Solução_Calculo_Stewart_6e

F.
448 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
11. r =cos3θ.Thisisa
three-leaved rose. The curve is
traced twice.
13. r =1+cos2θ. Thecurveis
symmetric about the pole and
both the horizontal and vertical
axes.
3
15. r =
⇒ e =2> 1, so the conic is a hyperbola. de =3 ⇒
1+2sinθ
d = 3 2
and the form “+2 sin θ” imply that the directrix is above the focus at
the origin and has equation y = 3 . The vertices are
1, π 2 2 and −3,
3π
2 .
2
17. x + y =2 ⇔ r cos θ + r sin θ =2 ⇔ r(cos θ +sinθ) =2 ⇔ r =
cos θ +sinθ
19. r =(sinθ)/θ. Asθ → ±∞, r → 0.
As θ → 0, r → 1. Inthefirst figure,
there are an infinite number of
x-intercepts at x = πn, n a nonzero
integer. These correspond to pole
points in the second figure.
21. x =lnt, y =1+t 2 ; t =1. dy dx
=2t and
dt dt = 1 dy
,so
t dx = dy/dt
dx/dt = 2t
1/t =2t2 .
When t =1, (x, y) =(0, 2) and dy/dx =2.
23. r = e −θ ⇒ y = r sin θ = e −θ sin θ and x = r cos θ = e −θ cos θ ⇒
dy
dx = dy/dθ dr
dx/dθ = sin θ + r cos θ
dθ
dr
cos θ − r sin θ = −e−θ sin θ + e −θ cos θ
−e −θ cos θ − e −θ sin θ · −eθ sin θ − cos θ
=
−eθ cos θ +sinθ .
dθ
When θ = π, dy
dx = 0 − (−1)
−1+0 = 1 −1 = −1.