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📚 Topic 3.14 · Unit 3

Polar Function
Graphs

r = f(θ) in the polar coordinate system · graphing by table of values · identifying polar functions from graphs · domain restrictions and portions of polar curves.

📋 Notes 🃏 Flashcards 🧠 Practice Quiz
🧭 Part 1 — Polar Functions and Notation
📐
The Basics of Polar Functions
θ is input (independent), r is output (dependent) · typical domain [0, 2π]
📌 Standard Phrasing on AP Exam
"The graph of the polar function r = f(θ), where f(θ) = 3 − 3sin(θ), is shown in the polar coordinate system for 0 ≤ θ ≤ 2π."
🔑 Key Variables
θ = input (angle — the independent variable)
r = f(θ) = output (radius — the dependent variable)
Domain: usually [0, 2π] for a complete curve
🎯

AP Exam fact: polar functions are MULTIPLE CHOICE only

On the AP Precalculus Exam, you will never need to sketch a polar function by hand. All polar function questions are multiple choice. The key skills are: reading r values from a graph, evaluating f at key θ values, and understanding what happens when the domain is restricted.

🔄 Part 1B — Negative r and Opposite Quadrant Reflection
⬅️
What Negative r Does to a Polar Point
Negative r reflects the point across the origin into the opposite quadrant
Opposite Quadrant
If θ is in Q1 and r < 0, the point appears in Q3. If θ is in Q2, the point appears in Q4. Always the opposite quadrant.
Equivalent Form
(−r, θ) = (r, θ+π)
Adding π to the angle has the same effect as negating r.
Physical Meaning
Face direction θ, then walk backwards distance |r|. The angle tells direction but negative r means you go the other way.
📊 f(θ) = sin(θ) on [0, 2π] — rectangular plot (left) alongside polar plot (right)
Rectangular: y = sin(θ)
Polar: r = sin(θ)
θ ∈ [0, π]: sin(θ) ≥ 0 (positive r) → points plot in the direction of θ, in Q1 and Q2 → upper half circle
θ ∈ [π, 2π]: sin(θ) < 0 (negative r) → points REFLECT across origin into the OPPOSITE quadrant → also traces the upper half circle!
📌

Key insight: r=sinθ traces the SAME circle twice

On θ∈[0,π]: r=sinθ is positive → points plot in the direction of θ (Q1 and Q2) → upper half circle.
On θ∈[π,2π]: r=sinθ is negative → points reflect across the origin → land in Q1 and Q2 again → same upper circle retraced!
This is why r=sinθ only needs [0,π] to complete one full circle.

📌 Tracing a specific negative-r point: θ = 3π/2, f(θ) = sin(3π/2) = −1
1
Angle direction: θ=3π/2 points straight down (negative y-axis).
2
r = −1: Negative radius means go in the opposite direction — opposite of down is up.
3
Result: The point plots at distance 1 upward from the origin = (0, 1) in rectangular = top of the circle. This is equivalent to (r=1, θ=π/2).
4
Verify with formula: (−1, 3π/2) = (1, 3π/2+π) = (1, 5π/2) = (1, π/2) ✓
(−1, 3π/2) = (1, π/2) → rectangular: (0, 1) → top of the unit circle
📋
Example 1 — The Cardioid f(θ) = 3 − 3sin(θ)
Build a table, then plot each (r, θ) point on the polar grid
θr = 3−3sin(θ)θr = 3−3sin(θ)
037π/64.5
π/61.55π/45.121
π/40.8794π/35.598
π/30.4023π/26
π/205π/35.598
2π/30.4027π/45.121
3π/40.87911π/64.5
5π/61.52π3
π3
Key values: r=0 at θ=π/2 (pinch point at origin), r=6 at θ=3π/2 (max radius at bottom), r=3 at θ=0, π, 2π.
Cardioid: f(θ)=3−3sin(θ) — pinches at origin (θ=π/2), max r=6 at θ=3π/2
🔑

Reading the cardioid table

r = 3 − 3sinθ. When sinθ = 1 (θ=π/2), r = 0 — the curve passes through the origin. When sinθ = −1 (θ=3π/2), r = 6 — the farthest point. The cardioid is symmetric about the vertical axis. The "pinch" always faces the direction where r = 0.

🔍 Part 3 — Identifying Polar Functions from Graphs
🧮
The Evaluation Strategy — Examples 2 & 3
Evaluate at θ=0, π/2, π to eliminate wrong choices
📌

Strategy: evaluate at key angles to eliminate choices

Step 1: Read r values from the graph at θ=0 and θ=π/2.
Step 2: Evaluate each answer choice at those θ values.
Step 3: Eliminate any choice that doesn't match.
The two most useful test values: θ=0 (distinguishes sin from cos) and θ=π/2 (confirms sign and amplitude).

