Solução_Calculo_Stewart_6e

F.
450 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
35. The curves intersect where 2sinθ =sinθ +cosθ ⇒
sin θ =cosθ ⇒ θ = π 4 , and also at the origin (at which θ = 3π 4
on the second curve).
A = π/4
0
1
(2 sin 2 θ)2 dθ + 3π/4 1
π/4
= π/4
0
(1 − cos 2θ) dθ + 1 2
= θ − 1 2 sin 2θ π/4
0
(sin θ 2 +cosθ)2 dθ
3π/4
(1 + sin 2θ) dθ
π/4
+ 1
θ − 1 cos 2θ 3π/4
= 1 (π − 1)
2 4 π/4 2
37. x =3t 2 , y =2t 3 .
L = 2
0
(dx/dt)2 +(dy/dt) 2 dt = 2
0
(6t)2 +(6t 2 ) 2 dt = 2
√
36t2 +36t
0
4 dt = 2
= 2
0 6 |t| √ 1+t 2 dt =6 2
0 t √ 1+t 2 dt =6 5
1 u1/2 1
2 du u =1+t 2 , du =2tdt
=6· 1 · 2
2 3
5
u 3/2 =2(5 3/2 − 1) = 2 5 √ 5 − 1
1
0
√
36t
2 √ 1+t 2 dt
39. L = 2π
π
24
=
−
r2 +(dr/dθ) 2 dθ =
2π
2π
θ (1/θ)2 +(−1/θ 2 2 +1
)
π
2 dθ =
π θ 2
√ √
θ 2 +1
π2 +1 4π2 +1
θ
= 2√ π 2 +1− √ 4π 2 +1
2π
+ln θ + 2π
θ 2 +1
π
=
√
2π + 4π2 +1
+ln
π + √ π 2 +1
π
−
2π
dθ
√
2π + 4π2 +1
+ln
π + √ π 2 +1
41. x =4 √ t, y = t3 3 + 1
2t 2 , 1 ≤ t ≤ 4
⇒
S = 4
2πy (dx/dt)
1 2 +(dy/dt) 2 dt = 4
2π 1
1 3 t3 + 1 t−2 √ 2
2 2/ t +(t2 − t −3 ) 2 dt
=2π 4
1
1
3 t3 + 1 2 t−2 (t 2 + t −3 ) 2 dt =2π 4
1
1
3 t5 + 5 6 + 1 2 t−5 dt =2π 1
18 t6 + 5 6 t − 1 8 t−4 4
1 = 471,295
1024 π
43. For all c except −1, the curve is asymptotic to the line x =1. For
c0, there is a loop to the left of the origin,
whose size and roundness increase as c increases. Note that the x-intercept
of the curve is always −c.