Watch a moth spiral around a porch light. At each moment, its position has an angle (which way from the light) and a radius (how far). The moth's radial story — closing in, backing off — is told entirely by whether r(θ) is increasing or decreasing as the angle sweeps.
Unit 1 taught you to read increasing/decreasing/concavity from f(x). CED 3.15 asks the same questions about r(θ). Nothing new is under the hood — the input is an angle and the output is a distance, and rates of change work exactly as they always have.
For a polar function r(θ) with r > 0 on the interval:
Read these from the trig part. For r = 3 + 2cos θ: cos θ decreases on (0, π), so r decreases from 5 to 1 — the curve spirals inward across the upper half-plane; then cos θ increases on (π, 2π) and the curve swells back out. Relative max of r at θ = 0 (r = 5), relative min at θ = π (r = 1).
(If r passes through 0 and goes negative, "distance from the pole" is |r| — the AP keeps rate questions on intervals where the sign is constant, typically r > 0.)
Same secant formula as Lesson 1, new letters:
AROC of r over [θ₁, θ₂] = ( r(θ₂) − r(θ₁) ) / ( θ₂ − θ₁ )
Units: distance per radian. Example: r = 4 sin θ over [0, π/2]: (4 − 0)/(π/2) = 8/π ≈ 2.55 units of distance per radian of sweep. Interpretation sentence: "on average, the curve moves about 2.55 units farther from the pole per radian as θ goes from 0 to π/2."
Everything from Unit 1 transfers: rates of change of r can themselves increase or decrease (concavity-style reasoning), AROC is a secant slope on the (θ, r) graph, and the (θ, r) graph — r plotted against θ like any function — is often the clearest way to see a polar function's behavior before wrapping it around the pole.
View 1 (function graph): plot r against θ on ordinary axes. For r = 2 + 2sin θ: a sinusoid, midline 2, amplitude 2, peak at θ = π/2. View 2 (polar graph): wrap that sinusoid around the pole: the cardioid.
Feature dictionary between views: sinusoid max ↔ farthest point from pole; sinusoid zero ↔ curve at the pole; sinusoid increasing ↔ spiraling outward. The exam moves between views without warning; practice both directions.
The unit circle defines sin/cos as coordinates (L17); unrolling them gives sinusoidal graphs with midline/amplitude/period (L18) that model periodic contexts via a·sin(b(x−c))+d (L19); tangent is the slope, inverses answer "which angle?" from restricted ranges (L20); equations are solved by inverse + symmetry + period (L21); identities rewrite expressions to make those solves possible (L22); polar coordinates reuse cos/sin as a positioning system with signature curves (L23); and rates of change read those curves' radial motion (L24). One system, eight lessons.
Problem: For r = 3 + 2cos θ, does the curve move toward or away from the pole as θ increases from 0 to π? What are the extreme distances?
Solution: On (0, π), cos θ falls from 1 to −1, so r falls from 5 to 1: the curve moves toward the pole throughout. Farthest point: r = 5 (θ = 0); closest on this sweep: r = 1 (θ = π).
Interpretation: The trig factor's monotonicity is the whole answer. No plotting required.
Problem: For r = 4 sin θ, compute the average rate of change of r over [π/6, π/3], and interpret it.
Solution: r(π/6) = 4(1/2) = 2; r(π/3) = 4(√3/2) = 2√3.
AROC = (2√3 − 2)/(π/3 − π/6) = (2√3 − 2)/(π/6) = 12(√3 − 1)/π ≈ 2.797
On average, r grows ≈ 2.8 distance-units per radian over this sweep — the point is moving away from the pole (positive rate).
Interpretation: Exact form first (12(√3−1)/π), decimal second. The sign tells the direction of radial motion; the magnitude, how urgently.
Problem: For the cardioid r = 2 + 2 sin θ on [0, 2π), find where the distance to the pole is at a relative maximum, at a relative minimum, and where the curve touches the pole.
