Rates of Change in Polar Functions (College Board AP® Precalculus): Study Guide

A useful way to keep track of all four cases is the rule:

• Distance from origin increases when $f\left(\theta \right)$ and its direction of change have the same sign (both positive, or both negative)

• Distance from origin decreases when $f\left(\theta \right)$ and its direction of change have opposite signs (one positive, the other negative)

Another way to think about it is that the distance from the origin is $\left|f\right(\theta \left)\right|$, so it is increasing whenever $\left|f\right(\theta \left)\right|$ is increasing

• Because in that case the signed value $f\left(\theta \right)$ is moving away from zero

Consider the graph of the polar function $r=f\left(\theta \right)$, where $f\left(\theta \right)=1-2\cos \theta$, in the polar coordinate system for $0\le \theta \le 2\pi$. Which of the following statements is true about the distance between the point with polar coordinates $\left(f\right(\theta ), \theta )$ and the origin?

(A) The distance is increasing for $\frac{\pi }{3}\le \theta \le \pi$, because $f\left(\theta \right)$ is positive and increasing on the interval.

(B) The distance is increasing for $0\le \theta \le \frac{\pi }{3}$, because $f\left(\theta \right)$ is negative and increasing on the interval.

(C) The distance is decreasing for $\frac{\pi }{3}\le \theta \le \pi$, because $f\left(\theta \right)$ is positive and decreasing on the interval.

(D) The distance is decreasing for $0\le \theta \le \frac{\pi }{3}$, because $f\left(\theta \right)$ is negative and decreasing on the interval.

Answer:

First check where $f\left(\theta \right)=1-2\cos \theta$ is positive or negative

• $f\left(\theta \right)=0$ when $\cos \theta =\frac{1}{2}$

  • i.e. $\theta =\frac{\pi }{3}$ or $\theta =\frac{5\pi }{3}$

• $f\left(\theta \right)>0$ when $\cos \theta <\frac{1}{2}$

  • so $f$ is positive on $\left(\frac{\pi }{3}, \frac{5\pi }{3}\right)$

• $f\left(\theta \right)<0$ elsewhere

  • so $f$ is negative on $\left[0, \frac{\pi }{3}\right)\cup \left(\frac{5\pi }{3}, 2\pi \right]$

Now check where $f$ is increasing or decreasing

• Since $f\left(\theta \right)=1-2\cos \theta$

  • $f$ increases when $\cos \theta$ decreases, i.e. on $(0, \pi )$

  • and it decreases when $\cos \theta$ increases, i.e. on $(\pi , 2\pi )$.

This lets you divide $0\le \theta \le 2\pi$ into four regions

• $f$ is negative and increasing on $0\le \theta \le \frac{\pi }{3}$

  • sign ($-$) and direction of change ($+$) don't match, so distance is decreasing

• $f$ is positive and increasing on $\frac{\pi }{3}\le \theta \le \pi$

  • sign ($+$) and direction of change ($+$) match, so distance is increasing

• $f$ is positive and decreasing on $\pi \le \theta \le \frac{5\pi }{3}$

  • sign ($+$) and direction of change ($-$) don't match, so distance is decreasing

• $f$ is negative and decreasing on $\frac{5\pi }{3}\le \theta \le 2\pi$

  • sign ($-$) and direction of change ($-$) match, so distance is increasing

The only answer option which contains both a correct behavior (increasing or decreasing) and a correct explanation is (A)

(A) The distance is increasing for $\frac{\pi }{3}\le \theta \le \pi$, because $f\left(\theta \right)$ is positive and increasing on the interval

Don't forget to consider the sign of $f$ when analyzing a polar function for relative extrema.

When all the values of $f$ under consideration are positive, then there is a convenient shortcut:

• relative maxima of $f$ give farthest points

• relative minima of $f$ give closest points

A polar function $r=f\left(\theta \right)$ is defined on the interval $[0, \pi ]$, and the table above gives selected values of $f\left(\theta \right)$. Assume that $f$ is continuous and has no relative extrema between the values of $\theta$ listed in the table.

