Solução_Calculo_Stewart_6e

F.
188 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
5. w = xe y/z , x = t 2 , y =1− t, z =1+2t ⇒
dw
dt = ∂w dx
∂x dt + ∂w dy
∂y dt + ∂w
dz
1
∂z dt = ey/z · 2t + xe y/z · (−1) + xe y/z − y
· 2=e y/z 2t − x z
z 2 z − 2xy
z 2
7. z = x 2 y 3 , x = s cos t, y = s sin t ⇒
9. z =sinθ cos φ, θ = st 2 , φ = s 2 t ⇒
∂z
∂s = ∂z ∂x
∂x ∂s + ∂z ∂y
∂y ∂s =2xy3 cos t +3x 2 y 2 sin t
∂z
∂t = ∂z ∂x
∂x ∂t + ∂z ∂y
∂y ∂t =(2xy3 )(−s sin t)+(3x 2 y 2 )(s cos t) =−2sxy 3 sin t +3sx 2 y 2 cos t
∂z
∂s = ∂z ∂θ
∂θ ∂s + ∂z ∂φ
∂φ ∂s =(cosθ cos φ)(t2 )+(− sin θ sin φ)(2st) =t 2 cos θ cos φ − 2st sin θ sin φ
∂z
∂t = ∂z ∂θ
∂θ ∂t + ∂z ∂φ
∂φ ∂t =(cosθ cos φ)(2st)+(− sin θ sin φ)(s2 )=2st cos θ cos φ − s 2 sin θ sin φ
11. z = e r cos θ, r = st, θ = √ s 2 + t 2 ⇒
∂z
∂s = ∂z ∂r
∂r ∂s + ∂z ∂θ
∂θ ∂s = er cos θ · t + e r (− sin θ) · 1
2 (s2 + t 2 ) −1/2 (2s) =te r cos θ − e r sin θ ·
= e r t cos θ −
s
√
s2 + t sin θ 2
∂z
∂t = ∂z ∂r
∂r ∂t + ∂z ∂θ
∂θ ∂t = er cos θ · s + e r (− sin θ) · 1
2 (s2 + t 2 ) −1/2 (2t) =se r cos θ − e r sin θ ·
= e r s cos θ −
t
√
s2 + t sin θ 2
s
√
s2 + t 2
t
√
s2 + t 2
13. When t =3, x = g(3) = 2 and y = h(3) = 7. BytheChainRule(2),
dz
dt = ∂f dx
∂x dt + ∂f dy
∂y dt = fx(2, 7)g0 (3) + f y(2, 7) h 0 (3) = (6)(5) + (−8)(−4) = 62.
15. g(u, v) =f(x(u, v),y(u, v)) where x = e u +sinv, y = e u +cosv ⇒
∂x
∂u = eu ,
∂x
∂v =cosv, ∂y
∂u = eu ,
∂y
∂g
= − sin v.BytheChainRule(3),
∂v ∂u = ∂f
∂x
∂x
∂u + ∂f
∂y
∂y
∂u .Then
g u(0, 0) = f x(x(0, 0),y(0, 0)) x u(0, 0) + f y(x(0, 0),y(0, 0)) y u(0, 0) = f x(1, 2)(e 0 )+f y(1, 2)(e 0 ) = 2(1) + 5(1) = 7.
Similarly, ∂g
∂v = ∂f ∂x
∂x ∂v + ∂f ∂y
∂y ∂v .Then
g v(0, 0) = f x(x(0, 0),y(0, 0)) x v(0, 0) + f y(x(0, 0),y(0, 0)) y v(0, 0) = f x(1, 2)(cos 0) + f y(1, 2)(− sin 0)
=2(1)+5(0)=2
17. u = f(x, y), x = x(r, s, t), y = y(r, s, t) ⇒
∂u
∂r = ∂u ∂x
∂x ∂r + ∂u ∂y
∂y ∂r , ∂u
∂s = ∂u
∂x
∂x
∂s + ∂u
∂y
∂y
∂s , ∂u
∂t = ∂u ∂x
∂x ∂t + ∂u
∂y
∂y
∂t