Why is the perpendicular got by means $\frac{dr}{rd\theta}\to -\frac{rd\theta}{dr}$ in polar form?

Question

At the current moment I'm reading a textbook, George F. Simmons' "Differential Equations with Applications and Historical Notes", and at first I had exactly the same question as was described here: Orthogonal Trajectories Using Polar Coordinates. Correct Calculations, Two Different Answers?

While that answer cleared some of my queries, this question bothers me;

In polar coordinates, why is the perpendicular is evaluated through: $\displaystyle{\frac{dr}{rd\theta}\to -\frac{rd\theta}{dr}}$ instead of $\displaystyle{\frac{dr}{d\theta}\to -\frac{d\theta}{dr}}$.

In my perspective, the differential equation of the family of original curves (in Cartesian coordinates) looks in the following manner:

If $\text{ }\displaystyle{\frac{dy}{dx} = f(x,y)}$, then the differential equation for the orthogonal trajectory of this family would

be, $\displaystyle{-\frac{dx}{dy} = f(x,y)}.$

Doing some research on my query, I found that answer. However, I did not understand the explanation.

I would really appreciate it if someone could explain the post in a more detailed way.

Answer 1

At some point $M(\theta_0)$ of a curve $\mathcal C$ given in polar coordinates by $r(\theta),$ the angle $V=(\overrightarrow{OM(\theta_0)},\overrightarrow{M'(\theta_0)})$ between the radial vector and the tangent is given by $$\tan V=\frac{r'(\theta_0)}{r(\theta_0)}.$$

Let $\rho(\theta)$ be a parametrization of some curve $\theta\mapsto N(\theta)$ normal to $\mathcal C$ at the same point, $N(\theta_0)=M(\theta_0).$ Similarly, the angle $W=(\overrightarrow{ON(\theta_0)},\overrightarrow{N'(\theta_0)})$ is given by $$\tan W=\frac{\rho'(\theta_0)}{\rho(\theta_0)}.$$

Now,$$W\equiv V+\frac\pi2\bmod\pi\implies\tan W=-\frac1{\tan V}\implies\frac{\rho'(\theta_0)}{\rho(\theta_0)}=-\frac{r(\theta_0)}{r'(\theta_0)}.$$