Submarine hunting
Question
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?
From Differential Equations with Applications and Historical Notes by George F. Simmons, second edition, page 34.
Answer
Setting up the problem
Using polar coordinates with the initial position of the submarine as the origin, the submarine's position at time \(t\) is \(r=vt/2\), where \(v\) is the speed of the destroyer, so clearly the submarine needs to follow an outgoing spiral path of some kind where the distance from the origin is proportional to the time.
The destroyer is three miles from the submarine at the start of the problem, so it cannot immediately locate the sub. Thus it should travel to the nearest possible location of the sub before beginning the spiral search pattern. Since the destroyer is at a location of three miles from the origin, and it travels at twice the speed of the sub, the earliest point it can locate the sub is if the sub goes directly towards the destroyer, which means the destroyer should go in a straight line for two miles towards the origin, then begin spiralling outwards.
The submarine hunt
In polar coordinates the destroyer's path is \((r(t), \theta(t))\). In cartesian coordinates this is \((x(t),y(t))=(r(t)\sin\theta(t),r(t)\cos\theta(t))\), and the speed of the destroyer, \(v=\sqrt{x'(t)^2+y'(t)^2}\) must be constantly the maximum possible for the best chance of finding the sub. Thus
\[ x'(t)^2+y'(t)^2=v^2. \] Substituting \(x'(t) = r'(t)\sin\theta(t)+r(t)\theta'(t)\cos\theta(t)\) and similar for \(y'(t)\) and doing some trigonometry gives \[ r'(t)^2+r(t)^2\theta'(t)^2=v^2. \] The sub's position at time \(t\) is \(vt/2\), so \(r(t)=vt/2\). Differentiating by \(t\) gives \(r'(t) = v/2\), and \[ (v/2)^2+(vt/2)^2(\theta'(t))^2=v^2 \]The \(v\)s all divide out, and this simplifies to \(t\theta'(t)=\sqrt3\), which has the solution \(\theta(t)=\sqrt3\log t+c\). If we put the start of the search as time \(t=1\) then we can discard the constant of integration \(c\).
Thus the radius of the search is proportional to the exponential of the angle relative to the origin, and the ideal search pattern for the destroyer is a logarithmic spiral.