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1988 Paper 3 Q10
D: 1700.0 B: 1500.0
differential equation polar coordinates pursuit curve symmetry exponential function regular polygon first order ODE

Four greyhounds \(A,B,C\) and \(D\) are held at positions such that \(ABCD\) is a large square. At a given instant, the dogs are released and \(A\) runs directly towards \(B\) at constant speed \(v\), \(B\) runs directly towards \(C\) at constant speed \(v\), and so on. Show that \(A\)'s path is given in polar coordinates (referred to an origin at the centre of the field and a suitable initial line) by \(r=\lambda\mathrm{e}^{-\theta},\) where \(\lambda\) is a constant. Generalise this result to the case of \(n\) dogs held at the vertices of a regular \(n\)-gon (\(n\geqslant3\)).

Solution:

TikZ diagram
It's straightforward to see that \(\dot{r} = -\frac{v}{\sqrt{2}}\) and \(\tan (\theta(t + \delta t) - \theta(t)) = \frac{v\delta t/\sqrt{2}}{r - v \delta t/\sqrt{2}} = \frac{v}{r\sqrt{2}} \delta t + o(\delta t^2) \Rightarrow \dot{\theta} = \frac{v}{r \sqrt{2}}\) Solving this system, we can see that \(r(t) = \frac{l - vt}{\sqrt{2}}\) and \(\frac{\d r}{\d \theta} = -r \Rightarrow r = \lambda e^{-\theta}\) where \(\lambda = \frac{l}{\sqrt{2}}\) where \(l\) is the initial square side length.
TikZ diagram
By the cosine rule, we can see that \(r(t + \delta t)^2 = r^2+(v\delta t)^2 - 2rv\delta t \cos (\frac12 (\pi - \frac{2\pi}{n}))\), ie: \(\frac{r(t + \delta t)^2 - r^2}{\delta t} = - 2r v \sin(\frac{\pi}{n}) \Rightarrow \dot{r} = - v \sin (\frac{\pi}{n})\). We can also observe that \(\tan (\theta(t + \delta t) - \theta(t)) = \frac{v \delta t \sin (\frac{\pi}{2} - \frac{\pi}{n})}{r - v \delta t \cos (\frac{\pi}{2} - \frac{\pi}{n})} \Rightarrow \dot{\theta} = \frac{v}{r} \cos (\frac{\pi}{n})\). Combining these, we can see that \(\frac{\d r}{\d \theta} = - r \tan (\frac{\pi}{n}) \Rightarrow r = \lambda e^{-\tan(\frac{\pi}{n})t}\) where \(\lambda\) is the initial radius of the circumscribed circle.

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