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Brachistochrone Racer

Learning goal

Watch the calculus of variations answer the question it was invented for. Five paths, same endpoints, five frictionless beads released together — and the cycloid wins every time, by a margin you can see.

Things to try

  1. Watch the race once without reading anything. The straight line is the shortest path and it loses badly. Minimising distance and minimising time are different problems, and the whole point of the subject is that "the best curve" depends entirely on what functional you are extremising.

  2. Notice the cycloid dives first. It drops steeply at the start to buy speed, then flattens to spend that speed on horizontal progress. The Euler–Lagrange equation is doing that trade-off optimally at every point, which is not something you could guess.

  3. Make the drop small and the run long (Y small, X large). The margin over the straight line grows to 25% or more — the steeper the required average slope, the less room there is to be clever, and conversely.

  4. Make the drop large and the run short. All five paths converge: when the motion is nearly vertical free-fall there is little to optimise, and the margin falls below a percent.

  5. Check the exact column. The cycloid's descent time is $\(T = \sqrt{a/g}\;\theta_1\)$ in closed form, where \(a\) and \(\theta_1\) come from solving \(x = a(\theta - \sin\theta)\), \(y = a(1-\cos\theta)\) through the endpoint. The numerically integrated time agrees to about 0.02% — the residual is quadrature error near the start, where the integrand \(\sqrt{(1+y'^2)/2gy}\) has an integrable singularity at \(y = 0\).

Why a cycloid

The functional to minimise is the descent time

\[T[y] = \int_0^X \sqrt{\frac{1 + y'^2}{2gy}}\,dx\]

The integrand has no explicit \(x\), so the Beltrami identity applies — \(f - y'\partial f/\partial y' = \text{const}\) — which drops the order by one and gives

\[y\left(1 + y'^2\right) = 2a = \text{const}\]

whose solution is the cycloid. That shortcut (one of the two "shortcut cases" in the notes) is what makes the problem tractable at all; attacking the full Euler–Lagrange equation directly is considerably worse.

The tautochrone, for free

The cycloid has a second remarkable property, discovered by Huygens before Bernoulli's challenge: a bead released anywhere on a cycloid reaches the bottom in the same time,

\[T = \pi\sqrt{a/g}\]

independent of the starting height. It is simultaneously the fastest descent curve and the curve of isochronous oscillation. Huygens tried to build pendulum clocks on this principle — cycloidal cheeks constraining the string — and it worked in theory and lost to friction in practice.

The history is worth knowing

Johann Bernoulli posed the problem publicly in 1696, giving Europe six months. Newton, then Master of the Mint, received it, solved it overnight, and published anonymously. Bernoulli is said to have recognised the author immediately — tanquam ex ungue leonem, "as the lion is known by its claw."

Solutions also came from Leibniz, l'Hôpital and Jakob Bernoulli, and Euler and Lagrange's generalisation of the technique became the calculus of variations — which then became Hamilton's principle, and hence all of mechanics, and eventually every field theory in physics.

Calculus of variations · Euler–Lagrange equation · Brachistochrone · Fermat's principle — the same logic for light · Hamilton's principle · Geodesics