# 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:

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

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

\(\displaystyle T\) | \(=\) | \(\displaystyle \int_{\theta_0}^\pi \sqrt {\frac {2 a^2 \paren {1 - \cos \theta} } {2 g a \paren {\cos \theta_0 - \cos \theta} } } \rd \theta\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sqrt {\frac a g} \int_{\theta_0}^\pi \sqrt {\frac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta\) |

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

- $\displaystyle T = \sqrt{\frac a g} \int_{\theta_0}^\pi \frac {\map \sin {\theta / 2} } {\sqrt {\map \cos {\theta_0 / 2} - \map \cos {\theta / 2} } } \rd \theta$

Now we make the substitution:

\(\displaystyle u\) | \(=\) | \(\displaystyle \frac {\map \cos {\theta / 2} } {\map \cos {\theta_0 / 2} }\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d u} {\d \theta}\) | \(=\) | \(\displaystyle -\frac {\map \sin {\theta / 2} } {2 \map \cos {\theta_0 / 2} }\) |

Recalculating the limits:

- when $\theta = \theta_0$ we have $u = 1$
- when $\theta = \pi$ we have $u = 0$.

So:

\(\displaystyle T\) | \(=\) | \(\displaystyle -2 \sqrt {\frac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \intlimits {2 \sqrt {\frac a g} \sin^{-1} u} 0 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \pi \sqrt {\frac a g}\) |

Thus the time to slide down a brachistochrone from any arbitrary point $\theta_0$ is:

- $T = \pi \sqrt {\dfrac a g}$

$\blacksquare$

## Sources

- 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $\S 1.6$: The Brachistochrone. Fermat and the Bernoullis: Problem $2$