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

240 0 CHAPTER 14 PARTIAL DERIVATIVES
linear approximation off at (2, 3, 4) is
f(x;y,z) ~ /(2,3,4) + J.,(2,3,4)(x ~ 2) + fv(2,3,4)(y - 3) + /z(2,3, 4)(z - 4)
= 40 +60(x- 2) + ¥CY - 3) + 3 2 5
(z- 4) = 60x+ 2 4
y+ 3 2
z - 120
5 5
Then (1.98) 3 J(3.01) 2 + (3.97) 2 = /(1.98,3.01, 3.97) ~ 60(1.98) + ¥(3.01) + ¥(3.97) -120 = 38.656.
du au dx au dy au dz 3 2 2 3
35. -d = ,---d +-;;--d +-;;--d =2xy (1+6p) + 3x y (per>+er>)+4z (pcosp+sinp)
p uX p uy p u z p
. az az ax az ay
37. By the Cham Rule, as = ax as + ay as. When s = 1 and t = 2, X = g(l, 2) =;:: 3 andy = h(1, 2) = 6, so
& . . & &~ &~
as = f,(3, 6)gs(1, 2) + fv (3, 6) h. (1, 2) = (7)( - 1) + (8)( -5) = - 47. Stmiiarly, Bt = ax 8t + ay at, so
az
at = f,(3, 6)gt(1, 2) + fv (3, 6) ht(1, 2) = (7)(4) + (8)(10) = 108.
az !'( 2 2) az !'( 2 2) [ h !' df ]
39. ax = 2x X - y ' ay = 1 - 2y X - y w ere = d( x2 - y2) . Then
az az 2 2 I 2 2
Y Bx + x {)y = 2xyf'(x - y ) + x - 2xyf (x - y ) = x.
az az az - y
41. -a = -a v+ -a 2 and
X U V X
a 2 z a (az). 2y az - y a (az) 2y az (a 2 z 8 2 z - y) - y (a 2 z - y a 2 z )
ax2 = Yax au +xa8v + ~ax av =xaav+Y au2 ~ + 8v8u~ +~ 8v2 ~+auavY
2y az 2 8 2 z 2y 2 a 2 z y 2 8 2 z
= x3 8v + Y &u2 - X2 &uav + x 4 av2
Thus
. uv 2
smcey = xv = - ory = uv.
y
® 2012 Ccngage l.e:lmin11- All Righi$ Re>en·ed. Moy not be scanned, copied, or duplicaled, or poSied 10 a publicly accessible wcbsiiC, in whole or in~"~" ·