12.3 Classic problems: brachistochrone, catenary, geodesics, Fermat

The Euler–Lagrange equation and its two first integrals (12.2) are now enough to solve the problems that built the subject. Each is a different integrand $F$ ; the method is identical every time.

The brachistochrone

A bead slides without friction from a high point $A$ to a lower point $B$ under gravity. Of all the curves joining them, which gives the shortest descent time? (Greek brákhistos, shortest; khrónos, time.) Not the straight line — the answer is a cycloid, and it is where the calculus of variations began.

▶ The fastest descent is a cycloid Derivation

Put $A$ at the origin with $y$ measured downward. A bead released from rest reaches depth $y$ with speed set by energy conservation, $\tfrac12 v^2 = g y$ , so $v = \sqrt{2gy}$ . The travel time is arc length over speed:

T[y] \;=\; \int \frac{ds}{v} \;=\; \int_0^{x_B} \frac{\sqrt{1+y'^2}}{\sqrt{2gy}}\; dx, \qquad F = \sqrt{\frac{1+y'^2}{2gy}}.

The integrand has no explicit $x$ , so use the Beltrami identity $F - y'\,\partial F/\partial y' = \text{const}$ . Computing $\partial F/\partial y' = \dfrac{1}{\sqrt{2gy}}\dfrac{y'}{\sqrt{1+y'^2}}$ and forming the combination, the $y'^2$ terms collapse:

F - y'\frac{\partial F}{\partial y'} \;=\; \frac{1}{\sqrt{2gy}}\cdot\frac{1}{\sqrt{1+y'^2}} \;=\; \text{const} \quad\Longrightarrow\quad y\,(1+y'^2) \;=\; k,

with $k$ a positive constant. This is the differential equation of a cycloid — the curve traced by a point on a rim of a wheel of radius $R = k/2$ rolling along the underside of the $x$ -axis. Its solution in parametric form is

x(\theta) = R(\theta - \sin\theta), \qquad y(\theta) = R(1 - \cos\theta),

as you can verify by substituting into $y(1+y'^2)=k$ using $y' = dy/dx = \sin\theta/(1-\cos\theta)$ . The single constant $R$ (and the range of $\theta$ ) is fixed by requiring the curve to pass through $B$ .

where

$T[y]$: descent time — the functional being minimised s
$v = \sqrt{2gy}$: bead speed at depth y (from rest, energy conservation) m/s
$R$: radius of the generating circle of the cycloid m
$\theta$: the rolling angle parametrising the cycloid rad

descent time: cycloid 0.80 s (fastest) · line 0.99 s · arc 0.82 s

x_B = 2.00 m depth y_B = 1.00 m

The straight line is the shortest path but not the fastest: the cycloid drops steeply at first to build speed, then flattens. Minimising travel time is a calculus-of-variations problem, and its extremal is this curve.

Race the three curves. The straight line is the shortest path but not the fastest: the cycloid dives steeply at the start to buy speed, then trades that speed for a gentler run-in to $B$ , and wins overall. Move $B$ and the winning margin changes but the cycloid always leads. That the fastest route is not the shortest is the lesson the brachistochrone taught the seventeenth century — and the first hint that “least” problems in physics rarely have the naïve answer.

⏳ The history — Bernoulli's challenge and the lion's claw

In June 1696 Johann Bernoulli posed the brachistochrone as a public challenge “to the sharpest mathematicians of the world,” allowing six months. Correct solutions came from Leibniz, from Johann’s brother and rival Jakob Bernoulli, from l’Hôpital, and — famously — from Isaac Newton, then Master of the Mint. Newton reportedly received the problem in the afternoon, stayed up through the night, and sent his solution anonymously. Bernoulli, recognising the author instantly from the power of the method, is said to have remarked “tanquam ex ungue leonem” — “I recognise the lion by his claw.” Johann’s own solution was a stroke of genius: he treated the bead as a ray of light and applied Fermat’s principle and Snell’s law layer by layer, uncovering the deep link between least-time optics and mechanics that Hamilton would later make into a single theory. The winning curve, the cycloid, is also Huygens’s tautochrone — the curve on which a bead’s period is independent of where it starts — so the fastest descent and the perfect pendulum are the same shape.

