Download A( 7 7.4 - Stewart Calculus

SECTION 7.4
7.4
■ Find the length of the arc of the given curve from point A to
point B.
A1, 0,
2. 9y xx 3 ;
2
3. y x 1 ;
2
3
B8, 3
B(4,
A0, 0,
2
3
■ Set up, but do not evaluate, an integral for the length of
the curve.
10 –12
10. y tan x,
)
11. y x 3,
A1, 0, B2, 1
4. 12xy 4y 4 3;
■
5–9
■
■
■
■
■
■
■
■
■
1
0x1
13. y x 3,
1x3
8. y ln1 x ,
0 x 12
2
9. y lncos x,
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■
■
■
0 x 4
■
■
■
■
■
■
■
■
■
■
■
■
0x1
14. y 1x,
6 x 3
7. y lnsin x,
■
0 x 2
13–16 ■ Use Simpson’s Rule with n 10 to estimate the arc
length of the curve.
4
1
x
6. y ,
4
8x 2
0x1
■
Find the length of the curve.
5. y 3 x 2 232,
0 x 4
12. y e x cos x,
A( 127 , 1), B( 67
24 , 2)
■
■
1
S Click here for solutions.
1– 4
2
■
ARC LENGTH
A Click here for answers.
1. y 1 x 23;
ARC LENGTH
15. y sin x,
0x
16. y tan x,
0 x 4
■
■
■
■
■
■
1x2
■
■
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■
■
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■
2
■
SECTION 7.4
ARC LENGTH
7.4
ANSWERS
E Click here for exercises.
√ √
1
1. 27
80 10 − 13 13
2. 14
3
√
13−8
3. 13 27
4. 59
24
5. 43
6. 181
9
7. ln 1 + √2
3
8. ln 3 −
1
2
√
2+1
π/4 √
10. 0
1 + sec4 x dx
1√
11. 0
1 + 9x4 dx
π/2 12. L = 0
1 + e2x (1 − sin 2x) dx
9. ln
13. 1.548
14. 1.132
15. 3.820
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16. 1.278
S Click here for solutions.
SECTION 7.4
7.4
ARC LENGTH
■
3
SOLUTIONS
E Click here for exercises.
2/3
⇒ dy/dx = − 23 x−1/3
1. y = 1 − x
x4
dy
1
1
+ 2 ⇒
= x3 − 3 ⇒
4
8x
dx
4x
2
1
1
dy
1
1
= 1 + x6 − +
= x6 + +
. So
1+
dx
2 16x6
2 16x6
3 3 1 −3 1 4 1 −2 3
L = 1 x + 4x
dx = 4 x − 8 x
1
81
1
1
= 4 − 72 − 4 − 18 = 181
9
6. y =
⇒
1 + (dy/dx)2 = 1 + 49 x−2/3 . So
8 √ 2/3
8
9x
+4
4
1 + 2/3 dx =
dx
L=
9x
3x1/3
1
1
40 √
1
= 18
u du
u = 9x2/3 + 4, du = 6x−1/3 dx
13
40
√ √
1
1
= 27
80 10 − 13 13
= 27
u3/2
13
Or: Let u = x1/3 .
7. y = ln (sin x)
2
2
1/2
2. 9y = x (x − 3) , 3y = x
⇒ y = 12 x1/2 − 12 x−1/2
(x − 3), y = 13 x3/2 − x1/2
2
⇒ (y ) = 14 x −
1
2
+ 12 x−1
2
⇒ 1 + (y ) = 14 x + 12 + 14 x−1 ⇒
1 + (y )2 = 12 x1/2 + 12 x−1/2 . So
4
4
L = 0 12 x1/2 + 12 x−1/2 dx = 12 23 x3/2 + 2x1/2
0
14
= 12 16
+
4
=
3
3
3. y
2
= (x − 1)3 , y = (x − 1)3/2
3
2
⇒ dy/dx =
(x − 1)1/2
2
⇒
9
4
1 + (dy/dx) = 1 + (x − 1). So
2 9
2
x−
L = 1 1 + 94 (x − 1) dx = 1
4
2
√
3/2
13−8
= 13 27
= 49 · 23 94 x − 54
5
4
dx
4
4. 12xy = 4y + 3, x =
−1
y
y
+
3
4
⇒
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2
5. y = 13 x + 2
3/2
⇒
√
1/2
(2x) = x x2 + 2 ⇒
dy/dx = 2 x + 2
2
1 + (dy/dx)2 = 1 + x2 x2 + 2 = x2 + 1 . So
1
1
L = 0 x2 + 1 dx = 13 x3 + x 0 = 43 .
