System of equations, limit points This is a worked out example in my book, but I am having a little trouble understanding it:
Consider the system of equations:
$$x'=y+x(1-x^2-y^2)$$
$$y'=-x+y(1-x^2-y^2)$$
The orbits and limit sets of this example can be easily determined by using polar coordinates. (My question: what is the motivation for that thinking? What should clue me in to thinking that I should use polar coordinates?)
The polar coordinate satisfies $r^2=x^2+y^2$ so by differentiating with respect to t and using the differential equations we get:
$r\cdot r'=x\cdot x'+y\cdot y'$ (I am unclear about how the book even got this first equation from $r^2=x^2+y^2$) 
$=x\cdot y+x^2(1-r^2)-x\cdot y+y^2(1-r^2)$ Substitute in $x'$ and $y'$ and then multiple out and replace with $r$, I get this step
$=r^2(1-r^2)$ cancel terms, I get this step too
$r'=r(1-r)$
similarly, the angle variable $\theta$ satisfies $\tan\theta=\frac yx$, so the derivative with respect to $t$ yields $\sec^2(\theta)\theta'=x^{-2}[x^2+xy(1-r^2)-y^2-xy(1-r^2)]=-\frac{r^2}{x^2}$ so $\theta=1$
Thus the solution goes clockwise around the origin at unit angular speed.
I don't understand the $\theta$ step at all or how they reached the conclusion of clockwise around the origin with unit angular speed..
But then it just jumps to saying "the origin is a fixed point, so α(0)=ω(0)={0} but I have no idea how they reached this conclusion..

When you see an $x^2 + y^2$ in a problem that is probably going to be tractable, one good first thing to try is to see how the problem looks transformed to polar coordinates.  (Even in a problem that comes about in real life, where you don't know the solution will be possible to obtain, this is a good first shot.)
When $r^2 = x^2 + y^2$ it is valid to take the derivative of each side with respect to $t$.
For example, 
$${d \over dt} (r^2) = 2r {dr \over dt} = 2rr'
$$
The book got the equation you present by doing this on both sides of the equation, and dividing by 2.
I see another answer and that you are unclear on how $\tan \theta$ comes in.  That is easy:  In polar coordinates, $y/x = \tan \theta$.  Draw yourself a right triangle with one point at the origin, the right angle on the X axis and the third point at $(x,y)$ and ask yourself what $\tan \theta$ would be in that picture.
By the way, the solution that approaches $r=1$ is in fact a stable orbit, since if $r = 1+ \epsilon$, $r' = -\epsilon r$ so the solution will approach $r=1$ in an exponential fachion.
Then you have seen how to get from that to 
$$
r' = r(1-r) $$
The first thing to notice about that equation is that considering $r$ alone (without $\theta$) it will have fixed points wherever $r' = 0$, that is, at $r=0$ and at $r=1$.  But then $r=0$ is a fixed point of the whole system since at $r=0$, $\theta$ is irrelevant.
However, that fixed point is an unstable fixed point, since near the origin with $r(t=0) = \epsilon$, the solution behaves like $r = \epsilon e^t$, exploding away from the origin.
