# Cycloid has Tautochrone Property

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 It has been suggested that this page or section be merged into Time of Travel down Brachistochrone/Corollary. (Discuss)

## 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

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 $\tuple {x, y}$, we have:

$(1): \quad v = \dfrac {\d s} {\d t} = \sqrt {2 g y}$

This can be written:

 $\displaystyle \d t$ $=$ $\displaystyle \frac {\d s} {\sqrt {2 g y} }$ $\displaystyle$ $=$ $\displaystyle \frac {\sqrt {\d x^2 + \d y^2} } {\sqrt {2 g y} }$

Thus the time taken for the bead to slide down the wire is given by:

$\displaystyle T_1 = \int \sqrt {\dfrac {\d x^2 + \d y^2} {2 g y} }$

From Equation of Cycloid, we have:

$x = a \paren {\theta - \sin \theta}$
$y = a \paren {1 - \cos \theta}$

Substituting these in the above integral:

$\displaystyle T_1 = \int_0^{\theta_1} \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {1 - \cos \theta} } } \rd \theta = \theta_1 \sqrt {\dfrac a g}$

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 $\tuple {x_0, y_0}$.

Take equation $(1)$ and replace it with:

$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$

Thus the total time to reach the bottom is:

 $\displaystyle T$ $=$ $\displaystyle \int_{\theta_0}^\pi \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {\cos \theta_0 - \cos \theta} } } \rd \theta$ $\displaystyle$ $=$ $\displaystyle \sqrt {\dfrac a g} \int_{\theta_0}^\pi \sqrt {\dfrac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta$ $\text {(2)}: \quad$ $\displaystyle$ $=$ $\displaystyle \sqrt {\dfrac a g} \int_{\theta_0}^\pi \dfrac {\sin \frac 1 2 \theta \rd \theta} {\sqrt {\cos^2 \frac 1 2 \theta_0 - \cos^2 \frac 1 2 \theta} } \rd \theta$ Double Angle Formula for Cosine

Setting:

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

and so:

$\d u = -\dfrac 1 2 \dfrac {\sin \frac 1 2 \theta \rd \theta} {\cos \frac 1 2 \theta_0}$

Then $(2)$ becomes:

 $\displaystyle T$ $=$ $\displaystyle -2 \sqrt {\dfrac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }$ $\displaystyle$ $=$ $\displaystyle 2 \sqrt {\dfrac a g} \Big [{\arcsin u}\Big]_0^1$ $\displaystyle$ $=$ $\displaystyle \pi \sqrt {\dfrac a g}$

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

Hence the result.

$\blacksquare$

## Also known as

This result is seen referred to as the pendulum property of the cycloid.

## Historical Note

The fact that Cycloid has Tautochrone Property was discovered by Christiaan Huygens in $1658$, during his work on developing a reliable and accurate pendulum clock.

The Tautochrone Problem was also solved independently by Niels Henrik Abel in $1823$, using the technique now known as Abel's integral equation.