# 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?

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.