Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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SECTION 14.5 THE CHAIN RULE 0 207 39. First we find :~ implicitly by taking partial derivatives of both sides with respect to R1 : oR . R 2 oR 1 = Ri. Then by symmetry, oR R 2 2 oR = R2 . When RI = 25, R2 = 40 and Ra = 50, -R 1 = 2 1 0 7 0 ¢::} R = 2 1°7° n. Since the possible error oR2 - ~· oR3 R 3 for each Ri is 0.5%, the maximum error of R is attained by setting t:J.R, = 0.005R.. So oR oR oR 2 ( 1 1 1 ) 1 !:l.R ~=:::: dR = oR !:l.R1 1 + oR 2 !:l.R2 + oR 3 !:l.R3 = (0.005)R R 1 + R 2 + Ra = (0.005)R = l7 ~=:::: 0.059 st. ~ l!:l.wl l!:l.hl . . . 41 . The errors in measurement are at ~ost 27o, so w $ 0 .. 02 and T $ 0.02. The relative error m the calculated surface area is !:l.S dS 0.1091(0.425w 0·425 - 1 )h 0·725 dw + 0.1091w 0 .4 25 (0.725h 0·725 - 1 } dh dw dh s 1':::: s= 0.1091w0.426h0.725 . = 0.425-:;:; + 0.725/i: To estimate the maximum relative error, we use : = I ~w I = 0.02 and ~~i = I ~hI = 0.02 => dS = 0.425 (0.02) + 0. 725 (0.02} = 0.023. Thus the maximum percentage error is approximately 2.3%. s 43. !:l.z = f(a + !:l.x, b + !:l.y) - f(a, b) =(a+ !:l.x? + (b + !:l.y) 2 - (a 2 + b 2 ) = a 2 + 2a!:l.x + (!:l.x) 2 + b 2 + 2b!:l.y + (!:l.y?- a 2 - b 2 = 2a !:l.x + (!:l.x? + 2b!:l.y + (!:l.y? But f,(a, b) = 2a and /y(a, b) = 2b and so !:l.z = f x(a, b) !:l.x + /y(a, b) !:l.y + !:l.x !:l.x + !:l.y !:l.y, which is Definition 7 with e:1 = !:l.x and e: 2 = !:l.y. Hence f is differentiable. 45. To show that f is continuous at (a, b) we need to show that lim f(x, y) = f(a, b) or (z,y)~(a,b) equivalently lim f(a + !:l.x, b + !:l.y) = f(a, b). Since f is differentiable at (a, b), (6x,6y)-(O,O ) f(a + !:l.x, b + !:l.y) - f(a, b)= !:l.z = f,(a, b) 6-x + /y(a, b) !:l.y + e:1 !:l.x + e:2 !:l.y, where €1 and €2 ---+ 0 as (!:l.x, !:l.y} --> (0, 0). Thus f(a + !:l.x, b + !:l.y) = f(a, b)+ f,(a, b) !:l.x + / 11 (a, b) !:l.y + e:1 !:l.x + e:2 !:l.y. Taking the limit of both sides as (.C:..x, .C:..y) ___.. (0, 0) gives lim f (a + .C:..x, b + .C:..y) = f (a , b). Thus f is continuous at (a, b). (6 x,61J)- (0 ,0) . 14.5 The Chain Rule 1. z = x? + y 2 + xy, x =sin t, y = et => ~ &~ &~ t -d =,.--d +,.--d =(2x+ y}cost+(2y +x)e t uX t uy t 3. z =· .J1 + x 2 + y 2 , x = In t, y = cost => dz oz dx 8z dy 1 2 2 -1/2 1 1 2 2 -1/ 2 . 1 (x . ) - =-- + --= 2 (1+x + y) (2x)·- + 2 (1+x + y) (2y)(-smt) = - - ysmt dt ox dt 8y dt t · .J1 + x2 + y2 t © 2012 Cc:ngagc Lc:aming. All Righrs Reserved. Mny not be sc.mncd, copied. or duplicalcd, or posrOO 10 a publi~ly occcsslblc website, in whole or in part.
208 D CHAPTER 14 PARTIAL DERIVATIVES 5. w=xevl =, x = t2, y = l-t, z=1+2t => dw = OW dx + ow dy + ow dz = ey/ z. 2t + xevl= (.!) .(- 1) + xeYI= (-J!...) . 2 = evl= (2t - ~ - 2xy) dt ox dt /Jy dt oz dt z z 2 z z2 7. z = x 2 y 3 , x = scost, y = ssint => az oz ax oz 8y 3 2 2 • !i"'" = !i"'"' + -a -a = 2xy cost+ 3x y sm t us uxus y s . az . 8z Ox az 8y 3 . 2 2 s . 2 ot = ox at+ oy at = (2xy )(- ssmt) + (3x y )(scost) = -2sxy smt + 3sx 2 y cost 9. z = sin() cos, () = st 2 , ¢ = s 2 t => ~; = ~; ~~ + ~~ ~! =(cos 8 cos q))(t~) + (- sinO sin q.,)(2st) = t 2 cosO cos¢ - 2stsin0 sin )(2st) + (-sinO sin )(s 2 ) = 2stcos0 cos¢- s 2 sinO sin
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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D ABBREVIATIONS AND SYMBOLS CD cu D
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viii o CONTENTS 12.4 The Cross Prod
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10 D PARAMETRIC EQUATIONS AND POLAR
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.2 CALCULUS WITH PARAMETR
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SECTION 10.3 POLAR COORDINATES 0 13
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SECTION 10.5 CONIC SECTIONS 0 27 5.
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x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1
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SECTION 10.6 CONIC SECTIONS IN POLA
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CHAPTER 10 REVIEW 0 35 the length o
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CHAPTER 10 REVIEW 0 37 EXERCISES 1.
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CHAPTER 10 REVIEW 0 39 25. x = t +
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0 PROBLEMS PLUS l lt sin u dx cost
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11 . D INFINITE SEQUENCES AND SERIE
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SECTION 11.1 SEQUENCES 0 47 35. a,.
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SECTION 11.1 SEQUENCES D 49 71. Sin
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SECTION 11.2 SERIES 0 51 ak+l- a1.:
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, SECTION 11.3 THE INTEGRAL TEST AN
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SECTION 11.5 ALTERNATING SERIES 0 6
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SECTION 11.5 ALTERNATING SERIES D 6
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SECTION 11.6 ABSOLUTE CONVERGENCE A
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17 lim I an+l I= SECTION 11.7 STRAT
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61 _ x_ x · sin x - x- tx 3 + 1~
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CHAPTER 11 REVIEW 0 97 J'(xn)(xn -
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CHAPTER 11 REVIEW 0 99 oo f (n) (O)
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CHAPTER 11 REVIEW 0 101 l 23. Consi
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49./- 1 - dx = -ln{4- x) + C and 4-
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D PROBLEMS PLUS 1. It would be far
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112 0 CHAPTER 12 VECTORS AND THE GE
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13 D VECTOR FUNCTIONS 13.1 Vector F
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SECTION 13.1 VECTOR FUNCTIONS AND S
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SECTION 15.5 APPLICATIONS OF DOUBLE
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SECTION 1505 APPLICATIONS OF DOUBLE
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-I SECTION 15.6 SURFACE AREA 0 267
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Therefore E = { (x, y, z) I -2 ~ X~
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43. I,. = foL foL foL k(y2 + z2)dz.
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SECTION 15.8 TRIPLE INTEGRALS IN CY
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M xv = I~1f I: I:2 6 - 3 r 2 (zK) r
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(b) The wedge in question is the sh
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8 8 8 (b)clivF = 'V ·iF=- (x+yz) +
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that is, D = {( x, y) I x 2 + y 2 :
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CHAPTER 16 REVIEW 0 337 27. JI 5 cu
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CHAPTER 16 REVIEW 0 339 TRUE-FALSE
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CHAPTER 16 REVIEW D 341 Alternate s
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3.c6 0 CHAPTER 17 SECOND-ORDER DIFF
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0 APPENDIX Appendix H Complex Numbe
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APPENDIX H COMPLEX NUMBERS 0 361 43
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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