Topic 3.15

Rates of Change in Polar Functions

For the bicycle pedal, $r = 17$ no matter what $\theta$ is: the pedal's distance from the crank center never changes, so its polar graph is a circle. Most polar curves are not like that. As $\theta$ increases, $r$ changes, and the point moves toward or away from the pole. This last note of Unit 3 asks two questions: which way is the point moving, and how fast is $r$ changing?

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01
Distance from the pole is increasing when $r$ and its change have the same sign.

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Toward or Away from the Pole?

Picture a dog on a leash tied to a post at the pole. The distance from the post is the length of leash that is out — that is $|r|$ , never negative. The sign of $r$ only tells you which SIDE of the post the dog is on (3.13A). So to know whether the dog is getting farther away, you have to look at $|r|$ , not just $r$ .

Four cases. Gray dot: before. Colored dot: after. Only the distance to the pole (the dark center dot) matters.

	r is increasing	r is decreasing
r > 0	\|r\| grows: moving AWAY	\|r\| shrinks: moving TOWARD
r < 0	\|r\| shrinks: moving TOWARD	\|r\| grows: moving AWAY

KEY RULE

Distance from the pole is increasing when $r$ and its change have the SAME sign.

Another way to see it: draw the DISTANCE $|r|$ as its own graph. Whenever that graph rises, the point is moving away from the pole; whenever it falls, the point is coming back. Compare the two pictures below — they carry exactly the same information.

Left: the polar curve. Right: the distance $|r|$ from the pole. Red means the distance is growing; teal means it is shrinking.

Two details worth noticing in the right-hand graph. It touches 0 at $\theta = 7\pi /6$ and $11\pi /6$ — those are the moments the curve passes through the pole. And it has TWO humps: a big one of height 3 (the outer curve) and a small one of height 1 (the inner loop). The inner loop never gets farther than 1 unit from the pole, even though it is drawn while $r$ is as low as −1.

CONCEPT

Three questions to ask every time

Is $r$ positive or negative on the interval?
Is $r$ increasing or decreasing on the interval?
Same signs (positive & increasing, or negative & decreasing) → moving AWAY. Different signs → moving TOWARD.

If $r$ changes sign inside the interval, split the interval at $r = 0$ and answer each piece separately.

Worked example

Example 1. (Warm-up) For $r = 4 + 2 \cos \theta$ , when is the point moving toward the pole and when is it moving away, for $0 \le \theta \le 2\pi$ ?

01
Question 1: the smallest $r$ is $4 - 2 = 2$ , so $r > 0$ for every $\theta$ . The sign never changes — this makes the problem easy.
02
Question 2: $\cos \theta$ decreases on $(0, \pi )$ and increases on $(\pi , 2\pi )$ , so $r$ does the same.
03
Question 3: on $(0, \pi )$ , positive and decreasing → moving TOWARD the pole. On $(\pi , 2\pi )$ , positive and increasing → moving AWAY.
04
Justification sentence (the way AP wants it): "On $(0, \pi )$ , $r(\theta ) > 0$ and $r$ is decreasing, so the distance $|r|$ between the point and the pole is decreasing."

CONCEPT

Try it yourself

Decide: moving toward or away?

$r$ goes from 5 to 2. 2. $r$ goes from −1 to −4. 3. $r$ goes from −6 to −3. 4. $r$ goes from 0.5 to 3.

Answers: 1. Toward (distance 5 → 2). 2. Away (distance 1 → 4). 3. Toward (distance 6 → 3). 4. Away (distance 0.5 → 3).

Worked example

Example 2. For $r = 1 + 2 \sin \theta$ (3.14B), find where the point moves away from the pole and where it moves toward it, for $0 \le \theta \le 2\pi$ .

01
Key angles from the rectangular graph: maximum $r = 3$ at $\pi /2$ , zeros at $7\pi /6$ and $11\pi /6$ , minimum $r = -1$ at $3\pi /2$ .
02
0 to $\pi /2$ : $r > 0$ and increasing → away. $\pi /2$ to $7\pi /6$ : $r > 0$ and decreasing → toward, reaching the pole.
03
$7\pi /6$ to $3\pi /2$ : $r < 0$ and decreasing (from 0 to −1) → AWAY. This is the first half of the inner loop.
04
$3\pi /2$ to $11\pi /6$ : $r < 0$ and increasing (from −1 to 0) → toward. $11\pi /6$ to $2\pi$ : $r > 0$ and increasing → away.

Red: moving away from the pole. Teal: moving toward it. The colors match in both graphs.

COMMON MISTAKE

"On $7\pi /6 < \theta < 3\pi /2$ , $r$ is decreasing, so the point is moving toward the pole."

That rule only works when $r > 0$ . Here $r < 0$ : it goes from 0 to −1, so $|r|$ goes from 0 to 1 and the point moves AWAY. In the polar graph this is the part of the inner loop leaving the pole. Always check the sign of $r$ first.

Worked example

Example 3. The table gives values of $r = f(\theta ) = 1 + 3 \cos \theta$ . Is the point moving toward or away from the pole on $[\pi /2, 3\pi /4]$ ? On $[3\pi /4, \pi ]$ ?

θ	0	π/4	π/2	3π/4	π
r	4	3.121	1	−1.121	−2

01
On $[\pi /2, 3\pi /4]$ , $r$ goes from 1 to −1.121: it CHANGES SIGN, so it passes through 0 somewhere in between (at $\theta \approx 1.911$ ). Split the interval there.
02
Before the zero: $r > 0$ and decreasing → moving toward the pole, all the way to the pole itself. After the zero: $r < 0$ and still decreasing → moving away again. So the answer is "toward, then away".
03
On $[3\pi /4, \pi ]$ , $r$ goes from −1.121 to −2: negative AND decreasing → same signs → moving AWAY. The distance grows from 1.121 to 2.

$r = 1 + 3 \cos \theta$ crosses 0 between $\pi /2$ and $3\pi /4$ : the point touches the pole there.

Quick check

On an interval, $r$ is negative and decreasing. Is the point moving toward or away from the pole?

Average Rate of Change of r

Remember from Unit 1: the average rate of change of a function over an interval is the change in output divided by the change in input. For a polar function, the input is $\theta$ and the output is $r$ .

\text{average rate of change of }r\text{ on }[\theta_1,\theta_2]=\frac{f(\theta_2)-f(\theta_1)}{\theta_2-\theta_1}

Its units are units of $r$ per radian. It tells you how fast, on average, $r$ changes as $\theta$ increases. Its sign tells you whether $r$ increased or decreased — but, as in Section 1, not by itself whether the point moved toward or away from the pole.

CONCEPT

Saying it in words

An average rate of change of −3.308 on $[0, \pi /3]$ means: "On average, $r$ decreases by about 3.308 units for each radian that $\theta$ increases, from $\theta = 0$ to $\theta = \pi /3$ ."

Always include the direction (increases/decreases), the amount, the units (per radian), and the interval.

Worked example

Example 4. The table gives values of $r = f(\theta ) = 2 - 4 \sin \theta$ . (a) Where is $f$ increasing or decreasing on $[0, \pi ]$ ? (b) Is the point moving toward or away from the pole on $(\pi /6, \pi /2)$ ? (c) Find the average rate of change of $f$ on $[0, \pi /3]$ . (d) Use an average rate of change to estimate $f(\pi /4)$ .

θ	0	π/6	π/3	π/2	2π/3	5π/6	π
r	2	0	−1.464	−2	−1.464	0	2

01
(a) The values fall from 2 to −2 on $[0, \pi /2]$ and rise back to 2 on $[\pi /2, \pi ]$ . So $f$ is decreasing on $[0, \pi /2]$ and increasing on $[\pi /2, \pi ]$ .
02
(b) On $(\pi /6, \pi /2)$ , $r$ is negative AND decreasing (0 → −2). Same signs, so $|r|$ grows: the point moves away from the pole.
03
$> (c) (f(\pi /3) - f(0)) \div (\pi /3 - 0) = (-1.464 - 2) \div (\pi /3) \approx -3.308$ units per radian. (Exactly, $-6\sqrt{3}/\pi$ .)
04
(d) Use the interval around $\pi /4$ from the table, $[\pi /6, \pi /3]$ : the rate is $(-1.464 - 0) \div (\pi /6) \approx -2.796$ . Start at $f(\pi /6) = 0$ and move $\pi /12$ : $f(\pi /4) \approx 0 + (\pi /12)(-2.796) \approx -0.732$ .

The secant line on $[\pi /6, \pi /3]$ gives the estimate. The actual value is $2 - 4 \sin (\pi /4) \approx -0.828$ .

Why is the estimate too high? On $[\pi /6, \pi /3]$ the graph of $r$ is concave up (its rate of change is increasing, Unit 1), so the curve bends BELOW the straight secant line. Any estimate read off that line is therefore too high. If the graph had been concave down, the curve would bend above the line and the estimate would be too low.

The straight dashed line is the secant; the square is the estimate and the dark dot is the true value.

Shape of r on the interval Secant line sits… Your estimate is…
concave up above the curve too high (overestimate)
concave down below the curve too low (underestimate)

Two more habits that keep the error small: use the SHORTEST interval in the table that contains the value you want, and keep the interval centered on that value when you can.

Shape of r on the interval	Secant line sits…	Your estimate is…
concave up	above the curve	too high (overestimate)
concave down	below the curve	too low (underestimate)

Quick check

For $r = f(\theta)$ with $f(0) = 4$ and $f\left(\frac{\pi}{2}\right) = 1$ , what is the average rate of change of $r$ on $\left[0, \frac{\pi}{2}\right]$ ?

Comparing Rates and Finding the Farthest Point

Worked example

Example 5. One petal of $r = 4 \sin 3\theta$ is drawn for $0 \le \theta \le \pi /3$ . Compare how fast $r$ changes on $[0, \pi /12]$ and $[\pi /12, \pi /6]$ , and find the point of the petal farthest from the pole.

01
Values: $r(0) = 0$ , $r(\pi /12) = 4 \sin (\pi /4) = 2\sqrt{2} \approx 2.828$ , $r(\pi /6) = 4 \sin (\pi /2) = 4$ .
02
Average rates: on $[0, \pi /12]$ , $2.828 \div (\pi /12) \approx 10.80$ ; on $[\pi /12, \pi /6]$ , $(4 - 2.828) \div (\pi /12) \approx 4.48$ units per radian. $r$ is still increasing, but more slowly.
03
$r > 0$ and increasing on $[0, \pi /6]$ , so the point moves away from the pole; $r > 0$ and decreasing on $[\pi /6, \pi /3]$ , so it moves back. $r$ changes from increasing to decreasing at $\theta = \pi /6$ , a relative maximum of $r$ .
04
The farthest point is $(4, \pi /6)$ , the tip of the petal. By symmetry, the average rates on $[\pi /6, \pi /4]$ and $[\pi /4, \pi /3]$ are −4.48 and −10.80.

Left: the petal, red while moving away, teal while returning. Right: the secant slopes shrink as $r$ nears its maximum.

CONCEPT

Justification sentences AP asks for

Moving away: "On this interval $r(\theta ) < 0$ and $r$ is decreasing, so $|r|$ is increasing and the point moves away from the pole."

Moving toward: "On this interval $r(\theta ) > 0$ and $r$ is decreasing, so $|r|$ is decreasing and the point moves toward the pole."

Comparing two points: "Because $|r(\theta _{1})| > |r(\theta _{2})|$ , the point at θ₁ is farther from the pole."

Always name the sign of $r$ AND the direction $r$ is changing. One of them alone is not a complete reason.

CONCEPT

Extreme distances

Where $r$ changes from increasing to decreasing (or back), $r$ has a relative maximum (or minimum).

The point is farthest from the pole where $|r|$ is largest. That can happen at a maximum of $r$ OR at a minimum of $r$ , if the minimum is negative.

