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MathidiaMathematics Academy
Polar coordinates
Lesson 30 of 75
Review

Tangents to polar curves

Objective. Determine gradient from parametric derivatives.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

To find the gradient of the tangent to a polar curve r = f(θ), the curve is treated parametrically with x = r cos θ and y = r sin θ, and the gradient dy/dx is calculated as (dy/dθ)/(dx/dθ) using the product rule on each. This connects polar curves back to parametric differentiation, since both x and y depend on the parameter θ. For example, horizontal tangents occur where dy/dθ = 0 (and dx/dθ ≠ 0), while vertical tangents occur where dx/dθ = 0.

Worked example

For the polar curve r = 2θ, find the gradient dy/dx at θ = π/2.

For the polar curve r = 2θ
Coordinate grid with the line For the polar curve r = 2θ
  1. 1x = r cos θ = 2θ cos θ, y = r sin θ = 2θ sin θ
  2. 2dx/dθ = 2cos θ − 2θ sin θ; dy/dθ = 2sin θ + 2θ cos θ
  3. 3At θ=π/2: dx/dθ = 2(0) − 2(π/2)(1) = −π; dy/dθ = 2(1) + 2(π/2)(0) = 2
  4. 4Gradient dy/dx = (dy/dθ)/(dx/dθ) = 2/(−π) = −2/π

Answer: −2/π

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Common mistake. Students try to differentiate r with respect to θ and call that the gradient, forgetting that dy/dx requires both x and y in terms of θ via the product rule, not just dr/dθ. Always form dx/dθ and dy/dθ separately before dividing.

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