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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).

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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$$x=\frac{k^2}{2}[\theta-\sin\theta+\mu (1-\cos\theta)],\,y=\frac{k^2}{2}[1-\cos\theta+\mu (\theta-\sin\theta)].$$

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  • $\begingroup$ 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? $\endgroup$ – Akaash Jan 19 at 8:27
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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.

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