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.

$y=\sqrt{x}$

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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 $t>0$.

Solution 5:

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.