8.4Graphing Equations in Polar Form
Graph cardioids (r = a ± a cosθ), limaçons, rose curves (r = a cos nθ), lemniscates, and spirals. Test for symmetry about the polar axis, θ = π/2, and the pole.
Polar curves like cardioids and rose curves produce beautiful and complex shapes from simple equations. Understanding their symmetry and structure is foundational for polar calculus (area between polar curves).
Essential Question
What are the characteristic shapes of polar curves — circles, limaçons, cardioids, roses, and lemniscates — and how do you identify and sketch them from their equations?
Lesson Overview
Basic Polar Curves
- •r = a — circle of radius |a| centered at the pole.
- •r = a cosθ or r = a sinθ — circle of diameter |a| passing through the pole.
- •θ = k — line through the pole at angle k.
Limaçons: r = a ± b cosθ or r = a ± b sinθ
- •|a/b| > 1: no inner loop (convex if a/b ≥ 2, dimpled if 1 < a/b < 2).
- •|a/b| = 1: cardioid (heart shape).
- •|a/b| < 1: limaçon with an inner loop.
Rose Curves: r = a cos(nθ) or r = a sin(nθ)
- •n even: 2n petals; n odd: n petals.
- •Petal length = |a|.
Lemniscates: r² = a² cos(2θ) or r² = a² sin(2θ)
Figure-eight shape symmetric about the pole. The curve only exists where the right-hand side is ≥ 0.
Symmetry Tests
- •Polar axis (x-axis): replace θ with −θ. If the equation is unchanged, the curve is symmetric about the polar axis.
- •Line θ = π/2 (y-axis): replace θ with π − θ. If unchanged, symmetric about θ = π/2.
- •Pole: replace r with −r. If unchanged, symmetric about the pole.
Worked Examples
Identify and sketch r = 3cosθ.
Recognize the form r = a cosθ with a = 3.
This is a circle of diameter |a| = 3 passing through the pole.
Multiply both sides by r: r² = 3r cosθ → x² + y² = 3x.
Complete the square: (x − 3/2)² + y² = (3/2)².
Center: (3/2, 0) in rectangular; radius = 3/2.
Identify and sketch r = 2 + 2sinθ.
Form: r = a + b sinθ with a = 2, b = 2.
Compute a/b = 2/2 = 1, so this is a cardioid.
Maximum r: at θ = π/2, r = 2 + 2(1) = 4.
Minimum r: at θ = 3π/2, r = 2 + 2(−1) = 0 (passes through pole).
Symmetric about the line θ = π/2 (y-axis) because sinθ is unchanged when θ → π − θ.
Identify and sketch r = 1 + 3cosθ. Find max r, min r, and where the inner loop occurs.
Form: r = a + b cosθ with a = 1, b = 3.
a/b = 1/3 < 1, so this is a limaçon with an inner loop.
Max r: at θ = 0, r = 1 + 3 = 4.
Min r: at θ = π, r = 1 − 3 = −2 (|r| = 2, plotted in opposite direction).
Inner loop: set r = 0 → 1 + 3cosθ = 0 → cosθ = −1/3 → θ = arccos(−1/3) ≈ 109.5° and θ ≈ 250.5°.
The inner loop is traced for θ between these two values.
Identify and sketch r = 4cos(3θ). Find the petal tips.
Form: r = a cos(nθ) with a = 4, n = 3 (odd).
n is odd → 3 petals; petal length = |a| = 4.
Petal tips occur where r = 4, i.e., cos(3θ) = 1 → 3θ = 0, 2π, 4π → θ = 0, 2π/3, 4π/3.
Petal tips: (4, 0°), (4, 120°), (4, 240°).
Petals lie along θ = 0, 2π/3, 4π/3.
Identify and sketch r² = 9cos(2θ). Find the extent of the curve.
Form: r² = a² cos(2θ) with a² = 9, a = 3. This is a lemniscate.
The curve exists only where cos(2θ) ≥ 0.
cos(2θ) ≥ 0 when 2θ ∈ [−π/2, π/2] ∪ [3π/2, 5π/2], i.e., θ ∈ [−π/4, π/4] ∪ [3π/4, 5π/4].
Maximum r: r² = 9 → r = 3, at θ = 0 and θ = π.
The curve is a figure-eight symmetric about the polar axis.
Guided Practice
Identify the curve r = −4sinθ and find its center and radius in rectangular form.
Hint: Multiply both sides by r to get r² = −4r sinθ, then use x² + y² = r² and y = r sinθ. Complete the square.
