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

CHAPTER 10 REVIEW 0 39
25. x = t + sin t, y = t - cost ~
d (dy)
dt dx
dxjdt
dy dyjdt 1 + sint
dx = d.xjdt = 1 +cos t
{1+cost) cost- {1 +sint)(- sint)
___ _ ___:(,_1,...-+.:..__c_os_t-'-)- 2 _____ = cost+ cos 2 t +sin t + sin 2 t =
l +cost
(l+cost)3
1 + cost + sin t
(1 + cost)3
3 2 .
27. We graph the curve x = t - 3t, y = t + t + 1 for -2.2 :=::; t :=::; 1.2.
By zooming in or using a cursor, we find that the lowest point is about
(1.4, 0. 75). To find the exact values, we find the t-value at which
dyjdt=2t+1 = 0 ¢=> t=~~ , (x,y),;,(lj,~).
29. x·= 2acost- acos2t ~ dx = -2asin t + 2asin2t = 2asint(2cost - 1) = 0 ¢:>
dt
sint = 0 or cost = ~ ~ t = O,·t, 1r, orr;.
y = 2asin t - a sin 2t ~ dy = 2a cost - 2a cos 2t = 2a(l +cost- 2 cos 2 t) = 2a(1 -cos t)(1 + 2 cost) = 0 =}
dt
271" 4"
t = 0 '3,or3.
Thus the graph has vertical tangents where t = t. 1r and ~"> 3 " , and horizontal tangents where t = 2 ; and 4 ;. To determine
what the slope is where t = 0, we use !'Hospital's Ru;e to evaluate lim ddy/jddt = 0, so there is a. horizontal tangent there.
t->0 X t
t X y y
0 a 0
3 " ~a :.::]:a
2
2rr
- ~a ~a
3 2
7r - 3a 0
-~a
4 7r
3 -~a 2
51r
~a
_ :.::]:a
3 2
(-3a,O)
(a,O)
)C
31. The curve r 2 = 9 cos 50 has 10 "petals." For instance, for - -fu :=::; 8 :=::; {' 0
, there are two petals, one with r > 0 and one
with r < 0.
33. The curves intersect when 4 cos 8 = 2 ~ cos 8 = ~ ~ 8 = ± i
for -1r :=::; 8 :=::; 1r. The points of intersection are (2, i) and (2,- ~).
r = 4cos 0
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