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

5. f~(t 2 i +tcos 1rtj +sin 1rt k ) dt = (! 0
1 t 2 dt) i + (I~ tcos 1rtdt) j +(I; sin 1rtdt) k
where we integrated by parts in the y-component.
I
= [ ~t 3 ]~ i + (~sin 1rt]~ - f 0
1 ~sin 1rtdt) j +[-~cos 1rt]~ k
1 • [ 1 t]l . 2 k 1 • 2 • 2 k
= ]j l + ;1fCOS1l" 0 J+ ;r ;= 3 1 - ;:lfJ+ ;r
1. r(t) = (t 2 , t 3 , t 4 ) . =? r 1 (t) = ( 2t,3e,4t 3 ) =? Jr 1 (t) J = J 4t 2 + 9t4 + 16t 6 and
CHAPTER 13 REVIEW· 0 175
. L = J; Jr 1 (t)J dt = J; J4t2 + 9t 4 + 16t 6 dt. Using Simpson's Rule with f(t) = J4t 2 + 9t4 + 16t 6 and n = 6 we
have tlt = 3 (i 0 = ~ and
L ~ ';' [f(O) + 4f(~) + 2f{1) + 4Ja) + 2/(2) + 4f(~) + /{3)]
= i [vo +o +o + 4 . ..j4(~/ + 9 (~) 4 + 16( ~) 6 + 2 . .J4(1)2 + 9{1) 4 + 16(1)6
~ 86.631
+ 4. J4G) 2 + 9G) 4 + 16Gt + 2 . .J4(2)2 + 9(2)4 + 16{2)6
+4· 4(~) 2 +9(~) 4 +16(~t+.J4(3) 2 +9{3)4+ 16(3)6 ]
9. The angle of intersection of the two curves, 8, is the angle between their respective tangents at the point of intersection.
For both curves the point {1, 0, 0) occurs when t = 0.
r W) = - sin t i +cos t j + k =? rHO) = j + k and r 2(t) = i + 2tj + 3t 2 k =} r~(O) = i.
r i {0) · r~ (0) = (j + k ) · i = 0. Therefore, the curves intersect in a right angle, that is, 8 = ~.
(b) T'(t) = -Ht 4 +t 2 + 1)- 3 1 2 (4t 3 +2t) (t 2 ,t, 1) + (t 4 + t 2 + 1)- 1 1 2 (2t, 1, 0)
- 2t 3 - t 2 1
(t4 + t2 + 1)3/2 (t , t , 1) + (t4 + t2 + 1)1/2 (2t, 1, 0)
- ( -2t 5 - t 3 , - 2t 4 - t 2 , - 2t 3 - t) + (2t 5 + 2t 3 + 2t, t 4 + t 2 + 1, 0) - (t 3 + 2t, - t 4 + 1, -2t 3 - t)
- (t4 + t2 + 1)3/2 - (t4 + t2 + 1)3/2
1 Jt 6 + 4t 4 + 4t 2 + t 8 - 2t 4 + 1 + 4t6 + 4t 4 + t 2 JtB + 5t 6 + 6t 4 + 5t2 + 1
IT (t)l = (t4 + t2 + 1)3/2 = (t4 + t2 + 1)3/2
and
(t 3 + 2t 1 - t 4 - 2t 3 - t)
N (t) - ' '
- Jt!!. + 5t6 + 6t4 + 5t2 + 1 .
JT 1 (t)\ J t!!. + 5t 6 + 6t 4 + 5t2 + 1 Jt
(c) "'(t) = Jrl(t)l = 4 + 4t 2 + 1
(t4 + t2 + 1)2 or (t4 + t2 + 1)3/2
I
13 4 3 II 12 2 d ( ) lv"l
. y = X , y = X an It X = ( 1 + (y')2)3/ 2
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