Solution: r follows sin θ: max where sin θ = 1 → θ = π/2, r = 4 (relative — in fact global — max). Min where sin θ = −1 → θ = 3π/2, r = 0: the relative minimum is the pole touch (the cardioid's cusp). r = 0 only there.
Interpretation: For limaçons/cardioids, extrema of r sit where the trig part peaks/troughs. When the minimum value is exactly 0, the "closest approach" and "pole crossing" coincide — the cusp.
Problem: The function r(θ) = 5 − 3cos θ. (a) Describe its (θ, r) graph. (b) Use that to describe the polar curve's radial behavior on [0, 2π), including extremes.
Solution: (a) A sinusoid in θ: midline 5, amplitude 3, period 2π, reflected cosine (−3cos θ): starts at its minimum (0, 2), peaks at (π, 8), returns to (2π, 2). (b) So the polar curve starts at distance 2 (θ = 0, on the positive x-axis), moves steadily away from the pole through the upper half (r increasing on (0, π)), reaches its far point 8 at θ = π (negative x-axis), then draws back toward the pole through the lower half, returning to distance 2. Since min r = 2 > 0, the curve never touches the pole: a dimpled limaçon (a = 5 > b = 3, and a < 2b... check: 2b = 6 > 5 → dimpled ✓).
Interpretation: Analyze the sinusoid first — it's familiar territory — then wrap. Every radial-motion sentence about the polar view is a monotonicity sentence about the sinusoid view.
1. (A). cos θ falls on (0, π) → r falls 5 → 1 → toward the pole.
2. (B). (4 − 0)/(π/2) = 8/π. (A) forgets to divide by the half-π properly (4/(π) misses the 2).
3. (A). r tracks sin θ, which increases on (0, π/2) and again on (3π/2, 2π). (B) is where sin is positive — value, not direction.
4. (C). −3cos θ peaks when cos θ = −1: r = 5 + 3 = 8 at θ = π. (A) is the midline; (B) the minimum.
5. (C). (2√3 − 2)/(π/6) = 12(√3 − 1)/π ≈ 2.80 (Example 2). (A) halves it; (D) inverts the division.
6. (B). r = 1 there — the smallest value r takes (cos at its trough): relative minimum (and not zero, since 3 − 2 = 1 > 0, so (C) fails).
7. (D). (1/cos θ)·cos θ = 1 wherever defined.
8. (C). T = 2π/(π/4) = 8. (B) drops the 2; (D) ignores b.
9. (B). cos 0 = 1 → arccos(1) = 0. (A) is arccos(−1).
10. (B). cos θ falls from 1 to 0 on (0, π/2): r falls 6 → 0 — toward the pole throughout (all the way onto it). (A) invents a turnaround.
11. (D). tan's asymptotes at argument ±π/2: x/2 = ±π/2 → x = ±π. (A) forgets the horizontal stretch by 2.
12. (FRQ-style, 6 points) (i) [2 pts] Max r = 4 + 3 = 7 at θ = π/2; min r = 4 − 3 = 1 at θ = 3π/2. Since min r = 1 > 0, the curve never passes through the pole (a dimpled limaçon: a = 4, b = 3, a > b). (ii) [2 pts] Toward the pole where r decreases, i.e. where sin θ decreases: (π/2, 3π/2). (iii) [2 pts] AROC = (r(π) − r(π/2))/(π − π/2) = (4 − 7)/(π/2) = −6/π ≈ −1.91 per radian. Negative → on average the curve is moving toward the pole on this interval (distance shrinking from 7 to 4).
12. (FRQ-style) 🚫 Let r(θ) = 4 + 3 sin θ for 0 ≤ θ < 2π. (i) Find the maximum and minimum values of r and the θ-values where they occur. Does the curve pass through the pole? (ii) On which θ-intervals is the curve moving toward the pole? (iii) Compute the average rate of change of r over [π/2, π], exactly, and interpret its sign.
The polar function r(θ) = 6 cos θ is graphed for −π/2 ≤ θ ≤ π/2 (which traces the full circle of radius 3 centered at (3, 0)).
