The Fastest Slide: Why the Cycloid Beats the Straight Line

Which slide gets a bead from the same high point to the same low point in the least time? The straight line is the shortest route, so it is the obvious favorite. Let us race it before trusting that intuition. Every candidate begins at A and ends at B. The blue track is the straight ramp. The green track is a circular arc. The red track is a cycloid, a curve traced by a point on the rim of a rolling circle. The three beads start from rest. Gravity is the same for all of them, the track is frictionless, and only the shape of the route changes. Ready. Go. The red bead dives sharply, gathers speed early, and reaches the bottom first. The green bead follows close behind. The blue bead on the direct ramp arrives last. The measured times put the cycloid first at about one point one four seconds. The circular arc takes about one point one seven. The straight ramp takes about one point three zero. So shortest does not mean fastest. The straight ramp saves distance, but the curved tracks spend more of their journey moving quickly. A steep initial drop can repay the extra distance later. The circular arc already exploits that bargain, but the cycloid balances early acceleration against later distance even better. Our task is to turn that visual bargain into a condition that selects one curve from every possible curve joining A to B.

Let y measure vertical depth below the release point, down toward the bottom. At depth y, energy conservation says the lost gravitational potential energy has become kinetic energy. The mass cancels, and a bead released from rest has speed square root of two g y. Deeper parts of the track are therefore traversed faster. Now inspect one tiny piece of path. Its horizontal change is d x, its vertical change is d y, and Pythagoras gives the arc length d s. Time is distance divided by speed, so this small piece costs d s over v. Replace d s by square root of one plus y prime squared, times d x. Adding the costs from the start to the finish gives this integral. It depends on the height y and the slope y prime at every x along the route. This notation T of y matters. The input is not one number. The input is the entire function y of x, and the output is one travel time. Ordinary calculus might minimize f of alpha. We move one number left or right and compare the resulting values. Here the choice is a curve. Moving one part can leave another part alone, and there are infinitely many places where the route can be nudged. A single slider can display one family of curves, but it cannot represent every possible deformation. The mathematical variation needs an arbitrary function to describe all those independent nudges. Write a nearby path as y plus epsilon times eta. Epsilon controls the overall size of the change. Eta of x controls its shape from place to place. The endpoints are prescribed, so the variation vanishes at both ends. Between them, eta is otherwise arbitrary. That freedom is what will turn one integral statement into a differential equation for the fastest path.

Call the integrand F of y and y prime. A fastest path must have zero first-order change in travel time for every allowed deformation eta. Differentiate under the integral sign. Changing y contributes F sub y times eta. Changing the slope contributes F sub y prime times eta prime. The eta-prime term is inconvenient because the arbitrary function is differentiated. Integrate that term by parts, moving the derivative onto F sub y prime. The boundary term vanishes because eta is zero at both fixed endpoints. What remains is one integral of eta times a coefficient. Now use the crucial freedom. Eta can be a small smooth bump wherever we choose. If the coefficient were positive or negative anywhere, a bump placed there would make the integral nonzero. Therefore the coefficient itself must vanish. This is the Euler-Lagrange equation, the differential condition satisfied by every stationary path. It is the analogue of setting an ordinary derivative equal to zero, but now the derivative tests every possible direction in a space of functions rather than the two directions on a number line. For this particular problem there is a useful shortcut. The integrand depends on y and y prime, but it has no explicit x in it. Differentiate F minus y prime times F sub y prime. The chain rule and the Euler-Lagrange equation reduce that derivative to F sub x. But F sub x is zero, so the expression is constant. This conserved quantity is often called the Beltrami identity. Substitute our square-root integrand. After the y-prime terms combine, F minus y prime F sub y prime becomes the reciprocal of these two square roots. Set that expression equal to C, square both sides, and rearrange. The fastest curve must satisfy y times one plus y prime squared equals a positive constant. We name the constant two a because that choice will make the geometry transparent. This equation is already the answer to the variational problem in differential form.

We now solve y times one plus y prime squared equals two a. The constant is positive, and y is the downward depth below the release point. The combination two a minus y suggests the substitution y equals a times one minus cosine theta. It automatically starts at y equals zero when theta equals zero. Solve the condition for y prime squared, then insert the substitution. The ratio becomes one plus cosine over one minus cosine, which is cotangent squared of theta over two. Differentiate y with respect to theta. That gives a sine theta. The slope equation also says d x over d y is tangent of theta over two. Multiply d y by d x over d y. The result is a sine theta times tangent of theta over two, which simplifies to a times one minus cosine theta. Integrate with respect to theta and choose x equals zero at the cusp. We obtain x equals a times theta minus sine theta. Together with the substitution for y, these are the parametric equations of a cycloid. Let theta grow. The point traces the curve from its cusp, initially almost straight downward, then gradually turns toward the finish. The two endpoint equations determine the two remaining constants. For the race, a is about one point five three and the finishing parameter is about two point eight eight. Near the release point, the vertical depth y is tiny, so the curve begins extremely steep. Farther down, the bead already has speed, and the curve spends that speed moving toward the destination. The straight line minimizes distance alone. The cycloid minimizes the integral of distance divided by speed, so it deliberately buys early speed with extra length. That is why it wins the race.

The fastest curve has another remarkable property. Put three beads at different heights on the same cycloid, and release all three from rest. The red bead has the longest arc to travel. The yellow bead starts lower. The green bead begins lower still, close to the bottom. Release them together. Their speeds and remaining distances are different throughout the descent, yet all three reach the bottom at the same instant. This is the tautochrone property. Tauto means the same, and chronos means time. On an ideal cycloid, the travel time to the bottom is independent of the release point. Let theta measure position from the cusp. The downward depth is a times one minus cosine theta. The bottom occurs at theta equals pi and depth two a. Let h be vertical height above the bottom. Subtracting the depth from two a gives a times one plus cosine theta, or two a cosine squared of theta over two. Now let s be arc length measured upward from the bottom. The cycloid's arc element is two a sine of theta over two, d theta. Integrating back to pi gives four a cosine of theta over two. Eliminate the cosine between the two formulas. The height above the bottom is exactly s squared divided by eight a. That quadratic relation is the secret. Gravitational potential energy is m g h. Because h is quadratic in s, the potential energy is a constant times s squared. Differentiate the potential with respect to s and reverse the sign to get the force. Newton's law gives an acceleration proportional to minus s. After cancelling the mass, this is the equation of simple harmonic motion. Its angular frequency is square root of g over four a. A bead released from rest begins at one extreme of that oscillation. The bottom is the equilibrium point, one quarter of a period later. One quarter period is pi over two omega, which simplifies to pi times square root of a over g. Notice what is missing: the starting arc length s zero. The higher bead travels farther, but it also begins with a larger restoring acceleration. The lower bead has less distance but receives a weaker pull. The cycloid balances those effects exactly. Return the beads to their three starting places. Their different amplitudes are now visible again. Release them once more. One shared harmonic clock carries every amplitude to zero at the same quarter period, so the three beads meet at the bottom together. The brachistochrone and tautochrone are therefore two faces of the same curve. Between fixed endpoints, the cycloid minimizes gravitational travel time. Along one cycloidal bowl, every release height shares one arrival time.