The Brachistochrone Problem I would like to acquire a good enough understanding of a proof of the Brachistochrone Problem in order that I may explain it as simply as possible to others. Can anyone recommend their favorite proof that would be discernable to someone with a mathematics background say, up to differential equations, and some coursework in general physics and mechanics?  
 A: Brachistochrone Problem: Find the curve connecting two given points $P$ and $Q$ traversed by a particle sliding from $O$ to $Q$ in the shortest time where friction and resistance of the medium are ignore.
Solution: 

Let the particle of mass $~m~$ starting from the point $O$ with zero velocity along the curve $OPQ$.
At time $~t~$, let the position of the particle be  $~P(x,y)~$, then by the principle of work and energy we have
$$\text{Kinetic Energy  at}\quad P\quad-\text{Kinetic Energy at}\quad O\quad=\quad \text{work done in moving the particle from}\quad O\quad \text{ to}\quad P$$
or, $$\frac{1}{2}m(\frac{ds}{dt})^2-0=mgy\implies \frac{ds}{dt}=\sqrt{2~g~y}$$
Therefore the time taking to reach the point $Q$ from $O$  is given by $$T=\int_0^T dt=\int_0^{x_1}\frac{ds}{\sqrt{2~g~y}}=\frac{1}{\sqrt{2~g}}\int_0^{x_1}\sqrt{\frac{1+y'^2}{y}}dx$$
Now $~T~$ will be minimum requires that Euler-Lagrange Equation $$\frac{\partial f}{\partial y}-\frac{d}{dx}(\frac{\partial f}{\partial y'})=0\qquad \text{with}\quad f(x,y,y')=\sqrt{\frac{1+y'^2}{y}}\quad \text{must be satisfied.}$$
Here $~f~$ is independent of $~x~$.
Then the Euler-Lagrange Equation reduces to $$f-y'\frac{\partial f}{\partial y'}=\text{constant} =c_1, (\text{say})$$
$$\sqrt{\frac{1+y'^2}{y}}-{\frac{y'^2}{\sqrt{y(1+y'^2)}}}=c_1$$
$$\implies\frac{1}{\sqrt{y(1+y'^2)}}=c_1$$
$$\implies y(1+y'^2)=\frac{1}{c_1^2}=a(\text{say})$$
$$\implies y'=\frac{dy}{dx}=\sqrt{\frac{a-y}{y}}\qquad .......(1)$$
Putting $~y=a~\sin^2 \theta~$ and integrating $(1)$ from $O$ to $P$, we have 
$$x=\int_0^x dx=\int_0^y \sqrt{\frac{y}{a-y}}dy=\int_0^{\theta} a~\sin^2{\theta}~d\theta=\frac{a}{2}(2~\theta-\sin 2\theta)=b(\phi-\sin \phi)$$where $~b=\frac{a}{2}~,~\phi=2~\theta$.
Substituting $~b=\frac{a}{2}~,~\phi=2~\theta$ in $~y=a~\sin^2 \theta~$ we have $$y=b~(1-\cos \phi)$$
Hence the required curve is $$x=b(\phi-\sin \phi)$$ $$y=b~(1-\cos \phi)$$which is a cycloid.
