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

214 0 CHAPTER 14 PARTIAL DERIVATIVES
using the values given in the table·: h (-20, 30) ~ f ( - 15 • 30 ) ~ · f (- 20 • 30 ) = - 26 -
5
( - 33) = 1.4,
!( -25 30) - f( -20 30) -39 - ( - 33)
fr( -20, 30) ~ ' _ 5
' = _ 5
= 1.2. Averaging these values gives fr( - 20, 30) ~ 1.3.
Similarly, f,( -20, 30) = lim /( - 20 • 30 + h~- !( - 20 • JO), so we can approximate f,( --120, 30) with h = ±10:
h-0
f (-20 30) ~ !(- 20, 40) - J( - 20, 30) = -34 - (- 33) = - 0 1
1J , 10 10 . '
!( - 20 20) - J( -20 30) - 30- ( - 33)
f,( - 20, 30) ~ ' _ 10
' = _ 10
= -0.3. Averaging these values gives f,( -20, 30) ~ -0.2.
Then Duf(- 20,30) ~ 1.3(-}z) + (-0.2>(-}z) ~ 0.778.
5. f(x, y) = ye- "' :::> fx (x, y) = -ye_., and fu(x, y) = e-"'. If u is a unit vector in the direction of 8 = 21r / 3, then
from Equation 6, Du /(0, 4) = f x (0, 4) co~ en + fv(O, 4) sin en = -4 · (- ~) + 1 · {1 = 2 + 4.
7. f(x, y) = sin(2x + 3y).
• I
(a) 'il f (x, y) = . 8
8 1 i + ~! j = [cos(2x + 3y) · 2] i + [cos(2x + 3y) · 3]j = 2 cos {2x + 3y) i + 3 cos (2x + 3y)j
X ~J •
(b) 'il f( -6,4) = (2cos0) i + (3cosO)j = 2 i + 3j
(c) By Equation 9, Du f( - 6, 4) = 'il f( -6, 4) · u = (2 i + 3j ) · H v'3 i - j ) = ~ (2v'3- 3) = v'3 - ~·
9. f(x, y, z) = x 2 yz- xyz 3
(a) 'il f(x, y, z) = (f,. (x, y, z), fv(x, y, z), f:(x, y, z)) = (2xyz - yz 3 , x 2 z- xz 3 , x 2 y - 3xyz 2 )
(b) 'il/(2, -1, 1) = (-4 + 1,4- 2,-4 + 6) = (-3, 2, 2)
(c) By Equation 14, Duf(2, - 1, 1) = 'il /(2, - 1, 1) · u = (-3, 2, 2) · ( 0, ~. - ~) = 0 + ~ - ~ = ~·
11. f(x, y) = e"' si.ny => 'il f(x, y) = (e"' siny, e"' cosy}, '\1 f(O , Tr/3) = ( 4, ~ ) ,and a
unit vector in the direction.ofv is u = J 2
+8 2
(- 6, 8) = 1~ (-6, 8) = ( -i, V. so
- '