In a 3 Blue 1 Brown video (https://www.youtube.com/watch?v=Cld0p3a43fU), 3b1b gave the following challenge:
Give an intuitive reason why time-minimizing curves going from one point to another under the effect of gravity look like straight lines in the t-$\theta$ space, which maps each time to the change in angle of the curve.
Which is equivalent to saying that the if the Brachistochrone curve is modelled as $f(x)$, then the function that measures the angle of the tangent line of $f$ at $x$, which we will call $g(x)=\tan^{-1}f'(x)$, then the function $g'(x)$ looks like a line, that is, $g''(x)$ is a constant. (Intuitively, $g(x)=0$.)
Did anyone make any discoveries concerning this? A scroll through the comments didn't reveal anything either.
We can view the sliding motion as a non-linear pendulum of variable length and variable center of oscillation on an instantaneous pivot I..
$$ L \ddot\theta + g \sin \theta=0$$
Angle $\theta $ is made to vertical line and $\phi= \theta-\pi/2 $ is to made to horizontal by path tangent.
Qualitatively at least
$$-y \ddot \phi+ g \cos \phi=0$$
Acceleration or the path curvature $1/R_1$ must be proportional to $\cos \phi/ y= 1/R_2$ which is in fact a geometrical property of the cycloid:
$$\frac{R_2}{R_1}=\frac{1}{2}= \frac{PC}{PI}$$
The locus of I is another cycloid. The answer could be improved after any specific aspects in the question from the video are addressed.