Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 15.5 THE CHAIN RULE ET SECTION 14.5 ¤ 191
turn functions of time t. We are given that dx
dt
dA
dt = ∂A dx
∂x dt + ∂A dy
∂y dt + ∂A dθ
∂θ dt
and θ = π/6 we have
⇒
0= 1 2 (30) sin π 6
=3,
dy
dt
= −2, and because A is constant,
dA
dt
=0. By the Chain Rule,
dA
dt = 1 dx
2
y sin θ ·
dt + 1 dy
2
x sin θ ·
dt + 1 dθ
2
xy cos θ · .Whenx =20, y =30,
dt
(3) +
1
(20)
2
sin π 6 (−2) +
1
(20)(30)
2
cos π dθ
6
dt
=45· 1
2 − 20 · 1
2 + 300 · √
3
2 · dθ
dt = 25 2
+150√ 3 dθ
dt
Solving for dθ dθ
gives
dt dt = −25/2
150 √ 3 = − 1
12 √ , so the angle between the sides is decreasing at a rate of
3
1/ 12 √ 3 ≈ 0.048 rad/s.
45. (a)BytheChainRule, ∂z
∂r = ∂z ∂z
cos θ +
∂x ∂y sin θ, ∂z
∂θ = ∂z
∂z
(−r sin θ)+ r cos θ.
∂x ∂y
2 2 ∂z ∂z
(b) = cos 2 θ +2 ∂z
2
∂z
∂z
∂r ∂x
∂x ∂y cos θ sin θ + sin 2 θ,
∂y
2 2 ∂z ∂z
= r 2 sin 2 θ − 2 ∂z ∂z
∂θ ∂x
∂x ∂y r2 cos θ sin θ +
2 ∂z
+ 1
2 2
2
∂z ∂z ∂z
= + (cos 2 θ +sin 2 θ)=
∂r r 2 ∂θ ∂x ∂y
47. Let u = x − y. Then ∂z
∂x = dz ∂u
du ∂x = dz
du
2 ∂z
r 2 cos 2 θ.Thus
∂y
2 ∂z
+
∂x
∂z
and
∂y = dz (−1). Thus∂z
du ∂x + ∂z
∂y =0.
2 ∂z
.
∂y
49. Let u = x + at, v = x − at. Thenz = f(u)+g(v),so∂z/∂u = f 0 (u) and ∂z/∂v = g 0 (v).
51.
Thus ∂z
∂t = ∂z ∂u
∂u ∂t + ∂z ∂v
∂v ∂t = af 0 (u) − ag 0 (v) and
∂ 2 z
∂t 2
= a ∂ df
∂t [f 0 (u) − g 0 0 (u) ∂u
(v)] = a
du ∂t − dg0 (v)
dv
∂v
= a 2 f 00 (u)+a 2 g 00 (v).
∂t
Similarly ∂z
∂x = f 0 (u)+g 0 (v) and ∂2 z
∂x = f 00 (u)+g 00 (v). Thus ∂2 z
2 ∂t = ∂ 2 z
2 a2
∂x . 2
∂z
∂s = ∂z ∂z
2s + 2r. Then
∂x ∂y
∂ 2 z
∂r ∂s = ∂ ∂r
∂z
∂x 2s + ∂ ∂z
∂r ∂y 2r
= ∂2 z ∂x
∂x 2 ∂r 2s + ∂ ∂z ∂y ∂z ∂
2s +
∂y ∂x ∂r ∂x ∂r 2s + ∂2 z ∂y
∂y 2 ∂r 2r + ∂ ∂z ∂x ∂z
2r +
∂x ∂y ∂r ∂y 2
=4rs ∂2 z
∂x 2 +
By the continuity of the partials,
∂2 z
∂y ∂x 4s2 +0+4rs ∂2 z
∂y + 2
∂2 z
∂x∂y 4r2 +2 ∂z
∂y
∂ 2 z
∂r∂s =4rs ∂2 z
∂x 2 +4rs ∂2 z
∂y 2 +(4r2 +4s 2 ) ∂2 z
∂x∂y +2∂z ∂y .