4.9 Arc Length
4.9
Brian E. Veitch
Arc Length
Recall the length of a line segment:
The distance of this line segment is the distance between (x1 , y2 ) and (x2 , y2 ). We use
the distance formula,
d=
p
(x2 − x1 )2 + (y2 − y1 )2
But what about something like this? How can we find how long this curve is?
1. The idea is to divide the interval into n equal subintervals each with width ∆x.
2. Find the length between the lines connecting the y-values (P1 P2 , P2 P3 , etc.).
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4.9 Arc Length
Brian E. Veitch
3. The length of the curve in each subinterval is approximately the length of each line
segment.
4. Add all those lengths together to get an approximate arc length.
L≈
n
X
Pi−1 Pi
i=1
5. The larger n gets, the better the approximation.
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4.9 Arc Length
Brian E. Veitch
6. Therefore, the exact length is
L = lim
n→∞
n
X
Pi−1 Pi
i=1
So how do we go from a limit to using an integral to calculate arc length.
Consider the following picture that contains the length of one segment.
1. Find the length again
d=
q
p
(xi − xi−1 )2 + (yi − yi−1 )2 = ∆x2i + ∆yi2
2. By the Mean Value Theorem, we know the interval [xi−1 , xi ], there is a point x∗i such
that
f (xi ) − f (xi−1 ) = f 0 (x∗i )(xi − xi−1 )
∆yi = f 0 (x∗i )∆xi
We can now write the length of the line segment as
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4.9 Arc Length
Brian E. Veitch
p
(xi − xi−1 )2 + (yi − yi−1 )2
q
∆x2i + ∆yi2
q
∆x2i + [f 0 (x∗i )]2 ∆x2i
q
∆x2i 1 + [f 0 (x∗i )]2
q
1 + [f 0 (x∗i )]2 · ∆xi
Pi−1 Pi =
=
=
=
=
3. The total length,
n
X
p
1 + f 0 (x∗i )2 · ∆x
L = lim
n→∞
i=0
which we rewrite as
Z bp
1 + f 0 (x)2 dx
L=
a
or
Z
b
L=
s
1+
a
dy
dx
2
dx
If you’re given a function in terms of y, x = h(y), the formula would be
Z
d
s
1+
L=
c
dx
dy
2
dy
Example 4.27. Find the length of the arc of the semicubical parabola y 2 = x3 between the
points (1,1) and (4,8).
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4.9 Arc Length
Brian E. Veitch
1. To graph it, rewrite it as y = x3/2 .
2. Find
dy
dx
2
dy
= x1/2
dx
3
3. Use the Arc Length Formula
4
Z
s
L =
1+
1
4
Z
r
=
1
dy
dx
2
dx
9
1 + x dx
4
We need to use substitution to finish integrating.
9
(a) Let u = 1 + x
4
4
9
(b) du = x → du = dx
4
9
(c) Change the limits of integration
If x = 1, u = 13/4
If x = 4, u = 10
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4.9 Arc Length
Brian E. Veitch
(d) Integrate
Z
1
4
r
Z
9
4 10 1/2
1 + x dx =
u du
4
9 13/4
4 2 3/2 10
· u 13/4
=
9 3
= 7.63371
Example 4.28. Find the exact length of the curve x =
1√
y(y − 3) over the interval [1,9].
3
1. This is a function in terms of y. In order to graph this, you’ll have to use the techniques
from Chapter 1.
2. The graph
3. Find
dx
dy
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4.9 Arc Length
Brian E. Veitch
dx
1 1/2
1
=
y · (1) + (y − 3) · y −1/2
dy
3
6
1 1/2
1
=
y + √ (y − 3)
3
6 y
1
= √ (2y + (y − 3))
6 y
1
= √ (3y − 3)
6 y
(y − 1)
=
√
2 y
4. Find 1 +
dx
dy
2
1+
y−1
√
2 y
2
=
=
=
=
206
(y − 1)2
1+
4y
4y y 2 − 2y + 1
+
4y
4y
2
y + 2y + 1
4y
(y + 1)2
4y
4.9 Arc Length
Brian E. Veitch
5. Set up the integral
Z
9
L =
s
1+
1
Z
9
=
1
Z
s
y−3
√
2 y
2
dy
(y + 1)2
dy
4y
9
y+1
dy
1/2
1 2y
Z 9
1 1/2 1 −1/2
y + y
dy
=
2
1 2
9
1 3/2
=
y + y 1/2 3
=
1
1
= (9 + 3) − ( + 1)
3
= 10.6667
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