Differential equations application problem I am studying differential equations, and I saw this interesting problem in another question (here):

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? 

The problem gives a hint: establish a polar coordinate system with the origin at the point where the submarine was sighted.
I honestly have no inkling as to how you can solve this problem. I am thinking the path must be some sort of spiral around the submarine's location (pursuit curve?) but I'm not sure.
 A: time $t$, velocity $v$, submarine subscript $s$, destroyer subscript $d$
At any given time, sub will be $r_s=tv$ away from the origin at a constant angle $\theta _0$
Destroyer will need to match this radial distance, so $r_d=tv$. We need to find $\theta _d (t)$. Vectoral velocity of the destroyer in polar coordinates is $\bar v_d = r_d'\hat r+r_d \theta _d '\hat \theta$, where $^$ denotes unit vector.
The magnitude of $|\bar v_d|=2v=\sqrt{ r_d'^2+r_d^2 \theta _d '^2}$
This gives us
$$\theta _d '= \pm \frac{\sqrt{3}}{t}$$
The $\pm$ makes sense because we can choose to wrap around from any direction we want.
The rest of the question can be completed with initial conditions, velocity, etc.
Example: Lets assume the destroyer starts at $(3,0)$ with velocity $2$ and the sub at $(0,0)$ with velocity $1$ (Cartesian). The destroyer first directly moves to $(1,0)$, where it will meet the destroyer in best cast scenario. At that point, he will be on the $r_d = vt$ position and he will follow the path in the image:

