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

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48 D CHAPTER 11 INFINITE SEQUENCES AND SERIES ' 59. From the graph, it appears that the sequence converges to t. As n--+ oo, 3/ n .. ... 2 +2 => 8 + 1/ n so lim an =~ · n -+oo 0 '----~--~--"--'----./ 21 61. 2.~-------------.. From the. graph, it appears 'that the sequence {an} ,;_ { ~ 2 ~o~: } is .. .. divergent, since it osci llates between 1 and -1 (approximately). To n2 prove this, suppose that {a,.} converges to L. If bn = --, 2 then 1+n {bn} converges to 1, and lim abn =!::. = L. But an = cosn, so n-+oo n 1 bn 63. - 2 0~-~-~~-----~-J IO lim abn does not exist. This contradiction shows that {an} diverges. -oon • From the graph, it appears that the sequence approaches 0. 1 · 3 · 5 · · · · · (2n- 1) 1 3 5 2n- 1 O < a,.= n = - · - · - · · .. ·-- (2n) 2n 2n 2n 2n 1 1 : at = 1060, a2 = 1123.60, a3 = 1191.02, a4 = 1262.48, and as = 1338.23. (b) lim a,. = 1000 lim (1.06)", so the sequence diverges by (9) with r = 1.06 > 1. n - oo n-+oo 67. (a) We are given that the init!al population is 5000, so Po = 5000. The number of catfish increases by 8% per month and is decreased by 300 per month, so Pt = Po + 8%Po - 300 = 1.08Po - 300, H = l.08P1 - 300, and so on. Thus, P,. = 1.08Pn-t - 300. (b) Using the recursive formula with Po = 5000, we get P 1 = 5100, P 2 = 520.8, P 3 = 5325 (rounding any portion of a catfish), P,. = 5451, P 5 = 5587, and P6 = 5734, which is lhe number of catfish in the pond after six monlhs. 69. If lrl ~ 1, then {r"} diverges by (9), so {nrn} diverges also, since lnr''l = n lr"l ~ lr''l· If lrl < 1 then = lim ___!__I ."' :::: 0, so lim nr" = 0, and hence { nr"} converges. :C-tOO :c-oo r-:Z: X-+00 - T T-% %-+00 - n T n-oo . lim xr"' = lim -=..._ :lb ~ ( In\ whenever lrl < 1. ® 2012 Ccngogc Lcoming. All Rights Rc:sen'Ctl. May not be S
SECTION 11.1 SEQUENCES D 49 71. Since {a,.} is a decreasing sequence, an > an+l for all n ~ 1. Because all of its terms lie between 5 and 8, {an} is a bounded sequence. By the Monotonic Sequence Theorem,'{ an} is convergent; that is, {an} has a limit L. L must be less than 8 since {an} is decreasing, so 5 $ L < 8. 1 ' d . . 1 1 1 " 73. a,. = 2 n + 3 IS ecreasmg smce an+1 = 2 (n + 1 ) + 3 = 2 n + 5 < 2 n + 3 = a1, tOr each n ~ 1. The sequence is bounded since 0
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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Page 7 and 8: viii o CONTENTS 12.4 The Cross Prod
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Page 11 and 12: SECTION 10.1 CURVES DEFINED BY PARA
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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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CHAPTER 11 PROBLEMS PLUS D 107 l 9.
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CHAPTER 11 PROBLEMS PLUS 0 109 x x
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112 0 CHAPTER 12 VECTORS AND THE GE
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120 D CHAPTER 12 VECTORS AND THE GE
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136 D CHA~TER 12 VECTORS AND THE GE
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D PROBLEMS PLUS 1. Since three-dime
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l CHAPTER 12 PROBLEMS PLUS 0 149 Eq
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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 13.1 VECTOR FUNCTIONS AND S
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SECTION 13.2 DERIVATIVES AND INTEGR
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SECTION 13.2 DERIVATIVES AND INTEGR
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SECTION 13.3 ARC LENGTH AND CURVATU
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SECTION 13.4 MOTION IN SPACE: VELOC
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2 SECTION 13.4 MOTION IN SPACE: VEL
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CHAPTER 13 REVIEW D 173 43. The tan
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5. f~(t 2 i +tcos 1rtj +sin 1rt k )
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CHAPTER 13 REVIEW 0 177 23. (a) r (
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180 0 CHAPTER 13 PROBLEMS PLUS (-2a
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182 0 CHAPTER 13 PROBLEMS PLUS vect
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184 0 CHAPTER 14 PARTIAL DERIVATIVE
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196 D CHAPTER 14 PARTIAL DERIVATIVE
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D PROBLEMS PLUS 1. The areas of the
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15 D MULTIPLE INTEGRALS 15.1 Double
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SECTION 15.2 ITERATED INTEGRALS D 2
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l SECTION 15.3 DOUBLE INTEGRALS OVE
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SECTION 15.3 DOUBLE INTEGRALS OVER
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SECTION 15.3 DOUBLE INTEGRALS OVER
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61 . Since m :S j(x, y) :S M, ffD m
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SECTION 15.4 DOUBLE INTEGRALS IN PO
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SECTION 15.5 APPLICATIONS OF DOUBLE
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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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SECTION 15.7 TRIPLE INTEGRALS 0 269
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SECTION 15.7 TRIPLE INTEGRALS D 271
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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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SECTION 15.9 TRIPLE INTEGRALS IN SP
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(b) The wedge in question is the sh
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SECTION 15.10 CHANGE OF VARIABLES I
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CHAPTER 15 REVIEW 0 289 15 Review C
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CHAPTER 15 REVIEW 0 291 l 9. The vo
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CHAPTER 15 REVIEW D 293 33. Using t
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49. Since u = x- y and v = x + y, x
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300 0 CHAPTER 15 PROBLEMS PLUS 13.
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16 0 VECTOR CALCULUS 16.1 Vector Fi
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SECTION 16.2 LINE INTEGRALS 0 305 2
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SECTION 16.2 LINE INTEGRALS D 309 3
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SECTION 16.3 THE FUNDAMENTAL THEORE
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SECTION 16.4 GREEN'S THEOREM D 313
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8 8 8 (b)clivF = 'V ·iF=- (x+yz) +
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SECTION 16.5 CURL AND DIVERGENCE 0
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SECTION 16.6 PARAMETRIC SURFACES AN
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that is, D = {( x, y) I x 2 + y 2 :
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SECTION 16.7 SURFACE INTEGRALS 0 32
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SECTION 16.8 STOKES' THEOREM 0 333
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dS SECTION 16.9 THE DIVERGENCE THEO
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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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344 0 CHAPTER 16 PROBLEMS PLUS Simi
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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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