The Brachistochrone, the Fastest Descent, math that surprises you lesson

Two wires run from the same starting point down to the same finishing point. One is straight, and the other plunges steeply first and then levels out, so it is clearly longer. A bead slides down each wire without friction, both released from rest at the same moment, and the bead on the longer wire can arrive first. This page explains how speed gained early can make up for extra distance, which curve is the fastest of all, and why one race cannot prove that on its own.

Key terms

Brachistochrone curve

Definition Fix a starting point $A$ and a finishing point $B$ that is lower than $A$ and not directly below it. A bead, treated as a single point, slides along a fixed wire from $A$ to $B$ under uniform gravity, starting from rest, with no friction and no air resistance. The brachistochrone problem asks which shape of wire brings the bead to $B$ in the least time. The answer, the brachistochrone curve, is an arc of a cycloid, the curve traced by a point on the rim of a circle that rolls without slipping along the underside of the horizontal line through $A$ , with the traced point starting at $A$ . For exactly one radius of the circle, the first arch of this cycloid passes through $B$ , and the brachistochrone is the part of that arch from $A$ to $B$ . The travel time along it is smaller than along any other wire from $A$ to $B$ . The arc starts vertically downward at $A$ . When the horizontal distance from $A$ to $B$ is more than $\pi/2$ times the drop, the arc dips below $B$ and climbs back up to it. A race in which the cycloid beats a straight wire only shows that it beats that one rival. That it beats every possible wire is a theorem, proved with the calculus of variations. The answer depends on the model. With friction, air resistance or a push at the start, the fastest curve is a different one. Johann Bernoulli posed the problem in 1696, and the name comes from the Greek words for shortest and time.

In plain words The fastest wire spends some extra length on a steep plunge at the start. The speed gained there carries the bead along the rest of the route quickly enough to make up for the longer path.

Example Put $B$ a horizontal distance of $\pi \approx 3.14$ meters from $A$ and $2$ meters lower, and take $g = 9.81$ meters per second squared. The cycloid through these points comes from a circle of radius $1$ meter turning half a turn. Its arc is $4$ meters long, while the straight wire has length $L = \sqrt{\pi^2 + 4} \approx 3.72$ meters. The straight wire drops $h = 2$ meters over its length $L$ , so the bead on it has the constant acceleration $gh/L$ along the wire. Covering $L$ from rest takes the time $t$ with $L = \tfrac12 (gh/L)\,t^2$ , so $t = L\sqrt{2/(gh)}$ . The cycloid card below shows that the cycloid takes $\pi\sqrt{r/g}$ with $r = 1$ meter. In seconds,

\begin{aligned} T_{\text{cycloid}} &\approx 1.003, \\[0.4em] T_{\text{straight}} &\approx 1.189. \end{aligned}

The straight wire takes about $18.5$ percent longer. Both beads finish at the same speed, about $6.26$ meters per second, because both have dropped $2$ meters. These figures hold for these endpoints and this model only, and other endpoints give other times and another percentage.

Conservation of energy

Definition A body of mass $m$ moving at speed $v$ has kinetic energy $\tfrac12 m v^2$ , and at height $h$ in uniform gravity $g$ it has potential energy $mgh$ . Conservation of energy says that when no force other than gravity does work on the body, the sum of the two energies stays constant. Friction and air resistance would do work, so the model leaves them out. A fixed wire without friction pushes on the bead only at right angles to its motion, so that push does no work. A bead released from rest that is a vertical distance $y$ below its start therefore satisfies $\tfrac12 m v^2 = m g y$ , so

v = \sqrt{2 g y},

whatever the shape of the wire and whatever the mass of the bead.

In plain words The speed of the bead depends only on how far it has fallen, not on the route it took. Two beads at the same depth move equally fast, so a wire gains time by reaching depth early, where the extra speed is used for most of the trip.

