Let $A$ be the origin and $B$ some lower point not directly below $A$. Let $\left(x\,y\right)$ be the horizontal and vertical distances of the bead from $A$, so that $y\>0$ as the bead falls ($g\>0$ as well). The problem stated is to minimize the time $t$ that it takes a bead to go from $A$ to $B$ under gravity alone. The travel time can be expressed as the integral of the inverse of the speed $v$ with respect to arclength $s$ along the bead's path:

$t\={\int }_A^B \frac{\mathrm{ds}}{v\left(s\right)}\,$

To express the speed in terms of arclength, notice that the potential energy of the bead is $V\left(y\right) \= -\mathrm{mgy}$ where the convention $V\left(0\right)\=0$ is used,  and the kinetic energy is $T\=\frac{1}{2}{\mathrm{mv}}^2$ the bead starts from rest, $E\left(0\right) \= 0$ as well. Using conservation of energy $\left(E\= T \+ V\right)\,$you get:

$$$0\=$$\frac{1}{2}{\mathrm{mv}}^2$ -$\mathrm{mgy}\.$

Rearranging terms gives $v\left(y\right) \= \sqrt{2 g y}\.$ Changing variables using:

$\frac{{\mathrm{ds}}^2}{{\mathrm{dx}}^2} \= 1 \+ \frac{{\mathrm{dy}}^2}{{\mathrm{dx}}^2}$,

where $\mathrm{ds}$ is the increment of length on the curve, the integral for minimization becomes:

$t\=\frac{1}{\sqrt{2g}}{\int }_0^{x_B} \sqrt{\frac{1\+y{\'}^2}{y}} \mathrm{dx}\,$

where $y\' \= \frac{\mathrm{dy}}{\mathrm{dx}}\.$ It remains to find the function $y\left(x\right)$ which minimizes the time.

This problem can be approached using the calculus of variations. In particular, the function $F \= \sqrt{\frac{1\+y{\'}^2}{ y}}$is varied until an extreme value for the integral is found. In most cases the Euler-Lagrange equation would be applied next:

$\frac{\partial F}{\partial y}-\frac{\ⅆ}{\ⅆx}\left(\frac{\partial F}{\partial y\'}\right)\=0.$

However, $F$ does not explicitly depend on $x$, and thus the simpler Beltrami identity is used instead:

$F-y\'\frac{\partial F}{\partial y\'}\=C\,$

where $C$ is an integration constant. Applying $F\,$ you get:

$\sqrt{\frac{1\+y{\'}^2}{y}}-\frac{y{\'}^2}{\sqrt{ y \left(1\+y{\'}^2\right)}} \=C\,$

Squaring both sides yields:

$y \left(1\+y{\'}^2\right) \=C^{-2}\.$

This differential equation is separable and can be integrated to give:

$x\=-\sqrt{\left(2a-y\right)y}\+a \left(\frac{\mathrm{\π}}{2}- \arcsin \left(1-\frac{y}{a}\right)\right)$,

where $a\=\frac{1}{2}{C}^2$, using $y\left(0\right)\=0$ and $x\left(0\right)\=0$ as initial conditions. Using the following definition:

$\mathrm{\θ}\=\frac{\mathrm{\π}}{2}-\arcsin \left(1-\frac{y}{a}\right)$,

the solution can be rewritten parametrically as:

$$\begin{gathered} x\left(\mathrm{\θ}\right) \= a \left(\mathrm{\θ}-\sin \mathrm{\θ}\right)\, \\ y\left(\mathrm{\θ}\right) \= a \left(1- \cos \mathrm{\θ}\right)\. \end{gathered}$$

This is a parameterization of a cycloid! To fully solve the problem, $a$ and ${\mathrm{\θ}}_B$ must be determined by satisfying the boundary conditions

$$$$\begin{gathered} x\left({\mathrm{\θ}}_B\right) \=x_B \, \\ y\left({\mathrm{\θ}}_B\right) \=y_B \end{gathered}$$,

which in general must be solved numerically.

The actual time can be then expressed in terms of $\mathrm{\θ}$ as follows:

You can simplify the Euler-Lagrange equation for the case that $F$ does not explicitly depend on$x$, that is, $\frac{\partial F}{\partial x}\=0$. Consider the total derivative of $F$ with respect to $x\:$

$\frac{\ⅆF}{\ⅆx}\=\frac{\partial F}{\partial y}y\'\+\frac{\partial F}{\partial y\'}y\'\'\+\frac{\partial F}{\partial x}\.$

The last term must be zero since $F$ does not explicitly depend on $x$ by assumption. Solving for $\frac{\partial F}{\partial y}y\'$ gives:

$\frac{\partial F}{\partial y}y\' \=\frac{\ⅆF}{\ⅆx}-\frac{\partial F}{\partial y\'}y\'\'\.$

Multiplying the Euler-Lagrange equation by $y\'$ and substituting the previous equation into it gives:

$\frac{\ⅆF}{\ⅆx}-\frac{\partial F}{\partial y\'}y\'\'-y\'\frac{\ⅆ}{\ⅆx}\left(\frac{\partial F}{\partial y\'}\right)\=0\,$

where left hand side can be written as:

$\frac{\ⅆ}{\ⅆx}\left(F-y\'\frac{\partial F}{\partial y\'}\right)\=0.$

Integrating this gives the Beltrami identity:

$F-y\'\frac{\partial F}{\partial y\'}\=C\,$

where $C$ is the integration constant.