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 {{WRT|Integration}} $x \in \closedint a b$ 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: {{begin-eqn}} {{eqn | l = x       | r = -\int \sqrt{\frac {-y} {y + C} } \rd y }} {{eqn | r = \int \frac {\sqrt {C u - 1} } {u^2} \rd u       | c = substituting $u = \dfrac 1 {C + y}$ }} {{eqn | r = -\frac {\sqrt {C u - 1 } } u + \frac C 2 \int \dfrac {\d u} {u \sqrt {C u - 1} }       | c = Primitive of $\dfrac {\sqrt {a x + b} } {x^2}$ }} {{eqn | r = -\frac {\sqrt {C u - 1} } u + \frac C 2 2 \, \map \arctan {C u - 1} + C_1       | c = Primitive of $\dfrac 1 {x \sqrt {a x + b} }$ }} {{eqn | r = -\sqrt {-y \paren {C + y} } + C \, \map \arctan {\frac {-y} {C + y} } + C_1 }} {{end-eqn}} 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 $y$: {{begin-eqn}} {{eqn | l = \sqrt {\frac {-y} {C + y} }       | r = \tan \theta }} {{eqn | ll= \leadsto       | l = \frac {-y} {C + y}       | r = \tan^2 \theta }} {{eqn | ll= \leadsto       | l = - 1 - \frac C y       | r = \frac 1 {\tan^2 \theta} }} {{eqn | ll= \leadsto       | l = y       | r = -C \sin^2 \theta }} {{eqn | r = -\frac C 2 \paren {1 - \map \cos {2 \theta} }       | c = Square of Sine }} {{eqn | ll= \leadsto        | r = -\frac C 2 \paren {1 - \map \cos \phi}       | c = Introduce a new parameter $\phi = 2 \theta$ }} {{end-eqn}} Substitution into the expression for $x$ results in: {{begin-eqn}} {{eqn | l = x       | r = C \paren {\theta - \frac {\map \sin {2 \theta} } 2 } }} {{eqn | r = C \paren {\frac \phi 2 - \frac {\sin \phi} 2}       | c = Introduce a new parameter $\phi = 2 \theta$ }} {{eqn | r = \frac C 2 \paren {\phi - {\sin \phi} } }} {{end-eqn}} 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. {{qed}}