Brachistochrone is Cycloid/Proof 2

Proof
Throughout this proof, we use the standard alignment of coordinate axes:
 * $X$-axis pointing rightwards
 * $Y$-axis is pointing 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 $\tuple {\map x t, \map y t}$:


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

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

In Euclidean space the arc length element is:


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

Hence:


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

Due to the symmetries of Euclidean space, the Principle of Conservation of Energy holds:


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

where $E$ is a constant of motion.

To determine $E$ use the following initial conditions:


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


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

Then it follows that:


 * $E = 0$

and:


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

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:


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

or


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

where


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

and $c$ is a real constant.

The aforementioned differential equation can be rearranged to:


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

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 differential equation can be solved for $\map x y$ in the following way:

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:


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

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:


 * $\intlimits {\dfrac {\d y} {\d x} } {x = b} {} = 0$

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:


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

and this derivative vanishes if:


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

Similarly:


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

which vanishes if:


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

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:


 * $b = \dfrac C 2 \pi$

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


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


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

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