Solução_Calculo_Stewart_6e

F.
144 ¤ CHAPTER 14 VECTOR FUNCTIONS ET CHAPTER 13
TX.10
3. Since (x +2) 2 = t 2 = y − 1 ⇒
y =(x +2) 2 − 1,thecurveisa
parabola.
(a), (c)
(b) r 0 (t) =h1, 2ti,
r 0 (−1) = h1, −2i
5. x =sint, y =2cost so
x 2 +(y/2) 2 =1and the curve is
an ellipse.
(a), (c) (b) r 0 (t) =cost i − 2sint j,
π
√
2
r 0 =
4 2 i − √ 2 j
7. Since y = e 3t =(e t ) 3 = x 3 ,the
curve is part of a cubic cuve. Note
that here, x>0.
(a), (c) (b) r 0 (t) =e t i +3e 3t j,
r 0 (0) = i +3j
d
9. r 0 (t)=
dt [t sin t] , d dt
t
2 , d dt [t cos 2t]
= ht cos t +sint, 2t, t(− sin 2t) · 2+cos2ti
= ht cos t +sint, 2t, cos 2t − 2t sin 2ti
11. r(t) =i − j + e 4t k ⇒ r 0 (t) =0i +0j +4e 4t k =4e 4t k
13. r(t) =e t2 i − j +ln(1+3t) k ⇒ r 0 (t) =2te t2 i + 3
1+3t k
15. r 0 (t) =0 + b +2t c = b +2t c by Formulas 1 and 3 of Theorem 3.
17. r 0 (t) = −te −t + e −t , 2/(1 + t 2 ), 2e t ⇒ r 0 (0) = h1, 2, 2i. So|r 0 (0)| = √ 1 2 +2 2 +2 2 = √ 9=3and
T(0) = r0 (0)
|r 0 (0)| = 1 h1, 2, 2i = 1
2
3 3 3 3 .
19. r 0 (t) =− sin t i +3j +4cos2t k ⇒ r 0 (0) = 3 j +4k. Thus
T(0) = r0 (0)
|r 0 (0)| = 1
√
02 +3 2 +4 (3 j +4k) = 1 (3 j +4k) = 3 j + 4 k.
2 5 5 5
21. r(t) = t, t 2 ,t 3 ⇒ r 0 (t) = 1, 2t, 3t 2 .Thenr 0 (1) = h1, 2, 3i and |r 0 (1)| = √ 1 2 +2 2 +3 2 = √ 14, so
T(1) = r0 (1)
|r 0 (1)| = √ 1
1
14
h1, 2, 3i = √
2
14
, √
3
14
, √
14
. r 00 (t) =h0, 2, 6ti,so
i j k
r 0 (t) × r 00 (t) =
1 2t 3t 2
2t 3t 2
=
0 2 6t
2 6t i − 1 3t 2
0 6t j + 1 2t
0 2 k
=(12t 2 − 6t 2 ) i − (6t − 0) j +(2− 0) k = 6t 2 , −6t, 2