Solução_Calculo_Stewart_6e

F.
418 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
23. We graph the curve x = t 4 − 2t 3 − 2t 2 , y = t 3 − t in the viewing rectangle [−2, 1.1] by [−0.5, 0.5]. This rectangle
corresponds approximately to t ∈ [−1, 0.8].
We estimate that the curve has horizontal tangents at about (−1, −0.4) and (−0.17, 0.39) and vertical tangents at
about (0, 0) and (−0.19, 0.37). We calculate dy
dx = dy/dt
dx/dt = 3t 2 − 1
. The horizontal tangents occur when
4t 3 − 6t 2 − 4t
dy/dt =3t 2 − 1=0 ⇔ t = ± 1 √
3
, so both horizontal tangents are shown in our graph. The vertical tangents occur when
dx/dt =2t(2t 2 − 3t − 2) = 0 ⇔ 2t(2t +1)(t − 2) = 0 ⇔ t =0, − 1 or 2. It seems that we have missed one vertical
2
tangent, and indeed if we plot the curve on the t-interval [−1.2, 2.2] we see that there is another vertical tangent at (−8, 6).
25. x =cost, y =sint cos t. dx/dt = − sin t, dy/dt = − sin 2 t +cos 2 t =cos2t.
(x, y) =(0, 0) ⇔ cos t =0 ⇔ t is an odd multiple of π 2 .Whent = π 2 ,
dx/dt = −1 and dy/dt = −1,sody/dx =1. Whent = 3π , dx/dt =1and
2
dy/dt = −1. Sody/dx = −1. Thus, y = x and y = −x are both tangent to the
curve at (0, 0).
27. x = rθ − d sin θ, y = r − d cos θ.
(a) dx
dy
dy
= r − d cos θ, = d sin θ,so
dθ dθ dx = d sin θ
r − d cos θ .
(b) If 0