[Math] Brachistochrone – Solution of a Cycloid – Parametric Equations

calculus-of-variationscycloidparametric

I am trying to understand the math behind the Brachistochrone.

I could understand all the technical intricacies of the mathematical treatment of the topic found at Wolfram-Mathworld|Brachistochrone Problem.

At the last part, they say,
$$
\boxed{\left[1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\right]y = \frac{1}{2gC^2} = k^2} \tag*{(1)}
$$
and then they just say, This equation is solved by the parametric equations and write two equations,
$$
x = \frac{1}{2}k^2(\theta-\sin \theta)\\ \tag*{(2)}
y=\frac{1}{2}k^2(1-\cos \theta)
$$
How did this come?

I plotted them on Desmos and I could clearly see a cycloid (Can be see here).

Note:

  1. I have not shown my steps here because, I am just studying the problem straight from Wolfram. If you wish to have a look at the steps, I have mentioned the link above.

  2. Learning Calculus of Variations on my own. I will be grateful if you could direct me towards some good resources.

  3. I know how to derive the parametric equation of a cycloid, I learnt it from Math.Stackexchange|How to find the parametric equation of a cycloid?. I just don't know how to solve $(1)$ using the two equations in $(2)$.

Thank you.

Best Answer

Note that we can use the trigonometric identities, $\sin(\theta)=2\sin(\theta/2)\cos(\theta/2)$ and $\frac12(1-\cos(\theta))=\sin^2(\theta/2)$, to write $x(\theta)$ and $y(\theta)$ as

$$\begin{align} &x(\theta)=k^2(\theta/2 -\sin(\theta/2)\cos(\theta/2))\tag1\\\\ &y(\theta)=k^2 \sin^2(\theta/2)\tag2 \end{align}$$

Then, it is easy to see that we have

$$\frac{dy}{dx}=\cot(\theta/2) \tag3$$

Finally, using $(2)$ and $(3)$ it is straightforward to show that

$$\left(1+\left(\frac{dy}{dx}\right)^2\right)y=k^2$$

as was to be shown!

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