📌 Example 2 — 3-petal rose, r=0 at θ=0 on the polar axis, r=3 at θ=π/6. Find f(θ).
1
Read from graph: r(0) = 0 → eliminates (C) 3cos(3θ) [cos(0)=1≠0] and (D) −3cos(3θ) [−cos(0)≠0].
2
Check θ=π/6: f(π/6) should = 3 (petal tip on vertical axis).
(A) 3sin(3·π/6) = 3sin(π/2) = 3 ✓
(B) −3sin(π/2) = −3 ✗ → eliminates (B).
✓ (A)
3sin(3θ)
f(0)=0 ✓ · f(π/6)=3sin(π/2)=3 ✓
(B)
−3sin(3θ)
f(π/6)=−3 ✗
(C)
3cos(3θ)
f(0)=3≠0 ✗
(D)
−3cos(3θ)
f(0)=−3≠0 ✗
📌 Example 3 — Limaçon, r=2 at θ=0, r=−2 at θ=π/2 (inner loop going opposite direction). Find f(θ).
1
Read: f(0) = 2 → evaluate each: (A)=2+4(0)=2 ✓, (B)=2−4(0)=2 ✓, (C)=2+4(1)=6 ✗, (D)=2−4(1)=−2 ✗ → eliminates (C) and (D).
2
Read: f(π/2) = −2 → (A) 2+4sin(π/2) = 2+4 = 6 ✗, (B) 2−4sin(π/2) = 2−4 = −2 ✓ → eliminates (A).
(A)
2+4sin(θ)
f(π/2)=6≠−2 ✗
✓ (B)
2−4sin(θ)
f(0)=2 ✓ · f(π/2)=2−4=−2 ✓
(C)
2+4cos(θ)
f(0)=6≠2 ✗
(D)
2−4cos(θ)
f(0)=−2≠2 ✗
✂️ Part 4 — Domain Restrictions on Polar Curves
🎯
Which Piece of the Curve? — Examples 4, 5, 6
Restrict domain → only part of the curve remains · trace r value behavior
🔑

Strategy for domain restriction questions

Step 1: Evaluate r at the endpoints and midpoint of [a, b].
Step 2: Note whether r is positive or negative (negative r → point plots in opposite direction).
Step 3: Determine if r is increasing or decreasing on the interval.
Step 4: Match to the portion of the graph in the correct quadrant with the correct r behavior.

📌 Example 4 — f(θ)=−3sin(θ). The graph shows a portion for a≤θ≤b. The visible arc has r<0 (reflection through origin) and is decreasing then increasing. Find a and b.
1
Evaluate: f(0)=0, f(π/2)=−3 (minimum), f(π)=0.
2
On [0, π/2]: f goes 0→−3 (negative, decreasing). On [π/2, π]: f goes −3→0 (negative, increasing).
3
The shown arc starts and ends at the origin (r=0), dips to r=−3 in the middle → this is [π/2, π]. f(π/2)=−3 and f(π)=0.
(A)
a=0, b=π/2
f goes 0→−3, decreasing — but shown arc has r increasing from −3 ✗
✓ (B)
a=π/2, b=π
f(π/2)=−3 (start), f(π)=0 (end). r is negative and increasing (less negative). ✓
(C)
a=0, b=π
Full half — includes both decreasing and increasing, too wide ✗
(D)
a=π, b=2π
On [π,2π] f is positive (sin negative → r positive) ✗
📌 Example 5 — f(θ)=3cos(3θ). The shown arc has r positive and decreasing. Find a and b.
1
Period of cos(3θ) = 2π/3, so each "petal" takes θ-width of π/3. One petal: cos(3θ) starts at max and returns to 0.
2
f(0) = 3cos(0) = 3 (max, positive). f(π/6) = 3cos(π/2) = 0 (zero). So on [0, π/6]: r positive and decreasing from 3→0.
3
This is the ONLY interval where f is positive and decreasing starting from the maximum.
(A)
a=π, b=3π/2
f(π)=3cos(3π)=−3 (negative) ✗
(B)
a=0, b=π/2
Too wide — f goes negative on (π/6, π/2) ✗
(C)
a=0, b=π/3
Too wide — includes negative r region ✗
✓ (D)
a=0, b=π/6
f(0)=3 (max), f(π/6)=0. Positive and decreasing throughout. ✓
📌 Example 6 — f(θ)=6cos(3θ), domain restricted to π/3≤θ≤2π/3. Points A(Q2), B(polar axis), C(Q3), D(origin) are labeled. One piece is "C to D" (Q3). What is the other piece?
1
Evaluate on [π/3, 2π/3]:
f(π/3) = 6cos(π) = −6 → negative r at θ=π/3 → plots in opposite direction → Q3 (near C)
f(π/2) = 6cos(3π/2) = 0 → origin (point D)
f(2π/3) = 6cos(2π) = 6 → positive r at θ=2π/3 (pointing into Q2) → near A
2
Piece 1: [π/3, π/2] = C→D (negative r, Q3, returning to origin). This is given.
3
Piece 2: [π/2, 2π/3] = D→A. Starts at origin (D), r positive and increasing, θ in Q2 direction → ends at A in Q2.
(A)
Q1 from D to B
B is on polar axis (Q1 boundary), f(2π/3)=6 points into Q2, not Q1 ✗
(B)
Q1 from B to D
Wrong quadrant and wrong direction ✗
✓ (C)
Q2 from D to A
D=origin at θ=π/2, A in Q2. On [π/2, 2π/3]: r=0→6, positive, increasing. θ=2π/3 points into Q2. ✓
(D)
Q3 from D to C
That's Piece 1 reversed, and in Q3 ✗
🌹 Part 5 — Polar Graph Families
📐
1. Basics — Points, Lines & Circles at the Pole
Simple polar equations for circles, radial lines, vertical and horizontal lines
⭕ Circles centered at the pole
r = a
Circle of radius |a| centered at the origin. Simple constant function.
📏 Lines through the pole
θ = α
A straight radial line at constant angle α through the origin.
↕️ Vertical lines
r = a sec θ
Equivalent to x = a in rectangular coordinates.
↔️ Horizontal lines
r = a csc θ
Equivalent to y = a in rectangular coordinates.
⭕
2. Off-Center Circles — Pass Through the Pole
r = 2a sinθ or r = 2a cosθ · radius = a · complete in [0, π]
Equation Radius Center (rect.) Direction Color →
r = 2a sinθ a (0, a) ⬆ Up
r = −2a sinθ a (0, −a) ⬇ Down
r = 2a cosθ a (a, 0) ➡ Right
r = −2a cosθ a (−a, 0) ⬅ Left
💡 Key: All pass through the origin (pole). Diameter = 2a. Complete circle traced in [0, π] — the second half [π, 2π] retraces the same circle due to negative r.
All 4 off-center circles (a=1)
● r=2sinθ ● r=−2sinθ ● r=2cosθ ● r=−2cosθ
💡

Why all 4 circles pass through the origin

At θ=0: r=2a·sin(0)=0 or r=2a·cos(0)=2a — one endpoint is always at the origin. The complete circle is traced in just [0, π] because the second half [π, 2π] has negative r values that reflect back, retracing the exact same arc.

🐌
3. Limaçons — r = a ± b cosθ ; r = a ± b sinθ
Shape determined by ratio a/b · four types · bulge direction from ± and cos/sin
Inner Loop a/b < 1
Example: r = 1 + 2cosθ
(a=1, b=2, a/b=0.5)
Passes through the pole.
Inner loop on far side.
Cardioid ❤️ a/b = 1
Example: r = 1 + cosθ
(a=1, b=1, a/b=1)
Touches pole exactly once.
Heart-shaped.
Dimpled 1 < a/b < 2
Example: r = 3 + 2cosθ
(a=3, b=2, a/b=1.5)
Never reaches the pole.
Pushed in on one side.
Convex a/b ≥ 2
Example: r = 4 + 2cosθ
(a=4, b=2, a/b=2)
Oval/circle-like.
No dent, no loop.
🔑

Direction of the limaçon bulge

All four examples above use + cosθ → bulge faces right (toward the polar axis).
Replace cosθ with sinθ → bulge rotates up. Use − instead of + → bulge flips to left (cos) or down (sin).
The a/b ratio controls the shape; the ± and cos/sin control the direction.

