Differential Equation Math Puzzle

Solution 1:

For part (a), you don't have to parameterize anything, really, or set up any complicated differential equations. All you have to do is show that, thinking of yourself as one of the dogs, the dog you're chasing is always running at right angles to your line of sight and therefore, since your speed is constant and directed straight at the target, you are getting closer at a constant rate -- in short, as far as the time to overtake it is concerned, the dog you're chasing may as well not be moving at all, which is to say you'll catch it in the time it takes to travel $1$ meter.

Solution 2:

Here is a solution for (b):

Let $$t\mapsto z(t)=r(t)e^{i\phi(t)}\qquad(t\geq0)$$ be the orbit of the dog starting at ${1\over2}(1+i)$. Then the orbit of the dog starting at ${1\over2}(-1+i)$ is simply $t\mapsto iz(t)$. It follows that at any moment the velocity vector $$\dot z=(\dot r+i r\dot\phi)e^{i\phi}$$ is parallel to $$iz-z=(-1+i)z=(-1+i)r e^{i\phi}\ .$$ This amounts to $$\dot r+ir\dot\phi=\lambda(-r+ir)\tag{1}$$ for some $\lambda>0$ changing with time. Comparing the imaginary parts in $(1)$ we see that in fact $\lambda=\dot\phi>0$, so that from looking at the real parts we obtain $\dot r=-r\dot\phi$, or $${dr\over d\phi}={\dot r\over\dot\phi}=-r\ .$$ It follows that the function $\phi\mapsto r(\phi)$ satisfies the differential equation $$r'=-r\ ,$$ with the solutions $r(\phi)=r_0 e^{-(\phi-\phi_0)}$. This implies that we see four logarithmic spirals.