Example Three tenths of a second after release in the race above, the bead on the cycloid has dropped about $0.410$ meters and moves at about $2.84$ meters per second. The bead on the straight wire has dropped only about $0.127$ meters and moves at about $1.58$ meters per second. Both pairs fit the formula, since $2g = 19.62$ and

\begin{aligned} \sqrt{19.62 \cdot 0.410} &\approx 2.84, \\[0.4em] \sqrt{19.62 \cdot 0.127} &\approx 1.58. \end{aligned}

At the finish both beads reach $\sqrt{2 \cdot 9.81 \cdot 2} \approx 6.26$ meters per second, but at different moments. Equal speeds at the finish do not mean equal travel times, because the time depends on the speed all along the way.

Cycloid

Definition A cycloid is the curve traced by a point on the rim of a circle as the circle rolls without slipping along a straight line. For the descent, let a circle of radius $r$ roll along the underside of a horizontal line, with the traced point starting at the point of contact. Measure $x$ to the right and $y$ downward from that start, and let $\theta$ be the angle in radians through which the circle has turned. Then

\begin{aligned} x &= r(\theta - \sin\theta), \\[0.4em] y &= r(1 - \cos\theta). \end{aligned}

One arch, for $0 \le \theta \le 2\pi$ , is $2\pi r$ wide, $2r$ deep at its middle and $8r$ long, and it hangs below the line like a bowl. A bead that slides from rest along such an arch, starting at its top end, reaches the point with angle $\theta$ at the time

t = \theta \sqrt{r/g}.

The reason is that a small turn $\Delta\theta$ moves the point along the curve by about $2r \sin(\theta/2)\,\Delta\theta$ , while the speed there is $\sqrt{2gy} = 2\sqrt{gr}\,\sin(\theta/2)$ , so every small turn of the circle takes the same time $\sqrt{r/g}\,\Delta\theta$ .

In plain words Roll a coin along the underside of a ruler and follow one point on its rim. The point drops straight down at first and then swings forward along a smooth bowl, which is the shape of the fastest track.

Example With $r = 1$ meter and half a turn, $\theta = \pi$ , the point ends at $x = \pi$ and $y = 2$ , the finishing point of the race. That half arch is $4r = 4$ meters long and the bead needs $\pi\sqrt{1/9.81} \approx 1.003$ seconds for it. The cycloid has a second surprise. A bead released from rest anywhere on the arch, not only at its top, reaches the lowest point after the same time $\pi\sqrt{r/g}$ , a property discovered by Christiaan Huygens.

Calculus of variations

Definition The calculus of variations studies problems in which the unknown is a whole function and the quantity to be made as small or as large as possible is an integral that depends on it. A typical problem fixes the two endpoints of a graph $y(x)$ for $x_0 \le x \le x_1$ and asks for the smallest value of

J[y] = \int_{x_0}^{x_1} F(x, y, y')\, dx .

Suppose $F$ has continuous second partial derivatives and a twice continuously differentiable function $y$ gives the smallest value among all such functions with the same endpoints. Then $y$ satisfies the Euler-Lagrange equation

\frac{d}{dx}\,\frac{\partial F}{\partial y'} = \frac{\partial F}{\partial y}.

Like setting a derivative equal to zero in ordinary calculus, this is a necessary condition only. A curve that satisfies it is a candidate, and proving that it really gives the smallest value takes a further argument. For the brachistochrone, measure $y$ downward from $A$ . A piece of wire of length $\Delta s$ takes about $\Delta s / \sqrt{2gy}$ seconds, and the travel time is

T[y] = \int_{0}^{x_B} \frac{\sqrt{1 + y'^2}}{\sqrt{2 g y}}\, dx .

Its integrand is infinite at the start, where $y = 0$ , so the equation is applied to the stretches of the curve away from the start. The further argument has been carried out for this problem, so the cycloid is the true fastest curve and not only a candidate.

In plain words Ordinary calculus finds the best number, and the calculus of variations finds the best curve. Its main equation narrows the search to a few candidates, but a candidate still has to be checked against every other curve.

