Solução_Calculo_Stewart_6e

F.
156 ¤ CHAPTER 14 VECTOR FUNCTIONS ET CHAPTER 13
TX.10
reaches the east bank after 8 s, and it is located at r(8) = 5(8)i + 3
4 (8)2 − 1 16 (8)3 j =40i +16j. Thus the boat is 16 m
downstream.
(b) Let α be the angle north of east that the boat heads. Then the velocity of the boat in still water is given by
5(cos α) i +5(sinα) j. Att seconds, the boat is 5(cos α)t meters from the west bank, at which point the velocity
of the water is
3
[5(cos α)t][40 − 5(cos α)t] j. The resultant velocity of the boat is given by
400
v(t) =5(cosα) i + 5sinα + 3
400 (5t cos α)(40 − 5t cos α) j =(5cosα) i + 5sinα + 3 2 t cos α − 3 16 t2 cos 2 α j.
Integrating, r(t) =(5t cos α) i + 5t sin α + 3 4 t2 cos α − 1 16 t3 cos 2 α j (where we have again placed
the origin at A). The boat will reach the east bank when 5t cos α =40 ⇒ t = 40
5cosα = 8
cos α .
In order to land at point B(40, 0) we need 5t sin α + 3 4 t2 cos α − 1 16 t3 cos 2 α =0
2 3 8
8
8
5 sin α + 3 cos α − 1 cos 2 α =0 ⇒ 1 (40 sin α +48− 32) = 0
4
16
cos α
cos α
cos α
cos α ⇒
⇒
40 sin α +16=0 ⇒ sin α = − 2 . Thus α =sin−1
− 2 5 5 ≈−23.6 ◦ , so the boat should head 23.6 ◦ south of
east (upstream). The path does seem realistic. The boat initially heads
upstream to counteract the effect of the current. Near the center of the river,
the current is stronger and the boat is pushed downstream. When the boat
nears the eastern bank, the current is slower and the boat is able to progress
upstream to arrive at point B.
33. r(t) =(3t − t 3 ) i +3t 2 j ⇒ r 0 (t) =(3− 3t 2 ) i +6t j,
|r 0 (t)| = (3 − 3t 2 ) 2 +(6t) 2 = √ 9+18t 2 +9t 4 = (3 − 3t 2 ) 2 =3+3t 2 ,
r 00 (t) =−6t i +6j, r 0 (t) × r 00 (t) =(18+18t 2 ) k. Then Equation 9 gives
a T = r0 (t) · r 00 (t)
|r 0 (t)|
a T = v 0 = d 3+3t
2
=6t
dt
= (3 − 3t2 )(−6t)+(6t)(6) 18t +18t3
= = 18t(1 + t2 )
3+3t 2 3+3t 2 3(1 + t 2 )
and Equation 10 gives a N = |r0 (t) × r 00 (t)|
=
|r 0 (t)|
=6t
or by Equation 8,
18 + 18t2
3+3t 2 = 18(1 + t2 )
3(1 + t 2 ) =6.
35. r(t) =cost i +sint j + t k ⇒ r 0 (t) =− sin t i +cost j + k, |r 0 (t)| = sin 2 t +cos 2 t +1= √ 2,
r 00 (t) =− cos t i − sin t j, r 0 (t) × r 00 (t) =sint i − cos t j + k.
Then a T = r0 (t) · r 00 (t)
|r 0 (t)|
=
sin t cos t − sin t cos t
√
2
=0and a N = |r0 (t) × r 00 (t)|
|r 0 (t)|
=
√
sin 2 t +cos 2 t +1 2
√ = √ =1.
2 2
37. r(t) =e t i + √ 2 t j + e −t k ⇒ r 0 (t) =e t i + √ 2 j − e −t k, |r(t)| = √ e 2t +2+e −2t = (e t + e −t ) 2 = e t + e −t ,
r 00 (t) =e t i + e −t k.Thena T = e2t − e −2t
= (et + e −t )(e t − e −t )
= e t − e −t =2sinht
e t + e −t e t + e −t
√ 2e −t i − 2 j − √ 2e t k 2(e
−2t
+2+e
and a N =
=
2t )
= √ 2 et + e −t
e t + e −t e t + e −t
e t + e = √ 2.
−t