Brachistochrone is Cycloid

Curve
Let a point $A$ be joined by a wire to a lower point $B$.

Suppose we are allowed to bend the wire into whatever shape we want.

Suppose that a bead is allowed to slide down without friction from $A$ to $B$.

The shape of the wire so that the bead takes least time to descend from $A$ to $B$ is a cycloid.

Proof

 * Brachistochrone.png

We invoke a generalization of Snell's Law.

This is justified, as we are attempting to demonstrate the curve that takes the smallest time.

Thus we have $\dfrac {\sin \alpha} v = k$, where $k$ is some constant.

By the Principle of Conservation of Energy, the speed of the bead at a particular height is determined by its loss in potential energy in getting there.

Thus, at the point $\left({x, y}\right)$, we have:
 * $v = \sqrt {2 g y}$

We have:

Combining all the above equations, we get:

where $c$ is another (more convenient) constant.

This is the differential equation which defines the brachistochrone.

Now we solve it:

Now we introduce a change of variable:
 * $\sqrt {\dfrac y {c - y}} = \tan \phi$

Thus:

Also:

Thus:

As the curve goes through the origin, we have $x = y = 0$ when $\phi = 0$ and so $c_1 = 0$.

Now we can look again at our expression for $y$:

To simplify the constants, we can substitute $a = c / 2$ and $\theta = 2 \phi$, and thus we get:

which are the parametric eqns of the cycloid.