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 $P_2 = (x_2, y_2)$?

  • $\begingroup$ Note the sqrt in your first line is not covering the denominator. In your second line equation y=(k^2)cosP, shouldn't the cos be squared? These small typos are making it hard for me to follow your process, because I feel I have to double check every line. $\endgroup$ Mar 29, 2017 at 17:20
  • $\begingroup$ Thanks @electronpusher: I've fixed these typos in my answer now! $\endgroup$
    – Tom
    Mar 29, 2017 at 17:22
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    $\begingroup$ Using your initial point in equation y(theta)=0 you can obtain theta=pi, and substituting into equation x(theta)=0 gives A=pi. $\endgroup$ Mar 29, 2017 at 17:33
  • $\begingroup$ $\theta_1 = \pi$ and $A=-\pi$ I think... I know I can then find $\theta_2$ by solving $y_2/x_2 = (1+\cos\theta_2)/[2(\sin\theta_2 + \theta_2 - \pi)]$ and then put $\theta_2$ back in to the expression for $y$. Is there an easier way? $\endgroup$
    – Tom
    Mar 29, 2017 at 21:57
  • $\begingroup$ Ah yes, $A=-\pi$. It's not clear to me what you're trying to do; you've found the arbitrary constant. You're calling $(x_2,y_2)$ the endpoint, but you haven't specified what endpoint means. Do you want the cycloid to be traced by a single rotation of a circle? $\endgroup$ Mar 29, 2017 at 22:44

1 Answer 1


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:


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.

  • $\begingroup$ Thank you, this answers the main part of my question. My interest is in finding the equation and least time of travel for the bead between $(0,0)$ and a fixed point $(x_2, y_2)$, so I don't know $R$ (or $k$): as I understand it, I need to solve the equation in my comment above for the parameters final point $\theta_2$ and then substitute it back in to get $k$. $\endgroup$
    – Tom
    Mar 30, 2017 at 7:32
  • $\begingroup$ Updated. This may be the end of the line for me, good luck with your project. $\endgroup$ Mar 30, 2017 at 16:14
  • $\begingroup$ Thanks very much – I'm on the right track now. $\endgroup$
    – Tom
    Mar 30, 2017 at 19:22

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