Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

162 0 CHAPTER 13 VECTOR FUNCTIONS
9. r (t) = (sint,cost,tant) =} r'(t) = (cost, - sint,sec 2 t) =}
lr'(t)l = ..jcos 2 t + (-sint)2 + (sec 2 t)2 = v'l + sec 4 t and L = J 0
,. 14 lr'(t)l dt = J~"' 14 v'1 +sec 4 t dt ~ 1.2780.
11. The projection of the curve C onto the xy-plane is the c6rve x 2 = 2y or y = ~ x 2 , z = 0. Then we can choose the parameter
x = t =} y = ~t 2 . Since C also lies on the surface 3z = xy, we have z .= ~xy = ~(t)(~t2) = i t 3 . Then paramen:ic
equations for C are x = t, y = ~ t2, z = *t 3 and the corresponding vector equation is r (t) = (t, ~ t 2 , ~ t 3 ). The origin
corresponds to t = 0 and the point (6, 18,_36) corresponds tot = 6, so
L = f 0
6
lr'(t)l dt = ]~ 6
1(1, t , lt 2 )1 dt = J~ V1 2 + t 2 + (~t 2 ) 2 dt = f 0
6
J1 + t 2 + i t4. dt
= f 0
6
J(1 + ~ t2 )2 dt = j~( 1 + tt 2 ) dt = [t + ~t 3 )~ = 6 + 36 = 42
13. r (t) = 2t i + (1 - 3t) j + (5 + 4t) k =} r'(t) = 2i- 3j + 4 k and "* = lr'(t)l :::::: v'4 +9 + 16 = .;29. Then
8 = 8( t) = J~ lr' ( u) I du = J; .;29 du = .;29 t. Therefore, t = ~ 8, and substituting for t in the original equation, we
have r (t(s)) = ~8 i + (1- ~8)j + (5+ ~8) k.
15. Here r (t) = (3 sin t , 4t, 3 cos t), so r' (t) = (3cos t, 4, -3 sin t) and lr'(t)l = ..)9 cos 2 t + 16 + 9 sin 2 t = v'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 8( t) = J; lr' ( u) I du = J; 5 du = 5t. 8( t) = 5 =} 5t = 5 . =} t = 1, thus your location after
moving 5 units along the curve is (3 sin 1, 4, 3 cos 1).
17. (a) r (t) = (t, 3 cost, 3 sin t) =} r' (t) = (1, - 3 sin t, 3 cost) =} lr' (t)l = ..)i + 9sin 2 t + 9 cos 2 t = v'IO.
· ( ) r' (t) 1 ( . ) / 1 ;l • 3 )
ThenT t = lr'{t)l = "710 1, - 3 sm t,3cost or \"Tlii,-'710smt,'711icost.
T'(t) = fto (0, - 3cost, -3sint) =} IT '(t)l = 7io ..)o + 9cos 2 t + 9sin 2 t = #o· Thus
N (t) = ~ ~:~~~ ~ = ~~~ (0,-3cos t,-3sin t) = (0,-cost,-sint).
(b) ~t(t) = IT '(t)l = 3/v'W = ~
lr'(t)l v'W 10
19. (a)r(t) = (v'2t,et,e- t) =} r'(t) = (v'2,et,-e-t) =} Jr'(t)J=v'2 +e2L + e 2 t =..J(et+e-t)2= et+e-t.
Then
[after multiplying by ::]
and
T '(t) = _1_ (v'2e' 2e2t 0)- 2e2t (V'fe' e2t - 1)
e 2 '+ 1 ' ' (e 2 ' + 1) 2 ''
~ (e 2 ' ~ 1
)2 ((e 2 ' + 1) (v'2e' ,2e 2 ',0) - 2e 2 ' ( .J2e',e 2 ', - 1)) = (e2' ~ 1
) 2 (v'2e' (1 .:... e 2 ') ,2e 2 ',2e 2 ')
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