Consider the graph of a function for . The purpose of this module is to find the length of this piece of the curve, known as the arclength of the function from to .

As in previous modules, the basic method is to find the arclength element and then integrate it:

By zooming in on a portion of the curve, it begins to look like a straight line. Then one can express in terms of the infinitesimal horizontal change and vertical change :

Now, by the Pythagorean theorem one finds that . A little algebra and the chain rule gives that

So the arclength of the function from to is given by


Find the arclength of the curve

Hint: recall the facts that

Computing the arclength element from the above formula gives

Therefore, we find that the arclength is


Find the arclength of the curve

First, one finds . So, with some careful algebra one sees that

Now note that by reversing the cancellation done in an earlier step when simplifying , one finds that . And so, continuing the computation, one finds

Thus, , and it follows that

Parametric curves

If a curve is defined parametrically, i.e. and for , then the arclength element can be written as

So the arclength of a parametric curve for is given by


Find the arclength for a circle of radius .

A simple parametrization for the circle of radius is

Note that ranges from 0 to . Using the above formula, we find that the arclength element is

(we used the Pythagorean identity from line three to line four). Therefore,

as desired.


Find the arclength for the spiral , for .

First, compute and . Then

Thus, . So one finds that

This integral was computed in the Trigonometric Substitution module. The answer becomes

Additional Examples


Compute the arclength of the curve

Computing the arclength element, we find

Therefore, the arclength is


A catenary is the curve that is formed by hanging a cable between two towers. It is a fact that the rate of change of the slope of a hanging cable is proportional to the rate of change of arclength with respect to . Mathematically,

for some constant . Use this fact to find the equation of the catenary. Then find the length of the catenary for .

Using the formula for the arclength element, the fact tells us that

Now, making a substitution of

simplifies the equation to become

This is a separable differential equation. Separating and integrating gives

The left side can be handled with either a trigonometric or hyperbolic trigonometric substitution. We take the latter approach, and let

So we have (remembering the Pythagorean identity for hyperbolic trigonometric functions from the trigonometric substitution module)

(we leave the constant of integration off for now since we will be integrating on the right side as well). On the right side, we have

Putting it together, we have

If we pick our coordinates so that occurs at the low point of the catenary, then note that at this point, we have

since the slope of the catenary is 0 at the low point. Using this fact and plugging in into the earlier equation gives

and so . This gives

Now integrating both sides gives

where is the value of the low point of the catenary.

To find the length of the catenary, we have

since hyperbolic sine is an odd function. This grows very quickly as increases, because


Show that the spiral

for has infinite arclength.


Plugging these into the formula for the arclength of a parametric curve and noting the cancellation of cross terms, we have

Therefore, the arclength is

This integral is difficult to compute exactly, but we only want to show it diverges, which is not as difficult. Note that

And so by the dominance of definite integrals,

but the integral on the right diverges to infinity by our earlier discussions of p-integrals. Thus, our integral on the left, being larger, also diverges to infinity.


  • Compute the arclength of , from to .