[Math] Parametric solution of the Brachistochrone problem

calculusordinary differential equationsparametric

In my attempt to find the parametric solution to the Brachistochrone problem
I write the differential equation resulting from the calculus of variations treatment as
$$y(1+y'^2) = k^2 \Rightarrow y' = \sqrt{\frac{k^2 – y}{y}}$$
Then attempt a parameterization of the form
$$\tan\phi = \frac{\mathrm{d}y}{\mathrm{d}x} = y' = \frac{\sqrt{k^2-y}}{\sqrt{y}}$$
and so
$$\cos\phi = \frac{\sqrt{y}}{k} \Rightarrow y = k^2\cos^2\phi \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}\phi} = -2k^2\cos\phi\sin\phi,$$
and
$$\frac{\mathrm{d}x}{\mathrm{d}\phi} = \frac{\mathrm{d}x}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}\phi} = \cot\phi\cdot(-2k^2\cos\phi\sin\phi) = -k^2(\cos 2\phi + 1)$$
which I integrate to
$$
x = -k^2(\frac{1}{2}\sin 2\phi + \phi) + A.
$$
Now I'm stuck because this doesn't look like the equation of a cycloid that I recognise, and I don't know what range of values $\phi$ should take.
The substitution $\theta = -2\phi$ almost works:
$$
x = \frac{k^2}{2}(\sin\theta + \theta + A), y = \frac{k^2}{2}(1+\cos\theta),
$$
but how do I find $A$ given the start and end points $P_1 = (0,0)$ and $P_2 = (x_2, y_2)$?

Best Answer

Note that substituting $t=\theta-\pi$ (or originally, $t=-(2\phi+\pi)$) will yield the traditional parametric form of a cycloid:

$$x=R(t-\sin t), y=R(1-\cos t)$$

where $R=\frac{k^2}{2}$ is the radius of the associated tracing circle.

Update: To find the angle value "$t_2$" yielding some point $(x_2,y_2)$, one might rearrange the y-component:

$$t_2=\arccos\Big(1-\frac{y_2}{R}\Big)$$

When the parameter $t_2$ is determined, the $x_2$ coordinate can be evaluated.

Note that it is possible to eliminate the $t$ parameter entirely and write x in terms of y. Interestingly, it is not possible in closed form to write $y$ in terms of $x$.

Update 2: A way to find $t$ that is especially effective for physical scenarios is related to your original approach:

$$\frac{y}{x} = \frac{1-\cos t}{t-\sin t}$$

where $\frac{y}{x}=\frac{y-0}{x-0}$ can be interpreted as the slope between points $(0,0)$ and $(x,y)$. In a physical situation, one may easily measure the slope from the origin to any arbitrary point $(x_i,y_i)$ and then determine the parameter value $t_i$ at that point, via the above equality. Once you have the physical dimensions and parameter value for point $i$, you can determine $R$ (from the cycloid equations). An experimentalist might measure several points $\{(x_i,y_i)\}$ and determine an experimental value of the radius from each, $R_i$, then average them to find a more accurate value for $R$ (other statistics may be performed as well, to quantify how closely the physical model matches the mathematical model of the cycloid).

As for finding the least time of travel, a few key steps that may interest you are presented in this video. Note that for this purpose it will probably make more sense to have a variable initial point (rather than the original) and a fixed endpoint (at the bottom of the inverted cycloid). See if you can determine your parameter value at the extreme of the cycloid.