Arc Length Formula

This lesson will be a short derivation of the formula for the arc length of a function on a given interval. If you've learned about the arc length formula before, you were probably shown something along the lines of this

and told to go along with it. However, this isn't the most intuitive thing in the world at first glance and all it does is beg the question of why it's true at all.

To start, let's try to divide the interval [a, b] into subintervals of width Δx and approximate the function by using secant line segments. For each subinterval, there is also a different Δyi denoting the change in y within that subinterval.

Notice that as we decrease Δx, the number of subintervals increases and the linear approximations approach y.

To approximate the actual arc length of y on the interval, we simply add together each of the yellow line segments, which we will call Aᵢ.

One useful observation before moving forward is that each Δx, Δyᵢ , and Aᵢ forms a right triangle, letting us use the Pythagorean theorem to show the following:

Putting everything together (literally), we have

Now come the vital steps. From inside the radical, we can factor out a Δx from the summand. Meaning that, if A is the arc length of y on the interval,

Now, we let Δx approach 0. Recall that as Δx approaches 0, the number of subintervals (n) approaches infinity, and so our approximation will approach the actual arc length.

Also notice that as Δx approaches 0, each Δyi also approaches 0, meaning that the quantity

will converge to the derivative of y at the point its subinterval shrinks to.

If this rings a few bells, it should, as we're now simply taking a Riemann integral of the function

on the designated interval [a, b].

So, if you ever happen to forget the arc length formula, it's quite straightforward to recall that you're simply summing up lots of hypotenuses on the interval [a, b] from which you can factor out a Δx. After doing that, letting Δx approach 0 to approach the actual arc just turns the sum into an integral from a to b.