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

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SECTION 14.5 THE CHAIN RULE D 211 (b) S = 2(£w + £h + wh), so by the Chain Rule, ~~ = ~~: + ~! :~ + ~~ ~~ = 2(w +h):+ 2(£ + h):~ + 2(£+ w) ~~ = 2(2 + 2)2 + 2(1 + 2)2 + 2(1 + 2)( - 3) = 10 m 2 fs (c) L 2 = £ 2 + w 2 + h 2 ::::? 2£ ~~ = 2f. ~ + 2w :~ + 2h ~: = 2(1)(2) + 2(2)(2) + 2(2)(-3) = 0 ::::? dL/ dt = 0 mjs. dP dT T dV 8.31 dT T dP 41. dt = 0.05, dt = 0.15, V = 8.31? and dt = Pdt - 8.31 p 2 Yt· Thus when P = 20 andT = 320, dV = 8 31 [ 0.15 _ (0.05)(320)] ~ _ 0 . 27 Lf s. dt . 20 400 43. Let x be the length of the first side of the triangle andy the lengtl1 of the second side. The area A of the triangle is given by A= ~xy sin B where 0 is the angle between the two sides. Thus A is a function of x, y, and (}, and x, y, and 0 are each in tum functions of timet. We are given that ~~ = 3, ~~ ·= -2, and because A is constant, :~ = 0. By the Chain Rule, dA 8A dx 8A dy 8A dO dt = ax dt + ay dt + ao dt dA 1 . 0 dx 1 . 0 dy 1 d8 ::::? dt = 2YSI.Il · dt + 2xsrn · dt + 2xycos(} · dt. When x = 20, y = 30, and 0 = 1r /6 we have 0 = ~(30)(sin ~)(3) + ~(20)(sin ~)( - 2) + ~(20)(30)(cos ~) ~~ · = 45 · l - 20 · l + 300 · vf3 · dO = ll + 150 '3 d(} 2 2 2 dt 2 v,) dt S o I vmg . .or ., -d dO g1ves . -d dO = ---;;; - 25/2 = -------:r.i· 1 sot h e ang I e be tween t h e s1 ·d es 1s · d ecreasmg · at a rate o f t t 150 v 3 . 12 v 3 1/ (12 V3) ~ 0.048 radj s. ( 45. (a) By the Chain Rule, : = ~= cos(} + ~; sin B, ~; = ~= - r sin B) + ~~ r cos 0. ( {Jz)2 ({)z ) 2 2 {)z {)z . ( {)z ) 2 • 2 (b) or = ax cos 0 +.2 8 x 8 y cosO ~mO + By sm 0, ( 8z ) 2 2 1 (az ) 2 [(az ) 2 (az) ] 2 (az ) 2 (oz ) 2 2 8r + r2 80 = 8x + 8y (cos 0 + sm . 0) = 8x + 8y . 8z dz 8u dz 8z dz 8z 8z 47. Letu = x - y. Then - 8 = d-- 8 = -d and -a = -d (-1). Thus -a + - 8 = 0. X UX U y U X y ® 2012 Ccngogc lc:uning. All Rights Res
212 D CHAPTER 14 PARTIAL DERIVATIVES 49. Let u =X + at, v =X - at. Then z = f(u) + g(v), so azjau ~ f'(u) and azjav ~ g 1 (v) . az f}z aU az av 1 1 Thus at = au at + av at = af (u)- ag (v) and 8 2 z = 2._ [/1( ) - I( )] = (df'(u) 8u - dg 1 (v ) au ) = 2/"( ) 2 " ( ) [)t2 a at u g v a du 8t dv 8t a u + a g v . S .. , 8z fl( ) I( ) 8 2 z !"( ) "( )' 8 2 z 2 8 2 z tmt arly .,-- = u + g v and
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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 .~ POLAR COORDINATES 0 1
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SECTION 10.4 A~~S AND LENGTHS IN PO
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SECTION 10.4 AREAS AND LENGTHS IN P
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SECTION 10.4 AREAS AND LENGTHS IN P
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SECTION 10.5 CONIC SECTIONS 0 27 5.
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SECTION 10.5 CONIC SECTIONS 0 29 35
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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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CHAPTER 10 REVIEW 0 41 2 2 . 45. ~
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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 11.2 SERIES 0 53 13. n s.,
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SECTION 11.2 SERIES 0 55 47. F~r th
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. . (1+ c)- 2 scn es and set 1t equ
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SECTION 11.3 THE INTEGRAL TEST AND
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, SECTION 11.3 THE INTEGRAL TEST AN
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SECTION 11.4 THE COMPARISON TESTS D
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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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SECTION 11.6 ABSOLUTE CONVERGENCE A
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17 lim I an+l I= SECTION 11.7 STRAT
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SECTION 11.8 POWER SERIES 0 75 l 00
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SECTION 11.8 POWER SERIES 0 n (b) I
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SECTION 11.9 REPRESENTATIONS OF FUN
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SECTION 11.9 REPRESENTATIONS OF FUN
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SECTION 11.10 TAYLOR AND MACLAURIN
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61 _ x_ x · sin x - x- tx 3 + 1~
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SECTION 11.11 APPLICATIONS OF TAYLO
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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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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 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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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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298 D CHAPTER 15 PROBLEMS PLUS To e
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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 0 307 (
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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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SECTION 16.4 GREEN'S THEOREM 0 315
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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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348 0 CHAPTER 17 SECOND-ORDER DIFFE
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352 D CHAPTER 17 SECOND-ORDER DIFFE
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356 D CHAPTER 17 SECOND-ORDER DIFFE
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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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