Example For the brachistochrone, the Euler-Lagrange equation reduces to

y\left(1 + y'^2\right) = c

for a constant $c$ . The cycloid with circle radius $r$ has slope $y' = \sin\theta / (1 - \cos\theta)$ , and

1 + y'^2 = \frac{2}{1 - \cos\theta}.

Multiplying by $y = r(1 - \cos\theta)$ gives $y(1 + y'^2) = 2r$ for every $\theta$ , a constant. Every cycloid of this kind is therefore a candidate, and the endpoints choose the one that passes through $B$ . For $B$ at $x = \pi$ and $y = 2$ this is the circle of radius $1$ turned through half a turn.

Check yourself

Scale up the race. Use a rolling circle of radius $4$ meters, so that $B$ is $4\pi \approx 12.57$ meters across and $8$ meters below $A$ . Find the time on the cycloid, the time on the straight wire and the speed at the finish. Which comparison does not change?

Show the answer

The bead on the cycloid needs half a turn, $\theta = \pi$ , so its time is $\pi\sqrt{4/9.81} \approx 2.006$ seconds. The straight wire has length $L = 4\sqrt{\pi^2 + 4} \approx 14.90$ meters and drop $h = 8$ meters, so its time is

t = L\sqrt{\frac{2}{9.81 \cdot 8}} \approx 2.378

seconds. Both beads finish at $\sqrt{2 \cdot 9.81 \cdot 8} \approx 12.53$ meters per second. Both times are twice the times for radius $1$ , because making every length $4$ times as large multiplies every time by $\sqrt{4} = 2$ . The ratio of the two times stays about $1.185$ , so the straight wire still takes about $18.5$ percent longer.

Test the steepest start. Keep $B$ at $\pi$ meters across and $2$ meters down. A third wire drops straight down $2$ meters, turns through a very small smooth bend, and then runs level for $\pi$ meters to $B$ . Treat the time spent in the bend as zero. Does the steepest possible start beat the cycloid?

Show the answer

It does not. The free fall of $2$ meters takes $\sqrt{2 \cdot 2/9.81} \approx 0.639$ seconds and ends at $\sqrt{2 \cdot 9.81 \cdot 2} \approx 6.26$ meters per second. Without friction the bend does not change the speed, so the level run takes $\pi / 6.26 \approx 0.502$ seconds. The total is about $1.140$ seconds. That beats the straight wire, at $1.189$ seconds, but loses to the cycloid, at $1.003$ seconds. The vertical drop gains speed as fast as possible but makes no horizontal progress during its first $0.64$ seconds. The cycloid balances the two, falling steeply at first and turning forward soon after.

Extend the pattern. Let the circle of radius $1$ meter roll through three quarters of a turn, $\theta = 3\pi/2$ , and let $B$ be the point it reaches. Where is $B$ , how deep does the fastest wire go, and how do the times on the cycloid and on the straight wire compare?

Show the answer

Since $\sin(3\pi/2) = -1$ and $\cos(3\pi/2) = 0$ , the point $B$ is at

\begin{aligned} x &= 3\pi/2 + 1 \approx 5.71, \\[0.4em] y &= 1, \end{aligned}

so $B$ is $5.71$ meters across and only $1$ meter down. The ratio $5.71/1$ is more than $\pi/2$ , so the fastest wire dips below $B$ . Its lowest point, at $\theta = \pi$ , is $2$ meters below $A$ and $1$ meter below $B$ , and from there the bead climbs back up to $B$ . The cycloid takes $(3\pi/2)\sqrt{1/9.81} \approx 1.50$ seconds. The straight wire has length $\sqrt{5.71^2 + 1} \approx 5.80$ meters and drop $1$ , so its acceleration is $g/5.80$ and its time is

5.80\sqrt{2/9.81} \approx 2.62

seconds, about $1.74$ times as long. The bead on the cycloid reaches $6.26$ meters per second at the bottom of the dip and slows to $\sqrt{2 \cdot 9.81 \cdot 1} \approx 4.43$ meters per second at $B$ . Going lower than the finish and climbing back is faster than any route that stays above it.