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

200 0 CHAPTER 14 PARTIAL DERIVATIVES
1 (1)(1-xy)-(x+y)(-x) _ 1+x 2 1 +x 2 1
zu = 1 + ( x+u )2 . (1- xy)2 (1- xy)2 +(x + y)2 (1 + x2)(1 + y2) = 1 + y2 ·
1-xy
2x 2 -2
( 1
+ x 2 )2, Zxy = 0, Zyx = 0, Zyy = -(1 + y ) . 2y =
2y
Thus U:cy = Uy:c.
61. u = cos(x 2 y) => U:c = -sin(x 2 y) · 2xy = -2xysin(x 2 y),
Uxy = -2xy · cos(x 2 y) · x 2 +sin(x 2 y) · (-2x) = -2x 3 ycos(x 2 y)- 2xsin(x 2 y) and
uu = -sin(x 2 y) ·x 2 = - x 2 sin(x 2 y), 1t 11
x = - x 2 ·cos(x 2 y) ·2xy+sin(x 2 y) · (- 2x) = -2x 3 ycos(x 2 y) -2xsin(x 2 y).
Thus Uxy = Uux.
63. f(x, y) = x 4 y 2 - x 3 y => fx = 4x 3 y 2 - 3x 2 y, fxx = 12x 2 y 2 - 6xy, fxxx = 24xy 2 - 6y and'
f xv = 8x 3 y- 3x 2 , fxvx = 24x 2 y- 6x.
67. u=er 8 sin(J => :~ . =er 8 cosfJ+sinfJ·er 8 (r)=e'. 0 (cosO+rsinO),
;
2
;
0 = er8 (sinfJ) +(cosO+ rsinO) erO (0) = ero (sinO+ OcosB + rOsinO),
{)~:~O = ero (0 sin 0) +(sinO+ 0 cosO+ rOsinO) · e'' 8 (0) = Oero (2sin0 + Ocos 0 + rO sinO).
X OW 8 2 w .
69.w=y+2z=x(y+2z)- 1 => ax=(y+2z)- I, 8yax = -(y+2z)- 2 (1)=-(y+2z)- 2 ,
8 3 w . . 4 ow
{)z{)y{)x = -(-2)(y + 2z)- 3 (2) = 4(y + 2z)- 3 = (y + 2
z)3 and {)y = x(-1)(y + 2z)- 2 (1) = -x(y + 2z)- 2 ,
71. Assuming that the third partial derivatives off are continuous (easily verified), we can write fxzy = fyxz· Then
f(x, y, z) = :J;y 2 z 3 + arCSin ( X vz) =?
fu = 2xyz 3 + 0, fvx = 2yz 3 , and fyxz = 6yz 2 = fxzy·
73. By Definition 4, fx(3, 2) = lim !( 3 + h, 2 l-!( 3 , 2 ) which we can approximate by considering h = 0.5 and h = -0.5:
h-0
f ( 3 2
) ~ !(3.5, 2) - f(3, 2) _ 22.4 - 17.5 _ 9 8
f ( 3 2
) ~ !(2.5, 2) - f(3, 2) _ 10.2- 17.5 _ 14 6
A .
"' ' ~ 0.5 - 0.5 - · ' "' ' ~ -0.5 - -0.5 - · · veragmg
these values, we estimate f,.(3, 2) to be approximately 12.2. Similarly, f x(3, 2.2) = lim !( 3 + h , 2·2}-!( 3 , 2·2) which
_ h"""""+O L
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