Parametric Differentiation

When a curve is defined parametrically as x = f(t), y = g(t), you often need to find the gradient dy/dx at a particular point. The Edexcel 9MA0 specification requires you to differentiate parametric equations using the chain rule.

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The Chain Rule for Parametric Differentiation

If x = f(t) and y = g(t), then:

dy/dx = (dy/dt) ÷ (dx/dt)

provided dx/dt ≠ 0.

This follows from the chain rule: dy/dx = (dy/dt) × (dt/dx) = (dy/dt) / (dx/dt).

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Worked Examples

Example 1: Simple Polynomial

A curve is defined by x = t², y = t³. Find dy/dx in terms of t.

dx/dt = 2t, dy/dt = 3t².

dy/dx = 3t² / 2t = 3t/2 (provided t ≠ 0).

Example 2: Trigonometric Parametrisation

A curve is defined by x = 5cos(θ), y = 5sin(θ). Find dy/dx.

dx/dθ = −5sin(θ), dy/dθ = 5cos(θ).

dy/dx = 5cos(θ) / (−5sin(θ)) = −cos(θ)/sin(θ) = −cot(θ).

This makes geometric sense: the circle has gradient −cot(θ) at the point with parameter θ.

Example 3: Finding the Gradient at a Point