Differential Calculus: Single Variable (8hrs) · Engineering Mathematics - I

Polar Curves

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Polar Curves

1. Polar Coordinates

A point PP is given by (r,θ)(r,θ) where rr = signed distance from pole OO and θθ = angle from polar axis (OxOx). Conversion:

x=rcosθ,y=rsinθ,r=√(x2+y2),tanθ=y/xx = r cosθ , y = r sinθ , r = √(x²+y²) , tanθ = y/x

A polar curve is r=f(θ)r = f(θ) (or F(r,θ)=0F(r,θ)=0). Examples: r=a(1−cosθ)r = a(1-cosθ) (cardioid), r=acos2θr = a cos 2θ (four-petalled rose), r=aθr = aθ (spiral of Archimedes), r=a(1+cosθ)r = a(1+cosθ) variations.

2. Symmetry — Quick Tests

For r=f(θ)r = f(θ):

  • Symmetric about initial line (θ=0θ=0): f(−θ)=f(θ)f(-θ)=f(θ) or f(2π−θ)=f(θ)f(2π-θ)=f(θ) — replace θ→−θθ→-θ, equation unchanged.
  • Symmetric about pole (OO): f(θ+π)=−f(θ)f(θ+π)= -f(θ) or equation in r2r².
  • Symmetric about θ=π/2θ=π/2: f(π−θ)=f(θ)f(π-θ)=f(θ).
    Symmetry halves sketching work; examiners accept a symmetry statement before plotting.

3. Tangent in Polar Form

In Cartesian terms x(θ)=f(θ)cosθ,y(θ)=f(θ)sinθx(θ)=f(θ)cosθ, y(θ)=f(θ)sinθ. The slope is:

dy/dx=[f′(θ)sinθ+f(θ)cosθ]/[f′(θ)cosθ−f(θ)sinθ]dy/dx = [f'(θ) sinθ + f(θ) cosθ] / [f'(θ) cosθ - f(θ) sinθ]

Let φφ be angle between radius vector OPOP and tangent at PP:

tanφ=r/[dr/dθ]=r/r′tan φ = r / [dr/dθ] = r / r'
  • φφ is measured from OPOP (produced) towards tangent. Then the angle the tangent makes with initial line is ψ=θ+φψ = θ + φ.
  • At the pole (r=0r=0): tangents are θ=αθ = α where f(α)=0f(α)=0 — set r=0r=0 and solve for θθ; these are the pole tangents (e.g., rose r=acos2θr=a cos2θ gives 44 tangents at 45°45° intervals).

4. Angle Between Radius Vector and Tangent — Derivation Sketch

In ΔΔ with r,r+Δr,Δθr, r+Δr, Δθ, tanφ=rΔθ/Δr→rdθ/drtan φ = r Δθ/Δr → r dθ/dr as Δθ→0Δθ→0.

5. Polar Subtangent / Subnormal and Pedal Equation

  • Polar subtangent (perp to OPOP): =r2/[dr/dθ]=r⋅tanφ= r² / [dr/dθ] = r · tan φ
  • Polar subnormal (along OPOP): =dr/dθ=rcotφ= dr/dθ = r cot φ
  • Pedal distance pp (perp from pole to tangent): 1/p2=1/r2+1/r4(dr/dθ)21/p² = 1/r² + 1/r⁴ (dr/dθ)² or p=rsinφp = r sinφ. The pedal equation p=p(r)p = p(r) eliminates θθ.

6. Worked Examples

(a) Cardioid r = a(1 - cosθ):
    r' = a sinθ ⇒ tan φ = r/r' = (1-cosθ)/sinθ = tan(θ/2) ⇒ φ = θ/2
    At θ=π/2 : r=a, φ=π/4, ψ=3π/4 ; at pole r=0 ⇒ 1-cosθ=0 ⇒ θ=0 (cusp) ✓

(b) Circle r = 2a cosθ (diameter on initial line):
    r' = -2a sinθ ⇒ tan φ = 2a cosθ/(-2a sinθ)= -cotθ = tan(π/2+θ) ⇒ φ=π/2+θ
    ψ = θ+φ = π/2+2θ ; subtangent = r²/r' = -2a cos²θ/sinθ                ✓

(c) Rose r = a cos 2θ — pole tangents: r=0 ⇒ cos2θ=0 ⇒ 2θ=±π/2, ±3π/2 ⇒ θ=±π/4, ±3π/4 ✓

(d) Spiral r = aθ : r' = a ⇒ tan φ = aθ/a = θ ⇒ φ = arctan θ → grows with radius ✓

(e) Find φ for r = a e^{θ cotα} (equiangular spiral):
    r' = a cotα e^{θ cotα}= r cotα ⇒ tan φ = r/r' = tanα ⇒ φ = α (constant!) ⇒ spiral cuts every radius vector at same angle α ✓

7. Sketching Tip for Exams

Make table θθ: 0,π/6,π/3,π/2,…0, π/6, π/3, π/2,… → rr → plot; mark pole tangents and axis symmetry; the curve is closed if r(θ+2π)=r(θ)r(θ+2π)=r(θ).

8. Exam Points

  • Derive tanφ=r/[dr/dθ]tan φ = r / [dr/dθ] and find φφ, ψψ at given θθ — 5 marks, very frequent.
  • Find pole tangents (r=0r=0) — 3 marks, frequent.
  • Polar subtangent / subnormal / pedal equation — 5 marks, frequent.
  • Symmetry and sketch r=a(1±cosθ),r=acosnθ,r=aθr=a(1±cosθ), r=a cos nθ, r=aθ — 5-7 marks.
  • Angle between two polar curves = ∣φ1−φ2∣|φ_1-φ_2| or ∣ψ1−ψ2∣|ψ_1-ψ_2| — 5 marks.
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