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

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
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