A tautochrone or isochrone curve is the curve for which the time taken by a frictionless particle sliding down it under uniform gravity to its lowest point is independent of its starting point. The curve is a cycloid, and the time is equal to π times the square root of the radius over the acceleration of gravity.

The tautochrone problem, the attempt to identify this curve, was solved by Christiaan Huygens in 1659. He proved geometrically in his Horologium oscillatorium (The Pendulum Clock, 1673) that the curve was a cycloid. This solution was later used to attack the problem of the brachistochrone curve. Jakob Bernoulli solved the problem using calculus in a paper (Acta Eruditorum, 1690) that saw the first published use of the term integral.

Later mathematicians such as Joseph Louis Lagrange and Leonhard Euler looked for an analytical solution to the problem.

"Virtual gravity" solution

Perhaps the simplest solution to the tautochrone problem is to note a direct relation between the angle of an incline and the gravity felt by a particle on the incline. A particle on a 90° vertical incline feels the full effect of gravity, while a particle on a horizontal plane feels effectively no gravity. At intermediate angles, the "virtual gravity" felt by the particle is $g\; \backslash sin\; \backslash theta\backslash ,$. The first step is to find a "virtual gravity" that produces the desired behavior.

The "virtual gravity" required for the tautochrone is simply proportional to the distance remaining to be traveled, which admits a simple solution:

$$

\frac{d^2s} \\ T & = & \pi \sqrt{\frac{r}{g}} \end{matrix}

(Based loosely on Proctor, pp. 135-139)

Abel's solution

Abel attacked a generalized version of the tautochrone problem (Abel's mechanical problem), namely, given a function $T(y)$ that specifies the total time of descent for a given starting height, find an equation of the curve that yields this result. The tautochrone problem is a special case of Abel's mechanical problem when $T(y)$ is a constant.

Abel's solution begins with the principle of conservation of energy — since the particle is frictionless, and thus losses no energy to heat, its kinetic energy at any point is exactly equal to the difference in potential energy from its starting point. The kinetic energy is $\backslash frac\{1\}\{2\}\; mv^2$, and since the particle is constrained to move along a curve, its velocity is simply $\backslash frac\{ds\}\{dt\}$, where $s$ is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height $y\_0\backslash ,$ to a height $y\backslash ,$ is $mg(y\_0-y)\backslash ,$, thus:

$$

\begin{matrix} \frac{1}{2} m \left ( \frac{ds}{dt} \right ) ^2 & = & mg(y_0-y) \\ \frac{ds}{dt} & = & \pm \sqrt{2g(y_0-y)} \\ dt & = & \pm \frac{ds}{\sqrt{2g(y_0-y)}} \\ dt & = & - \frac{1}{\sqrt{2g(y_0-y)}} \frac{ds}{dy} dy\\ \end{matrix}

In the last equation, we've anticipated writing the distance remaining along the curve as a function of height ($s(y)\backslash ,$), recognized that the distance remaining must decrease as time increases (thus the minus sign), and used the chain rule in the form $ds\; =\; \backslash frac\{ds\}\{dv\}\; dv$.

Now we integrate from $y=y\_0$ to $y=0$ to get the total time required for the particle to fall:

$$

T(y_0) = \int_{y=y_0}^{y=0} \, dt = \frac{1}{\sqrt{2g}} \int_{0}^{y} \frac{1}{\sqrt{y_0-y}} \frac{ds}{dy} dy\,

This is called Abel's integral equation and allows us to compute the total time required for a particle to fall along a given curve (for which $\backslash frac\{ds\}\{dy\}$ would be easy to calculate). But Abel's mechanical problem requires the converse — given $T(y\_0)\backslash ,$, we wish to find $\backslash frac\{ds\}\{dy\}$, from which an equation for the curve would follow in a straightforward manner. To proceed, we note that the integral on the right is the convolution of $\backslash frac\{ds\}\{dy\}$ with $\backslash frac\{1\}\{\backslash sqrt\{y\}\}$ and thus take the Laplace transform of both sides:

$$

\mathcal{L}= \frac{1}{\sqrt{2g}} \mathcal{L} \left [ \frac{1}{\sqrt{y}} \right \mathcal{L} \left \frac{ds}{dy} \right

Since $\backslash mathcal\{L\}\; \backslash left\backslash frac\{1\}\{\backslash sqrt\{y\}\}\; \backslash right=\; \backslash sqrt\{\backslash pi\}z^\backslash frac\{1\}\{2\}$, we now have an expression for the Laplace transform of $\backslash frac\{ds\}\{dy\}$ in terms of $T(y\_0)\backslash ,$'s Laplace transform:

$$

\mathcal{L}\left \frac{ds}{dy} \right = \sqrt{\frac{2g}{\pi}} z^{\frac{1}{2}} \mathcal{L}^*

This is as far as we can go without specifying $T(y\_0)\backslash ,$. Once $T(y\_0)\backslash ,$ is known, we can compute its Laplace transform, calculate the Laplace transform of $\backslash frac\{ds\}\{dy\}$ and then take the inverse transform (or try to) to find $\backslash frac\{ds\}\{dy\}$.

For the tautochrone problem, $T(y\_0)\; =\; T\_0\backslash ,$ is constant. Since the Laplace transform of 1 is $\backslash frac\{1\}\{z\}$, we continue:

$$

\begin{matrix} \mathcal{L}\left \frac{ds}{dy} \right & = & \sqrt{\frac{2g}{\pi}} z^{\frac{1}{2}} \mathcal{L}^* \\ & = & \sqrt{\frac{2g}{\pi}} T_0 z^{-\frac{1}{2}} \\ \end{matrix}

Making use again of the Laplace transform above, we invert the transform and conclude:

$$

\frac{ds}{dy} = T_0 \frac{\sqrt{2g}}{\pi}\frac{1}{\sqrt{y}}

It can be shown that the cycloid obeys this equation.

(Simmons, Section 54).

Bibliography

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• Simmons, George, Differential Equations with Applications and Historical Notes, McGraw-Hill, 1972. ISBN 0070575401.
• Proctor, Richard, A Treatise on The Cycloid and all forms of Cycloidal Curves (1878), posted by Cornell University Library.

Curves | courbe tautochrone