Solução_Calculo_Stewart_6e

F.
SECTION 15.6 DIRECTIONAL TX.10 DERIVATIVES AND THE GRADIENT VECTOR ET SECTION 14.6 ¤ 197
2x0
51. ∇F (x 0,y 0,z 0)=
a , 2y 0
2 b , −1
, so an equation of the tangent plane is 2x 0
2 c
a x + 2y 0
2 b y − 1 2 c z = 2x2 0
a + 2y2 0
2 b − z 0
2 c
or 2x 0
a x + 2y 0
2 b y = z x
2
2 c +2 0
a + y2 0
− z 0
2 b 2 c .Butz 0
c = x2 0
a + y2 0
, so the equation can be written as
2 b2 2x 0
a x + 2y 0
2 b y = z + z 0
.
2 c
53. The hyperboloid x 2 − y 2 − z 2 =1is a level surface of F (x, y, z) =x 2 − y 2 − z 2 and ∇F (x, y, z) =h2x, −2y, −2zi is a
normal vector to the surface and hence a normal vector for the tangent plane at (x, y, z). The tangent plane is parallel to the
plane z = x + y or x + y − z =0if and only if the corresponding normal vectors are parallel, so we need a point (x 0 ,y 0 ,z 0 )
on the hyperboloid where h2x 0, −2y 0, −2z 0i = c h1, 1, −1i or equivalently hx 0, −y 0, −z 0i = k h1, 1, −1i for some k 6= 0.
Then we must have x 0 = k, y 0 = −k, z 0 = k and substituting into the equation of the hyperboloid gives
k 2 − (−k) 2 − k 2 =1 ⇔ −k 2 =1, an impossibility. Thus there is no such point on the hyperboloid.
55. Let (x 0 ,y 0 ,z 0 ) be a point on the cone [other than (0, 0, 0)]. Then an equation of the tangent plane to the cone at this point is
2x 0 x +2y 0 y − 2z 0 z =2 x 2 0 + y 2 0 − z 2 0 .Butx
2
0 + y 2 0 = z 2 0 so the tangent plane is given by x 0 x + y 0 y − z 0 z =0,aplane
which always contains the origin.
57. Let (x 0,y 0,z 0) be a point on the surface. Then an equation of the tangent plane at the point is
x
2 √ + y
x 0 2 + z
√
y 0 2 √ x0 + y 0 + √ z 0
=
.But √ x 0 + y 0 + √ z 0 = √ c, so the equation is
z 0 2
x
√
x0
+
y
y0
+
z √
z0
= √ c.Thex-, y-, and z-intercepts are √ cx 0 , cy 0 and √ cz 0 respectively. (The x-intercept is found
by setting y = z =0and solving the resulting equation for x,andthey-andz-intercepts are found similarly.) So the sum of
the intercepts is √ √x0
c + y 0 + √
z 0 = c, a constant.
59. If f(x, y, z) =z − x 2 − y 2 and g(x, y, z) =4x 2 + y 2 + z 2 , then the tangent line is perpendicular to both ∇f and ∇g
at (−1, 1, 2). The vector v = ∇f × ∇g will therefore be parallel to the tangent line.
We have ∇f(x, y, z) =h−2x, −2y, 1i ⇒ ∇f(−1, 1, 2) = h2, −2, 1i,and∇g(x, y, z) =h8x, 2y, 2zi ⇒
i j k
∇g(−1, 1, 2) = h−8, 2, 4i. Hence v = ∇f × ∇g =
2 −2 1
= −10 i − 16 j − 12 k.
−8 2 4
Parametric equations are: x = −1 − 10t, y =1− 16t, z =2− 12t.
61. (a) The direction of the normal line of F is given by ∇F ,andthatofG by ∇G. Assuming that
∇F 6= 06=∇G, the two normal lines are perpendicular at P if ∇F · ∇G =0at P
⇔
h∂F/∂x,∂F/∂y,∂F/∂zi·h∂G/∂x, ∂G/∂y, ∂G/∂zi =0at P ⇔ F x G x + F y G y + F z G z =0at P .
(b) Here F = x 2 + y 2 − z 2 and G = x 2 + y 2 + z 2 − r 2 ,so
∇F · ∇G = h2x, 2y, −2zi·h2x, 2y, 2zi =4x 2 +4y 2 − 4z 2 =4F =0, since the point (x, y, z) lies on the graph of