# Problems approaching the Brachistochrone problem

I am having trouble understanding some of what the author mentions in this question regarding the Brachistochrone problem:

If a point-like mass is rolling down a hill from a point $$A$$ to a point $$B$$, the time it takes the mass to get from $$A$$ to $$B$$ depends on the profile of the hill. The Brachistochrone Curve is the profile that minimizes the time and we can find this profile using a Lagrangian extremization procedure. You can easily convince yourself that one can always find profiles which would make the time taken arbitrarily long, so if one finds a finite curve that extremizes the time taken, that curve should necessarily be a minimum.

Before jumping to the core of this problem, let us first look at the ‘cycloid’ curve, which can be thought of as the curve a fixed point on a circle of radius $$R$$ draws as the circle ‘rolls’ along a straight line (see left of Fig. 4). With $$\theta$$ as the angle parameterizing the circle’s rotation, the coordinates on the curve are $$x=R(\theta-\sin\theta)\quad\text{and}\quad y=R(1-\cos\theta)$$ The derivatives along that curve are $$x'(\theta) = R(1-\cos\theta)$$ and $$y'(\theta)=R \sin\theta$$ and bearing in mind that $$\cos\theta=1-y/R$$, ie $$x'(\theta)=y$$, we have $$\left(x'(y)\right)^{-1}=y'(x)=\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y'(\theta)}{x'(\theta)}=\sqrt\frac{2R-y}{y}\tag{1}$$

Firstly, I don't understand how the leftmost side of $$(1)$$ is true, is this a typo, or is $$x'\equiv\frac{\mathrm{d}x}{\mathrm{d}\theta}$$ really a function of $$y$$, ($$x'=x'(y)$$)? Now looking at the central equality in $$(1)$$, then, by the chain rule $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}x}=\frac{R\sin\theta}{y}\ne \sqrt\frac{2R-y}{y}$$

But even taking the given equality in $$(1)$$, $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y'(\theta)}{x'(\theta)}=\frac{R\sin\theta}{y}\ne \sqrt\frac{2R-y}{y}$$ This was done by simple insertion of the equations given in the question, but either way, I still cannot seem to arrive at the correct answer of $$\sqrt\frac{2R-y}{y}$$.

I am almost certain that the chain-rule must be being used somewhere, and at a wild guess perhaps there is some trick to it whereby in $$(1)$$ $$\frac{y'(\theta)}{x'(\theta)}{\stackrel{?}{=}}\frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}}$$

I don't doubt that I am missing something very simple, so any hints or tips that lead me to the answer would be greatly appreciated.

$$\require{begingroup}\begingroup\renewcommand{\dd}[1]{\,\mathrm{d}#1}$$Regarding the leftmost side of $$(1)$$, it's just the notation and there's no typo.

Here the author is not adopting the conventions that $$x'\equiv\dd{x}/\dd{\theta}$$ (implicitly to distinguish from, for example, $$\dot{x} = \dd{x}/\dd{t}$$) but the other convention. The derivative is denoted with a prime for anything, and it is written explicitly as a function with an argument, as in $$x'(u)\equiv\dd{x}/\dd{u}$$.

Namely, merely by definition of notation, $$x'(y)\equiv \frac{\dd{x}}{\dd{y}}~,\quad x'(\theta)\equiv \frac{\dd{x}}{\dd{\theta}}~,\quad y'(x)\equiv \frac{\dd{y}}{\dd{x}}~,\quad x'(\theta)\equiv \frac{\dd{x}}{\dd{\theta}}$$ The fact that $$\bigl(x'(y)\bigr)^{-1}=y'(x)$$ is the basic formula for the derivative of the inverse function.

As for the rightmost of $$(1)$$, note that

$$y=R(1-\cos\theta) \implies \cos\theta=1-\frac{y}R \implies \sin\theta=\sqrt{\frac{2y}R-\frac{y^2}{R^2}}$$ Plug it back in we get $$\frac{\dd{y}}{\dd{x}}=\frac{R\sin\theta}{y}= \frac{\sqrt{2yR-y^2} }y = \sqrt{\frac{2Ry-y^2}{y^2}}= \sqrt{\frac{2R-y}y}$$ as desired.$$\endgroup$$

• Thanks for your answer, I am now stuck on another part of the question and have asked a follow-up question here Dec 13, 2021 at 22:41