Brachistochrone racer

Sample project writeup — replace with a real one. Its physics + cs tags land it between chalkboard and terminal: dark, with dust and a faint grid.

Given two points, what curve gets a frictionless bead from the top one to the bottom one fastest under gravity? Not the straight line — Bernoulli's 1696 challenge, and the answer that launched the calculus of variations: the cycloid, the path traced by a point on a rolling wheel's rim.

This project is a small browser simulator that races beads down a straight line, a circular arc, a user-drawn curve, and the cycloid, side by side. Watching the cycloid bead dive steeply to build speed early — and still arrive first despite the longer path — does more for intuition than the derivation ever did for me.

The one equation that matters

Energy conservation fixes the speed at any depth: $v = \sqrt{2gy}$. So the total time along a curve $y(x)$ is the functional

and minimizing it with the Euler–Lagrange machinery (the integrand has no explicit $x$, so the Beltrami identity applies) gives $y\,(1 + y'^2) = 2r$ — the differential equation of a cycloid of rolling radius $r$.

Simulation notes

The integrator is a plain RK4 on arc length, not a physics engine — each curve is arc-length parameterized, and the bead's state is $(s, \dot{s})$ with $\ddot{s} = g \sin\theta(s)$ from the local slope. Two details bit me:

1. The cycloid through two arbitrary points requires solving a transcendental equation for the rolling radius; Newton's method on $\frac{y_1}{x_1} = \frac{1 - \cos\phi}{\phi - \sin\phi}$ converges in a handful of iterations from a decent bracket.
2. The straight-line bead must not win at shallow angles due to integration error — a good end-to-end test is checking the analytic straight-line time $\sqrt{2L^2 / (g\,y_1)}$ against the simulated one.