Transformation to polar coordinates I know this is very simple and I'm missing something trivial here...
I'm having trouble converting this set of equations to polar form:
$$
\dot{x_1}=x_2-x_1 (x_1^2+x_2^2-1)\\
\dot{x_2}=-x_1-x_2 (x_1^2+x_2^2-1)
$$
where
$$
r= (x_1^2+x_2^2)^{1/2}\\
\theta=\arctan\left(\frac{x_2}{x_1}\right)
$$
The book I'm going through has these converted to the following equations:
$$
\dot{r}=-r(r^2-1)\\
\dot{\theta}=-1
$$
Here are the steps I've taken...
$$
\frac{dr}{dt}=(x_1\dot{x_1}+x_2\dot{x_2})(x_1^2+x_2^2)^{-1/2}\\
\dot{x_1}+\dot{x_2}=x_2-x_1-(x_1+x_2)(x_1^2+x_2^2-1)\\
\dot{x_1}+\dot{x_2}=x_2-x_1-(x_1+x_2)(r^2-1)
$$
Now I'm not sure what the next step to take would be... I've tried a few things and none of them got me to the correct result. Any help would be appreciated! :)
 A: Instead of directly finding $\dot{r}$, I think it might be helpful to first notice that
$$
x_1=r\cos \theta, x_2=r\sin \theta
$$
Differentiate above with respect to $t$, you get
$$
\dot{x_1}=\dot{r} \cos \theta - r\dot{\theta}\sin \theta 
\quad
\dot{x_2}=\dot{r} \sin \theta + r\dot{\theta}\cos \theta 
$$
How would you get $\dot{r}$ from above two equations? One simple way is to do as follows:
$$
\dot{x_1} \cos \theta + \dot{x_2} \sin \theta = \dot{r} \left( \cos^2 \theta + \sin^2 \theta \right)=\dot{r}
$$
so you basically eliminate $\dot{\theta}$ term. Similarly, you get
$$
\dot{x_2} \cos \theta - \dot{x_1} \sin \theta = \dot{\theta} r
$$
and hopefully you can carry on from here: just substitute $\dot{x_1},\dot{x_2}$ to equations, and bear in mind that you have $x_1=r\cos \theta, x_2=r\sin \theta$.
A: Since $r^2=x^2+y^2, \;\;\;2r\dot{r}=2x\dot{x}+2y\dot{y}=2x[y-x(r^2-1)]+2y[-x-y(r^2-1)]$
$\hspace{1.7 in}=-2x^2(r^2-1)-2y^2(r^2-1)=-2r^2(r^2-1)$, 
$\hspace{.4 in}$so $\;\;\dot{r}=-r(r^2-1)$
Since  $\displaystyle\tan\theta=\frac{y}{x}, \;\;(\sec^{2}\theta)\;\dot{\theta}=\frac{x\dot{y}-y\dot{x}}{x^2}$ so
$\displaystyle\;\;\dot{\theta}=\frac{x\dot{y}-y\dot{x}}{(1+\tan^{2}\theta)(x^2)}=\frac{x[-x-y(r^2-1)]-y[y-x(r^2-1)]}{x^2+y^2}=\frac{-x^2-y^2}{x^2+y^2}=-1$
