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

206 0 CHAPTER 14 PARTIAL DERIVATIVES
Thus when T = 95 and H = 78, !(95, 78) ~ 127 + 4(95- 94) + 1(78 - 80) = 129, so we estimate the heat index to be
approximately 129°F.
25. z = e- 2 "' cos 27rt =>
az f)z -2 ( ) . -2 . 2 2 .
dz = -a dx + - 0
dt = e "' -2 cos 21rt dx + e "' (- sm2?rt)(27r) dt = -2e- "'cos 27rt dx- 21re- x sin 21rt dt
. X t
"3 am om 43 "2
ap oq
27. m = p 0 q · => dm = - dp + - dq = 5p q dp + 3pa q dq
29. R = o:(3 2 cos 1 =>
aR oR oR 2 2 .
dR = oo: do: + 0
(3 d(3 + O"f d"f = (3 cos '"Y do: + 2o:(3 cos 1 d(3 - o:(3 sm 1 d1
• 2 2
31. dx = ll.x = 0.05, dy = ll.y = 0.1, z = 5x + y , z., = lOx, Zy = 2y . . Thus when x. = 1 and y = 2,
dz = zx(1, 2) dx + z 11 (1, 2) dy = (10)(0.05) + (4)(0.1) = 0.9 while
ll.z = ! ,(1.05, 2.1)- /(1, 2) = 5(1.05? + (2.1) 2 - 5 -4 = 0.9225.
8A 8A .
33. dA = ox dx + oy dy = y dx + x dy and Jll.xJ :::; 0.1, Jll.yJ :::; ~.1. We use dx = 0.1, dy = 0.1 w1th x = 30, y = 24; then
the maximum error in the area is about dA = 24(0.1) + 30(0.1) = 5.4 cm 2 •
35. The volume of a can is V = 1rr 2 h and ll. V ~ dV is an estimate of the amount of tin. Here dV = 27rr h dr + 1rr~ dh, so put .
dr = 0.04, dh = 0.08 (0.04 on top, 0.04 on bottom) and then ll. V ~ dV = 27r( 48)(0.04) + 7r(16)(0.08) ~ 16.08 cm 3 .
Thus the amount of tin is about 16 cm 3 .
37. T = ";'gR 2
, so the differential ofT is
2r +R
dT = aT dR
8T d = (2r 2 + R 2 )(mg) - mgR(2R) dR + (2r 2 +R 2 )(0) - mgR(4r) d
oR + or . r (2r 2 + R 2 ) 2 (2r2 + R2)2 r
mg(2r 2 - R 2 ) 4mgRr
= (2r2 + R2)2 dR- (2r2 + R2)2 dr
Here we have ll.R = 0.1 and ll.r = 0.1, so we take dR = 0.1, dr = 0.1 with R = 3, r = 0.7. Then the change in the
tension T is approximately
mg[2(0.7?- (3) 2 ]
4mg(3)(0.7)
dT = (2(0.7)2 + (3) 2]2 (O.l) - (2(0.7) 2 + (3)2)2 (0. 1 )
0.802mg 0.84mg 1.642
(9.98)2 - (9.98) 2 = - 99.6004 mg ~ - 0.0 165 mg
Because the change is negative, tension· decreases.
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