Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 421
(b)WeusetheCAStofind the derivatives dx/dt and dy/dt, and then use Formula 1 to find the arc length. Recent versions
of Maple express the integral 4π
(dx/dt)2 +(dy/dt)
0
2 dt as 88E 2 √ 2 i ,whereE(x) is the elliptic integral
1
√
1 − x2 t
√ 2
dt and i is the imaginary number √ −1.
1 − t
2
0
Some earlier versions of Maple (as well as Mathematica) cannot do the integral exactly, so we use the command
evalf(Int(sqrt(diff(x,t)ˆ2+diff(y,t)ˆ2),t=0..4*Pi)); to estimate the length, and find that the arc
length is approximately 294.03. Derive’sPara_arc_length function in the utility file Int_apps simplifies the
integral to 11
4π
−4cost cos
11t
0
2 − 4sint sin 11t
2 +5dt.
57. x =1+te t , y =(t 2 +1)e t , 0 ≤ t ≤ 1.
dx
dt
2
+
dy
2
dt =(te t + e t ) 2 +[(t 2 +1)e t + e t (2t)] 2 =[e t (t +1)] 2 +[e t (t 2 +2t +1)] 2
= e 2t (t +1) 2 + e 2t (t +1) 4 = e 2t (t +1) 2 [1 + (t +1) 2 ], so
S = 2πy ds = 1
0 2π(t2 +1)e t e 2t (t +1) 2 (t 2 +2t +2)dt = 1
0 2π(t2 +1)e 2t (t +1) √ t 2 +2t +2dt ≈ 103.5999
59. x = t 3 , y = t 2 , 0 ≤ t ≤ 1.
dx
2
dt + dy
2
dt = 3t
2 2 +(2t) 2 =9t 4 +4t 2 .
S =
1
=2π
= π 81
2πy dx
0
13
= 2π
1215
4
dt
2
+
dy
dt
u − 4 √u 1
18
9
du
1
2
dt = 2πt 2 9t 4 +4t 2 dt =2π
0
u =9t 2 +4, t 2 =(u − 4)/9,
du =18tdt,sotdt = 1 18 du
13
13
2
5 u5/2 − 8 3 u3/2 =
3u π · 2 5/2 − 20u 3/2 81 15
4
4
√ 3 · 13
2
13 − 20 · 13 √ 13 − (3 · 32 − 20 · 8) =
1
0
t 2 t 2 (9t 2 +4)dt
= 2π
9 · 18
13
√
2π
1215 247 13 + 64
4
(u 3/2 − 4u 1/2 ) du
61. x = a cos 3 θ, y = a sin 3 θ, 0 ≤ θ ≤ π . dx
2
2 dθ + dy
2
dθ =(−3a cos 2 θ sin θ) 2 +(3a sin 2 θ cos θ) 2 =9a 2 sin 2 θ cos 2 θ.
S = π/2
2π · a sin 3 θ · 3a sin θ cos θdθ=6πa 2 π/2
sin 4 θ cos θdθ= 6 πa2 sin 5 0 0 5
θ π/2
= 6 0 5 πa2
63. x = t + t 3 , y = t − 1 t , 1 ≤ t ≤ 2. dx
2 dt =1+3t2 and dy =1+ 2 dt
t ,so
dx 2 3 dt + dy
2
dt =(1+3t 2 ) 2 +
1+ 2 2
t 3
and S =
2πy ds =
2
1
2π t − 1
(1 + 3t
t 2 ) 2 + 1+ 2 2
dt ≈ 59.101.
2 t 3
65. x =3t 2 , y =2t 3 , 0 ≤ t ≤ 5 ⇒
dx 2
dt + dy
2
dt =(6t) 2 +(6t 2 ) 2 =36t 2 (1 + t 2 ) ⇒
S = 5
2πx (dx/dt)
0 2 +(dy/dt) 2 dt = 5
0 2π(3t2 )6t √ 1+t 2 dt =18π 5
t2√ 1+t
0 2 2tdt
=18π 26
1
(u − 1) √ udu
u =1+t 2 ,
du =2tdt
=18π 26
1 (u3/2 − u 1/2 ) du =18π
=18π 2 · 676 √ 5
26 − 2 · 26 √ 3
26 − 2
− 2
5 3 =
24
π 5
949 √ 26 + 1
26
2
5 u5/2 − 2 3 u3/2
1