Identify r = 3 − 3cosθ and describe its symmetry.
Hint: Compute a/b = 3/3 = 1. What type of curve has a/b = 1? Test symmetry by replacing θ with −θ.
How many petals does r = 5sin(4θ) have? Find the length of each petal.
Hint: Use the rule: n even → 2n petals. Here n = 4. Petal length = |a|.
For r = 2 + 5cosθ, find the maximum and minimum values of r and describe the curve type.
Hint: Max r when cosθ = 1; min r when cosθ = −1. Then compute a/b = 2/5 to classify.
Test r = 4cos(2θ) for symmetry about the polar axis, the line θ = π/2, and the pole.
Hint: Polar axis: replace θ with −θ. Line θ = π/2: replace θ with π − θ. Pole: replace r with −r. Use cos(2θ) = cos(−2θ) and cos(2(π−θ)) = cos(2π−2θ) = cos(2θ).
Key Vocabulary
Polar curve
A curve defined by an equation in polar coordinates (r, θ).
Example: r = 1 + cosθ is a cardioid.
Limaçon
A polar curve of the form r = a ± b cosθ or r = a ± b sinθ. Shape depends on the ratio a/b.
Example: r = 2 + 3cosθ (inner loop since a/b < 1)
Cardioid
A special limaçon where a = b (i.e., a/b = 1). Heart-shaped curve that passes through the pole.
Example: r = 3 + 3sinθ
Rose curve
A polar curve of the form r = a cos(nθ) or r = a sin(nθ). Has n petals if n is odd, 2n petals if n is even.
Example: r = 2cos(3θ) has 3 petals
Lemniscate
A figure-eight shaped polar curve of the form r² = a² cos(2θ) or r² = a² sin(2θ).
Example: r² = 16cos(2θ), max r = 4
Inner loop
The smaller loop of a limaçon that appears when a/b < 1. It is traced when r is negative.
Example: r = 1 + 2cosθ has an inner loop
Petal
One lobe of a rose curve. The tip of a petal is at distance |a| from the pole.
Example: r = 5cos(2θ) has 4 petals each of length 5
Symmetry (polar)
A polar curve can be symmetric about the polar axis, the line θ = π/2, or the pole. Each is tested by substituting into the equation.
Example: r = cos(2θ) is symmetric about all three
Quick Check
Interactive Practice — 5 Questions
The equation r = 5 describes which curve?
r = 2 + 2cosθ is a:
How many petals does r = 3sin(5θ) have?
r² = 16sin(2θ) is a:
For r = 1 + 3cosθ, the maximum value of r is:
Common Mistakes
Saying r = a cos(nθ) always has n petals regardless of whether n is odd or even.
If n is odd, the rose has n petals. If n is even, the rose has 2n petals. For example, r = cos(4θ) has 8 petals.
Plotting r² = a² cos(2θ) for all values of θ, including where cos(2θ) < 0.
r² must be ≥ 0, so the lemniscate only exists where cos(2θ) ≥ 0, i.e., θ ∈ [−π/4, π/4] ∪ [3π/4, 5π/4].
Calling r = 2 + 2cosθ a limaçon with inner loop because it looks like one.
A cardioid has a/b = 1 exactly. r = 2 + 2cosθ has a = b = 2, so a/b = 1 — it is a cardioid, not a limaçon with inner loop.
Plotting only θ ∈ [0, π] for a rose curve and thinking you have the full curve.
Rose curves require θ ∈ [0, 2π] (or [0, π] for odd n). For even n, plotting only half the interval misses petals.
Math Tips
For limaçons r = a ± b cosθ: the ratio a/b determines the shape. a/b = 1 → cardioid; a/b < 1 → inner loop; 1 < a/b < 2 → dimpled; a/b ≥ 2 → convex.
Rose curves: odd n → n petals; even n → 2n petals. The petals of r = a cos(nθ) lie along the polar axis when n is odd.
To convert r = a cosθ to rectangular: multiply both sides by r to get r² = ar cosθ, then x² + y² = ax. Completing the square gives (x − a/2)² + y² = (a/2)², a circle.
Symmetry shortcut: if the equation is unchanged when θ → −θ (i.e., only even powers of sinθ appear, or sinθ is absent), the curve is symmetric about the polar axis.
For lemniscates r² = a² cos(2θ), the curve only exists where cos(2θ) ≥ 0, i.e., −π/4 ≤ θ ≤ π/4 and 3π/4 ≤ θ ≤ 5π/4.