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3.14 · Polar Function Notation
In r=f(θ), which is the input (independent variable) and which is the output?
θ is angle, r is radius
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3.14 · AP Exam Fact
On the AP Precalculus Exam, are polar function questions free-response or multiple choice?
Important for strategy!
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3.14 · Cardioid Key Values
For f(θ)=3−3sinθ, find r at θ=0, π/2, π, 3π/2, 2π.
Where is the pinch? Where is the max?
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3.14 · Evaluation Strategy
What is the two-step evaluation strategy for identifying f(θ) from a polar graph?
Which θ values to test first?
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3.14 · sin vs cos at θ=0
If the polar graph passes through the ORIGIN at θ=0, does f use sin or cos?
Evaluate sin(0) and cos(0)
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3.14 · Example 2 — Rose
Graph shows 3-petal rose, r=0 at θ=0, r=3 at θ=π/6. Which is f(θ)?
f(0)=0 eliminates two choices
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3.14 · Example 3 — Limaçon
Graph shows f(0)=2 and f(π/2)=−2. Which is f(θ)?
Check θ=0 first, then θ=π/2
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3.14 · Example 4 — Domain
f(θ)=−3sinθ. The shown arc starts at r=−3 and increases to 0. What is [a,b]?
f(π/2)=−3, f(π)=0
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✓ Polar Function Notation
θ = input (independent/angle). r = output (dependent/radius).
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✓ AP Exam Fact
MULTIPLE CHOICE only. No sketching by hand required. Focus on evaluating at key θ values and reading the graph.
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✓ Cardioid Key Values
f(0)=3 · f(π/2)=0 (pinch) · f(π)=3 · f(3π/2)=6 (max) · f(2π)=3
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✓ Evaluation Strategy
Step 1: read r at θ=0 and θ=π/2 from graph. Step 2: evaluate each choice, eliminate mismatches.
θ=0 distinguishes sin (gives 0) from cos (gives max). θ=π/2 confirms sign.
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✓ sin vs cos at θ=0
Uses sin — sin(0)=0 → r=0 at θ=0. cos(0)=1 so cos-based curves don't pass through origin at θ=0.
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✓ Example 2 — Rose
f(θ) = 3sin(3θ)
f(0)=0 eliminates cos choices. f(π/6)=3sin(π/2)=3 ✓.
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✓ Example 4 — Domain
[a,b] = [π/2, π]
f(π/2)=−3, f(π)=0. Negative and increasing.
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✓ Example 3 — Limaçon
f(θ) = 2−4sin(θ)
f(0)=2: C(=6)✗, D(=−2)✗. f(π/2): 2−4=−2 ✓.
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3.14 · Example 5 — Petal
f(θ)=3cos(3θ). One petal has r positive and decreasing. What is [a,b]?
Period of cos(3θ)=2π/3. One petal=π/3.
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3.14 · Example 6 — Two Pieces
f(θ)=6cos(3θ), domain π/3≤θ≤2π/3. One piece is C→D (Q3). What is the other?
f(π/2)=0 (D), f(2π/3)=6 (A in Q2)
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3.14 · Rose Petals Rule
r=a·sin(nθ) or r=a·cos(nθ). How many petals if n is odd? If n is even?
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3.14 · Negative r — Core Rule
Where does a polar point (r, θ) plot when r is NEGATIVE?
Think: opposite direction of θ
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3.14 · Negative r — Formula
State the equivalence formula: (−r, θ) = ?
Add something to θ
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3.14 · Negative r — Quadrant
θ is in Q2 and r < 0. In which quadrant does the point actually appear?
Opposite of Q2 is…
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3.14 · Negative r — sinθ trace
r = sinθ. On θ∈[π, 2π], r is negative. Where do those points plot on the polar graph?
sin(3π/2)=−1. The angle 3π/2 points down. r=−1 flips it…
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3.14 · Negative r — Example
Point (−1, 3π/2). Step by step: what is the equivalent positive-r form, and where does it plot?
(−r, θ) = (r, θ+π)
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✓ Example 6 — Two Pieces
Q2 from D to A
D=origin at θ=π/2. On [π/2,2π/3]: r→6, positive increasing, pointing Q2.
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✓ Example 5 — Petal
[a,b] = [0, π/6]
f(0)=3 (max), f(π/6)=0. Positive, decreasing.
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✓ Negative r — Core Rule
The point (r, θ) with r<0 plots in the OPPOSITE direction of θ — it is reflected across the origin into the opposite quadrant.
Face angle θ, then walk BACKWARDS distance |r|. The angle tells direction but negative r means you go behind the origin.
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✓ Rose Petals Rule
n odd → n petals. n even → 2n petals. Petal length = |a|.
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✓ Negative r — Formula
(−r, θ) = (r, θ+π)
Adding π to the angle is equivalent to negating r. Both describe the same physical point. Example: (−2, π/4) = (2, 5π/4).
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✓ Negative r — Quadrant
Opposite of Q2 is Q4. If θ is in Q2 and r<0, the point appears in Q4.
Rule: opposite quadrant. Q1↔Q3 · Q2↔Q4. The reflection is exactly 180° (π radians) across the origin.
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✓ Negative r — sinθ trace
On θ∈[π,2π], sinθ<0 → negative r → points REFLECT into Q1/Q2 → traces the SAME upper semicircle as θ∈[0,π]. That's why r=sinθ completes in [0,π]!
The negative r on the second half doesn't create new points — it retraces the circle. This is why off-center circles only need [0,π] to complete.
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✓ Negative r — Example
(−1, 3π/2) = (1, π/2) → plots at rectangular (0,1) = top of circle
sin(3π/2)=−1 so r=−1. θ=3π/2 points DOWN. Negative r flips UP. So: (−1, 3π/2) = (1, 3π/2+π) = (1, 5π/2) = (1, π/2) ✓
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3.14 · Basics — r=a
What shape does r=a describe? What about θ=α?
These are the simplest polar graphs
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3.14 · Off-Center Circles
State the four off-center circle equations and each one's center in rectangular coordinates.
r=2asinθ, r=−2asinθ, r=2acosθ, r=−2acosθ
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3.14 · Circle Cycle
Off-center circles r=a sinθ and r=a cosθ: over what interval is the circle completely traced?
Shorter than [0,2π]!
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3.14 · Limaçon Types
Name all four types of limaçon and the a/b ratio that produces each.
r = a ± b cosθ or a ± b sinθ
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3.14 · Limaçon Direction
r = a + b sinθ: which direction does the bulge face? What about r = a − b cosθ?
+sin=up, +cos=right, −sin=down, −cos=left
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3.14 · cos vs sin Rose
How does a cos rose differ from a sin rose visually? Where does the first petal start?
Hint: evaluate at θ=0
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3.14 · Cardioid Direction
State the direction (up/down/left/right) for each of the four cardioid forms.
r=a(1±cosθ) and r=a(1±sinθ)
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3.14 · Cardioid Max/Pinch
For r=a(1+cosθ): where is the maximum r, and where does the cardioid pinch (r=0)?
Check θ=0 and θ=π
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✓ Basics — r=a, θ=α
r=a → circle of radius |a| at origin. θ=α → radial line through origin. r=a secθ → vertical line x=a. r=a cscθ → horizontal line y=a.
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✓ Off-Center Circles
r=2asinθ: center (0,a) · r=−2asinθ: center (0,−a) · r=2acosθ: center (a,0) · r=−2acosθ: center (−a,0). All have radius a.
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✓ Circle Cycle
[0, π]
Off-center circles complete in [0,π]. The second half [π,2π] retraces the same circle due to negative r reflecting back.
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✓ Limaçon Types
a/b<1: inner loop · a/b=1: cardioid · 1
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✓ Limaçon Direction
+cosθ: RIGHT · −cosθ: LEFT · +sinθ: UP · −sinθ: DOWN
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✓ cos vs sin Rose
cos: petal starts ON polar axis (f(0)=a). sin: rotated — f(0)=0, no petal at axis.
sin rose = cos rose rotated by π/(2n).
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✓ Cardioid Direction
r=a(1+cosθ)→RIGHT · r=a(1−cosθ)→LEFT · r=a(1+sinθ)→UP · r=a(1−sinθ)→DOWN
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✓ Cardioid Max/Pinch
r=a(1+cosθ): max r=2a at θ=0 (right), pinch r=0 at θ=π.
Opposite for (1−cosθ): max LEFT, pinch at origin.