A-Level Further Mathematics revision: Polar Coordinates. Learning objectives, key points, worked examples and practice questions across AQA, Edexcel, OCR, WJEC and CCEA.
📌 Key Points
Key Fact: x = r cos θ, y = r sin θ; r = sqrt(x^2+y^2), tan θ = y/x (check quadrant)
Key Fact: r = a (circle centred at origin), r = 2a cos θ (circle on x-axis), r = 2a sin θ (circle on y-axis)
Key Fact: r = a(1 + cos θ) (cardioid), r = a cos(nθ) (n-petal rose if n odd, 2n-petal if n even)
Key Fact: r = aθ (Archimedean spiral), r^2 = a^2 cos 2θ (lemniscate)
Key Fact: Area = ½∫_{α}^{β} r^2 dθ; Area between curves = ½∫(r₁^2 - r₂^2) dθ
Key Fact: Tangents: dy/dθ = sin θ dr/dθ + r cos θ, dx/dθ = cos θ dr/dθ - r sin θ; dy/dx = (dy/dθ)/(dx/dθ)
Key Fact: Horizontal tangent: dy/dθ = 0 (r cos θ + sin θ dr/dθ = 0); Vertical: dx/dθ = 0
🎯 Learning Objectives
Convert between polar (r,θ) and Cartesian (x,y) coordinates
Sketch standard polar curves: circles, cardioids, roses, spirals, lemniscates
Find area enclosed by polar curves
Find tangents to polar curves (horizontal/vertical)
Understand symmetry in polar graphs
💡 Worked Example
Exam-Style Question
Question: Find the area enclosed by one petal of r = 2 cos 3θ
Model Answer:
r = 0 when cos 3θ = 0 -> 3θ = pi/2 -> θ = pi/6. One petal from -pi/6 to pi/6. Area = ½∫_{-pi/6}^{pi/6} (2 cos 3θ)^2 dθ = 2∫_{-pi/6}^{pi/6} cos^2 3θ dθ = ∫(1+cos 6θ) dθ = [θ + ⅙ sin 6θ]_{-pi/6}^{pi/6} = pi/3
❓ Practice Questions
Model answers are being added progressively - questions marked ✗ don't have one yet. Cross-check with your teacher or the official mark scheme.
Questions:
Convert r = 2 sin θ to Cartesian, identify the curve✗ answer coming soon
Sketch r = 1 + 2 cos θ✗ answer coming soon
Area inside r = 2 and outside r = 1 + cos θ✗ answer coming soon
Find horizontal tangents of r = 2 sin 2θ✗ answer coming soon
Area of one loop of r^2 = 4 cos 2θ✗ answer coming soon
Worked example (10 min): Model the example question: Find the area enclosed by one petal of r = 2 cos 3θ. Solution: r = 0 when cos 3θ = 0 -> 3θ = pi/2 -> θ = pi/6. One petal from -pi/6 to pi/6. Area = ½∫_{-pi/6}^{pi/6} (2 cos 3θ)^2 dθ = 2∫_{-pi/6}^{pi/6} cos^2 3θ dθ = ∫(1+cos 6θ) dθ = [θ + ⅙ sin 6θ]_{-pi/6}^{pi/6} = pi/3
Practice (10 min): Students attempt the practice questions independently; circulate and support.
Plenary (5 min): Review answers and address misconceptions.
🏠 Homework
Convert r = 2 sin θ to Cartesian, identify the curve
Sketch r = 1 + 2 cos θ
Area inside r = 2 and outside r = 1 + cos θ
Find horizontal tangents of r = 2 sin 2θ
Area of one loop of r^2 = 4 cos 2θ
🧾 Assessment
Check practice answers against the model answer; use the built-in practice questions as formative assessment.