Parametric Equations

Parametric equations define x and y separately in terms of a third variable (parameter), usually t or θ. This allows us to describe curves that are difficult or impossible to express as y = f(x).

Introduction

Instead of y = f(x), a curve may be defined by:

x = f(t)
y = g(t)

As t varies, the point (x, y) traces out a curve.

Worked Example 1

A curve is defined by x = 2t, y = t² − 1.

t	x	y
−2	−4	3
−1	−2	0
0	0	−1
1	2	0
2	4	3

This traces out a parabola. To find the Cartesian equation: t = x/2, so y = (x/2)² − 1 = x²/4 − 1.

Converting from Parametric to Cartesian

Method

Express t in terms of x (or y).
Substitute into the other equation.
Simplify.

Worked Example 2

x = 3 cos t, y = 3 sin t.

Use the identity cos²t + sin²t = 1:

(x/3)² + (y/3)² = 1

= x² + y² = 9 — a circle of radius 3, centre the origin.

Worked Example 3

x = 2t + 1, y = 4t² − 3.

t = (x − 1)/2, so y = 4((x − 1)/2)² − 3 = 4(x − 1)²/4 − 3

= (x − 1)² − 3

Parametric Equations of Common Curves

Curve	Parametric form
Circle (radius r)	x = r cos t, y = r sin t
Ellipse (a, b)	x = a cos t, y = b sin t
Parabola	x = at², y = 2at
Line	x = a + bt, y = c + dt

Differentiation with Parametric Equations

The chain rule gives:

dy/dx = (dy/dt) ÷ (dx/dt)

Worked Example 4

x = t³, y = t² − t. Find dy/dx.

dx/dt = 3t², dy/dt = 2t − 1

dy/dx = (2t − 1)/(3t²)

Worked Example 5

Find the equation of the tangent to the curve x = 2t, y = t² at the point where t = 3.

At t = 3: x = 6, y = 9.

dx/dt = 2, dy/dt = 2t = 6

dy/dx = 6/2 = 3

Tangent: y − 9 = 3(x − 6)

= y = 3x − 9

Finding Stationary Points

Stationary points occur where dy/dx = 0, i.e. where dy/dt = 0 (provided dx/dt ≠ 0).

Worked Example 6

x = t², y = t³ − 3t. Find the stationary points.

dy/dt = 3t² − 3 = 3(t² − 1) = 3(t − 1)(t + 1)

dy/dt = 0 when t = 1 or t = −1.

At t = 1: x = 1, y = 1 − 3 = −2
At t = −1: x = 1, y = −1 + 3 = 2

Stationary points: (1, −2) and (1, 2)

Areas Under Parametric Curves

The area under a parametric curve from t = t₁ to t = t₂ is:

Area = ∫ₜ₁ᵗ² y (dx/dt) dt

Worked Example 7

Find the area enclosed by the curve x = 3 cos t, y = 3 sin t for 0 ≤ t ≤ 2π.

Area = ∫₀²π (3 sin t)(−3 sin t) dt = −9 ∫₀²π sin²t dt

Using sin²t = (1 − cos 2t)/2:

= −9 × [t/2 − sin 2t/4]₀²π = −9 × (π − 0) = −9π

Taking the absolute value: Area = 9π (consistent with πr² for r = 3).

Second Derivatives

d²y/dx² = d/dt(dy/dx) ÷ (dx/dt)

Worked Example 8

x = t², y = t³. Find d²y/dx².

dy/dx = (3t²)/(2t) = 3t/2

d/dt(dy/dx) = d/dt(3t/2) = 3/2

d²y/dx² = (3/2)/(2t)

= 3/(4t)

Exam Tips

To convert to Cartesian form, eliminate the parameter using algebra or trigonometric identities.
For dy/dx, use the chain rule: (dy/dt)/(dx/dt). Do not attempt to differentiate x and y directly with respect to each other.
When finding tangent or normal equations, always find the coordinates at the given parameter value first.
For area calculations, adjust the limits from x-values to t-values and include dx/dt.
Watch out for the direction of traversal — if x decreases as t increases, you may get a negative area.

