Four-Dogs Pursuit 
Four dogs start at the corners of square $ABCD$ (labelled anti-clockwise). Running anti-clockwise, the dog starting at $A$ pursues the dog starting at $B$, which pursues the dog starting at $C$, which pursues the dog starting at $D$, which pursues the dog starting at $A$. They run at the constant speed of $7$ meters per second and the sides of the square are $30$ meters long. The pursuit stops when at least one dog has reached the centre of the square.

*

*Use a system of ordinary differential equations to model the
trajectories of the dogs.

*Make a plot of the paths followed by the dogs.

*Determine how long the pursuit lasts.


 A: $(x_a(t), y_a(t))$ represents the position of dog a at time t
$(x_a', y_a') = 7 (\frac {x_b-x_a}{(x_b-x_a)^2 + (y_b-y_a)^2},\frac{y_b-y_a}{(x_b-x_a)^2 + (y_b-y_a)^2})$
This factor of the distance between the dogs influencing the velocity vector is going to be messing.  So, lets simplify this a bit. 
$(x_a', y_a') = (x_b-x_a,y_b-y_a)$ will give the same trajectory of the dogs, but they won't meet in the middle until $t = \infty.$  That's okay, we can worry about how to scale that later.
$x_a' = x_b-x_a\\
x_b' = x_c-x_b\\
x_c' = x_d-x_c\\
x_d' = x_a-x_d$
Now we have a system of ordinary diff eq. that shouldn't be too hard to solve
$\mathbf x' = \begin{bmatrix}
-1&1& 0&0\\
0&-1&1& 0\\
0&0&-1&1\\
1&0&0&-1\end{bmatrix}\mathbf x$
$\mathbf x = e^{At}\mathbf x_0$
Diagonalize the matrix, solve for $\mathbf x$ I am sure it will be something along the lines of $x_a = x_a(0) e^{-t} \cos t$
Do the same thing for $y_a$
This will give you the solutions to i)ii).
Ian Miller as an elegant solution to iii) above.  Before I had read it, I was thinking somethign along the lines of $\int_0^\infty \sqrt {x_a^{'2} +y_a^{'2}} dt$ would give you the length of the path,  and then you could divide that by the speed of the dog. 
A: It turns out that you don't need to use differential equations to solve part 3.  Consider the view of dog A.  Since the setup is symmetric, the angle between the the velocity vectors of each dog and their target dog is the same, since there are four dogs we see that the angle must remain at 2pi/4 = pi/2, i.e. dog B will always be running perpendicular to dog A, thus B's running will never alter the distance between A and B, the only thing decreasing the distance between A and B is the running of A.  Therefore, it will take 30/7 seconds.
