I'm trying to solve the Brachistochrone problem and the first step is to get the functional which we are trying to minimise. In this case, it will be the time taken. Using vectors, I am able to get the acceleration at a point in the curve to be
$$
a =\frac{d^2s}{dt^2} = \frac{g}{\sqrt{1+(y')^2}}
$$
Also,
$$
ds = dx\sqrt{1+(y')^2}
$$
However, I am not able to relate these two and form the functional which is,
$$
t = \frac{1}{\sqrt{2g}}\int\sqrt{\frac{1+(y')^2}{y}}
$$
The derivations which I have found uses energy conservation to get the functional. But, I cannot find how to get the functional by just using the acceleration at a point on the curve.
Best Answer
The acceleration is given by $$\frac{d^2s}{dt^2}=\frac{g\color{red}{y'}}{\sqrt{1+y'^2}}$$ and using $ds/dx=\sqrt{1+y'^2}$ we obtain $$\frac{d^2s}{dt^2}=g\frac{dx}{ds}\frac{dy}{dx}=g\frac{dy}{ds}$$ and multiplying both sides by $ds/dt$ it follows that $$\frac{ds}{dt}\frac d{dt}\left(\frac{ds}{dt}\right)=g\frac{dy}{dt}\implies \int\frac{ds}{dt}\,d\left(\frac{ds}{dt}\right)=\int g\frac{dy}{dt}\,dt$$ so that $$\frac12\left(\frac{ds}{dt}\right)^2=gy+c\implies\frac{ds}{dt}=\sqrt{2gy+2c}.$$ Since $(ds/dt)\mid_{y=0}=0$ we have $c=0$, so $ds/dt=\sqrt{2gy}$. Finally, the time is given by $$t=\int dt=\int\frac{ds}{ds/dt}=\frac1{\sqrt{2g}}\int\sqrt{\frac{1+y'^2}y}\,dx$$ as desired.