AP®MathsCollege BoardPrecalculusRevision NotesTrigonometric & Polar FunctionsPolar Coordinates & Polar FunctionsRates of Change in Polar Functions

Rates of Change in Polar Functions (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 20 April 2026

Increasing & decreasing polar functions

For a polar function $r = f (θ)$ , the point on the graph corresponding to input $θ$ has polar coordinates $(f (θ), θ)$
The distance from the origin to this point is $| f (θ) |$
- the absolute value is needed, as output values can be signed
  - i.e. $f (θ)$ can be positive or negative
As $θ$ changes, this distance can increase or decrease
- depending on both the sign of $f (θ)$
- and whether $f (θ)$ is increasing or decreasing

When is the distance from the origin increasing or decreasing?

The distance from the origin to the point $(f (θ), θ)$ is increasing on an interval when either of the following is true
- $f (θ)$ is positive and increasing on the interval
  - $r$ is moving further from $0$ in the positive direction
  - so $| r |$ grows
- $f (θ)$ is negative and decreasing on the interval
  - $r$ is moving further from $0$ in the negative direction
  - so $| r |$ also grows
- E.g. if $f (θ)$ goes from $- 1$ to $- 3$ as $θ$ increases
  - $f$ is decreasing
  - but the distance from the origin has grown from $1$ to $3$

The distance from the origin to the point $(f (θ), θ)$ is decreasing on an interval when either of the following is true
- $f (θ)$ is positive and decreasing on the interval
  - $r$ is moving toward $0$ from the positive side
  - so $| r |$ shrinks
- $f (θ)$ is negative and increasing on the interval
  - $r$ is moving toward $0$ from the negative side
  - so $| r |$ shrinks
- E.g. if $f (θ)$ goes from $- 4$ to $- 1$ as $θ$ increases
  - $f$ is increasing
  - but the distance from the origin has shrunk from $4$ to $1$

Examiner Tips and Tricks

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

Distance from origin increases when $f (θ)$ and its direction of change have the same sign (both positive, or both negative)
Distance from origin decreases when $f (θ)$ 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 $| f (θ) |$ , so it is increasing whenever $| f (θ) |$ is increasing

Because in that case the signed value $f (θ)$ is moving away from zero

Worked Example

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

(A) The distance is increasing for $\frac{π}{3} \leq θ \leq π$ , because $f (θ)$ is positive and increasing on the interval.

(B) The distance is increasing for $0 \leq θ \leq \frac{π}{3}$ , because $f (θ)$ is negative and increasing on the interval.

(C) The distance is decreasing for $\frac{π}{3} \leq θ \leq π$ , because $f (θ)$ is positive and decreasing on the interval.

(D) The distance is decreasing for $0 \leq θ \leq \frac{π}{3}$ , because $f (θ)$ is negative and decreasing on the interval.

Answer:

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

$f (θ) = 0$ when $\cos θ = \frac{1}{2}$
- i.e. $θ = \frac{π}{3}$ or $θ = \frac{5 π}{3}$
$f (θ) > 0$ when $\cos θ < \frac{1}{2}$
- so $f$ is positive on $(\frac{π}{3}, \frac{5 π}{3})$
$f (θ) < 0$ elsewhere
- so $f$ is negative on $[0, \frac{π}{3}) \cup (\frac{5 π}{3}, 2 π]$

Now check where $f$ is increasing or decreasing

Since $f (θ) = 1 - 2 \cos θ$
- $f$ increases when $\cos θ$ decreases, i.e. on $(0, π)$
- and it decreases when $\cos θ$ increases, i.e. on $(π, 2 π)$ .

This lets you divide $0 \leq θ \leq 2 π$ into four regions

$f$ is negative and increasing on $0 \leq θ \leq \frac{π}{3}$
- sign ( $-$ ) and direction of change ( $+$ ) don't match, so distance is decreasing
$f$ is positive and increasing on $\frac{π}{3} \leq θ \leq π$
- sign ( $+$ ) and direction of change ( $+$ ) match, so distance is increasing
$f$ is positive and decreasing on $π \leq θ \leq \frac{5 π}{3}$
- sign ( $+$ ) and direction of change ( $-$ ) don't match, so distance is decreasing
$f$ is negative and decreasing on $\frac{5 π}{3} \leq θ \leq 2 π$
- 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 $\frac{π}{3} \leq θ \leq π$ , because $f (θ)$ is positive and increasing on the interval

Relative extrema of polar functions

What does a relative extremum of a polar function mean?

