Brachistochrone is Cycloid

Theorem
The shape of the brachistochrone is a cycloid.

Proof 1

 * Brachistochrone.png

We invoke a generalization of the Snell-Descartes 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 equations of the cycloid.

Proof 2
Through this solution, we use the standard alignment of coordinate axes: $x$ axis pointing rightwards and $y$ axis is pointint upwards.

Suppose that the curve passes through the point $\tuple {x, y}$ for some value of variable $t$.

Due to smoothness of the curve, one can define velocity $v$ at a point $\tuple {\map x t, \map y t}$:


 * $\displaystyle v = \dfrac {\d s} {\d t}$

where $\d s$ is an infinitesimal length element.

In Euclidean space we have:


 * $\displaystyle \d s = \sqrt{1 + y'^2} \d x$

Hence:


 * $\displaystyle v = \sqrt{1 + y'^2} \dfrac {\d x} {\d t}$

Due to the symmetries of Euclidean space, the following energy conservation identity holds:


 * $\displaystyle \frac {m v^2} 2 + m g y = E$

where $E$ is a constant of motion.

To determine $E$ use the following initial conditions:


 * $\displaystyle \tuple {\map x 0, \map y 0} = \mathbf 0$


 * $\displaystyle \tuple {\map {\dfrac {\d x} {\d t}} 0, \map {\dfrac {\d y} {\d t}} 0} = \mathbf 0$

Then it follows that:


 * $E = 0$

and:


 * $\displaystyle v = \sqrt{- 2 g y}$

Then the total travel time, integrated $x \in \closedint a b$ is:


 * $\displaystyle T = \int_a^b \frac {\sqrt{1 + y'^2}}{\sqrt{- 2 g y}} \rd x$

Application of Euler's Equation yields:


 * $\displaystyle \frac {\sqrt{1 + y'^2}} {\sqrt{- 2 g y}} - y' \frac {2 y'} {2 \sqrt {- 2 g y} \sqrt {1 + y'^2}} = c$

or


 * $\displaystyle \sqrt C = \sqrt {- y \paren {1 + y'^2}}$

where


 * $\displaystyle C = \frac 1 {2 c^2 g}$

and $c$ is a real constant.

The aforementioned differential equation can be rearranged to:


 * $\dfrac {\d x} {\d y} = \pm \sqrt{\frac {-y} {y + C} }$

Since we want to describe a downwards sliding bead, we have:


 * $\dfrac {\d y} {\d x} \le 0$

and we choose the minus sign.

This equation can be solved for $\map x y$:


 * $\displaystyle x = - \sqrt {- y \paren {C + y} } + C \arctan {\sqrt {\frac{- y}{C + y}} } + C_1$

From the initial condition $\tuple {\map x 0, \map y 0} = \mathbf 0$ it follows that:


 * $C_1 = 0$

To bring the solution to parametric form, introduce the following parametric dependence:


 * $\displaystyle \sqrt{\frac {-y} {C + y}} = \tan \theta$

which can be solved for $y$:

Substitution into the expression for $x$ results in:

To determine $C$ we use the boundary condition for the final point:


 * $\displaystyle \dfrac {\d y} {\d x} \Bigg|_{x = b} = 0$

For parametric equations we can rewrite this as:


 * $\dfrac {\d y} {\d x} = \dfrac {\d y} {\d \phi} \paren {\dfrac {\d x} {\d \phi} }^{-1}$

We need to find to which $\phi$ the point $x = b$ corresponds.

Notice that:


 * $\paren {\dfrac {\d y} {\d \phi} = 0 \land \dfrac {\d x} {\d \phi} \ne 0 } \implies \paren { \dfrac {\d y} {\d x} = 0 }$

Therefore:


 * $\displaystyle \dfrac {\d y} {\d \phi} = - \frac C 2 \map {\sin} \phi$

and this derivative vanishes if:


 * $\phi = \pi n, n \in \Z$

Similarly:


 * $\displaystyle \dfrac {\d x} {\d \phi} = \frac C 2 \paren {\map {\cos} \phi - 1}$

which vanishes if:


 * $\displaystyle \phi = 2 \pi n, n \in \Z$

By comparing both conditions on $\phi$ we limit the set of solutions to $\phi = \pi + 2 \pi n, n \in \Z$.

We choose the nearest appropriate value corresponding to $x = b > 0$.

Then substitution into the expression for $x$ results in:


 * $\displaystyle b = \frac C 2 \pi$

Finally, the parametric form of this curve is described by such a parametric solution:


 * $x = \frac b \pi \paren {\phi - {\sin \phi} }$


 * $y = - \frac b \pi \paren {1 - \map {\cos} {\phi} }$

This is the form of a cycloid as portraied upside down.

Also see

 * Cycloid has Tautochrone Property, in which it is shown that a cycloid is also the shape for which it takes the same time for the bead to reach the bottom from wherever it is released.