Ant on a rubber rope

Although solving the problem appears to require analytical techniques, it can actually be answered by a combinatorial argument by considering a variation in which the rope stretches suddenly and instantaneously each second rather than stretching continuously. Indeed, the problem is sometimes stated in these terms, and the following argument is a generalisation of one set out by Martin Gardner, originally in Scientific American and later reprinted.[1]

Consider a variation in which the rope stretches suddenly and instantaneously before each second, so that the target-point moves from ${\displaystyle x=c}$ to ${\displaystyle x=c+v}$ at time ${\displaystyle t=0}$, and from ${\displaystyle x=c+v}$ to ${\displaystyle x=c+2v}$ at time ${\displaystyle t=1}$, etc. Many versions of the problem have the rope stretch at the end of each second, but by having the rope stretch before each second we have disadvantaged the ant in its goal, so we can be sure that if the ant can reach the target point in this variation then it certainly can in the original problem, or indeed in variants where the rope stretches at the end of each second.

Let ${\displaystyle \theta (t)}$ be the proportion of the distance from the starting-point to the target point which the ant has covered at time t. So ${\displaystyle \theta (0)=0}$. In the first second the ant travels distance ${\displaystyle \alpha }$, which is ${\displaystyle {\frac {\alpha }{c+v}}}$ of the distance from the starting-point to the target-point (which is ${\displaystyle c+v}$ throughout the first second). When the rope stretches suddenly and instantaneously, ${\displaystyle \theta (t)}$ remains unchanged, because the ant moves along with the rubber where it is at that moment. So ${\displaystyle \theta (1)={\frac {\alpha }{c+v}}}$. In the next second the ant travels distance ${\displaystyle \alpha }$ again, which is ${\displaystyle {\frac {\alpha }{c+2v}}}$ of the distance from the starting-point to the target-point (which is ${\displaystyle c+2v}$ throughout that second). So ${\displaystyle \theta (2)={\frac {\alpha }{c+v}}+{\frac {\alpha }{c+2v}}}$. Similarly, for any ${\displaystyle n\in \mathbb {N} }$, ${\displaystyle \theta (n)={\frac {\alpha }{c+v}}+{\frac {\alpha }{c+2v}}+\cdots +{\frac {\alpha }{c+nv}}}$.

Notice that for any ${\displaystyle k\in \mathbb {N} }$, ${\displaystyle {\frac {\alpha }{c+kv}}\geqslant {\frac {\alpha }{kc+kv}}=\left({\frac {\alpha }{c+v}}\right)\left({\frac {1}{k}}\right)}$, so we can write

$${\displaystyle \theta (n)\geqslant \left({\frac {\alpha }{c+v}}\right)\left(1+{\frac {1}{2}}+\cdots +{\frac {1}{n}}\right)}$$

.

The term ${\displaystyle \left(1+{\frac {1}{2}}+\cdots +{\frac {1}{n}}\right)}$ is a partial harmonic series, which diverges, so we can find ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle 1+{\frac {1}{2}}+\cdots +{\frac {1}{N}}\geqslant {\frac {c+v}{\alpha }}}$, which means that ${\displaystyle \theta (N)\geqslant 1}$.

Therefore, given sufficient time, the ant will complete the journey to the target point. This solution could be used to obtain an upper bound for the time required, but does not give an exact answer for the time it will take.

Absolute position x vs time t graph of the animation above: an ant crawls at 1 cm/s (red) relative to and along an elastic rope initially 4 cm long, stretching at 2 cm/s, painted in eights (shaded background). The asymptote (dashed purple) shows the position of the ant if the rope were not stretching. As the ant's position has an exponential component, it eventually catches up with the end of the rope (green) regardless of the ratio of their initial speeds. The boundaries between the background stripes can be visualised as trajectories that the ant would take should it suddenly stop when its reaches the stripe. However, since the ant is crawling, it traverses the stripes until it reaches the final one at the end.

