Why does $\frac{1}{r}\frac{dr}{d\theta} = \cot \psi$?

In the discussion of linear fractional equations in Birkhoff and Rota's Ordinary Differential Equations, the authors assert that if we convert a DE of the form $y' = F\left(\frac{y}{x}\right)$ to polar coordinates, then we have \begin{align} \frac{1}{r}\frac{dr}{d\theta} = \cot \psi, \end{align} where $\psi = \gamma - \theta$, with $\gamma$ being the tangent direction and $\theta$ the radial direction. I'm afraid this has me entirely buffaloed -- why on earth is this true? I have neither an analytic nor geometric intuition as to how this could possibly be. I'm sure there's some elementaryish fact about $\frac{dr}{d\theta}$ which makes the answer obvious, but I have no clue what said fact is.

I do note that \begin{align} \frac{dy}{d\theta} &= r\cos\theta + \frac{dr}{d\theta}\sin\theta\Rightarrow\\ \frac{dr}{d\theta} & = \frac{dy}{d\theta}\csc\theta - r\cot\theta\Rightarrow\\ \frac{1}{r}\frac{dr}{d\theta} & = \frac{1}{y}\frac{dy}{d\theta} - \cot\theta, \end{align} but this is as close as I can come to getting a cotangent anywhere near the expression (and, of course, $\theta \neq \psi$ in general).

What gives?

Solution 1:

The prove of the relationship $\frac{1}{r}\frac{dr}{d\theta} = \cot (\gamma -\theta )$ is shown below :

On the figure, the relationship is easy to intuitively understand in the triangle PQT where PQ$=dr$ , QT$=r d\theta$ and angle$=(\gamma-\theta)$, of course infinitely magnified.