Solução_Calculo_Stewart_6e

F.
420 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
45. x = e t cos t, y = e t sin t, 0 ≤ t ≤ π.
dx
dt
Thus, L = π
0
2
+
dy
2
dt =[e t (cos t − sin t)] 2 +[e t (sin t +cost)] 2
=(e t ) 2 (cos 2 t − 2cost sin t +sin 2 t)
+(e t ) 2 (sin 2 t +2sint cos t +cos 2 t
= e 2t (2 cos 2 t +2sin 2 t)=2e 2t
√
2e
2t
dt = π
√
0 2 e t dt = √ 2 e t π
= √ 2(e π − 1).
0
47. x = e t − t, y =4e t/2 , −8 ≤ t ≤ 3
dx
dt
2 + dy
2
dt =(e t − 1) 2 +(2e t/2 ) 2 = e 2t − 2e t +1+4e t
= e 2t +2e t +1=(e t +1) 2
L = 3
(et +1)
−8
2 dt = 3
−8 (et +1)dt = e t + t 3t
−8
=(e 3 +3)− (e −8 − 8) = e 3 − e −8 +11
49. x = t − e t , y = t + e t , −6 ≤ t ≤ 6.
dx
2
dt + dy
2
dt =(1− e t ) 2 +(1+e t ) 2 =(1− 2e t + e 2t )+(1+2e t + e 2t )=2+2e 2t ,soL = 6
√
−6 2+2e
2t
dt.
Set f(t) = √ 2+2e 2t . Then by Simpson’s Rule with n =6and ∆t = 6−(−6) =2,weget
6
L ≈ 2 [f(−6) + 4f(−4) + 2f(−2) + 4f(0) + 2f(2) + 4f(4) + f(6)] ≈ 612.3053.
3
51. x =sin 2 t, y =cos 2 t, 0 ≤ t ≤ 3π.
(dx/dt) 2 +(dy/dt) 2 =(2sint cos t) 2 +(−2cost sin t) 2 =8sin 2 t cos 2 t =2sin 2 2t ⇒
Distance = 3π
√ √ π/2
0 2 |sin 2t| dt =6 2 sin 2tdt [by symmetry] = −3 √ π/2
2 cos 2t = −3 √ 2(−1 − 1) = 6 √ 2.
0
0
The full curve is traversed as t goes from 0 to π 2
, because the curve is the segment of x + y =1that lies in the first quadrant
(since x, y ≥ 0), and this segment is completely traversed as t goes from 0 to π 2 . Thus, L = π/2
0
sin 2tdt= √ 2,asabove.
53. x = a sin θ, y = b cos θ, 0 ≤ θ ≤ 2π.
dx
2
dt + dy
2
dt =(a cos θ) 2 +(−b sin θ) 2 = a 2 cos 2 θ + b 2 sin 2 θ = a 2 (1 − sin 2 θ)+b 2 sin 2 θ
= a 2 − (a 2 − b 2 )sin 2 θ = a 2 − c 2 sin 2 θ = a 2 1 − c2
a 2 sin2 θ = a 2 (1 − e 2 sin 2 θ)
1 − e2 sin 2 θdθ.
So L =4
π/2
0
a 2 1 − e 2 sin 2 θ dθ [by symmetry] =4a π/2
0
55. (a) x =11cost − 4cos(11t/2), y =11sint − 4sin(11t/2).
Notice that 0 ≤ t ≤ 2π does not give the complete curve because
x(0) 6=x(2π). In fact, we must take t ∈ [0, 4π] in order to obtain the
complete curve, since the first term in each of the parametric equations has
period 2π and the second has period 2π
= 4π
11/2 11
integer multiple of these two numbers is 4π.
, and the least common