AP Precalculus 3.15: Rates of Change in Polar Functions

In $r=f(\theta )$ the output $r$ is a signed radius. When $r$ is positive, the point $(r,\theta )$ sits on the terminal ray of $\theta$. When $r$ is negative, it sits on the opposite ray. Either way, its distance from the origin is $|r|$.

So to tell whether the point is moving toward the origin or away from it as $\theta$ increases, watch $|r|$, not $r$. That gives four cases.

• $r$ positive and increasing: $|r|$ grows, so the point moves away from the origin.
• $r$ positive and decreasing: $|r|$ shrinks, so it moves toward the origin.
• $r$ negative and decreasing: $r$ goes from $-1$ to $-2$ to $-3$, so $|r|$ grows and the point moves away.
• $r$ negative and increasing: $|r|$ shrinks, so it moves toward the origin.

The third case is the one students get wrong. Decreasing sounds like shrinking, but a negative number that decreases gets bigger in size.

Look at $f(\theta )=1-2\sin \theta$ in the diagram. From $\theta =\frac{\pi }{6}$ to $\theta =\frac{\pi }{2}$, $f$ falls from 0 to $-1$. It is negative and decreasing there, and the point swings out from the origin to the far end of the small inner loop.

Where $r$ changes from increasing to decreasing, or from decreasing to increasing, it has a relative extremum. At that angle the point is relatively closest to the origin or relatively farthest from it. Which one depends on the sign of $r$ there, because the distance is $|r|$.

Take $f(\theta )=1-2\sin \theta$ again. It has a relative maximum of 3 at $\theta =\frac{3\pi }{2}$. That value is positive and larger than the values beside it, so the point $(3,\frac{3\pi }{2})$ is relatively farthest from the origin: the bottom of the big loop.

The same function has a relative minimum of $-1$ at $\theta =\frac{\pi }{2}$. A minimum sounds like the closest point, but $|f|=1$ there, and $|f|$ is smaller on either side, down to 0 at $\frac{\pi }{6}$ and $\frac{5\pi }{6}$. So that point is also relatively farthest from the origin: the far end of the small loop.

Closest points come from extrema too. For $g(\theta )=3+2\cos \theta$, $g$ has a relative minimum of 1 at $\theta =\pi$, and $g$ is positive there, so that point is relatively closest to the origin.

So find where $r$ turns, then read $|r|$ to tell closest from farthest.

The average rate of change of $r$ with respect to $\theta$ over an interval from $\theta =a$ to $\theta =b$ is the change in the signed radius divided by the change in $\theta$:

It tells you how fast the signed radius is changing per radian, on average. On the graph of $r$ against $\theta$, it is the slope of the secant line between the two points.

For $f(\theta )=1-2\sin \theta$ from $\theta =0$ to $\theta =\frac{\pi }{2}$:

On average, the signed radius drops about 1.27 for each radian $\theta$ turns.

Use the signed values, signs and all. The point is 1 unit from the origin at both ends of this interval, so a calculation with distances would give 0, as if nothing happened. Something did: $r$ went from 1 to $-1$ and passed through the origin on the way. Keep $\theta$ in radians, since the rate is per radian.

The average rate of change gives you a line to estimate with. Start at a value you know and move along the secant line:

for an angle $\theta$ inside the interval from $a$ to $b$.

With $f(\theta )=1-2\sin \theta$ and the rate $-\frac{4}{\pi }$ from 0 to $\frac{\pi }{2}$, estimate $f\left(\frac{\pi }{6}\right)$:

The actual value is $1-2\cdot \frac{1}{2}=0$. The estimate is off by a third, because the graph of $f$ bends away from the secant line. It is still the best you can do when all you have is two values, as in a table, and a shorter interval around the angle usually gives a closer estimate.

Two slips to avoid. Multiply the rate by how far $\theta$ is from your starting angle, not by $\theta$ itself. And keep the sign of the rate: a negative rate lowers the estimate as $\theta$ moves forward.