(a) (i) Compute r at θ = −π/2, 0, π/2. (ii) Identify where r attains its maximum on the interval and the corresponding point's rectangular coordinates.
(b) (i) On which subinterval is r increasing, and on which decreasing? (ii) Translate each into a statement about the curve's distance from the pole.
(c) The average rate of change of r over [0, π/3] is (6cos(π/3) − 6cos 0)/(π/3) = (3 − 6)/(π/3) = −9/π. (i) Interpret this value. (ii) A student says a negative rate means the curve is "below the x-axis" on that interval. Correct the misinterpretation.
(a) [2 pts] (i) [1 pt] r(−π/2) = 0, r(0) = 6, r(π/2) = 0. (ii) [1 pt] Max r = 6 at θ = 0: rectangular point (x, y) = (r cos θ, r sin θ) = (6·1, 6·0) = (6, 0) — the far side of the circle from the pole.
(b) [2 pts] (i) [1 pt] Increasing on (−π/2, 0) (cos θ rising toward 1); decreasing on (0, π/2). (ii) [1 pt] From θ = −π/2 to 0 the curve moves away from the pole (distance grows 0 → 6); from 0 to π/2 it returns toward the pole (6 → 0).
(c) [2 pts] (i) [1 pt] On average, between θ = 0 and θ = π/3, the curve's distance from the pole shrinks by 9/π ≈ 2.86 units per radian of sweep — the point is closing in on the pole. (ii) [1 pt] The sign of r's rate of change describes radial motion (toward/away from the pole), not vertical position. In fact the curve is above the x-axis for 0 < θ < π/2 (y = r sin θ > 0 there) even while r decreases. "Negative rate" = approaching the pole.
1. (A). cos θ falls on (0, π) → r falls 5 → 1 → toward the pole.
2. (B). (4 − 0)/(π/2) = 8/π. (A) forgets to divide by the half-π properly (4/(π) misses the 2).
3. (A). r tracks sin θ, which increases on (0, π/2) and again on (3π/2, 2π). (B) is where sin is positive — value, not direction.
4. (C). −3cos θ peaks when cos θ = −1: r = 5 + 3 = 8 at θ = π. (A) is the midline; (B) the minimum.
5. (C). (2√3 − 2)/(π/6) = 12(√3 − 1)/π ≈ 2.80 (Example 2). (A) halves it; (D) inverts the division.
6. (B). r = 1 there — the smallest value r takes (cos at its trough): relative minimum (and not zero, since 3 − 2 = 1 > 0, so (C) fails).
7. (D). (1/cos θ)·cos θ = 1 wherever defined.
8. (C). T = 2π/(π/4) = 8. (B) drops the 2; (D) ignores b.
9. (B). cos 0 = 1 → arccos(1) = 0. (A) is arccos(−1).
10. (B). cos θ falls from 1 to 0 on (0, π/2): r falls 6 → 0 — toward the pole throughout (all the way onto it). (A) invents a turnaround.
11. (D). tan's asymptotes at argument ±π/2: x/2 = ±π/2 → x = ±π. (A) forgets the horizontal stretch by 2.
12. (FRQ-style, 6 points) (i) [2 pts] Max r = 4 + 3 = 7 at θ = π/2; min r = 4 − 3 = 1 at θ = 3π/2. Since min r = 1 > 0, the curve never passes through the pole (a dimpled limaçon: a = 4, b = 3, a > b). (ii) [2 pts] Toward the pole where r decreases, i.e. where sin θ decreases: (π/2, 3π/2). (iii) [2 pts] AROC = (r(π) − r(π/2))/(π − π/2) = (4 − 7)/(π/2) = −6/π ≈ −1.91 per radian. Negative → on average the curve is moving toward the pole on this interval (distance shrinking from 7 to 4).
🎯 Exam tip: For any polar rate question, sketch the humble (θ, r) sinusoid in the margin — increasing/decreasing/extrema read off it instantly, and each fact translates one-for-one into radial motion. The wrapped polar picture is for checking, not for reasoning.