# Solving the Euler-Lagrange equation for the brachistochrone problem with friction

This Wolfram Alpha Page contains a derivation of the parametric form of the brachistochrone curve that result from either assuming friction or its absence.

I am asking for help understanding how the solution to the differential equation obtained from applying the Euler-Lagrange equation to the integrand of the the integral representing the total time of descent is obtained. This differential equation can be found on step (30) of the page. I am asking for help in understanding the next step, how setting $$\frac{dy}{dx} = \cot(1/2\cdot\theta)$$ allows for the equation to be solved, obtaining the parametric equations for $$x$$ and $$y$$, shown in steps (32) and (33).

## 2 Answers

You want to know how $$x,\,y$$ are obtained in terms of $$\theta$$ from $$\frac{1+y^{\prime 2}}{(1+\mu y^\prime)^2}=\frac{C}{y-\mu x},\,y^\prime=\cot\frac{\theta}{2}$$. The former equation reduces to $$y-\mu x=C\frac{(1+\mu y^\prime)^2}{1+y^{\prime 2}}=C\left(\sin\frac{\theta}{2}+\mu\cos\frac{\theta}{2}\right)^2,$$because $$1+y^{\prime 2}=\csc^2\frac{\theta}{2}$$. Because of the forms of rotation matrices and compound-angle formulae, the effect of $$\mu$$ in this equation is clearly that of a rotation, thereby mixing $$x$$ and $$y$$ and adding a constant to $$\theta$$. So the frictionless result $$x=\frac{k^2}{2}[\theta-\sin\theta],\,y=\frac{k^2}{2}[1-\cos\theta]$$must generalise to$$x=\frac{k^2}{2}[\theta-\sin\theta+\mu (1-\cos\theta)],\,y=\frac{k^2}{2}[1-\cos\theta+\mu (\theta-\sin\theta)].$$

• Hi, thanks for the quick reply. I did not quite follow the argument, however. Would you be able to provide a derivation that a high school student could follow, in terms just of integration? Is there a way to 'extract' y (and/or) x directly from the substitution (and equation) given? – Akaash Jan 19 at 8:27

The friction effect should be included as a dissipation force $$-\mu y'$$

$$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'} = -\mu y'$$

then from here

$$f-y'\frac{\partial f}{\partial y'}=-\mu y + C_0$$

or

$$\frac{1}{\sqrt{2gy}\sqrt{1+y'^2}}=-\mu y + C_0$$

or

$$y'=\pm \sqrt{\frac{1}{\sqrt{2gy}(\mu y - C_0)^2}-1}$$

etc.