The Brachistochrone Curve

Let B to be a ball positioned at some point A. Considering that only gravitational force acts on B, what is the trajectory the ball should follow in order to reach a point C in the shortest time? The solution of this problem is known as the brachistochrone curve.

Although similar, the brachistochrone problem is different from the tautochrone problem. The amusing fact is that the solutions of both problems are the same. The brilliant argument exposed next is due to Johann Bernoulli.

The principle of least time

Light has different speeds at different mediums. That’s why we perceive the phenomenon of refraction when observing objects underwater while ourselves stand out of it. Willebrord Snellius was the first to give a relation between the speed of light in different mediums and its angle of refraction. By denoting n_1 and n_2 constants related with the speed of light at mediums M_1 and M_2 (also called refraction indexes of mediums M_1 and M_2); by \theta_1 the entry ray’s angle with the plane P separating M_1 and M_2; and by \theta_2 the out ray’s angle, the relation is written as:

\displaystyle \frac{n_1}{n_2} = \frac{ \sin \theta_1 }{\sin \theta_2}

Snellius probably based its relation on experimental evidence, but we can derive it from the principle of least time. It was argued by Pierre de Fermat that nature behaves in an optimum way. In particular, the light would follow the path of least time between any two points A and B. In the following, we are going to use Fermat’s principle to derive Snell’s law.


Assume points A and B are separated from each other by a height of H and a length of W. Moreover, assume the plane P separating mediums M_1 and M_2 is at a height h from A. It will be sufficient to discover the point Q, distant of a length x from A, in which the entry ray touches the plane P. The time light will take to travel from A to Q is denoted t_1 and time light takes from travel from Q to B is denoted t_2. We can easily compute t_1 and t_2:

\displaystyle t_1 = \frac{ (x^2 + h^2)^{1/2} }{v_1} \quad \quad t_2 = \frac{ \left( (W-x)^2 + (H-h)^2\right)^{1/2}}{v_2}.

The value \hat{x} that minimizes t = t_1 + t_2 is then

\displaystyle \frac{dt}{dx}(\hat{x}) = 0 \Rightarrow \frac{\hat{x}}{v_1(h^2+\hat{x}^2)^{1/2}} - \frac{W-x^\star}{v_2( (H-h)^2+(W-x^\star)^{2})^{1/2}} = 0.


\displaystyle \frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}.

In other words, the ratio \displaystyle \frac{ \sin \theta }{v} is constant. We are going to call this constant b.

Bernoulli’s Analogy

If the path traveled by light is the one of least time, maybe we can use this fact to derive the brachistochrone curve. That was precisely the idea of Johann Bernoulli. Imagine that a ball traveling from points A to B passes through many different mediums. In order to follow the path of least time, the ball’s trajectory should suffer the same changes of orientation that light would suffer by Snell’s law. If we consider thinner and thinner mediums we hope that we can find the brachistochrone curve using calculus:

\begin{array}{rl} b &= \displaystyle \frac{\sin \theta}{v} =\frac{dx/ds}{v}\\[0.25in] bvds &= dx \\[0.25in] \displaystyle b^2v^2(dx^2 + dy^2) &=dx^2 \\[0.25in] \displaystyle dx^2(1-b^2v^2) &=dy^2(b^2v^2) \\[0.25in] \displaystyle \left(\frac{dy}{dx}\right)^2 &= \displaystyle \frac{1-b^2v^2}{b^2v^2}\\[0.25in] \end{array}.

Using v = \sqrt{-2gy}

\begin{array}{rl} \displaystyle \left(\frac{dy}{dx}\right)^2 &= \displaystyle \frac{1+2b^2gy}{-2b^2gy}\\[0.25in] &= \displaystyle \frac{\frac{1}{-2b^2g} - y}{y}. \end{array}

By denoting \displaystyle 2a = -\frac{1}{2b^2g} we recover the equation of the tautochrone curve. Surprisingly, the tautochrone and the brachistochrone curves are the same: an arch of a cycloid!

\displaystyle \left(\frac{dy}{dx}\right) = \displaystyle \sqrt{\frac{2a - y}{y}}.

Further Reading

[1]Kunkel, Paul.The Brachistochrone. [Link]
[2] – Nishiyama, Yutaka. The Brachistochrone Curve: The Problem of Quickest Descent. [Link]
[3]Freire, Alex.The Brachistochrone Problem. [Link]


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s