Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 15 REVIEW ET CHAPTER 14 ¤ 217
Thus
x 2 ∂ 2 z
∂x − ∂ 2 z
2 y2
∂y = 2y ∂z
2 x ∂v + x2 y 2 ∂ 2 z
∂u − 2 2y2
= 2y x
∂z
∂v − 4y2
since y = xv = uv
y or y2 = uv.
43. ∇f =
z 2 √ ye
x √y , xz2 e x√ y
2 √ y , 2zex√ y
∂ 2 z ∂z
=2v
∂u∂v ∂v − 4uv
= ze x√ y
z √ y,
∂ 2 z
∂u∂v + y2
x 2 ∂ 2 z
∂v 2 − x2 y 2 ∂ 2 z
∂u 2 − 2y2
xz
2 √ y , 2
∂2 z
∂u∂v
45. ∇f = h1/ √ x, −2yi, ∇f(1, 5) = h1, −10i, u = 1 5 h3, −4i. ThenD u f(1, 5) = 43
5 .
47. ∇f = 2xy, x 2 +1/ 2 √ y , |∇f(2, 1)| = 4, 9 2
direction 4, 9 2
.
∂ 2 z
∂u∂v − y2 ∂ 2 z
x 2 ∂v 2
. Thus the maximum rate of change of f at (2, 1) is
√
145
2
in the
49. First we draw a line passing through Homestead and the eye of the hurricane. We can approximate the directional derivative at
Homestead in the direction of the eye of the hurricane by the average rate of change of wind speed between the points where
this line intersects the contour lines closest to Homestead. In the direction of the eye of the hurricane, the wind speed changes
from 45 to 50 knots. We estimate the distance between these two points to be approximately 8 miles, so the rate of change of
wind speed in the direction given is approximately
50 − 45
8
= 5 8
=0.625 knot/mi.
51. f(x, y) =x 2 − xy + y 2 +9x − 6y +10 ⇒ f x =2x − y +9,
f y = −x +2y − 6, f xx =2=f yy , f xy = −1. Thenf x =0and f y =0
imply y =1, x = −4. Thus the only critical point is (−4, 1) and
f xx(−4, 1) > 0, D(−4, 1) = 3 > 0,sof(−4, 1) = −11 is a local minimum.
53. f(x, y) =3xy − x 2 y − xy 2 ⇒ f x =3y − 2xy − y 2 ,
f y =3x − x 2 − 2xy, f xx = −2y, f yy = −2x, f xy =3− 2x − 2y. Then
f x =0implies y(3 − 2x − y) =0so y =0or y =3− 2x. Substituting into
f y =0implies x(3 − x) =0or 3x(−1+x) =0. Hence the critical points are
(0, 0), (3, 0), (0, 3) and (1, 1). D(0, 0) = D(3, 0) = D(0, 3) = −9 < 0 so
(0, 0), (3, 0),and(0, 3) are saddle points. D(1, 1) = 3 > 0 and
f xx (1, 1) = −2 < 0,sof(1, 1) = 1 is a local maximum.
55. First solve inside D. Heref x =4y 2 − 2xy 2 − y 3 , f y =8xy − 2x 2 y − 3xy 2 .
Then f x =0implies y =0or y =4− 2x,buty =0isn’t inside D. Substituting
y =4− 2x into f y =0implies x =0, x =2or x =1,butx =0isn’t inside D,
and when x =2, y =0but (2, 0) isn’t inside D. Thus the only critical point inside
D is (1, 2) and f(1, 2) = 4. Secondly we consider the boundary of D.
On L 1 : f(x, 0) = 0 and so f =0on L 1 .OnL 2 : x = −y +6and