Near an extreme value of $r$ , the average rates of change get close to 0: $r$ levels off, just like a sinusoid at its peak (3.4).

COMMON MISTAKE

"The point of $r = 2 - 4 \sin \theta$ farthest from the pole on $[0, \pi ]$ is where $r$ is largest, $r = 2$ ."

The minimum $r = -2$ at $\theta = \pi /2$ is just as far from the pole: $|-2| = 2$ . Distance uses $|r|$ , so a large negative $r$ counts. On $[0, \pi ]$ the farthest distance, 2, is reached three times: at $\theta = 0$ , $\pi /2$ , and $\pi$ .

Worked example

Example 6. For $r = 1 + 2 \sin \theta$ on $0 \le \theta \le 2\pi$ , find the point farthest from the pole, and the farthest point ON THE INNER LOOP.

01
Distance is $|r|$ , so list the candidates: the largest $r$ and the most negative $r$ . Here $r$ runs from a maximum of 3 (at $\theta = \pi /2$ ) down to a minimum of −1 (at $\theta = 3\pi /2$ ).
02
Compare distances: $|3| = 3$ and $|-1| = 1$ . The farthest point is at $\theta = \pi /2$ , distance 3, which is the point $(0, 3)$ .
03
The inner loop is drawn while $7\pi /6 < \theta < 11\pi /6$ , where $r < 0$ . On that interval the most negative value is $r = -1$ at $\theta = 3\pi /2$ , so the loop reaches only 1 unit from the pole — at the point $(0, 1)$ , directly above it.
04
Both answers are the two humps of the $|r|$ graph in Section 1 ✓.

CONCEPT

Try it yourself

For $r = 3 + 5 \cos \theta$ :

What is the largest value of $r$ , and at which $\theta$ ?
What is the most negative value of $r$ , and at which $\theta$ ?
Which of those two points is farther from the pole?

Answers: 1. $r = 8$ at $\theta = 0$ . 2. $r = -2$ at $\theta = \pi$ . 3. The $\theta = 0$ point, because $8 > |-2| = 2$ .

Common slips

"On $7\pi /6 < \theta < 3\pi /2$ , $r$ is decreasing, so the point is moving toward the pole."

That rule only works when $r > 0$ . Here $r < 0$ : it goes from 0 to −1, so $|r|$ goes from 0 to 1 and the point moves away. In the polar graph this is the part of the inner loop leaving the pole. Always check the sign of $r$ first.
"The point of $r = 2 - 4 \sin \theta$ farthest from the pole on $[0, \pi ]$ is where $r$ is largest, $r = 2$ ."

The minimum $r = -2$ at $\theta = \pi /2$ is just as far from the pole: $|-2| = 2$ . Distance uses $|r|$ , so a large negative $r$ counts. On $[0, \pi ]$ the farthest distance, 2, is reached three times: at $\theta = 0$ , $\pi /2$ , and $\pi$ .

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6 cards · Polar rates of change

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Recap card

5 lines to re-read the night before.

01
The distance from the pole is $|r|$ . Ask: is $r$ positive or negative? increasing or decreasing? Same signs → away; different signs → toward. Split the interval where $r = 0$ .
02
Average rate of change of $r$ on $[\theta _{1}, \theta _{2}] = (f(\theta _{2}) - f(\theta _{1})) \div (\theta _{2} - \theta _{1})$ , in units of $r$ per radian.
03
Estimate an unknown value using the average rate of change of the smallest interval available; concavity tells you if the estimate is too high (concave up) or too low (concave down).
04
Relative extrema of $r$ occur where $r$ switches between increasing and decreasing; the farthest point uses the largest $|r|$ .
05
Unit 3 complete: from periodic phenomena and the unit circle, through sinusoids, inverses, equations, and identities, to polar coordinates and polar functions.

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Rates of Change in Polar Functions

Toward or Away from the Pole?

Average Rate of Change of r

Comparing Rates and Finding the Farthest Point

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