A polar function $r = f (θ)$ has
- a relative maximum at a value of $θ$
  - where $f$ changes from increasing to decreasing
- and a relative minimum at a value of $θ$
  - where $f$ changes from decreasing to increasing
These relative extrema correspond to points on the polar graph that are relatively closest to or relatively farthest from the origin
- compared with nearby points on the graph
Because the distance from the origin is $| f (θ) |$ rather than $f (θ)$ itself
- the link between extrema of $f$ and extrema of the distance depends on the sign of $f$
At a relative maximum of $f$ where $f$ is positive
- $f$ reaches a local peak in the positive direction
  - so the point is relatively farthest from the origin compared to nearby points
At a relative minimum of $f$ where $f$ is positive
- $f$ reaches a local low point (still positive)
  - so the point is relatively closest to the origin compared to nearby points
At a relative maximum of $f$ where $f$ is negative
- $f$ reaches a local peak (still negative)
  - the value is closest to zero, so the point is relatively closest to the origin
At a relative minimum of $f$ where $f$ is negative
- $f$ reaches a local low point (most negative)
  - the value is furthest from zero, so the point is relatively farthest from the origin

Polar graph with a red curve in the shape of a sideways heart, showing a relative minimum at -1 on the polar axis and a relative maximum at 5 on the polar axis, marked with labels and arrows. — **Relative minimum and maximum points on the graph of a polar function**

Examiner Tips and Tricks

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

Average rate of change of polar functions

How is the average rate of change of a polar function defined?

The average rate of change of a polar function $r = f (θ)$ over an interval $[θ_{1}, θ_{2}]$ is defined in the usual way
- $average rate of change = \frac{f (θ_{2}) - f (θ_{1})}{θ_{2} - θ_{1}}$

This is the ratio of
- the change in the radius values
- to the change in the angle over the interval
Graphically, it represents the rate at which the radius is changing per radian (assuming $θ$ is measured in radians)
- The units are therefore "units of $r$ per radian"

How can the average rate of change be used to estimate values of a polar function?

The average rate of change over an interval $[θ_{1}, θ_{2}]$ can be used
- to estimate the value of $f (θ)$ at any $θ$ between $θ_{1}$ and $θ_{2}$
- using a linear approximation
  - $f (θ) \approx f (θ_{1}) + (average rate of change) \cdot (θ - θ_{1})$

This approximation treats $f$ as though it changes at a constant rate across the interval
- It will be most accurate
  - when the interval is small
  - and when $f$ does not change behavior drastically within the interval
The linear approximation formula can also be written starting from the right endpoint
- $f (θ) \approx f (θ_{2}) + (average rate of change) \cdot (θ - θ_{2})$

Worked Example

$θ$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$
$f (θ)$	$4.8$	$6.0$	$4.5$	$3.2$	$3.8$

A polar function $r = f (θ)$ is defined on the interval $[0, π]$ , and the table above gives selected values of $f (θ)$ . Assume that $f$ is continuous and has no relative extrema between the values of $θ$ 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.

Answer:

Reading across the table

From $θ = 0$ to $θ = \frac{π}{4}$ , $f$ increases from $4.8$ to $6.0$
From $θ = \frac{π}{4}$ to $θ = \frac{π}{2}$ , $f$ decreases from $6.0$ to $4.5$
From $θ = \frac{π}{2}$ to $θ = \frac{3 π}{4}$ , $f$ decreases from $4.5$ to $3.2$
From $θ = \frac{3 π}{4}$ to $θ = π$ , $f$ increases from $3.2$ to $3.8$

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

This is a relative maximum of $f$
and since $f$ is positive at this value, this means the point on the graph at $θ = \frac{π}{4}$ is relatively farthest from the origin compared with nearby points

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

This is a relative minimum of $f$
and since $f$ is positive at this value, this means the point on the graph at $θ = \frac{3 π}{4}$ is relatively closest to the origin compared with nearby points

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

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

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

Answer:

Apply the average rate of change formula:

$average rate of change = \frac{f (\frac{π}{2}) - f (\frac{π}{4})}{\frac{π}{2} - \frac{π}{4}} = \frac{4.5 - 6.0}{\frac{π}{4}} = \frac{- 1.5}{\frac{π}{4}} = - \frac{6}{π}$

So the average rate of change is

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

Answer:

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

$f (\frac{5 π}{12}) \approx f (\frac{π}{4}) + (- \frac{6}{π}) (\frac{5 π}{12} - \frac{π}{4})$

First simplify the angle difference

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

Then

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

$f (\frac{5 π}{12}) \approx 5.0$

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Rates of Change in Polar Functions (College Board AP® Precalculus): Revision Note

Increasing & decreasing polar functions

When is the distance from the origin increasing or decreasing?

Examiner Tips and Tricks

Worked Example

Relative extrema of polar functions

What does a relative extremum of a polar function mean?

Examiner Tips and Tricks

Average rate of change of polar functions

How is the average rate of change of a polar function defined?

How can the average rate of change be used to estimate values of a polar function?

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis

Rates of Change in Polar Functions (College Board AP® Precalculus): Revision Note

Increasing & decreasing polar functions

How is the distance from the origin related to a polar function?

When is the distance from the origin increasing or decreasing?

Relative extrema of polar functions

What does a relative extremum of a polar function mean?

Average rate of change of polar functions

How is the average rate of change of a polar function defined?

How can the average rate of change be used to estimate values of a polar function?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis