Time of Travel down Brachistochrone

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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:

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


Differentiating with respect to theta:

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


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:

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

$\blacksquare$


Sources