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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.

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If the derivative is negative then the value is decreasing. If $r$ is decreasing, what does that mean? Where is $r$ large and where is $r$ small? –  Rahul Jan 8 '13 at 4:47
    
I can see it now, I am thinking of a string attached to me and if its pulled away then the derivative is increasing and if I pull it then the derivative is decreasing. –  yiyi Jan 8 '13 at 4:50

2 Answers 2

up vote 3 down vote accepted

First, let's visualize the curve traced out in the $x$-$y$ plane by $r=\theta+\sin(2\theta)$, as $\theta$ varies over $[0,\pi]$:

enter image description here

Now, for the question posed in (c), let's focus on the part of the curve traced out for ${\pi\over 3}<\theta<{2\pi\over 3}$:

enter image description here

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 ${\pi\over 3}<\theta<{2\pi\over 3}$:

Mathematica graphics

$r$ 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 ${\pi\over 3}<\theta<{2\pi\over 3}$.

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Nice graphics (or perhaps I am just showing my age...)! –  copper.hat Jan 8 '13 at 6:25
    
@JohnD What tool did you use to generate the graphics? –  hondaman Jul 27 '13 at 7:06
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@hondaman: I used Mathematica. –  JohnD Jul 27 '13 at 17:33

The key here is that the particle gets closer to the origin as it moves counterclockwise along the curve. As $\theta$ increases, $r$ decreases. That is the meaning of the derivative. If the particle moved clockwise instead, $r$ would be increasing. Without a specific parameterization, you cannot know which is the true behavior of the particle, but it's safe to say that as the angle increases, the distance decreases. One tends not to worry about whether that behavior is forwards or backwards in time.

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