Fun Puzzle: The Ant Racing an Elastic Rope

趣题：与橡皮绳赛跑的蚂蚁 苏剑林 · 2014-04-09 · Questions fun

This is a widely circulated puzzle that many readers have probably already heard of. However, "widely known" doesn't necessarily mean "widely solved." So let me state the problem here, write down my own answer, and share my thoughts on that answer. The problem goes like this:

The Ant Racing an Elastic Rope
An ant crawls at a constant speed of 1 centimeter per second from one end of an elastic rope of length $l=100$ meters toward the other end. Every second, the rope stretches by 100 meters—for instance, after 10 seconds the rope will have lengthened by 1000 meters. Assume the rope can be stretched indefinitely and that the stretching is uniform. The ant, too, keeps crawling forward tirelessly. Since the stretching is uniform, the ant's position naturally gets carried forward at a proportional rate as well. The question is: can the ant ever reach the other end of the rope?

Solution

If you already have some basic calculus background, I don't think this problem is too hard. Can it be solved within the scope of high school mathematics? I don't think so, because although the process below can be discretized to get an approximate answer, estimating the time still requires computing the approximate value of the series

$$S(n)=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$$

which again requires calculus.

Suppose that at time $t$ the ant is located at position $x$. Its velocity at this moment is $\dot{x}$, which is its own crawling speed ($v_0 =1 cm/s$) plus the speed at which the rope is stretching at that point. The total stretching speed of the rope is $v_1=100 m/s$; at this moment the rope has length $l+v_1 t$, and the stretching speed at position $x$ is $v_2=\frac{v_1 x}{l+v_1 t}$ (allocated proportionally). So we have

$$\dot{x}=\frac{v_1 x}{l+v_1 t}+v_0$$

Let $v_1 x=y,l+v_1 t=\tau$, and the equation above becomes

$$\frac{dy}{d\tau}=\frac{y}{\tau}+v_0$$

Letting $\frac{y}{\tau}=u,\ln \tau =v$ and substituting, we get

$$\frac{du}{dv}=v_0$$

Therefore

$$u=v_0 v-v_0 \ln l;\frac{y}{\tau}=v_0 \ln \tau-v_0 \ln l$$

which finally gives

$$\ln\left(1+\frac{v_1 t}{l}\right)=\frac{v_1 x}{v_0 (l+v_1 t)}$$

For the ant to cross the entire rope, we need at least $x=l+v_1 t$, so

$$\ln\left(1+\frac{v_1 t}{l}\right)=\frac{v_1}{v_0}$$

For the specific numbers in this problem, that is

$$\ln\left(1+t\right)=10000$$

Solving this gives $t\approx 10^{4343} s$, which is $10^{4335}$ years!

My Take on the Answer

The answer shows that, in a finite amount of time, the ant can indeed cross to the other side. However, the time required is $10^{4335}$ years! This not only far exceeds the ant's lifespan, but also exceeds the current known age of the universe! From this perspective—or rather, from a physical point of view—this is simply not achievable in practice. That is to say, the answer a physicist would give should be that it's impossible to cross!

Physics is a discipline that combines experiment and theory, and experiments inevitably come with errors. Within the precision of an experiment, if the error is below some threshold it's considered equal, and if it's above some threshold it's considered infinite, and so on. For example, given a square with side length 1.0, its diagonal length would be measured as 1.4, not $\sqrt{2}$. Similarly, for the answer to this problem, $10^{4335}$ years, from the point of view of (current) physicists, is effectively an infinite amount of time—so, practically speaking, it is unachievable.