## The Tautochrone Curve

The tautochrone is the curve in which a ball, positioned anywhere on the curve, will take the same time to arrive at its lowest point.

In a physical model where uniform gravity is the only force acting on the ball, the tautochrone curve is a cycloid (the curve described by a fixed point in a rolling circle), as correctly proven by Huygens in 1673 [BOYDE] using pure geometric arguments. We are going to present two different ways to arrive at this result.

## A wild guess.

Let $\mathcal{C}$ be the curve of interest and consider any partition of it (finite or infinite) in points $p_i$ such that $p_0$ is the highest point in the partition, $p_1$ the second highest and so on. Let $b_i$ be a ball positioned at point $p_i$ of the partition. Assume $(0,0)$ is the highest point of the curve, as in the figure. Further, every ball $b_i$ arrives at the lowest point at time $t^\star$.

$\begin{array}{ll} s_{i}(t): & \text{Distance left to ball } b_i \text{ to arrive at lowest point.}\\[0.25in] v_i(t): & \text{Speed of the ball } b_i \text{ at time } t.\\[0.25in] a_i(t): & \text{Acceleration of the ball } b_i \text{ at time } t. \end{array}$

We also have the following equalities:

$v_i(0) = 0 \quad a_i(t^\star) = 0 \quad s_i(t^\star) = 0 \quad \forall i.$

The very first observation one can come up with is that the acceleration function must be non-negative and strictly decreasing. We are ready to make a wild guess.

Let $b_p$ and $b_q$ two balls positioned on the curve with such that:

$s_p(0) = \alpha \cdot s_q(0).$

One way to impose $b_p$ and $b_q$ to arrive at the lowest point at the same time is to write:

$v_p(t) = \alpha \cdot v_q(t), \: \forall t \in [0,t^\star].$

So we have that the distance traversed by $b_p$ will always be $\alpha$ times the distance traversed by $b_q$. Since both balls start with zero speed, we can also derive the expression of its acceleration:

$a_p(t) = \alpha \cdot a_q(t).$

It is reasonable to guess that the ball’s acceleration will depend on its position on the curve. Let’s assume we can write the equation for the acceleration in terms of the $x$-coordinate

$a(x) = w(x) \cdot s(x),$

where $w(\cdot)$ is a weight function. The simplest possibility for $w(\cdot)$ is a constant

$a(x) = A \cdot s(x).$

Substituting with the actually acceleration function of our model

$g\sin(\theta(x)) = A \cdot s(x).$

Derivating both sides

$\begin{array}{ll} \displaystyle g\cos(\theta)\frac{d\theta}{dx} &= \displaystyle A \frac{ds}{dx} \\[0.25in] &= \displaystyle A \sqrt{1 + (\frac{dy}{dx})^2} \\[0.25in] &= \displaystyle A \sqrt{1 + \tan(\theta)^2} = A\sec(\theta). \end{array}$

Then we have the following separable differential equation

$\displaystyle \frac{d\theta}{dx} = \frac{A}{g\cos(\theta)^2}. \quad \quad (1)$

Using the identity $\displaystyle \cos(\theta)^2 = \frac{1+\cos(2\theta)}{2}.$

$\begin{array}{ll} \displaystyle \frac{g}{2}\int{1+cos(2\theta)} &= \displaystyle \frac{g}{2} \left( \theta + \frac{\sin(2\theta)}{2} \right)\\[0.25in] &= \displaystyle Ax + C_x. \end{array}$

Finally,

$\displaystyle x(\theta) = \frac{g}{4A}(2\theta + \sin(2\theta) + C_x.$

We still have to find an equation for $y$.

$\displaystyle \frac{dy}{dx} = \tan(\theta).$

From equation (1):

$\begin{array}{ll} \displaystyle dy &= \displaystyle \frac{g}{A}\cos(\theta)^2d\theta \cdot \tan(\theta) \\[0.25in] \displaystyle &= \displaystyle \frac{g}{A}\sin(\theta)\cos(\theta)d\theta. \end{array}$

Integrating on both sides

$\begin{array}{ll} y &= \displaystyle \frac{g}{A}\frac{\sin(\theta)^2}{2} + C_y \\[0.25in] &= \displaystyle \frac{g}{4A}\left ( 1 - cos(2\theta) \right) + C_y. \end{array}$

In summary

$\displaystyle \mathbf{x(\theta)} = \frac{g}{4A}(2\theta + \sin(2\theta) ) + C_x \quad \quad \mathbf{y(\theta)} = \frac{g}{4A}\left ( 1 - cos(2\theta) \right) + C_y,$

the equation of a cycloid.

A different source of inspiration could come from harmonic oscillations. For small angles, the movement of a pendulum can be modeled as

$\displaystyle \ddot{\theta}(t) = -\frac{g}{l}\theta(t),$

where $l$ is the length of the chord. The solution is given by

$\displaystyle \theta(t) = \theta_0\cos( \frac{\sqrt{g}}{l} t + \phi),$

where $\theta_0$ is the amplitude, i.e., the maximum angle the pendulum forms with the vertical axis. Notice the period of the harmonic pendulum does not depend on the amplitude. This is exactly the property we want for the tautochrone. Hence, the solution of the differential equation:

$\displaystyle \ddot{s} = A \cdot s,$

is a tautochrone curve.

## Derivation with no guesses.

Although correct, the given solution may not satisfy everyone. It was more a guess than a derivation. In particular, we are not able to ask if there exists a different solution from a cycloid or not.

Consider the model in the next figure. A ball is positioned on curve $\mathcal{C}$ at height $d$. Define

$f(y) = \displaystyle \frac{ds}{dy} = \left( 1 + \frac{dx}{dy}^2 \right)^{\frac{1}{2}}.$

From the principle of conservation of energy we know that:

$\begin{array}{ll} & mg(d-y(t)) = \frac{mv^2}{2} \\[0.25in] \Rightarrow & v = \sqrt{2g(d-y)} \end{array}.$

Therefore, the time for a ball positioned at height $d$ is given by:

$\displaystyle T(d) = \frac{1}{\sqrt{2g}} \int_0^d{ \frac{f(y)}{\sqrt{(d-y)}} dy }.$

We want that the time a ball takes to arrive at the lowest point is independently of its starting height. Therefore $T(d)=T_0, \, \forall d$. We can solve the problem by applying the Laplace Transform.

$\begin{array}{ll} \mathcal{L} \{ T_0 \} &= \displaystyle \mathcal{L} \left\{ \frac{1}{\sqrt{2g}} \int_0^d{\frac{f(y)}{\sqrt{d-y}} dy} \right \}\\[0.25in] &= \displaystyle \frac{1}{\sqrt{2g}} \cdot \mathcal{L} \left\{ \int_0^d{f(y)g(d-y)dy} \right \}\\[0.25in] &= \displaystyle \frac{1}{\sqrt{2g}} \cdot \mathcal{L} \left\{ f(y) \right \} \cdot \mathcal{L} \left \{ g(y) \right \}, \end{array}$

where $\displaystyle g(y) = \frac{1}{\sqrt{y}}$. The Laplace Transform of $g(y)$ is easily computed.

$\displaystyle \mathcal{L} \left\{ g(y) \right\} = \int_0^{\infty}{e^{-sy}y^{-1/2}dy}.$

By using the substitution $u = (st)^{-1/2}$ one derives

$\displaystyle \frac{2}{\sqrt{s}}\int_0^{\infty}{e^{-u^2} du}.$

The last integral can be computed and gives the value $\displaystyle \frac{\sqrt{\pi}}{2}$. Therefore,

$\displaystyle \mathcal{L} \left\{ g(y) \right\} = \sqrt{\frac{\pi}{s}}.$

Hence,

$\displaystyle \mathcal{L} \left\{ f(y) \right \} = \sqrt{\frac{2g}{\pi}}\frac{T_0}{\sqrt{s}}.$

By observing $\displaystyle \mathcal{L} \left\{ y^{-1/2} \right \} = \frac{\sqrt{\pi}}{\sqrt{s}}$.

$\displaystyle f(y) = \frac{T_0 \sqrt{2g}}{\pi}\frac{1}{\sqrt{y}}.$

And now it rests to us to solve the differential equation:

$\begin{array}{ll} & f(y) = \displaystyle \left( 1 + \frac{dx}{dy}^2 \right)^{1/2}\\[0.25in] \Rightarrow & \displaystyle \frac{2T_0^2g}{\pi^2}\frac{1}{y} = 1 + \frac{dx}{dy}^2 \\[0.25in] \Rightarrow & \displaystyle \frac{dx}{dy} = \sqrt{ \frac{2\alpha - y}{y}}. \end{array}$

Using the substitution $y=2 \alpha \sin^2(\theta/2)$.

$\displaystyle x = \alpha(\theta + \sin(\theta)), \quad y = \alpha(1 - \cos(\theta)).$