Understanding shift to polar coordinates in the newtonian central force system of ODE's This is from Hirsch, Smale and Devaney chapter 13. The larger context is moving towards blowing up the singularity at the origin of the system.
The second order ODE is defined, $X:t\rightarrow \mathbb{R}^{3}$ as:
$X''=\frac{X}{|X|^{3}}=F(X)$, with the central force field. Or trading order for dimension:
$X'=V$, $V'=\frac{X}{|X|^{3}}$.
My book then ignores a dimension of movement (or of the configuration space) and just looks at the particles moving in the plane of $\mathbb{R}^2$, yielding a phase space in $\mathbb{R}^4$, composed of the configuration space of 2d position and tangent space of velocities. 
They then define the energy function on the phase space as: 
$E(V,X)=\frac{1}{2}|V|^2+\frac{1}{|X|}$, and restrict attention to negative values of $E=h<0$ (I think for geometric reasons relating to wanting the level sets of this function to be toroidal, but I am not sure). 
Finally, they introduce the polar change of coordinates in configuration space, for the new variables $(v_r,v_{\theta})$ in the tangent planes as:
$V=v_r\begin{bmatrix}\cos(\theta)\\\sin(\theta) \end{bmatrix}+
v_{\theta}\begin{bmatrix}-\sin(\theta)\\\cos(\theta) \end{bmatrix}$
Where did this come from? Why is the second vector the derivative of the first? 
The book then uses the identity $X'=V$ to conclude that
$X'=r'\begin{bmatrix}\cos(\theta)\\\sin(\theta) \end{bmatrix}+
r\theta'\begin{bmatrix}-\sin(\theta)\\\cos(\theta) \end{bmatrix}$ and so $v_r=\theta'$ which is clear, but then also $r\theta'$, which I do not understand, shouldn't this also be $\theta'$?
I can provide further context and more of the development in the book, but I am hoping understanding the above will help with the later parts. Also, the derivation of the energy function is a bit of a blackbox to me, as is ignoring a dimension of position, so any clarification on this would be appreciated.
 A: It's just a coincidence that the second vector is the derivative of the first with respect to $\theta$. These vectors are the basis vectors $e_r$ and $e_\theta$ of the orthonormal basis of the polar coordinate system, so this is $V=v_re_r+v_\theta e_\theta$. You can derive these basis vectors by differentiating the position $r\pmatrix{\cos\theta\\\sin\theta}$ with respect to $r$ and $\theta$ and then normalising.
Regarding your other questions:
The energy is the sum of kinetic energy and potential energy; the mass in the kinetic energy has been set to $1$, and the potential energy is the line integral of the force $X/|X|^3$ from a point at infinity, which you can verify by taking its gradient to recover the force.
That the problem can be reduced to two dimensions is due to the fact that there is no force that would accelerate the body outside the plane spanned by the origin, the original position and the original velocity. You can also see this in the context of preservation of angular momentum: The angular momentum is perpendicular to this plane, and there is no force that could change it, so the movement stays within that plane. Then it's just a matter of convenience to choose that plane to be some copy of $\mathbb R^2$ embedded in $\mathbb R^3$.
