Arc Length
Suppose f is continuously differentiable on the interval [a,b].
Let's derive a formula for the length L of the curve on the
interval, called the arc length over [a,b].
We'll start by subdividing the interval [a,b] into n subintervals
[x0, x1],
[x1, x2],
... ,
[xn-1,xn] where
a = x0 < x1 < ¼ < xn-1 < xn = b.
Introduce the line segments between
(x0, f(x0)) and
(x1,f(x1)),
(x1, f(x1)) and
(x2, f(x2)),
...,
(xn-1, f(xn-1)) and
(xn, f(xn)).
The resulting polygonal path approximates the curve given by y = f(x),
and its length approximates the arc length of f(x) over [a,b].
Let's find the length of the polygonal path by adding up the lengths
of the individual line segments. The kth line segment is the
hypotenuse of a triangle with base Dxk and height
f(xk)-f(xk-1) and so has length
Lk = |
___________________
Ö(Dxk)2+[f(xk)-f(xk-1)]2. |
By the
Mean Value Theorem,
there exists xk* Î [xk-1,xk] such that
f(xk)-f(xk-1)
xk-xk-1
|
= f¢(xk*) |
so
|
f(xk)-f(xk-1) = f¢(xk*)(xk-xk-1) =
f¢(xk*)Dxk. |
Thus,
Lk = |
_________________
Ö
(Dxk)2+[f¢(xk*)Dxk]2 |
= |
__________
Ö
1+[f¢(xk*)]2 |
Dxk. |
Finally, the length of the entire polygonal path is
n
å
k = 1
|
Lk = |
n
å
k = 1
|
__________
Ö
1+[f¢(xk*)]2 |
Dxk |
which has the form of a
Riemann sum.
Increasing the number of subintervals such that max Dxk ® 0, the summation åkn= 1 Lk
® L.
That is,
| L | = |
lim
max Dxk® 0
|
|
n
å
k = 1
|
__________
Ö
1+[f¢(xk*)]2 |
Dxk |
|
|
= |
ó
õ |
b
a
|
__________
Ö
1+[f¢(xk*)]2 |
dx |
|
by the definition of the
definite integral
as a limit of Riemann sums. Thus, we have proved the following:
Arc Length
Let f(x) be continuously differentiable on [a,b]. Then the arc
length L of f(x) over [a,b] is given by
|
L = |
ó
õ
|
b
a
|
_________
Ö
1+[f¢(x)]2 |
dx |
Similarly, if x = g(y) with g continuously differentiable on
[c,d], then the arc length L of g(y) over [c,d] is given by
|
L = |
ó
õ
|
d
c
|
_________
Ö
1+[g¢(y)]2 |
dy |
These integrals often can only be computed using numerical methods.
Example
We can compute the arc length of the graph of f(x) = x3/2 over
[0,1] as follows:
| L | = |
ó
õ
|
1
0
|
_________
Ö
1+[f¢(x)]2 |
dx |
|
| = |
ó
õ
|
1
0
|
___________
Ö
1+[3x1/2/2]2 |
dx |
|
| = |
ó
õ
|
1
0
|
_______
Ö
1+9x/4 |
dx |
|
| = |
|
| = |
(1+9/4)3/2-(1)3/2
|
| = |
(13/4)3/2-1
|
| » |
1.44. |
