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

158 0 CHAPTER 13 VECTOR FUNCTIONS
· 7. Since x = e 2 t = (et) 2 = y 2 , the
(a), (c)
y
(b) r'(t) = 2e 2 t i + et j ,
curve is part of a parabola. Note
r '(O) = 2 i + j
that here x > 0, y > 0.
X
9. r ' (t) = \! [t sin t], :t [e) ·, ! [t cos 2t]) = (t c~s t +sin t, 2t, t(- sin 2t) · 2 +cos 2t)
= (t cost+ sin t , 2t, cos 2t - 2t sin 2t)
11. r(t) =ti+j + 2Vt k => r'(t)=1 i + O j +2(~t - 1 1 2 )k = i+ ~ k
13. r (t) = et 2 i - j + ln(1 + 3t) k => r'(t) = 2tet 2 3
i + - - k
1 +3t
15. r ' (t) = 0 + b + 2t c = b + 2t c by Formulas 1 and 3 of Theorem 3.
r'(O) = (1, 2, . 2). So lr'(O)I = .JP + . 2 2 + 2 2 = .J§ = 3 and
19. r '(t) = - sin ti + 3j + 4cos 2tk => r'(O) = 3j + 4k. Thus
T (~) = 1::~~~~ = .J02 +~2 +42 (3 j +4k) = t(3j +4k) = ~ j + ~ k.
21. r(t) = (t, t 2 , t 3 ) => r' (t) = (1, 2t, 3t 2 ). Then r ' (1) = (1, 2, 3) and lr' (1)1 = \,/1 2 + 2 2 + 3 2 = v'l4, so
( ) r ' ( 1) 1 ( ) 1 1 2 a ) " ( ) ( )
T 1 = ir'( 1
)i = V'i4 1,2,3 = \V'i4'V'i4 ' V'i4 . r t = 0,2,6t ,so
j k
r '(t) x r"(t) = 1 2t 3t2
0 2 6t
= 12t 3t21 i - 11 3t2 1· + 11 2t I k
2 6t 0 6t J 0 2
= {12f- 6t~ ) i - (6t- O) j + (2 - 0) k = (6f, -6t, 2)
23. The vector equation for the curve is r(t) = ( 1 + 2 Jt, t 3 - t, t 3 + t ), so r '(t) = ( 1/ Vt, 3f - 1, 3t 2 + 1). The point
(3, 0, 2) corresponds tot= 1, so the tangent vector there is r ' (1) = (1, 2, 4). Thus, the tangent line goes through the point
(3, 0, 2) and is parallel to the vector (1, 2, 4). Parametric equations are x = 3 + t, y = 2t, z ,;,. 2 + 4t.
25. The vector equation for the curve is r (t) = ( e- t cost, e- t sin t, e-t), so
r' (t) = (e-t(-sin t) +(cost)( -e-t), e-t cost+ (sin t)( -e-t), (-e- t))
= ( - e-t(cos t +sin t), e- t(cos t - sin t), -e-')
The point (1, 0, 1) corresponds to t = 0, so the tangent vector there is
r '(O) = (-e 0 (cos 0 +sin 0), e 0 (cos0- sin 0), -e 0 ) = (-1, 1, -1}. Thus, the tangent line is parallel to the. vector
(- 1, 1, - 1} and parametric equations are x = 1 + ( -1)t = 1 - t, y = 0 + 1 · t = t, z = 1 + ( - 1)t = 1 - t.
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