Calculo.Solucoes.JS.Ed.6.Ingles part 01

Prévia do material em texto

23/11/2010.TX
TX.10
F.
"O que sabemos é uma gota, o que ignoramos é um oceano." Isaac Newton
Solução detalhada de todos os exercícios ímpares do livro: Cálculo, James Stewart - VOLUME I e VOLUME II
152 ¤ CHAPTER 14 VECTOR FUNCTIONS ET CHAPTER 13
55. (a) r0 = s0T ⇒ r00 = s00T+ s0T0 = s00T+ s0 dT
ds
s0 = s00T+ κ(s0)2N by the first Serret-Frenet formula.
(b) Using part (a), we have
r0 × r00 = (s0T)× [s00T+ κ(s0)2N]
= [(s0T)× (s00T)] + �(s0T)× (κ(s0)2N)� (by Property 3 of Theorem 13.4.8 [ ET 12.4.8])
= (s0s00)(T×T) + κ(s0)3(T×N) = 0+ κ(s0)3B = κ(s0)3B
(c) Using part (a), we have
r000 = [s00T+ κ(s0)2N]0 = s000T+ s00T0 + κ0(s0)2N+ 2κs0s00N+ κ(s0)2N0
= s000T+ s00
dT
ds
s0 + κ0(s0)2N+ 2κs0s00N+ κ(s0)2
dN
ds
s0
= s000T+ s00s0κN+ κ0(s0)2N+ 2κs0s00N+ κ(s0)3(−κT+ τ B) [by the second formula]
= [s000 − κ2(s0)3]T+ [3κs0s00 + κ0(s0)2]N+ κτ(s0)3B
(d) Using parts (b) and (c) and the facts thatB ·T = 0,B ·N = 0, andB ·B = 1, we get
(r0 × r00) · r000
|r0 × r00|2 =
κ(s0)3B · �[s000 − κ2(s0)3]T+ [3κs0s00 + κ0(s0)2]N+ κτ(s0)3B�
|κ(s0)3B|2 =
κ(s0)3κτ(s0)3
[κ(s0)3]2
= τ .
57. r =
t, 1
2
t2, 1
3
t3
�
⇒ r0 =
1, t, t2
�
, r00 = h0, 1, 2ti, r000 = h0, 0, 2i ⇒ r0 × r00 = 
t2,−2t, 1� ⇒
τ =
(r0 × r00) · r000
|r0 × r00|2 =
t2,−2t, 1
� · h0, 0, 2i
t4 + 4t2 + 1
=
2
t4 + 4t2 + 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× 108 × 2π, we find the approximate length of each helix to be
L=
U 2.9×108×2π
0
|r0(t)| dt = U 2.9×108×2π
0
t
(−10 sin t)2 + (10 cos t)2 +
�
34
2π
�2 dt =
t
100 +
�
34
2π
�2
�2.9×108×2π
= 2.9× 108 × 2π
t
100 +
�
34
2π
�2 ≈ 2.07× 1010 Å—more than two meters!
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
vave =
r(1)− r(0)
1− 0 =
(4.5 i+ 6.0 j+ 3.0k)− (2.7 i+ 9.8 j+ 3.7k)
1
= 1.8 i− 3.8 j− 0.7k. Similarly, over the other
intervals we have
[0.5, 1] : vave =
r(1)− r(0.5)
1− 0.5 =
(4.5 i+ 6.0 j+ 3.0k)− (3.5 i+ 7.2 j+ 3.3k)
0.5
= 2.0 i− 2.4 j− 0.6k
[1, 2] : vave =
r(2)− r(1)
2− 1 =
(7.3 i+ 7.8 j+ 2.7k)− (4.5 i+ 6.0 j+ 3.0k)
1
= 2.8 i+ 1.8 j− 0.3k
[1, 1.5] : vave =
r(1.5)− r(1)
1.5− 1 =
(5.9 i+ 6.4 j+ 2.8k)− (4.5 i+ 6.0 j+ 3.0k)
0.5
= 2.8 i+ 0.8 j− 0.4k
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