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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.

Community
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Tdonut
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  • That sounds like the right idea, although I don't know how to actually do that formally. Maybe this will help: http://www.math.cornell.edu/~numb3rs/blanco/Spree.html – Chee Han May 26 '15 at 18:32
  • Is the "full speed" of the submarine a known number? – Ian May 26 '15 at 19:11
  • @Ian No, it isn't, only that it is half that of the destroyer. – Tdonut May 27 '15 at 02:37
  • That's actually quite important to the problem, and is enough to use to solve it. Specifically it means that if the destroyer initially moves radially inward, then its radial position (for a while) will be $3-2vt$ while the submarine's radial position will be $vt$ forever. The destroyer should follow this course until they are equal i.e. until the destroyer has moved 1 mile, then it should follow the spiral path as in grdgfgr's answer. – Ian May 27 '15 at 02:44

1 Answers1

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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:

enter image description here

Gappy Hilmore
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  • I'm not sure this is quite right. Why is it enough for the radial velocity of the destroyer to be the same as the radial velocity of the submarine, when the destroyer starts out at $r_d=3$ and the submarine starts out at $r_s=0$? I would think that the destroyer needs to get to $r_s$ first, and then follow the spiral path that you're describing. Alternately the destroyer could simply wait for $r_s$ to be $3$ and then follow the path you're describing. – Ian May 26 '15 at 19:09
  • Yes, the destroyer should first arrive at 1 possible position before starting the path described above. – Gappy Hilmore May 26 '15 at 19:14
  • @Ian I have given an example. I'd appreciate if you have any input. – Gappy Hilmore May 26 '15 at 19:22
  • How did you obtain the vector velocity of the destroyer? – wjmolina Jul 30 '16 at 00:51
  • @EsX_Raptor That was explained. Which part do you not understand. – Gappy Hilmore Aug 01 '16 at 16:02