Cycloid has Tautochrone Property

Theorem
Consider a wire bent into the shape of an arc of a cycloid $C$ and inverted so that its cusps are uppermost and on the same horizontal line.

Let a bead $B$ be released from some point on the wire.

The time taken for $B$ to reach the lowest point of $C$ is:
 * $T = \pi \sqrt {\dfrac a g}$

independently of the point at which $B$ is released from.

That is, a cycloid is a tautochrone.

Proof

 * Brachistochrone.png

By the Principle of Conservation of Energy, the speed of the bead at a particular height is determined by its loss in potential energy in getting there.

Thus, at the point $\left({x, y}\right)$, we have:
 * $(1): \quad v = \dfrac {\mathrm d s} {\mathrm d t} = \sqrt {2 g y}$

This can be written:

Thus the time taken for the bead to slide down the wire is given by:
 * $\displaystyle T_1 = \int \sqrt {\dfrac {\mathrm d x^2 + \mathrm d y^2} {2 g y} }$

From Equation of Cycloid, we have:
 * $x = a \left({\theta - \sin \theta}\right)$
 * $y = a \left({1 - \cos \theta}\right)$

Substituting these in the above integral:


 * $\displaystyle T_1 = \int_0^{\theta_1} \sqrt {\dfrac {2 a^2 \left({1 - \cos \theta}\right)} {2 a g \left({1 - \cos \theta}\right)} } \, \mathrm d \theta = \theta_1 \sqrt {\dfrac a g}$

This is the time needed for the bead to reach the bottom when released when $\theta_1 = \pi$, and so:


 * $T_1 = \pi \sqrt {\dfrac a g}$

Now suppose the bead is released at any intermediate point $\left({x_0, y_0}\right)$.

Take equation $(1)$ and replace it with:
 * $v = \dfrac {\mathrm d s} {\mathrm d t} = \sqrt {2 g \left({y - y_0}\right)}$

Thus the total time to reach the bottom is:

Setting:
 * $u = \dfrac {\cos \frac 1 2 \theta} {\cos \frac 1 2 \theta_0}$

and so:
 * $\mathrm d u = -\dfrac 1 2 \dfrac {\sin \frac 1 2 \theta \, \mathrm d \theta} {\cos \frac 1 2 \theta_0}$

Then $(2)$ becomes:

That is, wherever the bead is released from, it takes that same time to reach the bottom.

Hence the result.

Also see

 * Brachistochrone is Cycloid, in which it is shown that a cycloid is also the shape for which the time is shortest.