You'll have to excuse me if this questions is extremely trivial; it's been years since I went back to elementary calculus and I humbly accept that I haven't really gotten deep into polar equations/coordinates during my college years.
That, or I'm going coocoo studying this at 1 AM on a Saturday night.
Suppose I'm given a polar equation $(r,\theta)$ where $r$ is a function of $\theta$. I want to know the geometric interpretation of $\frac{dr}{d\theta}$ to the slope of the curve generated by the equation.
I'm aware that $x = r(\theta)*Cos(\theta)$ and $y = r(\theta)*Sin(\theta)$ such that we can find the slope of the curve from: $$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}*Sin(\theta)+r*Cos(\theta)}{\frac{dr}{d\theta}*Cos(\theta)-r*Sin(\theta)}$$
Are there more geometric significance from $\frac{dr}{d\theta}$ to the curve beyond just how "fast" $r(\theta)$ shinks or expands from the origin?
Knowing $\frac{dy}{dx}$ just from looking at the curve, are there methods to estimate $\frac{dr}{d\theta}$ beyond plugging in the equation above or converting $r(\theta)*Cos(\theta) \rightarrow x$ and $r(\theta)*Sin(\theta) \rightarrow y$
Draw a particle in the plane at a point with polar coordinates $(r_0,\theta_0)$, and draw the circle $x^2+y^2=r_0^2$. Now allow the particle to move along a curve $(r(t),\theta(t))$ for a small time $\Delta t$ such that from $t\rightarrow t+\Delta t$, the curve is approximately linear and the polar coordinates are now $(r_0+\Delta r,\theta_0+\Delta \theta)$.
If you do this the curve makes an angle $\alpha$ with the circle $x^2+y^2=r_0^2$ and if you draw the picture you will see that
$$\tan\alpha=\frac{\Delta r}{r_0\cdot \Delta \theta}\Rightarrow \frac{dr}{d\theta}\approx r\tan \alpha.$$
If you can't draw this picture get back to me.