Solução_Calculo_Stewart_6e

F.
SECTION 14.2 DERIVATIVES TX.10 AND INTEGRALS OF VECTOR FUNCTIONS ET SECTION 13.2 ¤ 145
23. The vector equation for the curve is r(t) = 1+2 √ t, t 3 − t, t 3 + t ,sor 0 (t) = 1/ √ t, 3t 2 − 1, 3t 2 +1 .Thepoint
(3, 0, 2) corresponds to t =1, so the tangent vector there is r 0 (1) = h1, 2, 4i. Thus, the tangent line goes through the point
(3, 0, 2) and is parallel to the vector h1, 2, 4i. Parametric equations are x =3+t, y =2t, z =2+4t.
25. The vector equation for the curve is r(t) = e −t cos t, e −t sin t, e −t ,so
r 0 (t) = e −t (− sin t)+(cost)(−e −t ), e −t cos t +(sint)(−e −t ), (−e −t )
= −e −t (cos t +sint),e −t (cos t − sin t), −e −t
The point (1, 0, 1) corresponds to t =0, so the tangent vector there is
r 0 (0) = −e 0 (cos 0 + sin 0),e 0 (cos 0 − sin 0), −e 0 = h−1, 1, −1i. Thus, the tangent line is parallel to the vector
h−1, 1, −1i and parametric equations are x =1+(−1)t =1− t, y =0+1· t = t, z =1+(−1)t =1− t.
27. r(t) = t, e −t , 2t − t 2 ⇒ r 0 (t) = 1, −e −t , 2 − 2t .At(0, 1, 0),
t =0and r 0 (0) = h1, −1, 2i. Thus, parametric equations of the tangent
line are x = t, y =1− t, z =2t.
29. r(t) =ht cos t, t, t sin ti ⇒ r 0 (t) =hcos t − t sin t, 1,tcos t +sinti.
At (−π, π, 0), t = π and r 0 (π) =h−1, 1, −πi. Thus, parametric equations
of the tangent line are x = −π − t, y = π + t, z = −πt.
31. The angle of intersection of the two curves is the angle between the two tangent vectors to the curves at the point of
intersection. Since r 0 1(t) = 1, 2t, 3t 2 and t =0at (0, 0, 0), r 0 1(0) = h1, 0, 0i is a tangent vector to r 1 at (0, 0, 0). Similarly,
r 0 2(t) =hcos t, 2cos2t, 1i and since r 2 (0) = h0, 0, 0i, r 0 2 (0) = h1, 2, 1i is a tangent vector to r 2 at (0, 0, 0). Ifθ is the angle
between these two tangent vectors, then cos θ = √ 1 √
1 6
h1, 0, 0i·h1, 2, 1i = √ 1
6
and θ =cos −1 √
1
6
≈ 66 ◦ .
1
33.
0 (16t3 i − 9t 2 j +25t 4 k) dt =
π/2
1
1
1
dt
0 16t3 i −
0 9t2 dt j +
0 25t4 dt k
= 4t 4 1
i − 3t 3 1
j + 5t 5 1
k =4i − 3 j +5k
0 0 0
35. (3 sin 2 t cos t i +3sint cos 2 t j +2sint cos t k) dt
0
π/2
= 3sin 2 π/2
t cos tdt i + 3sint cos 2 π/2
tdt j + 2sint cos tdt k
0 0 0
= sin 3 t π/2
0
i + − cos 3 t π/2
0
j+ sin 2 t π/2
0
k =(1− 0) i +(0+1)j +(1− 0) k = i + j + k