The catenary and the soap film

Hang a flexible chain from two points and it settles into a curve that minimises its gravitational potential energy — the catenary. The identical curve arises, for a different reason, as the shape of a soap film spanning two rings: the film minimises its area, and the surface of revolution of least area has a catenary profile.

▶ Least area of revolution gives y = a cosh(x/a) Derivation

Revolving $y(x)$ about the $x$ -axis sweeps a surface of area $A = \int 2\pi y\,\sqrt{1+y'^2}\;dx$ . Drop the constant $2\pi$ ; the integrand $F = y\sqrt{1+y'^2}$ has no explicit $x$ , so Beltrami again:

F - y'\frac{\partial F}{\partial y'} = y\sqrt{1+y'^2} - \frac{y\,y'^2}{\sqrt{1+y'^2}} = \frac{y}{\sqrt{1+y'^2}} = a \;(\text{const}).

Rearranging, $y'^2 = (y/a)^2 - 1$ , a separable first-order equation whose solution is the hyperbolic cosine:

\boxed{\;y(x) = a\,\cosh\!\left(\frac{x-b}{a}\right).\;}

The constants $a$ and $b$ are set by the two endpoints. The hanging chain gives the same $\cosh$ for a subtly different problem — minimise potential energy $\int y\,ds$ subject to fixed chain length — a constrained problem we solve with a Lagrange multiplier in 12.4.

Geodesics

The geodesic problem — the shortest path between two points on a surface — is the arc-length functional restricted to the surface. On a plane the extremal is a straight line (12.2); on a sphere it is a great circle; on a cylinder a helix. Writing the surface’s own distance element $ds$ in surface coordinates and applying Euler–Lagrange yields the geodesic equation, whose solutions are the “straightest possible” curves a surface allows. This is the exact object that, for the curved spacetime of general relativity, becomes the path of a freely falling body: gravity as geometry is a variational principle, $\delta\int ds = 0$ .

Fermat’s principle and Snell’s law

Light travels between two points along the path of least time — Fermat’s principle. In a medium of refractive index $n$ the local speed is $c/n$ , so the time is proportional to the optical path length $\int n\,ds$ . From this one functional, the law of refraction falls out.

▶ Snell's law from least time Derivation

Take layers stacked in $y$ , with index $n(y)$ , and describe the ray by $x(y)$ . The optical path is $\int n(y)\sqrt{1 + x'^2}\;dy$ with $x' = dx/dy$ ; the integrand $F = n(y)\sqrt{1+x'^2}$ has no explicit $x$ , so the coordinate $x$ is cyclic and its conjugate quantity is conserved:

\frac{\partial F}{\partial x'} = \frac{n(y)\,x'}{\sqrt{1+x'^2}} = \text{const}.

The fraction $x'/\sqrt{1+x'^2}$ is $\sin\theta$ , where $\theta$ is the angle the ray makes with the vertical (the normal to the layers). Hence

\boxed{\;n(y)\,\sin\theta \;=\; \text{const}\;}

along the ray — Snell’s law in its general form. Across a sharp interface between two media it becomes $n_1\sin\theta_1 = n_2\sin\theta_2$ . The same first-integral logic gives the acoustic Snell’s law for sound crossing a sound-speed gradient (Sound 7.2) and underlies ray (geometric) acoustics.

These four problems — one integrand each — show the pattern the whole subject runs on: write the quantity to be extremised as $\int F\,dx$ , apply Euler–Lagrange or a first integral, and read off the curve. The next lesson handles the case the catenary flagged — extremising when the competitors must also satisfy a constraint.