Solução_Calculo_Stewart_6e

F.
TX.10SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ¤ 415
41. It is apparent that x = |OQ| and y = |QP | = |ST|. From the diagram,
x = |OQ| = a cos θ and y = |ST| = b sin θ. Thus, the parametric equations are
x = a cos θ and y = b sin θ. To eliminate θ we rearrange: sin θ = y/b
⇒
sin 2 θ =(y/b) 2 and cos θ = x/a ⇒ cos 2 θ =(x/a) 2 . Adding the two
equations: sin 2 θ +cos 2 θ =1=x 2 /a 2 + y 2 /b 2 . Thus, we have an ellipse.
43. C =(2a cot θ, 2a),sothex-coordinate of P is x =2a cot θ. LetB =(0, 2a).
Then ∠OAB isarightangleand∠OBA = θ,so|OA| =2a sin θ and
A =((2a sin θ)cosθ, (2a sin θ)sinθ). Thus,they-coordinate of P
is y =2a sin 2 θ.
45. (a) There are 2 points of intersection:
(−3, 0) and approximately (−2.1, 1.4).
(b) A collision point occurs when x 1 = x 2 and y 1 = y 2 for the same t. So solve the equations:
3sint = −3+cost (1)
2cost =1+sint (2)
From (2), sin t =2cost − 1. Substituting into (1),weget3(2 cos t − 1) = −3+cost ⇒ 5cost =0 () ⇒
cos t =0 ⇒ t = π or 3π . We check that t = 3π satisfies (1) and (2) but t = π 2 2 2 2
does not. So the only collision point
occurs when t = 3π 2
, and this gives the point (−3, 0). [We could check our work by graphing x1 and x2 together as
functions of t and, on another plot, y 1 and y 2 as functions of t. If we do so, we see that the only value of t for which both
pairs of graphs intersect is t = 3π .] 2
(c) The circle is centered at (3, 1) instead of (−3, 1). There are still 2 intersection points: (3, 0) and (2.1, 1.4),butthereare
no collision points, since ()inpart(b)becomes5cost =6 ⇒ cos t = 6 > 1. 5
47. x = t 2 ,y = t 3 − ct.Weuseagraphingdevicetoproducethegraphsforvariousvaluesofc with −π ≤ t ≤ π. Note that all
the members of the family are symmetric about the x-axis. For c0 the graph crosses itself at x = c, so the loop grows larger as c increases.