ∫θ0 2 2a – r2 π /22∫ θ θ0= 4 a (1− cos ) d = 2 a ( π–2)Following the hint, treat this as a surfacex = f ( y,z)= ay− yz .a−2yf y =, f z = 02 y a−y( )2 2 afy+ fz+ 1 =.2 y( a−y)The region S in the yz-plane is a quarter circle.aA( G)= 4∫∫dzdy2 y a−yS= 2aa∫2 2a − y∫( )0 0a 2 2( − y)1( − y)y aa − y a+y= 2a∫dy = 2a∫dyy ay0 0Make the substitution:2y = atanu( )2 2a+ y = a 1+ tan u = asecua+y= cscuy2dy = 2a tanu sec u du( )π /4∫0π /42 3( ( ))dzdyAG = 2a cscu⋅2atan usecudu∫= 4asec udu02 1 1π /4⎡⎤⎣2 2 ⎦02= 4a secutan u+ ln secu+tan u= 2a2+ ln 1+2a213.14.Let f ( x,y)2 2( x − y )= .a2x−2yfx( x, y) = ; fy( x, y)=aa2π a2 24r + aA( G)= ∫ rdrdθ0 ∫ 0 a22πaπa5 5 –12 2 1/2= (4 r a ) rdra∫ + =06−xfx( x, y) = ; fy( x, y) = 02 2a − x2( fxx y ) ( fyx y )( , ) + ( , ) + 1=A(all sides) = 8( )aa2 2−π /2 a0 ∫0 2 2 2∫2π /2 a2 2dθa a0∫= 8 = 8 (1) = 81+sinθaxa=2 2 2a − r cos θadθrdr− r cos θInstructor’s Resource Manual Section 13.6 831

15. f =− 2; x f =− 2yx∫∫y2 2SA = 4x + 4y + 1dAR2π2 2= ∫ 4 10 ∫ r + rdrdθ021 2π2= +8∫03⎡ 3/22 ⎤⎢( 4r1)⎥ dθ⎣ ⎦03/2− ( 17 −1)2π3/217 1= ∫ dθ= ≈36.180 12 6π17. ( )z = f x, y = ( D− Ax−By)/C2 2( ) ∫∫AG = ( AC / ) + ( B/ C) + 1dARD/ A D/ B−( A/ B) x 2 2= ∫( A / C) + ( B/ C) + 1dydx0 ∫02 2= ( A/ C) + ( B / C) + 1 × Area( Triangle)1 DD 2 2= ( A / C ) + ( B / C ) + 12 AB2D 2 2 2= A + B + C2ABC16. Using Formula 44 from the table of integrals,f =− 2 x; f = 0xy20 3 2AG ( ) = ∫0 ∫ 4x + 1dxdy−320 3 2= 2∫0 ∫ 4x+ 1dxdy0320 ⎡ 2 12 ⎤= ∫ 4 1 ln 2 4 10 ⎢x x + + x+ x +2⎥ dy⎣⎦020 ⎡ 1⎤= ∫ 3 37 ln 6 37 ( 0 ln1)dy0 ⎢ + + − +2⎥⎣⎦⎡ 1⎤= 20 ⎢3 37 + ln ( 6 + 37 ) ≈389.882⎥⎣⎦18. Let z = C− x be the equation of the plane thatdefines the roof, where C is a constant. Thus,f = − 1 and f = 0 .xy( ) ∫∫ ( )AG∫∫2 2= − 1 + 0 + 1dAR( )= 2 dA = π 18 2 ≈1440sq.ft.R219. x = y = 0 (by symmetry)h1+hLet h = 2 . Planes z = h1and z = h cut out2the same surface area as planes z = h and z = h2.Therefore, z = h,the arithmetic average of h 1and h 2.20.Area = 2πah= 2 π aa ( – acos φ) = 2 π a (1– cos φ)2832 Section 13.6 Instructor’s Resource Manual

21. a.A =π b2d.b.c.2B = 2 π a (1–cos φ)(Problem 20)2 ⎡ ⎛2 a 1–cos b ⎞⎤= π ⎢ ⎜ ⎟a⎥⎣ ⎝ ⎠⎦2 4 62⎡ b b b ⎤= 2 π a ⎢ – + +…2 4 4⎥⎢⎣2! a 4! a 6! a ⎥⎦⎡⎤=π b + … ≤π⎢⎣12a360a⎥⎦2 42 b b2⎢1– – b2 4⎥22.D = 2πah2 2( )2 2 22 πaa [ –( a – b )]= 2 π a a– a – b =2 2a+a – b22πab2= >πb2 2a+a – bTherefore, B < A = C < D.2 2 2yz + xz + xy[ AS ( )] [ AS ( )] [ AS ( )]2 2 2= [ AS ( )cos α] + [ AS ( )cos β] + [ AS ( )cos γ](where α, β, and γ are direction angles for anormal to S.)2 2 2 2 2= [ A( S)] (cos α + cos β + cos γ) = [ A( S)]23. In the following, each double integral is over S xyA( S ) f( x, y) = A( S )( ax + by+c)xy∫∫⎡ xdA ydA ⎤= ∫∫ dA ⎢a + b + c⎥⎢ dA dA ⎥⎣ ∫∫ ∫∫ ⎦= a x dA + b y dA + c dAxy∫∫∫∫ ∫∫ ∫∫∫∫= ( ax + by + c)dA= Volume of solid cylinder under Sxy2 2 2 2a – ( a– h) = b – h , soThus, C = 2πah⎛ 2b ⎞2= 2 π a=πb.⎜2a⎟⎝ ⎠2bh = .2a24. Because the slopes of both roofs are the same, thearea of Tmwill be the same for both roofs.(Essentially we will be integrating over aconstant). Therefore, the area of the roofs will bethe same.25. Let G denote the surface of that part of the planez = Ax+ By+ C over the region S . First,suppose that S is the rectanglea≤ x≤b,c≤ y ≤ d . Then the vectors u and vthat form the edge of the parallelogram G areu = ( b− a) i+ 0 j+ A( b−a)k andv = 0 i+ ( d − c) j+ B( d −c)k. The surface area ofG is thusu× v =−A( b−a)( d −c) i− B( b−a)( d −c) j+( b−a)( d −c)k2 2= ( b−a)( d − c) A + B + 1A normal vector to the plane is n = −Ai− Bj+k .Thus,Instructor’s Resource Manual Section 13.6 833

nk ⋅ −A, −B,1 ⋅ 0,0,1cosγ= =nk 2 2A + B + 11 ⋅1=2 2A + B + 11 2 2 u×v A( G)secγ= = A + B + 1 = = .cosγA( S) A( S)If S is not a rectangle, then make a partition ofS with rectangles R1, R2, …, Rn. The Riemannsum will ben∑m=1A( G ) = sec γ ∑ A( R ) .mnm=1As we take the limit as P → 0 the sumconverges to the area of S . Thus the surfacearea will benA( G) = lim sec γ∑ A( Rm) = sec γ A( S).P → 0 m=126. Let γ = γ( x, y, f( x, y))be the acute anglebetween a unit vector n that is normal to thesurface and makes an acute angle with the z-axis.Let F( xyz , , ) = z− f( xy , ). Then the normalvector to the surface F( xyz , , ) = 0 = z−f( xy , )is parallel to the gradient∇ F( x, y, z) =−fxi− fyj+1k. The unit normalvector is thus2 2n = ( −fxi− fyj+ 1 k )/ fx + fy+ 1The cosine of the angle γ is thusnk ⋅ −fxi − fyj + 1kcosγ= = ⋅knk 2 2fx+ fy+ 11=2 2f + f + 1xHence, secγ = f + f + 1 .y2 2x ymb.f = 2 x, f = 2yx( )y3 3 x 2 20 0∫∫A G = 4x + 4y + 1dydxParabolic rule with n = 10 givesSA ≈ 15.423328. a. f = 2 x, f =− 2yb.x( )∫∫y2 2AG = 4x + 4y + 1dAS3/2( )π= 37 −1 ≈ 29.329724(same integral as problem 27a)f = 2 x, f =−2yx( )3 3−x2 20 0∫∫yA G = 4x + 4y + 1dydxParabolic rule with n = 10 givesSA ≈ 15.4233(same integral as problem 27b)29. The surface area of a paraboloid and a hyperbolicparaboloid are the same over identical regions.So, the areas depend on the regions.E = F < A= B< C = D27. a. fx= 2 x, fy= 2y( )∫∫2 2AG = 4x + 4y + 1dASπ /2 3 2∫ r0 ∫0= 4 + 1rdrdθ2( r )π ⎡ 1= 4 12⎢ +⎣123/2( )3/2⎤⎥⎦π= 37 −1 ≈ 29.32972430834 Section 13.6 Instructor’s Resource Manual

13.7 Concepts Review1. volume2.∫∫∫Sxyz dV3. y; y4. 0Problem Set 13.71.2.3.7 2 7x( x –1– y ) dy dx = –2 x dx = –40∫ ∫ ∫–3 0 –3⎛45⎞(3 y + x) dy dx = ⎜ + 5x ⎟dx= 55⎝ 2 ⎠2 4 2∫∫ ∫0 –1 04 2z y+2z 4 2z∫∫ ∫ ∫∫dx dy dz = ( y + 2 z ) dy dz1 z–1 0 1 z–1 24 ⎡y⎤= ∫ ⎢ + 2yz⎥1⎢⎣2 ⎥⎦2zy=z–1dz4⎛27z1⎞= ∫+ 3 z–dz1 ⎜2 2⎟⎝⎠⎡ 3 27z 3z z⎤189= ⎢ + – ⎥ = = 94.5⎢⎣6 2 2⎥⎦2416.7.8.9.10.95 3⎡ 2 5yzx ⎤ 5 3⎡81yz − yz ⎤∫∫ ⎢ ⎥ dz dy =0 0 2∫∫ ⎢ ⎥dz dy0 0⎢⎣ ⎥⎦ 22z⎢⎣ ⎥⎦3⎡ 2 45 −z ( z −243)y⎤5243y=⎢⎥∫dy = dy0 ⎢ 12 ⎥ ∫02⎢⎣⎥⎦05⎡2432 ⎤= ⎢ y = 1518.754⎥⎣ ⎦0z ⎛1⎞2xdxdz = ( z –1) dz =⎝3⎠32 2 2 30 1 0⎜ ⎟∫∫ ∫π /2z∫ ∫ ∫∫ ∫ysin( x + y + z ) dx dy dz0 0 0π /2 z= [– cos(2 y+ z) + cos( y+z)]dydz0 0/20⎜sin 3zsin z 1= ∫π ⎛ – + sin2 z–⎟⎞ dz =⎝ 2 2 ⎠ 34 x+1 2 4 2–2 x–1 –2∫ ∫ ∫3 y dydx = (6x + 2) dx = 156π /2 0y0 sin2z π /2⎛ 1 ⎞ 2∫ 0∫ ∫(1 – cos 2 zdydz )= ⎜– ⎟(sin 2 z)(1–cos2 z)⎝ 2 ⎠π=− ≈− 0.392784.5 3 4 2 2 ⎛625⎞6 ∫ zdz (72)(3)0 ∫ ydy xdx–2 ∫ =1 ⎜ ⎟⎝ 4 ⎠= 33,75011.1 3 (12–3 x–2 y)/6∫∫∫0 0 0f ( x, y, z)dzdydx5.24−x−y24 24−x⎡1⎛1 2 ⎞⎤yz z dy dx4 0 ⎢x⎜ + ⎥2⎟⎣ ⎝ ⎠⎦0∫ ∫( x+ y−24)( x− y−24)24 24−x⎡⎤= ∫ dy dx4 ∫ ⎢ ⎥0 ⎢⎣2x⎥⎦24−x⎡ 22y( y −3 24( x−24⎤⎢) ) ⎥= ∫4⎢ −dx6x⎥⎢⎥⎣⎦0324 ( x − 24)= ∫ − dx4 3x24⎡32 x⎤= ⎢− 576x+ 12x − + 4608ln x⎥≈1927.54⎢⎣9 ⎥⎦412.23 2 4– y∫∫∫0 0 0f ( x, y, z)dx dy dzInstructor’s Resource Manual Section 13.7 835

13.2 4 y /2∫∫∫0 0 0f ( x, y, z)dx dy dz17.14.4 y 3 x/2∫∫ ∫0 0 0f ( x, y, z)dzdxdyUsing the cross product of vectors along edges, itis easy to show that 2,6,9 is normal to theupward face. Then obtain that its equation is2x + 6y + 9z = 18.3 (9– x)/3 (18–2 x–6 y)/9f ( x, y, z)dzdydx∫∫ ∫0 2 x /3 018.2 2 23 9– x 9– x – y∫∫ ∫0 0 0f ( x, y, z)dzdydx15.2 3 4− −2∫∫ ∫z x zf ( x , y , z ) dy dx dz0 0 019.24 1 1– y∫∫∫1 0 0f ( x, y, z)dzdydxAlternate:( − )12 / 5 4 x / 2 4−x−2z∫ ∫ ∫0 x /3 0f ( x, y, z)dy dz dx16.1z∫∫ ∫0 0 02yf ( x, y, z)dxdydz20.23 1 2 y–y∫∫∫0 0yf ( x, y, z)dxdydz836 Section 13.7 Instructor’s Resource Manual

21.22 8 2– y /4∫∫ ∫0 2x01281dz dy dx =1525. Let δ(x, y, z) = x + y + z. (See note with nextproblem.)22.22 y 4– y /8∫∫∫0 0 011dz dx dy =31 1– x 1– x–y1m = ∫∫ ( x y z)dz dy dx0 0 ∫ + + =081 1– x 1– x–y1Myz= ∫∫ x( x y z)dzdydx0 0 ∫ + + =03044x = ; Then y = z = (symmetry).151523.1 y y 1 y∫∫ ∫ ∫∫V = 4 1dz dx dy = 4 ydx dy0 0 0 0 01 21y ydy y00∫= 4 = [2 ] = 2or1 1 y1 1∫∫ ∫ ∫∫V = 4 1dz dy dx = 4 ydy dx2 20 x 0 0 x1 11⎡23/2⎤8 33 2 ( )0⎣⎦x∫0∫= 4 y dy = 1−x dx38 411 8⎛3⎞= ⎡x− x ⎤ = = 23⎣ 4 ⎦ ⎜ ⎟0 3⎝4⎠26.2 2 2( x, y, z) = k( x + y + z )In evaluating the coordinates of the center ofmass, k is a factor of the numerator anddenominator and so may be canceled. Hence, forsake of convenience we may just let k = 1 whendetermining the center of mass. Note that this isnot valid if we are concerned with values ofmoments or mass.2 4 3 y /4224. 2∫ 21 32 9.05100 ∫ dz dy dx = ≈x + 2∫0523 9– x 4 2 2 2∫∫ ∫m = 4 ( x + y + z ) dz dy dx0 0 023 9– x ⎡ 2 2 64⎤= 4∫∫4( x + y ) + dydx0 0 ⎢3 ⎥⎣⎦π /2 3⎛ 2 64 ⎞= 4 4rrdrdθ0 0⎜+ ⎟⎝ 3 ⎠∫ ∫ (change to polar)23π/2 ⎡4 32r⎤π/2⎢ ⎥ θ0 30⎢⎣⎥⎦0π /2 3 4 2 2xy = 4 ∫ ( )0 ∫ +0∫0∫ ∫ 354= 4 r + d = 4 177dθ= πM z r z r dz dr dθ(polar coordinates)π /2 3 3∫ ∫= 4 (8r+ 64 r)drdθ0 0π /2∫= 4 450 dθ = 900π0900π150z = = ≈ 2.5425; x = y = 0 (by354π59symmetry)Instructor’s Resource Manual Section 13.7 837

27. Let δ(x, y, z) = 1. (See note with previousproblem.)29.⎛1 ⎞ ⎛π⎞m= ⎜ ⎟ (volume of sphere) = ⎜ ⎟a⎝8⎠ ⎝6⎠M xy2 2 2 2 2a a – x a – x – y0 0 0= ∫∫ ∫3zdzdydxπ= ∫ ∫ ∫= ⎜ ⎟⎝ ⎠⎛3⎞z = ⎜ ⎟ a⎝8⎠⎛3⎞x = y = ⎜ ⎟ a (by symmetry)⎝8⎠2 2π /2 a a – r⎛ ⎞ 4zr dz dr dθa0 0 0 1630.The limits of integration are those for the firstoctant part of a sphere of radius 1.2 2 21 1– x 1– x – y0 0 0∫∫ ∫f ( x, y, z)dzdydx28.2 2y + z is the distance of (x, y, z) from the x-axis.4 2 − /2 4− −2∫∫ ∫x x yf ( x , y , z ) dzdydx0 0 031.2 2– z29– x0 0 0∫∫ ∫f ( x, y, z)dydxdzFigure is same as for Problem 32 except that thesolid doesn’t need to be divided into two parts.2 2I = ( y + z ) δ ( x, y, z)dVx∫∫∫S22 4– z y 2 216= ∫∫ ( y z ) zdxdydz0 0 ∫ + =0332.5 2 2– x 9 9– y 2– x∫∫∫ ∫∫ ∫f ( x, y, zdzdxdy ) +f( x, y, zdzdxdy )0 0 0 5 0 0838 Section 13.7 Instructor’s Resource Manual

33.a.b.1 4–2x 1 1 6–2x 3– x– y/2∫∫ ∫ ∫∫ ∫dz dy dx + dz dy dx = 3+ 1=40 0 0 0 4–2x01 1 6–2 x–2z∫∫∫0 0 01dy dx dz = 4c. A( Sxz) f( x, z ) ( S xz is the unit square in the corner of xz-plane; and⎡ ⎛1⎞ ⎛1⎞⎤= (1) ⎢6 – 2 ⎜ ⎟– 2 ⎜ ⎟ = 42 2⎥⎣ ⎝ ⎠ ⎝ ⎠⎦⎛1 1⎞( x, z) = ⎜ , ⎟⎝2 2⎠ is the centroid of S xz.)34. The moment of inertia with respect to the y-axis is the integral (over the solid) of the function which gives thesquare of the distance of each point in the solid from the y-axis.1 1 6–2 –2 2 2 7∫∫∫x z kx ( + z ) dydxdz = k0 0 0335.1 1 6–2 x–2z1 1=0 0 0 ∫∫ 0 0∫∫∫1 2 10x=02 31(30 – z) dy dx dz (30 – z)(6 – 2 x – 2 z)dx dz = ∫ ([(30 – z)(6 x– x – 2 xz)] ) dz1 1 2⎡ 65z2z⎤ 709= ∫ (30– z)(5–2 z) dz = (150–65z + 2 z ) dz0 ∫= ⎢150 z – + ⎥ =0⎢⎣2 3 ⎥⎦60The volume of the solid is 4 (from Problem 33).7096 709Hence, the average temperature of the solid is = ≈ 29.54 ° .4 2436. T( x y z), , = 30 − z = 29.54 . The set of all points whose temperature is the average temperature is the planez = 0.46.1 1 6 − 2x−2z1 137. yz = ∫∫∫ = ∫∫ ( 6−2 −2)M xdydxdz x x z dxdz0 0 0 0 011⎛7 ⎞ ⎡7 1 2 ⎤ 11= ∫ z dz z z0⎜ −3⎟ = ⎢ − =3 2⎥⎝ ⎠ ⎣ ⎦0611⎡2 2 3 2 ⎤= ∫ 3x x x z dz0 ⎢ − −3⎥⎣⎦x=01 1 6−2x−2z1 1⎡1 2 ⎤M xz = ∫∫∫ ydydxdz= ( 6 2 x 2 z)dxdz0 0 0 ∫∫ − −0 0⎢2⎥⎣⎦1⎛38 2 ⎞ 25= ∫ 10z 2z dz0⎜− + =3⎟⎝⎠ 311⎡2 2 3 2 2⎤= ∫ 18x 6x x 12xz 2x z 2xz0 ⎢ − + − + +3⎥⎣⎦x=01 1 6–2 –2 1 1∫∫∫ =0 0 0 ∫∫ 0 02 311 2⎡5z2z⎤ 11(5 z– 2 z ) dz ⎢ ⎥0⎢⎣2 3 ⎥⎦60M xy =∫x zz dy dx dz z (6–2 x –2 z ) dx dz= = =Hence, ( x y z)⎛11/ 6 25 / 3 11/ 6 ⎞ ⎛11 25 11 ⎞, , = ⎜ , , ⎟=⎜ , , ⎟⎝ 4 4 4 ⎠ ⎝24 12 24⎠1 2 10x=0= ∫ ([ z(6 x– x – 2 xz)] ) dzInstructor’s Resource Manual Section 13.7 839

38. a. It will be helpful to first label the corner points at the top of the region.z(1, 0, 3)(0, 0, 4)(0, 1, 3)z4z = 4 − y(1, 1, 2)y3R 2z = 3 − yR 1x1yFixing z and y, we will be looking at the figure along the x-axis. The resulting projection is shown in the figureabove and to the right. The possible values of x depends on where we are in the yz-plane. Therefore, we split upthe solid into two parts. The volume of the solid will be the sum of these two smaller volumes. In the lowerportion, x goes from 0 to 1, while in the upper portion, x goes from 0 to 4 − y − z (the plane that bounds the topof the square cylinder).V 1 3−y 1 1 4 45 11 1 30 0 0 dxdzdy −y −y−z= ∫∫ ∫ + ∫∫ 0 3−y ∫ 0dxdzdy = + =2 2b.1 1 4−y−z1 dx dy dz = 30 0 0∫∫∫c. A( Sxy) f( x, y ) ( S xy is the unit square in the corner of xy-plane; and⎡ 1 1⎤= () 1 ⎢4− − = 32 2⎥⎣ ⎦⎛1 1⎞( x, y) = ⎜ , ⎟⎝2 2⎠ is the centroid of S xz.)39.1 1 4−x−y1 1 1⎛7⎞m = ∫∫∫ k dz dy dx = k(4 x y) dy dx k x dx 3k0 0 0 ∫∫ − − = − =0 0 ∫ 0⎜2⎟⎝ ⎠( 4 )1 1 4 −x−y1 1 2 1 ⎛7 2⎞17kM yz = ∫∫∫ kx dz dy dx = k x x xy dy dx k x x dx0 0 0 ∫∫ − − =0 0 ∫ − =0⎜2⎟⎝ ⎠ 122( 4 )1 1 4 −x−y1 1 1 ⎛5 x ⎞ 17kMxz= ∫∫∫ kydzdydx= k y xy y dydx k x dx0 0 0 ∫∫ − − =0 0 ∫ − =0⎜3 2⎟⎝ ⎠ 121 1 4−x−y1 1⎛ 2 212x y ⎞ ⎛37 7x x ⎞ 55kMxy= ∫∫∫ kzdzdydx= k 8 4x 4y xy dydx k dx0 0 0 ∫∫ − + − + + = − + =0 0⎜ 2 2 ⎟ ∫ 0⎜ 6 2 2 ⎟⎝ ⎠ ⎝ ⎠12Myz 17 k /12 17x = m= 3k= 36Mxz 17 k /12 17y = m= 3k= 36M xy 55 k /12 55z = m= 3k= 36840 Section 13.7 Instructor’s Resource Manual

40. The temperature, as a function of ( x, yz , ) is T( x, y, z) = 40+ 5z. The average temperature is41.1 1 1 4−x−y1 1 1 4−x−yVolume∫∫∫∫∫∫0 0 0 0 0 0( )T ( x, y, z) dz dy dx = 40+5z dz dy dx31 121 ⎛5 25y⎞= 200 60x x 60y 5xy dy dx3∫∫− + − + +0 0⎜2 2 ⎟⎝⎠=⎛− +⎝1715= ≈ 47.6436121 1025 115x5x3∫0⎜18 6 6⎞dx⎟⎠Figure 1: When the center of mass is in this position, it will go lower when a little more soda leaks out since massabove the center of mass is being removed.Figure 2: When the center of mass is in this position, it was lower moments before since mass that was below thecenter of mass was removed, causing the center of mass to rise.Therefore, the center of mass is lowest when it is at the height of the soda, as in Figure 3. The same argumentwould hold for a soda bottle.42. The result obtained from a CAS is:2 2 2 2 2 2c b 1 −z / c a 1 −y / b −z / c8 2 2 8 2 2 8 2 2 8∫∫ 8( ) ( )0 0 ∫xy + xz + yz dxdydz = a b c + a bc + ab c = acb ca + cb + ab015 15 15 1543. a.12 x12 k 21 = ∫0 ∫ ky dy dx =0 ∫ x dx0 212⎡k 3 ⎤1= ⎢ x = 288 k ⇒ k =6⎥⎣ ⎦028812 x 1> = ∫4 ∫ ydydx4 288x12 ⎡ 1 2⎤12 1 2= ∫4 ⎢ y4( 16)576⎥ dx = ∫ x − dx⎣ ⎦457612⎡ 1 3 ⎤ 20= ⎢ ( x − 16x)=576⎥⎣⎦427b. PY ( 4)12 x 1E X = ∫0 ∫ x ydydx0 288x12 ⎡ 1 2⎤12 1 3= ∫0 ⎢ xy576⎥ dx = ∫ x dx⎣ ⎦ 00 57612⎡ 1 4 ⎤= ⎢ x = 92304⎥⎣ ⎦0c. [ ]44. a.2 4 y1 = ∫∫∫ kxy dx dy dz0 0 0y2 4⎡k 2 ⎤= ∫∫ 0 0 ⎢ x y2⎥ dy dz⎣ ⎦042 4k3 2⎡k4⎤= ∫∫ ydydz= y dz0 0 2∫0⎢8⎥⎣ ⎦021= ∫ 32kdz= 64k⇒ k=0642 4 4 1> = ∫∫∫ xydydxdz0 2 x 6442 4⎡1 2 ⎤= ∫∫ 0 2 ⎢ xy128⎥ dx dz⎣ ⎦x2 4 1 3= ∫∫ 0 2( 16 x −x ) dx dz12842 1 ⎡ 2 1 4⎤2 9 9= ∫ 8x x dz dz0 128⎢ − = =4⎥ ∫⎣ ⎦ 02 32 16b. P( X 2)Instructor’s Resource Manual Section 13.7 841

2 4 y 1 2= ∫∫∫ x y dx dy dz0 0 0 64y2 4⎡1 3 ⎤= ∫∫ 0 0 ⎢ x y192⎥ dydz⎣ ⎦042 4⎡ 1 4⎤ 2⎡ 1 5⎤∫∫ y dydz y dz0 0 ⎢192⎥ = ∫0⎢960⎥⎣ ⎦ ⎣ ⎦0216 32= ∫ dz =0 15 15c. E[ X ]4 4 3 2 245. a. ( > 2) = ( + )∫∫P X x y dy dx2 x 2564 ⎡ 3 ⎛ 2 1 3⎞⎤= ∫ x y y dx2⎢ + ⎥256⎜3⎟⎣ ⎝ ⎠⎦( 3 16)4 1 3 2= ∫ − x + x + dx2 64⎡ 1 ⎛ 1 4 3 ⎞⎤7= ⎢ x x 16x64⎜− + + ⎥ =4⎟⎣ ⎝⎠⎦162 4−x3 2 2b. ( + < 4) = ( + )∫∫P X Y x y dydx0 x 256( 3 6 8)4x424−x2 ⎡ 3 ⎛ 2 1 3⎞⎤= ∫ x y y dx0⎢ + ⎥256⎜3⎟⎣ ⎝ ⎠⎦2 1 3 2= ∫ − x + x − x+dx0 321 ⎡ 1 4 3 2 ⎤ 1= x x 3x 8x32⎢− + − + =4⎥⎣⎦ 4c. E[ X + Y]3( ) ( )4 y2 2= ∫∫ x + y x + y dx dy0 0 256y4 ⎡ 3 ⎛1 4 1 2 2 1 3 3⎞⎤= ∫0⎢256⎜ x + x y + x y+xy dx4 2 3⎟⎥⎣ ⎝⎠⎦044 25 4 ⎡ 5 5⎤= ∫ y dy = y = 50 1024⎢1024⎥⎣ ⎦0b ∞P a X b f x y dydxa −∞46. a. ( < < ) = ∫∫ ( , )b ∞=⎛f ( x,y)dy⎞∫dxa ⎜ −∞ ⎟⎝∫⎠b= ∫ fX( x)dxax20∞ ∞E X xf x y dy dxb. [ ] = ∫ ( , )−∞∫−∞∞ ⎛ ∞∫ ∫ ( , )47. X ( ) = ∫ 0= x f x y dy⎞dx−∞ ⎜ −∞ ⎟⎝⎠∞= ∫ xf X ( x)dx−∞x 1f x ydy288x⎡ 1 2⎤1 2= ⎢ y x ;0 x 12576⎥ = ≤ ≤⎣ ⎦05761E X x f x dx x dx12 12 30 576 0121 ⎡14 ⎤( ) = ∫ ⋅ X ( ) = ∫= x 9576 ⎢ =4 ⎥⎣ ⎦d bf y f x y z dxdzc ay2 y 1 2⎡1 2 ⎤= ∫∫ xydxdz= 0 0 64∫0⎢ x y128⎥ dz⎣ ⎦0231 3 y= ∫ ydz= ;0≤ y≤40 128 6448. Y ( ) = ∫ ∫ ( , , )13.8 Concepts Review1. r dz dr dθ;2.3.4.0ρ2 sinφdρdθ dφπ /2 1 3 3∫ r cosθ sinθ dzdrdθ0 ∫0∫0π 2π1 4cos 2sin d d d0 0 0∫∫ ∫4π15Problem Set 13.8ρ φ φ ρ θ φ1. The region is a right circular cylinder about the z-axis with radius 3 and height 12.2π3 122π3∫ [ ]0 ∫ rz dr dθ= 120 0 ∫0 ∫ r dr d02π322π= ⎡6r ⎤∫ d = 54d= 1080 ⎣ ⎦ ∫0 0θ θ πθ842 Section 13.8 Instructor’s Resource Manual

2. The region is a hollow right circular cylinderabout the z-axis with inner radius 1, outer radius3, and height 12.2π3 122π3∫ [ ]0 ∫ rz dr dθ= 121 0 ∫0 ∫ r dr d12π322π= ⎡6r ⎤∫ d = 48d= 960 ⎣ ⎦ ∫1 0θ θ π3. The region is the region under the paraboloid2z = 9 − r above the xy-plane in that part of theπfirst quadrant satisfying 0 ≤θ≤ .49−r2π /4 3⎡12 ⎤∫0 ∫0⎢ rz2⎥ dr dθ⎣ ⎦0π /4 3⎡12 ⎤= ∫0 ∫0⎢ r( 81− 18r+r ) drd2⎥⎣⎦3π /41 ⎡81 2 3 1 4⎤= ∫ r 6r r d0 2⎢ − +2 4⎥⎣⎦0π /4 243 243π= ∫ dθ=0 4 164. The region is a right circular cylinder about the z-⎛ 1 ⎞axis through the point ⎜0, , 02⎟⎝ ⎠ with radius 1 2and height 2.π sinθ 2π sinθrz dr dθ= 2r dr dθ∫ ∫[ ]∫ ∫0 0 00 0π sinθ2 π 2= ⎡r ⎤∫ dθ = sin θ dθ0 ⎣ ⎦ ∫00ππ ⎛1 1 ⎞ ⎡1 1 ⎤= ∫ cos 2θ dθ θ sin 2θ0 ⎜ −2 2⎟ = ⎢ −2 4⎥⎝ ⎠ ⎣ ⎦01= π25. The region is a sphere centered at the origin withradius a.aπ 2π⎡13 ⎤∫0 ∫0⎢ ρ sin ϕ3⎥ dθ dϕ⎣ ⎦0π 2π1 3= ∫ a sin d d0 ∫ ϕ θ ϕ0 3π 2π3= ∫ a sin ϕdϕ0 3π 3⎡ 2π3 ⎤ 4πa= ⎢− a cosϕ=3⎥⎣⎦ 30θθθ6. The region is one-eighth of a sphere in the firstoctant of radius a, centered at the origin.7.8.9.10.11.aπ /2 π /2⎡13 2 ⎤∫0 ∫ cos sin0 ⎢3⎥ d d⎣⎦0π /2 π /2 1 3 2= ∫a cos sin d d0 ∫03πρ ϕ ϕ θ ϕπ /2 3 2= ∫ a cos sin d0 6π /2 3⎡ π 3 ⎤ πa= ⎢− a cosϕ=6⎥⎣⎦018ϕ ϕ ϕϕ ϕ θ ϕ2π2 40 02 rdzdrdθ = 8π≈25.1327r∫ ∫ ∫2π 2 9– r2⎛2⎞ 3/2rdzdrdθ = π0 0 0⎜ ⎟∫ ∫ ∫V=2π3 25−r2= ∫ ∫ ∫0 0 42π3 25−r2[ rz]0 ∫04∫( )(27 – 5 )⎝3⎠≈ 33.1326r dz dr dθ2π3 2=⎡r 25 r 4r ⎤∫ − − drdθ0 ∫0⎢⎣⎥⎦2π⎡ 1 3/2252 22⎤= ∫ − −r − r dθ0 ⎢3⎥⎣⎦2π7 14π= ∫ dθ=0 3 3V ==∫2 π 4 4+r sinθ∫ ∫0 0 02 π 4 4+r sinθ[ rz]0 ∫00∫dr dθr dz dr dθdr dθ2π4 2= ⎡4r r sinθ⎤∫0 ∫ + drd0 ⎣ ⎦42π⎡ 2 1 3 ⎤= ∫ 2r r sin d0 ⎢ + θ3⎥⎣⎦042π⎡ 64 ⎤= ∫ 32 + sinθdθ0 ⎢3⎥⎣⎦0= 64π2π2 5– r20 0 r2 rdzdrdθ/4∫ ∫ ∫2π 2⎡ 321/2 r ⎤= ∫ r(5 r ) drdθ0 ∫ ⎢ − − ⎥0⎢⎣4 ⎥⎦2π3/2 3/25 –4 2 π(5 –4)= ∫ dθ = ≈15.03850 3 3θ30θInstructor’s Resource Manual Section 13.8 843

12.π / 2 2cosθ r2sinθcosθ42⋅ ∫ rdzdrd0 ∫ θ =0 ∫ 0313. Let δ(x, y, z) = 1.(See write-up of Problem 26, Section 13.7.)2π2 12–2r2m = ∫0 ∫0∫2 r dz dr dθ = 24πr2π2 12–2r2Mxy= ∫0 ∫0∫2 zrdzdrdθ = 128πr16z =3x = y = 0 (by symmetry)14. Let δ(x, y, z) = 1.(See comment at beginning of write-up ofProblem 26 of the previous section.)2π2 12– r22π2 2∫ ∫ ∫ ∫ ∫m = r dz dr dθ= r (12 – r ) dr dθ0 1 0 0 0422π⎡ 2 r ⎤ 2π⎛57 ⎞ 57π= ∫ ⎢6 r – ⎥ dθ= ∫ dθ=0 4 0 ⎜ ⎟⎢⎣⎥⎦⎝ 4 ⎠ 212π2 12– r2M xy = ∫ z r dz dr dθ0 ∫1 ∫ 02π22 2(12 – r ) (–2 r)= ∫ dr dθ0 ∫ 1 –42π3 311 – 8 273π= ∫ dθ =0 12 2273πTherefore,2 91z = = ≈ 4.7895.57π192x y 0= = (by symmetry)15. Let δ ( x, y, z)= kρπ 2πb 2 4 4m= ∫∫ kρρ sin φ dρ dθ dφ= kπ( b – a )0 0∫a16.π/2 π/2 2a2 2π /6 0 a csc∫ ∫ ∫8 k ρ ρ sin φ d ρ d θ d φφ⎛56⎞ 5= ⎜ ⎟kπa3⎝ 5 ⎠17. Assume that the hemisphere lies above the xyplane.Let δ(x, y, z) = ρ.(Letting k = 1 - - see comment at the beginning ofthe write-up of Problem 26 of the previoussection.)π/2 2πa 30 0 0π/2 2π4a0 0m ρ sinφdρdθ dφ= ∫ ∫ ∫= ∫ ∫sin4φ d θ d φ4 4π /2πasinφπa= ∫d φ =0 2 2π/2 2πa 40 0 0M ρ sinφcosφdρdθ dφxy= ∫ ∫ ∫(z = ρ cos φ)π/2 2π0 0= ∫ ∫a5sin 2 φ d θ d φ 10π /25πasin 2φ⎛π⎞5d φ a0∫= = ⎜ ⎟5 ⎝5⎠zπa55 2a ; xπa4 5y 02= = = = (by symmetry)18. Assume that the hemisphere lies above the xyplane.δ(x, y, z) = ρ sin φ (letting k = 1)π π a⎛1⎞m = ∫ ρ sin φdρdθ dφa0 ∫ = π0 ∫ 0⎜ ⎟⎝8⎠19.20.21.22.23./2 2 3 2 2 4π/2 2πa 4 2M xy = ∫ sin0 ∫0 ∫ 0cos d d d⎛ 2 ⎞ 5= ⎜ ⎟πa⎝15⎠16az = ≈0.3395a15πx y 0= = (by symmetry)∫∫ρ φ φ ρ θ φ2 2 2 2 1/2Iz= ( x + y ) k( x + y ) dVSπ π ⎛ k ⎞= ∫ ∫ ∫= π⎝16⎠/2 2 a 5 4 2 6kρ sin φdρdθ dφa0 0 0⎜ ⎟π/2 2π4 2Volume = ∫ ρ sinφdρdθ dφπ /4∫0 ∫ 0π/2 2 π64sinφπ/2128πsinφ= ∫ d d dπ/4∫ =0 3∫ π/4364 2π= ≈ 94.78153π π/6 1 2πρ φdρdθ dφ0 0 0∫∫ ∫θ φ φsin = ≈0.34919π 2π3ρ 3ρ 2sinφdρdφdθ= 486π≈1526.810 0 0∫∫ ∫π sinθr sinθVolume = ∫∫ 0 0 ∫ r2 rdzdrdθπ sinθ42πsinθ= ∫∫ rr ( sin θ – r ) drdθ = dθ0 0 ∫ 0 121 π ⎡1+cos4θ⎤= 1–2cos2θdθ48∫+0 ⎢2 ⎥⎣⎦π= ≈ 0.098232844 Section 13.8 Instructor’s Resource Manual

24. Consider the following diagram:IIIIIIMethod 1: (direct, requires 2 integrals)V = I + IIπ /2 2π 2 2cos ϕ 2 π /4 2π2 2sinπ /4 0 0 0 0 0∫ ∫ ∫ ∫ ∫ ∫= ρ ϕdρdθ dϕ+ρ sinϕdρdθ dϕ( − ) π ( − )2 2π82 2 28 3 2π= + = ≈7.86943 3 3Method 2: (indirect, requires 1 integral)V = upper sphere volume −III8 2ππ /4 2π 2 2cosϕ2= −sin3∫0 ∫0 ∫ ρ ϕdρdθ dϕ28 2π27 ( 2−8) π 28 ( −3 2)π= − = ≈7.86943 3 32 2 2 1/225. a. Position the ball with its center at the origin. The distance of (x, y, z) from the origin is ( x + y + z ) = ρ.2 2 2 1/2 /2 /2 a 2 4( ) 8 (sin )S x y z dV π π+ + =0 0 0ρ θρ dρdθ dφ=π a4πa3 a= .⎡⎛4⎞ 3 ⎤ 4⎢⎜⎟ π a3⎥⎣⎝⎠ ⎦∫∫∫ ∫ ∫ ∫Then the average distance from the center isb. Position the ball with its center at the origin and consider the diameter along the z-axis. The distance of (x, y, z)2 2 1/2from the z-axis is ( x + y ) = ρ sin φ./2 /24 22 2 1/2 π π a2a π∫∫∫ ( x + y ) dV = 8 ( ρsin φ)( ρ sin θ)dρdθ dφS∫0 ∫ =0 ∫ 04⎡a4π2⎤⎢ 4 3Then the average distance from a diameter is⎣ ⎥⎦πa= .⎡ 4 3( 3 ) πa⎤ 16⎣ ⎦c. Position the sphere above and tangent to the xy-plane at the origin and consider the point on the boundary to bethe origin. The equation of the sphere is ρ = 2a cos φ, and the distance of (x, y, z) from the origin is ρ./2 2 2 cos42 2 2 1/2 π π a φ 28πa∫∫∫ ( x + y + z ) dV = ρρ ( sin θ ) d ρ d θ d φS∫0 ∫ =0 ∫ 05⎡8πa4 ⎤⎢ 5 6Then the average distance from the origin is⎣ ⎥⎦a= .⎡ 4 3( 3 ) πa⎤ 5⎣ ⎦Instructor’s Resource Manual Section 13.8 845

26. Average value of ax + by + cz + d on S is∫∫∫ ∫∫∫ ∫∫∫ ∫∫∫ ∫∫∫( ax + by + cz + d)dV a kx dV + b ky dV + c kz dV + d k dVS S S S S=∫∫∫ dVk dVS∫∫∫SaM yz + bM xz + cM xy + dm= = ax + by + cz + d = f ( x , y , z ).m27. a. M yz= ∫∫∫kx dVS4 (sin α)ρ φ θ ρ φ ρ θ φ = ka π4π /2 α a24 k ( sin cos )( sin ) d d d0 0 0= ∫ ∫ ∫4akαρ φ ρ θ φ =3π /2 α a 2m = ∫∫∫ kdV = 4k sin d d dS ∫0 ∫0 ∫0⎡ka4π (sin α ) ⎤⎢ 4 ⎥3 aπ(sin α) Therefore, x =⎣ ⎦= .⎡4a 3kα⎤ 16α⎢⎣3 ⎥⎦3b.3πa16(See Problem 25b.)28. a.b.2 2 1/2 2I = k[( x + y ) ] dVz∫∫∫Sπ/2 π/2 a 2 28a πk 2a m8 k ( ρ sin φ) ( ρ sin φ)dρdθ dφ= = (since0 0 015 5= ∫ ∫ ∫2 22 2a m 2 7a mI′ = I + d m= + a m=5 55 24 3m= ⎛⎜⎞⎟πa k ⎞⎟⎝3⎠ ⎠c.⎡ 22a m ⎤2I = 2 ⎢ + ( a+b)m⎥⎢⎣5⎥⎦2 22 m(7a + 10ab+5 b )=529. Let m 1 and m 2 be the masses of the left and right balls, respectively. Thenm2 = cm1.m1(– a– b) + m2( a+b)y =m1+m2m1(– a– b) + cm1( a+ b) – a– b+ c( a+b)= =m + cm 1+c1 1( a+ b)(–1 + c) c–1 = = ( a+b )1+ c c+1my(Analogue)1 1+m2y2 m1 my = = y21 + y2m + m m + m m + m1 2 1 2 1 24 3m1= π a k and34 3m2= π a ( ck),so3846 Section 13.8 Instructor’s Resource Manual

13.9 Concepts Review4.1. u-curve; v-curve2. the integrand; the differential of dx dy; the regionof integration3. Jacobian4. J( u,v )Problem Set 13.95.1.2.6.3.7. x = u+ 2; v y = u−2vG(0,0) = (0,0); G(2,0) = (2, 2)G(2,1) = (4,0); G(0,1) = (2, −2)The image is the square with corners (0,0) ,(2,2) , (4,0) , and (2, − 2) . The Jacobian isJ =1 2=−4.1 −2Instructor’s Resource Manual Section 13.9 847

8. x = 2u+ 3 v;y = u−v9.G(0,0) = (0,0); G(3,0) = (6,3)G(3,1) = (9, 2); G(0,1) = (3, −1)The image is the parallelogram with vertices(0,0) , (6,3), (9,2) , and (3, − 1) . The Jacobianis J =2 3=−51 −12x = u + v 2 ; y = vG(0,0) = (0,0); G(1,0) = (1,0)G(1,1) = (2,1); G(0,1) = (1,1)12. u = 2x− 3 y; v = 3x− 2ySolving for x and y2 3 3 2gives x =− u+ v;y =− u+ v. The5 5 5 52 3−4 9 1Jacobian is J = 5 5 =− + = .3 2− 25 25 55 513.2u = x + y 2 ; v = x. Solving for x and y gives2x v;y u v= = − . The Jacobian is0 1J = 1 −v= −22u−v 2u−v212u−v10.Solving for u and v gives2u = x− y and v = yThe u = 0 curve is2 20 = x − y ⇒ x = y .2 2The u = 1 curve is 1= x− y ⇒ x = y + 1.The v = 0 curve is y = 0 , and the v = 1 curve isy = 1. The image is the set of ( x, y ) that satisfy2 2y ≤ x≤ y + 1, 0 ≤ y ≤ 1 . The Jacobian isJ =2u2v= 2u0 12 2x = u;y = u − vSolving for u and v gives2u = x and v = x − yThe u = 0 curve is x = 0 , and the u = 3 curve isx = 3 . The v = 0 curve is2 20 = x − y ⇒ y = x and the v = 1 curve is2 21= x − y ⇒ y = x − 1. The image istherefore the set of ( x, y ) that satisfy2 2x −1 ≤ y ≤ x ; 0≤ x≤ 3. The Jacobian isJ = 1 02v2u−2v=− .14.x and y givesJacobian isu = x2 − y2 ; v = x+ y.Solving forv 2 2+ u ;v −x yu2v2v1 1 u−2 2 22 1J =v v =1 1 u− + 2v2v2 22v= = . The15. u = xy;v = x. Solving for x and y gives16.x = v; y = u/v. The Jacobian is0 11J = 1 u−=−v 2 vv.u = x2 ; v = xy . Solving for x and y givesvx = u;y = . The Jacobian isu101J = 2 u = .−v1 2u3/22uu11. u = x+ 2 y; v = x− 2ySolving for x and ygives x = u/2 + v/2; y = u/4 − v/4. The1 1 1Jacobian is J =1/2 1/2= − − =− .1/4 −1/4 8 8 4848 Section 13.9 Instructor's Resource Manual

17. Let u = x+ y,v = x− y . Solving for x and ygives x = u/2 + v/2and y = u/2 − v/2. The1 11Jacobian is J = 2 2 =− .1 1− 22 2The region of integration gets transformed to thetriangle in the uv-plane with vertices (1,1), (4,4),and (7,1). The integral in the uv-plane is moreeasily done by holding v fixed and integrating u.Thus,x+y 4 8−vu 1∫∫ ln dA = ln du dvx−y∫1∫vv 2R8−v1 4 ⎡ u ⎤= ln2∫1⎢− u+u dvv⎥⎣ ⎦v1 4 ⎡8−v ⎤= 8 2 v (8 v)lndv2∫ 1 ⎢− + + −v⎥⎣⎦1 ⎡ 49⎤= 3− 64ln4+ ln7+16ln162⎢2⎥⎣⎦≈ 3.1566918. Let u = x+ y,v = x− y . Solving for x and ygives x = u/ 2 + v/ 2 and y = u/2 − v/2. The1 11Jacobian is J = 2 2 =− .1 1− 22 2The region of integration gets transformed to thetriangle in the uv-plane with vertices (1,1), (4,4),and (7,1). The integral in the uv-plane is moreeasily done by holding v fixed and integrating u.Thus, with the help of a CAS for the outerintegral, we have∫∫Rx+y 4 8−vudA =x−y∫1∫vv4 ⎡ 3/21 u ⎤= ⎢ ⎥∫128−vdu dvdv1 1/2⎢⎣3 v ⎥⎦v43/2 3/21 ⎡(8 − v)v∫ ⎢1 1/2 1/2⎤= − ⎥dv3 ⎢⎣v v ⎥⎦49 19 7−1= − − 4π+ 16tan 76 6≈ 6.5729519. Let u = 2x− y and v = y . Thenx = u/2 + v/2and y = v . The Jacobian is1/2 1/2 1J = = . Thus,0 1 2sin π(2 x−y) cos π( y−2 x)dA∫∫( ) ( )R3 8−v1= ∫∫ sin ( πu) cos ( π( −u)) du dv0 v+221 3 8−v1= 2sin ( ) cos ( )2∫∫πuπu dudv0 v+2 21 3 8−v= sin ( 2 )4∫∫ πu dudv0 v+28−v1 3⎡cos2πu ⎤=4∫0⎢ −dv2π⎥⎣ ⎦v+21 3= cos( 2 π( v 2) ) cos( 2 π(8 v)) dv8π∫⎡0 ⎣ + − − ⎤⎦1 3= ⎡cos( 2πv+ 4π) −cos( 16π −2πv)⎤8π∫0⎣⎦ dv1 3= ⎡cos( 2πv) − cos( 2πv)⎤dv= 08π∫0⎣⎦20. Let u = 2x− y and v = y . Thenx = u/2 + v/2and y = v . The Jacobian is1/2 1/2 1J = = . Thus,0 1 2(2 x−y)cos( y−2 x)dA∫∫R1= u −u dudv3 8−vcos( )0 v+2 23 8−v∫ [ sin cos ]0v+2∫∫1= u u+u dv21= ( − 2cos2 + 10cos5 − 8cos82+ 2sin2− 4sin5+2sin8)≈ 6.23296Instructor’s Resource Manual Section 13.9 849

21. The transformation to spherical coordinates isx = ρ sinφcosθy = ρ sinφsinθz = ρcosφThe Jacobian is∂x ∂x ∂x∂ρ ∂θ ∂φ∂y ∂y ∂ysinφ cosθ −ρsinφsinθ ρcosφcosθJ = = sinφ sinθ ρsinφcosθ ρcosφsinθ∂ ρ ∂ θ ∂ φ cosφ 0 −ρsinφ∂z ∂z ∂z∂ρ ∂θ ∂φ= cosφ −ρ sinφsinθ ρcosφcosθ 0sinφcosθ ρcosφcosθ sin cos sin sinsin cos cos sin sin sin cos sin( sinφ θ ρ φ θρ φ θ ρ φ θ−φ θ ρ φ θ+ − ρ φ−)sinφsinθ ρsinφcosθ2 2 2 2 2 2 2= cosφ −ρ sinφsinθcosφsinθ −ρ cosφsinφcos θ − ρsinφ ρsin φcos θ + ρsin φsinθ( ) ( )cos sin ( sin cos ) sinsin ( cos sin )2 2 2 2 2 3=− ρ φ φ θ + θ −ρ φ2 2 2=− ρ φ φ + φ2=−ρsinφa 0 022. Let x = ua, y = vb,z = wc . Then the Jacobian is J = 0 b 0 = abc and the region of integration becomes the0 0 csphere with radius 1 centered at the origin. Thus,4V = ∫∫∫ 1dV = ∫∫∫ 1dz dy dx = ∫∫∫ 1abc dwdv du = π abc3ellipsoid ellipsoid unitsphereThe moment of inertia about the z-axis is2 2M = x + y dzdydxz∫∫∫unitsphere∫∫∫ ( )2 2 2 2( )ellipsoid= au + bv abcdwdvdu( )π∫ ( cos sin sin sin )π∫ sin φ( a cos θ b sin θ)dφdθ2( 3aπ cos θ + 3bsin θ)dθ2 2( a b )2ππ 1 2 2 2 2 2 2 2 2 20 ∫0 ∫ sin cos sin sin sin02π2 2 4 2 2 44∫ a θ φ b θ φ dφdθ0 02π4 2 2 2 24∫0 02π2 2 232∫02∫= a ρ φ θ + b ρ φ θ ρ φabc d ρdφdθabc= +abc= +abc=3abcπ= +32850 Section 13.9 Instructor's Resource Manual

23. Let X = xUV ( , ) and Y = y( U, V). If R is a region in the xy -plane with preimage S in the uv -plane, then(( , ) ∈ ) = (( , ) ∈ )P X Y R P U V SWriting each of these as a double integral over the appropriate PDF and region givesf ( x, y) dy dx = g( u, v)dv du∫∫ ∫∫RSNow, make the change of variable x = xuv ( , ) and y = y( u, v)in the integral on the left. Therefore,∫∫ ∫∫ ∫∫( )g( uvdvdu , ) = f( xydydx , ) = f xuv ( , ), yuv ( , ) Juv ( , ) dvduS R SThus probabilities involving ( UV , ) can be obtained by integrating ( )f ( xuv ( , ), yuv ( , ) Juv ( , ) is the joint PDF of ( UV , ).24. Let u = x+ y,v = x− y.a. Solving for x and y gives x = u/ 2 + v/ 2 and y = u/2 − v/2. The Jacobian is1/2 1/2 1 1 1J = =− − =−1/2 −1/2 4 4 2The joint PDF for ( UV , ) is therefore⎛u v u v⎞guv ( , ) = f⎜+ , −2 2 2 2⎟⎝⎠⎧1 ⎪ , if 0 ≤ ( u+ v )/2 ≤ 2,0 ≤ ( u−v)/2 ≤ 280, otherwise= ⎨⎪⎩⎧ ⎪ ≤ + ≤ ≤ − ≤f xuv ( , ), yuv ( , Juv ( , ) . Thus,1= , if 0 u v 4,0 u v 4⎨8⎪⎩ 0, otherwiseTo find the marginal of U , we fix a u and integrate over all possible v . For 0≤v≤ 2,u 1 u4−u1 8−2uugU( u)= ∫ dv = and for 2< u ≤ 4,g ( ) 1−uU u =8 4∫ dv = = − .− 4+u 8 8 4Therefore,⎧⎪u/4, if0≤u≤ 2gU( u) = ⎨1 − u/4, if 2< u ≤ 4⎪⎩ 0 otherwiseInstructor’s Resource Manual Section 13.9 851

25. a. Let u = x+ y,v = x. Solving for x and ygives x = v,y = u− v. The Jacobian isJ =0 1=−1.1 −1Thus,guv ( , ) = f( vu , −v) −1⎧ −v−( u−v)= e , if 0 ≤v,0≤u−v⎨ ⎩ 0, otherwise⎧ −u= e , if 0≤v≤u⎨ ⎩ 0, otherwiseb. The marginal PDF for U is obtained byintegrating over all possible v for a fixedu . If u ≥ 0 , thenu − u −u0g ( u)= e dv = ueU∫⎧ −uThus, g ( ) ue , if 0 uU u = ≤⎨ ⎩ 0, otherwise13.10 Chapter ReviewConcepts Test1. True: Use result of Problem 33, Section13.2, and then change dummyvariable y to dummy variable x.7. False: Let f(x, y) = x, g( x, y) = x ,R = {(x, y): x in [0, 2], y in [0, 1]}.The inequality holds for the integralsbut f(0.5, 0) > g(0.5, 0).8. False: Let f(0, 0) = 1, f(x, y) = 0 elsewhere2 2for x + y ≤ 1.9. True: See the write-up of Problem 26,Section 13.7.10. True: For each x, the density increases as yincreases, so the top half of R is moredense than the bottom half. For eachy, the density decreases as the xincreases, so the right half of R is lessdense than the left half.11. True: The integral is the volume betweenconcentric spheres of radii 4 and 1.That volume is 84π .12. True: See Section 13.6.A(T) = (Area of base)(sec 30°)2 ⎛ 2 ⎞ 2 3π=π (1) ⎜ ⎟=⎝ 3 ⎠ 313. False: There are 6.14. False: The integrand should be r.22. False: Let f(x, y) = x. 1st integral is 1 ;3 2nd15. True:f2 2x fy+ + 1≤ 9=3is 1 .63. True: Inside integral is 0 sincean odd function in x.3 3sin( x y ) is4. True: Use Problem 33, Section 13.2. Each2x2 y2integrand, e and e , determinesand even function.5. True: It is less than or equal to6. True:which equals 2.2 21 0∫∫1dx dyf ( x0, y0)f( x, y)≥ in some2neighborhood N of ( x0, y 0)due tothe continuity. Then⎛1⎞∫∫ f ( x, y) dA≥ f( x0, y0)dAR ∫∫ N⎜ ⎟⎝2⎠⎛1⎞= ⎜ ⎟ ( 0, 0)(Area ) 0.2 f x y N >⎝ ⎠16. True: J ( r,θ )17. False: ( )=cosθ−rsinθ= rsinθr cosθJ u, v =2 0= 40 2Sample Test Problems1.2.3.∫1 2 30⎛1⎞1⎜ ⎟( x – x ) dx = ≈0.0417⎝2⎠2420dy = 0–2∫ (Integrand determines an oddfunction in x.)22sinθ∫π/2 ⎡rcosθ⎤π/2 2⎢ ⎥ dθ = 2sin θ cosθ dθ0 2∫ 0⎢⎣⎥⎦r= 03/2⎡2sin θ ⎤π2= ⎢ ⎥ =⎢⎣3 ⎥⎦30852 Section 13.10 Instructor's Resource Manual

4.2 x⎛ π ⎞ 2⎛ dy dx π ⎞( x –3) dx –π1 3 ⎜ ⎟ = =3 1 ⎜ ⎟⎝ ⎠ ⎝3⎠2∫∫ ∫c.π/2 π/2 a 28∫ sin0 ∫0 ∫ 0ρ φdρdθ dφ5.1 yf ( x , y ) dxdy0 0∫∫10.π sin x 5π( x + y ) dydx = ≈3.92700 04∫∫6.π /2 cosx f ( x , y ) dydx0 0∫ ∫11.1 x 1−x22 18∫∫∫z dzdydx =0 0 037.1/2 1–2y1–2 y–zf ( x , y , z ) dxdzdy0 0 0∫ ∫ ∫12.2π3 –2 2π∫ ( r ) rdrd [ln r]3 0 ∫ θ =2 ∫ 0 2 dθ2π ⎛3⎞ ⎛3⎞= ∫ ln dθ 2 ln 2.54760 ⎜ ⎟ = π ⎜ ⎟≈⎝2⎠ ⎝2⎠13.2 3 20 1∫∫m= xy dxdy =3238.4 4– z∫∫ ∫0 0 0yf ( x , y , z ) dxdydz2 3 3Mx= ∫∫ xy dxdy = 160 12 3 2 2 208My= ∫∫ x y dxdy =0 19⎛13 3 ⎞( x, y) = ⎜ , ⎟⎝ 6 2⎠14.Ix2 3 4 128= ∫∫ xy dxdy = = 25.60 159. a.b.2–2 2–2–2∫∫ ∫8 a a x a x y dz dy dx0 0 0π /2 a a2– r28∫0 ∫0∫0rdzdrdθ15.21/2= ( , ) = (9– ) ; x ( , ) = 0;2 –1/2y ( , ) = – (9– )z f x y y f x yf x y y y3 y 2 2 –1Area = ∫∫ (9 – ) + 10 y /33 y3(9 2 ) −1/2= ∫∫ − y0 y /3y y dx dydxdy= 3 2 –1/2 2 1/2 3∫ (9– y ) (2 y ) dy = [–2(9– y ) ]0 0 = 6Instructor’s Resource Manual Section 13.10 853

16. a.b.π /2 2 3rrdrdzdθ = 9π≈28.27430 0 0∫ ∫ ∫π/2 2 4−r221/2∫ (4 )0 ∫ z r rdzdrdθ0∫−0⎛8⎞= ⎜ ⎟π≈5.0265⎝5⎠17. δ(x, y, z) = kρ18.19.20.π 2π3 2m = ∫∫ kρρ sinφdρdθdφ= 80πk0 0∫12π 4(1+sin θ )m = ∫∫ 1dA=rdrdθR ∫0 ∫ 02π⎛1−cos2θ⎞= 8∫1+ 2sinθ+ dθ= 24π0 ⎜⎟⎝2 ⎠2π 4(1+sin θ )M x = ∫∫ ydA=( rsin θ ) rdrdR ∫0 ∫ 0= 80π80 πy = = 10 ; x = 0 (by symmetry)24π3a ( b / a)( a– x) ( c / ab)( ab– bx– ay)m = ∫∫ kxdzdydx0 0 ∫ 0⎛ k ⎞ 2= ⎜ ⎟ abc⎝24⎠π 2sinθr23π∫∫ rdzdrdθ4.71240 0 ∫= ≈02u+v v−u21. Let x = and y = . Then we have,22sin x − y cos x+ y = sin ucosv and( ) ( )1 12 21 12 21 ⎛ 1⎞1J( u,v)= = −⎜− ⎟=.− 4 ⎝ 4⎠2Thus,∫∫Rsinπ0( − ) cos( + )πx y x y dxdy= 1∫∫ sin u cos vdudv0 0 21 ππ= cos v[ −cosu]dv002∫∫π= cos vdv=0θReview and Preview Problems1. Answers may vary. One solution isx = 3cos t, y = 3sin t, 0 ≤ t < 2π2 2 2 2Then: x + y = 9cos t+ 9sin t = 9 asrequired.2. Answers may vary. One solution isx = cost+ 2, y = sin t+ 1, 0 ≤ t < 2π2 2 2 2Then: ( x − 2) + ( y− 1) = cos t+ sin t = 1 asrequired.3. Answers may vary. One solution for the circle isx = 2cos t,y = 2sintTo have the semicircle where y > 0 , we needsin t > 0 , so we restrict the domain of t to0 < t < π .4. Answers may vary. Considerx = acos( − t), y = asin( −t), 0 ≤t ≤ π ; thisis a semicircle.a. Since sin( − t) =− sin t , and since sin t ≥ 0 on[ 0,π ] , 0y ≤ on [ ]0,π .b. As t goes from 0 to π , - t goes from 0 to− π so the orientation is clockwise.5. Answers may vary. One solution isx = − 2+ 5 t, y = 2 for t ∈ [0,1] .6. Note that x+ y = 9 , so a simpleparameterization is to let one variable be t andthe other be 9 – t . Since we are restricted to thefirst quadrant, we must have t > 0 and 9− t > 0;hence the domain is t ∈ (0,9) . Finally, sinceorientation is to be down and to the right, wewant y to decrease and x to increase as tincreases. Thus we use x = t and y = 9 − t .7. Note that x+ y = 9 , so a simpleparameterization is to let one variable be t andthe other be 9 – t . Since we are restricted to thefirst quadrant, we must have t > 0 and 9− t > 0;hence the domain is t ∈ (0,9) . Finally, sinceorientation is to be up and to the left, we want yto increase and x to decrease as t increases.Thus we use x = 9 − t and y = t .8. Since we are restricting the parabola to the pointswhere y > 0 , a simple parameterization is2x = t y = −t t∈ −, 9 , [ 3,3]Note that the orientation is left to right.854 Review and Preview Instructor’s Resource Manual

9. Since we are restricting the parabola to the pointswhere y > 0 , a simple parameterization is10.2x =− t y = −t t∈ −, 9 , [ 3,3]Note that the orientation is right to left.2 2b ⎛dx⎞ ⎛dy⎞L = ∫dta ⎜ ⎟ + ⎜ ⎟⎝ dt ⎠ ⎝ dt ⎠. Using theparameterization in problem 6,dx dyx = t, y = 9 − t, a = 0, b = 9, 1, 1dt= dt= − and9 2 29so L = ∫ 1 + 1 dt = ⎡ 2t⎤= 9 20 ⎣ ⎦.0(Note: this can be verified by finding thedistance between the points (0,9) and (9,0) ).11. ∇ f ( xy , ) = fx( xy , ) i+fy( xy , ) j. Now iff( x, y) = xsinx+ ycosy, thenf ( xy , ) = xcosx+ sin x, f ( xy , ) = cosy−ysinyxThus:∇ f( x, y) = ( xcos x+ sin x) i+ (cos y−ysin y)j12. ∇ f ( xy , ) = fx( xy , ) i+fy( xy , ) j. Now if−xyf( x, y)= xe + ye , then−xy −xy 2 xyf ( x, y) = ( − yxe + e ) + ( y e )xxy−xy2 xy= − xy e + y e(1 )and2 −xy xy xyf ( xy , ) =− ( xe ) + ( xye + e )y2 −xyxy=− xe + (1 + xye )Thus,−xy2 xy∇ f( x, y) = [(1 − xy) e + y e ] i2 −xyxy+− [ xe + (1 + xye ) ] j13. ∇ f ( xyz , , ) = fx i+ fy j+fzk . Now if2 2 2f( x, y, z)x y z= + + , thenfx( xyz , , ) = 2 x, fy( xyz , , ) = 2 y, fz( xyz , , ) = 2zso∇ f( x, y, z) = 2xi+ 2yj+2zk14. ∇ f ( xyz , , ) = fx i+ fy j+fzk . Now iff( x, y, z)=12 2 2x + y + z−2xfx=,2 2 2 2( x + y + z )andfz−2z=( x + y + z )2 2 2 2y, then−2yf y =,2 2 2 2( x + y + z ),so−2x−2y∇ f( x, y, z)= i+j2 2 2 2 2 2 2 2( x + y + z ) ( x + y + z )−2z+k2 2 2 2( x + y + z )15. ∇ f ( xyz , , ) = fx i+ fy j+fzk . Now iff ( x, y, z)= xy + xz + yz , then fx( x, y, z)= y+ z,f y ( xyz , , ) = x+ z, and fz( x, y, z)= x+ yso( ) ( ) ( )∇ f ( xyz , , ) = y+ zi+ x+ zj+ x+yk16. ∇ f ( xyz , , ) = fx i+ fy j+fzk . Now if17.1f( x, y, z)=, then2 2 2x + y + z−xfx=,32 2 2( x + y + z ) 2− yf y =, and32 2 2( x + y + z ) 2−zfz=,32 2 2( x + y + z ) 2so−x−y∇ f( x, y, z)= i+j3 32 2 2 2 2 2 2( x + y + z ) ( x + y + z ) 2−z+k32 2 2( x + y + z ) 2π 2 π1−cos2t∫ sin tdt=dt0 ∫ 0 21 π= ( 1 cos2 )2∫ − t dt0π1⎡1 ⎤= sin 22⎢t−t2 ⎥⎣ ⎦01⎡⎛ 1 ⎞ ⎛ 1 ⎞⎤= π sin ( 2π) 0 sin ( 2 0)2⎢⎜ − ⎟−⎜ − ⋅ ⎟2 2⎥⎣⎝ ⎠ ⎝ ⎠⎦π=2ππt tdt 1 t dt 2π1 ⎤⎣ 4 ⎦ 01=− cos 2 − cos 041=− ( 1 − 1 ) = 0418. ∫ sin cos =0 ∫ 0sin ( 2 )= ⎡−cos( 2t)( ( π ) ( ))Instructor’s Resource Manual Review and Preview 855

19.20.21.22.12 1 2 1 2y=2⎡ ⎤ ⎡xy⎤xy dydx= ⎢ xydy⎥dx= ⎢ ⎥ dx⎢ ⎥ ⎢ 2⎣ ⎦ ⎣ ⎥⎦∫∫ ∫ ∫ ∫01 0 1 0 y=11 2⎛3x⎞⎡3x⎤ 3= ∫⎜⎟dx= ⎢ ⎥ =⎝ 2 ⎠ ⎢⎣4 ⎥⎦410 0⎡⎤( x + 2 y) dydx = ∫ ⎢∫( x + 2 y)dy⎥dx=⎢⎣⎥⎦1 4 1 42 2∫∫−11 −1 11412 2y=2∫y=1−1 −131⎡x⎣+ 15x⎤= 16 + 16 = 32⎦−1( 3 15)⎡x y+ y ⎤ dx = x + dx =⎣ ⎦2π 2 2π 2 2π3r=2⎡ ⎤ ⎡2 2r ⎤∫∫ ∫ ∫ ∫0 1 0 1 0 r=12π2π∫00∫r drdθ = ⎢ r dr⎥dθ = ⎢ ⎥ dθ=⎢ ⎥ ⎢ 3⎣ ⎦ ⎣ ⎥⎦7 7 14πdθ= θ = ≈14.663 3 32ππ2∫∫∫0 0 12ρ sinφdρdφdθ =2π π⎡2⎤2∫∫ ⎢∫ρ sinφdρ ⎥ dφdθ=0 0 ⎢⎣1⎥⎦2π πρ=2⎡ 3ρ ⎤∫∫ ⎢ sinφ⎥dφdθ=30 0 ⎢⎣⎥⎦ρ= 12π π 2πφ=π⎡ 7 ⎤⎡ 7 ⎤∫ ⎢∫sinφ dφ⎥ dθ = cosφ dθ3∫ ⎢− 3 ⎥ =0 ⎢⎣0 ⎥⎦0 ⎣ ⎦φ=02π14 28π∫ dθ= ≈ 29.323 3023. Note that2π π R2ρ sinφdρdφdθ=∫∫∫0 0 r2π π R0 0 r2π π 30 0ρ=r2π⎡π1 (3 3∫ ⎢∫)sin30⎢⎣02π⎡ 3 30 φ=02π⎡2 3 3 ⎤ 4 3 4 30⎡⎤2⎢ ρ sinφdρ ⎥ dφdθ=⎢⎣⎥⎦ρ=R⎡ρ⎤⎢ sinφ⎥dφdθ=⎢⎣3 ⎥⎦⎤R − r φdφ⎥dθ=⎥⎦φ=π1 ⎤⎢− ( R − r )cos φ dθ=3⎥⎣⎦∫∫ ∫∫∫∫∫⎢ ( R − r ) dθ πR πr3 ⎥ = −⎣ ⎦ 3 3which we recognize as the difference between thevolume of a sphere with radius R and the volumeof a sphere with radius r . Thus the volume inproblem 22 is that of a spherical shell with centerat (0,0,0) and, in this case, outer radius = 2 andinner radius = 1 .2 224. Let f ( xy , ) = z= 144−x − y ; then:f ( xy , ) = − 2 x f ( xy , ) = − 2yxy2 2We note that z = 36 when x + y = 108 ; thusthe surface area we seek projects onto theS = ( x, y)| x 2 + y2 = 108circular region inside { }Hence∫∫2 2A = 4x + 4y + 1 dA; or, convertingS2 2 2to polar coordinates ( r = x + y and r = 6 32when r = 108) ,2π2π6 3A= ( 4r + 1) rdrdθ=∫0 02r=6 3⎡ 1 3(42r 1) 2⎤⎢ + dθ=12⎥⎣⎦0 r=02π1 3 212∫0∫ ∫[433 −1] dθ≈1501.7π≈ 4717.7856 Review and Preview Instructor’s Resource Manual

25. This will be the unit vector, at the point (3, 4,12), in the direction of ∇ F , where2 2 2F( xyz , , ) = x + y + z .Now∇ F( xyz , , ) = Fi+ F j+ Fk = 2xi+ 2yj+2zkx y zso that ∇ F(3, 4,12) = 6,8,24 and the unit vectorin the same direction is∇ F 6 8 24 3 4 12= , , = , , .∇F26 26 26 13 13 13This agrees with our geometric intuition, since2 2 2x + y + z = 169 is the surface of a sphere withcenter at O = (0,0,0) and radius = 13. Now theplane tangent to a sphere (center at (0,0,0) andradius r ) at any point P= ( abc , , ) isperpendicular to the radius at that point; so itwould follow that the vector OP = a, b,c isperpendicular to the tangent plane and hencenormal to the surface. The unit normal in the 1direction of OP is simply abc , , , or in ourrcase 1 3, 4,12 .13Instructor’s Resource Manual Review and Preview 857

CHAPTER 14Vector Calculus14.1 Concepts Review6.1. vector-valued function of three real variables or avector field2. gradient field3. gravitational fields; electric fields4. ∇⋅F,∇× FProblem Set 14.11.7. 2x −3 y, − 3 x,28. (cos xyz) yz, xz,xy9. f ( x, y, z) = ln x + ln y + ln z ;–1 –1 –1∇ f ( x, y, z) = x , y , z2.10. x, yz ,y11. e cos z, xcos z, – xsinz12.–2z2 –2z∇ f( x, y, z) = 0, 2 ye , –2y e3.–2z= 2ye0,1, – y13. div F = 2x – 2x + 2yz = 2yz2curl F = z ,0,–2y14. div F = 2x + 2y + 2zcurl F = 0, 0, 0 = 04.15. div F= ∇ ⋅ F = 0+ 0+ 0=0curl F= ∇× F= x−x, y− y,z− z = 016. div F = –sin x + cos y + 0curl F = 0,0,0 = 05.17. div F = e x cos y+ e x cos y+ 1 = 2e x cos y+1curl F = 0, 0, 2esin yx18. div 0 0 0 0= ∇⋅ = + + =F Fcurl F = ∇ × F = 1-1,1-1,1-1 = 0858 Section 14.1 Instructor’s Resource Manual

19. a. meaninglessb. vector fieldc. vector fieldd. scalar fielde. vector fieldg. vector fieldh. meaninglessi. meaninglessj. scalar fieldk. meaninglessf. vector field20. a. div(curl F ) = div ⋅ Py – Nz, Mz – Px, Nx – My= ( P – N ) + ( M – P ) + ( N – M ) = 0b. curl(grad f ) = curl f , f , f = f – f , f – f , f – f = 0x y zzy yz xz zx yx xyc. div( fF)=div fM , fN, fP = ( fM + f M ) + ( fN + f N) + ( fP + f P)x x y y z z= ( f)( M + N + P ) + ( f M + f N + f P) = ( f)(div F) + (grad f)⋅Fx y z x y zyx zx zy xy xz yzd. curl( fF)=curl fM , fN,fP= ( fP + f P) – ( fN + f N), ( fM + f M) – ( fP + f P), ( fN + f N) – ( fM + f M)y y z z z z x x x x y y= ( f ) P – N , M – P , N – M + f , f , f × M, N,P = (f)(curl F) + (grad f) × Fy z z x x y x y z21. Let–3f( x, y, z) = – c r , so–4 r –5grad( f) = 3c r = 3 c r r.r–3Then curl F = curl ⎡ (–c⎤⎢⎣r ) r ⎥⎦( – c –3 ) (curl ) ( 3 c –5 r)–3 –5( c r ) 0 ( c r ) r r–3div F =div ⎡( –c ⎤⎢⎣r ) r ⎥⎦–3 –5= ( – c ) (div ) + ( 3 c ) ⋅= ( – c )(1+ 1+ 1) + ( 3c)–3 3= ( –3cr ) + 3cr = 0= r r + r × r (by 20d)= – ( ) + 3 ( × ) = 0+ 0=0r r r r r (by 20c)–3 –5 2r r rcurl ⎡⎢⎣r r⎤⎥⎦r (0) r ( 0)= 0−m −m −m−222. − c = ( − c ) + mc( )div ⎡– c ⎤⎢ ⎥ = – c (3) + mc⎣r r ⎦r r r= ( m–3)c r– m – m – m–2 2– m23.–1/2 –1/2 –1/2grad f=f′ () rxr , f′ () ryr , f′() rzr(if r ≠ 0)–1/2 –1/2= f′ () r r x, y, z = f′() r r r–1/ 2curl F = [ f( r)][curl r] + [ f′( r) r r]× r–1/ 2= [ f( r)][curl r] + [ f′( r) r r]× r= 0 + 0 = 024. div F = div[f(r)r] = [f(r)](div r) + grad[f(r)] ⋅ r′ –1= + ⋅[ f( r)](div r) [ f ( r) r r]r′ –1= + ⋅[ f( r)](3) [ f ( r) r ]( r r )′ –1 2= + = + ′3 f () r [ f () r r ]( r ) 3 f() r rf () rNow if div F = 0, and we let y = f(r), we have thedydifferential equation 3y+ r = 0, which can bedrsolved as follows:dy dr–3= –3 ; ln y = –3ln r + ln C = ln Cr ,y rfor each C ≠ 0. Then y = Cr –3 , or–3f () r = Cr , is a solution (even for C = 0).Instructor’s Resource Manual Section 14.1 859

25. a. Let P = ( x0, y0).div F = div H = 0 since there is no tendencytoward P except along the line x = x0,andalong that line the tendencies toward andaway from P are balanced; div G < 0 sincethere is no tendency toward P except alongthe line x = x0,and along that line there ismore tendency toward than away from P;div L > 0 since the tendency away from P isgreater than the tendency toward P.b. No rotation for F, G, L; clockwise rotationfor H since the magnitudes of the forces tothe right of P are less than those to the left.c. div F = 0; curl F = 02–div G = –2yey < 0 since y > 0 at P; curlG = 02 2 –1/2= +div L ( x y ) ; curl L = 0div H = 0; curl H = 0, 0, – 2xewhich2– xpoints downward at P, so the rotation isclockwise in a right-hand system.26. F( x, yz , ) = Mi+ Nj + Pk ,whereM( x, y, z) = y, N( x, y, z) = 0, P( x, y, z) = 021.510.51.2 1.4 1.6 1.8 2a. Since the velocity into (1, 1, 0) equals thevelocity out, there is no tendency to divergefrom or accumulate to the point.Geometrically, it appears thatdiv F (1,1,0) = 0 . Calculating,∂M ∂N ∂Pdiv F ( xyz , , ) = + + = 0 + 0 + 0 = 0∂x ∂y ∂zb. If a paddle wheel is placed at the point (withits axis perpendicular to the plane), thevelocities over the top half of the wheel willexceed those over the bottom, resulting in anet clockwise motion. Using the right-handrule, we would expect curlF to point into theplane (negative z). By calculatingcurl F = (0 − 0) i+ (0 − 0) j+ (0 − 1) k = −k27. F( x, yz , ) = Mi+ Nj + Pk , wherexyM =− , N =− ,2 2 3 2(1 + x +2 2 3 2y ) (1 + x + y )P = 010.5-1 -0.5 0.5 1-0.5-1a. Since all the vectors are directed toward theorigin, we would expect accumulation at thatpoint; thus div F (0,0,0) .should be negative.Calculating,2 23( x + y )div F ( xyz , , ) =2 2 5 2(1 + x + y )2− + 02 2 3 2(1 + x + y )so that div F (0,0,0) = −2b. If a paddle wheel is placed at the origin (withits axis perpendicular to the plane), the forcevectors all act radially along the wheel and sowill have no component acting tangentiallyalong the wheel. Thus the wheel will not turnat all, and we would expect curl F = 0. Bycalculatingcurl F = (0 − 0) i+ (0 −0)j⎛ 3yx3xy⎞+ ⎜−⎟k⎜(1 + x2+ y2)32(1 + x2+ y2)32⎟⎝⎠= 0860 Section 14.1 Instructor’s Resource Manual

28. div v = 0 + 0 + 0 = 0;curl v = 0,0, w+ w = 2ωk29.1 2 2∇ f ( x, y, z) = mω2 x,2 y,2 z = mωx, y,z2= F(x, y, z)30.2∇ f = div(grad f) = div fx , fy,fz= fxx + fyy + fzza.b.c.2∇ f = 4–2–2=02∇ f = 0+ 0+ 0=02∇ f = 6 x–6x+ 0=02–1d. ∇ f = div(grad f)= div( grad r )–3= div ( – r r ) = 0 (by problem 21)Hence, each is harmonic.31. a. F× G = ( fygz − fzgy) i−( fxgz − fzgx) j+ ( fxgz − fzgx)k . Therefore,∂ ∂ ∂div( F× G ) = ( fygz − fzgy) − ( fxgz − fzgx) + ( fxgy − fygx).∂x ∂y ∂zUsing the product rule for partials and some algebra gives⎡∂f∂fz y ⎤ ⎡∂fx∂fz⎤ ⎡∂fy∂fx⎤div( F× G)= gx ⎢ − ⎥+ g y gzy z⎢ −z x⎥+ ⎢ − ⎥⎣ ∂ ∂ ⎦ ⎣ ∂ ∂ ⎦ ⎣ ∂x ∂y⎦⎡∂g∂gz y ⎤ ⎡∂ggx ∂gz⎤ ⎡∂y ∂gx⎤+ fx ⎢ − ⎥+ fy fzy z⎢ −z x⎥+ ⎢ − ⎥⎣ ∂ ∂ ⎦ ⎣ ∂ ∂ ⎦ ⎣ ∂x ∂y⎦= G⋅curl( F) −F⋅curl( G)b.⎛∂f ∂g ∂f ∂g ⎞ ⎛∂f ∂g ∂f ∂g ⎞ ⎛∂f ∂g ∂f ∂g⎞∇ f ×∇ g = ⎜ − ⎟i−⎜− ⎟j+ ⎜ − ⎟k⎝∂y ∂z ∂z ∂y ⎠ ⎝∂x ∂z ∂z ∂x ⎠ ⎝∂x ∂y ∂y ∂x⎠∂ ⎛∂f ∂g ∂f ∂g ⎞ ∂ ⎛∂f ∂g ∂f ∂g ⎞ ∂ ⎛∂f ∂g ∂f ∂g⎞Therefore, div( ∇ f ×∇ g)= ⎜ − ⎟− ⎜ − ⎟+ ⎜ − ⎟.∂x ⎝∂y ∂z ∂z ∂y ⎠ ∂y⎝∂x ∂z ∂z ∂x ⎠ ∂z⎝∂x ∂y ∂y ∂x⎠Using the product rule for partials and some algebra will yield the resultdiv( ∇ f ×∇ g) = 032.lim F ( x , y , z )a b c( x, y, z) →( , , )F( x, y, z)– L < ε.= L if for each ε > 0 there is a δ > 0 such that 0 < x, y, z – a, b,c < δ implies thatF is continuous at (a, b, c) if and only if lim = F ( a, b, c).( x , y , z ) → ( a , b , c )14.2 Concepts Review1. Increasing values of t2.3.4.n∑ f ( xi, yi)Δsii=12 2f ( xt (), yt ()) [ x′ ()] t + [ y′()] tdrF ⋅dtInstructor’s Resource Manual Section 14.2 861

Problem Set 14.21. 1 (27 t 3 + t 3 )(9 + 9 t 4 ) 1/2 dt = 14( 2 2 –1)∫ ≈ 25.598002.31/2∫1⎛ t ⎞ ⎛1 25t⎞ ⎛ 1 ⎞ 3/2() tdt (26 –1)0⎜ ⎟ + = ⎜ ⎟ ≈2 ⎜4 4 ⎟ 4500.2924⎝ ⎠ ⎝ ⎠ ⎝ ⎠3. Let x = t, y = 2t, t in [0, π ] .π(sin + cos ) = (sin + cos 2 ) 1+4C0Then ∫ x yds ∫ t t dt= 2 5 ≈ 4.47214. Vector equation of the segment isx,y = − 1, 2 + t 2, − 1 , t in [0, 1].1 2– t 1/2 2 –1∫ (–1+ 2 te ) (4 + 1) dt = 5 e (1– 3 e ) ≈− 1.712405.6.7.8.∫ 1 3 2 4 1/2 ⎛1⎞ 3/2(2t+ 9 t )(1 + 4t + 9 t ) dt = (14 –1)0⎜ ⎟ ≈⎝6⎠8.56392π2 2 2 2 2 1/2 2π2(16cos t+ 16sin t+ 9 t )(16sin t + 16cos t+ 9) dt = (16 + 9 t )(5) dt0 032π3= ⎡80t+ 15t⎤ = 160π+ 120π ≈ 4878.11⎣ ⎦0∫ ∫2 2 2[( t –1)(2) + (4 t )(2 t)]dt =0∫4 3 2dx + dy =0 –1∫ ∫(–1) (4) 6010039.10.∫ ∫ ∫3 3 3 3 3 3ydx+ xdy= ydx+ xdy+ ydx+xdyC C1 C2∫1 2 3 3–2828[( t – 3) (2) + (2 t) (2 t)] dt = ≈ 23.657135–2 3 2 3= ∫ dy + dx = + =1 ∫ –4(–4) (–2) 192 (–48) 14411. y = –x + 23∫ ([ 2(– 2)](1) [ – 2(– 2)](–1)) 01 x+ x+ + x x+ dx =12.1 2 1 2[ x + ( x)2 x] dx = 3x dx = 10 0∫ ∫(letting x be the parameter; i.e., x = x,y = x2 )13. x, yz , = 1, 2,1 + t 1, -1, 01 2 17∫ [(4 – t)(1) + (1 + t)(–1) – (2 – 3 t+ t )(–1)] dt = ≈ 2.833306141 3 – 2 – 2[( e t )( e t ) + ( e t + e t )(–3 t ) + ( e t )(2 e t )] dt0∫⎛1⎞ 4 ⎛2⎞ 3 ⎛1⎞–2 5= ⎜ ⎟e + ⎜ ⎟e – e+⎜ ⎟e–⎝4⎠ ⎝3⎠ ⎝2⎠12≈ 23.9726862 Section 14.2 Instructor’s Resource Manual

15. On C 1 : y = z = dy= dz = 0On C 2 : x = 2, z = dx = dz = 0On C 3 : x = 2, y = 3, dx = dy = 02 3 4x dx + (2 – 2 y) dy + (4 + 3 – z)dz0 0 0∫ ∫ ∫2224⎡ x ⎤[2 – 2 ] 3 ⎡ z ⎤= ⎢ ⎥ + y y 0 + ⎢7 z–⎥⎢⎣ 2 ⎥⎦ 20 ⎢⎣ ⎥⎦0= 2 + (–3) + 20 = 1916. x, yz , = t 2,3, 4 , t in [0, 1].1[(9 )(2) (8 )(3) (3 )(4)] 270 t + t + t dt =∫17.2 21/2m = ∫ k x ds = k x (1+4 x ) dxC ∫–2⎛k⎞ 3/2= ⎜ ⎟(17 –1) ≈11.6821k⎝6⎠18. Let δ(x, y, z) = k (a constant).3 2 2 2 2 2 1/2m k 1ds k π2 2 1/2= ∫ = 1( a sin t + a cos t + b ) dtC ∫= 3 π ka ( + b)02 2 1/2 3π2 2 2 1/29 π bk( a + b )M xy = k∫zds = k( a + b ) btdtC∫ =022 2 1/2 3π2 2 1/2 2 2 1/2M xz = k∫yds = k( a + b ) asintdtC∫ = ak( a + b ) (2) = 2 ak( a + b )02 2 1/2 3π2 2 1/2M yz = k∫xds = k( a + b ) acostdtC∫ = ak( a + b ) (0) = 00M yz M xz 2aTherefore, x = 0; y;m= = m= M xy 3 πbz = = .3 π m 219∫3 3 2( x – y ) dx xy dyC += 0 6 9 2 6 2∫ [( – )(2 ) ( )( )(3 )]–1 t t t + t t t dt7=− ≈ − 0.15914420.Cx– – y 5 ⎡ 3 3 1 11( ) ⎛ ⎞edx e dy t – ⎛ ⎞⎛ ⎞⎤= ⎢ ⎜ ⎟ ⎜ ⎟⎜ ⎟ dtt 2 t t⎥⎣ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎦∫ ∫ = 123.621. W = ∫ F⋅ ( ) ( – )Cd r = ∫Cx + y dx + x y dy π /2= ∫ [( acost+ bsin t)(– asin t) + ( acos t – bsin t)( bcos t)]dt0π /2 2 2 2 2 π /22 2–( a + b )sin 2t= ∫ [–( a + b )sin tcos t+ab(cos t – sin t)]dt = abcos 2t dt0∫+0 22 2π /2⎡2 2( a + b )cos2t absin2t⎤a + b= ⎢+ ⎥ =⎢⎣4 2 ⎥⎦–2022. x, yz , = t 1,1,1 , t in [0, 1].1(2 x – y) dx + 2 z dy + ( y – z) dz = ( t + 2t + 0) dtC0∫ ∫ = 1.5Instructor’s Resource Manual Section 14.2 863

23.π ⎡ ⎛ π ⎞ ⎛ 2sin π t⎞ ⎛cos π t⎞ ⎛ t ttcos π ⎞ ⎛sin π ⎞– t ⎤ dt 2 – 1.36340 ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟+π ⎜ ⎟+ ⎜ ⎟ = ≈2 2 2 2 2⎥⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ π∫24.∫∫W = F⋅ Cd r =Cydx + zdy + xdz64 8 412= + 12 + = ≈ 27.46675 3 15= 2 2 3 2∫ [( )(1) ( )(2 ) ( )(3 )]0 t + t t + t t dt= 2 4 3 2∫ 0 (2t + 3 t + t ) dt25. The line integralCi, i = 1,2,3,∫ F • dr represents the work done in moving a particle through the force field F along the curveC ia. In the first quadrant, the tangential component to C 1 of each force vector is in the positive y-direction , the samedirection as the object moves along C 1 . Thus the line integral (work) should be positive.b. The force vector at each point on C 2 appears to be tangential to the curve, but in the opposite direction as theobject moves along C 2 . Thus the line integral (work) should be negative.c. The force vector at each point on C 3 appears to be perpendicular to the curve, and hence has no component inthe direction the object is moving. Thus the line integral (work) should be zero26. The line integralCi, i = 1,2,3,∫ F • dr represents the work done in moving a particle through the force field F along the curveC ia. In the first quadrant, the tangential component to C 1 of each force vector is in the positive y-direction , the samedirection as the object moves along C 1 . Thus the line integral (work) should be positive.b. The force vector at each point on C 2 appears to be perpendicular to the curve, and hence has no component inthe direction the object is moving. Thus the line integral (work) should be zeroc. The force vector at each point on C 3 is along the curve, and in the same direction as the movement of theobject. Thus the line integral (work) should be positive.27.⎛ y ⎞ 2 3 2 2 2 2 1/21 ds (1 10sin t )[( 90cos t sin t ) (90sin t cos t ) ] dtC ⎜ + ⎟ = + − +⎝ 3 ⎠ 0∫ ∫ = 225Christy needs 450 = 2.25 gal of paint.2008π28. ∫ 0, 0, 1.2 ⋅ dx, dy, dz = 1.2 dzC∫ = 1.2(4) 38.4 120.64C ∫ 0 dt = π≈ ft-lbTrivial way: The squirrel ends up 32π ft immediately above where it started.( 32π ft)(1.2 lb) ≈ 120.64 ft-lb29. C: x + y = aLet x = t, y = a – t, t in [0, a].2 2Cylinder: x + y = a; ( x + y) = a ;2 2 2 2Sphere: x + y + z = aThe curve of intersection satisfies:2 2 2x + 2xy + y = a2z = 2 xy; z = 2xy.864 Section 14.2 Instructor’s Resource Manual

Area 8 2 8 a2 2xyds02 t ( aC– t ) (1) (–1)a 2= ∫ = ∫+ dt = 16 ∫ at – t dt0( ) 2a⎡– aa2 2 2 – a ⎤⎢t⎛t⎞–116 at – t sin 2 ⎥⎛⎡ ⎛ 2 2a ⎞= +⎛π⎞⎤ ⎡ ⎛a⎞⎛−π⎞⎤⎞⎢ 2 2 ⎜ a ⎟⎥= 16⎜⎢0 + ⎥− ⎢0+⎥⎟⎜ ⎜ 8 ⎟⎜ ⎟2 ⎜ 8 ⎟⎜ ⎟2 ⎟⎢⎝ 2⎣⎠⎥⎦⎝⎢⎣ ⎝ ⎠⎝ ⎠⎥⎦ ⎢⎣ ⎝ ⎠⎝ ⎠⎥⎦⎠02= 2aπTrivial way: Each side of the cylinder is part of a plane that intersects the sphere in a circle. The radius of eacha⎛a⎞⎛a⎞ a 2circle is the value of z in z = 2xywhen x = y = . That is, the radius is 2 ⎜ ⎟⎜ ⎟ = . Therefore, the2⎝2⎠⎝2⎠2⎡ 2⎛a2 ⎞ ⎤2total area of the part cut out is r⎢π ⎥ = 2 a π.⎢ ⎜ 2 ⎟⎝ ⎠ ⎥⎣ ⎦30.32 a 2kaIy= ∫ kx ds = 4k t 2dt4 2c ∫ =03(using same parametric equations as in Problem 29)I = I (symmetry)xy3kaIz = Ix + Iy= 8 2 32 2 231. C: x + y = a⎡ π⎤Let x = a cos θ, y = a sin θ, θ in ⎢0, .2 ⎥⎣ ⎦2 2Area = 8 ∫ –C a x dsπ /2 2 2= 8 ∫ ( asin ) (– asin ) + ( acos ) d0θ θ θ θπ /2 2 2 π /2= 8 ∫ ( asin θ) a dθ = 8 a [–cos θ]002= 8aInstructor’s Resource Manual Section 14.2 865

32. Note that r = a cos θ along C.2 2 2 1/2 2 2 1/2Then ( a – x – y ) = ( a – r ) = acos θ.⎧x = rcosθ = ( asin θ)cosθ⎫⎡ π⎤Let ⎨⎬,θ in 0, .⎩y = rsinθ = ( asin θ)sinθ⎢⎭ 2 ⎥⎣ ⎦Therefore, x′ ( θ ) = acos2 θ; y′( θ) = asin2 θ.Then2 2 2 1/2Area = 4 ∫ ( a – x – y ) dsCπ /2 2 2 1/2= 4 ∫ ( acos )[( asin 2 ) + ( acos 2 ) ] d02θ θ θ θ = 4a .33. a.2 π /2 2 2 2 1/2x yds= (3sin t) (3cos t)[(3cos t) (–3sin t) ] dtC ∫+0∫π /2π /2 2 ⎡⎛1⎞ 3 ⎤= 81∫sin tcostdt = 81 sin t 270⎢⎜⎟ =3⎥⎣⎝⎠ ⎦0b.C42 2 3 2 32xy dx + xy dy = (3 – t)(5 – t) (–1) dt + (3 – t)(5 – t) (–1) dt0 0∫ ∫ ∫3= 2 ( 3–13 2∫ t t + 55 t – 75) dt = –148.5014.3 Concepts Review1. f(b) – f(a)2. gradient; ∇f() r3. 0; 04. F is conservative.Problem Set 14.31. M y = –7 = Nx,so F is conservative.2 2f ( x, y) = 5 x –7xy+ y + C2. M y = 6y+ 5 = Nx,so F is conservative.3 2 3f ( x, y) = 4x + 3xy + 5 xy–y + C3.4.4 5M y 90 x y– 36y Nx= ≠ since4 5Nx90 x y–12 y ,= so F is not conservative.2 3 8M y –12x y 9 y Nx,= + = so F isconservative.5 3 4 9f ( x, y) = 7 x – x y + xy + Cyx7. M = 2 e – e = N so F is conservative.8.yf ( x, y) = 2 xe – ye + C– x –1M y – e y Nx,yxx= = so F is conservative.– xf ( x, y) = e ln y+C9. M y = 0 = Nx, Mz = 0 = Px,and Nz= 0 = Py,so F is conservative. f satisfies22fx ( x, y, z) = 3 x , f y ( x , y , z ) = 6 y , and2fz( x, y, z) = 9 z .Therefore, f satisfies31. f ( x, y, z) = x + C1( y, z),32. f ( x, y, z) = 2 y + C2( x, z),and33. f ( x, y, z) = 3 z + C ( x, y).A function with an arbitrary constant thatsatisfies 1, 2, and 3 is3 3 3f ( x, y, z) = x + 2y + 3 z + C.10. M = 2 x = N , M = 2 z = P , andNzy y z x= 0 = P , so F is conservative.y2 2f ( x, y, z) = x y+ xz + sinπ z+C35.12 2 –3M ⎛ ⎞y = ⎜– ⎟x y = Nx,⎝ 5 ⎠2 3 –2f ( x, y)⎛ ⎞= ⎜ ⎟x y + C⎝5⎠so F is conservative.6.2 2 2M y = (4 y )(–2xysin xy ) + (8 y)(cos xy ) ≠ Nx2 2 2since Nx= (8 x)(– y sin xy ) + (8)(cos xy ), so Fis not conservative.866 Section 14.3 Instructor’s Resource Manual

11. Writing F in the formF( x, yz , ) = M( xyz , , ) i+ Nxyz ( , , ) j+Pxyz ( , , ) k−2xwe have M( xyz , , ) = , Nxyz ( , , ) = 0,2 2( x + z )−2zPxyz ( , , ) =2 2( x + z )so that∂M ∂N ∂M 4xz ∂P= 0 = , = = ,∂y ∂x ∂z 2 2 2( x + z ) ∂x∂N∂P= 0 = . Thus F is conservative by Thm. D.∂z∂yWe must now find a function f ( xyz , , ) such that∂f −2x ∂f ∂f −2z= , = 0, =∂x 2 2 2 2( x + z ) ∂y ∂z( x + z )Note that F is a function of x and z alone so fwill be a function of x and z alone.a. Applying the first condition gives−2xf( x, y, z)= ∫ dx2 2x + z⎛ 1 ⎞= ln ⎜ C2 2 ⎟+1( z)⎝ x + z ⎠b. Applying the second condition,−2z∂f∂ ⎛ 1 ⎞ ∂C1= = ln2 2 ⎜ 2 2 ⎟+ =( x + z ) ∂z ∂z ⎝ x + z ⎠ ∂z−2z∂C1 ∂C+ which requires 1 = 02 2( x + z ) ∂z∂zHence⎛ 1 ⎞f( x, y, z) = ln⎜ 2 2 ⎟⎝ x + z ⎠ + C12. Writing F in the formF( x, yz , ) = M( xyz , , ) i+ Nxyz ( , , ) j+Pxyz ( , , ) k2we have M( xyz , , ) = 0, Nxyz ( , , ) = 1+2 yz ,2Pxyz ( , , ) = 1+2y zso that∂ M N M P N P= 0 = ∂ , ∂ = 0 = ∂ , ∂ = 4 yz =∂ .∂y ∂x ∂z ∂x ∂z ∂yThus F is conservative by Thm. D.We must now find a function f ( xyz , , ) such that∂ f 2 f f2= 1+ 2 yz , ∂ = 0, ∂ = 1+2yz∂y ∂x ∂zNote that f is a function of y and z only.a. Applying the first condition gives∫2 2 2f ( xyz , , ) = (1+ 2 yz) dy= y+ yz + C1( z)..b. Applying the second condition,2 ∂f∂ 2 2 ∂C1+ 2 y z = = ( y+ y z ) + 1∂z ∂z ∂z2 ∂C1= 2y z+ ∂ z∂C1which requires = 1or C1( z)= z+c.∂z2 2Hence f ( xyz , , ) = y+ z+ yz + C13. M y = 2y+ 2 x = Nx,so the integral is2 2independent of the path. f ( x, y)= xy + x y(3,1) 2 2∫ ( y + 2 xy) dx+ ( x + 2 xy)dy( −1,2)2 2 (3,1)= [ xy + x y] ( −1,2)= 14x14. M y = e cos y = Nx,so the line integral isindependent of the path.xLet fx( x, y) = e sin y and f y ( x, y) = e x cos y.xThen f ( x, y) = e sin y+ C1( y)andf ( x, y) = e x sin y+C2( x).xChoose f ( x, y) = e sin y.By Theorem A,(1, π / 2) xxe sin ydx+e cos ydy∫(0,0)x (1, π / 2)= [ e sin y](0,0) = e(Or use line segments (0, 0) to (1, 0), then (1, 0)⎛ π ⎞ ⎞to ⎜ 1, ⎟ . ⎟⎝ 2 ⎠ ⎠15. For this problem, we will restrict ourconsideration to the setD = {( x, y) | x > 0, y > 0}(that is, the first quadrant), which is open andsimply connected.a. Now, F( x, y)= M i+N j where3 3xyM( x, y) = , N( x, y)=4 4 2 4 4 2( x + y ) ( x + y )3 3∂M −8x y ∂Nthus = = so F is∂y4 4 3( x + y ) ∂xconservative by Thm. D, and hence ∫ Fr ()dCindependent of path in D by Theorem C.;i r isInstructor’s Resource Manual Section 14.3 867

