Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 15.5 THE CHAIN RULE ET SECTION 14.5 ¤ 189
19. w = f(r, s, t), r = r(x, y), s = s(x, y), t = t(x, y) ⇒
∂w
∂x = ∂w ∂r
∂r ∂x + ∂w ∂s
∂s ∂x + ∂w ∂t
∂t ∂x ,
∂w
∂y = ∂w ∂r
∂r ∂y + ∂w ∂s
∂s ∂y + ∂w ∂t
∂t ∂y
21. z = x 2 + xy 3 , x = uv 2 + w 3 , y = u + ve w ⇒
∂z
∂u = ∂z ∂x
∂x ∂u + ∂z ∂y
∂y ∂u =(2x + y3 )(v 2 )+(3xy 2 )(1),
∂z
∂v = ∂z ∂x
∂x ∂v + ∂z ∂y
∂y ∂v =(2x + y3 )(2uv)+(3xy 2 )(e w ),
∂z
∂w = ∂z ∂x
∂x ∂w + ∂z ∂y
∂y ∂w =(2x + y3 )(3w 2 )+(3xy 2 )(ve w ).
When u =2, v =1,andw =0,wehavex =2, y =3,
so ∂z
∂u = (31)(1) + (54)(1) = 85, ∂z
∂v = (31) (4) + (54)(1) = 178, ∂z
= (31)(0) + (54)(1) = 54.
∂w
23. R =ln(u 2 + v 2 + w 2 ), u = x +2y, v =2x − y, w =2xy ⇒
∂R
∂x = ∂R ∂u
∂u ∂x + ∂R ∂v
∂v ∂x + ∂R
∂w
=
2u +4v +4wy
u 2 + v 2 + w 2 ,
∂R
∂y = ∂R ∂u
∂u ∂y + ∂R ∂v
∂v ∂y + ∂R
∂w
=
4u − 2v +4wx
u 2 + v 2 + w 2 .
∂w
∂x =
∂w
∂y =
2u
u 2 + v 2 + w 2 (1) +
2u
u 2 + v 2 + w 2 (2) +
2v
u 2 + v 2 + w 2 (2) +
2v
u 2 + v 2 + w 2 (−1) +
When x = y =1we have u =3, v =1,andw =2,so ∂R
∂x = 9 ∂R
and
7 ∂y = 9 7 .
25. u = x 2 + yz, x = pr cos θ, y = pr sin θ, z = p + r ⇒
2w
u 2 + v 2 + w 2 (2y)
2w
u 2 + v 2 + w 2 (2x)
∂u
∂p = ∂u ∂x
∂x ∂p + ∂u ∂y
∂y ∂p + ∂u ∂z
=(2x)(r cos θ)+(z)(r sin θ)+(y)(1) = 2xr cos θ + zr sin θ + y,
∂z ∂p
∂u
∂r = ∂u ∂x
∂x ∂r + ∂u ∂y
∂y ∂r + ∂u ∂z
=(2x)(p cos θ)+(z)(p sin θ)+(y)(1) = 2xp cos θ + zp sin θ + y,
∂z ∂r
∂u
∂θ = ∂u ∂x
∂x ∂θ + ∂u ∂y
∂y ∂θ + ∂u ∂z
=(2x)(−pr sin θ)+(z)(pr cos θ)+(y)(0) = −2xpr sin θ + zpr cos θ.
∂z ∂θ
When p =2, r =3,andθ =0we have x =6, y =0,andz =5,so ∂u ∂u ∂u
=36, =24,and
∂p ∂r ∂θ =30.
27.
xy =1+x 2 y,soletF (x, y) =(xy) 1/2 − 1 − x 2 y =0.ThenbyEquation6
dy
dx = − Fx
F y
1
2
= −
(xy)−1/2 (y) − 2xy
1
2 (xy)−1/2 (x) − x = − y − 4xy xy
2 x − 2x 2 xy = 4(xy)3/2 − y
x − 2x 2 xy .
29. cos(x − y) =xe y ,soletF (x, y) =cos(x − y) − xe y =0.
Then dy
dx = −F x − sin(x − y) − ey sin(x − y)+ey
= − =
F y − sin(x − y)(−1) − xey sin(x − y) − xe . y