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

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204 D CHAPTER 14 PARTIAL DERIVATIVES<br />

tangent plane become almost indistinguishable. (Here, the tangent plane is below the surface.) If we zoom in farther, the<br />

surface and the tangent plane will appear to coincide.<br />

_ xy sin {X - y)<br />

9 · !( x,y ) - 2 2<br />

1+x +y<br />

ACAS . f ( ' ) ysin(x - y) + xycos(x-y) 2x2 ysin{x- y) d<br />

g1ves "' x,y = - an<br />

1 + x2 + y2 (1 + x2 + y2)2<br />

fv(x, y) =<br />

x sin ( x - y) - xy cos ( x - y) 2xy 2 sin ( x - y) ·<br />

- ( 2 . We use the CAS to evaluate these at (1 , 1), and then<br />

+ x + y 1 + x2 + y2)<br />

1<br />

2 2<br />

substitute the results into Equation 2 to compute an equation oqhe tangent plane: z = tx - tY· The surface and tangent<br />

plane are shown in the first graph below. After zooming in, the surface and the tangent plane become almost indistinguishable,<br />

as shown in the second graph. (Here, the tangent plane is shown with fewer traces than the surface.) If we zoom in farther, the<br />

surface and the tangent plane will appear to coincide.<br />

z 0<br />

-1<br />

11. f(x, y) = 1 + x Jn(xy - 5): The partlal derivatives are f, (x, y) = x · -<br />

and fv(x, y) = x · -<br />

xy-5 xy- ,<br />

1 - (y) + Jn (xy - 5) · 1 = xy 5<br />

+ln(xy- 5)<br />

. xy - 5 xy -<br />

1 - (x) = __£_ 5<br />

, so f x(2 , 3) = 6 and / 11 (2, 3) = 4. Both f , and / 11 are continuous functions for<br />

xy > 5, so by Theorem 8, f is differentiable at (2, 3). By Equation 3, the linearization off at (2, 3) is given· by<br />

L(x, y) = /(2, 3) + f x (2, 3)(x- 2) + f v (2, 3)(y - 3) = 1 + 6(x - 2) + 4(y - 3) = 6x + 4y - 23.<br />

13. f(x, y) ~ _ x _ . The partial !;lerivatives are f x(x, y) = l(x t y) ) ~'t( 1 ) = y/ (x + y) 2 and<br />

x + y x +y<br />

fv(x, y) = x ( -1){x + y)- 2 • 1 = -x/ (x + y) 2 , so f x(2, 1) = ~and / 11<br />

(2; 1) = - ~.<br />

Both f, and j 11 are continuous<br />

functions for y =I - x, so f is differentiable at (2, 1) by Theorem 8. The linearization of f at (2, 1) is given by<br />

L(x, y) = !(2, 1) + f x(2 , 1)(x - 2) + / 11 (2;1)(y - 1) = 1 + ~ (x - 2) - ~(y - 1) = ~ X - ~ y + ~·<br />

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