Arc Length

Using Calculus to find the length of a curve.
(Please read about Derivatives and Integrals first)

Imagine we want to find the length of a curve between two points. And the curve is smooth (the derivative is continuous).

arc length curve

First we break the curve into small lengths and use the Distance Between 2 Points formula on each length to come up with an approximate answer:

arc length between points

The distance from x0 to x1 is:

S1 = (x1 − x0)² + (y1 − y0

And let's use  Δ (delta) to mean the difference between values, so it becomes:

S1 = (Δx1)² + (Δy1

Now we just need lots more:

S2 = (Δx2)² + (Δy2
S3 = (Δx3)² + (Δy3
...
...
Sn = (Δxn)² + (Δyn

 

We can write all those many lines in just one line using a Sum:

S ≈
n
Σ
i=1
(Δxi)² + (Δyi

But we are still doomed to a large number of calculations!

Maybe we can make a big spreadsheet, or write a program to do the calculations ... but lets try something else.

We have a cunning plan:

Let's go:

First, divide and multiply Δyi by Δxi:

S ≈
n
Σ
i=1
(Δxi)² + (Δxi)²(Δyi/Δxi

Now factor out (Δxi:

S ≈
n
Σ
i=1
(Δxi)²(1 + (Δyi/Δxi)²)

Take (Δxi out of the square root:

S ≈
n
Σ
i=1
1 + (Δyi/Δxi  Δxi

Now, as n approaches infinity (as we head towards an infinite number of slices, and each slice gets smaller) we get:

S =
lim
n→∞
n
Σ
i=1
1 + (Δyi/Δxi  Δxi

We now have an integral  and we write dx to mean the Δx slices are approaching zero in width (likewise for dy):

S =
    b
a
1 + (dy/dx dx

And dy/dx is the derivative of the function f(x), which can also be written f’(x):

S =
    b
a
1 + (f’(x) dx

The Arc Length Formula

And now suddenly we are in a much better place, we don't need to add up lots of slices, we can calculate an exact answer (if we can solve the differential and integral).

Note: the integral also works with respect to y, useful if we happen to know x=g(y):

S =
    d
c
1 + (g’(y) dy

So our steps are:

Some simple examples to begin with:

arc length constant

Example: Find the length of f(x) = 2 between x=2 and x=3

f(x) is just a horizontal line, so its derivative is f’(x) = 0

Start with:  
S =
    3
  2
1 + (f’(x) dx
Put in f’(x) = 0:  
S =
    3
  2
1 + (0 dx
Simplify:  
S =
    3
  2
dx
Calculate the Integral:  
S = (3-2) = 1

So the arc length between 2 and 3 is 1. Well of course it is, but it's nice that we came up with the right answer!

Interesting point: the "(1 + ...)" part of the Arc Length Formula guarantees we get at least the distance between x values, such as this case where f’(x) is zero.

arc length slope

Example: Find the length of f(x) = x between x=2 and x=3

The derivative f’(x) = 1

Start with:  
S =
    3
  2
1 + (f’(x) dx
Put in f’(x) = 1:  
S =
    3
  2
1 + (1 dx
Simplify:  
S =
    3
  2
2 dx
Simplify more:  
S = 2
    3
  2
dx
Calculate the Integral:  
S = 2(3-2) = 2

And the diagonal across a unit square really is the square root of 2, right?

OK, now for the harder stuff. And this is useful, too:

rope bridge

Example: Metal posts have been installed 6m apart across a gorge.

Find the length for the hanging bridge that follows the curve:

f(x) = 5 cosh(x/5)

Here is the actual curve:

catenary graph

Let us solve the general case first!

A hanging cable forms a curve called a catenary:

f(x) = a cosh(x/a)

Larger values of a have less sag in the middle
And "cosh" is the hyperbolic cosine function.

The derivative is: f’(x) = sinh(x/a)

The curve is symmetrical, so it is easier to work on half of the catenary, from the center to an end at "b":

Start with:  
S =
    b
  0
1 + (f’(x) dx
Put in f’(x) = sinh(x/a):  
S =
    b
  0
1 + sinh²(x/a) dx
Use the identity  
1 + sinh²(x/a) = cosh²(x/a):
 
S =
    b
  0
cosh²(x/a) dx
Simplify:  
S =
    b
  0
cosh(x/a) dx
Calculate the Integral:   S = a sinh(b/a)

Now, remembering the symmetry, let's go from −b to +b:

S = 2a sinh(b/a)

In our specific case a=5 and the 6m span goes from −3 to +3

S = 2×5 sinh(3/5)
= 6.367 m
(to nearest mm)

This is important to know! If we build it exactly 6m in length there is no way we could pull it hard enough for it to meet the posts. But at 6.367m it will work nicely.

 

arc length graph

Example: Find the length of y = x(3/2) from x = 0 to x = 4.

 

The derivative is y’ = (3/2)x(1/2)

Start with:  
S =
    4
  0
1 + (f’(x) dx
Put in (3/2)x(1/2):  
S =
    4
  0
1 + ((3/2)x(1/2) dx
Simplify:  
S =
    4
  0
1 + (9/4)x dx

We can use integration by substitution:

And we get:  
S =
    10
  1
(4/9)u du
Integrate:   S = (8/27) u(3/2) from 1 to 10
Calculate:   S = (8/27) (10(3/2) - 1(3/2)) = 9.073...

Conclusion

The Arc Length Formula is:

S =
    b
a
1 + (f’(x) dx

Steps: