Ancora Imparo

28 February 2014

Snail Problem, Part 5

Filed under: mathematics — Darmok @ 03:04 UTC
Tags: , ,

I hope no one is tired of snails yet. In the last several posts, I tackled the following problem: a snail crawls along a twig of length L0 at 1 cm/d, and the twig grows (along its entire length) at 2 cm/d. Will the snail reach the end, and when? In parts 1, 2, and 3, I solved this problem, and then in part 4 I turned to the general case, leaving the snail’s velocity and twig growth velocity unspecified.

We found that the snail’s position, x, can be given by the following equation, where g is the growth rate of the twig, s is the (inherent) velocity of the snail, and t is the time:

$\displaystyle x = \frac{s}{g}(gt+L_0)\ln\frac{gt+L_0}{L_0}.$

We want to find the time t where the snail reaches the end of the twig. Since the length of the twig is gt+L0, we set x equal to this.

$\displaystyle x = gt+L_0$

Substituting our equation for x gives

$\displaystyle \frac{s}{g}(gt+L_0)\ln\frac{gt+L_0}{L_0}=gt+L_0$.

To solve this for t, we start by dividing both sides by gt+L0 (note that this is the twig length, and is only applicable for nonzero twig lengths).

$\displaystyle \frac{s}{g}\ln\frac{gt+L_0}{L_0}=1$

We can multiply both sides by g/s (note that the snail velocity cannot be zero).

$\displaystyle \ln\frac{gt+L_0}{L_0}=\frac{g}{s}$

Exponentiating both sides gives

$\displaystyle e^{\ln\frac{gt+L_0}{L_0}}=e^{\frac{g}{s}}$

$\displaystyle \frac{gt+L_0}{L_0}=e^{\frac{g}{s}}$

$\displaystyle gt+L_0 = L_0 e^{\frac{g}{s}}$

$\displaystyle gt = L_0 e^{\frac{g}{s}}-L_0$

$\displaystyle gt = L_0(e^{\frac{g}{s}}-1)$

And finally, the time is

$\displaystyle t_f = \frac{L_0}{g}\left(e^{g/s}-1\right) \blacksquare$

So the time it takes is the initial length of the twig divided by its growth rate (which incidentally is the amount of time it would have taken to grow to its start position) times the quantity e to the power of the ratio of the twig and snail rates minus 1.

Interestingly, this will be finite no matter how small s is. If the snail is extremely slow, then the exponent g/s will be very large, but as long as s is positive, the snail will eventually reach the end of the twig — no matter how fast the twig grows!

Let’s also see how long the twig will be when the snail reaches the end. Recall that the length of the twig is

$\displaystyle L(t) = gt+L_0$

We can substitute our time above in for t:

$\displaystyle L(t_f) = g\left[\frac{L_0}{g}\left(e^{g/s}-1\right)\right] +L_0$

$\displaystyle L(t_f) = L_0\left(e^{g/s}-1\right)+L_0$

$\displaystyle L(t_f) = L_0\left[(e^{g/s}-1)+1\right]$

And finally, the length is

$\displaystyle L(t_f) = L_0 e^{g/s}\, \blacksquare$

1 Comment

1. Hi Darmok … I see that comments are possible now. Great! I’ve checked your calculations in parts 4 and 5 and agree with everything. I’ve found some info about the background of the snail problem, and will post that info later today as a reblog of your post. Ken R, 01-March-2014

Comment by krobe8 — 1 March 2014 @ 12:18 UTC