1
2
−2
y
dx
= y2 −
,
dy
4
2
1
dx
y −4
= y4 − +
so
⇒
dy
2
16
2
dx
1
y −4
= y4 + +
⇒
1+
dy
2
16
2
dx
y −2
1+
= y2 +
. So
dy
4
3
2
2
y
y −2
1
dy =
−
y2 +
L=
4
3
4y 1
1
8
1
1
1
59
= 3 − 8 − 3 − 4 = 24
dy
dx
2
dy
cos x
=
= cot x ⇒
dx
sin x
= 1 + cot2 x = csc2 x. So
π/3
π/3
csc x dx = [ln (csc x − cot x)]π/6
√ = ln √23 − √13 − ln 2 − 3
√
2
√ 3 = ln 1 + √
= ln √3 2 1− √3 = ln 2 +
3
3
(
)
L=
π/6
dy
−2x
⇒
=
dx
1 − x2
2
2
1 + x2
dy
4x2
=1+
=
. So
1+
dx
(1 − x2 )2
(1 − x2 )2
1/2 1/2
1 + x2
2
dx
=
L=
−1
+
dx
1 − x2
(1 − x) (1 + x)
0
0
1/2 1
1
dx
−1 +
+
=
1+x
1−x
0
2
8. y = ln 1 − x
⇒
= [−x + ln (1 + x) − ln (1 − x)]1/2
0
1
3
1+
⇒
= − 12 + ln 32 − ln 12 − 0 = ln 3 −
1
2
1
(− sin x) = − tan x
cos x
2
⇒ 1 + (y ) = 1 + tan2 x = sec2 x. So
√
π/4
= ln 2 + 1
L = 0 sec xdx = ln (sec x + tan x)|π/4
0
9. y = ln (cos x)
⇒ y =
2
⇒ 1 + (y ) = 1 + sec4 x. So
π/4 √
1 + sec4 x dx.
L= 0
10. y = tan x
3
⇒ y = 3x2
1√
L = 0 1 + 9x4 dx.
11. y = x
2
⇒ 1 + (y ) = 1 + 9x4 . So
x
⇒ y = ex (cos x − sin x) ⇒
2
1 + (y ) = 1 + e2x cos2 x − 2 cos x sin x + sin2 x
12. y = e cos x
= 1 + e2x (1 − sin 2x)
π/2 1 + e2x (1 − sin 2x) dx.
So L = 0
4
■
SECTION 7.4
ARC LENGTH
2
2
⇒ 1 + (y ) = 1 + 3x2 = 1 + 9x4 ⇒
√
1√
L = 0 1 + 9x4 dx. Let f (x) = 1 + 9x4 . Then by
Simpson’s Rule with n = 10,
3
13. y = x
L≈
1/10
3
[f (0) + 4f (0.1) + 2f (0.2) + 4f (0.3)
+ · · · + 2f (0.8) + 4f (0.9) + f (1)] ≈ 1.548
2
dy
1
1
dy
1
14. y =
= 1+ 4.
⇒
=− 2 ⇒ 1+
x
dx
x
dx
x
1
Therefore, with f (x) = 1 + 4 ,
x
2
L = 1 1 + 1/x4 dx ≈ S10
=
2−1
10 · 3
[f (1) + 4f (1.1) + 2f (1.2) + 4f (1.3)
+ 2f (1.4) + 4f (1.5) + 2f (1.6) + 4f (1.7)
+ 2f (1.8) + 4f (1.9) + f (2)]
≈ 1.132104
2
2
15. y = sin x, 1 + (dy/dx) = 1 + cos x,
√
π√
L = 0 1 + cos2 x dx. Let g (x) = 1 + cos2 x. Then
π
L ≈ π/10
g (0) + 4g 10
+ 2g π5 + 4g 3π
3
π
3π 10
+ 2g 2π
+
4g
+
2g
+ 4g 7π
5
2
5
10
+ 2g 4π
+ 4g 9π
+ g (π)
5
10
≈ 3.820
2
⇒ 1 + (y ) = 1 + sec4 x. So
√
π/4 √
L= 0
1 + sec4 x dx. Let g (x) = 1 + sec4 x. Then
π
L ≈ π/40
g (0) + 4g 40
+ 2g 2π
+ 4g 3π
3
40
5π 6π 40 7π + 2g 4π
+
4g
+
2g
+ 4g
40
40
8π 40 9π 40 π + 2g 40 + 4g 40 + g 4
≈ 1.278
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16. y = tan x