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

CHAPTER 14 REVIEW 0 239
21. f(x, y , z) = xky 1 z'" => f x = kxk-lylzm., ! 11 = lxky 1 - 1 z m, fz = mxky 1 z""'- 1 , f:"x = k(k- 1)xk- 2 y 1 z"",
fvv = l(l- 1)xky 1 - 2 zm, f zz = m(m -1)xky 1 z"'- 2 , f xv = fvcx: = klx k- 1 y 1 - 1 z"', ! ex::= f ::;r; = kmx"- 1 y 1 zm-t,
fvz = f ::v = lmxkyl-1 zm-1
23. z = x y + xe 11 1"' =>
25. (a) zx = 6x + 2 => zx (l ,-2) = 8 and z 11 = - 2y => zv(1, -2) = 4, so an equation of the tangent plane is
z - 1 = 8(x - 1) + 4(y + 2) or z = 8x + 4y + 1.
(b) A normal vector to the tangent plane (imd the surface) at (1, - 2, 1) is (8, 4, - 1). Then parametric equations for the normal
·
· d . . x-1 y + 2 z- 1
I me there are x = 1 + 8t, y = - 2 + 4 t, z = 1 - t, an symmetnc equations are - - = - - = 8 4
--=t·
27. (a) Let F(x , y ,z) = x 2 + 2y 2 - 3z 2 • Then F.,= 2x, Fy = 4y, Fz = - 6z, so F,.,(2, - 1, 1) = 4, F 11 (2, - 1, 1) = - 4,
F: (2, - 1, 1) = - 6. From Equation 14.6.1 9, an equation ofthetangent planeis 4(x - 2)- 4(y + 1) - 6(z - 1) = 0
or, equivalently, 2x - 2y - 3z = 3.
· 14620 · · fi h II' X- 2 Y + 1 z- 1
(b) From Equatwns . . , symmetnc equatwns or t e norma 1 ~e are - 4
- = --=4 = --=fl·
29. (a) Let F (x, y , z ) = x + 2y + 3z- sin(xyz ) .. Then F,., = 1-yz cos(xyz ), Fv = 2 - xz cos(xyz), F: = 3 - x y cos(xyz),
so F :z: (2, -1, 0) = 1, Fy{2, - 1, 0) = 2,·F: (2, - 1, 0) = 5. From Equation 14.6. 19, an equation of the tangent plane is
1(x - 2) + 2(y + 1) + 5(z - 0) = 0 or x + 2y + 5z = 0.
' . 14 6 20 . .