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

SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS 0 159
27. First we parametrize the curve C of intersection. The projection of C onto the xy-plane is contained in the circle·
x 2 + y 2 = 25, z = 0, so we can write x = 5 cost, y = 5 sin t. C also lies on the cylinder y 2 • + z 2 = 20, and z ~ 0
near the point (3, 4, 2), so we can write z = )20 - y2 = V20 - 25 sin 2 t. A vector equation then for C is
r (t) = (5c~st,5sint, )20- 25sin 2 t) =} r'(t) = ( -5sint,5cost, ~(20- 25sin 2 t)- 1 1 2 (-50sintcost)).
The point (3, 4, 2) corresponds tot = cos- 1 (~),so the tangent vector there is
The tangent line is parallel to this vector and passes through (3, 4, 2), so a vector equation for the line
is r (t) = (3 - 4t)i + ( 4 + 3t)j + (2 - 6t)k.
29. r(t) =(t,e- t, 2t - e ) =* r'(t)=(1,-e- t,2 - 2t). At(0,1,0),
t = 0 and r '(O) = {1, - 1, 2). Thus, parametric equations of the tangent
.line are x = t, y = 1 - t, z = 2t.
31 . r(t) = (tcost,t,tsint) =* r'(t) = (cost-tsint, 1,tcost+sint).
At ( -1r, 1r, 0), t = 1r and r ' (1r) = ( -1, 1, -1r). Thus, parametric equations
of the tangent line are x = -1r -
t, y = 1r + t, z = -1rt.
33. The angle of intersection of the two curves is the angle between the two tangent vectors to the curves at the point of
intersection. Since r; (t) = (1, 2t, 3t 2 ) and t = 0 at (0, 0, 0), r} (0) = (1, 0, 0) is a tangent vector to r 1 at (0, 0, 0). Similarly,
r~(t) = (cost, 2cos2t, 1) and since r 2(0) = (0, 0, 0), r~ (0) = (1, 2, 1) is a tangent vectortor2 at (0, 0,0). IfO is the angle
between these two tangent vectors, then cos 0 = 7f7r, (1, 0, 0 ~ · (1, 2, 1) = -js and 9 = col:l- 1 ( -js) ~ 66°.
35. J; (t i - t 3 j + 3t 5 k ) dt = (J; t dt). i- (J; t 3 dt) j + (! 0
2
3t
5
dt) k
= (~t 2 ]~ i - [tt 4 ]~ j + [~t 0 ] ~ k
= H4- O)i - t{16 - O)j + ~(64- O) k = .2i -4j +32k
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