# 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 $AB$.

### Corollary

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:

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

## 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.

We have the equations of the cycloid:

 $\displaystyle x$ $=$ $\displaystyle a \left({\theta - \sin \theta}\right)$ $\displaystyle y$ $=$ $\displaystyle a \left({1 - \cos \theta}\right)$
 $\displaystyle \frac {\mathrm d x} {\mathrm d \theta}$ $=$ $\displaystyle \left({1 - \cos \theta}\right)$ $\displaystyle \frac {\mathrm d y} {\mathrm d \theta}$ $=$ $\displaystyle a \sin \theta$

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

Then:

 $\displaystyle \left({\frac {\mathrm d s} {\mathrm d \theta} }\right)^2$ $=$ $\displaystyle \left({\frac {\mathrm d x} {\mathrm d \theta} }\right)^2 + \left({\frac {\mathrm d y} {\mathrm d \theta} }\right)^2$ $\displaystyle$ $=$ $\displaystyle a^2 \left({1 - \cos \theta}\right)^2 + a^2 \sin^2 \theta$ $\displaystyle$ $=$ $\displaystyle a^2 \left({1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta}\right)$ $\displaystyle$ $=$ $\displaystyle 2 a^2 \left({1 - \cos \theta}\right)$ $\displaystyle$ $=$ $\displaystyle 4 a^2 \sin^2 \left({\frac \theta 2}\right)$

Thus:

$\dfrac {\mathrm d s} {\mathrm d \theta} = 2 a \sin \left({\dfrac \theta 2}\right)$

Now:

 $\displaystyle v$ $=$ $\displaystyle \dfrac {\mathrm d s} {\mathrm d t}$ $\displaystyle \therefore \ \$ $\displaystyle \int \mathrm d t$ $=$ $\displaystyle t$ $\displaystyle$ $=$ $\displaystyle \int \frac {\mathrm d s} v$

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

 $\displaystyle m g y$ $=$ $\displaystyle \frac {m v^2} 2$ $\displaystyle \therefore \ \$ $\displaystyle v$ $=$ $\displaystyle \dfrac {\mathrm d s} {\mathrm d t}$ $\displaystyle$ $=$ $\displaystyle \sqrt {2 g y}$ $\displaystyle$ $=$ $\displaystyle \sqrt {2 g a \left({1 - \cos \theta}\right)}$

Thus:

 $\displaystyle \frac {\mathrm d t} {\mathrm d \theta}$ $=$ $\displaystyle \frac {\mathrm d t} {\mathrm d s} \theta$ $\displaystyle$ $=$ $\displaystyle \frac {a \sqrt {2 \left({1 - \cos \theta}\right)} } {\sqrt{2 g a \left({1 - \cos \theta}\right)} }$ $\displaystyle$ $=$ $\displaystyle \sqrt {\frac a g}$

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

At the top:

$\theta = 0$

and at the bottom:

$\theta = \pi$

Then:

 $\displaystyle T$ $=$ $\displaystyle \int \, \mathrm d t$ $\displaystyle$ $=$ $\displaystyle \sqrt {\frac a g} \int_0^\pi \, \mathrm d \theta$ $\displaystyle$ $=$ $\displaystyle \pi \sqrt {\frac a g}$

So the time to slide down this brachistochrone from the top to the bottom is:

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

$\blacksquare$