Calculating The Length of a Path

June 20, 2006 – 2:46 am

When I learned calculus in high-school, integrals were introduced as a way of measuring the area below a function’s graph. I was quite struck by this concept, and I tried to find other geometric properties of functions I could calculate. The first thing that came to mind was measuring the length of the function’s path.

I remember thinking about how to go about this until I almost fell asleep. Then the solution came to me, and I immediately got up and wrote it down. Here it is: I first looked at a very small interval dx. In a sufficiently small interval, I assumed, the function would be almost linear, so it would look like this:

The length (dl) of this path element is then:

dl = \frac{dx}{\cos \theta}
\cos \theta = \frac{1}{\sqrt{1+\tan^2\theta}} = \frac{1}{ \sqrt{1+ f \prime (x)^2} }
\Rightarrow dl = dx \sqrt{1+ f \prime (x)^2}

And the path’s length is the sum of all the path elements:

length = \int_a^b dl = \int_a^b dx \sqrt{1+ f \prime (x)^2}

I verified this result by using it to calculate the circumference of a circle by setting f(x)=\sqrt{R^2-x^2} and a=0, b=2R. I didn’t know how to do many integrals back then so I used an integrals table, and I was thrilled to get the correct result of \pi R for the half-circle.

In retrospect it seems quite trivial, but back then I was very proud of this accomplishment. I started thinking about whether this might be a new result, and whether I should try to publish it. So I picked up a calculus book I had on my shelf to see if, by any chance, it mentioned this problem. And sure enough it was there: A section on measuring the length of a path. The proof was a bit involved (much longer than mine) and I couldn’t follow it, but the same result was obtained. And so after the thrill of discovery, I felt the pain of rediscovery… :)

Today I’m familiar with other ways of getting this result. The quickest (and least rigorous, I guess) is this lovely one-liner:

length = \int_a^b \sqrt{dx^2+dy^2} = \int_a^b dx \sqrt{1+ (\frac{dy}{dx})^2} = \int_a^b dx \sqrt{1+ f\prime(x)^2}

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