Solução_Calculo_Stewart_6e

F.
194 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
17. g(x, y, z) =(x +2y +3z) 3/2 ⇒
3
∇g(x, y, z) = (x +2y 2 +3z)1/2 (1), 3 (x +2y 2 +3z)1/2 (2), 3 (x +2y 2 +3z)1/2 (3)
= 3
√ √
2 x +2y +3z, 3 x +2y +3z,
9
√
2 x +2y +3z , ∇g(1, 1, 2) = 9 , 9, 27
2 2 ,
and a unit vector in the direction of v =2j − k is u = 2 √
5
j − 1 √
5
k,so
D u g(1, 1, 2) = 9
, 9, 27
2 2 ·
2
0, √
5
, − √ 1 5
= √ 18
5
− 27
2 √ = 9
5 2 √ . 5
19. f(x, y) = xy ⇒ ∇f(x, y) =
1
2 (xy)−1/2 (y), 1 y
2 (xy)−1/2 (x) =
2 xy , x
2 ,so∇f(2, 8) =
1, 1 4 .
xy
The unit vector in the direction of −→
PQ = h5 − 2, 4 − 8i = h3, −4i is u =
3 , − 4
5 5 ,so
D u f(2, 8) = ∇f(2, 8) · u =
1, 1 4 · 3 , − 4
5 5 =
2
. 5
21. f(x, y) =y 2 /x = y 2 x −1 ⇒ ∇f(x, y) = −y 2 x −2 , 2yx −1 = −y 2 /x 2 , 2y/x .
∇f(2, 4) = h−4, 4i, or equivalently h−1, 1i, is the direction of maximum rate of change, and the maximum rate
is |∇f(2, 4)| = √ 16 + 16 = 4 √ 2.
23. f(x, y) =sin(xy) ⇒ ∇f(x, y) =hy cos(xy),xcos(xy)i, ∇f(1, 0) = h0, 1i. Thus the maximum rate of change is
|∇f(1, 0)| =1in the direction h0, 1i.
25. f(x, y, z) = x 2 + y 2 + z 2 ⇒
1
∇f(x, y, z)=
2 (x2 + y 2 + z 2 ) −1/2 · 2x, 1 2 (x2 + y 2 + z 2 ) −1/2 · 2y, 1 2 (x2 + y 2 + z 2 ) −1/2 · 2z
x
=
x2 + y 2 + z , y
2 x2 + y 2 + z , z
,
2 x2 + y 2 + z 2
∇f(3, 6, −2) =
√
3
49
,
|∇f(3, 6, −2)| =
3
7
√
6
49
, √ −2
49
= 3
, 6 , − 2
7 7 7 . Thus the maximum rate of change is
2
+
6 2
+
− 2 2 9+36+4
=
=1in the direction 3
, 6 , − 2
7 7 49 7 7 7 or equivalently h3, 6, −2i.
27. (a) As in the proof of Theorem 15, D u f = |∇f| cos θ. Since the minimum value of cos θ is −1 occurring when θ = π,the
minimum value of D u f is − |∇f| occurring when θ = π, that is when u is in the opposite direction of ∇f
(assuming ∇f 6=0).
(b) f(x, y) =x 4 y − x 2 y 3 ⇒ ∇f(x, y) = 4x 3 y − 2xy 3 ,x 4 − 3x 2 y 2 ,sof decreases fastest at the point (2, −3) in the
direction −∇f(2, −3) = − h12, −92i = h−12, 92i.
29. The direction of fastest change is ∇f(x, y) =(2x − 2) i +(2y − 4) j,soweneedtofind all points (x, y) where ∇f(x, y) is
parallel to i + j ⇔ (2x − 2) i +(2y − 4) j = k (i + j) ⇔ k =2x − 2 and k =2y − 4. Then2x − 2=2y − 4 ⇒
y = x +1, so the direction of fastest change is i + j at all points on the line y = x +1.