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

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 0 219
61. Let (xo, y 0 , zo) be a point on the surface. Then an equation of the tangent plane at the point is
X y Z JXO+Jy0+ /Zo r;. . · ·
--+ --+ r.;:: = . But JXo + ..;Yo + v'Zo = v c, so the equation IS
2y'Xo 2..;Yo 2yZo 2
-=-- + _]f_ + ~ = .;c. The x-, y-, and z-intercepts are foO, -ICYo and .jCZO respectively. (The X-intercept is found by
y'Xo ..;Yo v zo
setting y = z = 0 and solving the resulting equation for x, and they- and z-intercepts are found similarly.) So the sum of the
intercepts is .jC ( JXo + ffo + v'Zo) = c, a constant.
63. 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 perpenoicular to both "il f and "ilg
at ( - 1, 1, 2). The vector v = "il f x "ilg will therefore be parallel to the tangent line.
We have "il f(x, y, z) = ( - 2x, -2y, 1) =? "il f ( - 1, 1, 2) = (2, - 2, 1), and "il g(x, y , z ) = (8x, 2y, 2z) =?
"Vg(- 1, 1, 2)= (-8, 2,4).Hencev="Vf x "Vg =
j k
2 - 2 1 = - 10 i-16j-12k.
- 8 2 4
Parametric equations are: x = -1 - lOt, y = 1 - 16t, z = 2 - 12t.
65. (a) The direction of the normal line ofF is given by "il F, and that of G by "VG. Assuming that
"il F -1= 0 -1= "VG, the two normal lines are perpendicular at P if "il F · "VG = 0 at P -¢:?
(8Fj8x, 8F/8y,8F/8z) · (8G j8x, 8G/ 8y, 8G/ 8z) = 0 at P # F_..,G, + FvGv + FzGz = 0 at P.
(b) Here F = x 2 + y 2 - z 2 and G = x 2 + y 2 + z 2 - r 2 , so
"il F · "VG = (2x, 2y, - 2z) · (2x, 2y, 2z) = 4x 2 + 4y 2 -
4z 2 = 4F = 0, since the point (x, y, z ) lies on the graph of
F = 0. To see that this is true without using calculus, note that G = 0 is the equation of a sphere centered at the origin and
F = 0 is the equation of a right circular cone with vertex at the origin (which is generated by lines through the origin). At
any point of intersection, the sphere's normal line (which passes through the origin) lies on the cone, and thus is
perpendicular to tl1e cone's normal line. So the surfaces with equations F = 0 and G = 0 are everywhere orthogonal.
67. Let u = (a, b) and v = (c, d). Then we know that at the given point, Du f = "il f · u = afx + bfv and
Dv I = "il I · v = clx + dfv· But these are just two linear equations in the two unknowns fx and fv, and since u and v are
not parallel, we can solve the equations to find "V f = (f.,, f v) at the given point. In fact,
"V f = I d Du f - b Dv f , a Dv f - c D u f ) .
\ ad - be ad - be
@) 2012 CcnJ:!ISC Lc:uning. All Rights Rosel"\·ed. May nol be SC4Med, cop k-d, or duplicated, or postod lo n publicly accessible wobsite, in wbolc or in port.