. Since F is conservative, we can find a function( , ) f xysuch that3 3∂f x ∂f y= , =∂x4 4 2 4 4 2( x + y ) ∂y( x + y )Applying the first condition gives3x−1f ( xy , ) = ∫ dx= + Cy ( )4 4 2 4 4( x + y ) 4( x + y )Appling the second condition gives3y ∂f ∂ ⎛ −1⎞ ∂C= = + =4 4 2 4 4( x + y ) ∂y ∂y⎜4( x y )⎟⎝ + ⎠ ∂y3y ∂C+ ;4 4 2( x + y ) ∂ y∂Chence = 0 and C(y) = constant. Therefore∂y−1f ( xy , ) = + c4 44( x + y ), and so, by Thm. A,(6,3) 3 3xy∫ 4 4 2 4 4 2(2,1) ( ) dx +( )dyx + y x + y⎛ 1 ⎞ ⎛ 1 ⎞= f (6,3) − f(2,1)= ⎜− + c⎟−⎜− + c⎟⎝ 5508 ⎠ ⎝ 68 ⎠20=1377xc. Consider the linear path C: y = , 2≤ x≤ 6 in2D which connects the points (2, 1) and (6, 3);1then dy = dx . Thus2(6,3) 3 3xy∫dx + dy =4 4 2 4 4 2(2,1)( x + y ) ( x + y )6 3x 3x( )2 1∫dx + ( dx)=4 x 4 2 4 x 4 2 22 ( x + ( ) ) ( x + ( ) )2 26616 ⎡ −4 ⎤ 1 1∫ dx =5 ⎢ 4⎥=− + =217x⎣17x⎦ 25508 68201377.16. For this problem, we can use the whole real planeas D.a. F( x, y)= M i+N j where2 2M ( xy , ) = 3x −2xy− y ,2 2N( x, y) = 3y −2xy−x∂M∂Nthus =−2x− 2y= so F is∂y∂xconservative by Thm. D, and hence∫ Fr ()d i r is independent of path in D byCTheorem C.b. Since F is conservative, we can find a functionf ( xysuch , ) that∂f2 2 ∂f2 2= 3x −2 xy − y , = 3y −2xy −x.∂x∂yApplying the first condition gives2 2f ( xy , ) = 3x −2xy−y dx∫3 2 2= x −x y− xy + C( y)Appling the second condition gives2 2 ∂f∂ 3 2 2 ∂C3y −2xy− x = = ( x −x y− xy ) +∂y ∂y ∂y2 ∂C=−x− 2 xy+ ; ∂ yhence∂C2= 3y∂yandC( y)3 2 2 3f ( xy , ) x xy xy y c3= y + c. Therefore= − − + + , and so, byThm. A,(4,2)∫2 2 2 2(3x −2 xy− y ) dx+ (3x −2 xy−y ) dy( −1,1)= f(4,2) − f( − 1,1) = 24 + c − 0 + c = 24( ) ( )c. Consider the simple linear pathC: x = 5y−6,1≤ y ≤ 2 in D which connects thepoints ( 1,1)− and (4, 2); then dx = 5dy.Thus(4,2)2 2 2 2∫ (3x −2 xy − y ) dx + (3y −2 xy − x ) dy =( −1,1)22 2∫ (64y − 168y+ 108)(5 dy) + ( − 32y + 72y−36)dy122= ∫ (288y − 768y+504) dy13 22= ⎡96y − 384y + 504y⎤⎣⎦1= 240 + ( − 216) = 24868 Section 14.3 Instructor’s Resource Manual

217. M y = 18 xy = Nx, M z = 4 x = Px,Nz= 0 = Py.By paths (0, 0, 0) to (1, 0, 0); (1, 0, 0) to (1, 1, 0);(1, 1, 0) to (1, 1, 1)1 1 2 10dx + 9 y dy + (4z + 1) dz = 60 0 02 3 2f ( x, y) = 3x y + 2 xz + z.)∫ ∫ ∫(Or use18. M y = z = Nx, Mz = y = Px,Nz= x = Py.f(x, y) = xyz + x + y + zThus, the integral equals(1, 1, 1)[ xyz + x + y + z] (0.1, 0)= 3.19. M y = 1 = Nx, Mz = 1 = Px,Nz= 1 = Py(so pathindependent). From inspection observe thatf ( x, y, z)= xy+ xz+ yz satisfiesf = y+ z, x+ z,x+ y , so the integral equals(–1, 0, π)[ xy + xz + yz] (0, 0, 0) = – π . (Or use linesegments (0,1, 0) to (1,1, 0) , then (1,1, 0) to(1,1,1) .)20. M y = 2 z = Nx, Mz = 2 y = Px,Nz= 2x = Pybypaths (0,0,0) to ( π ,0,0), ( π ,0,0) to ( π , π ,0).ππ∫ cos xdx+ sin ydy = 20 ∫ 02zOr use f ( x, y, z) = sinx + 2 xyz –cos y + .221. fx = M, fy = N,fz= P22.fxy M y,fxzf yz= and f = N , so M = N .M zNzyx= and f = P , so M = P .zxfzyxPyxyz xNz= Py= and = , so .– kxfx ( x, y, z) =,2 2 2x + y + zso– k 2 2 2f ( x, y, z) = ln( x + y + z ) + C1( y, z).2Similarly,– k 2 2 2f ( x, y, z) = ln( x + y + z ) + C2( y, z),2using ; yf and– k 2 2 2f ( x, y, z) = ln( x + y + z ) + C3( y, z),2using f z.Thus, one potential function for F is– k 2 2 2f ( x, y, z) = ln( x + y + z ).2xr23. F( x, y, z) = k r = kr= k x, y,zrk 2 2 2f ( x, y, z) ⎛ ⎞= ⎜ ⎟( x + y + z ) works.⎝2⎠⎛1⎞24. Let f = ⎜ ⎟ hu ( ) where u = x2 + y2 + z2 .⎝2⎠1 1Then fx= ⎜ ⎛ ⎟ ⎞ h′( u) ux= ⎜ ⎛ ⎟⎞ g( u)(2 x) = xg( u).⎝2⎠ ⎝2⎠Similarly, f y = yg( u)and fz= zg( u).Therefore, f ( x, y, z) = g( u) x, y,z2 2 2= g( x + y + z ) x, y, z =F ( x, y, z).b25. ∫ F⋅ dr = ( m ′′ ⋅ ′)dtC ∫ r rab= m∫( x′′ x′ + y′′′ y + z′′′z ) dta2 2 2b( x ) ( y ) ( z )m ⎡ ′ ′ ′= ⎢ + +⎤⎥⎢⎣2 2 2 ⎥⎦am 2b m 2 2= ⎡ ′() t ⎤ ⎡ ′( b) – ′( a)⎤2 ⎢=⎣ r ⎥⎦a2 ⎢⎣ r r ⎥⎦26. The force exerted by Matt is not the only forceacting on the object. There is also an equal butopposite force due to friction. The work done bythe sum of the (equal but opposite) forces is zerosince the sum of the forces is zero.27. f ( xyz , , ) = - gmzsatisfies∇ f( x, y, z) = 0, 0, – gm =F . Then, assumingthe path is piecewise smooth,Work ( 2, 2, 2)= ∫ [– ] x y zCF ⋅ d r = gmz( x1, y1, z1)= – gm( z2 – z1) = gmz ( 1 – z2).28. a. Place the earth at the origin.44GMm ≈ 7.92(10 )– GMmf () r = is a potential function ofrF(r). (See Example 1.)147.1(109)⎡–GMm ⎤Work = ∫ Fr ( ) ⋅ dr= C ⎢⎣r ⎥⎦r= 152.1(109)32≈ –1.77(10 ) joulesb. ZeroInstructor’s Resource Manual Section 14.3 869

29. a.2 2( – )M = y ; M2 2 y =x y2 2 2( x + y ) ( x + y )2 2( – )N = – x ; N2 2 x =x y2 2 2( x + y ) ( x + y )Problem Set 14.41.b.y (sin t)M = = = sin t2 2 2 2( x + y ) (cos t+sin t)x (– cos t)N = – = = −cost2 2 2 2( x + y ) (cos t+sin t)∫CF r∫⋅ d = Mdx+NdyC2π= ∫ [(sin t )(– sin t ) + (– cos t )(cos t )] dt02π= – ∫ 1 dt = –2π≠002. ∫2∫∫2 xydx+ y dy= (0−2 x)dACS2 2y642–2x dx dy = – ≈ –4.26670 y15= ∫∫30. f is not continuously differentiable on C since f isundefined at two points of C (where x is 0).31. Assume the basic hypotheses of Theorem C aresatisfied and assume ∫ Fr ( ) i d r = 0 for everyCclosed path in D . Choose any two distinct pointsA and B in D and let C1, C2be arbitrarypositively oriented paths from A to B in D. Wemust show thatFr () idr=Fr () i dr∫C∫C1 2Let − C2be the curve C 2 with oppositeorientation; then −C2is a positively oriented pathfrom B to A in D. Thus the curve C = C1∪−C2is a closed path (in D) between A and B and so,by our assumption,0 = Fr ( ) idr= Fr ( ) idr+ Fr ( ) idr=∫∫ ∫ ∫C C1 −C2Fr () idr−Fr () idr∫C1 C2Thus we have∫∫Fr () idr=Fr () i drwhichC1 C2proves independence of path.14.4 Concepts Review∂ N ∂ M1. –∂ x ∂ y2. –2; –23. source; sink4. rotate; irrotational3.−( 1/2 −–1/2)1 ∫ ydx+ xdy=x y dAC ∫∫S22⎛1⎞ 2 x /2 –1/ 2 –1/ 2= ⎜ ⎟ ( x – y ) dydx2∫∫ ⎝ ⎠ 0 03 2= – ≈ –0.84855 ∫2 2(2 x + y ) dx+ ( x + 2 y) dy = (2 x– 2 y)dACS32 x /4 2⎡4 6(2 – 2 ) x –x ⎤= ∫∫ x ydydx= dx0 0 ∫ ⎢ ⎥0⎢⎣2 16⎥⎦16 8 72= – = ≈ 2.05715 7 35∫∫870 Section 14.4 Instructor’s Resource Manual

4.8.5. ∫∫∫xydx+ ( x+ y) dy=(1– x)dACS1 –2y+21= ∫∫ (1 – xdxdy ) =0 03⎛1⎞A( S) = ⎜ ⎟ xdy–ydx2 ∫ ⎝ ⎠ C⎛1⎞ 2⎡⎛3⎞ 3 ⎛1⎞ 3⎤⎛1⎞0 2 2= ⎜ ⎟ x – x dx [2 x – x ] dx2∫0 ⎢⎜ ⎟ ⎜ ⎟ −⎜ ⎟2 2⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝2∫⎣⎦ ⎠ 22=39. ∫C= 02 2( x + 4 xy) dx + (2x + 3 y) dy = (4 x –4 x)dA∫∫S6.7. ∫3x2( e + 2 y) dx + ( x + sin y) dy = (2 x –2) dASC4 6 4∫∫ ∫1 2 1∫∫(2 x– 2) dxdy = 24dy= 24(3) = 72∫∫∫∫∫∫a. div F dA = ( M x + N y)dASS= (0 + 0) dA = 0S∫∫b. (curl F) ⋅ k dA = ( Nx– M y)dASS∫∫1 1∫∫ ∫ ∫= (2 x – 2 ydA ) = (2 x– 2 ydxdy )S0 0∫1= (1 – 2 ydy ) = 00∫∫10. a. (0 + 0) dA = 0Sb.1 1∫∫ ∫ ∫∫∫( b – a) dA ( – ) –S = b a dx dy = b a0 011. a. (0 + 0) dA=0Sb.2 2∫∫ (3 x – 3 y ) dA= 0, since for theSintegrand, f(y, x) = –f(x, y).⎛1⎞A( S) = ⎜ ⎟ xdy–ydx2 ∫ ⎝ ⎠ C⎛1⎞ 2 2 2 ⎛1⎞0= ⎜ ⎟ [4 x – 2 x ] dx [4 x– 4 x]dx2∫+0⎜ ⎟⎝ ⎠ ⎝2∫⎠ 28=3Instructor’s Resource Manual Section 14.4 871

12.15.13.14.∫∫∫∫∫∫a. div F dA = ( M x + N y)dASS= (1+ 1) dA = 2[ A( S)] = 2πS∫∫∫∫b. (curl F) ⋅ k dA = ( Nx– M y)dASS= (0 – 0) dA = 0S∫∫∫∫ ∫ ∫(curl F) ⋅ kdA = F⋅Tds – F⋅TdsS C1 C2= 30 – (–20) = 501 1F⋅ nds = (2x + 2 x) dA = 4x dx dy = 2CS0 0 ∫ ∫∫ ∫ ∫ ∫∫∫W = F⋅ T ( x – y)Cds =SN M dA1 1= ∫∫ (–2 y – 2 y) dA = –4ydxdyS ∫0∫0= 1∫ –4 –20 ydy= ∫∫∫16. F⋅ Tds = (2 y – 2 y) dA = 0CS17. F is a constant, so Nx= My= 0.F⋅ Tds = ( N – M ) dA = 0 ∫ ∫C∫∫S18. Mdx+ Ndy= ( Nx– My) dA=0S∫∫xyTherefore, ∫ F ⋅ dr is independent of path sinceC∫ ∫ ∫C1 C2CF⋅ dr– F⋅ dr = F⋅ dr= 0(Where C is the loop C 1 followed by – C 2.)∫Therefore, F⋅ dr = F⋅dr , so F isC1 C2conservative.∫19. a. Each equals2 2 2 2 –2( x – y )( x + y ) .b.∫2 2 –1 2 2 –1 2 2 2y( x + y ) dx– x( x + y ) dy =π (–sin t –cos t)dtC ∫02π= ∫ 0 –1dtc. M and N are discontinuous at (0, 0).20. a. Parameterization of the ellipse: x = 3 cos t, y = 2 sin t, t in [0,2 π ].∫2π ⎡02sint3cost⎤⎢(–3sin t) – (2cos t) –22 2 2 2 ⎥dt= π⎣9cos t+ 4sin t 9cos t+4sin t ⎦872 Section 14.4 Instructor’s Resource Manual

.1 2 –1 –1 2 –1 –1 2 –1 1 2 –1–(1 + y ) dy+ ( x + 1) dx+ (1 + y ) dy+ –( x + 1) dx = –2π–1 1 1 –1∫ ∫ ∫ ∫c. Green’s Theorem applies here. The integral is 0 since Nx– M y.21. Use Green’s Theorem with M(x, y) = –y and N(x, y) = 0.(– ydx ) = [0 – (–1)] dA=AS ( ) ∫C∫∫SNow use Green’s Theorem with M(x, y) = 0 and N(x, y) = x.x dy = (1 – 0) dA = A( S) ∫C∫∫S22.⎛ 1⎞ 2 ⎛1⎞2– (0 )C⎜ ⎟y dx = + y dA= Mx⋅ x dy2 S C⎜ ⎟⎝ ⎠ ⎝2⎠∫ ∫∫ ∫ = ∫∫ ( –0) = yS x dA M23.⎛1⎞A( S) = ⎜ ⎟ xdy–ydx2 ∫ ⎝ ⎠ C⎛1⎞ 2π3 2 3 2= ⎜ ⎟ [( acos t)(3asin t)(cos t) – ( asin t)(3acos t)(– sin t)]dt2∫ ⎝ ⎠ 0⎛3⎞ 2= ⎜ ⎟8 a π⎝ ⎠24. W = ∫ F⋅ T (curl )Cds = ∫∫ F ⋅kSdA⎛ 2 23aπ⎞15aπ= ∫∫ ( Nx– M y) dA = (–3–2) dAS ∫∫= –5[ AS ( )] = –5 = – ,S⎜ 8 ⎟⎝ ⎠15the result of Problem 23.using25. a.b.2 2x + y1 1⋅ = = =2 2 3/2 2 2 1/2( x + y ) ( x + y ) aFnTherefore,∫C1 1Fn ⋅ ds = 1 ds (2 a) 2 .a∫= C aπ = π2 2 2 2( x + y )(1)–( x)(2 x) ( x + y )(1)–( y)(2 y)div F = + = 02 2 2 2 2 2( x + y ) ( x + y )c.Mx=2 2( x + y )is not defined at (0, 0) which is inside C. ∫ ∫∫ ∫∫d. If origin is outside C, then Fn ⋅ ds = div FdA = 0 dA = 0.C S SIf origin is inside C, let C′ be a circle (centered at the origin) inside C and oriented clockwise. Let S be theregion between C and C′ . Then 0= ∫∫ div F dA (by “origin outide C” case)S= ∫ Fn ⋅ ds – ⋅ dsC ∫ Fn (by Green’s Theorem) = ds – 2C′∫ Fn ⋅ π (by part a), so 2 .C∫ ⋅ ds = πCFn26. a. Equation of C:x, y = x0, y0 + t x1 – x0, y1 – y0,t in [0, 1].Thus;∫ ∫= 1[ 0 + ( 01 – 0 )]( 1 – 0 ) ,x dy x t x x y y dtCwhich equals the desired result.b. Area( P)= ∫ xdy whereCC = C1∪C2 ∪…∪ Cnand C i is the ithedge. (by Problem 21)∫ ∫ ∫x dy x dy x dy= + +…+C C Ci=11 2 n( xi – xi–1 )( yi – yi–1)= ∑ (by part a)2c. Immediate result of part b if each x i andeach y i is an integer.Instructor’s Resource Manual Section 14.4 873

d. Formula gives 40 which is correct for thepolygon in the figure below.b.1⎡ ⎛ x⎞ ⎛ y⎞⎤– sec ⎜ ⎟+ sec ⎜ ⎟⎥dy dx = 0⎣ ⎝ ⎠ ⎝ ⎠⎦3 3 2 29∫–3 ∫ –3⎢3 329. a. div F = –2 sin x sin ydiv F < 0 in quadrants I and IIIdiv F > 0 in quadrants II and IV27. a. div F = 4sin(x)sin(y)x2 2+y28. a.b. 4(36) = 1441 2⎛ ⎞ 1 2⎛ ⎞div F = – sec ⎜ ⎟+sec ⎜ ⎟9 ⎝3⎠ 9 ⎝ 3⎠xy30.b. Flux across boundary of S is 0.2Flux across boundary T is –2(1– cos3) .1 2 2–( x + y ) / 4 2 2div F = e ( x + y – 4)42 2so div F < 0 when x + y < 4 and div F > 02 2when x + y > 4.⎛ ⎛ x ⎞⎞ ⎛ ⎛ y ⎞⎞ln ⎜cos⎜ ⎟⎟– ln ⎜cos⎜ ⎟⎟⎝ ⎝3⎠⎠ ⎝ ⎝ 3⎠⎠⎛ 2 2( x + y ) ⎞exp –⎜ 4 ⎟⎝ ⎠874 Section 14.4 Instructor’s Resource Manual

14.5 Concepts Review1. surface integral2.3.n∑ g( xi, yi, zi)ΔSii=14. 2; 18π2 2fx+ fy+ 1Problem Set 14.51.2.34.∫∫2 2 1/2[ x + y + ( x+ y+ 1)](1 + 1 + 1) dAR1 1 2 20 0∫∫= 3( x + y + x+ y+1) dxdy =R1/28 33⎛1 1 ⎞ 1 1⎛6 ⎞x⎜ + + 1⎟ dA= xdxdy⎝4 4 ⎠ 0 0⎜ 2 ⎟⎝ ⎠∫∫ ∫ ∫=∫∫642 –1/2 2( + ) [– (4– ) ] + 0 + 1R x y x x dA3 1 2( x+y)dy dx0 0 21/2(4 – x )= ∫ ∫= ∫032x+ 1(4 – x )21/2dx⎡ 21/2 –1⎛x ⎞⎤= ⎢–2(4 – x ) + sin ⎜ ⎟2⎥⎣⎝ ⎠⎦π+ 6 π= = 2+ ≈ 3.04723 32π1 2 2 1/2 ⎛ π ⎞∫ r (4r 1) rdrdθ (25 5 1)0 ∫ + = +0⎜ ⎟⎝60⎠≈ 2.9794037.2 1/2 3 2 2 1/2∫∫ y(4 y + 1) dA = (4 y 1) ydydxR ∫ +0∫033/2 3/2(17 – 1) 17 – 1= ∫dx = ≈ 17.27320 12 4∫∫1/2( x + y )(0 + 0 + 1) dAR1 1∫∫ ( x+y) dxdy = 10 0Bottom (z = 0):Top (z = 1): Same integral1 11Left side (y = 0): ∫∫ ( x+ 0) dxdz =0 021 13Right side (y = 1): ∫∫ ( x+ 1) dxdz =0 021 11Back (x = 0): ∫∫ (0 + ydydz ) =0 021 13Front (x = 1): ∫∫ (1 + ydydz ) =0 02Therefore, the integral equals1 3 1 31+ 1+ + + + = 6.2 2 2 28. Bottom (z = 0): The integrand is 0 so the integralis 0.4 8–2x 128Left face (y = 0): ∫∫ z 1 dzdx =0 03Right face (z = 8 – 2x – 4y):2 4–2y1/2∫∫ (8–2 x –4 y)(4+ 16+1) dxdy0 0⎛32⎞= ⎜ ⎟ 21⎝ 3 ⎠2 8–4y 64Back face (x = 0): ∫∫ z 1 dzdy =0 03⎛32⎞Therefore, integral = 64 + ⎜ ⎟ 21 ≈112.88.⎝ 3 ⎠9.5.π ⎛5⎞(4 + 1) rdrdθ= ⎜ ⎟π≈1.9635⎝8⎠sinθ2r0 0∫∫6.∫∫∫∫Fn ⋅ ds = (– Mfx– Nf y + P)dAGR1 1– y= ∫∫ (8 + 4 + 0)0 0y x dxdy= 1 2∫ [8(1 – ) 2(1 – ) ]0 +y y y dy= 1 2∫ (–6 4 2) 20 y + y+ dy =Instructor’s Resource Manual Section 14.5 875

10.11.12.13.14.3 (6–2 x)/3 2 ⎛ 1 ⎞( x – 9) – dydx 11.250 0⎜ ⎟ =⎝ 2 ⎠∫∫5 1 2 –1/2[– xy(1– y ) 2] dy dx 200 –1 + =∫∫(In the inside integral, note that the first term isodd in y.)∫∫[– Mfx– Nfy+ P]dAR2 2 –1/2 2 2 –1/2= ∫∫ [–2 x( x + y ) – 5 y( x + y ) + 3] dAR2π1 –1= ∫ [(–2 cos – 5 sin ) 3]0 ∫ 0 r θ r θ r + rdrdθ2π1= ∫ (–2cos θ – 5sinθ + 3)dθrdr0 ∫ 0⎛1⎞= (6 π ) ⎜ ⎟= 3π≈9.4248⎝2⎠∫∫ ∫∫m 2= kx ds = kx 2 3 dAGRa a– x 2 ⎛ 3k⎞ 4= 3k∫∫x dydx =⎜ a0 0⎜ 12 ⎟⎝ ⎠∫∫ ∫∫2 2 1/2m = kxyds = kxy ( x + y + 1) dAGR1 1 2 2 1/2( 1)0 0∫∫= kxy x + y + dx dy⎛ k ⎞= ⎜ ⎟(9 3 – 8 2 + 1) ≈0.3516k⎝15⎠M = zds = ( a – x – y ) 3 dAxy∫∫ ∫∫Sa a y3 ( a – x – y ) dxdy–0 0= ∫∫Ra ⎡2( a– y)⎤= 3 ∫ ⎢aa ( – y)– – ya ( – y)⎥ dy0⎢⎣2⎥⎦2 2 333 a ⎛ a –y ⎞= ∫ay + dy =a0 ⎜ 2 2 ⎟⎝⎠6M xy aaz = = ; then x ym 33= = (by symmetry).16. By using the points ( a,0,0), (0, b,0), (0,0, c)wecan conclude that the triangular surface is ax y zportion of the plane + + = 1, ora b c⎛z f( x, y) c 1 x y ⎞= = ⎜ − − ⎟, over the region⎝ a b⎠ x( 1 )Rxy= {( x, y) | 0 ≤ x≤ a, 0 ≤ y ≤ b − } . Sinceawe are assuming a homogeneous surface, we willassume δ ( xyz , , ) = 1.2 2x yS Rc2c2c2∫∫+ + dAa2b2c2Rb2c2+ a2c2+ a2b21ab ∫∫Ra. m = 1dS = f + f + 1dA==∫∫= (∫∫ab b22c + a22c + a22b)2 abdA2 2 2 2 2 2Let w =bc + ac + abab; thenabwm =215.Let δ = 1.∫∫ ∫∫1/2m = 1 ds = (1+ 1+1) dASRa a–y a∫∫ ∫= 3 dx dy = 3 ( a – y)dy0 0 02 3a=2b. M = zdS = zwdAxySa vx⎛ c c ⎞= w∫∫⎜c − x − y ⎟ dy dx⎝ a b ⎠00a⎛2cb cb cbx ⎞ wabc= w∫− x+ dx =⎜ 2 a 22a⎟⎝⎠60∫∫∫∫Rab−xvx =a876 Section 14.5 Instructor’s Resource Manual

