Translate linear motion to polar coordinates I'm a math noob, so even though I want to understand a general solution to my problem, I don't even know how to articulate it except as an example.  So:
The setup:
Suppose I want to draw a segment from (1,2) to (0,1) on the Cartesian plane.
Suppose I get to do so by tracing a vector along said segment.  That is, I have a stepper motor at (0,0) that rotates about the z axis, and a pen parallel to the z axis, tip facing the xy plane, attached to the tip of a rod that can extend linearly from the axle of the stepper motor.  Effectively, I can represent arbitrary vectors from the polar origin (ignoring length limitations, lets assume I can manage sqrt(5).
My problem is thus:
If I run the stepper motor at constant velocity, this represents constant angular velocity.  This is not a problem, except I would need to know how to compensate for this by moving the pen tip (vector magnitude) such that the end result is a smooth -y-ward motion.  I have no idea where to begin.  I would even appreciate answers that point me toward being able to ask this question more succinctly.  

The transformation from Cartesian coordinates to polar coordinates is
$$
x = r\cos\theta, \quad y=r\sin\theta.
$$
Since the equation of the line on which your two points lie is $y=x+1$, its polar equation is
$$
r\sin\theta=r\cos\theta+1,
$$
which you can solve for $r$ to obtain
$$
r = \frac1{\sin\theta - \cos\theta}.
$$
That tells you what you want the radial position of the pen to be, given the current angle of the stepper motor.
If the constant angular velocity, which is $\frac{d\theta}{dt}$, is equal to $a$ (make sure you use radians per second, not degrees per second!), then the required velocity of your pen tip in the radial direction should then be
$$
\frac{dr}{dt} = \frac{dr}{d\theta} \frac{d\theta}{dt} = \frac{\sin\theta+\cos\theta}{\sin2\theta-1} \cdot a.
$$