(a) Based on the information in the table, on which interval between consecutive table values does $f$ have a relative maximum at the right endpoint, and on which interval does $f$ have a relative minimum at the right endpoint? For each relative maximum or minimum point, state whether the corresponding point on the polar graph is relatively closest to or relatively farthest from the origin.

(b) Find the average rate of change of $f$ with respect to $\theta$ on the interval $\left[\frac{\pi }{4}, \frac{\pi }{2}\right]$. Give an exact value and include the appropriate units.

(c) Use the average rate of change found in part (b) to estimate the value of $f\left(\frac{5\pi }{12}\right)$.

Answer:

(a)

Reading across the table

• From $\theta =0$ to $\theta =\frac{\pi }{4}$, $f$ increases from $4.8$ to $6.0$

• From $\theta =\frac{\pi }{4}$ to $\theta =\frac{\pi }{2}$, $f$ decreases from $6.0$ to $4.5$

• From $\theta =\frac{\pi }{2}$ to $\theta =\frac{3\pi }{4}$, $f$ decreases from $4.5$ to $3.2$

• From $\theta =\frac{3\pi }{4}$ to $\theta =\pi$, $f$ increases from $3.2$ to $3.8$

So $f$ changes from increasing to decreasing at $\theta =\frac{\pi }{4}$, which is the right endpoint of the interval $\left[0, \frac{\pi }{4}\right]$

• This is a relative maximum of $f$

• and since $f$ is positive at this value, this means the point on the graph at $\theta =\frac{\pi }{4}$ is relatively farthest from the origin compared with nearby points

$f$ changes from decreasing to increasing at $\theta =\frac{3\pi }{4}$, which is the right endpoint of the interval $\left[\frac{\pi }{2}, \frac{3\pi }{4}\right]$

• This is a relative minimum of $f$

• and since $f$ is positive at this value, this means the point on the graph at $\theta =\frac{3\pi }{4}$ is relatively closest to the origin compared with nearby points

$f$ has a relative maximum at the right endpoint of $\left[0, \frac{\pi }{4}\right]$, and at $\theta =\frac{\pi }{4}$
the point on the polar graph is relatively farthest from the origin

$f$ has a relative minimum at the right endpoint of $\left[\frac{\pi }{2}, \frac{3\pi }{4}\right]$, and at $\theta =\frac{3\pi }{4}$
the point on the polar graph is relatively closest to the origin

(b)

Apply the average rate of change formula:

$\text{average rate of change}=\frac{f\left({\displaystyle \frac{\pi }{2}}\right)-f\left({\displaystyle \frac{\pi }{4}}\right)}{{\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{\pi }{4}}}=\frac{4.5-6.0}{{\displaystyle \frac{\pi }{4}}}=\frac{-1.5}{{\displaystyle \frac{\pi }{4}}}=-\frac{6}{\pi }$

So the average rate of change is

$-\frac{6}{\pi }$ units of $r$ per radian

(c)

Use the average rate of change to estimate $f\left(\frac{5\pi }{12}\right)$ by linear approximation starting from $\theta =\frac{\pi }{4}$

$f\left(\frac{5\pi }{12}\right)\approx f\left(\frac{\pi }{4}\right)+\left(-\frac{6}{\pi }\right)\left(\frac{5\pi }{12}-\frac{\pi }{4}\right)$

First simplify the angle difference

$\frac{5\pi }{12}-\frac{\pi }{4}=\frac{5\pi }{12}-\frac{3\pi }{12}=\frac{2\pi }{12}=\frac{\pi }{6}$

Then

$f\left(\frac{5\pi }{12}\right)\approx 6.0+\left(-\frac{6}{\pi }\right)\left(\frac{\pi }{6}\right)=6.0-1=5.0$

So

$f\left(\frac{5\pi }{12}\right)\approx 5.0$