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

SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 0 217
41. Let F(x, y, z) = 2(x - 2) 2 + (y - 1? + (z - 3) 2 . Then 2(x - 2) 2 + (y - 1f + (z - 3) 2 = 10 is a level surface of F.
F:.(x, y,z) = 4(x - ·2) =? F,(3, 3, 5) = 4, F 11 (x,y,z) = 2(y- 1) =? F 11 (3,3,5) = 4, and
Fz(x,y, z) =2{z - 3) =? Fz{3,3, 5) = 4.
(a) Equation 19 gives an equation of the tangent plane at {3, 3, 5) as 4{x - 3) + 4{y - 3) + 4{z- 5) = 0
4x + 4y + 4z = 44 or equivalently x + y + z = 11.
. I I' h . . x - 3 y -
(b)
3 z - 5 . I l
By EquatiOn 20, the norma me as symmetnc equat10ns -
- or eqUiva ent y
4 - = - 4 - = - 4
x - 3 = y - 3 = z - 5. Corresponding parametric equations are x = 3 + t, y =; 3 + t, z = 5 + t.
43. Let F(x, y, z ) = xyz 2 • Then xyz 2· = 6 is a level surface ofF and 'V F(x, y, z ) = (yz 2 , xz 2 , 2xyz).
(a) 'V F(3, 2, 1) ~ (2,3, 12) is a normal vector for the tangent plane at {3, 2, 1), so an equation of the tangent plane
is 2(x - 3) + 3(y - 2) + 12(z - 1) = 0 or 2x + 3y + 12z = 24.
(b) The normal line has direction (2, 3, 12), so parametric equations are x = 3 + 2t, y = 2 + 3t, z = 1 + 12t, and
. . x-3 . y-2 z- 1
symmetriC equations are -
2 - = - 3 - = l2.
45. Let F (x, y, z) = x + y + z - e"'Yz. Then x + y + z = e"' 11 :; is the level surface F(x, y , z) = 0,
and 'V F(x, y, z) = (1 - yze"'Y:;, 1 - xze"' 11 "", 1 - x ye"' 11 z).
(a) 'V F(O, 0, 1) = (1, 1, 1) is a normal vector for the tangent plane at {0, 0, 1), so an equation of the tangent plane
is 1(x- 0) + 1{y - 0) + 1(z - 1) = 0 or x + y + z = 1.
(b) The normal line has direction (1, 1, 1), so parametric equations are x = t, y = t, z = 1 + t, and symmetric equations are
x = y =z- 1.
47. F(x, y , z) = xy + yz + zx, 'V F (x, y, z) = (y + z, x + z, y + x), 'V F{1, 1, 1) = (2, 2, 2), so an equation of the tangent
plane is 2:r; + 2y +2z = 6 or x + y + z = 3, and the normal line is given by x - 1 = y - 1 = z - 1 or x = y = z. To graph
3 - xy
the surface we solve for z : z = --- .
- x+y
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