M xy 2( wabc)cc. Thus z = .m= 6( abw) = 3In a like⎛manner, using y g( x, z) b 1 x z ⎞= = ⎜ − − ⎟⎝ a c⎠ overthe regionxRxz= {( x, z) | 0 ≤ x≤ a, 0 ≤ z ≤c( 1−)}anda⎛x h( y, z) a 1 y z ⎞= = ⎜ − − ⎟ over the region⎝ b c⎠ yRyz= {( y, z) | 0 ≤ y ≤b, 0 ≤ z ≤ c 1−},awe can show x = and3⎛a b c⎞of mass is ⎜ , , ⎟⎝3 3 3⎠ .17. ( uv , ) = u + 3v + ( 4−u 2 −v2)r i j k( )by = so the center3b19. ( )r uv , = 2cosvi+ 3sinvj+uk20. ( )r uv , = ui+ 3sinvj+5cosvk18. ( uv , ) = 2u + 3v + ( u 2 + v2)r i j k21.r (,) uv = sinvi+ cosvj+0 k,ur (,) uv = ucosvi− usinvj+1kvr × r = cos vi−sinvj−ukuuvv2 2 2 2r × r = cos v+ sin v+ u = 1+uUsing integration formula 44 in the back of thebook we get6 π62 2A = u + 1dvdu = π u + 1 du =π∫∫ ∫−60 −66⎡u 2 1 2 ⎤⎢ u + 1+ ln u + u + 12 2⎥ =⎣⎦−6⎡37 + 6 ⎤π ⎢6 37 + ln ⎥ ≈122.49⎢⎣37 − 6 ⎥⎦Instructor’s Resource Manual Section 14.5 877

22.ru( u, v) = cosusin vi− sin usin vj+0 k,rv( uv , ) = sin ucosvi+ cosucos vj+cosvkru× rv=−sinusin vcosvi−cos usin vcosvj+ sin vcosvk[ i j k]= sin vcosv −sin u − cosu+ 12 2ru× rv= sin vcosv sin u+ cos u+12= 2 sin vcos v = sin 2v2Thus2π 2π 2π2A = sin 2 2 sin 22∫ ∫ vdudv= π ∫ v dv=0 0 0⎡ π2 ⎤⎢ ⎥ π 22π⎢4∫sin2vdv⎥ = 2 2π[ − cos2v]0 =⎢ 0⎣⎥⎦4 2π≈ 17.77Using integration formula 48 in the back of thebook we get4ππ ⎡ w 2 2 625 2 ⎤A= ⎢ ( 2w + 25)w + 25 − ln w+ w + 25 ⎥4 ⎣ 8 8⎦0π ⎡π2 2 625=2⎢ ( 32π + 25)16π + 25 − ln 4π+ 16π+ 254 ⎣ 2 8625 ⎤+ ln 5 ≈ 5585.428 ⎥⎦23.10.50-0.510.5-1-10-0.50-0.50.51 -1ru(,) uv = 2cos u vi+ 2sin u vj+5 k,2 2rv(,) uv =− u sinvi+ u cosvj+0k2 2 3ru× rv=−5u cosvi− 5u sinvj + 2uk2=− u v + v − u[ 5cos i 5sin j 2 k]24.r ( uv , ) = −sinucos vi−sinusin vj−sin uk,ur ( uv , ) =− cosusin vi+ cosucosvj+0kvr × r = sin ucosucosvi+sin ucosusinvjuv−sinucosuk[ i j k]= sin ucosu cos v + sin v −12 2ru× rv= sinucos u cos v+ sin v+12= 2 sinucos u = sin 2u2Thus (see problem 22)π 2 2ππ 22A = sin 2 2 sin 22∫ ∫ udvdu=π ∫ udu0 0 02π 2= π[ − cos 2u]= 2π≈ 4.443202 2ru× rv= u 25 + 4uThus2π2π2 2A= u 4u + 25dvdu∫ ∫0 02π2 2= 2πu 4u + 25 du∫04π2w 2 2 1= 2 π ∫ w + (5) ( dw)=w=2u420dw=2 duπ44π2 2 2∫ w w + (5) dw.0878 Section 14.5 Instructor’s Resource Manual

25. δ ( xyz , , ) = kz= 5 ku( k> 0) . Thus26. a.2π2π2 2m= ( 5ku ⎛) u 4u 25⎞∫ ∫ ⎜ + ⎟dvdu=⎝⎠0 02π3 2∫5k u 4u + 25 du =016π2+ 255k⎛t−25⎞8∫ ⎜ ⎟⎝ 4 ⎠25t= 4u2+ 25dt=8u dutdt=16π2+ 255k⎡2 5 32502⎤32⎢ t − t5 3⎥ ≈⎣⎦255k[ 139760 + 833 ] ≈ 21968 k322 2δ ( x, yz , ) = k x + y = kcosuThusπ 2 2πA = ( k cos u)( 2sinu cos u)dv du∫ ∫0 0π 2= 2 2πk sinu cos u dut=cosudt=−sinu du∫0202= −2 2πk t2 2πk= ≈ 2.962k3b. δ ( x, yz , ) = kz=kcosuThus the density function is the same as inpart a. and hence so is the mass: ≈ 2.962 k27. r =− 5sin usin vi+ 5cosusin vj+0k andurv= 5cosucosvi+ 5sinucosvj+−5sinvk .Thus,i j kru× rv= −5sin usin v 5cosusin v 05cosucos v 5sin ucosv −5sinv2 2= ( − 25cosusin v) i+ ( − 25sinusin v)j+2 2( −25sin usin vcosv−25cos usin vcos v)k2 2= ( − 25cosusin v) i+ ( − 25sinusin v)j++ ( −25sinvcos v)k= ( − 25sin v) cosusin vi+sin usinvj+ cos vk∫1dt[ ]Thus:ru×rv=− 25sin v2 2 2(cosusin v) + (sin usin v) + cos v= 25 sin v2 2 2 2[(cos u+ sin u)sin v+cos v= 25 sin v2 2sin v+ cos v = 25 sin v∫∫ ∫∫28. m = z ds = 3 dAGR2( = 3 AR ( ) = 3 π (3) = 27 π , ignoring the subtlety)2π3−ε2∫ rdrdθεε 0 0 ∫→ 0ε→0= lim 3 = lim 3(3 − ) π= 27 π29. a. 0 (By symmetry, sinceg(x, y, –z) = –g(x, y, z).)b. 0 (By symmetry, sinceg(x, y, –z) = –g(x, y, z).)c.∫∫ ∫∫2 2 2 2( x + y + z ) dS = a dSGG2 2 2 4= a Area( G) = a (4 π a ) = 4πad. Note:2 2 2( x + y + z ) dS =2x dS +2y dSe.∫∫ ∫∫ ∫∫G G G2 2∫∫ z dS 3 x dSG ∫∫ G= =(due to symmetry of the sphere with respectto the origin.)Therefore,∫∫2 1 2 2 2x dS ⎛ ⎞= ( x y z ) dSG⎜ ⎟ + +⎝3∫∫ ⎠ G⎛1⎞4 4πa= ⎜ ⎟4 π a = .⎝3⎠3∫∫G42 2 ⎛2⎞4 8πa( x + y ) dS = ⎜ ⎟4π a =⎝3⎠330. a. Let the diameter be along the z-axis.I = kx ( + y)dSz∫∫2 2G2 2 2= =G G G1. ∫∫ x dS ∫∫ y dS ∫∫ z dS (bysymmetry of the sphere)2 2 2 22. ( x + y + z ) dS = a dS∫∫ ∫∫GG2 2 2 4= a (Area of sphere)= a (4 π a ) = 4πaThus,2 2 2 4 8πa kIz= ∫∫ k( x + y ) dS = k(4 π a ) = .G3 3(using 1 and 2)44Instructor’s Resource Manual Section 14.5 879

. Let the tangent line be parallel to the z-axis.Then42 8πa k 2 2I = Iz+ ma = + [ k(4 π a )] a3420πa k= .331. a. Place center of sphere at the origin.F = ka ( – zdS ) = ka 1 dS–k zdS∫∫ ∫∫ ∫∫G G G2 3= ka(4 π a ) – 0 = 4πa kb. Place hemisphere above xy-plane with centerat origin and circular base in xy-plane.F = Force on hemisphere + Force on circularbase= 2ka ( – zdS ) + ka( π a )∫∫G∫∫ ∫∫= ka 1 dS – k z dS +π a kGG22 a3∫∫ R 2 2 2= ka(2 π a ) – k z dA +πa ka – x – yaak k z dAz3= 3 π – ∫∫ R3 2 3= 3 πak– ka( π a) = 2πakc. Place the cylinder above xy-plane withcircular base in xy-plane with the center atthe origin.F = Force on top + Force on cylindrical side+ Force on base= 20 + kh ( − zdS ) + kh( π a)∫∫G∫∫ ∫∫= kh 1 dS – k z dS +π a hkGGa2= kh(2 π ah) – 4k∫∫z + 0 + 1dA +πa hkR 2 2a – y(where R is a region in the yz-plane:0 ≤ y = a,0≤ z ≤ h)2 2 a h az= 2 π +π –4 ∫∫ 0 0 2 2ah k a hk k dz dya – y2 2 2= 2 π ah k +πa hk – πkah2 2=π ah k +π a hk =π ahk( h + a)23232. x = y = 0Now let G′ be the 1st octant part of G.⎛a⎞M xy = ∫∫ kdS= 4 kzdS 4k z dAG ∫∫ =G′ ∫∫ R′⎜ ⎟⎝ z ⎠(See Problem 19b.)= 4ak [Area ( R′ )]2 2 2 2( a – h1 ) ( a – h2)= 4 akπ⎢ ⎡ –⎤⎥⎢⎣4 4 ⎥⎦2 2( 2 – 1 )= akπh h∫∫mG ( ) = kdS=k[Area( G)]G= k[2 π a( h2 – h1)] = 2 π ak( h2 – h1)2 2π ak( hTherefore,2 – h1 ) h1+hz = =2 .2 πak( h2 – h1) 214.6 Concepts Review1. boundary; ∂ S2. F⋅n3. div F4. flux; the shapeProblem Set 14.6∫∫∫1. (0+ 0+ 0) dV = 0S∫∫∫2. (1+ 2 + 3) dV = 6 V ( S) = 6S∫∫∫∫∫3. Fn ⋅ dS = ( M x + N y + Pz)dV∂ SS1 1 1(0 1 0) dx dy dz 8−1 −1 −1 ∫ ∫ ∫= + + =∫∫∫∫∫4. Fn ⋅ dS = ( M x + N y + Pz)dV∂ SS= 2 2 23 ∫∫∫ ( x + y + z ) dx dy dzSConverting to spherical coordinates we have880 Section 14.6 Instructor's Resource Manual

∫∫∫2 2 23 ( x + y + z ) dx dy dz===S∫∫∫2 23 ( sin )Sπ 2πa43 ( sin )∫∫∫ρ ρ φ dρdθ dφρ φ dρdθ dφ0 0 05 π 2π3asin5∫∫ φdθdφ0 05 π56πa12πasin d5∫ φ φ50= =∫∫∫∫∫5. Fn ⋅ dS = ( M x + N y + Pz)dV∂ SS2 2 23abc=4c b a= ∫∫∫ 0 0 0 (2 xyz + 2 xyz + 2 xyz ) dx dy dzc b 2 2233a yzdydzc abz0 0 0 2∫∫ ∫= =dz6.∫∫∫ ⎛4⎞ 3(3 – 2 + 4) dV = 5 V ( S) = 5 π(3)S⎢⎜⎟3⎥ = 180 π= 565.49⎡⎣⎝⎠⎤⎦∫∫∫7. 2 ( x + y+z ) dVS2π2 4– r264π= 2 ∫ ( r cosθ + r sin θ + z)r dz dr dθ0 ∫0∫= ≈ 67.02038.∫∫∫ ∫∫∫( M x + N y + Pz) dV = (2x + 1+2 z)dVSS2π2 2 2[(2 cos )(2 – cos ) (2 – cos ) ]0 02π⎛2 8cosθ⎞12 – 4cos θ – dθ200⎜⎟3∫ ∫2π2 2– r cosθ(2rcosθ1 2 )0 0 0∫ ∫ ∫= r θ + r r θ + r r θ drdθ∫= = π⎝⎠= + +z rdzdrdθ2π2 3 2 2(6 – cos – cos )0 0= ∫ ∫r r θ r θ drdθ∫∫∫9. (1 + 1+ 0) dV = 2(volume of cylinder)S2= 2 π (1) (2) = 4π≈12.566410.4 4– x 4– x–y(2 + 2 + 2 ) = (2 + 2 + 2 )S0 0 0∫∫∫ x y z dV ∫ ∫ ∫ x y z dzdydx = 64∫∫∫ ∫∫∫11. ( M x + Ny + Pz) dV = (2+ 3+4) dVSS= 9(Volume of spherical shell)π= = π≈⎝ ⎠⎛4⎞ 3 39 ⎜ ⎟(5 – 3 ) 1176 3694.51312.2π2 2(0 + 0 + 2 zdV ) = 2S0 1 012 37.6991∫∫∫ ∫ ∫ ∫= π≈zrdzdrdθInstructor’s Resource Manual Section 14.6 881

∫∫13. FindS = ( M x + N y + Pz)dS =∂SS(0 + 2y + 0) dV = 2 y dx dy dz14.∫∫∫S∫∫∫∫∫∫Using the change of variable (from ( x, yz , ) to(, r y, θ )) defined byx = rcos θ, y = y, z = rsinθyields theJacobian∂x ∂y ∂zS∂r ∂r ∂rcosθ0 sinθ∂x ∂y ∂zJry ( , , θ ) = = 0 1 0 =∂y ∂y ∂y−rsinθ0 rcosθ∂x ∂y ∂z∂θ ∂θ ∂θ2 2rcosθ + rsinθ = r . Further, the region S is2now defined by r ≤1, 0 ≤ y ≤ 10 . Hence, bythe change of variable formula in Section 13.9,2π10 12 ydxdydz= 2 yrdrdydθ=∫∫∫S0 0 02π10 2π∫∫∫ydydθ = 50dθ = 100π≈314.160 0 0∫∫∫ ∫ ∫dS = ( M + N + P ) dV =Finx y z∂SS2 2 2∫∫∫3 ( x + y + z ) dxdydzS∫∫∫Use the change of variable (basically sphericalcoordinates with the role of z and y interchangedand maintaining a right handed system):x = ρsinφsin θ, y = ρcos φ, z = ρsinφcosθThen the region S becomes2 2 2 2ρ ≤ 1 ( x + y + z ≤1)πφ2y2 1φ2x2z2 2yπ],40 ≤ ≤ ( ≥0)sin ≤ ( + ≤ ) so thatρ∈[0,1], φ∈[0, θ ∈ [0,2 π]. TheJacobian of the transformation is15.16.∂x ∂y ∂z∂ρ ∂ρ ∂ρ∂x ∂y ∂zJ =∂θ ∂θ ∂θ∂x ∂y ∂z∂φ ∂φ ∂φsinφsinθ cosφ sinφcosθ= ρsinφcosθ 0 −ρsinφsinθρcosφsinθ −ρsinφ ρcosφcosθ=−ρ 2sin3φsin 2θ −ρ 2sinφcos 2φcos2θ −ρ2cos2φsinφsin 2θ − ρ2sin3φcos2θ =− ρ2 ⎡sinφ sin2φ⎛sin2θ cos2θ⎞cos2⎜⎟ φ⎝⎠⎛sin2θcos2 ⎤θ ⎞⎢⎜ + ⎟⎝⎠⎥⎣⎦=− ρ2 sinφ.Thus,2 2 23 ∫∫∫ ( x + y + z ) dxdydzSπ4 2π1= 3 ρ2( ρ2sin φ)dρdθ dφ∫ ∫ ∫0 0 0ππ4 2π43 6π= sinsin5∫ ∫=5∫0 0 06π⎛2−2 ⎞= ≈1.1045 ⎜ 2 ⎟⎝ ⎠φ dθ dφ φdφ⎛1⎞⎜ ⎟ (1+ 1+ 1) dV = V ( S)3 S⎝ ⎠ ∫∫∫1V( S)= dS3∫∫ Fn ⋅ for F = x, yz ,∂ S12πa h= 33∫∫∫ dV = rdzdrdθS ∫0 ∫0∫02π2π2 2= ∫ rh dr dθdθ20 ∫ =0 ∫ = π0 2 22=π a ha a h a h882 Section 14.6 Instructor’s Resource Manual

17. Note:1. ∫∫ ( ax + by + cz)dS = d dS = dDR∫∫ (R is theRslanted face.)a, b,c2. n =(for slanted face)2 2 2 1/2( a + b + c )3. F⋅ n = 0 on each coordinate-plane face.⎛1⎞Volume = ⎜ ⎟ ⋅ dS where F = x, yz , .3 ∫∫ ⎝ ⎠ SFn ( )⎛1⎞= ⎜ ⎟ ⋅ dS3∫∫ Fn (by Note 3)⎝ ⎠ R⎛1 ⎞ ( ax + by + cz)= ⎜ ⎟dS3∫∫⎝ ⎠ R 2 2 2a + b + cdD=2 2 23 a + b + c∫∫∫18. div F dV = 0dV= 0 (“Nice” if there isS∫∫∫San outer normal vector at each point of ∂ S.)19. a. div F = 2 + 3 + 2z = 5 + 2z∫∫ Fn ⋅ dS = (5 + 2 z)dV∂ S ∫∫∫= 5dV+ 2 z dVSSS= S +⎛4π⎞20π20π= 5⎜⎟+2 z (Volume of S)= + 2(0)(Volume) =⎝ 3 ⎠3 3b.2 2 2 3/2Fn ⋅ = ( x + y + z ) x, y,z ⋅∫∫∫∫∫Fn ⋅ dS = 1dV= 4 π (1) = 4π∂ SSc. div F = 2x + 2y + 2z∫∫ Fn ⋅ dS = 2( x + y + z)dV∂ S ∫∫∫S2∫∫∫ ∫∫∫ 5 (Volume of ) 2Mxyx, y,z2 2 2x + y + z= 2∫∫∫ x dV (Since x = z = 0 as in a.)S⎛4π⎞ 16π= 2Myz= 2( x)(Volume of S) = 2(2) ⎜ ⎟=⎝ 3 ⎠ 3d. Fn ⋅ = 0 on each face except the face R in the plane x = 1.∫∫∂ S∫∫R∫∫ ∫∫2 2 2 2Fn ⋅ dS = Fn ⋅ dS = 1, 0, 0 ⋅ 1, 0, 0 dS = 1dSR= ( x + y + z ) = 1 on ∂ S.R2= (1) = 1e. div F = 1 + 1 + 1 = 3⎛1⎡1⎤ ⎞∫∫ Fn ⋅ dS = 3dV = 3(Volume of S)∂ S ∫∫∫= 3 (4)(3) (6) 36S⎜=3⎢ ⎟2⎥⎝ ⎣ ⎦ ⎠Instructor’s Resource Manual Section 14.6 883

f.2 2 2 2 2 2div F = 3x + 3y + 3z = 3( x + y + z ) = 3 on ∂ S.2 2 2 ⎛38π⎞ 12π∫∫ Fn ⋅ dS = 3 ( x + y + z ) dV∂ S ∫∫∫= 3S⎜ ⎟=⎝ 215 ⎠ 5(That answer can be obtained by making use of symmetry and a change to spherical coordinates. Or you couldgo to the solution for Problem 22, Section 13.9, and realize that the value of the integral in this problem is 3 2 .g.2 2Fn ⋅ = [ln( x + y )] x, y, 0 ⋅ 0, 0, 1 = 0 on top and bottom.x, y, 02 2F⋅ n = (ln 4) x, y, 0 ⋅ = (ln 4) x + y = (ln 4) 4 = 2ln 4 = 4ln 2 on the side.2 2x + y∫∫∫∫F⋅ n dS = 4ln 2 dS = (4ln 2)[2 π(2)(2)]= 32πln 2∂ SR20. a. div F = 0 (See Problem 21, Section 14.1.)Therefore, ∫∫ F⋅ ndS= div dV = 0.∂ S ∫∫∫ FSb. ∫∫ F⋅ n dS = 4π(by Gauss’s law with –cM = 1 as in Example 5).∂ Sc.r r 1 1Fn ⋅ = ⋅ = = on ∂ S.r r r a⎛1⎞Thus, ∫∫ F⋅ n dS = 1dS∂S⎜ ⎟⎝a∫∫⎠ ∂S⎛1 ⎞⎛= 1 ⎞⎜ ⎟(Surfacearea of sphere)2= ⎜ ⎟ (4 π a ) = 4 πa.⎝a⎠⎝2⎠rrd. Fn ⋅ = f ( r ) r⋅ = r f ( r ) = af( a) on ∂ S.∫∫∫∫2F⋅ ndS = af a dS = af a πa∂S∂S( ) 1 [ ( )](4 ) = 4 π a f( a)ae. The sphere is above the xy-plane, is tangent to the xy-plane at the origin, and has radius .2nn( f )div F = r div r+ grad ⋅r (See Problem 20c, Section 14.1.)nn–1= + + + n f ⋅∫∫r(1 1 1)∫∫∫rrF⋅ ndS = (3 + n)r dV∂ SSrnn n n= 3 r + n r = (3 + n)r32 π π/2 a cos n 2(3 n) φ 2πaρ ( ρ sin φ ) d ρ d φ d θ =0 0 0n + 4= + ∫ ∫ ∫n+3∫∫ ∫∫ ∫∫∫21. D f dS f dS div( f ) dV∂S n = ∇ ⋅ n = ∇∂S S2∫∫∫ f dV (See next problem.)S= ∇884 Section 14.6 Instructor’s Resource Manual

∫∫∫∫22. f( ∇f ⋅ n) dS = ( f∇f)⋅n dS = div( f∇f)dV∂S∂SS= ∫∫∫ div( ∇ f) + ( ∇f) ⋅( ∇f)dV (See Problem 20c, Section 14.1.)S2= ⎡( fxx fyy fzz) f ⎤∫∫∫ + + + ∇ dVS ⎢⎣⎥⎦2 2 2= [( ∇ f) + ∇ f ] dV = ∇f dVSS∫∫∫∫∫ ∫∫∫ (Since it is given that∫∫∫∫∫∫∫2f∇ = 0 on S.)23. fD g dS f ( g ) dS∂Sn = ∇ ⋅n = ( f ∇g) ⋅ n dS = div( f ∇g)dV (Gauss)∂S∂ SS= ∫∫∫ [ (div ) ( ) ( )]S f ∇ g + ∇f ⋅ ∇gdV2= ∫∫∫ ( f ∇ g +∇ f ⋅∇ g ) dV (See Problem 20c, Section 14.1.)S∫∫ n n fD g dS –24. ( fD g – gD f ) dS∂ S∫∫∫gD f dS∂Sn∂Sn2 2= ( f ∇ g +∇f ⋅∇g) dV – ( g∇ f +∇g ⋅∇f ) dVSS2 2= ∫∫ ∫∫∫∫∫ ∫∫∫ (by Green’s 1st identity)∫∫∫= ( f ∇ g – g∇f ) dVS14.7 Concepts Review1. (curl F)⋅n2. Möbius band3. ∫∫ (curl F) ⋅ndSs4. curl FProblem Set 14.7 ∫ ∫∫ ∫∫1. F⋅ TdS = ( N – M ) dA = 0dA= 02.∂ S Rx yR∫ ∫ ∫ ∫F⋅ Tds = 0dx + xy dx + yz dy + xz dz + yz dy∂ S C1 C2 C31 21= ∫ ( t + 7 t –4) dt = –06Instructor’s Resource Manual Section 14.7 885