A key observation is that the speed of the ant at a given time ${\displaystyle t>0}$ is its speed relative to the rope, i.e. ${\displaystyle \alpha }$, plus the speed of the rope at the point where the ant is. The target-point moves with speed ${\displaystyle v}$, so at time ${\displaystyle t}$ it is at ${\displaystyle x=c+vt}$. Other points along the rope move with proportional speed, so at time ${\displaystyle t}$ the point on the rope at ${\displaystyle x=X}$ is moving with speed ${\displaystyle {\frac {vX}{c+vt}}}$. So if we write the position of the ant at time ${\displaystyle t}$ as ${\displaystyle y(t)}$, and the speed of the ant at time ${\displaystyle t}$ as ${\displaystyle y'(t)}$, we can write:

$${\displaystyle y'(t)=\alpha +{\frac {v\,y(t)}{c+vt}}}$$

This is a first order linear differential equation, and it can be solved with standard methods. However, to do so requires some moderately advanced calculus. A much simpler approach considers the ant's position as a proportion of the distance from the starting-point to the target-point.[2]

Consider coordinates ${\displaystyle \psi }$ measured along the rope with the starting-point at ${\displaystyle \psi =0}$ and the target-point at ${\displaystyle \psi =1}$. In these coordinates, all points on the rope remain at a fixed position (in terms of ${\displaystyle \psi }$) as the rope stretches. At time ${\displaystyle t\geqslant 0}$, a point at ${\displaystyle x=X}$ is at ${\displaystyle \psi ={\frac {X}{c+vt}}}$, and a speed of ${\displaystyle \alpha }$ relative to the rope in terms of ${\displaystyle x}$, is equivalent to a speed ${\displaystyle {\frac {\alpha }{c+vt}}}$ in terms of ${\displaystyle \psi }$. So if we write the position of the ant in terms of ${\displaystyle \psi }$ at time ${\displaystyle t}$ as ${\displaystyle \phi (t)}$, and the speed of the ant in terms of ${\displaystyle \psi }$ at time ${\displaystyle t}$ as ${\displaystyle \phi '(t)}$, we can write:

$${\displaystyle \phi '(t)={\frac {\alpha }{c+vt}}}$$

$${\displaystyle \therefore \phi (t)=\int {{\frac {\alpha }{c+vt}}\,dt}={\frac {\alpha }{v}}\ln(c+vt)+\kappa }$$

where ${\displaystyle \kappa }$ is a constant of integration.

Now, ${\displaystyle \phi (0)=0}$ which gives ${\displaystyle \kappa =-{\frac {\alpha }{v}}\ln {c}}$, so ${\displaystyle \phi (t)={\frac {\alpha }{v}}\ln {\left({\frac {c+vt}{c}}\right)}}$.

If the ant reaches the target-point (which is at ${\displaystyle \psi =1}$) at time ${\displaystyle t=T}$, we must have ${\displaystyle \phi (T)=1}$ which gives us:

$${\displaystyle {\frac {\alpha }{v}}\ln {\left({\frac {c+vT}{c}}\right)}=1}$$

$${\displaystyle \therefore T={\frac {c}{v}}\left(e^{v/\alpha }-1\right)}$$

and hence for the length of the rubber band when the ant catches the target point (i.e. the distance travelled by the ant):

$${\displaystyle y\left(T\right)=c+vT=c\times e^{v/\alpha }.}$$

(For the simple case of v = 0, we can consider the limit ${\displaystyle \lim _{v\rightarrow 0}T(v)}$ and obtain the simple solution ${\displaystyle T={\tfrac {c}{\alpha }}}$.) As this gives a finite value ${\displaystyle T}$ for all finite ${\displaystyle c}$, ${\displaystyle v}$, ${\displaystyle \alpha }$ (${\displaystyle v>0}$, ${\displaystyle \alpha >0}$), this means that, given sufficient time, the ant will complete the journey to the target-point. This formula can be used to find out how much time is required.

For the problem as originally stated, ${\displaystyle c=1\,\mathrm {km} }$, ${\displaystyle v=1\,\mathrm {km} /\mathrm {s} }$ and ${\displaystyle \alpha =1\,\mathrm {cm} /\mathrm {s} }$, which gives ${\displaystyle T=(e^{100\,000}-1)\,\mathrm {s} \,\!\approx 2.8\times 10^{43\,429}\,\mathrm {s} }$. This is a vast timespan, even compared to the estimated age of the universe, which is only about 4×10¹⁷ s. Furthermore, the length of the rope after such a time is similarly huge, 2.8×10⁴³⁴²⁹ km, so it is only in a mathematical sense that the ant can ever reach the end of this particular rope.