MrBillyShears said:
Let me say from the beginning I'm not talking about the non-coordinate unit vectors for polar coordinates. I'm talking about basis vectors.
I don't understand what you're saying here. "Non-coordinate unit vectors" sounds like something that has nothing to do with the coordinate system. "for polar coordinates" sounds like the exact opposite. Then the second sentence suggests that the vectors you were talking about in the first sentence aren't basis vectors. Every linearly independent set with two vectors is a basis.
I'm still assuming that you're talking about these guys:
\begin{align}
&\hat r=(\cos\varphi,\sin\varphi)\\
&\hat\varphi =(-\sin\varphi,\cos\varphi)
\end{align} For all ##r,\varphi##, the vectors above are the ones that the polar coordidinate system associates with the tangent space at the point ##(r\cos\varphi,r\sin\varphi)##. Since that tangent space is an identical copy of the vector space that your ##\vec r## is an element of, you can also use these vectors as a basis for that space.
MrBillyShears said:
how does one use these basis vectors in order to describe a vector?
Same way you use any other basis.
MrBillyShears said:
I know they are different at every point, so which point do you use?
A point on a particle's trajectory at which you intend to calculate something, like that particle's centripetal acceleration.
MrBillyShears said:
Why is there different basis vectors at every point?
They're defined by
\begin{align}
\hat r=\frac{\frac{d\vec r}{dr}}{\left|\frac{d\vec r}{dr}\right|},\qquad \hat\varphi=\frac{\frac{d\vec r}{d\varphi}}{\left|\frac{d\vec r}{d\varphi}\right|}
\end{align} The right-hand sides have different values at each point. If you don't find this useful, then you can use some other basis. But you will certainly find these bases useful when you describe circular motion, where the particle's position and velocity are respectively ##r\hat r## and ##r\dot\varphi\hat\varphi##, at
every point.
MrBillyShears said:
if you pick \vec{r} to describe your vectors, would the "tails" of your vectors come from the origin, or from your point \vec{r}?
The position vector should be drawn from the origin, but the velocity vector from the particle's location. It makes sense to think of the position vector ##\vec r## as an element of your original copy of ##\mathbb R^2## ("the configuration space"), and the velocity vector as an element of a different copy of ##\mathbb R^2## ("the tangent space at ##\vec r##") with its origin attached to the point ##\vec r##. When you're dealing with a significantly less trivial manifold than ##\mathbb R^2##, you pretty much
have to think this way.