Solução_Calculo_Stewart_6e

F.
196 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14 TX.10
41. Let F (x, y, z) =x 2 − 2y 2 + z 2 + yz. Thenx 2 − 2y 2 + z 2 + yz =2isalevelsurfaceofF
and ∇F (x, y, z) =h2x, −4y + z, 2z + yi.
(a) ∇F (2, 1, −1) = h4, −5, −1i is a normal vector for the tangent plane at (2, 1, −1),soanequationofthetangentplane
is 4(x − 2) − 5(y − 1) − 1(z +1)=0or 4x − 5y − z =4.
(b) The normal line has direction h4, −5, −1i, so parametric equations are x =2+4t, y =1− 5t, z = −1 − t, and
symmetric equations are x − 2 = y − 1
4 −5 = z +1
−1 .
43. F (x, y, z) =−z + xe y cos z ⇒ ∇F (x, y, z) =he y cos z, xe y cos z, −1 − xe y sin zi and ∇F (1, 0, 0) = h1, 1, −1i.
(a) 1(x − 1) + 1(y − 0) − 1(z − 0) = 0 or x + y − z =1
(b) x − 1=y = −z
45. F (x, y, z) =xy + yz + zx, ∇F (x, y, z) =hy + z, x + z, y + xi, ∇F (1, 1, 1) = h2, 2, 2i,soanequationofthetangent
plane is 2x +2y +2z =6or x + y + z =3, and the normal line is given by x − 1=y − 1=z − 1 or x = y = z. Tograph
thesurfacewesolveforz: z = 3 − xy
x + y .
47. f(x, y) =xy ⇒ ∇f(x, y) =hy, xi, ∇f(3, 2) = h2, 3i. ∇f(3, 2)
is perpendicular to the tangent line, so the tangent line has equation
∇f(3, 2) ·hx − 3,y− 2i =0 ⇒ h2, 3i·hx − 3,x− 2i =0 ⇒
2(x − 3) + 3(y − 2) = 0 or 2x +3y =12.
2x0
49. ∇F (x 0 ,y 0 ,z 0 )=
a , 2y0
2 b , 2z0 . Thus an equation of the tangent plane at (x 2 c 2 0 ,y 0 ,z 0 ) is
2x 0
a x + 2y 0
2 b y + 2z
0 x
2
2 c z =2 0
2 a + y2 0
2 b + z2 0
=2(1)=2since (x 2 c 2 0 ,y 0 ,z 0 ) is a point on the ellipsoid. Hence
x 0
a x + y0
2 b y + z0
2 c z =1is an equation of the tangent plane. 2