Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 10.3 POLAR COORDINATES ¤ 427
53. To show that x =1is an asymptote we must prove lim
r→±∞ x =1.
x =(r)cosθ =(sinθ tan θ)cosθ =sin 2 θ.Now,r →∞ ⇒ sin θ tan θ →∞ ⇒
θ →
π −
,so lim x =
2 r→∞
θ →
π +
2
,so lim x =
r→−∞
lim sin2 θ =1.Also,r →−∞ ⇒ sin θ tan θ →−∞ ⇒
θ→π/2 −
lim sin2 θ =1. Therefore,
θ→π/2 +
lim
r→±∞
x =1 ⇒ x =1is
a vertical asymptote. Also notice that x =sin 2 θ ≥ 0 for all θ,andx =sin 2 θ ≤ 1 for all θ. Andx 6= 1, since the curve is not
defined at odd multiples of π . Therefore, the curve lies entirelywithintheverticalstrip0 ≤ x