Solução_Calculo_Stewart_6e

F.
148 ¤ CHAPTER 14 VECTOR FUNCTIONS ET CHAPTER 13
TX.10
13. r(t) =2t i +(1− 3t) j +(5+4t) k ⇒ r 0 (t) =2i − 3 j +4k and ds = dt |r0 (t)| = √ 4+9+16= √ 29. Then
√ √
29 du = 29 t. Therefore, t = √
1
29
s, and substituting for t in the original equation, we
s = s(t) = t
0 |r0 (u)| du = t
0
have r(t(s)) = 2 √
29
s i +
1 − 3 √
29
s
j + 5+ √ 4
29
s k.
15. Here r(t) =h3sint, 4t, 3costi,sor 0 (t) =h3cost, 4, −3sinti and |r 0 (t)| = 9cos 2 t +16+9sin 2 t = √ 25 = 5.
The point (0, 0, 3) corresponds to t =0, so the arc length function beginning at (0, 0, 3) and measuring in the positive
direction is given by s(t) = t
0 |r0 (u)| du = t
5 du =5t. s(t) =5 ⇒ 5t =5 ⇒ t =1, thus your location after
0
moving 5 units along the curve is (3 sin 1, 4, 3 cos 1).
17. (a) r(t) =h2sint, 5t, 2costi ⇒ r 0 (t) =h2cost, 5, −2sinti ⇒ |r 0 (t)| = 4cos 2 t +25+4sin 2 t = √ 29.
Then T(t) = r0 (t)
|r 0 (t)| = 1 √
29
h2cost, 5, −2sinti or
√
2
29
cos t,
√
5
29
, − √ 2
29
sin t .
T 0 (t) = 1 √
29
h−2sint, 0, −2costi ⇒ |T 0 (t)| = 1 √
29
4sin 2 t +0+4cos 2 t = 2 √
29
. Thus
N(t) = T0 (t)
|T 0 (t)| = 1/√ 29
2/ √ h−2sint, 0, −2costi = h− sin t, 0, − cos ti.
29
(b) κ(t) = |T0 (t)|
|r 0 (t)|
= 2/√ 29
√
29
= 2 29
19. (a) r(t) = √ 2 t, e t ,e −t ⇒ r 0 (t) = √ 2,e t , −e −t ⇒ |r 0 (t)| = √ 2+e 2t + e −2t = (e t + e −t ) 2 = e t + e −t .
Then
T(t) = r0 (t)
|r 0 (t)| = 1 √
2,e t , −e −t 1 √
=
2e t ,e 2t , −1
after multiplying by et
and
e t + e −t e 2t +1
e t
T 0 1 √
(t)=
2e t , 2e 2t , 0 2e 2t √
−
2e t
e 2t +1
(e 2t +1) 2 ,e 2t , −1
1
=
(e 2t +1) √ 2e t , 2e 2t , 0 − 2e √ 2t 2e t ,e 2t , −1 1 √
=
2e
t
1 − e 2t , 2e 2t , 2e 2t
(e 2t +1) 2 (e 2t +1) 2
Then
|T 0 1 1
(t)| =
2e
2t
(1 − 2e
(e 2t +1) 2t + e 4t )+4e 4t +4e 4t =
2e 2 (e 2t +1) 2 2t
(1 + 2e 2t + e 4t )
Therefore
(b) κ(t) = |T0 (t)|
|r 0 (t)|
=
1
2e
(e 2t +1) 2t (1 + e 2t ) 2 =
2
√
2e t (1 + e 2t )
(e 2t +1) 2 =
√
2 e
t
e 2t +1
N(t)= T0 (t)
|T 0 (t)| = e2t +1 1 √
√ 2 e t (1 − e 2t ), 2e 2t , 2e 2t
2 e
t (e 2t +1) 2
1 √
= √ 2 e t (1 − e 2t ), 2e 2t , 2e 2t 1
= 1 − e 2t , √ 2 e t , √ 2 e t
2 et (e 2t +1)
e 2t +1
=
√
2 e
t
e 2t +1 ·
1
e t + e −t =
√
2 e
t
e 3t +2e t + e −t =
√
2 e
2t
e 4t +2e 2t +1 =
√
2 e
2t
(e 2t +1) 2