📏 Part 1 — Distance from Origin: The Core Rule
The Signed Radius and Distance
r can be positive or negative — the "signed radius" tells distance AND direction
Key idea: distance from origin = |r|, not r
The actual distance from the curve to the origin is |r| (the absolute value). So even when r is negative, the curve is still some positive distance from the origin. What matters is whether |r| is growing or shrinking — which depends on the sign of r AND whether r is increasing or decreasing.
| r = f(θ) is… | Then the distance from the curve to the origin is… | Why |
|---|---|---|
| positive and increasing | INCREASING | r > 0 and growing → |r| grows |
| negative and decreasing | INCREASING | r < 0 and more negative → |r| grows |
| positive and decreasing | DECREASING | r > 0 and shrinking → |r| shrinks |
| negative and increasing | DECREASING | r < 0 and less negative → |r| shrinks |
Memory shortcut — same sign = distance increasing
If the sign of r and the direction of change match → distance increasing:
· Positive + increasing = same direction ✓ → farther
· Negative + decreasing = same direction ✓ → farther
If they oppose → distance decreasing:
· Positive + decreasing → closer · Negative + increasing → closer
📌 Example 1 — f(x)=1+2sinx on [0,2π]. Complete the table of intervals.
1
Find zeros: 1+2sinx=0 → sinx=−1/2 → x=7π/6 and x=11π/6.Max at x=π/2 (f=3), min at x=3π/2 (f=−1).
f(x)=1+2sinx — teal=positive region, pink=negative region. Zeros at 7π/6 and 11π/6.
| Description of f | Interval(s) | Distance |
|---|---|---|
| Positive and increasing | [0, π/2) and (11π/6, 2π] | Increasing |
| Positive and decreasing | (π/2, 7π/6) | Decreasing |
| Negative and decreasing | (7π/6, 3π/2) | Increasing |
| Negative and increasing | (3π/2, 11π/6) | Decreasing |
🔍 Part 2 — Applying the Rule (Examples 2 & 3)
Using a Rectangular Sketch to Analyze Polar Distance
AP Exam tip: sketch f as a rectangular function first, then determine distance behavior
AP Exam strategy: sketch in rectangular coordinates first
Even though the question is about the polar graph, it's much easier to see whether r is positive/negative and increasing/decreasing on a rectangular (Cartesian) y=f(θ) graph. Sketch f(θ) vs θ, identify the sign and slope in each interval, then apply the rules.
📌 Example 2 — f(θ)=2−4cosθ on [0,2π]. Which statement is true about the distance from (f(θ),θ) to the origin?
1
Key values: f(0)=−2, f(π/3)=0 (zero), f(π)=6 (max), f(5π/3)=0 (zero), f(2π)=−2.2
On (π, 5π/3): f goes from 6→0 → positive and decreasing → distance decreasing.3
On (5π/3, 2π): f goes from 0→−2 → negative and decreasing → distance increasing. Eliminates (A), (B), (D).(A)
π < θ < 5π/3: increasing
f is positive but decreasing → distance decreasing ✗
(B)
5π/3 < θ < 2π: increasing because f negative and increasing
On (5π/3,2π) f is negative and decreasing (going to −2), not increasing ✗
✓ (C)
π < θ < 5π/3: decreasing because f positive and decreasing
f(π)=6, f(5π/3)=0 → positive, decreasing → distance decreasing ✓
(D)
5π/3 < θ < 2π: decreasing because f negative and decreasing
f is negative and decreasing on this interval → distance INCREASING, not decreasing ✗
📌 Example 3 — f(θ)=3sin(2θ) on [0, 3π/2]. On which interval is distance decreasing?
