On the Length of Curves

I'm somewhat depressed for various reasons, so I shall discuss mathematics: the pastime of all famous depressives.

I was riding in a car to Seattle in the dark. One can't easily read in the dark, so I was forced into pure thought. Among other things, I entertained the problem of a finding the length of any curve. Let's say say you have some crazy curve represented by a the function y=f(x). Finding the distance along the curve over an interval [a,b] turned out to be an interesting problem of whose solution I was not aware. So I thought about it, and I had an epiphany. It was a very elementary epiphany, but an epiphany which led me to this solution to the problem:

Consider a function y=f(x). Let L be the length of the curve over an interval [a,b]. Now divide L in differential units dl. Since each unit dl approximates a straight line, the distance of each is given by the Pythagorean theorem. So dl=((x₁ - x₂)² + (y₁ - y₂)²)^1/2. Rewriting this gives dl=(Δx² + Δy²)^1/2, and finally dl=(dx² + dy²)^1/2 since Δx and Δy shrink to differentials for a dl. Integration gives L=∫(dx² + dy²)^1/2. So simple! It checks out for y=x. I tried it for y=x² doing the integration using trig substitutions and it is a pretty ungodly integral to solve. Numerical integration performed on a TI-89 leads me to suspect that I've got the correct solution.

This is the kind of thing that makes me love calculus. Without having the sacred knowledge, this problem is unbelievably daunting and would require somekind of brilliant insight. With the calculus, it just takes a few minutes of thinking and voila! The result is simple, elegant and practical.