31.03.2019 Views

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

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS 0 159<br />

27. First we parametrize the curve C of intersection. The projection of C onto the xy-plane is contain<strong>ed</strong> in the circle·<br />

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<br />

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<br />

r (t) = (5c~st,5sint, )20- 25sin 2 t) =} r'(t) = ( -5sint,5cost, ~(20- 25sin 2 t)- 1 1 2 (-50sintcost)).<br />

The point (3, 4, 2) corresponds tot = cos- 1 (~),so the tangent vector there is<br />

The tangent line is parallel to this vector and passes through (3, 4, 2), so a vector equation for the line<br />

is r (t) = (3 - 4t)i + ( 4 + 3t)j + (2 - 6t)k.<br />

29. r(t) =(t,e- t, 2t - e ) =* r'(t)=(1,-e- t,2 - 2t). At(0,1,0),<br />

t = 0 and r '(O) = {1, - 1, 2). Thus, parametric equations of the tangent<br />

.line are x = t, y = 1 - t, z = 2t.<br />

31 . r(t) = (tcost,t,tsint) =* r'(t) = (cost-tsint, 1,tcost+sint).<br />

At ( -1r, 1r, 0), t = 1r and r ' (1r) = ( -1, 1, -1r). Thus, parametric equations<br />

of the tangent line are x = -1r -<br />

t, y = 1r + t, z = -1rt.<br />

33. The angle of intersection of the two curves is the angle between the two tangent vectors to the curves at the point of<br />

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,<br />

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<br />

between these two tangent vectors, then cos 0 = 7f7r, (1, 0, 0 ~ · (1, 2, 1) = -js and 9 = col:l- 1 ( -js) ~ 66°.<br />

35. J; (t i - t 3 j + 3t 5 k ) dt = (J; t dt). i- (J; t 3 dt) j + (! 0<br />

2<br />

3t<br />

5<br />

dt) k<br />

= (~t 2 ]~ i - [tt 4 ]~ j + [~t 0 ] ~ k<br />

= H4- O)i - t{16 - O)j + ~(64- O) k = .2i -4j +32k<br />

® 2012 Ccngugc Lcnming. All Righls Reserv<strong>ed</strong>. Moy not he scann<strong>ed</strong>. copi<strong>ed</strong>. or duplicotcd. or post<strong>ed</strong> to a publicly acce.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!