# Demonstrating that a cycloid solves the Tautochrone Problem

I've become stuck on a problem from my calculus textbook, while attempting some revision.

The problem: I'm told that the following parametric equations describe an arc of a cycloid that is the solution to the Tautochrone Problem, that is to say that the time taken for an object to reach the bottom of the curve is independent of the objects starting point. (So long as we don't bring friction into the picture)

$$x(\theta) = a(\theta - Sin(\theta));\\ y(\theta) = a(Cos(\theta)-1).$$

The question asks that verify that the curve is a solution to the Tautochrone Problem and provides the following two hints.

1. The speed of an object (v) at $\theta$, that started from $\theta_{0}$, is given by $v = \sqrt{2g(y(\theta_{0})-y(\theta))}$
2. The transit time to the bottom of the curve (T) is given by the following integral,

$$T = \displaystyle\int_{\theta_{0}}^\pi \frac{1}{v} ds$$

Where 's' is the arc length parameter for the curve.

My attempt: So given the information above, and that this question is in a chapter on parametric curves, it seems that the only thing to do is determine a relationship between the arc-length parameter and the parameter given in the equations provided. So start by writing the position vector for any point on the curve as, $\boldsymbol{r}(\theta) = \left(x(t), y(t)\right)$

The arc-length (s) of the curve, on the interval $\left[\theta_{0}, \theta\right]$, is given by the integral.

\begin{align} s & = \int_{\theta_{0}}^{\theta} \ \left|\frac{d \ \boldsymbol{r}}{d \ \tau}\right|d \ \tau \\ \\ \therefore \frac{d \ s}{d \ \theta} & = \left|\frac{d \ \boldsymbol{r}}{d \ \theta}\right| \end{align}

Returning back to our integral for the Transit time, we use the above to change the variable of integration.

$$T = \displaystyle\int_{\theta_{0}}^{\pi} \frac{1}{v}\frac{ds}{d\theta}d\theta$$

$$\therefore T = \displaystyle\int_{\theta_{0}}^{\pi} \frac{1}{v}\left|\frac{d\boldsymbol{r}}{d\theta}\right|d\theta$$

Trouble is I get this horrible integral and i'm not sure that it would even show what I've set out to show. Or in other words i'm pretty sure i've gone wrong somewhere, or maybe i've just plain gone down the wrong track altogether.

$$T = \sqrt{\frac{1}{ga}}\displaystyle\int_{\theta_{0}}^{\pi} \frac{2\sin\left(\theta/2\right)}{\left(2\cos(\theta_{0}) - 2\cos(\theta)\right)^{1/2}}\ d\theta$$

Edit: So turns out the integral is totally solvable with a simple trig identity that i had no idea about...

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You are plugging in the solution - you will not prove anything with that. You need to write the integral in terms of an unknown curve and then use the Euler-Lagrange equation to find a differential equation for that curve. – Ron Gordon Nov 15 '13 at 13:52
The point of the exercise wasn't to find an equation for the curve that solves the Tautochrone problem, it was simply to verify that the given curve was a solution. ie. demonstrate that the transit time of a body moving along a cycloid is independent of the starting point. (Probably should have mentioned that this problem comes from a basic calculus text) – Marc Fletcher Nov 18 '13 at 9:10

Use the fact that: $$\cos \theta_0 -\cos \theta = \cos^2 \frac{\theta_0}{2} -\cos^2 \frac{\theta}{2}.$$ And change the variable to: $$u =\frac{ \cos \frac{\theta}{2}}{\cos \frac{\theta_0}{2}}.$$ You will get a very easy integral.