Polar radius and position vector: two-dimensional kinematics for high school students We consider for example this image,

which is a polar graph of a naval unit's on-board instrumentation with vector radius (or polar radius) $\rho$ and anomaly $\theta$ or polar angle.
We know that a point $P=(x,y)$ in an orthogonal Cartesian coordinate system may be identified in a polar diagram with coordinates $P\equiv(\rho,\theta)$ or viceversa.


If we consider the trajectory $\Gamma$ (the curve coloured in brown) of a target and $\mathbf r=\mathbf r(t)$ is your position vector is it possible to say that there is an analogy between the position vector $\mathbf r$ and the polar radius $\rho$? Or are the two quantities distinct because the first is a vector and the polar radius is a scalar?

 A: The position vector $\boldsymbol{r}(t)$ is the parametrization of the curve it rides on. But converting this to a polar form $\rho(\theta)$ is also a parameterization of the same curve.
For example an ellipse can be parameteized with
$$ \boldsymbol{r}(t) = \pmatrix{x(t) \\ y(t)} = \pmatrix{ a \cos t \\ b \sin t}  \tag{1}$$
This position vector objeys the equation of the ellipse $$ \left( \tfrac{x}{a} \right)^2 + \left( \tfrac{y}{b} \right)^2 = 1$$ where $a$ is the semi-major axis, and $b$ the semi-minor axis.
Now consider the polar coordinates
$$ \boldsymbol{r}(t) = \pmatrix{x(t) \\ y(t)} = \pmatrix{ \rho \cos \theta \\ \rho \sin \theta} $$
that yields the solution
$$ \rho(\theta) = \frac{ a b}{\sqrt{ a^2 - (a^2-b^2) \cos^2 \theta}} \tag{2}$$
Expressions (1) and (2) are equivalent to each other since both describe the same ellipse.
A: The polar radius is a scalar quantity telling you how far you are from the origin.
The position vector is a vector hence has two pieces of information, magnitude (the polar radius) and direction (the angle). Since you are in 2D, you need two coordinates to unequivocally determine your position: you need the vector.
