Solução_Calculo_Stewart_6e

F.
146 ¤ CHAPTER 14 VECTOR FUNCTIONS ET CHAPTER 13
TX.10
37.
e t i +2t j +lnt k dt = e t dt i + 2tdt j + ln tdt k
= e t i + t 2 j +(t ln t − t) k + C, whereC is a vector constant of integration.
39. r 0 (t) =2t i +3t 2 j + √ t k ⇒ r(t) =t 2 i + t 3 j + 2 3 t3/2 k + C, whereC is a constant vector.
But i + j = r (1) = i + j + 2 k + C. ThusC = − 2 k and r(t) 3 3 =t2 i + t 3 2
j +
3 t3/2 − 2 k.
3
For Exercises 41–44, let u(t) =hu 1 (t),u 2 (t),u 3 (t)i and v(t) =hv 1 (t),v 2 (t),v 3 (t)i. In each of these exercises, the procedure is to apply
Theorem 2 so that the corresponding properties of derivatives of real-valued functions can be used.
41.
d
dt [u(t)+v(t)] = d dt hu 1(t)+v 1 (t),u 2 (t)+v 2 (t),u 3 (t)+v 3 (t)i
d
=
dt [u1(t)+v1 (t)] , d dt [u2(t)+v2(t)] , d
dt [u3(t)+v3(t)]
= hu 0 1(t)+v 0 1(t),u 0 2(t)+v 0 2(t),u 0 3(t)+v 0 3(t)i
= hu 0 1(t),u 0 2 (t) ,u 0 3(t)i + hv 0 1(t),v 0 2(t),v 0 3(t)i = u 0 (t)+v 0 (t)
43.
d
dt [u(t) × v(t)] = d dt hu 2(t)v 3 (t) − u 3 (t)v 2 (t),u 3 (t)v 1 (t) − u 1 (t)v 3 (t),u 1 (t)v 2 (t) − u 2 (t)v 1 (t)i
= hu 0 2v 3 (t)+u 2 (t)v 0 3(t) − u 0 3(t)v 2 (t) − u 3 (t)v 0 2(t),
u 0 3(t)v 1(t)+u 3(t)v 0 1 (t) − u 0 1(t)v 3(t) − u 1(t)v 0 3(t),
45.
u 0 1(t)v 2 (t)+u 1 (t)v 0 2(t) − u 0 2(t)v 1 (t) − u 2 (t)v 0 1(t)i
= hu 0 2(t)v 3 (t) − u 0 3(t)v 2 (t) ,u 0 3(t)v 1 (t) − u 0 1(t)v 3 (t),u 0 1(t)v 2 (t) − u 0 2(t)v 1 (t)i
+ hu 2(t)v 0 3(t) − u 3(t)v 0 2(t),u 3(t)v 0 1 (t) − u 1(t)v 0 3(t),u 1(t)v 0 2(t) − u 2(t)v 0 1(t)i
= u 0 (t) × v(t)+u(t) × v 0 (t)
Alternate solution: Let r(t) =u(t) × v(t). Then
r(t + h) − r(t) =[u(t + h) × v(t + h)] − [u(t) × v(t)]
=[u(t + h) × v(t + h)] − [u(t) × v(t)] + [u(t + h) × v(t)] − [u(t + h) × v(t)]
= u(t + h) × [v(t + h) − v(t)] + [u(t + h) − u(t)] × v(t)
(Be careful of the order of the cross product.) Dividing through by h and taking the limit as h → 0 we have
r 0 u(t + h) × [v(t + h) − v(t)] [u(t + h) − u(t)] × v(t)
(t) = lim
+ lim
= u(t) × v 0 (t)+u 0 (t) × v(t)
h→0 h
h→0 h
by Exercise 14.1.43(a) [ET 13.1.43(a)] and Definition 1.
d
dt [u(t) · v(t)] = u0 (t) · v(t)+u(t) · v 0 (t) [by Formula 4 of Theorem 3]
= hcos t, − sin t, 1i·ht, cos t, sin ti + hsin t, cos t, ti·h1, − sin t, cos ti
= t cos t − cos t sin t +sint +sint − cos t sin t + t cos t
=2t cos t +2sint − 2cost sin t