Riddler Solution for August 19th, 2016

Let \(y(t)\) be the location of the farmer, \(r(t)\) be the location of the ram at time \(t\). Fixing the farmer's speed at unity let \(v\) be the speed of the ram. By considering the case where the ram heads directly for the exit we see that \(v\) must be at least \(\sqrt 2\) if it is to have any chance at catching the farmer.

The constraint that the ram always moves towards the farmer means that the ram's velocity \(\dot{r}\) is:

\[ \dot{r} = v \frac{ y-r }{ \| y - r\| } \]

Re-arranging terms and taking the derivatives of both sides yields:

\[ \frac{d}{dt} \Big\{\|y-r\| \dot{r} \Big\} = v(\dot{y} - \dot{r}) \]

and after applying the chain rule: \[ \Big(\frac{d}{dt}\|y-r\|\Big)\dot{r} + \|y-r\|\ddot{r} = v(\dot{y} - \dot{r}) \] Since the norm of \(\dot{r}\) is constant, \(\ddot{r}\) and \(\dot{r}\) are perpendicular, so taking the inner product of the last equation with \(\dot{r}\) results in:

\[ \Big(\frac{d}{dt}\|y-r\|\Big)v^2 = v(\dot{y}\cdot\dot{r} - v^2) \]

Now integrate both sides from \(t = 0\) to \(t = 1\) and let \(v\) be the exact speed where the ram just catches the farmer, i.e. \(r(1) = y(1) = (1,1)\). \[ \begin{align} \int_0^1\frac{d}{dt} \|y-r\| dt & = \|y(1)-r(1)\| - \|y(0)-r(0)\| \\ & = 0 - 1 \\ & = -1 \\ \end{align} \]

\[ \begin{align} \int_0^1 \dot{y} \cdot \dot{r} dt & = \dot{y} \cdot \int_0^1 \dot{r} dt \\ & = \dot{y} \cdot (r(1) - r(0)) \\ & = 1 \\ \end{align} \] Substituting these results gives the following equation for \(v\): \[ \begin{align} -v^2 & = v(1-v^2) \\ -v & = 1 - v^2 \\ v^2 - v - 1 & = 0 \\ \end{align} \] Clearly \(v\) must be \(> 1\), so the only sensible solution is \(v = \frac{1+\sqrt{5}}{2}\).

Extra Credit Variation

Allow the farmer's initial position to vary within the paddock. The ram still enter from the southwest corner and the farmer still heads for the northeast corner.

Because \(\dot{y}\) is still constant the above analysis holds (with a different upper limit of integration), and the equation for the critical ram velocity becomes:

\[ \|A\|v^2 - Bv - \dot{y}\cdot(1,1) = 0 \] where \[ \begin{align} A &= (1,1) - y(0) \\ B &= \|y(0)\| \\ \dot{y} & = A / \|A\| \end{align} \] Solving this for an arbitrary \(y(0) = (a,b)\) results in the interesting formula: \[ v = \frac{ \sqrt{a^2 + b^2} + \sqrt{ (a-2)^2 + (b-2)^2 }}{2\sqrt{(a-1)^2+(b-1)^2}} \]

This suggests that we can define the problem for starting points outside of the paddock, and that the velocity is symmetric about the point \((1,1)\), i.e. \(v(a,b) = v(2-a,2-b)\).

It also indicates that the following times are the same:

  1. The time it takes the farmer to run to \((1,1)\) and back to his original position.
  2. The time it takes the ram to run to the farmer's starting position and then to \((2,2)\).