Differential Equation Math Puzzle Dog race:
Edit 2:
I posted a possible answer below. However, I am unsure how the authors arrived at the solution.
Maybe someone can offer an explanation.
Four dogs are positioned at the corners of a square ($d= 1m$), chase each other in clockwise direction with the same constant speed . As their target is moving, they will follow a curved path, eventually colliding in the center of the square.

(a) Why is the total length of the path just $1 m$?
(b) Find and solve a differential equation for the radius $r(\theta)$ in polar coordinates.
This is a homework question! I just want hints, no solutions please.
The tough part is setting up an equation for the radius i.e for the motion of one of the dogs. I was thinking about an equation similar to that of an archimedean spiral.
$$r(\theta)=a+b\theta \space \space \space \text{or} \space \space \space r(\theta)=a\theta^{\frac{1}{n}}$$
However, I have no idea what values $a$ or $b$ should be. Any hints are appreciated.
Edit 1:
This is my second attempt at a solution:
If Dog 1 is positioned at $(r, \theta)$ $\implies$ Dog 2 is positioned at $(r, \theta+\frac{\pi}{2})$ 
Picture: 

$$x_1=r \cos (\theta) \\ y_1= r \sin (\theta) \\ \\ \\ x_2=r \cos (\theta+\frac{\pi}{2})=-r \sin(\theta) \\y_2=r \sin (\theta+\frac{\pi}{2})=r \cos (\theta)$$
If these are the two position vectors then the vector joining the two points is my velocity vector.
$$\implies \frac{dy}{dx}=\frac{y_2-y_1}{x_2-x_1}=\frac{r \sin (\theta+\frac{\pi}{2})-r \sin (\theta)}{r \cos (\theta+\frac{\pi}{2})-r \cos (\theta)}=\frac{ \sin (\theta+\frac{\pi}{2})- \sin (\theta)}{ \cos (\theta+\frac{\pi}{2})- \cos (\theta)}=\frac{\cos(\theta)-\sin(\theta)}{-  \sin(\theta)-\cos(\theta)}$$
Am I on the right path? But How do I deduce $\dfrac{dr}{d\theta}$?
 A: 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.  
A: 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.
A: First off: I don't know how to do this problem the way they are asking. But I think I can do this problem my own way. Maybe this will be a hint enough for you to do it the original way.
Let's track the top dog (starts at $(0,1)$). Let its position at time $t$ be $(x(t),y(t))$. By symmetry, the dog it is chasing is at position
$$(u(t),v(t)) = (\frac{x(t) + y(t)}{\sqrt{2}},\frac{-x(t) + y(t)}{\sqrt{2}})$$ which is just the original position rotated by 45 degrees.
The speed of chase is a constant $s$ but the direction is from $(x(t),y(t))$ to $(u(t),v(t))$ so the velocity vector is 
$$(\dot{x}(t),\dot{y}(t)) = s\frac{(u(t)-x(t),v(t)-y(t))}{|(u(t)-x(t),v(t)-y(t))|}$$
Can you change to polar coordinantes and take it from there?
A: This answer is taken from Calculus 10th Edition by Larson and Edwards
Link to solution: Solution
This is how they derive the differential equation:
If a dog is located at $(r, \theta)$ in the first quadrant, then its neighbor is at $(r, \theta+ \frac{\pi}{2})$
$$(x_1,y_1)=(r \cos\theta, r \sin\theta)
\\(x_2,y_2)=(-r\sin\theta,r\cos\theta)$$
The slope joining these points is
$$\frac{r\cos\theta-r\sin\theta}{-r\sin\theta-r\cos\theta}=\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\text{slope of tangent line at} (r,\theta)$$
$$\color{}{\frac{dy}{dx}=\frac{\frac{dy}{dr}}{\frac{dx}{dr}}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}=\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}}$$
$$\color{blue}{\implies \frac{dr}{d\theta}=-r}$$
