Solução_Calculo_Stewart_6e

F.
SECTION 15.4TX.10
TANGENT PLANES AND LINEAR APPROXIMATIONS ET SECTION 14.4 ¤ 185
as shown in the second graph. (Here, the tangent plane is shown with fewer traces than the surface.) If we zoom in farther, the
surface and the tangent plane will appear to coincide.
11. f(x, y) =x y. The partial derivatives are f x (x, y) = y and f y (x, y) = x
2 y ,sof x(1, 4) = 2 and f y (1, 4) = 1 4 .Both
f x and f y are continuous functions for y>0, so by Theorem 8, f is differentiable at (1, 4). By Equation 3, the linearization of
f at (1, 4) is given by L(x, y) =f(1, 4) + f x (1, 4)(x − 1) + f y (1, 4)(y − 4) = 2 + 2(x − 1) + 1 4 (y − 4) = 2x + 1 4 y − 1.
13. f(x, y) = x
x + y .
The partial derivatives are f x(x, y) =
1(x + y) − x(1)
(x + y) 2 = y/(x + y) 2 and
f y (x, y) =x(−1)(x + y) −2 · 1=−x/(x + y) 2 ,sof x (2, 1) = 1 9 and f y(2, 1) = − 2 9 .Bothf x and f y are continuous
functions for y 6=−x, sof is differentiable at (2, 1) by Theorem 8. The linearization of f at (2, 1) is given by
L (x, y) =f(2, 1) + f x(2, 1)(x − 2) + f y(2, 1)(y − 1) = 2 3 + 1 9 (x − 2) − 2 9 (y − 1) = 1 9 x − 2 9 y + 2 3 .
15. f(x, y) =e −xy cos y. The partial derivatives are f x (x, y) =e −xy (−y)cosy = −ye −xy cos y and
f y(x, y) =e −xy (− sin y)+(cosy)e −xy (−x) =−e −xy (sin y + x cos y),sof x(π, 0) = 0 and f y(π, 0) = −π.
Both f x and f y are continuous functions, so f is differentiable at (π, 0), and the linearization of f at (π, 0) is
L(x, y) =f(π, 0) + f x(π, 0)(x − π)+f y(π, 0)(y − 0) = 1 + 0(x − π) − π(y − 0) = 1 − πy.
17. Let f(x, y) =
2x +3
4y +1 .Thenf x(x, y) = 2
4y +1 and f y(x, y) =(2x +3)(−1)(4y +1) −2 −8x − 12
(4) =
(4y +1) .Bothf 2 x and f y
are continuous functions for y 6=− 1 4 , so by Theorem 8, f is differentiable at (0, 0). Wehavef x(0, 0) = 2, f y (0, 0) = −12
and the linear approximation of f at (0, 0) is f(x, y) ≈ f(0, 0) + f x (0, 0)(x − 0) + f y (0, 0)(y − 0) = 3 + 2x − 12y.
19. f(x, y) = x
20 − x 2 − 7y 2 ⇒ f x (x, y) =− 20 − x2 − 7y and f 7y
y(x, y) =− 2 20 − x2 − 7y ,
2
so f x (2, 1) = − 2 and f 3 y(2, 1) = − 7 . Then the linear approximation of f at (2, 1) is given by
3
f (x, y) ≈ f(2, 1) + f x(2, 1)(x − 2) + f y(2, 1)(y − 1) = 3 − 2 3 (x − 2) − 7 3 (y − 1) = − 2 3 x − 7 3 y + 20
3 .
Thus f(1.95, 1.08) ≈− 2 (1.95) − 7 20
(1.08) + =2.84¯6.
3 3 3
21. f(x, y, z) = x 2 + y 2 + z 2 ⇒ f x(x, y, z) =
f z (x, y, z) =
x
y
, fy(x, y, z) =
x2 + y 2 + z2 x2 + y 2 + z ,and
2
z
x2 + y 2 + z ,sof x(3, 2, 6) = 3 , f
2 7 y(3, 2, 6) = 2 , f 7 z(3, 2, 6) = 6 . Then the linear approximation of f at
7