1
f(x)=3sin(2x) has period π. Sketch: positive hump [0,π/2], negative trough [π/2,π], positive hump [π,3π/2].2
Key values: f(3π/4)=3sin(3π/2)=−3 (minimum), f(π)=0.3
On (3π/4, π): f goes from −3→0 → negative and increasing → distance decreasing.(A)
0 < θ < π/4
f positive and increasing → distance increasing ✗
(B)
π/2 < θ < 3π/4
f negative and decreasing → distance increasing ✗
✓ (C)
3π/4 < θ < π
f(3π/4)=−3, f(π)=0 → negative and increasing → distance decreasing ✓
(D)
π < θ < 5π/4
f positive and increasing → distance increasing ✗
📈 Part 3 — Relative Extrema of Polar Functions
Relative Maxima and Minima — Example 4
Max: r changes increasing→decreasing · Min: r changes decreasing→increasing
| r = f(θ) changes from… | Then the polar graph has a… | Meaning |
|---|---|---|
| increasing → decreasing | RELATIVE MAXIMUM | Point farthest from origin in that region |
| decreasing → increasing | RELATIVE MINIMUM | Point closest to origin in that region |
Trap: sign change ≠ relative extremum
If r changes from negative to positive (or positive to negative), that does NOT mean there's a relative extremum — it just means the curve crossed the origin. A relative extremum requires r to change from increasing to decreasing (max) or decreasing to increasing (min).
📌 Example 4 — f(θ)=1−2sin(2θ) on [0,π]. Which statement is true?
1
Key values: f(0)=1, f(π/4)=1−2(1)=−1 (local min of f), f(π/2)=1, f(3π/4)=1−2(−1)=3 (local max of f at r=3), f(π)=1.2
On (2π/3, π): f(θ) changes from increasing to decreasing (it peaks somewhere then starts down toward f(π)=1). The maximum of f occurs around θ≈3π/4 where f=3.3
On the interval (2π/3, π): r changes from increasing to decreasing → relative maximum on that interval.(A)
Rel. min on (π/3, 2π/3) because r changes neg→pos
Sign change ≠ relative extremum. r changing neg→pos does not give a minimum ✗
(B)
Rel. min on (2π/3, π) because r changes dec→inc
On (2π/3,π) r is actually changing increasing→decreasing ✗
(C)
Rel. max on (π/3, 2π/3) because r changes pos→neg
Sign change ≠ relative extremum ✗
✓ (D)
Rel. max on (2π/3, π) because r changes inc→dec
r increases then decreases on (2π/3,π) → relative maximum (farthest point from origin) ✓
📊 Part 4 — Average Rate of Change
ARC Formula + Examples 5, 6, 7
Change in r ÷ change in θ · units: radius per radian
📌 Average Rate of Change of a Polar Function
ARC = [f(b) − f(a)] / (b − a)
For r=f(θ) over the interval a≤θ≤b.
Geometrically: the average rate at which the radius r changes per radian of θ.
Same formula as rectangular ARC — just applied to polar r and θ instead of y and x.
Geometrically: the average rate at which the radius r changes per radian of θ.
Same formula as rectangular ARC — just applied to polar r and θ instead of y and x.
📌 Example 5 — f(θ)=3−3cosθ. Find the average rate of change on [π/2, π].
1
f(π) = 3−3cos(π) = 3−3(−1) = 6.2
f(π/2) = 3−3cos(π/2) = 3−3(0) = 3.3
ARC = (f(π)−f(π/2))/(π−π/2) = (6−3)/(π/2) = 3/(π/2) = 3·(2/π) = 6/π.ARC = 6/π ≈ 1.91 radians⁻¹ (r increases by 6/π per radian on this interval)
📌 Example 6 — f(θ)=3sin(2θ) on [0,2π]. On which interval is the ARC equal to zero?
