Tautochrone curve.html

Four points run over a cycloid from different positions, but they arrive at the bottom at the same time. The blue arrows show the points' acceleration. On the top is the time-position diagram.

A tautochrone or isochrone curve is the curve for which the time taken by an object sliding without friction in 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.

1 The tautochrone problem
2 Lagrangian solution
3 "Virtual gravity" solution
4 Abel's solution
5 External links
6 Bibliography

The tautochrone problem

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.

Lagrangian solution

If the particle's position is parametrized by the arclength s(t) from the lowest point, the kinetic energy is proportional to $\scriptstyle \dot{s}^2$ . The potential energy is proportional to the height y(s). In order to be an isochrone, the Lagrangian must be that of a simple harmonic oscillator: the height of the curve must be proportional to the arclength squared.

$y(s) = s^2 \,$

Where the constant of proportionality has been set to 1 by changing units of length. The differential form of this relation is

$dy = 2s ds \,$

$dy^2 = 4s^2 ds^2 = 4y (dx^2 + dy^2) \,$

Which eliminates s, and leaves a differential equation for dx and dy. To find the solution, integrate for x in terms of y:

${dx \over dy} = {\sqrt{1-4y}\over 2\sqrt{y}} \,$

$x = \int \sqrt{1-4u^2} du$

Where $\scriptstyle u=\sqrt{y}$ . This integral is the area under a circle, which can be naturally cut into a triangle and a circular wedge:

$x= {1\over 2} u \sqrt{1-4u^2} + {1\over 4} \sin^{-1}(2u) \,$

$y= u^2 \,$

To see that this is a strangely parametrized cycloid, change variables to disentangle the transcendental and algebraic parts: define the angle $θ = sin - 1 (2 u)$ .

$8x = 2\sin(\theta) cos(\theta) + 2\theta = \sin(2\theta) + 2\theta \,$

$8y = 2\sin(\theta)^2 = 1 - \cos(2\theta)\,$

Which is the standard parametrization, except for the scale of x, y and $θ$ .

"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 $g \sin \theta\,$ . 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:

$\frac{d^2s}{{dt}^2} = - k^2s$

$s = A \cos kt \,$

It can be easily verified both that this solution solves the differential equation and that a particle will reach $s=0\,$ at time $\frac{\pi}{2k}$ from any starting height $A\,$ . The problem is now to construct a curve that will produce a "virtual gravity" proportional to the distance remaining to travel, i.e, a curve that satisfies:

$g \sin \theta = - k^2 s \,$

The explicit appearance of the distance remaining is troublesome, but we can differentiate to obtain a more manageable form:

$g \cos \theta d\theta = - k^2 ds \,$

$ds = - \frac{g}{k^2} \cos \theta d\theta\,$

This equation relates the change in the curve's angle to the change in the distance along the curve. We now use the Pythagorean theorem, the fact that the slope of the curve is equal to the tangent of its angle, and some trigonometric identities to obtain $d s$ is terms of $d x$ :

$\begin{matrix} ds^2 & = & dx^2 + dy^2 \\ \ & = & \left ( 1 + \left ( \frac{dy}{dx} \right ) ^2 \right ) dx^2 \\ \ & = & ( 1 + \tan^2 \theta ) dx^2 \\ \ & = & \sec^2 \theta dx^2 \\ ds & = & \sec \theta dx \end{matrix}$

Substituting this into the first differential equation lets us solve for $x$ in terms of $θ$ :

$\begin{matrix} ds & = & - \frac{g}{k^2} \cos \theta d\theta \\ \sec\theta \ dx & = & - \frac{g}{k^2} \cos \theta d\theta \\ dx & = & - \frac{g}{k^2} \cos^2 \theta d\theta \\ & = & - \frac{g}{2 k^2} \left ( \cos 2 \theta + 1 \right ) d\theta \\ x & = & - \frac{g}{4 k^2} \left ( \sin 2 \theta + 2 \theta \right ) + C_x \end{matrix}$

Likewise, we can also express $d x$ in terms of $d y$ and solve for $y$ in terms of $θ$ :

$\begin{matrix} \frac{dy}{dx} & = & \tan \theta \\ dx & = & \cot \theta dy \\ \cot \theta dy & = & - \frac{g}{k^2} \cos^2 \theta d\theta \\ dy & = & - \frac{g}{k^2} \sin \theta \cos \theta d\theta \\ & = & - \frac{g}{2k^2} \sin 2 \theta d\theta \\ y & = & \frac{g}{4k^2} \cos 2 \theta + C_y \end{matrix}$

Substituting $\phi = - 2\theta\,$ and $r = \frac{g}{4k^2}\,$ , we see that these equations for $x$ and $y$ are those of a circle rolling along a horizontal line — a cycloid:

$\begin{matrix} x & = & r ( \sin \phi + \phi ) + C_x \\ y & = & r ( \cos \phi ) + C_y \end{matrix}$

Solving for $k$ and remembering that $T = \frac{\pi}{2k}$ is the time required for descent, we find the descent time in terms of the radius $r$ :

$\begin{matrix} r & = & \frac{g}{4k^2} \\ k & = & \frac{1}{2} \sqrt{\frac{g}{r}} \\ T & = & \pi \sqrt{\frac{r}{g}} \end{matrix}$

(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 loses 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 $\frac{1}{2} mv^2$ , and since the particle is constrained to move along a curve, its velocity is simply $\frac{ds}{dt}$ , where $s$ is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height $y_0\,$ to a height $y\,$ is $mg(y_0-y)\,$ , 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 ( $s(y)\,$ ), recognized that the distance remaining must decrease as time increases (thus the minus sign), and used the chain rule in the form $ds = \frac{ds}{dv} dv$ .

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

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

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 $\frac{ds}{dy}$ would be easy to calculate). But Abel's mechanical problem requires the converse — given $T(y_0)\,$ , we wish to find $\frac{ds}{dy}$ , 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 $\frac{ds}{dy}$ with $\frac{1}{\sqrt{y}}$ and thus take the Laplace transform of both sides:

$\mathcal{L}[T(y_0)] = \frac{1}{\sqrt{2g}} \mathcal{L} \left [ \frac{1}{\sqrt{y}} \right ] \mathcal{L} \left [ \frac{ds}{dy} \right ]$

Since $\mathcal{L} \left [ \frac{1}{\sqrt{y}} \right ] = \sqrt{\pi}z^{-\frac{1}{2}}$ , we now have an expression for the Laplace transform of $\frac{ds}{dy}$ in terms of $T(y_0)\,$ 's Laplace transform:

$\mathcal{L}\left [ \frac{ds}{dy} \right ] = \sqrt{\frac{2g}{\pi}} z^{\frac{1}{2}} \mathcal{L}[T(y_0)]$

This is as far as we can go without specifying $T(y_0)\,$ . Once $T(y_0)\,$ is known, we can compute its Laplace transform, calculate the Laplace transform of $\frac{ds}{dy}$ and then take the inverse transform (or try to) to find $\frac{ds}{dy}$ .

For the tautochrone problem, $T(y_0) = T_0\,$ is constant. Since the Laplace transform of 1 is $\frac{1}{z}$ , we continue:

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