Slope of the Orthogonal Trajectory in Polar Coordinates (Versus the Slope of the Orthogonal Trajectory in the $xy$-plane)

Question

When finding the orthogonal trajectories of a family of curves in the $xy$ plane, we do the following:

1. Differentiate the equation of the family of curves with respect to the independent variables, which gives us the slope of the family of curves;
2. Eliminate the parameter (if it didn't already get eliminated during differentiation);
3. Rearrange to get the form $\dfrac{dy}{dx} = f(x, y)$;
4. Take the negative reciprocal of $\dfrac{dy}{dx} \implies \dfrac{-dx}{dy} = f(x, y)$, which gives us the slope of the orthogonal trajectories;
5. Solve the differential equation using separation of variables or some other method. This gives us the equation of the orthogonal trajectories.

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In my previous (related) question, I mentioned a peculiar problem in my textbook, "Differential Equations with Applications and Historical Notes, 3rd edition", by Simmons and Finlay, where the authors used polar coordinates to solve the orthogonal trajectories problem. In the aforementioned question, I discovered that the reason for my confusion was because the slope of the orthogonal trajectories in polar form is different to the slope of the orthogonal trajectories in the xy plane: The slope of the orthogonal trajectories in the xy plane, as previously mentioned, is the negative reciprocal of $\dfrac{dy}{dx} \implies \dfrac{-dx}{dy} = f(x, y)$, whilst the slope of the orthogonal trajectories in polar form is the negative reciprocal of $\dfrac{1}{r}\dfrac{\mathrm dr}{\mathrm d\theta} \implies -r\dfrac{\mathrm d\theta}{\mathrm dr} = f(r, \theta)$. In other words, and more generally, we can see that the way in which we get the slope of the orthogonal (normal) trajectories (vectors) in polar form is different from that in the $xy$-plane.

My original confusion stemmed from this fact that, with polar coordinates, we do not only take the negative reciprocal of the operator $\dfrac{\mathrm dr}{\mathrm d\theta}$, but also the negative reciprocal of $\dfrac{1}{r}$ along with it. This difference was not mentioned in the textbook; the only case that was mentioned was that of dealing with the operator $\dfrac{dy}{dx} = f(x, y)$ -- coordinates of the xy-plane.

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I would greatly appreciate it if people could please take the time to post a step-by-step proof that clearly shows that, unlike the orthogonal trajectory in the $xy$-plane, the orthogonal trajectory in polar form is found by taking the negative reciprocal of $\dfrac{dr}{rd\theta} \implies -r\dfrac{\mathrm d\theta}{\mathrm dr} = f(r, \theta)$. My goal is to convince myself that this is true -- that the way we get the slope for the orthogonal trajectories is different between equations using coordinates in the xy-plane $\left( \dfrac{dy}{dx} \implies \dfrac{-dx}{dy} = f(x, y) \right)$ and those using coordinates in polar form $\left( \dfrac{1}{r}\dfrac{\mathrm dr}{\mathrm d\theta}\ \implies -r\dfrac{\mathrm d\theta}{\mathrm dr} = f(r, \theta) \right)$.

Answer 1

Suppose we have two curves that cross at right angles. We represent them in polar coordinates as functions $r_1$ and $r_2$ from angle to radius. Suppose the crossing happens at $(r, \theta)$, so that $r_1(\theta) = r_2(\theta) = r$.

For $i \in {1, 2}$, put $x_i(\theta) = r_i(\theta)\cos(\theta)$ and $y_i(\theta) = r_i(\theta)\sin(\theta)$. Then $\frac{dx_i}{d\theta} = \frac{dr_i}{d\theta}\cos(\theta) - r_i\sin(\theta)$, and $\frac{dy_i}{d\theta} = \frac{dr_i}{d\theta}\sin(\theta) + r_i\cos(\theta)$.

The curves cross at right angles, so we have $\frac{dx_1}{d\theta}\frac{dx_2}{d\theta} + \frac{dy_1}{d\theta}\frac{dy_2}{d\theta} = 0$. (This can be deduced from $\frac{dy_1}{dx_1} = -\frac{dx_2}{dy_2}$ when they are finite, but it is more general). Substitute:

$$ \left(\frac{dr_1}{d\theta}\cos(\theta)-r\sin(\theta)\right) \left(\frac{dr_2}{d\theta}\cos(\theta)-r\sin(\theta)\right) + \left(\frac{dr_1}{d\theta}\sin(\theta)+r\cos(\theta)\right) \left(\frac{dr_2}{d\theta}\sin(\theta)+r\cos(\theta)\right) = 0 $$

Multiply out the brackets, cancel the terms in $\cos(\theta)\sin(\theta)$, and use $\cos^2(\theta) + \sin^2(\theta) = 1$:

$$\frac{dr_1}{d\theta}\frac{dr_2}{d\theta} + r^2 = 0$$

Divide through by $r^2$ and subtract 1:

$$\left(\frac1r\frac{dr_1}{d\theta}\right) \left(\frac1r\frac{dr_2}{d\theta}\right) = -1$$

Answer 2

simple:)

Consider a curve in polar coordinate. The angle $\psi$ between the radial and tangent directions is given by $$\tan(\psi) = \frac{r \ d\theta}{dr}$$

Consider the curve with angle $\psi_1$. The curve that intersects it orthogonally has angle $\psi_2 = \psi_1 + \frac{\pi}{2}$. Now $$\tan(\psi_2) = \frac{−1}{\tan{\psi_1}}$$ Thus, at the point of orthogonal intersection, the value of $$\frac{r \ d\theta}{dr}$$