Day 5: Save the Five Golden Rings!

Stokes' Law gives drag force = $6\pi r \mu v = 8v$ after substituting in values.

The force going forwards is precisely $e^{-\alpha t}$ as given in the question. I will substitute $\alpha$ in at the end to make the solution clearer.

So resultant force: $F = e^{- \alpha t} - 8v$ , and using $F=ma$ , $\frac{\mathrm{d}v}{\mathrm{d}t} =\frac18 e^{- \alpha t} - v$

So we have a first order differential equation to solve. Multiply by the integrating factor $e^t$ to get an exact equation and solve (since $\frac{d}{dt}(e^tv) = e^t\frac{dv}{dt} + e^tv$ ): $e^tv = \int \frac18 e^{(1-\alpha)t} \mathrm{d}t = \\ \frac{1}{8(1-\alpha)}\left(e^{(1-\alpha)t} - 1 \right)$ since $v=0$ at $t=0$ . So: $v = \frac{1}{8(1-\alpha)}\left(e^{-\alpha t} - e^{-t} \right)$ Therefore integrating with respect to $t$ : $x = \frac{1}{8(1-\alpha)}\left(\frac{1}{-\alpha}e^{-\alpha t} + e^{-t} + \frac{1}{\alpha} - 1 \right)$ since $x=0$ at $t=0$ .

Simplifying and substituting in $\alpha = 0.001$ gives: $x = \frac{1}{8\times 0.999}\left(999 + e^{-t}-1000e^{-0.001t}\right)$

Finally we are looking for the maximum length so the elf will hold his breath up to the maximum; therefore substitute $t=100$ to give $x=11.78$ m, giving the final answer $\boxed{1178}$ cm as required.