Solução_Calculo_Stewart_6e

F.
TX.10
8 FURTHER APPLICATIONS OF INTEGRATION
8.1 Arc Length
1. y =2x − 5 ⇒ L = 3
1+(dy/dx)2 dx = 3
1+(2)2 dx = √ 5[3− (−1)] = 4 √ 5.
−1
−1
The arc length can be calculated using the distance formula, since the curve is a line segment, so
L =[distance from (−1, −7) to (3, 1)] = [3 − (−1)] 2 +[1− (−7)] 2 = √ 80 = 4 √ 5
3. y =cosx ⇒ dy/dx = − sin x ⇒ 1+(dy/dx) 2 =1+sin 2 x.SoL =
2π
0 1+sin 2 xdx.
5. x = y + y 3 ⇒ dx/dy =1+3y 2 ⇒ 1+(dx/dy) 2 =1+(1+3y 2 ) 2 =9y 4 +6y 2 +2.
So L = 4
1
9y4 +6y 2 +2dy.
7. y =1+6x 3/2 ⇒ dy/dx =9x 1/2 ⇒ 1+(dy/dx) 2 =1+81x. So
L = 1 √ 82
0 1+81xdx= u 1/2
1
du u =1+81x,
82
= 1
1 81 du =81dx · √
u 2 3/2 81 3
= 2
243 82 82 − 1
1
9. y = x5
6 + 1
10x 3 ⇒ dy
dx = 5 6 x4 − 3 10 x−4
⇒
1+(dy/dx) 2 =1+ 25
36 x8 − 1 + 9
2 100 x−8 = 25
36 x8 + 1 + 9
2 100 x−8 = 5
6 x4 + 3 10 x−42 .So
L =
2 5
1 6 x4 + 3
10 x−42 dx = 2
5
1 6 x4 + 3 x−4 dx = 1
10 6 x5 − 1 x−3 2
= 32
− 1
10 1 6 80 − 1 −
1
6 10
= 31 + 7 = 1261
6 80 240
11. x = 1 √
3 y (y − 3) =
1
3 y3/2 − y 1/2 ⇒ dx/dy = 1 2 y1/2 − 1 2 y−1/2 ⇒
1+(dx/dy) 2 =1+ 1 y − 1 + 1 4 2 4 y−1 = 1 y + 1 + 1 4 2 4 y−1 1
2.So
=
2 y1/2 + 1 2 y−1/2
L = 9
1
= 1 2
1
2 y1/2 + 1 2 y−1/2
dy = 1 2
24 −
8
3
=
1
2
64
3
=
32
3 .
9
2
3 y3/2 +2y 1/2 = 1 2 · 27 + 2 · 3 − 2 · 1+2· 1
2 3 3
1
13. y =ln(secx) ⇒ dy sec x tan x
= =tanx ⇒ 1+
dx sec x
2 dy
=1+tan 2 x =sec 2 x,so
dx
L = π/4
√
sec2 xdx= π/4
|sec x| dx =
π/4
π/4
sec xdx= ln(sec x +tanx)
0
0 0
0
=ln √ 2+1 − ln(1 + 0) = ln √ 2+1 355