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

SECTION 13.3 ARC LENGTH AND CURVATURE D 165
41. Using a CAS, we find (after simplifYing)
K(!)
6 v'4cos2 t - 12cost + 13
~~:(t) = ( 17
_ 12
cos t) 3 12 • (To compute cross
products in Maple, use the VectorCa1cu1us or
Li nearA1gebra package and the Cross Produc t (a, b)
command; in Mathematica, use Cross (a, b].) Curvature is
largest at integer multiples of2:rr.
0 21T 41T 61r I
43. X = t 2 => :i; = 2t => X = 2, y = t 3 => iJ = 3t 2 => ij = 6t.
\xy- iJx\ j(2t)(6t) - (3t 2 )(2) j j12t2- 6t 2 1 6t2
Then ~~:(t) = [:i;2 + y2)3/2 = [(2t)2 + (3t2)2j3/ 2 = (4t2 + W)3/2 = (4t2 + 9t4)3/2 ·
45. x = et cos t => :i; = e 1 (cos t- sint) => x = e 1 ( - sint- cost)+ et(cost- sin t) = - 2et sin t,
y = et sint => iJ = et (cos t + sin't) => y = e 1 (- sint +cost)+ et(cos t + sint) = 2et cost. Then
\:i;ij - 1ixl jet(cos t- sin t)(2et cost) - e 1 (cos t + sin t)( -2et sin t}j
x;( t) = . . = .:....__;..______;_;_---:-:_-:---:--------'-:-'-;;-;;;--~
[x 2 + y 2 ] 3 1 2 ([et(cos t- sin t)J2 + [et(cos t +sin t)j2) 3 / 2
j2e 21 (cos 2 t - sin t cost + sin t cos t + sin 2 t)j j2e 2 t(1) j 2e2t 1
= [e21(cos2t - 2costsint +sin 2 t + cos2 t + 2cos t sint +sin 2 t)] 3 2
' = [e2t(1 + 1)] 3 1 2 = e 31 (2) 3 / 2 = -..fiet
2 ) _ _ r'(t) _ (2t,2t 2 ,1) _ (2t,2t\ 1) _ ( 2 2 1
2 t 2 + 1
,soT(1) - 3•3•3)·
47. ( 1, 3 ,1 correspondstot - 1. T (t)-\r'(t)\ - v' 4
t 2 + 4
t 4
+ - 1
T '(t) = - 4t(2t 2 + 1)- 2 ( 2t, 2t 2 , 1) + (2t 2 + 1)- 1 (2,4t, 0) [by Formula 3 of Theorem 13.2.3]
= (2t 2 + 1)- 2 ( -8t 2 + 4t 2 + 2, -8t 3 + 8t 3 + 4t, -4t) = 2(2t2 + 1)- 2 (1- 2t 2 , 2t, -2t)
(1 - 2t 2 , 2t, -2t) - (1- 2t 2 , 2t, - 2t)
v'1 - 4t2 + 4t4 + 8t2 - 1 + 2t 2
49. (0,7r,-?) corresponds tot ='lr. r(t) = (2sin 3t,t,2cos3t) =>
T(t) = r' (t) = (6 cos 3t,1,-6sin3t) =: - 1- ( 6 cos 3 t, 1 ,- 6 sin 3 t ).
\r'(t)\ } 36 cos2 3t + 1 + 36sin 2 3t V37
T(1r) = ~ ( - 6, 1, 0) is a normal vector for the normal plane, and so (-6, 1, 0) is also normal. Thus an equation for the
plane ·is - 6(x - 0) + 1(y -1r) + O(z + 2) = 0 ory·- 6x = ~ ·
18
ffi=>
N (t) = ~~:m , = (-sin3t, O, - cos 3t). So N(1r) = (0, 0, 1) and B (1r) = ~ (-6, 1, 0) x (0,0, 1) = ~ {1, 6, 0).
Since B(1r) is a normal to the osculating plane, so is (1, 6, 0).
An equation for the plane is 1(x- 0) + 6(y -1r) + O(z + 2) = 0 or x + 6y = 61r.
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