Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS D 5
41. It is apparent that x = JOQJ andy= IQPI = JSTJ. From the diagram,
y
X = I OQI = a cos B and y = I STI = b sine. Thus, the parametric equations are
x = a cos B and y = b sin B. To eliminate B we rearrange: sin B = y / b =>
sin 2 B = (y/bl and cosO= x/a =>
cos 2 B = (x/a) 2 • Adding the two
X
equations: sin 2 B + cos 2 B = 1 = x 2 /a 2 + y 2 /b 2 .,Thus, we have an ellipse.
43. C = (2a cot B, 2a), so the x-coordinate of Pis x = 2a cot B. Let B = (0, 2a).
Then L.OAB is a rightangle and L.OBA = B, so JOAI '= 2asinB and
A = ( (2a sin B) cos B, (2a sin B) sin B). Thus, they-coordinate of P
is y = 2asin 2 B.
45. (a) 4 There are 2 points of intersection:
( -3, 0) and approximately ( - 2.1, 1.4).
~
2a
-4
(b) A collision point occurs when x1 =.xz and Yl = Y2 for the same t. So solve the equations:
3sint= - 3+cost (1)
2 cost = 1 + sin t (2)
From (2), sin t = 2 cost- 1. Substituting into (1), we get 3(2 cost - 1) = - 3 +cost => 5 cost = 0 (*) =>
cost = 0 => t = t or 3 ;. We check that t = 3 2..,. satisfies (1) imd (2) butt = t does not. So the only collision point
occurs when t = 3 2 ", and this gives the point (- 3, 0). [We could check our work by graphing Xt and x 2 together as
functions oft and, on another plot, y 1 and y2 as functions oft. If we do so, we see that the only value oft for which both
pairs of graphs intersect is t ~ 3 ;.]
(c) The circle is centered at (3, 1) instead of (- 3, 1). Th~re are still 2 intersection points: (3, 0) and (2.1, 1.4), but there are
no collision points, since(*) in part (b) becomes 5 cost = 6 => cost = ~ > 1.
47. x = t 2 , y = t 3 - ct. We use a graphing device to produce the graphs for various values of c with -7r ::; t ::; 7r. Note that all
the members of the family are symmetric about the x -axis. For c < 0, the graph does not cross itself, but for c = 0 it has a
cusp at (0, 0) and for c > 0 the graph crosses itself at x = c, so the loop grows larger as c increases.
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