🌹
4. Rose Curves — r = a cos(nθ) or r = a sin(nθ)
n odd → n petals · n even → 2n petals · petal length = |a| · cos vs sin = rotation
🌹 Petal Count Rule
n odd → n petalse.g. cos(3θ) → 3 petals
n even → 2n petalse.g. cos(2θ) → 4 petals
Petal length = |a|
🔄 cos vs sin
cos rose: petal ON the polar axis at θ=0 (f(0)=a).
sin rose: rotated — no petal at θ=0 (f(0)=0).
sin rose = cos rose rotated by π/(2n).
r = cos(3θ) n=3 odd → 3 petals
Petals: 3
Petal at θ=0? ✅ Yes — cos
Traces in: [0, π]
r = sin(3θ) n=3 odd → 3 petals
Petals: 3
Petal at θ=0? ❌ No — sin
Rotated by: π/6
r = cos(2θ) n=2 even → 4 petals
Petals: 2n = 4
Petal at θ=0? ✅ Yes — cos
Traces in: [0, 2π]
r = sin(2θ) n=2 even → 4 petals
Petals: 2n = 4
Petal at θ=0? ❌ No — sin
Rotated by: π/4
💡

Spot the difference between cos and sin roses instantly

Compare the top two graphs (cos(3θ) vs sin(3θ)) — same 3 petals, but sin is rotated 30° (= π/6 = π/(2·3)). Same for the bottom pair: cos(2θ) vs sin(2θ) are both 4-petal roses, but sin is rotated 45° (= π/4 = π/(2·2)). The rotation is always π/(2n).

❤️
5. Cardioids — r = a(1 ± cosθ) or r = a(1 ± sinθ)
Heart-shaped · touches pole once · max r = 2a · direction from ± and cos/sin
r = a(1 + cosθ) ➡ Right
Pinch: (0, π) — left
Max 2a: (2a, 0) — right
Bulge: toward polar axis →
r = a(1 − cosθ) ⬅ Left
Pinch: (0, 0) — right
Max 2a: (2a, π) — left
Bulge: away from polar axis ←
r = a(1 + sinθ) ⬆ Up
Pinch: (0, 3π/2) — bottom
Max 2a: (2a, π/2) — top
Bulge: upward ↑
r = a(1 − sinθ) ⬇ Down
Pinch: (0, π/2) — top
Max 2a: (2a, 3π/2) — bottom
Bulge: downward ↓
🔑

Reading cardioid direction instantly

cos cardioid: bulge along horizontal axis. +cosθ → right ❤️ bulges right, pinch at left. −cosθ → left ❤️ bulges left, pinch at right.
sin cardioid: bulge along vertical axis. +sinθ → up ❤️ bulges up, pinch at bottom. −sinθ → down ❤️ bulges down, pinch at top.
The colored dot on each graph marks the max r=2a. The pinch is always at the origin.

🔄
6. Common Polar Graph Cycles
How much of [0, 2π] is needed to trace each shape once
⭕ Off-center
Circle
r = a sinθ
r = a cosθ
[0, π]
🌹 Rose
(n odd)
r = a cos(nθ)   [n odd]
r = a sin(nθ)   [n odd]
[0, π]
🌹 Rose
(n even)
r = a cos(nθ)   [n even]
r = a sin(nθ)   [n even]
[0, 2π]
🐌 Cardioid
& Limaçon
r = a + b cosθ
r = a − b cosθ
r = a + b sinθ
r = a − b sinθ
[0, 2π]
∞ Lemniscate
r² = a²cos(2θ)
r² = a²sin(2θ)
[0, π]
⚡
Quick Reference
Screenshot and save this!
🔑 Key Strategies
To ID f(θ): evaluate at θ=0 and θ=π/2
f(0): sin→0, cos→max. Distinguishes type.
Negative r → point plots opposite to θ
Domain restriction: trace r from a to b
r=0 → curve passes through origin (pinch)
Limaçon: a/b<1=loop · =1=cardioid · 1-2=dimpled · ≥2=convex
Rose n odd→n petals · n even→2n petals · length=|a|
Cardioid: +cos→right · −cos→left · +sin→up · −sin→down
⚠️ Common Mistakes
❌ Confusing sin vs cos polar curves
f(0)=0 → must be sin-based. f(0)≠0 → must be cos-based. Check θ=0 first — it immediately eliminates half the choices.
❌ Ignoring negative r values
Negative r means the point plots in the OPPOSITE direction of θ. On [0,π/2], f(θ)=−3sinθ gives negative r, so the arc appears in Q3/Q4, not Q1.
❌ Confusing n-petal roses and period
r=a·sin(nθ) has n petals if n is odd, 2n petals if n is even. One petal spans θ-width of π/n. For 3cos(3θ): one petal width = π/3.
❌ Domain [a,b] gives the whole curve
Restricting domain cuts out portions. Identify what r does on the restricted interval — only that arc is drawn.
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