Translating Polar Functions

Question

How do I rewrite a polar function (expressed in a polar coordinate system $r = F(\theta)$ so the entire curve is shifted right or left $h$ units and up or down $k$ units?

Answer 1

Thank you for taking the time to address my question. My goal is to have the function rewritten with respect to the polar system so that $r$ values are mapped to $\theta$ values, but to shift the curve as if the Pole were at $(p_x, p_y)$ instead of $(0, 0)$. Although I suppose I could have a process where the function were rewitten as if the Pole were at $(r_p, \theta_p)$ with $\theta_p=\arctan{\frac{p_y}{p_x}}$ and $r_p=\sqrt{p_x^2 + p_y^2}$.

I took the previous suggestion and tried focusing on the specific case of a circle and after some further messing around I was able to get this to work for a polar circle function, rewriting $r = a$ (where $a$ is the radius) as $$r = r_p \cos(\theta - \theta_p) + \sqrt{a^2 - r_p^2 \sin(\theta - \theta_p)^2}$$ leading to $$r = \sqrt{p_x^2 + p_y^2} \cos\left(\theta - \arctan\frac{p_y}{p_x}\right) + \sqrt{a^2 - (p_x^2 + p_y^2) \sin \left(\theta - \arctan\frac{p_y}{p_x}\right)^2}.$$ This gave me a function with terms containing constants $p_x, p_y$ that I could adjust to make a Cartesian like translation of a circle rendered in the polar plane. But when I tried to reproduce this process for something like a Leminscate, $r = \pm \sqrt{a^2 \cos(n \theta)}$, I got stuck again and kept distorting the curve.

Answer 2

A simple trick is to get $r=F(\theta)$ in parametric equations, and then apply $\sqrt{x^2+y^2}=r$.

If you move the function to right in $\Delta x$ and to up in $\Delta y$, then $$ \begin{cases} x_{new}(\theta)=r\cos(\theta)-\Delta x\\ y_{new}(\theta)=r\sin(\theta)-\Delta y \end{cases} $$ Then $r_{new}$ can be expressed as $$ r_{new}=\sqrt{(r\cos(\theta)-\Delta x)^2+(r\sin(\theta)-\Delta y)^2} $$

Answer 3

If you are translating the pole by $\Delta x$ + $\Delta y$, it's simply:

$$ r_{new}= r + \Delta x.\cos(\theta) + \Delta y.\sin(\theta) $$