:500px 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 $\tuple {x, y}$, we have: :$v = \sqrt {2 g y}$ We have: {{begin-eqn}} {{eqn | l = \sin \alpha       | r = \cos \beta       | c =  }} {{eqn | r = \frac 1 {\sec \beta}       | c =  }} {{eqn | r = \frac 1 {\sqrt {1 + \tan^2 \beta} }       | c =  }} {{eqn | r = \frac 1 {\sqrt {1 + \paren {y'}^2} }       | c =  }} {{end-eqn}} Combining all the above equations, we get: {{begin-eqn}} {{eqn | l = \frac {\frac 1 {\sqrt {1 + \paren {y'}^2} } } {\sqrt {2 g y} }       | r = k       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 {1 + \paren {y'}^2}       | r = 2 k^2 g y       | c =  }} {{eqn | ll= \leadsto       | l = y \paren {1 + \paren {y'}^2}       | r = \frac 1 {2 k^2 g} = c       | c =  }} {{end-eqn}} where $c$ is another (more convenient) constant. This is the differential equation which defines the brachistochrone. Now we solve it: {{begin-eqn}} {{eqn | l = y \paren {1 + \paren {y'}^2}       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \sqrt {\frac {c - y} y}       | c =  }} {{eqn | ll= \leadsto       | l = \int \rd x       | r = \int \sqrt {\frac y {c - y} } \rd y       | c =  }} {{end-eqn}} Now we introduce a change of variable: :$\sqrt {\dfrac y {c - y} } = \tan \phi$ Thus: {{begin-eqn}} {{eqn | l = \sqrt {\frac y {c - y} }       | r = \tan \phi       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \paren {c - y} \tan^2 \phi       | c =  }} {{eqn | ll= \leadsto       | l = c       | r = y \paren {1 + \tan^2 \phi}       | c =  }} {{eqn | r = y \sec^2 \phi       | c =  }} {{eqn | r = \sin^2 \phi       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d \phi}       | r = 2 c \sin \phi \cos \phi       | c =  }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x}       | r = \sqrt {\frac {c - y} y}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac 1 {\tan \phi}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d \phi}       | r = \frac {\d y} {\d x} \frac {\d x} {\d \phi}       | c =  }} {{eqn | r = 2 c \sin \phi \cos \phi       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d \phi}       | r = \paren {\frac {\d y} {\d \phi} } / \paren {\frac {\d y} {\d x} }       | c =  }} {{eqn | r = 2 c \sin \phi \cos \phi \tan \phi       | c =  }} {{eqn | r = 2 c \sin^2 \phi       | c =  }} {{eqn | r = c \paren {1 - \cos 2 \phi}       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \phi}       | r = c \paren {1 - \cos 2 \phi}       | c =  }} {{eqn | ll= \leadsto       | l = \int \rd x       | r = \int c \paren {1 - \cos 2 \phi} \rd \phi       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \frac c 2 \paren {2 \phi - \sin 2 \phi} + c_1       | c =  }} {{end-eqn}} 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 $y$: {{begin-eqn}} {{eqn | l = y       | r = c \sin^2 \phi       | c =  }} {{eqn | r = \frac c 2 \paren {1 - \cos 2 \phi}       | c =  }} {{end-eqn}} To simplify the constants, we can substitute $a = c / 2$ and $\theta = 2 \phi$, and thus we get: {{begin-eqn}} {{eqn | l = x       | r = a \paren {\theta - \sin \theta}       | c =  }} {{eqn | l = y       | r = a \paren {1 - \cos \theta}       | c =  }} {{end-eqn}} which are the parametric equations of the cycloid. {{qed}}