Time of Travel down Brachistochrone

Theorem
Let a wire $AB$ be curved into the shape of a brachistochrone.

Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.

Let a bead $P$ be released at $A$ to slide down without friction to $B$.

Then the time taken for $P$ to slide from $A$ to $B$ is:
 * $T = \pi \sqrt {\dfrac a g}$

where $a$ is the radius of the generating circle of the cycloid which forms $AB$.

Proof
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid.

Let $A$ be located at the origin of a cartesian plane.

We have the equations of the cycloid:

Differentiating with respect to theta:

Let $s$ be the distance along $AB$ from the origin.

Then:

Thus:
 * $\dfrac {\d s} {\d \theta} = 2 a \map \sin {\dfrac \theta 2}$

Now:

We also have, from the Principle of Conservation of Energy, that:

Thus:

Let $T$ be the time taken to slide down the brachistochrone.

At the top:
 * $\theta = 0$

and at the bottom:
 * $\theta = \pi$

Then:

So the time to slide down this brachistochrone from the top to the bottom is:
 * $T = \pi \sqrt {\dfrac a g}$