Tangents to polar curves

x and y in terms of θ

For r = f(θ): x = r cos θ = f(θ)cos θ and y = r sin θ = f(θ)sin θ.

The gradient is dy/dx = (dy/dθ)/(dx/dθ), as for any parametric curve.

Parallel and perpendicular

• Tangent parallel to the initial line: dy/dθ = 0 (and dx/dθ ≠ 0).
• Tangent perpendicular to the initial line: dx/dθ = 0 (and dy/dθ ≠ 0).

Differentiate r sin θ or r cos θ with the product rule, then use double angle formulae to get a quadratic in cos θ or sin θ.

Checking the solutions

• Reject angles where r < 0.
• At the pole (r = 0) the curve may have a cusp, as the cardioid does at θ = π, so check the point is a genuine tangent.
• Give the points in polar form (r, θ), with exact values where possible.

At a point that is not special (such as r = 3 + 8sin θ at θ = π/6), just evaluate dy/dx there to get a tangent's gradient.

Worked example

Find the points on r = a(1 + cos θ), -π < θ ≤ π, where the tangent is parallel to the initial line.

1. y = a(1 + cos θ)sin θ = a(sin θ + ½sin 2θ).
2. dy/dθ = a(cos θ + cos 2θ) = a(2cos² θ + cos θ - 1) = a(2cos θ - 1)(cos θ + 1).
3. cos θ = 1/2 gives θ = ±π/3, r = 3a/2. cos θ = -1 gives θ = π, which is the pole, a cusp, so it is rejected.

Answer: (3a/2, π/3) and (3a/2, -π/3)

Check yourself

1. For a polar curve, which condition gives a tangent parallel to the initial line?
   • d(r cos θ)/dθ = 0
   • dr/dθ = 0
   • d(r sin θ)/dθ = 0
   • r = 0
   Show answer

   d(r sin θ)/dθ = 0. Parallel to the initial line means horizontal: dy/dθ = 0, with y = r sin θ.

2. On r = 2(1 + cos θ), find r at the points where the tangent is parallel to the initial line (θ ≠ π).
   Show answer

   3. From the worked example, cos θ = 1/2, so r = 2(1 + 1/2) = 3.

3. On r = a(1 + cos θ), the tangent is perpendicular to the initial line at θ = kπ, with 0 < k < 1. Find k. Give your answer as a fraction in its simplest form.
   Show answer

   2/3. x = a(cos θ + cos² θ). dx/dθ = -a(sin θ + 2sin θ cos θ) = -a sin θ(1 + 2cos θ) = 0. In 0 < θ < π: cos θ = -1/2, θ = 2π/3.

4. On r = 4sin 2θ, 0 < θ < π/2, the tangent is parallel to the initial line where tan² θ = 2. Find this θ in radians, to 3 decimal places.
   Show answer

   0.955. y = 4sin 2θ sin θ = 8sin² θ cos θ. dy/dθ = 8sin θ(2cos² θ - sin² θ) = 0, so tan² θ = 2 and θ = arctan √2 = 0.955.

5. At the same point of r = 4sin 2θ, find r to 3 decimal places.
   Show answer

   3.771. sin 2θ = 2tan θ/(1 + tan² θ) = 2√2/3, so r = 8√2/3 = 3.771.

6. Find the gradient of the tangent to r = 3 + 8sin θ at θ = π/6, to 3 decimal places.
   Show answer

   3.811. r = 7 and dr/dθ = 8cos(π/6) = 4√3. dy/dθ = r' sin θ + r cos θ = 2√3 + 7√3/2 = 11√3/2; dx/dθ = r' cos θ - r sin θ = 6 - 7/2 = 5/2. Gradient = (11√3/2)/(5/2) = 11√3/5 = 3.811.

7. On r = 4(1 + cos θ), the tangent is parallel to the initial line at (6, π/3) and (6, -π/3). Find the distance between these two points, to 3 decimal places.
   Show answer

   10.392. y = ±6sin(π/3) = ±3√3, and x is the same for both. Distance = 6√3 = 10.392.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit F2
• Pearson Edexcel IAL Further Pure Mathematics 2 scheme of work

Last updated 8 November 2026. Spotted a mistake? Tell us.