Submarine-Destroyer ODE question An answer to the question Best Book For Differential Equations? mentions the following problem:

A destroyer is hunting a submarine in a dense fog. The fog lifts for a moment, discloses the submarine on the surface 3 miles away, and immediately descends. The speed of the destroyer is twice that of the submarine, and it is known that the latter will at once dive and depart at full speed in a straight course of unknown direction. What path should the destroyer follow to be certain of passing directly over the submarine? Hint: Establish a polar coordinate system with the origin at the point where the submarine was sighted.

Can someone suggest an elegant solution to this problem?
 A: The strategy
For convenience, I'll assume the submarine travels 1 mile per minute. Suppose the destroyer starts out $R$ miles away (after potentially rescaling depending on the sub's speed). In my solution, the destroyer will sit still for R minutes, then take a slow outward spiraling path.
In the suggested polar coordinates system, the destroyer starts at position $(r, \theta) = (R, 0)$. The goal is for the destroyer to choose a spiraling route $\gamma(t) = (r(t), \theta(t))$ such that the destroyer will always be above a possible sub position. At time $t$, the possible sub positions are $(t, \theta)$ for any $\theta$ (just depends on the sub's initial starting direction). So basically we want to choose $\gamma(t)$ with $r(t) = t$ for all $t \ge R$, and $\theta(t)$ eventually covering the whole interval $[0, 2 \pi]$, and $\|\gamma'(t)\| = 2$ for all $t$.
The requirement $\|\gamma'(t)\| = 2$ is where the differential equation will come from. We have
$$\begin{align}
4 &= \|\gamma'(t)\|^2 \\
4 &= \left\|\frac{d}{dt} (r(t) \cos(\theta(t)), r(t) \sin(\theta(t))\right \|^2
\\
4 &= r'(t)^2 + r(t)^2 \theta'(t)^2 \\
4 &= 1^2 + t^2 \theta'(t)^2 \\
\theta'(t) &= \frac{\sqrt{3}}{t} \\
\theta(t) &= \sqrt{3} \ln(t) + c
\end{align}$$
where we have chosen the positive square root, which just means the destroyer's spiral will be counterclockwise.
We have the initial condition $\theta(R) = 0$, so we get $\theta(t) = \sqrt{3} \left( \ln(t) - \ln(R) \right)$. So in our final strategy, the destroyer will sit still until time $R$, and then follow the path $$\boxed{\begin{align}
r(t) &= t \\
\theta(t) &= \sqrt{3} \left( \ln(t) - \ln(R) \right)
\end{align}}$$
This path will catch the sub by the time we finish making one full revolution, i.e. when $\theta(t) = 2 \pi$. Solving the equation above, we see this will happen at time
$$t^* = e^{\frac{2 \pi}{\sqrt{3}} + \ln(R)}.$$
Strategy tweak: catching the sub as fast as possible
The strategy above is pretty good, but it wastes some time at the start. To catch the sub as fast as possible, we could instead start out by moving directly toward the point $(0,0)$ where the sub was sighted. Let $\tilde R = \frac{R}{3}$. Then when we reach the point $(\tilde R, 0)$, this will be the first moment when the sub COULD be directly under us, if it happened to flee directly toward us.
This puts us in basically the same situation as above, except with $R$ replaced by $\tilde R$. Following the same strategy, we'll end up catching the sub within a max time of
$$t^* = e^{\frac{2 \pi}{\sqrt{3}} + \ln(\tilde R)}$$
or
$$t^* = e^{\frac{2 \pi}{\sqrt{3}} + \ln(R/3)}.$$
A: Let $r(\theta_0)$ be a potential position for the submarine. This is where the destroyer begins his pursuit.  Then   $$r(\theta)=r(\theta_0) +\frac{1}{2} \int_{\theta_0}^{\theta} \sqrt{r(\theta)^2+r'(\theta)^2} d\theta$$ implies the appropriate differential equation for the destroyer's pursuit curve.
The idea is the destroyer's outward expansion occurs fast enough to account for the fact that he is twice as fast as the submarine.  That is the destroyer's radial position less his starting position must be half the arclength of the path that leads  it to that position for all possible polar angles.
