Searching for the Submarine

Let the submarine start at the origin and the aircraft at (6,0).

Hand-wavy bit: the optimal path involves flying directly toward the origin, to where it would find the submarine if it happened to flee directly to the right (easily seen to be at (2,0)), then flying around in a spiral such that at each point the aircraft would spot the submarine if it traveled at the appropriate angle.

Assuming that bit's true, let $\vec{v}(t)$ denote the position of the aircraft at time t, with $t=0$ corresponding to the point (2,0). At time t, the submarine is 2+20t units from the origin, so we need $|v(t)| = 2+20t.$ And we need $|v'(t)|=40$ for all t for the aircraft's speed.

Let's guess then that $v(t) = \big\langle (2+20t)\cos(f(t))\,,\, (2+20t)\sin(f(t)) \big\rangle$ for some function $f(t)$ . (This is a spiral with the correct distances, and tuning $f$ can enforce the speed requirement.)

From here things are straightforward; solving the differential equation $|v'(t)|=40$ with $f(0)=0$ for $f$ , we get $f(t)=\sqrt{3}\ln(1+10t)$ .

We need to follow the spiral through one revolution; solving $f(T)=2\pi$ gives $T=\frac{1}{10}(e^{2\pi/\sqrt{3}}-1)$ .

Finally, the distance traveled is $4+40T=4e^{2\pi/\sqrt{3}}\approx 150.489466$ .