3.∫∫ ∫ = ∫ ( + ) + ( + ) += ∫ dt + ∫ + t t + t dt + ∫ dt + ∫ t dt (*)2 2(curl F)⋅ ndS= F⋅TdsS∂ S∂ S1 π1 π 21 [(1 sin )(– sin ) cos ] –1 sin0 0 0 0π∫ (– sin t cos t ) dt –20y z dx x z dy ydz= + =The result at (*) was obtained by integrating along S by doing so along C1, C2, C3,C 4 in that order.Along C 1 : x = 1, y = t, z = 0, dx = dz = 0, dy = dt, t in [0, 1]Along C 2 : x = cos t, y = 1, z = sin t, dx = –sin dt, dy = 0, dz = cos t dt, t in [0, π ]Along C 3 : x = –1, y = 1 – t, z = 0, dx = dz = 0, dy = dt, t in [0, 1]Along C 4 : x = –cos t, y = 0, z = sin t, dx = sin t dt, dy = 0, dz = cos t dt, t in [0, π ]4. ∂ S is the circlexy-plane).∫ ∫∂Sx2 2+ y = 1, z = 0 (in the2 3F⋅ T ds = xy dx + x dy = (cos xz ) dz∂S3 2π 3⎛ 3 ⎞= ∫ xdy= (cos t)(– cos tdt ) = – πS ∫0⎜ ⎟⎝ 4 ⎠≈ –2.35622 25. ∂ S is the circle x + y = 12, z = 2.Parameterization of circle:x = 12 sin t, y = 12 cos t,z = 2, t in [0,2 π ]∫ ∫F⋅ T ds = yz dx + 3 xz dy + z dz∂S∂S2π2 t 2∫ 0= (24sin – 72cos t ) dt = –48 π≈ –150.806. ∂ S is the circlex2 2+ y = 1, z = 027.8.⎛ 1 ⎞(curl F) ⋅ n = 3, 2, 1 ⋅ ⎜ ⎟1, 0, –1 = 2⎝ 2 ⎠ ∫∂ S∫∫F⋅ Tds = 2dS = 2 A( S)S= 2[sec(45 ° )](Area of a circle)=8π≈25.1327⎡⎛1 ⎞ ⎤(curl F) ⋅ n = –1,–1,–1⋅ ⎢⎜⎟ 0,1,–1⎥= 0,⎣⎝2 ⎠ ⎦so the integral is 0.9. (curl F) = -1 + 1, 0 -1,1-1 = 0,-1,0The unit normal vector that is needed to applyStokes’ Theorem points downward. It is–1, – 2, – 1n =.6∫∫S∫ ∫(curl F)⋅ NdS= F⋅Tds∂ S= ( z – y) dx + ( z + x) dy + (– x – y)dzSx = cos t ; y = sin t; z = 0; t in [0,2 π ]∫2π= [( −sin t)( − sin t) + (cos t)(cos t)]dt0= 2π≈6.2832 ∫C∫∫F⋅ Tds= (curl F)⋅ndSS⎛ 2 ⎞ ⎛ 2 ⎞= ∫∫ dS(1 4 1)S⎜ ⎟ = ∫∫ + +6 R⎜ ⎟⎝ ⎠ ⎝ 6 ⎠∫∫1/2dA= 2 dA = 2(Area of triangle in xy − plane)R= 2(1) = 2886 Section 14.7 Instructor’s Resource Manual

10.2 2 ⎡⎛1 ⎞ ⎤(curl F) ⋅n= 0, 0, –4 x – 4 y ⋅ ⎢⎜⎟ –1, 0, 1 ⎥⎣⎝2 ⎠ ⎦2 2= –2 2( x + y )2 2 2∫∫ – ( x + y ) dSS 21 1 2 2 8= –4 ∫∫ ( x + y ) dx dy = – 0 0311. (curl F) ⋅ n = 0, 0, 1 ⋅ x, y,z = z∫∫ ∫∫SzdS = 1dA=Area of R2R⎛1⎞ ππ ⎜ ⎟ = ≈0.7854⎝2⎠412. (curl F ) = –1− 1, –1–1, –1–1 = –2 1,1,1 ,1, 0, 1n = , so (curl F) ⋅ n = –2 2.2 ∫C∫∫S∫∫ dS ∫∫F⋅ Tds= (curl F)⋅ndS= –2 2 1 = –2 2 (sec45 ° ) dAS= 2– [ ( )] –42 2 AR = π13. Let H(x, y, z) = z – g(x, y) = 0.∇H– gx, gy,1Then n = =points upward.∇ H 2 21+ gx+ g yThus,(curl F) ⋅ ndS= (curl F) ⋅nsecγdA14.∫∫∫∫S S xy– gx, – gy,12 2= ∫∫ (curl F ) gx+ gy+ 1dASxy2 2gx+ gy+ 1(Theorem A, Section 14.5)= (curl ) ⋅ – g , – g , 1 dA∫∫ FSxycurl F = z , 0, – 2y2x ∫ F⋅ Tds= (curl ) ⋅ dSC ∫∫ F n (Stoke’sSTheorem)= ∫∫ (curl F ) ⋅ – gx, – gy, 1 dA (Problem 13)SxyRy∫∫= z 2 ,0,–2 y ⋅ – y,– x,1dAS xy(where z = xy )1 1 2 313= ∫∫ (– x y – 2 y) dxdy = –0 01215. curl F = 0- x,0-0, z-0 = - x,0,z ∫C∫∫F⋅ Tds= (curl F)⋅ndSS= ∫∫ (curl F) ⋅ – gx, – gy, 1 dAwhereSxyz = g( x, y) = xy . (Problem 13)∫∫2= – x, 0, z ⋅ – y , –2 xy,1dAS xy1 1 2 2∫∫= ( xy + 0 + xy ) dxdy0 01 2 2 1 1 2 1= ∫ ([ x y ] 0 )0x=dy = ∫ y dy =0 3 ∫C∫∫∫∫ F16. F⋅ Tds= (curl F)⋅ndSS= (curl ) ⋅ – g , – g , 1 dA∫∫Sxyx2y2 2= – x, 0, z ⋅ –2 xy , –2 x y,1dAS xy(where= ∫∫S xy2 2 )z = x y2 23x ydAπ /2 a 2 2 πa12 ( r cos θ ) ( r sin θ ) rdrdθ=0 08= ∫ ∫ ∫∂S ∫∫= ∫∫ F ⋅17. F⋅ Tds= (curl F)⋅ndSS(curl ) – g , – g , 1 dASx yxy⎡ 2 2 2 –1/2x, y,( a − x – y ) ⎤= 2, 2, 0 ⎢⎥∫∫ ⋅ S ⎢ xy2 2 2 –1/2 ⎥∫∫⎢⎣( a – x – y )2 2 2 –1/2= 2 ( x + y)( a – x – y ) dA= ∫∫S xy2 2 2 –1/22 y( a – x – y ) dAS xyπ /2 a sinθ2 2 –1/2θ0 02= ∫ ∫4 ( rsin )( a – r ) drdθ4a= joules36dA⎥⎦18. curl F = 0 by Problem 23, Section 14.1. Theresult then follows from Stokes’ Theorem sincethe left-hand side of the equation in the theoremis the work and the integrand of the right-handside equals 0.Instructor’s Resource Manual Section 14.7 887

19. a. Let C be any piecewise smooth simpleclosed oriented curve C that separates the“nice” surface into two “nice” surfaces, S 1and S 2.∫∫ (curl F)⋅ndS∂ S∫∫= (curl F ) ⋅ ndS+ (curl F ) ⋅ndS∫∫S1 S2∫ ∫(–C is C with= F⋅ Tds+ F⋅ Tds= 0C– Copposite orientation.)b. div(curl F) = 0 (See Problem 20, Section14.1.) Result follows. ∫S∫∫S= ∫∫ [ (curl ∇ ) + ( ∇ ×∇ )] ⋅ n20. ( f ∇g) ⋅ Tds = curl( f∇g)⋅ndSf g f g dSS= ∫∫ ( ∇ f ×∇g) ⋅n dS,since curl ∇ g = 0.S(See 20b, Section 14.1.)Sample Test Problems1.2. div F = 2yz − 6y + 2ycurl F = 4 yz,2 xy, − 2xzgrad(div F) = 0, 2z− 6 + 4 y, 2ydiv(curl F) = 0(See 20a, Section 14.1.)3. curl( f ∇ f) = ( f)(curl ∇ f) + ( ∇ f ×∇ f)= (f)(0) + 0 = 0214.8 Chapter ReviewConcepts Test4. a.b.2f ( x, y) = x y+ xy+ sin y+C– xf ( x, y, z)= xyz + e + e + Cy1. True: See Example 4, Section 14.12. False: It is a scalar field.3. False: grad(curl F) is not defined since curlF is not a scalar field.4. True: See Problem 20b, Section 14.1.5. True: See the three equivalent conditions inSection 14.3.6. True: See the three equivalent conditions inSection 14.3.7. False:2N = 0 ≠ z = Pz8. True: See discussion on text page 750.9. True: It is the case in which the surface is ina plane.10. False: See the Mobius band in Figure 6,Section 14.5.11. True: See discussion on text page 752.y5. a. Parameterization is x = sin t, y = –cos t, t in⎡ π⎤⎢0, .2 ⎥⎣ ⎦π /2 2 2 2 1/2 π∫ (1 – cos t)(sin t+ cos t)dt =04≈ 0.7854b∫π /2 20[ tcos t – sin tcost+sin tcos t]dt(3 π – 5)= ≈ 0.737566. M = 2y = N so the integral is independent of7.xythe path. Find any function f(x, y) such thatfx ( x, y)y2= and f y ( x, y) = 2 xy.2f ( x, y) = xy + C ( y)and2f ( x, y) = xy + C ( x),so let122f ( x, y) = xy .2 (1, 2)(0, 0)Then the given integral equals [ xy ] = 4.2 (3, 4)(1, 1)[ xy ] = 4712. True: div F = 0, so by Gauss’s DivergenceTheorem, the integral given equals∫∫∫ 0 dV where D is the solid sphereDfor which S = ∂ D.8.– y (1, 1, 4) –1(0, 0, 0)[ xyz + e + e ] = 2 + e + e ≈ 5.0862888 Section 14.8 Instructor’s Resource Manual

9. a.4 1 10 dx + (1 + y ) dy + x dx + y dy = 0 + – –3 2 31 1 2 0 0 20 0 1 1∫ ∫ ∫ ∫1=2b.A vector equation of C 3 is x,y = 2,1 + t −2, − 1 for t in [0, 1], so let x = 2 – 2t, y = 1 – t for t in [0, 1]be parametric equations of C 3.2 1 2 1 2 213 4∫ 0 dx + (4 ) [2(1– ) (–2) 5(1– ) (–1)] 0 – 30 ∫ + y dy + t t dt0 ∫+ = + =03 3c.x = cos ty =− sin tt in [0,2 π ]2π2 2 2π2+ + =0 ∫0∫[(cos t)(sin t)(– sin t) (cos t sin t)(cos t)] dt (1– sin t)cost dt2π⎡ 3sin t ⎤= ⎢sin t – ⎥ = 0⎢⎣3 ⎥⎦0∫∫10. div F dA = 2dA = 2 A( S) = 8S11. Let∫∫S2 2f ( x, y) = (1– x – y ) andThen Flux = ∫∫ Fn ⋅ dSG2 2∫∫ ∫∫ 2g( x, y) = –(1– x – y ), the upper and lower hemispheres.= [– Mf – Nf + P] dA+ [– Mg – Ng + P]dA = ∫∫ PdA (since f = – g and f = – g )Rx yRx y2 2= ∫∫ 6 dA = 6 (Area of R, the circle xR= 6π ≈ 18.8496+ y = 1, z = 0)RxxyyInstructor’s Resource Manual Section 14.8 889

12.∫∫G1 –2x+2 2 2xyz dS = xy( x + y)(sec)dA = 3 ( x y+xy ) dydx∫∫R2 3 1 4 3 2= – ( x –6x 9 x –4 x)dx3∫ +0–1, –1, 1 ⋅ 0, 0, 1cosν=3Therefore, secν = 3.∫∫0 02 3 1 3 3= – – + 3 – 2 = ≈ 0.34643 5 2 52 2 31 4 x (1– x) 8 x(1– x)= 3∫+0 2 3dx2 213. ∂ S is the circle x + y = 1, z = 1.A parameterization of the circle is x = cos t,y = sin t, z = 1, t in [0,2 π ].⎡ 3 y ⎛ xyz ⎞ ⎤∫F⋅ Tds = x y dx + e dy + z tan dz∂S ∫∂S⎢⎜ ⎟4⎥⎣⎝ ⎠ ⎦2π3 sint= ∫ [(cos t) (sin t)(– sin t) + ( e )(cos t)]dt = 00∫3= ( x y + e dy )∂ Sy∫∫∫14. div F dv = [(cos x) + (1– cos x) + (4)] dVS∫∫∫= 90π ≈ 282.7433S∫∫∫5dV 5 V ( S) 5 ⎛1⎞⎡⎛4⎞(3) 3 ⎤S⎜ ⎟⎢⎜ ⎟ ⎥= = = π⎝2⎠⎣⎝3⎠⎦15. curl F = 3−0,0−0, −1− 1 = 3, 0, − 2n =a, b,12 2a + b + 13 a–2 3 a–2 ∫ F⋅ Tds= (curl ) ⋅ dSC ∫∫ F n =S ∫∫dS =[ A( S)]S 2 2 2 2a + b + 1 a + b + 1=3 a –2(9 π)2 2a + b + 1(S is a circle of radius 3.)=9 π(3 a –2)2 2a + b + 1890 Section 14.8 Instructor’s Resource Manual

CHAPTER 1515.1 Concepts Review1.2.2r a1r a2 0;+ + = complex conjugate roots– x xCe 1 + C2e3. ( C1+C2x) ex4. C1cosx+C2sinxProblem Set 15.11. Roots are 2 and 3. General solution is2x3xy = C1e + C2e.2. Roots are –6 and 1. General solution is–6xxy = C1e + C2 e .3. Auxiliary equation: r2 + 6 r –7 = 0,(r + 7)(r – 1) = 0 has roots –7, 1.–7xxGeneral solution: y = C1e + C2e–7xxy′ = –7C1e + C2eIf x = 0, y = 0, y′ = 4, then 0 = C1+ C2and1 14 = –7 C1+ C2,so C 1 = – and C 2 = .2 2x –7xe – eTherefore, y = .24. Roots are –2 and 5. General solution is–2x5xy C1e C2 e .⎛12 ⎞ 5 x ⎛5⎞ –2xy = ⎜ ⎟ e – ⎜ ⎟ e .⎝ 7 ⎠ ⎝7⎠= + Particular solution is5. Repeated root 2. General solution is2xy = ( C1+C2x) e .6. Auxiliary equation:2 2r 10r 25 0, ( r 5) 0root − 5 .–5 x –5xGeneral solution: y = C1e + C2xeor–5xy = ( C1+C2x)e+ + = + = has one repeatedDifferentialEquations7. Roots are 2± 3. General solution is2x 3 x – 3xy = e ( C1e + C2e).8. Roots are –3± 11. General solution is( )–3x 11 x – 11x1 2 .y = e C e + C e9. Auxiliary equation: r 2 + 4= 0 has roots ±2i.General solution: y = C1cos 2x+C2sin 2xπIf x = 0 and y = 2, then 2 = C1;if x = and4y = 3, then 3 = C2.Therefore, y = 2cos2x+ 3sin2x.10. Roots are ±3i. General solution isy = ( C cos3x+ C sin3 x).Particular solution is1 2y = −sin 3x− 3cos3x.11. Roots are –1 ± i. General solution is– xy = e ( C cosx+C sin x).1 212. Auxiliary equation: r2 + r+ 1= 0 has roots–1 3 .2 ± 2 iGeneral solution:(–1/ 2 ) x 3 (–1/ 2) xy = C 31e cos⎛ ⎞ x C2e sin⎛ ⎞⎜ +x2 ⎟ ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠⎡– x /2 ⎛ 3 ⎞ ⎛ 3 ⎞ ⎤y = e ⎢C1cos⎜ x+C2sinx⎥2 ⎟ ⎜ 2 ⎟⎢⎣⎝ ⎠ ⎝ ⎠ ⎥⎦13. Roots are 0, 0, –4, 1.General solution is–4xxy = C1+ C2x+ C3e + C4 e .14. Roots are –1, 1, ±i. General solution is– x xy = C1e + C2e + C3cos x+C4sin x.15. Auxiliary equation: r4 + 3 r2 –4=0,2( r 1)( r –1)( r 4) 0+ + = has roots –1, 1, ±2i.General solution:– x xy = C1e + C2e + C3cos 2x+C4sin 2xInstructor’s Resource Manual Section 15.1 891

16. Roots are –2, 3, ±i. General solution is–2x3xy = C1e + C2e + C3cos x+C4sin x.–2x2xy = C1e + C2 e .y = C1(cosh2 x–sinh2 x) + C2(sinh2x+ cosh2 x)= (– C1+ C2)sinh 2 x+ ( C1+C2)cosh 2x= D1sinh 2x+D2cosh 2x17. Roots are –2, 2. General solution isu– u18. e = cosh u+ sinh u and e = cosh u– sinh u.2 2Auxiliary equation: r – 2 br – c = 0Roots of auxiliary equation:General solution:2 22b± 4b + 4c2 2= b± b + c2( b+ b2+ c2) x ( b– b2+ c2) x1 2y = C e + C ebx ⎡ ⎛ 2 2 2 2 ⎞ ⎛ 2 2 2 2 ⎞⎤= e C1 cosh⎛b + c x⎞+ sinh⎛b + c x⎞+ C2cosh⎛b + c x⎞– sinh⎛b + c x⎞⎢ ⎜ ⎜ ⎟ ⎜ ⎟⎟ ⎜ ⎜ ⎟ ⎜ ⎟⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥⎣ ⎝ ⎠ ⎝ ⎠⎦bx ⎡2 2 2 2 ⎤= e ( C1+ C2) cosh⎛b + c x⎞+ ( C1+ C2)sin⎛b + c x⎞ bx ⎡2 2 2 2 ⎤⎢⎜ ⎟ ⎜ ⎟⎣⎝ ⎠ ⎝ ⎠⎥= e D1cosh⎛b c x⎞D2sinh⎛b c x⎞⎜ + ⎟+ ⎜ + ⎟⎦⎢⎣ ⎝ ⎠ ⎝ ⎠⎥⎦⎛19. Repeated roots – 1 ⎞ ⎛ 3 ⎞⎜ ⎟ ± i.⎝ 2⎠ ⎜ 2 ⎟⎝ ⎠⎡– x /2⎛ 3⎞ ⎛ 3⎞⎤General solution is y = e ⎢( C1+ C2x)cos ⎜ x+ ( C3 + C4x)sin x⎥.2 ⎟ ⎜ 2 ⎟⎢⎣⎝ ⎠ ⎝ ⎠ ⎥⎦20. Roots 1 ± i. General solution is21.xy = e ( C1cosx+C2sin x)xx= e ( csin cos x+ ccos sin x) = ce sin( x+).γ γ γ2(*) x y′′ + 5xy′+ 4y= 0zx eLet = . Then z = ln x;dy dy dz dy 1y′ = = = ;dx dz dx dz x2dy′ d ⎛dy 1 ⎞ dy –1 1 d y dzy′′ = = ⎜ ⎟= +dx dx ⎝dz x ⎠ dz 2 x 2x dz dx2dy –1 1 d y 1= +dz 2 x 2x dz x⎛ 2dy d y ⎞ ⎛ dy ⎞– + + 5 + 4y= 0⎜ dz 2 ⎜ ⎟dz ⎟⎝ ⎠ ⎝ dz ⎠(Substituting y′ and y′′ into (*))2d y dy+ 4 + 4y= 02dz dz22Auxiliary equation: r + 4r+ 4=0, ( r + 2) = 0has roots –2, –2.–2zGeneral solution: y = ( C + C z) e ,1 2–2lnxy = ( C1+C2ln x)e–2y = ( C1+C2ln x)x22. As done in Problem 21,⎡2⎛dy ⎞ ⎛d y ⎞⎤⎛dy⎞⎢– a⎜ ⎟+ a ⎥+ b + cy = 0.dz 2 ⎜ ⎟⎢ ⎝ ⎠ ⎜ dx ⎟⎣ ⎝ ⎠⎥⎦⎝dz⎠⎛ 2d y⎞ ⎛dy⎞Therefore, a + ( b– a) + cy = 0.⎜ 2 ⎜ ⎟dz ⎟⎝ ⎠ ⎝dz⎠23. We need to show that y'' + a1y' + a2y= 0 if r 1 andr2are distinct real roots of the auxiliary equation.We have,rx 1 r2xy'= C11 re + C2r2e.2 rx 12 rx 2y''= C11 r e + C2r2eWhen put into the differential equation, weobtain2 rx 12 rx 2y''+ a y'+ a y = C r e + C r e1 2 1 1 2 2rx 1 rx 2 rx 1 r2x( ) ( )+ a1 C1re 1 + C2r2e + a2 C1e + C2eThe solutions to the auxiliary equation are givenby12r1 = ( −a1− a1 −4 a2) and212r2 = ( − a1+ a1 −4 a2).2Putting these values into (*) and simplifyingyields the desired result: y'' + a1y' + a2y= 0 .(*)892 Section 15.1 Instructor’s Resource Manual

24. We need to show that y'' + a1y' + a2y= 0 if α ± βiare complex conjugate roots of the auxiliary equation. Wehave,α x(( α 1 β 2) ( β ) ( α 2 β 1) ( β ))y' = e C + C cos x + C − C sin xα x2 2 2 2(( α ) ( ) ( ) ( ))1 β 1 αβ 2 β α 2 β 2 αβ 1 βy'' = e C − C + 2 C cos x + C − C −2 C sin xWhen put into the differential equation, we obtain2 2 2 2(( α β αβ ) ( β ) ( α β αβ ) ( β ))x x x(( α β ) ( β ) ( α β ) ( β )) ( ( β ) ( β ))α x1 2 1 1 2 2 2 1y'' + a y' + a y = e C − C + 2 C cos x + C − C −2 C sin xα α α1 1 2 2 1 2 1 2+ ae C + C cos x + C − C sin x + a Ce cos x + C e sin x (*)From the solutions to the auxiliary equation, we find that−a11 2α = and β =− i a1 − 4 a2.2 2Putting these values into (*) and simplifying yields the desired result: y'' + a1y' + a2y= 0 ..25. a.bi 2 3 4 5( ) ( ) ( ) ( )1 ( )bi bi bi bi ⎛ 2 4 6 3 5 7b b b ⎞ ⎛ b b b ⎞e = + bi + + + + +… = 1– + – + …+ i b– + – +…2! 3! 4! 5! ⎜ 2! 4! 6! ⎟ ⎜ 3! 5! 7! ⎟⎝ ⎠ ⎝ ⎠= cos (b) + i sin (b)a+bi a bi ab. e = e e = e [cos( b) + isin( b)]c.( ix )xD ⎡ α+βx e ⎤ ααxαx= Dx[ e (cosβx+isin βx)]= αe (cos βx+ isin βx) + e (– iβsin βx+iβ cos βx)⎣ ⎦α x= e [( + i)cos x+( i– )sin x]α β β α β β( α+β )( ) ixαxα x+ e = ( + i)[ e (cos x+ isin x)]= e [( + i)cos x+( i– )sin x]α β α β β βTherefore,( α+ βix) ( α+βix)Dx[ e ] = ( α + iβ)eα β β α β β26.27.28.29.30.( α+ βix) ( α+βix)ce 1 + c2e[c 1 and c 2 are complex constants.]αxαxα x= ce 1 [cos β x+ isin βx] + c2e [cos(– βx) + isin(– βx)]= e [( c1+ c2)cos β x+( c1 – c2) isin β x]α x= e [ C1cosβx+ C2sin β x],where C1 = c1+ c2,and C2 = c1 – c2.Note: If c 1 and c 2 are complex conjugates, then C 1 and C 2 are real.5.16228 x –1.162278xy = 0.5e + 0.5e–2.5 x –2.5xy = 3.5xe + 2e–0.25xy = 1.29099e sin(0.968246 x)0.333333 xy = e [2.5cos(0.471405 x ) – 4.94975sin(0.471405 x )]Instructor’s Resource Manual Section 15.1 893

15.2 Concepts Review1. particular solution to the nonhomogeneous equation; homogeneous equation2.3.–2x3x–6 + Ce 1 + C2e2y = Ax + Bx+C4.1 xy = Bxe 3Problem Set 15.21.2.–3x3xyh= C1e + C2e⎛ 1 ⎞yp= ⎜– ⎟x+0⎝ 9 ⎠1 –3x 3xy ⎛ ⎞= ⎜– ⎟x+ C1e + C2e⎝ 9 ⎠–3x2xyh= C1e + C2e⎛ 1⎞ 2 ⎛ 1⎞ ⎛ 7 ⎞yp= ⎜– ⎟x + ⎜– ⎟x+⎜–⎟⎝ 3⎠ ⎝ 9⎠ ⎝ 54⎠⎛ 1⎞ 2 ⎛1⎞ ⎛ 7 ⎞ –3x2xy = ⎜– ⎟x – ⎜ ⎟x–⎜ ⎟+ C1e + C2e⎝ 3⎠ ⎝9⎠ ⎝54⎠3. Auxiliary equation: r 2 – 2r+ 1= 0 has roots 1, 1.yh= ( C1+C2x) ex2Let yp= Ax + Bx+ C; y′p = 2 Ax+B;y′′ p = 2A.2 2Then (2 A) – 2(2 Ax + B) + ( Ax + Bx + C) = x + x.2 2Ax + (–4 A+ B) x+ (2 A– 2 B+ C)= x + xThus, A = 1, –4A + B = 1, 2A – 2B + C = 0, soA = 1, B = 5, C = 8.2General solution: y = x + 5x+ 8 + ( C + C x) ex1 24.5.– x2yh= C1e + C2⋅ yp= 2 x + (–4) x2 – xy = 2 x –4x+ C1e + C22x 3x 1 xyhC1e C2e y ⎛ ⎞= + ⋅ p = ⎜ ⎟e ⋅y⎝2⎠⎛1⎞ x 2x 3x= ⎜ ⎟e + C1e + C2e⎝2⎠894 Section 15.2 Instructor’s Resource Manual

6. Auxiliary equation: r 2 + 6r+ 9= 0, ( r+ 3) 2 = 0 has roots –3, –3.–3xyh= ( C1+C2x)e– x – x – xyp = Be ; y′ p = – Be ; y′′p = Be .LetThen ( Be ) + 6(– Be ) + 9( Be ) = 2 e ;General solution:– x – x – x – x – x – x 14Be = 2 e ; B =21 – x–3xy ⎛ ⎞= ⎜ ⎟e + ( C1+C2x)e⎝2⎠7.8.–3 x – xyh= C1e + C2e⎛ 1 ⎞ –3xyp= ⎜ – ⎟ xe⎝ 2 ⎠1 –3 x –3 x – xy ⎛ ⎞= ⎜– ⎟xe + C1e + C2e⎝ 2 ⎠– xyh= e ( C1cosx+C2sin x)⎛3⎞ –2xy p = ⎜ ⎟ e⎝2⎠3 –2 x – xy ⎛ ⎞= ⎜ ⎟e + e ( C1cosx+C2sin x)⎝2⎠9. Auxiliary equation: r 2 – r –2=0,(r + 1)(r – 2) = 0 has roots –1, 2.– x 2xyh= C1e + C2eLet y p = Bcos x+ Csin x;y′ p = – Bsin x+ Ccos x;y′′ p = – Bcos x– Csin x.Then ( −Bcosx−Csin x) −( − Bsinx+Ccos x)− 2( Bcos x+ Csin x) = 2sin x.( −3 B− C)cos x+ ( B− 3 C)sin x = 2sin x, so − 3B− C = 0 so –3B – C = 0 and B− 3C= 2;⎛1⎞ ⎛3⎞xGeneral solution: ⎜ ⎟cos x – ⎜ ⎟sinx+ C1e + C2e⎝5⎠ ⎝5⎠2 – x1 –3B = ; C = .5 510.–4xyh= C1e + C2⎛ 1 ⎞ ⎛ 4 ⎞yp= ⎜– ⎟cos x+⎜ ⎟sinx⎝ 17 ⎠ ⎝17⎠⎛ 1 ⎞ ⎛ 4 ⎞–4xy = ⎜– ⎟cosx+ ⎜ ⎟sinx+ C1e + C2⎝ 17 ⎠ ⎝17⎠11. yh= C1cos 2x+C2sin 2x⎛1⎞yp= (0) xcos 2x+ ⎜ ⎟ xsin 2x⎝2⎠1y ⎛ ⎞= ⎜ ⎟xsin 2x+ C1cos 2x+C2sin 2x⎝2⎠Instructor’s Resource Manual Section 15.2 895

12. Auxiliary equation: r 2 + 9= 0 has roots ±3i, so yh= C1cos3x+C2sin 3 x.Let y = Bx cos3x + Cxsin 3 x;y′ = (–3 bx + C)sin 3 x + ( B + 3 Cx)cos3x;ppy′′ p = (–9Bx + 6 C)cos3 x + (–9 Cx – 6 B)sin 3x.Then substituting into the original equation and simplifying, obtain 6Ccos3x- 6Bsin 3x = sin 3x, so C = 0 and1B = – .61General solution: y ⎛ ⎞= ⎜– ⎟xcos3x+ C1cos3 x+C2sin 3x⎝ 6 ⎠13. yh= C1cos3x+C2sin 3x⎛1⎞ ⎛ 1 ⎞ 2xyp= (0)cos x+ ⎜ ⎟sinx+⎜ ⎟e⎝8⎠ ⎝13⎠⎛1⎞ ⎛ 1 ⎞ 2xy = ⎜ ⎟sin x+ ⎜ ⎟e + C1cos3x+C2sin 3x⎝8⎠ ⎝13⎠14.h– x1 2y = C e + C⎛1⎞ x ⎛3⎞2yp= ⎜ ⎟e + ⎜ ⎟x + (–3) x⎝2⎠ ⎝2⎠⎛1⎞ x ⎛3⎞2 – xy = ⎜ ⎟e + ⎜ ⎟x –3x+ C e + C⎝2⎠ ⎝2⎠1 215. Auxiliary equation:2 xr – 5r+ 6= 0 has roots 2 and 3, so yh= C e + C eLet y = Be x ; y′ = Be x ; y′′= Bex .p p px x x xBe Be + Be = e 2 x xBe = 2 e ; B = 1.x 2x 3xy = e + C1e + C2ex 2x 3x2 1 3 2Then ( )–5( ) 6( ) 2 ;General solution:2 3x1 2 .y′ = e + C e + C eIf x = 0, y = 1, y′ = 0, then 1= 1+ C1+ C2and 0= 1+ 2C1+ 3 C2;C1 = 1, C2= –1.x 2x 3xTherefore, y = e + e – e .16.17.h–2x2x1 2y = C e + C e⎛ 4 ⎞yp= (0)cos x+ ⎜ – ⎟ sin x⎝ 5 ⎠4–2x2xy ⎛ ⎞= ⎜– ⎟sinx+ C1e + C2e⎝ 5 ⎠⎛ 4⎞ ⎛9⎞ x ⎛11⎞y = ⎜– ⎟sinx+ ⎜ ⎟e + ⎜ ⎟e⎝ 5⎠ ⎝5⎠ ⎝ 5 ⎠hx 2x1 2y = C e + C e–2 2x⎛1 ⎞yp= ⎜ ⎟ (10 x+19)⎝4⎠1 (10 19)xy ⎛ ⎞= ⎜ ⎟ x+ + C1e + C2e⎝4⎠2xsatisfies the conditions.896 Section 15.2 Instructor’s Resource Manual

2 x –2x18. Auxiliary equation: r 2 – 4= 0 has roots 2, –2, so yh= C1e + C2 e .2 x –2x2 x –2xLet y = v e + v e subject to ve ′ v′2 x –2x 2x+ e = and v′ (2 e ) + v′(–2 e ) = e .p1 2 ,1 2 0,1 22x2xThen v1(4 ′–2x2x1e ) = e and v2(–4 ′4 x /4 xee ) = e ; v′ 1 = and v2 ′ = – e ; v 1 = and v 2 = – .4416General solution:2x2xxe ey = – + C e + C e4 1619. yh= C1cosx+C2sinxy = – cosln sin x – cos x– xsinxp2 x –2x1 2y – cos xln sin x – xsin x C3cos x C2sinx= + + (combined cos x terms)20. yh= C1cosx+C2sinxy = – sin xln csc x+cot xpy = – sin xln csc x+ cot x + C cos x+C sin x1 221. Auxiliary equation: r 2 – 3r+ 2= 0 has roots 1, 2, so yh= C1e + C2 e .x 2xx 2xLet yp= v1e + v2esubject to ve 1′ v2′x 2 x x –1+ e = 0, and v1′ ( e ) + v2′(2 e ) = e ( ex+1) .xx– e– e–1Then v1′ = so vx x 1 = ∫ dx =x x ∫ due ( e + 1) e ( e + 1) uu ( + 1)⎛ –1 1 ⎞ ⎛u+ 1⎞e + 1 – x= ∫ ⎜ + ⎟du = –lnu + ln( u + 1) = ln⎜ ⎟ = ln = ln(1 + e )⎝ u u+1 ⎠ ⎝ u ⎠xexev2′ – x – x=2 so vx x 2 = – e + ln(1 + e )e ( e + 1)(similar to finding v 1 )x – x x 2 x – x x 2xGeneral solution: y = e ln(1 + e ) – e + e ln(1 + e ) + C e + C ex 2 x – x x 2x1 2y = ( e + e )ln(1 + e ) + D e + D exx1 22x4x22.h2x3x1 2y = C e + C e ;ypx 2x 3x1 2y = e + C e + C e= ex( ) ( )′′ ( )′( )23. Lyp= vu 1 1+ vu 2 2 + bvu 1 1+ vu 2 2 + cvu 1 1+vu 2 2= ( vu 11 ′ + vu 11′ + v2′ u2+ v2u2′ ) + bvu ( 11 ′ + vu 11′ + v2′ u2+ v2u2′) + cvu ( 11+v2u2)= ( vu 11 ′′ + vu 11 ′ ′ + vu 11 ′ ′ + vu 11′′ + v2′′ u2+ v2′ u2′ + v2′ u2′ + v2u2′′ ) + bvu ( 11 ′ + vu 11′ + v2′ u2+ v2u2′) + cvu ( 11+v2u2)= v1( u1′′ + bu1′ + cu1) + v2( u2′′ + bu2′ + cu2) + b( v11 ′ u + v2′ u2) + ( v11 ′′ u + v11 ′ u′ + v2′′ u2 + v2′ u2) + ( v11 ′ u′ + v2′ u2′)= v ( u′′ + bu′ + cu ) + v ( u′′ + bu′ + cu ) + b( v′ u + v′ u ) + ( v′ u + v′ u )′ + ( v′ u′ + v′ u′)1 1 1 1 2 2 2 2 11 2 2 11 2 2 11 2 2= v (0) + v (0) + b(0) + (0) + k( x) = k( x)1 2Instructor’s Resource Manual Section 15.2 897

24. Auxiliary equation: r 2 + 4= 0 has roots ±2i.yh= C1cos 2x+C2sin 2x3Now write sin x in a form involving sin βx’s orcos βx’s.3 3 1sin x = sin x– sin 3x4 4(C.R.C. Standard Mathematical Tables, or derive it using half-angle and product identities.)Let y = Asin x+ Bcos x+ Csin 3x+ Dcos3x;py′ = Acos x– Bsin x+ 3Ccos3 x– 3Dsin 3x;py′′ p = – Asin x– Bcos x– 9Csin 3 x– 9Dcos3x.Then3 1y′′ p + 4yp = 3Asinx+3Bcos x–5Csin3 x–5Dcos3x= sin x – sin 3 x,so4 41 1A = , B = 0, C = , D = 0.4201 1General solution: y = sin x+ sin 3x+ C1cos 2x+C2sin 2x4 2015.3 Concepts Review1. 3; π2. π ; decreases3. 04. electric circuitProblem Set 15.31.2k = 250, m = 10, B = k/ m = 250 /10 = 25, B = 5(the problem gives the mass as m = 10kg )Thus, y '' =− 25 y.The general solution is y = C1cos5t+ C2sin 5 t.Apply the initial condition to get y = 0.1cos5 t.The period is 2 π seconds.52. k = 100 lb/ft, w = 1 lb, g = 32 ft/s 2 1, y 0 = ft,12⎛ 1 ⎞B = 40 2 . Then y = ⎜ ⎟cos(402) t .⎝12⎠Amplitude is 1 ft = 1 in.122Period isπ ≈ 0.1111 s.40 2ππ π3. y = 0.1cos5t= 0 whenever 5t= + π k or t = + k .210 5⎛ π π ⎞ ⎛ π π ⎞ ⎛π ⎞y' ⎜ + k⎟ = 0.5 sin 5⎜ + k⎟ = 0.5 sin ⎜ + π k⎟= 0.5 meters per second⎝10 5 ⎠ ⎝10 5 ⎠ ⎝ 2 ⎠898 Section 15.3 Instructor’s Resource Manual

4.110 = k ⎛ ⎞⎜ ⎟ , so k = 30 lb/ft, w = 20 lb,⎝3⎠g = 32 ft/s 2 , y 0 = –1 ft, v 0 = 2 ft/s, B = 4 3Then y = C1cos(4 3 t) + C2sin(4 3 t).⎛ 3t⎞y = cos(4 3 t) +⎜ ⎜sin(4 3 t)6 ⎟satisfies the initial conditions.⎝ ⎠5. k = 20 lb/ft; w = 10 lb; y 0 = 1 ft;1q = s-lb/ft, B = 8, E = 0.32102 2E – 4B < 0, so there is damped motion. Roots of auxiliary equation are approximately –0.16 ± 8i.–0.16t–0.16 tGeneral solution is y ≈ e ( C cos8t+C sin8 t).y ≈ e (cos8 t+ 0.02sin 8 t ) satisfies the initial conditions.1 26. k = 20 lb/ft; w = 10 lb; y 0 = 1 ft; q = 4 s-lb/ft(20)(32) (4)(32)2 2B = = 8; E = = 12.8; E – 4B < 0, so damped motion.10102 2– E±E –4BRoots of auxiliary equation are= –6.4 ± 4.8 i.2–6.4tGeneral solution is y = e ( C1cos 4.8t+C2sin 4.8 t).–6.4 t–6.4ty′ = e (–4.8C1sin 4.8t+ 4.8C2cos 4.8 t) – 6.4 e ( C1cos 4.8t+C2sin 4.8 t)4If t = 0, y = 1, y′ = 0, then 1 = C1and 0= 4.8 C2 –6.4 C1,so C 1 = 1 and C 2 = .3Therefore,–6.4t⎡ ⎛4⎞ ⎤y = e ⎢ cos 4.8t+ ⎜ ⎟ sin 4.8 t .3 ⎥⎣ ⎝ ⎠ ⎦7. Original amplitude is 1 ft. Considering the contribution of the sine term to be negligible due to the 0.02 coefficient,–0.16 tthe amplitude is approximately e .–0.16te ≈ 0.1 if t ≈ 14.39 , so amplitude will be about one-tenth of original in about 14.4 s.8. C 1 = 1 and C 2 = –0.105, soQ9. LQ′′ RQ′6 6+ + = E();t 10 Q′ + 10 Q = 1;CtIntegrating factor: e–6 –6DQe [ t ] = 10 e t ; Qe t = 10 e t + C;–6 – tQ = 10 + Ce–0.16 ty = e (cos8 t+0.105sin 8 t ).If t = 0, Q = 0, then C = –10 .–6 –6 – t –6 – tTherefore, Qt () = 10 –10 e = 10 (1– e ).–6–6Q′ + Q = 10–6 –6 – t–610. Same as Problem 9, except C = 4–10 , so Qt () = 10 + (4–10 ) e .–6 – tThen I() t = Q′() t = –(4–10 ) e .11.Q= 120sin 377t–6[2(10 )]a. Q(t) = 0.00024 sin 377tb. I( t) = Q′( t) = 0.09048cos377tInstructor’s Resource Manual Section 15.3 899

Q12. LQ′′ RQ′–2 Q9+ + = E;10 Q′′ + = 20; Q′′ + 10 Q = 2000C–7102 99/2The auxiliary equation, r + 10 = 0, has roots ± 10 i.9/2 9/2Qh= C1cos10t+C2sin10t–9 –6Q p = 2000(10 ) = 2(10 ) is a particular solution (by inspection).–6 9/2 9/2General solution: Qt ( ) = 2(10 ) + C1cos10 t+C2sin10tThen I() t Q′9/2 9/2 9/2 9/2= () t = –10 C1sin10 t+10 C2cos10 t.–6If t = 0, Q = 0, I = 0, then 0= 2(10 ) + C 1 and 0 = C2.9/2 –6 9/2 –3/2 9/2I( t) = –10 (–2[10 ])sin10 t = 2(10 )sin10 t.Therefore,13.Q3.5Q′′ + 1000Q+ = 120sin 377t–6[2(10 )](Values are approximated to 6 significant figures for the remainder of the problem.)Q′′ + 285.714Q′+ 142857Q = 34.2857sin 377tRoots of the auxiliary equation are–142.857 ± 349.927i.–142.857tQh= e ( C1cos349.927t+C2sin 349.927 t)–4 –6Qp= –3.18288(10 )cos377t+2.15119(10 )sin 377t–4 –6Then, Q = –3.18288(10 )cos377t+ 2.15119(10 )sin 377 t+Q h .I = Q′ = 0.119995sin 377t+ 0.000810998cos377t+Qh′0.000888 cos 377t is small and Q′ h → 0 as t →∞ , so the steady-state current is I ≈ 0.12sin 377t.14. a. Roots of the auxiliary equation are ±Bi.y = C cos Bt+C sin Bt.h1 2⎡ c ⎤yp= ⎢ ⎥sinAt2 2⎢⎣( B – A ) ⎥⎦The desired result follows.⎛ c ⎞b. yp= ⎜ – ⎟ tcos Bt,so⎝ 2B⎠⎛ c ⎞y = C1cos Bt + C2sin Bt – ⎜ ⎟t cos Bt.⎝2B⎠c. Due to the t factor in the last term, itincreases without bound.15. Asin( βt+ γ) = A(sin βtcosγ + cos βtsin γ)= ( Acos γ )sin βt+( Asin γ)cosβt= β + β where C1 = AcosγandC1sin t C2cos t,C2 = Asin γ .[Note that2 2 2 2 2 2 2C1 + C2 = A cos γ + A sin γ = A .)16. The first two terms have period 2 π and the lastBhas period 2 π. Then the sum of the three termsA⎛2π⎞ ⎛2π⎞is periodic if m⎜ ⎟=n⎜ ⎟ for some integers⎝ B ⎠ ⎝ B ⎠B mm, n; equivalently, if = , a rational number.A n900 Section 15.3 Instructor’s Resource Manual

17. The magnitudes of the tangential components ofthe forces acting on the pendulum bob must beequal.15.4 Chapter ReviewConcepts Test1. False:2y is not linear in y.2. True: y and y′′ are linear in y and y′′ ,respectively.Therefore,s = Lθ, soTherefore,Hence,2d s– m = mgsin θ.2dt2 2d s d θ= L2 2dt dt.2d θ– mL = mg sin θ.2dt2d θ g= – sin θ.2dt L18. a. Since the roots of the auxiliary equation areg g± i , the solution of θ′′ ⎛ ⎞() t + ⎜ ⎟θ= 0L⎝ L ⎠g gC1cos t C2sin t,L L⎛ g ⎞θ = C⎜t+γL ⎟⎝ ⎠θ = + which can bewritten as(by Problem 15).The period of this function is22π L LR L= 2 π = 2 π = 2 π R .g G GM GMLL1p 2πR 1 1 GM R1 L1= = .p2 L2πR2 R22 L2GMTherefore,b. To keep perfect time at both places, requireR2 p1 = p2.Then 1 = 80.85 , so3960 81R2 ≈ 3963.67.The height of the mountain is about3963.67 - 3960 = 3.67 mi (about 19,387 ft).is3. True:2y′ = sec x+sec xtanx22 y′ 2– y = (2sec x + 2sec xtan x)2 2–(tan x + 2sec xtan x+sec x)2 2= sec x – tan x = 14. False: It should involve 6.5. True:2D adheres to the conditions forlinear operators.2 2D ( kf) = kD ( f)2 2 2D ( f + g)= D f + D g6. False: Replacing y by Cu 11( x) + C2u2( x)would yield, on the left side,Cf 1 ( x) + C2f( x) = ( C1+C2) f( x)which is f(x) only if C1+ C2 = 1 orf(x) = 0.7. True: –1 is a repeated root, with multiplicity3, of the auxiliary equation.8. True: Lu ( 1 – u2) = Lu ( 1)– Lu ( 2)= f( x)– f( x) = 09. False: That is the form of y h.y p shouldhave the form Bx cos 3x + Cx sin 3x.10. True: See Problem 15, Section 15.3.Sample Test Problemsx1. u′ + 3 u = e . Integrating factor is3x4xDue [ ] = e1 x –3xy ⎛ ⎞= ⎜ ⎟e + C1e⎝4⎠1 x –3xy′ ⎛ ⎞= ⎜ ⎟e + C1e⎝4⎠1 x –3xy ⎛ ⎞= ⎜ ⎟e + C3e + C2⎝4⎠3 xe .2. Roots are –1, 1.– x xy = C1e + C2eInstructor’s Resource Manual Section 15.4 901

3. (Second order homogeneous)2The auxiliary equation, r – 3r+ 2= 0, has rootsx 2x1, 2. The general solution is y = C e + C ex 2xy′ = C1e + 2C2e1 2 .If x = 0, y = 0, y′ = 3, then 0 = C1+ C2and3= C1+ 2 C2,so C1 = –3, C2= 3.x 2xTherefore, y = –3e + 3 e .4. Repeated root 3 – .25.– x xyhC1e C2e– x xyp= –1+ C1e + C2e(–3/ 2) xy = ( C1+C2x)e= + (Problem 2)6. (Second-order nonhomogeneous) The auxiliary2equation, r + 4r+ 4= 0, has roots –2, –2.7.–2 x –2 x –2xyh= C1e + C2xe = ( C1+C2x)eLet yp = Be x ; y′ p = Be x ; y′′p = Bex .x x x x( Be ) + 4( Be ) + 4( Be ) = 3 e , soGeneral solution:–2xyh( C1 C2x)e⎛1⎞ 2 –2xyp= ⎜ ⎟ x e⎝2⎠1B = .3xe–2xy = + ( C1+C2x)e3= + (Problem 12)⎡⎛1⎞ 2 ⎤ –2xy = ⎢⎜⎟x + C1+C2x e2⎥⎣⎝⎠⎦8. Roots are ±2i.y = C1cos 2x+C2sin 2xy = sin 2x satisfies the conditions.9. (Second-order homogeneous)2The auxiliary equation, r + 6r+ 25= 0, hasroots –3 ± 4i. General solution:–3xy = e ( C1cos4x+C2sin4 x)10. Roots are ±i. yh= C1cosx+C2sinxy = xcos x– sin x+sin xln cos xpy = xcos x− sin xln cos x + C1cos x+C3sinx(combining the sine terms)12. (Fourth-order homogeneous)4 2The auxiliary equation, r – 3 r –10= 0 or2 2( r –5)( r + 2) = 0, has roots – 5, 5, ± 2 i.General solution:5 x – 5xy = C1e + C2e + C3cos 2x+C4sin 2x13. Repeated roots ± 2– 2x2xy = ( C1+ C2x) e + ( C3 + C4x)e14. a. Q′ ( t) = 3 – 0.02Qb. Q′ ( t) + 0.02Q= 30.02tIntegrating factor is e0.02t0.02tDQe [ ] = 3e–0.02tQt () = 150+Ce–0.02tQt () 150–30e= goes through (0, 120).c. Q → 150 g, as t →∞.15. (Simple harmonic motion)k = 5; w = 10; y 0 = –1(5)(32)B = = 410Then the equation of motion is y = –cos 4t.The amplitude is –1 = 1; the period is 2 π π= .4 216. It is at equilibrium when y = 0 or –cos 4t = 0, orπ 3πt = , ,….8 8y′ () t = 4sin4, t so at equilibrium y′ =± 4 = 4.17. Q′′ + 2Q′+ 2Q= 1Roots are –1 ± i.– tQhe ( C1cost C2sin t)= + and1Q p = ;2– t1Q = e ( C1cost+ C2sin t)+2– tI() t = Q′() t = – e [( C1 – C2)cos t+ ( C1+C2)sin t]– tI() t e sint= satisfies the initial conditions.11. Roots are –4, 0, 2.–4x2xy = C1e + C2 + C3e902 Section 15.4 Instructor’s Resource Manual

Calculus 9e Purcell-Varberg-Rigdon (Solution).

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