Solução_Calculo_Stewart_6e

F.
416 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
49. Note that all the Lissajous figures are symmetric about the x-axis. The parameters a and b simplystretchthegraphinthe
x-andy-directions respectively. For a = b = n =1the graph is simply a circle with radius 1. Forn =2the graph crosses
itself at the origin and there are loops above and below 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.
a = b =1 n =2 n =3
10.2 Calculus with Parametric Curves
1. x = t sin t, y = t 2 + t ⇒ dy dx
dy
=2t +1, = t cos t +sint,and
dt dt dx = dy/dt
dx/dt = 2t +1
t cos t +sint .
3. x = t 4 +1, y = t 3 + t; t = −1.
dy
dt =3t2 +1, dx
dt =4t3 ,and dy
dx = dy/dt
dx/dt = 3t2 +1
.Whent = −1,
4t 3
(x, y) =(2, −2) and dy/dx = 4 −4
= −1, so an equation of the tangent to the curve at the point corresponding to t = −1
is y − (−2) = (−1)(x − 2),ory = −x.
5. x = e √t , y = t − ln t 2 ; t =1.
dy 2t
=1−
dt t =1− 2 2 t , dx
dt = e√ t
2 √ dy
,and
t dx = dy/dt
dx/dt = 1 − 2/t
e √ t
/ 2 √ t · 2t
2t = √ 2t − 4
√
te
t
.
When t =1, (x, y) =(e, 1) and dy
dx = − 2 e , so an equation of the tangent line is y − 1=−2 e (x − e),ory = −2 e x +3.
7. (a) x =1+lnt, y = t 2 +2; (1, 3).
dy
dt
dx
=2t,
dt = 1 dy
, and
t dx = dy/dt
dx/dt = 2t
1/t =2t2 .
At (1, 3), x =1+lnt =1 ⇒ ln t =0 ⇒ t =1and dy =2,soanequationofthetangentisy − 3=2(x − 1),
dx
or y =2x +1.
(b) x =1+lnt ⇒ x − 1=lnt ⇒ t = e x−1 ,soy =(e x−1 ) 2 +2=e 2x−2 +2and dy
dx =2e2x−2 .
When x =1, dy
dx =2e0 =2, so an equation of the tangent is y =2x +1,asinpart(a).
9. x =6sint, y = t 2 + t; (0, 0).
dy
dx = dy/dt
dx/dt
2t +1
= .Thepoint(0, 0) corresponds to t =0,sothe
6cost
slope of the tangent at that point is 1 6
. An equation of the tangent is therefore
y − 0= 1 6 (x − 0),ory = 1 6 x.