Help with something physics teacher glossed over when deriving Kepler's Laws. While driving Kepler's Laws my physics teacher was talking about a position vector that points to an object in space orbiting another much more massive object. As he was going through his calculations, he said the following was true. 
$$\vec r \cdot \frac{d\vec r}{dt}= |\vec r|\left|\frac{d\vec r}{dt}\right|.$$
If a position vector is pointing to an object that is orbiting in an elliptical or a circular orbit, then the velocity vector must be tangential to position vector (maybe this is where I'm wrong?). If that is true then 
$$\vec r \cdot \frac{d\vec r}{dt}=|\vec r|\left|\frac{d\vec r}{dt}\right|\cos(90^\circ)=0.$$
I guess I just don't understand how that could be true for a position and velocity vector of an orbiting object? I wish I could track you through his entire train of thought but it took him about 30 minutes and he used vector calculus which I haven't taken yet so I kind of got lost in the math. Hopefully this is enough to go on.
 A: There may be some notational confusion between $\displaystyle\left|\frac{d\vec r}{dt}\right|$ and $\displaystyle\frac{d |\vec r|}{dt}$ which are not the same thing.
It is not always true that
$$\vec r \cdot \frac{d\vec r}{dt}= |\vec r|\left|\frac{d\vec r}{dt}\right|$$
However, it is true that
$$\vec r \cdot \frac{d\vec r}{dt}= |\vec r|\frac{d|\vec r|}{dt}$$
To see this, note that  the position vector is $\vec r = x \vec i + y \vec j + z \vec k$, in terms of Cartesian components $x$, $y$ and $z$ and  the constant Cartesian basis vectors $\vec i$, $\vec j $, and $\vec k$. The only "vector calculus" you need is that
$$\frac{d \vec r}{dt} = \frac{dx}{dt} \vec i + \frac{dy}{dt} \vec j + \frac{dz}{dt} \vec k$$
Since the magnitude of the position vector is $|\vec r| = \sqrt{x^2 + y^2 + z^2}$ it follows from ordinary calculus that
$$\frac{d|\vec r|}{dt} = \frac{1}{2\sqrt{x^2 + y^2 +z^2}}\left(2x \frac{dx}{dt} + 2y \frac{dy}{dt} + 2z \frac{dz}{dt} \right) = \frac{1}{|\vec r|}\left(x \frac{dx}{dt} + y \frac{dy}{dt} + z \frac{dz}{dt} \right) \\ =  \frac{1}{|\vec r|} \vec r \cdot \frac{d \vec  r}{dt},$$
and multiplying both sides by $|\vec r|$ we get $\displaystyle \vec r \cdot \frac{d\vec r}{dt}= |\vec r|\frac{d|\vec r|}{dt}$.
Alternatively using more "vector calculus":
We have $\vec r \cdot \vec r = |\vec r|^2$ and
$$2|\vec r| \frac{d |\vec r|}{dt}= \frac{d |\vec r|^2}{dt} = \frac{d}{dt} (\vec r \cdot \vec r) = \vec r \cdot \frac{d \vec r}{dt} +   \frac{d \vec r}{dt}\cdot\vec r = 2 \vec r \cdot \frac{d \vec r}{dt} $$
