Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 0 215
17. h(r, s, t) = ln(37· + 6s + 9t) => 'ilh(r, s, t) = (3/(3r + 6s + 9t), 6/(3r + 6s + 9t), 9/(3r + 6s + 9t)),
'ilh(1, 1, 1) = (k, ft, ~ ), and a unit vector in the direction ofv = 4 i + 12j + 6 k
is u = .; 16
+; 4 4+3& (4i + 12j + 6k) = ~ i .+ ~ j + ~ k, so
( ) ' 'il f(x, y) = (Hxy)- 1 1 2 (y), Hxy)- 112 (x)) = / yt=' ~) ·so 'il!(2, 8) = (1, ~
--
).
\2 yxy 2 yxy
. .
The unit vector in the direction of PQ = (5 - 2, 4 - 8) = (3, -4) is u = ( ~, - ~ ), so
Duf(2,8)='il/(2,8)·
.
u=
(
1,4)·
1 (35•-6")- -5.
2
21. f(~,y) = 4yy'X => 'ilf(x,y) = ( 4y · ~x - 1 1 2 ,4v'x) = (2yjy'x,4.,fi).'
· 'il f(4, 1) = {1, 8) is.the direction of maximum rat~ ofch~ge, and the maximum rate is·/'il /(4, 1)/ = v'1 + 64 = v'65.
23. J(x, y) = sin(xy) => 'il f(x, y) = (y cos(xy), x cos(xy)), 'il!(1, 0) = (0, 1). Thus the maximum rate of change is
/'il f(1, 0)/ = 1 in the direction {0, 1).
25. f(x,y,z) = Jx2 + y2 + .z2 =>
'il f(3, 6, -2) = ( -7,w. ~· ~) = ( ~. ¥, -¥ ). Th~ s the maximum rate ofchange is
27. (a) As in the proof of Theorem 15, Du f = /'il!I cos 0. Since the minimum value of cos(} is - 1 occurring when(} = 11', the
minimum value of Du f is - /'il f/ occurring when(} = 11', that is when u is in the opposite direction of 'il f.
(assuming \7 f "f 0).
(b) f(x, y) = x 4 y - x 2 y 3 => 'il f(x, y) = ( 4x 3 y - 2xy 3 , x 4 - 3x 2 y 2 ), so f decreases fastest at the point (2, -3) in the
direction-'il /(2, - 3) = - {12, -92) = (-12, 92).
29. The directionoffastestchange is 'ilf(x,y) = (2x- 2) i + (2y- 4)j , so we need to find all points (x,y) where 'ilf(x,y) is
parallel to i + j ¢'> (2x - 2) i + (2y - 4)j = k (i + j) ¢'> k = 2x- 2 and k = 2y - 4. Then 2x- 2 = 2y- 4 =>
y = x + 1, so the direction of fastest change is i + j at all points on the line y = x + 1.
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