Answers · The physics underneath

Why is the fastest way down a hill not a straight line?

In shortBecause time is distance divided by speed, and a steeper path gives the bead more speed early even though it is longer.

The quickest descent is not the straight line because the bead starts from rest. The straight wire is the shortest route, but it keeps the bead slow for the first part of the drop, and time is accumulated as length divided by speed.

Speed knows only height

With no friction, the wire does no work on the bead. Its energy is motion plus height. Start the bead at rest at height zero. After it has descended a distance $h$ , the drop has released $mgh$ of potential energy, and all of that is motion: $mgh=\frac12 m v^2$ , so $v=\sqrt{2gh}$ . At $h=0.5$ m with $g=9.8$ m/s $^2$ , $v=\sqrt{9.8}=3.13$ m/s. At $h=1$ m, $v=\sqrt{19.6}=4.43$ m/s. Those speeds are the same on every smooth wire through those heights. The shape has no say in the speed at a given height.

v=\sqrt{2gh},\qquad T=\sum_{\text{path}}\frac{\Delta s}{v}

The time is therefore a sum of small pieces. A piece of wire of length $\Delta s$ where the speed is $v$ costs $\Delta s/v$ seconds. A longer path can win if it puts more of its length where $v$ is large, because each metre there is cheap in time. The straight line puts a large share of its length near the top, where $v$ is small.

FIG 1

path	length (m)	time (s)
straight	1.414	0.639
bent at (0.2,-0.5)	1.482	0.594

For the straight path, $v$ runs from $0$ to $4.43$ m/s, and $T=2(1.414)/4.43=0.639$ s. For the bent path, the first segment takes $2(0.5385)/3.13=0.344$ s and the second takes $2(0.9434)/(3.13+4.43)=0.250$ s, totalling $0.594$ s. The bent path is $0.068$ m longer and $0.045$ s quicker. The comparison is not a proof that this bent path is best; it is a proof that the straight line is not.

Why the straight line loses

The natural guess treats quickest as shortest. That would be right only if the bead moved at the same speed everywhere, because then time would be length divided by a constant. Here speed is not constant. At $h=0.25$ m it is $\sqrt{4.9}=2.21$ m/s, so one metre there costs about $1/2.21=0.45$ s; at $h=1$ m it is $4.43$ m/s, so one metre there costs about $0.23$ s. The bent path is longer by $0.068$ m, yet its early steep drop buys enough speed to save about $0.045$ s over the straight line. If speed were fixed at $4.43$ m/s, the straight line would take $1.414/4.43=0.319$ s and the bent path $1.482/4.43=0.335$ s, and the short path would win. Starting from rest reverses that comparison.

The true winner is a cycloid, the curve traced by a point on a rolling wheel. It is steeper than the straight line at the start and flatter at the finish, and it is longer than the straight line. The rule it satisfies at every point is $\sin\theta/\sqrt{h}=\text{constant}$ , where $\theta$ is the angle the wire makes with the vertical. Near the top $h$ is small, so $\sin\theta$ must be small and the wire is steep; near the bottom $h$ is larger, so the wire can flatten out. That is the same condition light obeys when it bends into a faster layer, with the bead's speed playing the part of the speed of light.

\frac{\sin\theta}{\sqrt{h}}=\text{constant along the quickest path}

A second fact falls out of the same curve. Release the bead from halfway up the cycloid, or from a hand's width above the bottom, and it reaches the bottom in the same time. The extra distance from higher up is exactly paid for by the extra speed. That property was not asked for; it is a consequence of the shape that already won the race.

Where it stops

This comparison assumes a smooth wire, no friction, and a bead released from rest. With friction, energy is not conserved, the speed at a given height depends on the path taken to get there, and the cycloid is no longer the quickest shape. If the bead starts with speed $v_0$ , replace $v=\sqrt{2gh}$ by $v=\sqrt{v_0^2+2gh}$ . The same length-over-speed sum still decides the race, but the winning curve changes.

Worth keeping

At any height on a smooth wire, the bead's speed is fixed by how far it has fallen, not by the shape of the wire.
The quickest path is the one that minimises the sum of small lengths divided by the speeds there, and that path is longer than the straight line.
Friction destroys the argument because speed then depends on the path and not only on height.

Where this goes next

This question is one step of Motion and What Is Conserved, a course that works through its subject from the beginning. The page that teaches this part of it is The Quickest Way Down, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that page The whole course