Solução_Calculo_Stewart_6e

F.
152 ¤ CHAPTER 14 VECTOR FUNCTIONS ET CHAPTER 13
TX.10
55. (a) r 0 = s 0 T ⇒ r 00 = s 00 T + s 0 T 0 = s 00 T + s 0 dT
ds s0 = s 00 T + κ(s 0 ) 2 N by the first Serret-Frenet formula.
(b) Using part (a), we have
r 0 × r 00 =(s 0 T) × [s 00 T + κ(s 0 ) 2 N]
(c) Using part (a), we have
=[(s 0 T) × (s 00 T)] + (s 0 T) × (κ(s 0 ) 2 N) (by Property 3 of Theorem 13.4.8 [ ET 12.4.8])
=(s 0 s 00 )(T × T)+κ(s 0 ) 3 (T × N) =0 + κ(s 0 ) 3 B = κ(s 0 ) 3 B
r 000 =[s 00 T + κ(s 0 ) 2 N] 0 = s 000 T + s 00 T 0 + κ 0 (s 0 ) 2 N +2κs 0 s 00 N + κ(s 0 ) 2 N 0
= s 000 T + s 00 dT
ds s0 + κ 0 (s 0 ) 2 N +2κs 0 s 00 N + κ(s 0 ) 2 dN
ds s0
= s 000 T + s 00 s 0 κ N + κ 0 (s 0 ) 2 N +2κs 0 s 00 N + κ(s 0 ) 3 (−κ T + τ B) [by the second formula]
=[s 000 − κ 2 (s 0 ) 3 ] T +[3κs 0 s 00 + κ 0 (s 0 ) 2 ] N + κτ(s 0 ) 3 B
(d) Using parts (b) and (c) and the facts that B · T =0, B · N =0,andB · B =1,weget
(r 0 × r 00 ) · r 000
|r 0 × r 00 | 2 = κ(s0 ) 3 B · [s 000 − κ 2 (s 0 ) 3 ] T +[3κs 0 s 00 + κ 0 (s 0 ) 2 ] N + κτ(s 0 ) 3 B
|κ(s 0 ) 3 B| 2 = κ(s0 ) 3 κτ(s 0 ) 3
[κ(s 0 ) 3 ] 2 = τ.
57. r = t, 1 2 t2 , 1 t3 ⇒ r 0 3
= 1,t,t 2 , r 00 = h0, 1, 2ti, r 000 = h0, 0, 2i ⇒ r 0 × r 00 = t 2 , −2t, 1 ⇒
τ = (r0 × r 00 ) · r 000 t 2 , −2t, 1 ·h0, 0, 2i 2
|r 0 × r 00 | 2 =
=
t 4 +4t 2 +1 t 4 +4t 2 +1
59. For one helix, the vector equation is r(t) =h10 cos t, 10 sin t, 34t/(2π)i (measuring in angstroms), because the radius of each
helix is 10 angstroms, and z increases by 34 angstroms for each increase of 2π in t. Using the arc length formula, letting t go
from 0 to 2.9 × 10 8 × 2π,wefind the approximate length of each helix to be
L = 2.9×10 8 ×2π
|r 0 (t)| dt =
2.9×10 8 ×2π
(−10 sin t)
0 0
2 +(10cost) 2 +
34 2
2π
dt = 100 +
2.9×10
8 ×2π
34 2
2π
=2.9 × 10 8 × 2π 100 +
34 2
≈ 2.07 × 10 10 Å — more than two meters!
2π
14.4 Motion in Space: Velocity and Acceleration ET 13.4
1. (a) If r(t) =x(t) i + y (t) j + z(t) k is the position vector of the particle at time t, then the average velocity over the time
interval [0, 1] is
v ave =
r(1) − r(0)
1 − 0
intervals we have
=
[0.5, 1] : v ave =
[1, 2] : v ave =
[1, 1.5] : v ave =
(4.5 i +6.0 j +3.0 k) − (2.7 i +9.8 j +3.7 k)
1
r(1) − r(0.5)
1 − 0.5
r(2) − r(1)
2 − 1
r(1.5) − r(1)
1.5 − 1
=
=
=
(4.5 i +6.0 j +3.0 k) − (3.5 i +7.2 j +3.3 k)
0.5
(7.3 i +7.8 j +2.7 k) − (4.5 i +6.0 j +3.0 k)
1
(5.9 i +6.4 j +2.8 k) − (4.5 i +6.0 j +3.0 k)
0.5
=1.8 i − 3.8 j − 0.7 k. Similarly, over the other
=2.0 i − 2.4 j − 0.6 k
=2.8 i +1.8 j − 0.3 k
=2.8 i +0.8 j − 0.4 k