Solução_Calculo_Stewart_6e

F.
31. T =
SECTION 15.6 DIRECTIONAL TX.10 DERIVATIVES AND THE GRADIENT VECTOR ET SECTION 14.6 ¤ 195
k
x2 + y 2 + z and 120 = T (1, 2, 2) = k so k =360.
2 3
h1, −1, 1i
(a) u = √ ,
3
D u T (1, 2, 2) = ∇T (1, 2, 2)·u = −360 x 2 + y 2 + z 2
−3/2
hx, y, zi ·u = − 40 h1, 2, 2i· √3
1
h1, −1, 1i = − 40
3
(1,2,2)
3 √ 3
(b) From (a), ∇T = −360 x 2 + y 2 + z 2 −3/2 hx, y, zi,andsincehx, y, zi is the position vector of the point (x, y, z),the
vector − hx, y, zi, and thus ∇T , always points toward the origin.
33. ∇V (x, y, z) =h10x − 3y + yz, xz − 3x, xyi, ∇V (3, 4, 5) = h38, 6, 12i
(a) D u V (3, 4, 5) = h38, 6, 12i·
√
1
3
h1, 1, −1i = √ 32
3
(b) ∇V (3, 4, 5) = h38, 6, 12i,orequivalently,h19, 3, 6i.
(c) |∇V (3, 4, 5)| = √ 38 2 +6 2 +12 2 = √ 1624 = 2 √ 406
35. A unit vector in the direction of −→
AB is i and a unit vector in the direction of −→
AC is j. ThusD −→
AB
f(1, 3) = f x(1, 3) = 3 and
D −→
AC
f(1, 3) = f y (1, 3) = 26. Therefore ∇f(1, 3) = hf x (1, 3),f y (1, 3)i = h3, 26i, and by definition,
D −→ f(1, 3) = ∇f · u where u is a unit vector in the direction of −→
AD, whichis
5 , 12
AD 13 13 . Therefore,
D −→
5
f (1, 3) = h3, 26i· , 12
AD 13 13 =3· 5
12
+26· = 327 .
13 13 13
∂(au + bv)
37. (a) ∇(au + bv) =
,
∂x
(b) ∇(uv) = v ∂u
= a ∇u + b ∇v
∂x + u ∂v
∂x
∂u
u
v
(c) ∇ = ∂x − u ∂v
∂x ,
v v 2
∂(au + bv)
= a ∂u
∂y
∂x + b ∂v
∂x
∂u
,a
∂y + b ∂v ∂u
= a
∂y ∂x , ∂u ∂v
+ b
∂y ∂x , ∂v
∂y
∂u
,v
∂y + u ∂v ∂u
= v
∂y ∂x , ∂u ∂v
+ u
∂y ∂x , ∂v
= v ∇u + u ∇v
∂y
v ∂u
∂y − u ∂v
∂u
v
∂y ∂x , ∂u
∂y
=
v 2
∂v
− u
∂(u
(d) ∇u n n )
=
∂x , ∂(un )
= nu n−1 ∂u ∂u
∂y
∂x ,nun−1 = nu n−1 ∇u
∂y
v 2
∂x , ∂v
∂y
=
v ∇u − u ∇v
v 2
39. Let F (x, y, z) =2(x − 2) 2 +(y − 1) 2 +(z − 3) 2 .Then2(x − 2) 2 +(y − 1) 2 +(z − 3) 2 =10isalevelsurfaceofF .
F x(x, y, z) =4(x − 2) ⇒ F x(3, 3, 5) = 4, F y(x, y, z) =2(y − 1) ⇒ F y(3, 3, 5) = 4,and
F z(x, y, z) =2(z − 3) ⇒ F z(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 =44or equivalently x + y + z =11.
(b) By Equation 20, the normal line has symmetric equations x − 3
4
= y − 3
4
= z − 5
4
or equivalently
x − 3=y − 3=z − 5. Corresponding parametric equations are x =3+t, y =3+t, z =5+t.
⇔