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The long walk

October 9, 2009 at 5:17 pm | In Puzzles | 2 Comments
Tags: expanding string, Mathematics, puzzle, the long walk

I thought that I’d follow up my recent post on The Monty Hall Problem with another puzzle, which in my opinion is a bit more interesting and somewhat more involved. As with most puzzles there are several variations (see for example [1-3]), however they are all more or less similar and can certainly be solved following the method detailed below. As a starting point we take the problem as formulated in [3].

An ant starts to crawl along a taut rubber rope 1km long at a speed of 1cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch uniformly by 1km per second (so that after 1 second it is 2km long, after 2 seconds it is 3km long, etc). Will the ant ever reach the end of the rope?

The key points to note here is that the ant is crawling at a constant speed relative to the rubber rope and that the rope is stretching in a uniform fashion. Intuition would perhaps suggest that the ant wouldn’t make it; however, as we shall see, intuition can often be wrong as was the case with The Monty Hall Problem.

Let us now make a more general formal statement of the problem in the same vein as [3].

Consider a thin elastic string, colinear with the x-axis and held taut between the points $x=0$ and $x=\alpha\left(t\right)$ , where $\alpha\left(t\right) > 0$ , $t > 0$ . A particle moves along the string with constant velocity $u > 0$ relative to the point on the string where the particle is currently located. The string stretches smoothly and uniformly such that the end point of the string obeys the differential equation $\alpha^\prime\left(t\right) = v$ with initial value $\alpha\left(0\right) = \alpha_0$ , where $v = \text{const.} > 0$ and $\alpha_0 > 0$ .

Will the particle reach the point $x=\alpha\left(t\right)$ in finite time?

It is obvious (or perhaps not so obvious) that the velocity of the particle, relative to the external coordinate system is the sum of its velocity relative to its current point on the string and the velocity of the same point on the string. To illustrate this point, imagine that you are walking down the aisle of a moving train carriage. To everyone else on the train you appear to be moving at walking pace down the aisle. Now imagine that you are stood at the side of the tracks watching a train go past whilst someone is walking down the aisle of the carriage. From your point of view, their speed is the sum of their walking speed and the speed of the train.

Although the above analogy illustrates the situation rather well, the problem in hand is somewhat more complicated. In the train-passenger example the each point in the carriage was moving at the same speed. However, in our problem the string is stretching, which means that any two points on the string are moving relative to each other. In other words, each point on the string moves with a different velocity. If you need convincing of this, take an elastic band and draw two lines across its width. If you stretch the elastic band you will see that the distance between the two lines increases.

So, how do we work out how fast each point is moving? Well the first step is to work out how fast the end point of the string is moving. We stated earlier that $\alpha^\prime\left(t\right) = v = \text{const.}$ and $\alpha\left(0\right) = \alpha_0$ . By integrating with respect to $t$ we obtain the position of the end point as a function of time

(1)

$\displaystyle\alpha\left(t\right) = vt + \alpha_0$

What can we say about the remaining points on the string? Since the string stretches uniformly, we know that the position of each point relative to the entire length of the string remains unchanged. In other words, the position of each point scaled with respect to the length of the string, $\alpha\left(t\right)$ remains constant. What does this mean? Well, let the string initially be 1m long and let us place a mark at 0.5m. If the string is then stretched to 2m, the mark will then be at 1m, or half the length of the string as it was originally. Let us consider a point $x=p\left(t\right)$ on the string. We may then write

(2)

$\displaystyle\frac{p\left(t\right)}{\alpha\left(t\right)} = A = \text{const.}$

(3)

$\displaystyle\Rightarrow p\left(t\right) = A\left(vt+\alpha_0\right)$

We take the derivative with respect to time

(4)

$\displaystyle p^\prime\left(t\right) = Av$

Eliminating $A$ using our original expression for $p\left(t\right)$ yields a the velocity of any point on the string as a function of time

(5)

$\displaystyle p^\prime\left(t\right) = \frac{v\cdot p\left(t\right)}{vt + \alpha_0}$

As we said earlier, the speed of the particle relative to the x-axis is the sum of the particle’s speed relative to its current point on the rope and the speed of the current point on the rope relative to the x-axis. We were given the former quantity and we have just derived an expression for the latter quantity. Therefore, denoting the position of the particle with respect to the x-axis as $\beta\left(t\right)$ we may write

(6)

$\displaystyle\beta^\prime\left(t\right) = u + \frac{v\cdot \beta\left(t\right)}{vt + \alpha_0}$

which may be rewritten in canonical form

(7)

$\displaystyle\beta^\prime\left(t\right) - \frac{v\cdot \beta\left(t\right)}{vt + \alpha_0} = u$

which may further be rewritten (making use of an integrating factor [4] of $\left(vt+\alpha_0\right)^{-1}$ )

(8)

$\displaystyle \frac{d}{dt}\left(\frac{\beta\left(t\right)}{vt+\alpha_0}\right) = \frac{u}{vt+\alpha_0}$

The integration is straightforward and yields

(9)

$\displaystyle\beta\left(t\right) = \frac{u}{v}\ln\left(\frac{vt+\alpha_0}{\beta_1}\right)$

where $\beta_1$ corresponds to the constant of integration. So we now have expressions for the position of both the end of the string (1) and the particle (9) in terms of time.

Recall that our objective was to determine if our particle will ever reach the end of string in finite time. The most obvious method to determine this is to solve the system of equations (1) and (9), in order to find an expression for time in terms of the other variables. Noting that the numerator in the argument of the logarithm in (3) is $\alpha\left(t\right)$ we may write

(10)

$\displaystyle \alpha\left(t\right) = \frac{u}{v}\ln\left(\frac{\alpha\left(t\right)}{\beta_0}\right)$

Unfortunately, the solution of the above expression cannot be written in terms of elementary functions (if you are interested the solution can be written in terms of the Lambert function [5]). However, we can find an explicit solution for the time taken for the particle to reach the point $x = \alpha\left(t\right)$ , by means of a coordinate transformation.

Let us make a change of coordinates $x \rightarrow \gamma$ such that the rope now extends from $\gamma = 0$ to $\gamma = \alpha_1 = \text{const.}$ . In other words, we move to a coordinate system which expands with the string such that in this coordinate system the length of the string remains constant. Therefore, each point on the string is now stationary with respect to to the new coordinate system. Formally, we define the transformation between $x$ and $\gamma$ as

(11)

$\displaystyle \gamma\left(t\right) = \frac{x}{\alpha\left(t\right)} = \frac{x}{vt+\alpha_0 }$

Hence, in changing coordinate system we have in effect removed the time dependence from the position of the points on the string. Instead, our coordinate system is now time dependent. The next question is “what is the speed of the particle in these new coordinates?” In order to determine how velocities transform between the two coordinate systems we simply need to differentiate (11) with respect to time

(12)

$\displaystyle \begin{aligned}\frac{d}{dt}\gamma\left(t\right) & = \frac{d}{dt}\left( \frac{x}{vt+\alpha_0 }\right) \\ & = \frac{dx}{dt}\frac{1}{vt+\alpha_0} - \frac{vx}{\left(vt + \alpha_0\right)^2}\end{aligned}$

Inserting equations (6) into equation (12) yields the desired transformation for velocities

(13)

$\displaystyle \eta^\prime\left(t\right) = \frac{u}{vt+\alpha_0}$

Where $\eta\left(t\right)$ denotes the position of the particle with respect to the coordinate system $\gamma\left(t\right)$ and $\eta^\prime\left(t\right)$ therefore denotes the velocity of particle. Integrating (13) with respect to time yields the position of the particle with respect to the coordinate system $\gamma\left(t\right)$ as a function of time

(14)

$\displaystyle \eta\left(t\right) = \frac{u}{v}\ln\left(vt+\alpha_0\right) + \eta_0$

Writing time as a function of position

(15)

$\displaystyle t\left(\eta\right) = \frac{1}{v}\left\{\exp\left[\frac{v}{u}\left(\eta-\eta_0\right)\right] - \alpha_0\right\}$

So, we have now obtained an expression for the time taken for the particle to reach a point $\eta$ , after starting at point $\eta_0$ in coordinates $\gamma$ . Now we are in a position to answer the question that was initially posed: “will the particle reach the point $\alpha\left(t\right)$ in finite time?”. As said previously, this corresponds to the case when $\beta\left(t\right) = \alpha\left(t\right)$ . Substituting $x = \alpha\left(t\right)$ into (11) we obtain the corresponding point in $\gamma$ coordinates: $\gamma\left(t\right) = 1$ . Hence, setting $\eta\left(t\right)$ equal to unity in (15) yields

(16)

$\displaystyle T = \frac{1}{v}\left\{\exp\left[\frac{v}{u}\left(1-\eta_0\right)\right] - \alpha_0\right\}$

And here we arrive at the desired result. We observe that $T$ is finite for all finite, positive values of $v,u,\eta_0, \alpha_0$ . Therefore, the ant will reach the end of the elastic string provided that the string is initially of finite length, the ant initially starts from a finite position and both the speed of expansion of the string and the speed of the ant is finite. Note that this is still true even when the speed of expansion far exceeds the speed of the ant.

To illustrate this, let us consider some specific cases. Let the ant initially start from the origin. In this case

(17)

$\displaystyle \eta_0 = -\frac{u}{v}\ln\left(\alpha_0\right)$

Equation (16) then becomes

(18)

$\displaystyle T = \frac{\alpha_0}{v}\left(e^{v/u} - 1\right)$

Inputting the values as originally stated in the problem ( $v = 1\times10^3\;m/s,\;u = 1\times10^{-2}\;m/s,\;\alpha_0 = 1\times10^{3}\;m$ ) we obtain a value of $T = e^{10^{5}}\;s$ , which is very large indeed, but still finite. As we would expect, for fixed $v$ (speed of expansion), the larger $u$ (speed of ant), the shorter the time taken to reach the end of the string. For the limiting case of $v \rightarrow 0$ (with fixed $u$ ), i.e. when the string is not expanding, we find that (making use of L’Hôpital’s rule)

(19)

$\displaystyle \begin{aligned}\lim_{v\to0} T & = \lim_{v\to0}\frac{\alpha_0}{v}\left(e^{v/u} - 1\right) \\ & = \lim_{v\to0}\frac{\alpha_0}{u}e^{v/u} \\ & = \frac{\alpha_0}{u}\end{aligned}$

which again is as expected.

To summarise, the answer to the original question as stated at the beginning of the post is: yes, the ant will always reach the end of the string provided that the string is finite and is expanding at a finite rate and the ant has finite speed.

References

[1] Su, Francis E., et al. “Inchworm on a Rubber Rope.” Mudd Math Fun Facts. http://www.math.hmc.edu/funfacts/.
[2] Sarcone G.A., Waeber M.J. “Puzzle # 114: Infinite beetle path”. Archimedes’ Lab. http://www.archimedes-lab.org/.
[3] Wikipedia. “Ant on a rubber rope”. Wikipedia. http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope
[4] Munkhammar, Joakim. “Integrating Factor.” From MathWorld–A Wolfram Web Resource, created by Eric W. Weisstein. http://mathworld.wolfram.com/IntegratingFactor.html
[5] Weisstein, Eric W. “Lambert W-Function.”
From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/LambertW-Function.html

$latex T < \infty \; \forall \; 0 < v,u,\eta_0, \alpha_0 < \infty$

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