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

(p+r)2
SECTION 14o5 THE CHAIN RULE D 209
l
19o
w
/ I~
r s r
/\ /\ 1\
X J X y X J
210 z = x 4 + x 2 y, x = s + 2t - u, y = stu 2 =>
w = f(r, s, t), r = r(x, y), s = s(x, y), t = t(x, y) =>
ow ow or ow as ow at aw aw ar aw as ow at
fu = &ofu+asax+8tax·~=aroy+asay + Bt~
oz = az ox+ az.ay = ( 4 xa + 2xy)( 1 ) + (x2)(tu2),
OS OX OS ay as ' 0
oz = oz ax+ az ay = (4x3 + 2xy)(2) + (x2)(su2)
m &m ~m ·
oz = az ax+ az ay = (4x3 + 2xy)( - 1} + (x2)(2stu).
au ax au ay a·u
- When s = 4, t = 2, and u = 1 we have x = 7 and y = 8,
so az = (1484)(1} + {49)(2} = 1582, aaz = {1484) (2) + (49){4) = 3164, aaz = (1484)(-1) + {49)(16} = -700.
as t u
23o w = xy + yz + zx, x = r cos 0, y = r sin 8, z = r8 =>
aw = awax + aawaay + aawaaz ~(y +z}(cos8} + (x+z)(sin8}+(y+x) (O), .
ar axar yr zr
aw = aw ax+ aaw aay + aaw aaz8 = (y + z)(- rsinO) + (~ + z )(rcosO) + (y + x}(r·)o
ao ox o8 . y 8 z
aw
When ro = 2 and()= ?r/2 we have x = 0, y = 2, and z = 1r, so ar = (2 + rr}(O) + (0 + rr){1) + (2 + 0)(1r / 2) = 27r and
~~ = (2 + 1r)( - 2) + (0 + 1r)(O) + (2 + 0)(2) = -2rro
..
25o N = p + q, p = u + vw, q = v + uw, r = w + uv =9
p+r
aN = aN ap + aN aq + aN ar.
au ap au aq au 8r au
'
= (p + r){1)- (p + q){1) ( 1
) + (p + r)(1) - (p + q)(O) (w) + (p + r)(O) - (p + q)(1) (v)
(p+r)2 (p + r)2 (p + r)2
=
(r- q) + (p+r)w - (p+q)v
(p+r)2
oN = aN ap + aN oq + aN or = r - q (w) + p + ro ( 1
) + -(p + q) (u) = (r - q)w + (p + r} - (p + q)u'
av op ov 8q 8v or ov 0 (p +r) 2 . (p+1')2 (p+ r)2
aN _ aN ap + 8N aq + 8N !!!.._ = ro - q (v) + p+r (u)+ -(p +q) ( 1
) = (r- q)v +(p+r)u- (p + q)
ow - ap ow aq ow or OW (p + r )2 (p + r) 2 (p + r) 2 (p + r )2
· aN -1 + (24)(4) - (25)(3) 20 5
When u = 2, v = 3, and w = 4 we have p = 14, q = 11, and r = ~0, so ou = . ( 24
)2 . = 576
= 144
,
aN _ (-1)(4)+24-(25)(2) _ -30 __ ~ d 8N = · (- 1 )(3) + (24)(2)-25 =~ = ~
av - (24)2 - 576 - 96' an aw (24)2 576 144 °
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