I thought I understood why the cycloid is both a brachistochrone (path of shortest time) and tautochrone (path of equal time for different starting points and the same end point) for the problem of a bead sliding, without friction, along a wire from $(x_1, y_1)$ to $(x_2, y_2)$, given $y_2 < y_1$ (with gravity pointing down along the negative $y$-axis).

But then someone I know asked about the case of $x_1=x_2$, where the cycloid becomes a vertical line. This must be a brachistochrone, but clearly isn't a tautochrone because objects take longer to fall to the same point from different heights. So where does the brachistochrone = tautochrone relation break down in this case?

What if $x_2 = x_1 + \epsilon$ for $\epsilon \ll y_2-y_1$? The brachistochrone would still be a cycloid, but that's practically straight down: would it still be a tautochrone?

  • $\begingroup$ I think it's a tautochrone when the end point is the minimum of the cycloid. I haven't checked it in any way, but it seems reasonable to me. $\endgroup$
    – Arthur
    Commented Jul 16, 2018 at 8:53

1 Answer 1


A brachistochrone always includes the cusp of the cycloid (not surprisingly, since the tangent becomes vertical there and this is the fastest way to accelerate initially), whereas the tautochrone always includes the minimum point (it is not isochronous to any other point, as can be seen by examining the integral for the descent time given on MathWorld with a more general angle than $\pi$).

Therefore only the totality of the first descending arc is both a brachistochrone between its endpoints and a tautochrone to its bottom endpoint. Such a cycloid exists joining $(x_1,y_1)$ to $(x_2,y_2)$ when $\frac{x_2-x_1}{y_1-y_2} = \frac{\pi}{2}$, as may be seen by examining the parametric equations $\frac{1}{2}k^2(\theta-\sin{\theta},\cos{\theta}-1)$. Any brachistochrone that contains a minimum is a tautochrone to its minimum (from either side, of course). On the other hand, any tautochrone that contains a cusp is a brachistochrone from that cusp to any other point on it.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .