Solução_Calculo_Stewart_6e

F.
412 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
9. x = √ t, y =1− t
(a)
t 0 1 2 3 4
x 0 1 1.414 1.732 2
y 1 0 −1 −2 −3
(b) x = √ t ⇒ t = x 2 ⇒ y =1− t =1− x 2 .Sincet ≥ 0, x ≥ 0.
So the curve is the right half of the parabola y =1− x 2 .
11. (a) x =sinθ, y =cosθ, 0 ≤ θ ≤ π. x 2 + y 2 =sin 2 θ +cos 2 θ =1.Since
0 ≤ θ ≤ π,wehavesin θ ≥ 0,sox ≥ 0. Thus, the curve is the right half of
(b)
the circle x 2 + y 2 =1.
13. (a) x =sint, y =csct, 0