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

166 0 CHAPTER 13 VECTOR FUNCTIONS
51. The ellipse is given by the parametric equations x = 2 cost, y = 3 sin t, so using the result from Exercise 42,
~~:(t) = l:i:ii- xyl = l(-2sint}(-3sint)- (3cost)(- 2cost)1 = 6
[:i: 2 + 1?] 3 / 2 ( 4 sin 2 t + 9 cos 2 t~ S/ 2 ( 4 sin 2 t + 9 cos 2 t)3/ 2 •
At (2, 0}, t = 0. N.ow ~~:(0} = 2
67
= ~.so the radius of the osculating circle is
1/ ~(0} = ~ and its center is ( -~, 0). Its equation is therefore (x + ~ ) 2 + y 2 = ¥·
s
At (0, 3}, t = ~.and~~:(~) = ~ = ~· So the radius of the osculating circle is~ and
its center is ( 0, ~). Hence its equation is x 2 + (y - ~) 2 = lf.
53. The tangent vector is normal to the normal plane, and the ve~ tor (6, 6, - 8) is normal to the given plane.
But T (t) II r'(t) and (6, 6, -8) II (3, 3, -4), so we need to find t such that r' (t) II (3, 3, - 4}.
r (t) = (t 3 1 3t, t4 ) :::::? r ' (t) = (3t 2 , 3, 4t 3 ) II (3, 3, - 4) when t = - 1. So the planes are parallel at the point ( - 1, -3, 1).
55. First we parametrize the curve of intersection. We can choose y = t; then x = y 2 = t 2 and z = x 2 = t 4 , and the curve is
given by r (t) = (t 2 , t, t 4 ). r ' (t) = (2t, 1, 4t 3 ) and the point (1, 1, 1) corresponds tot= 1, so r ' (1) := (2, 1, 4) is a normal
vector for the normal plane. Thus an equation of the normal plane is
r' (t) 1
2(x - 1) + 1(y - 1} + 4(z- 1) = 0 or 2x + y + 4z = 7. T (t) = -
1
'( )I = . ( 2t, 1, 4t 3 ) and
· r t J 4t 2 + 1 + 16t 6
T ' (t) = - H 4t2 + 1 + 1?t 6 ) - S/ 2 (8t + 96t 5 ) (2t, 1, 4t 3 ) + ( 4t 2 + 1 + 16t 6 ) - l/ 2 ( 2, 0, 12t 2 ). A normal vector for
the osculating plane is B(l) = T (1) x N (1), but r'(1) = (2, 1, 4) is parallel to T (1) and
T ' (1} = - ~ (21)- 3 : 2 ( 104} (2, 1, 4} + (21)- 1 1 2 (2, 0, 12) = 21
3TI (-31, -26, 22) is parallel to N (1) as is (-31, .:...26, 22),
so (2, 1, 4} x (- 31, - 26, 22} = (126, -168, -21) is normal to the osculating plane. Thus an equation for the osculating
plane is 126(x - 1) - 168(y - 1) - 21(z- 1) = 0 or 6x- 8y - z = -3.
dTI dT
57 - ldTI_IdT/dtl_ ldT / dtl d N dT / dt -N ldt dt _dT /dt_dT b h Ch' R l
. ~~:- ds- ds/dt - ds/dt an - l dT /dtl 'so~~: - ~dTids - ds/dt-ds yte am ue.
dt dt
59. (a) IB/ = 1 ~ B · B = 1
d
~ ds (B · B )= 0
dB J. B
ds
(b) B = T x N :::::?
: = ! (T x N) = :~ (T x N } ds~dt = :t (T x N } l r'~t)l = [(T ' x N ) + (T X N' ))lr' ~t )l
= [( T X IT ' I + T X N lr'(t)l = Tr'Tt)'l
1 T ' ) ( ')] 1 , T X N '
dB
:::::? -J. T
ds
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