Solução_Calculo_Stewart_6e

F.
200 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
15. f(x, y) =(x 2 + y 2 )e y2 −x 2 ⇒
f x =(x 2 + y 2 )e y2 −x 2 (−2x)+2xe y2 −x 2 =2xe y2 −x 2 (1 − x 2 − y 2 ),
f y =(x 2 + y 2 )e y2 −x 2 (2y)+2ye y2 −x 2 =2ye y2 −x 2 (1 + x 2 + y 2 ),
f xx =2xe y2 −x 2 (−2x)+(1− x 2 − y 2 ) 2x −2xe y2 −x 2 +2e y2 −x 2 =2e y2 −x 2 ((1 − x 2 − y 2 )(1 − 2x 2 ) − 2x 2 ),
f xy =2xe y2 −x 2 (−2y)+2x(2y)e y2 −x 2 (1 − x 2 − y 2 )=−4xye y2 −x 2 (x 2 + y 2 ),
f yy =2ye y2 −x 2 (2y)+(1+x 2 + y 2 ) 2y 2ye y2 −x 2 +2e y2 −x 2 =2e y2 −x 2 ((1 + x 2 + y 2 )(1 + 2y 2 )+2y 2 ).
f y =0implies y =0, and substituting into f x =0gives
2xe −x2 (1 − x 2 )=0 ⇒ x =0or x = ±1. Thus the critical points are
(0, 0) and (±1, 0). NowD(0, 0) = (2)(2) − 0 > 0 and f xx (0, 0) = 2 > 0,
so f(0, 0) = 0 is a local minimum. D(±1, 0) = (−4e −1 )(4e −1 ) − 0 < 0
so (±1, 0) are saddle points.
17. f(x, y) =y 2 − 2y cos x ⇒ f x =2y sin x, f y =2y − 2cosx,
f xx =2y cos x, f xy =2sinx, f yy =2.Thenf x =0implies y =0or
sin x =0 ⇒ x =0, π,or2π for −1 ≤ x ≤ 7. Substituting y =0into
f y =0gives cos x =0 ⇒ x = π or 3π , substituting x =0or x =2π
2 2
into f y =0gives y =1, and substituting x = π into f y =0gives y = −1.
Thus the critical points are (0, 1), π
2 , 0 , (π, −1), 3π
2 , 0 ,and(2π, 1).
D π
2 , 0 = D 3π
2 , 0 = −4 < 0 so π
2 , 0 and 3π
2 , 0 aresaddlepoints.D(0, 1) = D(π, −1) = D(2π, 1) = 4 > 0 and
f xx (0, 1) = f xx (π, −1) = f xx (2π, 1) = 2 > 0,sof(0, 1) = f(π, −1) = f(2π, 1) = −1 are local minima.
19. f(x, y) =x 2 +4y 2 − 4xy +2 ⇒ f x =2x − 4y, f y =8y − 4x, f xx =2, f xy = −4, f yy =8.Thenf x =0
and f y =0each implies y = 1 x, so all points of the form x
2 0 , 1 x 2 0
are critical points and for each of these we have
D x 0 , 1 x 2 0 = (2)(8) − (−4) 2 =0. The Second Derivatives Test gives no information, but
f(x, y) =x 2 +4y 2 − 4xy +2=(x − 2y) 2 +2≥ 2 with equality if and only if y = 1 x. Thus f
2
x 0, 1 2 x0 =2are all local
(and absolute) minima.