# Whether a square can be traversed in finite time

You are at point $$C$$ inside square $$ABCD$$ in the Cartesian plane, in which $$A=(0,0), B=(0,1), C=(1,1), D=(1,0)$$. You want to get to vertex $$A$$. However, your “speed” in $$\frac{\text{units}}{\text{sec}}$$ is everywhere equal to your y-coordinate. Can you get from $$C$$ to $$A$$ in finite time? If you can, what is the minimal time required for the journey?

A friend asked me this as a challenge recently out of what I think was a calculus textbook. I haven’t found any concrete way of resolving the question one way or another (or even modeling it properly), but most of my intuition says that the voyage should not be possible in finite time. Specifically, it seems to me that this problem is somehow related to the fact that the harmonic series, as well as the integral of the harmonic series, diverges. On the other hand, perhaps this problem is like Zeno’s paradoxes- an infinite number of decreasing steps adding up to something finite.

On solving the problem itself, I know that one can simplify whether it can be done in finite time to whether going straight down from $$B$$ to $$A$$ can be done in finite time. On minimizing the time taken (if it exists), I have no idea how to determine how to test infinite functions from $$(1,1)$$ to $$(0,0)$$ for their “speed”s, although I conjecture ones that are nowhere concave up should always be faster.

$$y=\sqrt{x}$$

• The minimum time trajectory (if it exists) could be found by variational calculus.
– user65203
Commented Apr 4, 2020 at 16:39
• If this is indeed from a textbook, it seems odd that it asks the follow-up question about minimum time if it's not possible at all. Did your friend maybe modify the question? Commented Apr 6, 2020 at 13:35
• He might have. It was definitely (in some form) from some textbook, at least. Commented Apr 6, 2020 at 19:56

Another answer: To get from height $$y$$ to height $$\frac y2$$, you need to travel a distance of at least $$\frac y2$$ at speeds less than $$y$$. This takes at least half a second. So no matter how close you are to 0, you still have more than half a second left.

• I think this can be rewritten as a divergent riemann sum... hmm Commented Apr 4, 2020 at 19:34
• Yeah, if you divide $[0,1]$ into intervals with boundaries $\frac1n$ and bound the time required to cross each interval, you get the divergent harmonic series $\frac12+\frac13+\frac14+\dots$. If you instead use only the boundaries $\frac1{2^n}$, you get the more obviously divergent series $\frac12+\frac12+\frac12+\dots$, which corresponds to my answer.
– Karl
Commented Apr 4, 2020 at 19:39
• This answer provides a good argument, but it could be improved by explicitly answering the question: no, it's not possible to get from C to A in finite time. Commented Apr 5, 2020 at 23:55
• @Dukeling: I think that's pretty obvious! Commented Apr 6, 2020 at 14:37
• So much resembles the Achilles and the tortoise paradox :) Commented Apr 6, 2020 at 18:06

You’ve already found that the existence question can be reduced to going straight from $$B$$ to $$A$$. On that leg, the $$y$$ coordinate $$y(t)$$ follows the ordinary first-order linear differentional equation $$y'=y$$, with solution $$y=c\,\mathrm e^{-t}$$. If you start at $$t=0$$, that determines the constant to be $$c=1$$, so your $$y$$ coordinate is $$y=\mathrm e^{-t}$$. This decays to $$0$$ exponentially and doesn’t reach $$0$$ in finite time.

• How does one prove that if finite time is not possible when going through B, then finite time is not possible at all ?
– user65203
Commented Apr 4, 2020 at 16:35
• @YvesDaoust: If you project the path onto the $y$ axis, you are always slower than you would have been at the corresponding point of the axis, since part of your speed is used for going horizontally; thus you take longer than you would have on the axis. Commented Apr 4, 2020 at 16:40
• Ok, I get it. Thanks.
– user65203
Commented Apr 4, 2020 at 16:42

Even easier...

When you are $$y$$ units away from the $$x$$-axis, you are one second away at current speed, and this speed can only decrease. No matter how much you approach, you are still always more than one second away.

• I think the pure fact that the speed can only decrease isn't sufficient to show that you can't reach the goal. For example, if your speed dropped from $y$ to $y/2$ over the course of the journey, you'd still hit the goal. Commented Apr 6, 2020 at 18:36
• @templatetypedef: Not what I said... At any moment, at current speed, you still have one more second to go. In other words, you never cut down the time left. Commented Apr 6, 2020 at 22:47

You have (ignoring horizontal movement) $$\tag1y'\ge -y.$$ Let $$f(t)=y(t)e^t$$. Then $$f'(t)= (\underbrace{y(t)+y'(t)}_{\ge 0})\,e^t\ge 0$$ so that $$f(t)\ge f(0)=y(0)\qquad \text{for all }t\ge 0.$$ In particular $$y(t)\ge y(0)e^{-t}>0$$ for all $$t>0$$.

This is the standard "half-plane" model of the hyperbolic plane.

Your “speed” in units/sec is everywhere equal to your y-coordinate.


This is equivalent to having constant unit speed in the standard half-plane model of the hyperbolic plane. "Half-plane" refers to the upper half plane, i.e. where $$y>0$$.

To get between any two points, the fastest path is along the circle centered on the x-axis that passes through the two points.

Unfortunately for anyone who wants to get from C to A in finite time, the x-axis itself is not part of the plane and cannot be reached.

From the point of view of the plane, the x-axis consists of the "points at infinity". Each of these points is a direction you can go in, like what "the +x direction" is for the standard Euclidean plane. You can walk in that direction as long as you want, but you will never arrive there.