Thursday, June 19, 2014

A Brachistochrone Problem

A dog running on a polished floor spies a bone out of the corner of its eye and begins scrabbling like mad to reach it.

Given the dog's initial speed and initial distance from the bone, and taking the dog's acceleration vector to be constant in magnitude, find the shortest time in which the dog may reach the bone.

Mathematically, the problem I'm posing is this: Given parameters \(d>0\), \(v>0\), and \(a>0\) (defined below), find the smallest time \(T>0\) so that there exists a piecewise-continuous direction function \(\theta(t)\) governing the motion \((x(t), y(t))\) as follows:

\[(x(0),y(0))=(0,d)\] \[(x(T),y(T))=(0,0)\]

There is no constraint on the final velocity.

I haven't solved the problem. I conjecture that the shortest time is

\[T=\sqrt{2}\cdot \frac{d}{v}\cdot\sigma\cdot\sqrt{1+\sqrt{1+\frac{1}{\sigma^2}}}\]
where \(\sigma\) is a "slipperiness factor" defined by \(\sigma = v^2/ad\). This value of \(T\) is achieved with a constant acceleration vector (i.e., \(\theta(t)\) constant). Here is a movie showing this motion:

This is better than a guess in that I've looked at some particular perturbations of the constant-acceleration trajectory and have also computed some specific non-nearby solutions as well. It'd be cool to see a faster, less obvious solution if there is one!

This problem is related to one solved a few years ago by Williams College faculty and students, concerning the optimal path for baserunning. More generally, both problems are about how best to navigate in space subject to a condition on the acceleration vector.

Be sure to let me know if you can improve upon my answer! And in the meantime, here's a funny video of dogs sliding on wood floors. (You can see the physics better if you mute the audio.)

No comments: