# Ancora Imparo

## 23 February 2014

### Snail Problem, Part 3

Filed under: mathematics — Darmok @ 20:28 UTC
Tags: , , ,

This is a continuation of my attempt to solve 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? You can see part 1 where I successfully set up the differential equation and unsuccessfully tried to solve it, and part 2 where I did manage to solve it.

At the end of the last post, I had solved for the equation that describes the position of the snail over time,

$\displaystyle x = \left(t+\frac{L_0}{2}\right)\ln\left(\frac{2t}{L_0}+1\right)$.

Also, recall that the length of the twig is

$\displaystyle L = 2t + L_0$

since it grows at 2 cm/d. We want to find the time t when these two are equal.

$\displaystyle 2t + L_0 = \left(t+\frac{L_0}{2}\right)\ln\left(\frac{2t}{L_0}+1\right)$

[Edit: I missed a very obvious step at this point, so the next few steps will be unhelpful. Free free to skip ahead to the next edit.] We need to isolate the natural logarithm.

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{2t+L_0}{t+\frac{L_0}{2}}$

Exponentiating both sides gives

$\displaystyle e^{\ln\left(\frac{2t}{L_0}+1\right)} = e^{\frac{2t+L_0}{t+\frac{L_0}{2}}}$

$\displaystyle \frac{2t}{L_0}+1 = e^{(2t+L_0)(t+\frac{L_0}{2})^{-1}}$

Well, the left side is great, but the right side is a mess.

$\displaystyle \frac{2t}{L_0}+1 = \left[e^{(t+\frac{L_0}{2})^{-1}}\right]^{(2t+L_0)}$

If I bring the 2tL0 term out, I could split it up.

$\displaystyle \frac{2t}{L_0}+1 = \left[e^{(t+\frac{L_0}{2})^{-1}}\right]^{(2t+L_0)}$

$\displaystyle \frac{2t}{L_0}+1 = \left[e^{(t+\frac{L_0}{2})^{-1}}\right]^{2t}\left[e^{(t+\frac{L_0}{2})^{-1}}\right]^{L_0}$

This was unhelpful and I have no idea how to simplify this.

[Edit: At this point I get back on track.] Let’s go back to the last simple equation.

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{2t+L_0}{t+\frac{L_0}{2}}$

I should get rid of this complex fraction, even if it means making the numerator more complex.

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{2t+L_0}{t+\frac{L_0}{2}}\cdot\frac{2}{2}$

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{4t+2L_0}{2t+L_0}$

Oh, it’s so obvious now. I don’t know how I missed realizing that this fraction would simplify.

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = \frac{2(2t+L_0)}{2t+L_0}$

$\displaystyle \ln\left(\frac{2t}{L_0}+1\right) = 2$

I feel a bit foolish. Now, let’s exponentiate both sides.

$\displaystyle e^{\ln\left(\frac{2t}{L_0}+1\right)} = e^2$

That’s much simpler.

$\displaystyle \frac{2t}{L_0}+1 = e^2$

$\displaystyle \frac{2t}{L_0} = e^2 - 1$

$\displaystyle t = (e^2-1)\cdot\frac{L_0}{2}$.

Here are my initial observations: It seems “elegant” enough to be a solution to this problem. I like that it depends on e. I also notice that the time it takes is directly proportional to the initial length of the twig. I’m a bit surprised, as the snail’s net speed is not constant but rather increases as it progresses. Also, it’s apparent in retrospect that the first factor, $\left(t+\frac{L_0}{2}\right)$ in the equation for the snail’s position over time,

$\displaystyle x = \left(t+\frac{L_0}{2}\right)\ln\left(\frac{2t}{L_0}+1\right)$,

is half the length of the twig at that time, L(t):

$\displaystyle x(t) = \frac{1}{2}L(t)\ln\left(\frac{2t}{L_0}+1\right)$.

I don’t have an intuitive understanding of what the other quantity could represent, nor do I know if the 2 and 1 have any connection to the twig growth and snail inherent velocity. It would be interesting to repeat this but leave all the quantities unspecified.

## 1 Comment

1. […] the last three posts (see Snail Problem parts 1, 2, and 3), I tackled the following problem: a snail crawls along a twig of length L0 at 1 cm/d, and the twig […]

Pingback by Snail Problem, Part 4 | Ancora Imparo — 27 February 2014 @ 06:29 UTC