Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 419
33. The curve x =1+e t , y = t − t 2 = t(1 − t) intersects the x-axis when y =0,
that is, when t =0and t =1. The corresponding values of x are 2 and 1+e.
The shaded area is given by
x=1+e
x=2
(y T − y B ) dx =
t=1
t=0
35. x = rθ − d sin θ, y = r − d cos θ.
[y(t) − 0] x 0 (t) dt = 1
0 (t − t2 )e t dt
= 1
0 tet dt − 1
0 t2 e t dt = 1
0 tet dt − t 2 e t 1
0 +2 1
0 tet dt [Formula 97 or parts]
=3 1
0 tet dt − (e − 0) = 3 (t − 1)e t 1
− e [Formula 96 or parts]
0
=3[0− (−1)] − e =3− e
A = 2πr
0
ydx= 2π
0 (r − d cos θ)(r − d cos θ) dθ = 2π
0 (r2 − 2dr cos θ + d 2 cos 2 θ) dθ
= r 2 θ − 2dr sin θ + 1 2 d2 θ + 1 2 sin 2θ 2π
0
=2πr 2 + πd 2
37. x = t − t 2 , y = 4 3 t3/2 , 1 ≤ t ≤ 2. dx/dt =1− 2t and dy/dt =2t 1/2 ,so
(dx/dt) 2 +(dy/dt) 2 =(1− 2t) 2 +(2t 1/2 ) 2 =1− 4t +4t 2 +4t =1+4t 2 .
Thus, L = b
(dx/dt)2 +(dy/dt)
a
2 dt = 2
√
1+4t2 dt ≈ 3.1678.
1
39. x = t +cost, y = t − sin t, 0 ≤ t ≤ 2π. dx/dt =1− sin t and dy/dt =1− cos t,so
dx
2
dt + dy
2
dt =(1− sin t) 2 +(1− cos t) 2 =(1− 2sint +sin 2 t)+(1− 2cost +cos 2 t)=3− 2sint − 2cost.
Thus, L = b
(dx/dt)2 +(dy/dt)
a
2 dt = 2π √
0 3 − 2sint − 2costdt≈ 10.0367.
41. x =1+3t 2 , y =4+2t 3 , 0 ≤ t ≤ 1.
dx/dt =6t and dy/dt =6t 2 ,so(dx/dt) 2 +(dy/dt) 2 =36t 2 +36t 4 .
Thus, L = 1
√
36t2 +36t
0
4 dt = 1
6t √ 1+t
0 2 dt
=6 2 √
1 u 1 du [u =1+t 2 , du =2tdt]
2
=3
2
3 u3/2 2
1
=2(2 3/2 − 1) = 2 2 √ 2 − 1
43. x = t
1+t , y =ln(1+t), 0 ≤ t ≤ 2. dx (1 + t) · 1 − t · 1 1 dy
= = and
dt (1 + t) 2 (1 + t)
2
dt = 1
1+t ,
2 2 dx dy 1
so + =
dt dt (1 + t) + 1
4 (1 + t) = 1 1+(1+t)
2
= t2 +2t +2
. Thus,
2 (1 + t) 4 (1 + t) 4
2
√
t2 +2t +2
3
√
u2 +1
L =
dt =
du
0 (1 + t) 2 1 u 2
= − √ 10
3
+ln 3+ √ 10 + √ 2 − ln 1+ √ 2
u = t +1,
du = dt
√
24 u2 +1
= − +ln u + 3
u
u
2 +1
1