Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 15 REVIEW ET CHAPTER 14 ¤ 215
T x (6, 2) ≈
T (8, 2) − Tx(6, 2)
2
=
90 − 87
2
=1.5, T x (6, 2) ≈
T (4, 2) − T (6, 2)
−2
Averaging these values, we estimate T x (6, 2) ≈ 4.0. Finally, we estimate T xy (6, 4):
T xy(6, 4) ≈ T x(6, 6) − T x (6, 4)
2
=
3.0 − 3.5
2
Averaging these values, we have T xy (6, 4) ≈−0.25.
13. f(x, y) = 2x + y 2 ⇒ f x = 1 2 (2x + y2 ) −1/2 (2) =
=
74 − 87
−2
= −0.25, T xy(6, 4) ≈ T x(6, 2) − T x (6, 4)
−2
=
=6.5.
4.0 − 3.5
−2
1
, 2x + y
2 fy = 1 (2x + 2 y2 ) −1/2 y
(2y) =
2x + y
2
= −0.25.
15. g(u, v) =u tan −1 v ⇒ g u =tan −1 v, g v = u
1+v 2
17. T (p, q, r) =p ln(q + e r ) ⇒ T p =ln(q + e r ), T q = p
q + e , T r r =
per
q + e r
19. f(x, y) =4x 3 − xy 2 ⇒ f x =12x 2 − y 2 , f y = −2xy, f xx =24x, f yy = −2x, f xy = f yx = −2y
21. f(x, y, z) =x k y l z m ⇒ f x = kx k−1 y l z m , f y = lx k y l−1 z m , f z = mx k y l z m−1 , f xx = k(k − 1)x k−2 y l z m ,
f yy = l(l − 1)x k y l−2 z m , f zz = m(m − 1)x k y l z m−2 , f xy = f yx = klx k−1 y l−1 z m , f xz = f zx = kmx k−1 y l z m−1 ,
f yz = f zy = lmx k y l−1 z m−1
23. z = xy + xe y/x ⇒ ∂z
∂x = y − y x ey/x + e y/x ,
x ∂z
∂x + y ∂z
∂y = x
y − y x ey/x + e y/x
+ y
∂z
∂y = x + ey/x and
x + e y/x = xy − ye y/x + xe y/x + xy + ye y/x = xy + xy + xe y/x = xy + z.
25. (a) z x =6x +2 ⇒ z x(1, −2) = 8 and z y = −2y ⇒ z y(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 (and the surface) at (1, −2, 1) is h8, 4, −1i. Then parametric equations for the normal
line there are x =1+8t, y = −2+4t, z =1− t, and symmetric equations are x − 1
8
= y +2
4
= z − 1
−1 .
27. (a) Let F (x, y, z) =x 2 +2y 2 − 3z 2 .ThenF x =2x, F y =4y, F z = −6z, soF x (2, −1, 1) = 4, F y (2, −1, 1) = −4,
F z (2, −1, 1) = −6. From Equation 15.6.19 [ET 14.6.19], an equation of the tangent plane is
4(x − 2) − 4(y +1)− 6(z − 1) = 0 or, equivalently, 2x − 2y − 3z =3.
(b) From Equations 15.6.20 [ET 14.6.20], symmetric equations for the normal line are x − 2
4
= y +1
−4 = z − 1
−6 .
29. (a) r(u, v) =(u + v) i + u 2 j + v 2 k and the point (3, 4, 1) corresponds to u =2, v =1.Thenr u = i +2u j ⇒
r u(2, 1) = i +4j and r v = i +2v k ⇒ r v(2, 1) = i +2j. A normal vector to the surface at (3, 4, 1) is
r u × r v =8i − 2 j − 4 k, so an equation of the tangent plane there is 8(x − 3) − 2(y − 4) − 4(z − 1) = 0 or equivalently
4x − y − 2z =6.
(b) A direction vector for the normal line through (3, 4, 1) is 8 i − 2 j − 4 k, so a vector equation is
r(t) =(3i +4j + k)+t (8 i − 2 j − 4 k), and the corresponding parametric equations are x =3+8t, y =4− 2t,
z =1− 4t.