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

216 0 CHAPTER 14 PARTIAL DERIVATIVES
k
Jx2 + y2 +z2 3
31. T = and 120 = T(1, 2, 2) = - so k = 360.
k
( )
- (1, - 1, 1)
au- v'3 ,
DuT(1, 2, 2) = 'VT(1, 2, 2) · u = [ .:._360( x 2 + y 2 + z 2 ) -S/ 2 (x, y, z)] . u = - .12 3 (1, 2, 2). -4.. (1, -1, 1) = -~
. . {1,2,2) v3 3v3
(b) From (a), 'VT = -360(x 2 + y 2 + z 2 ) - S/ 2 (x, y, z), and since (x, y, z) is the position vector of the point (x, y, z), the
vector - (x, y, z), and thus 'VT, always points toward the origin.
33. 'VV(x, Y •. z) = (lOx- 3y + yz, xz - 3x, xy), 'VV(3, 4, 5) = (38, 6, 12)
{a) Du V(3,4,5) = {38, 6, 12} · ~(1, 1, -1} = '7a
(b) 'VV(3, 4, 5) = (38, 6, 12), or equivalently, (19, 3, 6).
(c) IV'V(3, 4, 5)1 = v'382 + 6 2 + 12 2 = v'l624 = 2 v'406
-----t
35. A unit vector in the direction o(AB is i and a unit vector in the direction of AC isj. Thus D_..... f(1, 3) = f.,(1, 3) ~ 3 and
AB
D- f(l, 3) = / y(1, 3) = 26. Therefore \7 /(1, 3) = (!.,(1, 3), /y(l, 3)) = (3, 26}, and by definition,
AC
-----t
DAD f (1 , 3) = V' f · u where u is a unit vector in the direction of AD, which is (fa, H). Therefore,
37 ()
DAD f (1, 3) = (3, 26) . . ( 153, H) = 3. fa + 26. H = 312;.
. a v au + v ~ ' a a ~ + {) , a ~- + ~ a a , ~ + ~ ' a
'["7( b)=(o(au+bv) 8(au+bv)) =\ au bav au bav)= (au au; b\av au;
.= a 'Vu + b 'Vv
uX y ux X uy uy X uy ux y
-----t
(d)
'["7 n ( o(un) 8(u")) ( n - 1 au n - 1 au) n - 1 '["7
v u - , - nu , nu · - nu v u
ax oy ox {)y
39. f(x, y) = x 3 + 5x 2 y + y 3 . "*
Duf(x,y) = 'Vf(x, y) · u = (3x 2 + lOxy, 5x 2 + 3y 2 ) · (~, ~) = ~x 2 + 6xy+ 4x 2 + ¥v 2 = ~x 2 + 6xy + ¥y 2 . Then
D~.J(x ,y) = Du (Duf(x,y)] = \7 [Duf(x,y)] · u = (¥x + 6y, 6x + -\1Y) · (~, ~)
= l;s4x + Jfy + ¥x + ~y = 2is4x + ~ar,oY
and D~f(2 , 1) = 2iu4 (2) + 12866 (1) = 72754 .
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