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

SECTION 10.2 CALCULUS WITH PARAMETRIC-CURVES 0 7
itself at the origin and there are loops above and b~low the x-axis. In general, the figures have n - 1 points of intersection,
all of which are on the y-axis, and a total of n closed loops.
-1.1
1.1
t1 =3
n =2
n=l
- 2.1
\- ---+- (a,b) =( ~.2)'
(a, b) = (2, I)
-3.1
(a, b) = (2, 3)
(a, b) = (3, 2)
- 1.1
a=b = l
- 2.1 -3.1
n=2 n =3
10.2 Calculus with Parametric Curves
1. x = t sin t, y = t 2 + t ::}
dy . dx . dy dy I dt 2t + 1
-d = 2t + 1, -d = t cost+ sm t , and -d = d l d = . .
t t x x t t cos t + sm t
z . 3 dy = - 3t2 dx = 4- 2t d dy = dyldt = -3tz -Wh - 1
3. x = 1 + 4t - t , y = 2 - t ; t = 1. dt ' dt , an dx dx I dt 4 - 2t . en t '
( x, y) = ( 4, 1) and dy I dx = - ~, so an equa.tion of the tangent to the curve at the point corresponding to t = 1 is
y- 1 = -~(x- 4), oq/ = - ~x + 7.
5. x = t cos t, y = t sin t; t = 1r.
dy . dx . ' dy dy I dt t cos t + sin t
-d = tcost+smt, -d = t(-smt)+cost,and-d = d ld = . .
t t x x t -tsm t+cost
When t = 1r, (x, y) = ( -1r, 0) and dy I dx. = - 1r I( -1) = 1r, so an equation of the tangent to U1e curve at the point
corresponding to t = 1r is y - 0 = 1r[x -
( -1r) ) , or y = 1rx + 1r 2 •
7. (a) x = 1 + ln t, y = t 2 + 2; (1, 3).
dy dx 1 dy dyldt 2t 2
dt = 2 t, dt = t' and dx = dxldt = 1l t = 2 t · At( 1 • 3 ),
x=l+lnt = 1 ::} lnt = O =? t =1 an~ : = 2,so an equationofthetangentis y-3=2 (x - l),
ory = 2x + 1.
(b)x = 1 + ln t =? lnt=x -1 =? t=e"'-l,soy=t2.+2=(e"'- 1 ) 2 + 2 = e 2 " - 2 +2,andy'=e 2 x-z . 2.
At (1, 3), y' = e 2 (l)- z · 2 = 2, so an equation of the tangent is y - 3 = 2(x - 1), or y = 2x + 1.
9. x = 6sint, y = t2+t; (0,0).
dy dy 1 dt 2t + 1 .
- = d ld = --. The potnt (0, 0) corresponds tot = 0, so the
dx x t 6cost
slope of the tangent at that point is· lJ. An equation of the tangent is therefore
y - 0 = i(x- 0), or y = ix.
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