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

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CHAPTER 13 PROBLEMS PLUS 0 181<br />

The projectile hits the ground when (v sina)t- ~gt 2 = 0 => t = 2 ; sin a, so the distance travel<strong>ed</strong> by the projectile is<br />

(2v/g)sin o 1 (211/g)sin<br />

L(a) = lv(t)l dt = .9<br />

1 0 0<br />

v ) 2 v2<br />

( t - - sin a + 2 cos 2 a dt<br />

.9 9<br />

t - (v /g) sino<br />

=g 2<br />

[<br />

+ [(vl g) cosa]2 ln (t-~sino +<br />

2 g<br />

[using Formula 2 I in the Table of lntegrals]<br />

g<br />

=<br />

[v.<br />

2 (~.rna)' + (~ -gsma oo.o )' + (~'"'")' m ( ~ •ina+ (~ •mo)' + (~=a Y)<br />

+~sin o<br />

g (~ •ina )' + (~=o )'- (~oooo )' m( -~•ina+ (~•ina )' + (~=o Y)]<br />

2<br />

= g_ [~sino · ~+ v<br />

cos 2 2<br />

a ln(~ sino+~) +~ sino·~ - v cos 2 a ln(- ~ sino+~)]<br />

2 g g g2 g g g g g2 g g<br />

v 2 . + v 2 2 1n( (vl g)sina+v/g) v 2 . + v 2 2<br />

1 (1+sino)<br />

= - stno -cos a = -sma -cos an<br />

g 2f1 - (vI g) sin a +vI g g 2g 1 - sin a<br />

We want to maximize L( a) for 0 s; a s; 1r / 2.<br />

v<br />

2<br />

v<br />

2<br />

[ 2 1 - sin a 2 cos a . ( 1 + sin a)]<br />

L'(a) =-coso+ - cos a · . · . 2 1<br />

2 - 2cosa smo ln .<br />

g g +sma (1-smo) . 1 -sm o<br />

v 2 • v 2 [<br />

= - cos a + - cos 2 . ( 1 + sin<br />

2<br />

a)]<br />

2 a ·-- - 2 cos a sm a ln .<br />

g g coso , 1 -sma<br />

v 2 v 2 [ • (l+sina)] v 2 . = - COSO! [ 2- SIDQ . • ln (1+ sina . ) ]<br />

9 g 1 -SlDO! 9 1 -SJDO! ,<br />

= - COSO!+- COSO! 1-SlDO! ln<br />

L( a ) has critical points for 0 < a < 1r 12 when L' (a) = 0 => 2 - sin a ln ( ~ + s~ a) = 0 [since cos a =1= 0].<br />

- SIDQ<br />

Solving by graphing (or using a CAS) gives a~ 0.9855. Compare values at the critical point and the endpoints:<br />

L(O) = 0, L( 1r 12) = v 2 I g, and £(0.9855) ~ 1.20v 2 I g. Thus the distance travel<strong>ed</strong> by the projectile is maximiz<strong>ed</strong><br />

for a ~ 0.9855 or ~ 56°.<br />

7. We can write the vector equation as r (t) = ·at 2 + bt + c where a= (a1, a2, a a}, b = (b1 , b2 , ba }, and c = (c1, c2, c3).<br />

Then r ' (t) = 2t a+ b which says that each tangent vector is the sum of a scalar multiple of a and the vector b . Thus the<br />

tangent vectors are all parallel to the plane determin<strong>ed</strong> by a and b so the curve must be parallel to this plane. [Here we assume<br />

that a and bare nonparallel. Otherwise the tangent vectors are all parallel and the curve lies along a single line.] A normal<br />

© 2012 Ccngoge Lcoming. All RighLS Rescm :d. May oot be scanocd, copi<strong>ed</strong>, 0< duplicnt<strong>ed</strong>. 0< poSi<strong>ed</strong> to a publicly :10e

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