Rates of Change of Polar Functions

You’ll be able to

Determine where a polar function r(θ) is increasing or decreasing
Explain what an increasing r means about the distance from the origin
Compute the average rate of change of r with respect to θ and interpret it

r is a function of θ, and it has rates of change too

This is the part of polar work that AP Precalculus emphasizes and that most textbooks skip. Treat r = f(θ) as an ordinary function whose input happens to be an angle. Where f is increasing, the curve is moving away from the origin as the angle sweeps forward. Where f is decreasing, it is moving toward the origin. All the rate-of-change language from Unit 1 applies unchanged.

Average rate of change of a polar function

AROC of r on [θ₁, θ₂] = [f(θ₂) − f(θ₁)] / (θ₂ − θ₁)

Units are distance per radian. A positive value means the curve is receding from the origin on average over that interval.

Increasing r does not mean increasing y

The distinction matters. On r = 1 + cos θ over [0, π/2], the value of r decreases from 2 to 1 — the curve approaches the origin. But the point itself moves counterclockwise, so its y-coordinate increases over part of that stretch. Statements about r are statements about distance from the origin, not about height or about horizontal position. Read the question carefully to see which is being asked.

Worked example

For r = 3 + 2 sin θ, determine where r is increasing on [0, 2π] and find the average rate of change on [0, π/2].

1.r depends on θ only through sin θ, and the coefficient 2 is positive, so r increases exactly where sin θ increases.
2.sin θ increases on [0, π/2] and on [3π/2, 2π]; it decreases on [π/2, 3π/2].
3.So r increases on [0, π/2] ∪ [3π/2, 2π] and decreases on [π/2, 3π/2].
4.AROC on [0, π/2]: r(0) = 3 + 0 = 3, and r(π/2) = 3 + 2 = 5.
5.AROC = (5 − 3)/(π/2 − 0) = 2/(π/2) = 4/π ≈ 1.273.

Answer: r increases on [0, π/2] and [3π/2, 2π], and decreases on [π/2, 3π/2]. The average rate of change on [0, π/2] is 4/π ≈ 1.273 units per radian — positive, so over that quarter-turn the curve moves away from the origin at an average of about 1.27 units per radian.

On the exam

AP Precalculus asks about polar rates of change far more than about sketching polar curves. Expect to be given r = f(θ) and asked where the curve moves toward or away from the origin, with justification — and the justification must reference whether f is increasing or decreasing.

Checkpoint

For r = 4 cos θ on the interval [0, π/2], what happens to the distance from the origin?

Where r changes sign

When r = 0 the curve passes through the origin. When r changes sign, the curve crosses to the opposite side, since negative r is plotted backward. For r = 1 + 2 cos θ, setting 1 + 2 cos θ = 0 gives cos θ = −1/2, so θ = 2π/3 and 4π/3. Between those angles r is negative, and that negative stretch is exactly the inner loop of the limaçon. Zeros of r are therefore the key structural feature of a polar curve.

Worked example

For r = 2 − 4 cos θ, find where r = 0 and describe the behavior of the distance from the origin on [0, π].

1.r = 0 when 2 − 4 cos θ = 0, so cos θ = 1/2, giving θ = π/3 (within [0, π]).
2.r(0) = 2 − 4 = −2. Negative, so the curve starts 2 units out in the direction θ = π, opposite the polar axis.
3.On [0, π/3], cos θ decreases from 1 to 1/2, so −4 cos θ increases and r rises from −2 to 0.
4.On [π/3, π], cos θ continues to decrease to −1, so r continues rising to 2 − 4(−1) = 6.
5.So r increases throughout [0, π], from −2 to 6, passing through 0 at θ = π/3.

Answer: r = 0 at θ = π/3, and r increases monotonically from −2 to 6 across [0, π]. The distance from the origin, |r|, first decreases from 2 to 0 and then increases to 6 — which is why |r| and r must be distinguished when the question asks about distance.

Checkpoint

A polar function has r(θ) negative and decreasing on an interval. What is happening to the curve?

Answer the 2 checkpoints as you read.