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

SECTION 13.3 ARC LENGTH AND CURVATURE d 163
Then
Therefore
T '(t) e 2 t + 1 1 ( rn t( 2t) 2t 2t)
N (t) = !T '(t)! = J2 et (e2t + 1)2 v2e 1 - e , 2e , 2e
= 1 (J2et(1- e2t) 2e2t 2e2t) = _1_ (1- e2t J2et· J2et)
J2et(e2t+1) ' , e2t+ 1 , ,
!T'(t)! J2et 1 - J2et - J2e2t - .J2e2t
(b) ~~:(t) = lr'(t)l = e2t + 1 . et + e t - eSt+ 2et + e-t - e4t + 2e2t + 1 -:- (e2t + 1)2
21. r (t) = t 3 j + e k =} r'(t) = 3e j + 2t k, r"(t) = 6tj + 2 k, lr'(t)! = y'o 2 + (3t2)2 + (2t)2 = v'9t4 + 4t2,
, " _ 2 _ lr'(t) X r"(t)i _ ·6t 2 _ 6t 2
2 ) 312 .
r' (t) x r"(t) = -6t2 i, Jr (t) X r (t)J - 6t . Then ~~:(t)- I () 3 -
3 - ( 9 1 .
r' t I ( v'9t4 + 4t2 ) t' + 4t
23. r (t) = 3t i + 4sin t j + 4cost k =? r'(t) = 3 i + 4costj - 4sin tk, r"(t) = -4sintj - 4cos~k,
Jr '(t)! = Jg + 16 cos 2 t + 16sin 2 t = .J9+T6 = 5, r '(t) x r"(t) = -16 i + 12cos t j - 12 sin t k,
Jr' (t) x r " (t)! = }256 + 144 cos2 t + 144sin 2 t = J405 = 20. Then ~~:(t) = lr' (t) x r~ (t)J = 2 ~ = ..! .
, Jr'(t)[ · 5 25
25. r (t) = (t, t 2 , t 3 ) =? r' (t) = (1, 2t, 3t 2 ). The point (1, 1, 1) corresponds to t = 1, and 1) 1 (1) = (1, 2, 3) =?
Jr '(1)1 = v'1 + 4 +9 = Jf4. r"(t) = (0, 2,6t ) =?
r"(1) = (0, 2,6). r'(1) x r"(1) = (6, - 6,2),so
·lr'(1) x r"(1)! = yf36 + 36 + 4 = J76. Then ~~:(1_} = lr'(~;,~ 1 ~'~( 1 )1 = ::;; = ~ VSJ.
4 , 3 "( ) 2 ( ) lf"(x)! l12x 2 1 12x 2
27. f (x) = x , f (x) = 4x , f x = 12x • II: x = [l + (J'(x))2)3/2 = [ 1 + ( 4 x3)2]3/ 2 = ( 1
+ 16
x6)3/ 2
29. f(x) = xe"', J'(x) = xe"' + e"', f"(x) = xe"' + 2e"',
lf"(x)i !xe"'+2e"'l !x+2Je"'
~~:{x) = [1 + (/'(x))2J312 = [1 + (xe"' + e"')2)3/2 = [1 + (xe"' + e"')2 j3/2
Jy"(x)J e"' _ :r 2x -3/2
31. Since y' = y" = e"', the curvature is ~~:(x) = 312 = ( 1
+ e 2 "' ) 312 - e (1 + e ) .
. [1 + (y'(x))2]
To find the maximum curvature, we first find the critical numbers of ~~:(x):
1
2:r
+
3 2x 1 2 2x
'( ) _ "'( 2x)-3/2 "'(-;!)( 2x)- 5/2( 2 2x) _ x e - e _ x - e
II: x - e 1 + e + e . 2 1 + e e - e (1 + e2"')5/2 - e (1 + e2x)5/2.
~~:' (x) = 0 when 1 - 2e 2 "' = 0, so e 2 "' = ~ or x = - ~ ln 2. And since 1 - 2e 2 "' > 0 for x < - ~ ln 2 and 1 - 2e 2 "' < 0
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