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

SECTION 14.3 PARTIAL DERIVATIVES 0 201
· · · d I f ( 3 2 2
) !(3.5, 2.2) - !(3, 2.2) 26.1- 15.9
we can approximate by cons1denng h = 0.5 an 1 = - 0.5: "' , . ::::::
= = 20.4,
. . ' 0 . 5
0.5
f(2.5, 2.2) - f(3, 2.2) 9.3 - 15.9
f x(3, 2 A . I I h f ( )'
2.~) ~ _ _ = _ _ 0 5 0 5
= 13. . veragmg t 1ese va ues, we ave "' 3, 2.2 ~ 16.8.
To estimat~ f, 11 (3, 2), we first need an estimate for .fx(3, 1.8):
~ /(3.5, 1.8) - /(3, 1.8) - 20.0 - 18.1 - 3 8 f (3 1 8) ~ /(2:5, 1.8)- !(3, 1.8) - 12.5 - 18.1 -
.fx( 3 • 1. 8 ) ~ 0.5 - 0.5 - . ' x ' . ~ -0.5 - - 0.5 - 1 1. 2 .
. . a
Averaging these values, we get f,(3, 1.8) ~ 7.5. Now fx 11 (x, y) = -;::;- [f.,(x, y)) and fx(x, y) is itself a function of two
uy
.. ()a[ ( )) li f,(x,y+h)-fx(x,y)
variables, so Defimtton 4 says that fx 11 x, y = -;::;- fx x, Y = m I · =>
. uy h-o a
( )
f x(3, 2 +h)- fx(3, 2) W . I· · 1 · · k · 1 1
h- o h
f , 11
3, 2 = lim . e can estimate t liS va ue usmg our prev1ous wor Wit 1 a = 0.2 and h = -0.2:
) ~ f x(3, 1.8)- J,(3, 2) = 7.5 - 12.2 =
f ( 3 2 ) ~ j,(3, 2.2)- f x(3, 2) = 16.8- 12.2 = 23 f ( 3 2
xy • ~ 0.2 0.2 ' xy ' -0.2 -0.2
Av~raging these values, we estimate f, 11 (3, 2) to be appr~ximately 23.25.
23·5·
U:r:x = -(x2 + y2 + z2)- 3/2 - x( -~) (x2 + y2 + z2) - 5f2(2x) = 2x2 - y2- z2 .
. l . (x2 + y2 + z2)5/ 2
2y2 - x2 - z·2 2z 2 - x2 - y2
By symmetry, u uu = .(x2 + y2 + z2 )5/2 and U ::z = (x2 + y2 + z2)5/2 .
2x2 - y2 - z2 + 2y2 - x2 - z2 + 2z2 - x2 - y2
Thus Uxo:·+ u,,y + Uzz = ( 2 ? 2)"/2 = 0.
. X +y- + Z "
8[f(v) + g(w)] df(v) 8v dg(w) 8w ·
79. Let v = x +at, w = x - at. Then Ut = f)t = ~ ot +~7ft = af'(v) - ag'(w) and
uu = o[af'(v) a~ ag'(w)J = a[af"(v) + ag"(w)) = a 2 [f"(v) + y"(w)). Similarly, by using the Chain Rule we have
u, = f'(v) + g'(w) and Uxx = f"(v) + g"(w). Thus Utt = a 2 Ux:r.·
8z e"' 8z eY 8z 8z e" e·u e"' + eY
81. z = in(e"' + e!l) => -'- = --- and - = --- , so - +- = - -- + --- = --- = 1.
8x e"' + e!l 8y e"' + eY Dx 8y e"' + e11 e"' + eY e"' + eY
D 2 z e"'(e"' + eY) .- e"'(e'") e"'+ 11 8 2 z 0 - e 11 (e"') ex+y
8x2 = (e +eY) 2 = (e"' + e11 )2' 8x8y=(e"' + ev)2 = (e"' + eY)2' and
8 2 z e 11 (e"' + e 11 )-e 11 (e 11 )
8y2 (e"' + eY )2 ( ) 2 • Thus
e" +eY
e"'+Y )2 (e"'+u? (e"'+Y?
(e"' + eu) 2 = (e"' + eY)1 - (ex + eu )1 = 0
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