Ant on a stretchy rope puzzle

[Gavran]

In the general case, the time that the ant needs to reach the end of the rope is $$ T=\frac cv(e^\frac v\alpha-1) $$, where $c$ is the initial length of the rope, $v$ is the rate at which the rope stretches, and $\alpha$ is the speed of the ant relative to the rope.
Clearly, any positive real values of $c$, $v$, and $\alpha$ will always return the positive real value of $T$, which means that the ant will always reach the end of the rope no matter what positive real values are assigned to $c$, $v$, and $\alpha$.

 

[DaveC426913]

In the general case, the time that the ant needs to reach the end of the rope is $$ T=\frac cv(e^\frac v\alpha-1) $$, where $c$ is the initial length of the rope, $v$ is the rate at which the rope stretches, and $\alpha$ is the speed of the ant relative to the rope.
Clearly, any positive real values of $c$, $v$, and $\alpha$ will always return the positive real value of $T$, which means that the ant will always reach the end of the rope no matter what positive real values are assigned to $c$, $v$, and $\alpha$.

Google is too dumb to solve this. It returns undefined.

(1000 / 1000) * (e^(1000 / .01) - 1) = undefined

 

[Ibix]

Google is too dumb to solve this. It returns undefined.

(1000 / 1000) * (e^(1000 / .01) - 1) = undefined

It's obviously well approximated by $e^{100000}$. Noting that $e\approx 10^{0.434}$ this is $(10^{0.434})^{100000}\approx 10^{43400}$. That is rather larger than the maximum value of a double precision variable, which is what Google is probably using.

 

[A.T.]

Google is too dumb to solve this. It returns undefined.

Pfff... if you name yourself after a large number, you should do better on that.

Try:
https://www.mathsisfun.com/calculator-precision.html

 

[kochanskij]

Assuming I made no mistakes, here's a complete answer.

Spoiler: Complete solution

Let the band have initial length $L$, with one end at rest and the other moving at constant speed $V$. Thus at time $t$ the band has length $L+Vt$, and a point at distance $x$ from the stationary end has speed $\frac x{L+Vt}V$. The ant's speed relative to the rubber is $u$, so its speed relative to the stationary end is $$\frac{dx}{dt}=u+\frac{xV}{L+Vt}$$Maxima says that this is satisfied by $$\frac xL=\frac uV\left(1+\frac{Vt}L\right)\ln\left(1+\frac{Vt}L\right)$$meaning that the ant reaches $x=L+Vt$ when$$\begin{eqnarray*}
\frac Vu&=&\ln\left(1+\frac{Vt}L\right)\\
t&=&\frac LV\left(e^{V/u}-1\right)
\end{eqnarray*}$$Plugging in $L=10^3\,\mathrm{m}$, $V=10^3\,\mathrm{ms^{-1}}$, and $u=10^{-2}\,\mathrm{ms^{-1}}$ this becomes $t=e^{10^5}-1\approx 10^{43400}\,\mathrm{s}$.

Bear in mind that the universe is less than $10^{18}\,\mathrm{s}$ old. So there is a solution to the maths (assuming I didn't make any mistakes), but it's not realistic.

Yes, I got the same result. You gave an excellent complete solution.

 

[Roberto Pavani]

My 2 cents:
For the general case where the rope length grows as $L(t) = L_0 \cdot a(t)$, the ant reaches the end if and only if:

$\int_0^\infty \frac{dt}{a(t)} = \infty$

The critical threshold is $a(t) = t \ln(t)$: the ant still makes it (barely). Anything growing faster than $t \ln(t)$ and the ant never arrives.

 

[haruspex]

My 2 cents:
For the general case where the rope length grows as $L(t) = L_0 \cdot a(t)$, the ant reaches the end if and only if:

$\int_0^\infty \frac{dt}{a(t)} = \infty$

The critical threshold is $a(t) = t \ln(t)$: the ant still makes it (barely). Anything growing faster than $t \ln(t)$ and the ant never arrives.

I haven’t checked your iff condition, but taking that to be correct, wouldn’t $t\ln(t)\ln\ln(t)\ln\ln\ln(t)…$ also do?

 

[Roberto Pavani]

I haven’t checked your iff condition, but taking that to be correct, wouldn’t tln⁡(t)ln⁡ln⁡(t)ln⁡ln⁡ln⁡(t)… also do?

You're right! f(n) = n·ln(n)·ln(ln(n))·... works too, for any finite number of iterated logs (integral test: each step gives the next iterated logarithm, still → ∞). The ant always makes it. You'd need something like f(n) = n^(1+ε) to actually stop it.

My point was simply that even in the "accelerating rope" case, there are growth rates faster than linear where the ant still arrives.

 

[haruspex]

You'd need something like f(n) = n^(1+ε) to actually stop it.

What if we define $f(x)=1+\ln(x)$, $g_{n+1}(x)=f(g_n(x))$, $h_{n+1}(x)=h_n(x)g_n(x)$, with suitable settings for $g_0, h_0$, for $x>=1$?
Does $g_n$ converge for all x? If so, is the limiting function slower than $x^{1+\epsilon}$ for all $\epsilon>0$?