1
ARC=0 when f(a)=f(b). Check each option:2
(A) f(π/4)=3sin(π/2)=3, f(3π/4)=3sin(3π/2)=−3. ARC≠0 ✗3
(B) f(3π/4)=−3, f(5π/4)=3sin(5π/2)=3. ARC≠0 ✗4
(C) f(3π/4)=3sin(3π/2)=−3, f(7π/4)=3sin(7π/2)=3sin(−π/2+4π)=−3. f(3π/4)=f(7π/4)=−3 → ARC=0 ✓(A)
π/4 ≤ θ ≤ 3π/4
f(π/4)=3, f(3π/4)=−3. ARC=(−3−3)/(π/2)=−12/π≠0 ✗
(B)
3π/4 ≤ θ ≤ 5π/4
f(3π/4)=−3, f(5π/4)=3. ARC=6/(π/2)≠0 ✗
✓ (C)
3π/4 ≤ θ ≤ 7π/4
f(3π/4)=−3 and f(7π/4)=−3. Same values → ARC=0 ✓
(D)
π/4 ≤ θ ≤ 7π/4
f(π/4)=3, f(7π/4)=−3. ARC≠0 ✗
📌 Example 7 — Polar graph with points A, B, C, D in order as θ increases. On which interval is the ARC of r least (most negative)?
1
"Least" means most negative (smallest value). ARC is negative when r decreased over the interval.2
From the graph: A to B: r stays same (ARC=0). B to C: r increases (ARC>0). C to D: r stays same (ARC=0). D to A: r decreased (ARC<0).3
The only negative ARC is D to A → this is the least.(A)
A to B
ARC = 0 (r unchanged) ✗
(B)
B to C
ARC > 0 (r increased) ✗
(C)
C to D
ARC = 0 (r unchanged) ✗
✓ (D)
D to A
ARC < 0 (r decreased) → least (most negative) ✓
📐 Part 5 — Estimating Values Using Linear Approximation
Linear Approximation with ARC — Example 8
Use ARC as slope in point-slope form to estimate f at nearby θ values
📌 Linear Approximation Formula
f(θ) ≈ f(θ₁) + [f(b)−f(a)]/(b−a) · (θ − θ₁)
θ₁ is a known point, [a,b] is the interval, and (θ−θ₁) is how far from θ₁ you want to estimate.
This is just point-slope form with the ARC as the slope.
This is just point-slope form with the ARC as the slope.
📌 Example 8 — Table: f(π/6)=2, f(7π/6)=−1. Use ARC over [π/6, 7π/6] to approximate f(5π/6).
1
Compute ARC over [π/6, 7π/6]: ARC = (f(7π/6)−f(π/6))/(7π/6−π/6) = (−1−2)/(π) = −3/π.2
Use point-slope from θ₁=π/6: f(5π/6) ≈ f(π/6) + ARC·(5π/6−π/6) = 2 + (−3/π)(4π/6) = 2 + (−3/π)(2π/3).3
= 2 + (−3·2π)/(3π) = 2 + (−6π)/(3π) = 2 − 2 = 0.f(5π/6) ≈ 0
Quick Reference
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🔑 Key Rules
r+ inc OR r− dec → distance INCREASING
r+ dec OR r− inc → distance DECREASING
r inc→dec: relative MAX · r dec→inc: relative MIN
ARC = [f(b)−f(a)] / (b−a)
Linear approx: f(θ)≈f(θ₁)+ARC·(θ−θ₁)
⚠️ Common Mistakes
❌ r negative and decreasing → distance decreasing
When r is negative and decreasing (more negative), |r| is GROWING → distance is INCREASING. Think about |r|, not r.
❌ Sign change of r = relative extremum
r going from negative to positive (or vice versa) just means the curve crossed the origin. Relative extremum requires r to change from increasing to decreasing (or vice versa).
❌ ARC = least means smallest positive
"Least" means most negative. An ARC of −5 is less than an ARC of 0 or +3. The interval where r decreased the most has the least ARC.
❌ ARC = 0 means the function is constant
ARC=0 over [a,b] just means f(a)=f(b) — the function returns to the same value. The function can go up and down in between.