Time of Travel down Brachistochrone/Corollary

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 from anywhere on the wire between $A$ and $B$ to slide down without friction to $B$.

Then the time taken for $P$ to slide 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 coordinate plane.

Let the bead slide from an intermediate point $\theta_0$.

We have:
 * $v = \dfrac {\mathrm d s} {\mathrm d t} = \sqrt{2 g \left({y - y_0}\right)}$

which leads us, via the same route as for Time of Travel down Brachistochrone, to:

Using the Half Angle Formula for Cosine and Half Angle Formula for Sine, this gives:


 * $\displaystyle T = \sqrt{\frac a g} \int_{\theta_0}^\pi \frac {\sin \left({\theta / 2}\right)} {\sqrt{\cos \left({\theta_0 / 2}\right) - \cos \left({\theta / 2}\right)} } \, \mathrm d \theta$

Now we make the substitution:

Recalculating the limits:
 * when $\theta = \theta_0$ we have $u = 1$
 * when $\theta = \pi$ we have $u = 0$.

So:

Thus the time to slide down a brachistochrone from any arbitrary point $\theta_0$ is:
 * $T = \pi \sqrt {\dfrac a g}$