I have being doing problems from the released AP BC Calcululs Free-Response questions, and I have come to realize that I don't have a very good idea of explain or a deep understanding of how to tell if a particle is moving to the orgion (pole?).
Example 2005 AP BC Free-Response Question #2.
The curve above is drawn in the $xy$-plane and is described by the equation in polar coodinates $r =\theta +\sin\left(2\theta\right)$ for $0\leq \theta \leq \pi$, where $r$ is measured in meters and $\theta$ is measured in radians. The derivative of $r$ with respect to $\theta$ is given by $\frac{\mathrm{d}r}{\mathrm{d}\theta} = 1 + 2\cos\left(2\theta\right)$.
(c) for $\frac{\pi}{3} < \theta < \frac{2\pi}{3}$, $\frac{\mathrm{d}r}{\mathrm{d}\theta}$ is negative. What does this fact say about $r$? What does this fact say about the curve?
I have learned from doing problems and reading the answers that if the derivative is negative its getting closer to the origin. However, I have no clear understanding why?
I don't trust my justification:
Just as with the Cartesian coordinate system, when the 1st derivative is negative the particle is decreasing, moving downward, thus in the polar coordinate system that is moving to the origin.
First, let's visualize the curve traced out in the $x$-$y$ plane by $r=\theta+\sin(2\theta)$, as $\theta$ varies over $[0,\pi]$:
Now, for the question posed in (c), let's focus on the part of the curve traced out for ${\pi\over 3}<\theta<{2\pi\over 3}$:
The dashed line is showing the length of $r$ as $\theta$ varies. In particular, note that for these $\theta$ values, $r$ is decreasing as a function of $\theta$.
I drew these to help you "see" the geometry in this polar form, but we could also accomplish this much more simply by just plotting $r$ as a function of $\theta$ for ${\pi\over 3}<\theta<{2\pi\over 3}$:
$r$ is certainly looks like it decreases with $\theta$, and a little calculus verifies this since ${dr\over d\theta}=1+2\cos(2\theta)<0$ for ${\pi\over 3}<\theta<{2\pi\over 3}$.