The ant and the rubber string.

Question

We have an ant on the tip of a horizantal rubber string of length say $\text{10 cm}$. The ant moves $\text{5 cm}$ each second, and the rubber string is stretched $\text{100 cm}$ each second.

Will the ant ever reach the end of the rubber string?

(The numbers ofcourse aren't specific they are just an example)

As the first thought it is impossible, but note that the stretching of the rubber also affects the position of the ant, i.e. if the ant moves its first $\text{5 cm}$ it will be at the half way of the string, then we elongate the string till it becomes $\text{110 cm}$ and now the ant is still at the half way of the string and that is at $\text{55 cm}$.

For example, if $x_i$ represents the position of the ant at the $i^{th}$ second and $y_i$ represents the elongation of the string at the $i^{th}$ second.

So in our example

$\text{$x_1=5 \qquad y_1=10$}$

$\text{$x_2=60 \qquad y_2=110$}$

$\text{$x_3=119.55 \qquad y_3=210$}$ ....

I tried to work that into equations and sequences, and I came up with this $$x_i=\frac{x_{i-1}}{y_{i-1}} × y_i +5$$ and ofcourse $$y_i=y_{i-1}+100$$

So the question now becomes could $x_i=y_i$ at a certain $i$?

Is my work correct? Any other proved solution is also appreciated...

Actually the real paradox is that if we are elongating the rubber string $\text{1 km}$ each second and the ant is still moving at the rate $\text{5 cm}$ each second. Will the ant ever reach the end of the string?

Answer 1

The ant will reach the end, but if its speed is rather slower than the string stretches it will take a long time. The usual way to state the problem has the string stretching instantaneously once per second.

Just to be specific, let the ant walk at $1$ cm per second and the string start at $100$ cm and stretch $100$ cm per second. In the first second the ant covers $1$ cm, which is $1/100$ of the string. Then the string stretches to $200$ cm and the ant covers $1/200$ of the string. In the $n^{th}$ second the ant covers $\frac 1{100n}$ of the string. After the $n^{th}$ second the ant has covered $$\frac 1{100}\left(\frac 11+\frac 12+\frac 13 +\ldots +\frac 1n\right)=\frac 1{100}H_n$$ where $H_n$ is the $n^{th}$ harmonic number. We have $H_n \approx \log n + \gamma$, so we need $$\log n \approx 99.5\\n\approx e^{99.5}$$

Answer 2

Here's a different approach. Warning : the justification of the model is not very rigourous, but then again, we're trying to model some concrete physical phenomenon by maths, so it can't be fully rigourous. It's still a pretty good model though, and we get similar results to the ones we get with the discrete modelization

Ross Millikan's answer provides a solution when the process is discrete : you count seconds, and at each second, the rubber band is elongated of so much, and the ant moves of that much. This is what was being asked, but it's not very realistic.

I assume that we put the rubber band next to an infinite ruler, starting at $0$.

Concretely, the rubber band would move continuously. Since we want it to grow $100$ cm each seconds, and since the easiest model is the linear one, I'll be assuming that the length of the rubber band at time $t$ is $y(t) = 100(t+1)$.

Let $x(t)$ denote the position of the ant at time $t$ (position with respect to the infinite ruler next to which we are performing the experiment). I'm assuming the ant is walking at constant speed (on the rubber band) $v$.

Then in a really really small amount of time $\mathrm{d}t$, we can approximate the process by a discrete one (that's the non rigourous part), that is if $\mathrm{d}t$ is small enough, we can say that within this time, it's as if we had simply stretched the rubber band, then moved the ant (we could do it the other way around, the formula is a bit different, but the equation we get is the same); that is $x(t+\mathrm{d}t) \approx \frac{x(t)}{y(t)}y(t+\mathrm{d}t)+ v\mathrm{d}t$. To make this approximation precise (and go back to the rigourous world), I'll assume that for small $\mathrm{d}t$, the error between $x(t+\mathrm{d}t)$ is negligeable when compared to $\mathrm{d}t$.

So my assumption is that, as $\mathrm{d}t\to 0$, $x(t+\mathrm{d}t) = \frac{x(t)}{y(t)}y(t+\mathrm{d}t)+ v\mathrm{d}t + o(\mathrm{d}t)$. Hop, now we've finished modelling, and we can proceed in full rigor. Because what this implies, is that $x(t+\mathrm{d}t) = x(t)(1+\frac{\mathrm{d}t}{t+1})+ v\mathrm{d}t + o(\mathrm{d}t)$ (small computation with the explicit formula for $y(t)$), and so for $\mathrm{d}t\to 0$, we get $\frac{x(t+\mathrm{d}t)-x(t)}{\mathrm{d}t} = \frac{x(t)}{t+1} + v+o(1)$. Now we go to the limit and get that $x$ is differentiable, and for $t\geq 0$, $x'(t) = \frac{x(t)}{t+1} + v$.

So we have a differential equation telling us how $x$ behaves, and we know how $y$ behaves. The ant reaches the end of the rubber band if and only if $x(t) = y(t)$ for some $t\geq 0$.

Now it's a classical exercise in differential equations to solve this, and we get that $x(t) = v\ln(t+1)(t+1) + c(t+1) = (t+1)(v\ln(t+1)+c)$ for some constant $c$.

Except if we want a precise computation, the constant $c$ won't matter that much but we can say what it is : evaluate at $0$ to get $0=x(0) = c$. So $c=0$

So $x(t) = v(t+1)\ln(t+1)$. Note that we get a logarithm, just like in Ross Millikan's answer (at least in terms of order of magnitude of the growth), which tells us that the discrete model is not so far from the continuous model.

But now $x(t) = y(t)$ becomes an equation we can solve: $v(t+1)\ln(t+1) = 100(t+1)$. This tells us $t=e^\frac{100}{v}-1$, so in our situation where $v=5$, we get $t=e^{19}-1$; and in Ross's situation where $v=1$, we get $t=e^{100}-1$, which is pretty close to what they found : it tells us one more time that the two models aren't that different.

Of course the general formula is that the ant will get to the end of the rubber band at $t=e^\frac{d}{v}-1$, where $d$ is the amount by which the rubber band expands : the ant always gets there, but if $d>>v$, it will take a long time.