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

SECTION 13.3 ARC LENGTH AND CURVATURE 0 167
(c) B = T x N => T _L N , B _L T and B _LN. SoB, T and N form an orthogonal set of vectors in the threedimensional
space !R 3 . From parts (a) and (b), dBfds is perpendicular to both BandT, so dB/ ds is parallel to N .
Therefore, dB/ ds = -r(s)N , where r(s) is a scalar.
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(d) Since B = T x N , T _L N and both T and N are unit vectors, B is a unit vector mutually perpendicular to both T and
N. For a plane curve, T and N always lie in the plane of the curve, so that B is a constant unit vector always
perpendicular to the plane. Thus dB/ ds = 0, but dBfds = - r(s)N and N =/= 0, so r(s) = 0.
61. (a) r ' = s' T => r" = s" T + s' T ' = s" T + s' ~~ s' = s" T + ~(s') 2 N by the first Serret-Frenet formula.
(b) Using part (a), we have
(c) Using part (a), we have
r' X r" = (s' T ) X (s" T + ~cs') 2 N ]
= ((si T ) x (s" T )] + ((s'T ) x (~(s'? N)) [by Property 3 of Theorem 12.4.11)
= (s' s")(T x T ) + ~~;(s') 3 (T x N ) = 0 + ~~;(s'? B = ~~;(s') 3 B
r 111 = [s" T + ~~;(s') 2 N ]' = s 111 T + s" T ' ;t- ~'(s'? N + 2~~;s' s" N + ~~;(s'? N '
= s"' T + s"dT s' + ~'(s'? N + 2~s' s" N + ~(s') 2 dN s'
ds
ds
= s"' T + s" s'~ N + K 1 (s') 2 N + 2~s' s" N + ~~;(s') 3 (-~ T + rB) [by the second formula]
= [s"'- K 2 {s'?J T + (3~s' s" + ~~;'(s') 2 ] N + ~~;r(s') 3 B
(d) Using parts (b) and (c) and the facts that B · T = 0, B · N = 0, and B · B = 1, we get
(r ' x r ") . r 111 = ~~:(s') 3 B · { (s 111 - ~ 2 (s') 3 ] T + (3~s' s" + ~~ (s') 2 ] N + KT(s') 3 B} = ~~:(s') 3 ~r(s')3 = T
Jr ' x r " J 2 · J~~:(s')3 B J 2 (~~;(s')3] 2 •
63. r = (t, tt2, !t 3 ) => r' = (1, t, t 2 ), r" = (0, 1, 2t), r 111 = (0, 0, 2) => r ' x r" = (t 2 , - 2t, 1) =>
(r'x r")· r"' (t2, - 2t,1) ·(0,0,2) 2
T - - - --,---....,..--
- Jr' x r 11 J 2 - t 4 + 4t 2 + 1 - t 4 + 4t 2 + 1
65. For one helix, the vector equation is r (t) = (10 cost, 10 sin t, 34t/(2rr)) (measuring in angstroms), because the radius of each
helix is 10 angstroms, and z increases by 34 angstroms for each increase of 271' in t. Using the arc length formula, Jetting t go
from 0 to 2.9 x 10 8 x 21r, we find the appmximate length of each helix to be
B s . / ] 2.9xl0 8 x2rr
L = J;·9xlO
x zrr Jr'(t)J dt = j~ · 9 x 10 x zrr y (- 10 sint) 2 + {10cost) 2 + U!) 2 dt = ..j100 + (~!) 2 t
0
= 2.9 x 10 8 x 21r J10o + ( g! ) 2 ~ 2.07 x 10 10 A- more than two meters!
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