Solução_Calculo_Stewart_6e

F.
192 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
53.
∂z
∂r = ∂z ∂z ∂z
cos θ + sin θ and
∂x ∂y ∂θ = − ∂z ∂z
r sin θ + r cos θ. Then
∂x ∂y
and
∂ 2 z
∂r 2 =cosθ ∂ 2 z
∂x 2 cos θ +
∂2 z
∂y ∂x sin θ +sinθ
=cos 2 θ ∂2 z
+2cosθ sin θ
∂2 z
∂x2 ∂x∂y +sin2 θ ∂2 z
∂y 2
∂ 2 z
∂z
∂ 2
2
= −r cos θ
∂θ ∂x +(−r sin θ) z
∂x
2
(−r sin θ)+
∂2
∂ 2 z
∂y 2 sin θ +
z
∂y ∂x r cos θ
∂2 z
∂x∂y cos θ
−r sin θ ∂z
∂ 2
∂y + r cos θ z
∂y r cos θ +
∂2 z
2 ∂x∂y (−r sin θ)
= −r cos θ ∂z ∂z
− r sin θ
∂x ∂y + r2 sin 2 θ ∂2 z
∂x − 2 2r2 cos θ sin θ
∂2 z
∂x∂y + r2 cos 2 θ ∂2 z
∂y 2
Thus
∂ 2 z
∂r + 1 ∂ 2 z
2 r 2 ∂θ 2 + 1 ∂z
r ∂r =(cos2 θ +sin 2 θ) ∂2 z
− 1 r
= ∂2 z
∂x + ∂2 z
as desired.
2 ∂y2 ∂x + sin 2 θ +cos 2 θ ∂ 2 z
2 ∂y 2
∂z
cos θ
∂x − 1 ∂z
sin θ
r ∂y + 1 r
cos θ ∂z
∂x
∂z
+sinθ
∂y
55. (a) Since f is a polynomial, it has continuous second-order partial derivatives, and
f(tx, ty) =(tx) 2 (ty)+2(tx)(ty) 2 +5(ty) 3 = t 3 x 2 y +2t 3 xy 2 +5t 3 y 3 = t 3 (x 2 y +2xy 2 +5y 3 )=t 3 f (x, y).
Thus, f is homogeneous of degree 3.
(b) Differentiating both sides of f(tx, ty) =t n f(x, y) with respect to t using the Chain Rule, we get
∂
∂t f(tx, ty) = ∂ ∂t [tn f(x, y)]
⇔
∂
∂(tx)
f(tx, ty) · + ∂
∂(ty)
f(tx, ty) · = x ∂
∂
f(tx, ty)+y
∂(tx) ∂t ∂(ty) ∂t ∂(tx) ∂(ty) f(tx, ty) =ntn−1 f(x, y).
Setting t =1: x ∂
∂x f(x, y)+y ∂ f(x, y) =nf(x, y).
∂y
57. Differentiating both sides of f(tx, ty) =t n f(x, y) with respect to x using the Chain Rule, we get
∂
∂
f(tx, ty) =
∂x ∂x [tn f(x, y)]
⇔
∂
∂ (tx)
f(tx, ty) ·
∂ (tx) ∂x + ∂
∂ (ty)
f(tx, ty) ·
∂ (ty) ∂x = ∂
tn
∂x f(x, y) ⇔ tfx(tx, ty) =tn f x(x, y).
Thus f x(tx, ty) =t n−1 f x(x, y).
15.6 Directional Derivatives and the Gradient Vector ET 14.6
1. We can approximate the directional derivative of the pressure function at K in the direction of S by the average rate of change
of pressure between the points where the red line intersects the contour lines closest to K (extendtheredlineslightlyatthe
left). In the direction of S, the pressure changes from 1000 millibars to 996 millibars and we estimate the distance between
these two points to be approximately 50 km (using the fact that the distance from K to S is 300 km). Then the rate of change of
pressure in the direction given is approximately
996 − 1000
50
= −0.08 millibar/km.