Learn9709Paper 3 · Pure Mathematics 3DifferentiationParametric differentiation

Differentiation · Subtopic 4 of 5

Parametric Differentiation

Curves described by a third variable $t$ : $x=f(t)$ , $y=g(t)$ . How to find their gradient with $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ , and how to find tangents and special points.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Understand a curve given parametrically by $x=f(t)$ , $y=g(t)$ .
Find $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ in terms of $t$ .
Find gradients, tangents and normals at a given value of $t$ .
Find points where the tangent is horizontal or vertical, or has a given gradient.

Pause and think

A point moves so that at time $t$ it is at $(2t,\ t^2)$ . How fast is $x$ changing? How fast is $y$ changing? So how steep is its path?

Section 2 of 14

Core idea

Start hereDescribing a path by time

A ball's flight is easiest to describe with time: horizontally $x=20t$ , vertically $y=15t-5t^2$ . Two equations, one for each coordinate, with a parameter $t$ . To find the direction of travel at any moment, you need $\frac{dy}{dx}$ , and it comes straight from the two rates.

The big idea

For $x=f(t)$ , $y=g(t)$ : $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ . Find points by substituting $t$ ; eliminate $t$ for the Cartesian equation.

1. The formula

Tracing x = t², y = 2t

t = -2.80

\dfrac{dx}{dt}=2t

\dfrac{dy}{dt}=2

, so

\dfrac{dy}{dx}=\dfrac{2}{2t}=\dfrac1t

. The short line through the moving point is the tangent with that gradient.

Why

\frac{dy}{dx}=\frac{dy/dt}{dx/dt}

$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$
Chain rule.
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$
Divide (needs $\frac{dx}{dt}\neq0$ ).

Worked example $x=t^2$ , $y=2t$ : tangent at $t=2$

Find the tangent.

Goal: Line equation.
Strategy: Point from $t$ ; gradient from the formula.

$\frac{dx}{dt}=2t,\ \frac{dy}{dt}=2\Rightarrow\frac{dy}{dx}=\frac1t=\tfrac12$
At $t=2$ .
$(x,y)=(4,4)$
Substitute $t=2$ .
$y-4=\tfrac12(x-4)\Rightarrow y=\tfrac12x+2$
Tangent.

Answer

y=\frac12x+2

✓ CheckCartesian:

y^2=4x

; implicit

2yy'=4\Rightarrow y'=\frac24=\frac12

y=4

✓.

Worked exampleEllipse $x=3\cos\theta$ , $y=2\sin\theta$

Find $\frac{dy}{dx}$ , the horizontal tangents, and the Cartesian equation.

Goal: Three results.
Strategy: Formula; $\cos^2+\sin^2=1$ .

$\frac{dy}{dx}=\frac{2\cos\theta}{-3\sin\theta}=-\tfrac23\cot\theta$
Formula.
$\cos\theta=0\Rightarrow\theta=\tfrac\pi2,\tfrac{3\pi}2:\ (0,\pm2)$
Numerator zero.
$\left(\tfrac x3\right)^2+\left(\tfrac y2\right)^2=1$
$\cos\theta=\frac x3$ , $\sin\theta=\frac y2$ .

Answer

-\frac23\cot\theta

;

(0,\pm2)

;

\frac{x^2}9+\frac{y^2}4=1

✓ CheckTop and bottom of an ellipse have horizontal tangents ✓.

Common misconceptionUpside down

\frac{dy}{dx}

\frac{dy/dt}{dx/dt}

, not the other way round.

Pause and think

$x=e^{2t}$ , $y=e^t+t$ . Find $\frac{dy}{dx}$ .

Summary

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ .
Point from $t$ .
Eliminate $t$ for Cartesian form.

Parameter: The third variable both coordinates depend on.

HistoryCurves for cars. In the 1960s, the engineer Pierre Bézier at Renault needed a way to describe car body shapes to computers. He used parametric curves now called Bézier curves. Every font on your screen and every vector drawing is built from them.

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