# Differential equation of cycloid

derivativesordinary differential equationsphysics

I have a problem where I study the brachistochrone curve which is described by the following differential equation:

$$(1+(y')^2)y = k^2 ,$$ where $$k$$ is the appropriate constant

I have to explain why this is reasonable and where it comes from.

So I started from a cycloid. The coordinates that describe it are:

$$x = R(\theta-\sin\theta)$$

$$y = R(1-\cos\theta)$$

where $$\theta = \omega t$$, $$\omega$$ is the angular velocity of a point.

From that I derive the following differential equation

$$(1+(y')^2)y^2 = k^2$$

but I got an extra $$^2$$ on $$y$$. Is that ok? The only I can think of to get rid of it is to say that the brachistochrone curve is only a segment of the cycloid so it is ok to have different differential equations and we can get rid of $$^2$$.

Thanks.

You have $$\frac{dy}{dx}=\frac{\sin\theta}{1-\cos\theta}=\frac{2\sin (\frac12\theta)\cos (\frac12\theta)}{2\sin^2 (\frac12\theta)}=\cot(\frac12\theta)$$
Therefore, $$1+(y')^2=1+\cot^2(\frac12\theta)=\csc^2(\frac12\theta)$$ Meanwhile, $$y=R\cdot2\sin^2(\frac12\theta)$$
Hence the result $$(1+(y')^2)y=2R=k^2$$