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

194 0 CHAPTER 14 PARTIAL DERIVATIVES
25. h(x, y) = g(f(x, y)) = (2x + 3y - 6) 2 + y'2x + 3y - 6. Since f is a. polynomial, it is continu o~s on JR 2 and g· is
continuous on its domain { t I t ;::: 0}. Thus h is continuous on its domain.
D = {(x, y) I 2x + 3y-6;::: 0} = { (x, y) I y;::: - ~x + 2}, which consists of all points on or above the line y = - ~x + 2.
27.
From the graph, it appears that f is discontinuous along the line y = x.
3
2
z 1
0
-1
2
If we consider f(x, y) = e 1 /(a:-y) as a composition offun.ctions,
g(x, y) = 1/(x- y) is a rational function and therefore continuous except
where x- y = 0 =} y = x. Since the function h(t) = et is continuous
everywhere, the composition h(g(x,y)) = e 1 1(x-v) = f(x,y) is
continuous except along the line y = x, as we suspected.
29. The functions xy and 1 + ex-v are continuous everywhere, and 1 + ex- v is never zero, so F(x, y) = xy is continuous
. 1 + ex-y
on its domain JR 2 .
31. F(x, y) =
l +x2 +y2
1 - x 2 -y 2
is a rational function and thus is continuous on its domain
33. G(x, y) = ln(x 2 + y 2 - 4) = g(f(x, y)) where f(x, y) = x 2 + y 2 - 4, continuous on ]~_2, and g(t) = In t, continuous on its
domain {t ['t > 0}. Thus G is continuous on its domain { (x,_y) I x 2 + y 2 -
4 > 0} = { (x, y ) I x 2 + y 2 > 4 }, the_ exterior
ofthe circle x 2 + y 2 = 4.
35. f(x, y, z ) = h(g(x, y, z )) where g(x, y, z) = x 2 + y 2 + z 2 , a polynomial that is continuous
everywhere, and h(t) = arcsint, continuous on [- 1, 1). Thus f is continuous on its domain
{(x,y, z) l-1 S x 2 + y 2 + z 2 S 1} = {(x,y,z)! x 2 + y 2 + z 2 S 1 }, so f is continuous on the unit ball.
if (x, y) =I= (0, 0)
if (x,y) = (0,0)
The first piece off is a rational function defined everywhere except at the
origin, so f is continu~us on JR 2 except possibly at the origin. Since x 2 s 2x 2 + y 2 , we have jx 2 y 3 / (2x 2 + y 2 ) I. s jy 3 j. We
. 2 3
know that jy 3 \ --+ 0 as (x, y) --+ (0, 0). So, by the Squeeze Theorem, lim f(x, y) = lim
2 ~ y 2 = 0.
(a;,y)-(0,0) (x,y)- (0,0) X + Y
But f(O, 0) = 1, so f is discontinuous at (0, 0). Therefore, f is continuous on the set { (x, y) I (x, y) =/= (0, 0) }.
39 lim
- x 3 + y 3 1 . (r cos 8)3 + (1' sin8) 3 lim ( 3 . 3 B)
. - -- = tm = r· cos 8 + rsm =
( x ,y)-(0,0) X 2 -f- y 2 r-->O+ 1' 2 r~o+
0
-x2 - y 2
e - 1
. 41. lim
(a;,y)--> (0,0) x2 +y2
lim e- r (- 2r·)
2r
r - o+
= lim - e- r 2 = - e 0 = - 1
r·-o+
2
[using !'Hospital's Rule]
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