How to "vertically" scale an arbitrary polar equation? Given an arbitrary polar equation $r = P(\theta)$, which represents some graph, when plotted, is it possible to write a new $r = Q(\theta) = f(P(g(\theta)))$, such that plotting $Q(\theta)$ produces a new graph, which is exactly the same as $P(\theta)$, except that it is scaled along one axis?
 A: No, not in the way you describe. Your $f$ takes radial values as inputs, and it cannot distinguish between a radial value when $\theta$ is "horizontal" (a multiple of $\pi$) and when it is "vertical". For example, if $P$ is constant (so the graph is a circle) then $g$ is irrelevant, and $f$ would have no way to stretch vertically without also stretching horizontally.
If you want to scale vertically by $k$, then first assume we have an "old" point (before the stretch transformation) where the angle is in $[0,\pi/2)$. And after the stretch, it is at a "new" location, but still it would have angle in $[0,\pi/2)$.
$$\begin{align}
y_{\text{new}}&=k\cdot y_{\text{old}}&x_{\text{new}}=x_{\text{old}}\\
r_{\text{new}}\sin\left(\theta_{\text{new}}\right)&=k\cdot r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}\sin\left(\arctan\frac{y_{\text{new}}}{x_{\text{new}}}\right)&=k\cdot r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}\frac{\frac{y_{\text{new}}}{x_{\text{new}}}}{\sqrt{1+\frac{y_{\text{new}^2}}{x_{\text{new}^2}}}}&=k\cdot r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}\frac{k\frac{y_{\text{old}}}{x_{\text{old}}}}{\sqrt{1+k^2\frac{y_{\text{old}^2}}{x_{\text{old}^2}}}}&=k\cdot r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}\frac{\tan\left(\theta_{\text{old}}\right)}{\sqrt{1+k^2\tan^2\left(\theta_{\text{old}}\right)}}&= r_{\text{old}}\sin\left(\theta_{\text{old}}\right)\\
r_{\text{new}}&=P\left(\theta_{\text{old}}\right)\cos\left(\theta_{\text{old}}\right)\sqrt{1+k^2\tan^2\left(\theta_{\text{old}}\right)}\\
r_{\text{new}}&=P\left(\theta_{\text{old}}\right)\sqrt{\cos^2\left(\theta_{\text{old}}\right)+k^2\sin^2\left(\theta_{\text{old}}\right)}\\
\end{align}$$
So multiplying $P(\theta)$ by $\sqrt{\cos^2\left(\theta\right)+k^2\sin^2\left(\theta\right)}$ (not composing by some function) will have the desired effect.

To get a plottable curve for the full demonstration:
$$\begin{align}
\theta_{\text{old}}&=\arctan\left(\frac{y_{\text{old}}}{x_{\text{old}}}\right)\\
&=\arctan\left(\frac{y_{\text{new}}}{kx_{\text{new}}}\right)\\
&=\arctan\left(\frac{\tan\left(\theta_{\text{new}}\right)}{k}\right)\\
\end{align}$$
Compose this into the above for something like $r=P(g(\theta))\cdot f(\theta)$. Here is what it looks like in a Desmos plot.
I'll leave it to you to explore what happens for a point with angle outside of $[0,\pi/2)$. The details with $\arctan$ and the sign on trig functions need modifications.
