# Quick! Find a Solution to the Brachistochrone Problem

**Figure 1.**

Figure 2 shows the cycloid swept out by a point \(P\) on the rim of a circular wheel rolling on the ceiling. Let \(PC^\prime\) be the tangent at \(P\), with \(C^\prime\) lying on the circle. Note that \(PC^\prime \perp CP\): The velocity of a point on the rigid body is perpendicular to the point’s radius vector relative to the instantaneous center of rotation \(C\). We conclude that \(CC^\prime\) is a diameter. But in that case,

\[\begin{equation}\tag{1}

y = CP~\mathrm{sin}~\theta = D ~\mathrm{sin}^2 \theta,

\end{equation}\]

where \(\theta\) is the angle between the tangent and the vertical.

**Figure 2.**

Looking again at the bead, its sliding time along \(\gamma\) is \(\int ds/v\). Given that \(\upsilon = \sqrt{2gy}\) (using conservation of energy and the assumption \(\upsilon_A = 0)\), this time is

\[\begin{equation}\tag{2}

k \int_{\gamma} ~\frac{ds}{\sqrt{y}},

\end{equation}\] where \(k\) is a constant we don’t care about.

Now, \((2)\) is of the form \(\int F(y)ds\), and minimizers of such functionals satisfy (a short calculus-free derivation of this can be found in [1]); for \((2)\) this amounts to

\[\begin{equation}\tag{3}

\frac{\mathrm{sin}~\theta}{\sqrt{y}} = \mathrm{constant}.

\end{equation}\]

which is the same equation as \((1)\)!

The cycloid is thus a critical curve for the time functional \((2)\) (although this does not prove minimality; a proof can be found in almost any book on calculus of variations, e.g., [1]).

**Acknowledgments:** The work from which these columns are drawn is funded by NSF grant DMS-1412542.

**References**

[1] M. Levi, Classical Mechanics with Calculus of Variations and Optimal Control, AMS, Providence, Rhode Island, 2014.