Parametric Differentiation — Pure Mathematics 3 (9709) — Mathematics

Introduction

Many curves cannot be described conveniently as $y = f(x)$ . Instead, both $x$ and $y$ are expressed separately as functions of a parameter $t$ :

x = f(t), \qquad y = g(t).

This is common in mechanics (e.g. projectile paths) and in curve sketching where the Cartesian form is unwieldy. In the 9709 exam, parametric differentiation appears in Paper 3 and is routinely tested through questions asking for the gradient at a point, the equation of a tangent, or the equation of a normal — so mastering the chain-rule formula below is essential.

Core Concept

To find $\dfrac{dy}{dx}$ for a parametric curve, apply the chain rule:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt} \neq 0.

Why this works. By the chain rule, $\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt}$ . Rearranging gives the formula above. No elimination of $t$ is required.

Procedure.

Differentiate $y$ with respect to $t$ to get $\dfrac{dy}{dt}$ .
Differentiate $x$ with respect to $t$ to get $\dfrac{dx}{dt}$ .
Divide: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .
To find the gradient at a specific point, substitute the value of $t$ corresponding to that point.
For a tangent, use $y - y_1 = m(x - x_1)$ with $m = dy/dx$ .
For a normal, use gradient $= -1/m$ .

The product and quotient rules (assumed known) are often needed in steps 1–2 when $\dfrac{dy}{dt}$ or $\dfrac{dx}{dt}$ involves a product or quotient of functions of $t$ .

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Introduction

Core Concept

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