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

SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS D 203
(c) f (0 0) = lim f(h, 0) - f(O, 0) = lim {O/ h2) - 0 = 0 and jy(O, 0) = lim j(O, h) -
"' ' J•-O h h- o h h-o h
f(O, 0) = '0.
(d) B (3) f (0 0) = of, = lim f.,(O, h)- f , (O, O) = lim ( - h 5 - O) / h 4 = -1 while by (2),
Y ' "'Y ' oy h- o h h- o h
f ,(0, 0) = ofy = lim jy(h, 0) - jy(O, 0) = lim h5/ h4 = 1.
!I ox h ..... o h ,, ..... o h
(e) For (x,·y) i= (0, 0), we use a CAS to compute
f xy(x, y)=
x6 + gx4y2 _ gx2y4 _ y6 ·
(x2+y2)3
Now as (x, y) -> (0, 0) along the x-axis, fxv(x, y)-> 1 while~
(x, y) -> (0, 0) along they-axis, fxy(X, y) -> - 1. Thus fx 11 isn't
continuous at (0, 0) and Clairaut's Theorem doesn't apply, so there is
no contradiction. The graphs of f, 11 and fyx are identical except at the
origin, where we observe the discontinuity.
14.4 Tangent Planes and Linear Approximations
1. z = f(x, y) = 3y 2 - 2x 2 + x =? f .,(x, y) = -4x + 1, / 11 (x, y) = 6y, so f,(2, -1) = ~ 7, / y(2, -1) = -6.
By Equation 2, an equation of the tangent plane is z - ( -3) = f, (2 , - 1)(x- 2) + fv(2, -1)[y- ( -:-1)] =:­
z+3= - 7(x - 2)-6(y + 1) or z= - 7x - 6y + 5.
3. z = f(x, y) = .,fXY => f,(x, y) = ~(xy) - 112 · y = ~gx, fu(x, y) = ~(xy)- 1 1 2 · x = ~.../XlY, so f.,(1 , 1) = ~
and fv(1, 1) = ~-Thus an equation of the tangent plane is z - 1 = f,(1, 1)(x- 1) + jy(1, 1)(y- 1) =?
z- 1 = ~(x -1) + Hv - 1) or x + y- 2z = 0.
5. z = f(x, y) = xsin(x + y) =? f , (x, y) = x · cos(x + y) + sin(x + y) · 1 = x cos(x + y) + sin(x + y),
fv(x, y) = xcos(x + y), so f,( -1, 1) = ( -1) cosO+ sin 0 = - 1, / 11 ( -1, 1) = ( -1) cosO = - 1 and an equation of the
tangent plane is z - 0 = ( -1) ( x + 1) + ( -1) (y - 1) or x + y + z = 0.
7. z = f(x, y) = x 2 + xy + 3y 2 , so f,.(x, y) = 2x + y => f,(1, 1) = 3, fv(x, y) = x + 6y =? fv(1, 1) = 7 and an
equation of the tangent plane is z - 5 = 3(x- 1) + 1(y- 1) or z = 3x + 7y - 5. After zooming in, the surface and the
® 2012 Ccngogc Le;uning. All Rights Res