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

SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS 0 205
15. f(x, y) = e-"' 11 cosy. The partial derivatives are f,.(x , y) = e-"' 11 ( -y) cosy .= -ye-"'!1 cosy and
J 11 (x, y) = e- "'V(- siny) + (cosy)e-"' 11 ( - x) = -e-"' 11 (siny + xcosy), so f.,(1T, 0) = 0 and / 11(1T, 0) = -1r.
Both f., and fv are continuous func;tions, so I is differentiable at (1r, 0), and the linearization of I at (1r, 0) is
L(x, y) = j(1r, 0) + /:r(1r, O)(x- 1r) + j 11 (1r, O)(y - 0) = 1 + O(x -1r) - 1r(y - 0) = 1 -1ry.
2x + 3 2 . ( ) ( )( )( )_2 - 8x - 12
17.Letf(x,y) = --.Thenj.,(x,y)= - 4 1
andj11 x,y = 2x+3 -1 4y+1 (4)=( 4 1
)2 .Bothf.,andl 11
4y + 1 y + y +
are continuous functions for y =I= - ~. so by Theorem 8, f is differentiable at (0, 0) . We have f:,,(O, 0) = 2, f 11 (0, 0) = -12
and the linear approximation off at (0, 0) is f(x, y) ~ f(O, 0) + f:,(O, O)(x - 0) + / 11 (0, O)(y - 0) = 3 + 2x- 12y.
19. We can estimate /(2.2, 4.9) using a linear approximation off at (2, 5), given by.
f(x, y) ~ /(2, 5) + f,.(2, 5)(x - 2) + / 11 (2, 5)(y - 5) = 6 + 1(x - 2) + ( -1)(y- 5) = x- y + 9. Thus
!(2.2, 4.9) ~ 2.2- 4.9 + 9 = 6.3.
21.j(x,y,z)"=Jx 2 +y 2 +z 2 =>
f.,(x,y,z)= J
2 +x 2 + 2 ,f 11(x,y,z)= J 2
y 2 2
,and
X y Z X +y + Z
f:(x, y, z) = z , so f.,(3, 2, 6) = ~. fv(3, 2, 6) = ~; f,(3, 2, 6) = ~·Then the linear approximation off
,jx2 +y2 +z2
at (3, 2, 6) is given by
l(x, y, z) ~ /(3, 2, 6) + j .,(3, 2, 6)(x - 3) + / 11 (3, 2, 6) (y- 2) + f z(3, 2, 6)(z' - 6)
= 7 + ~(x- 3) + ~(y- 2) + ¥