A-Level Deep Dive: Parametric Equations

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section C: Coordinate geometry, sub-strand C2 (Year 2 content) covers the parametric equations of curves and conversion between Cartesian and parametric forms; use parametric equations in modelling in a variety of contexts (refer to the official specification document for exact wording). This Year 2 sub-strand layers on top of Year 1 coordinate geometry (lines, circles, Cartesian forms) and is examined in Paper 2 alongside trigonometry and differentiation. It connects directly to Section H — Differentiation (parametric differentiation $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ ), Section G — Trigonometry (the Pythagorean identity $\cos^2 t + \sin^2 t = 1$ is the workhorse for eliminating the parameter from trigonometric parametrisations), and Paper 3 — Mechanics, Section O (projectile motion is the canonical applied parametrisation, with $x = (u\cos\alpha)t$ and $y = (u\sin\alpha)t - \tfrac{1}{2}gt^2$ ). The AQA formula booklet does not list standard parametric forms — they must be recognised and manipulated from first principles.

Worked example with full mark scheme

Question (8 marks):

A curve $C$ is defined parametrically by $x = 3\cos t$ , $y = 2\sin t$ for $0 \leq t < 2\pi$ .

(a) Show that the Cartesian equation of $C$ can be written as $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ . (3)

(b) Find $\dfrac{dy}{dx}$ in terms of $t$ , and hence find the equation of the tangent to $C$ at the point where $t = \dfrac{\pi}{4}$ , giving your answer in the form $ax + by = c$ with $a, b, c$ exact. (5)

Solution with mark scheme:

(a) Step 1 — isolate the trigonometric functions.

From $x = 3\cos t$ , $\cos t = \dfrac{x}{3}$ . From $y = 2\sin t$ , $\sin t = \dfrac{y}{2}$ .

M1 — rearranging both parametric equations to express $\cos t$ and $\sin t$ in terms of $x$ and $y$ . A common error is to attempt to solve for $t$ directly ( $t = \arccos(x/3)$ ), which produces an unhelpful single-branch form.

Step 2 — apply the Pythagorean identity.

Using $\cos^2 t + \sin^2 t = 1$ :

$\left(\dfrac{x}{3}\right)^2 + \left(\dfrac{y}{2}\right)^2 = 1$

M1 — recognising and applying the identity $\cos^2 t + \sin^2 t = 1$ as the elimination tool for the parameter. This is the signature technique for trigonometric parametrisations.

Step 3 — write in the requested form.

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

A1 — printed answer obtained, with squaring of fractions performed correctly ( $(x/3)^2 = x^2/9$ , not $x^2/3$ ).

(b) Step 1 — differentiate parametrically.

$\dfrac{dx}{dt} = -3\sin t$ and $\dfrac{dy}{dt} = 2\cos t$ .

M1 — both derivatives correct. The negative sign on $\dfrac{dx}{dt}$ is the most-missed detail (chain-rule slip on $\cos t$ ).

Step 2 — form the gradient.

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2\cos t}{-3\sin t} = -\dfrac{2\cos t}{3\sin t}$

A1 — correct $\dfrac{dy}{dx}$ in terms of $t$ . Equivalent forms (e.g. $-\tfrac{2}{3}\cot t$ ) are fully accepted.

Step 3 — evaluate at $t = \pi/4$ .

At $t = \pi/4$ : $\cos t = \sin t = \dfrac{\sqrt{2}}{2}$ , so the gradient is $-\dfrac{2}{3}$ . The point on the curve is $\left(3 \cdot \dfrac{\sqrt{2}}{2}, 2 \cdot \dfrac{\sqrt{2}}{2}\right) = \left(\dfrac{3\sqrt{2}}{2}, \sqrt{2}\right)$ .

M1 — both the gradient value $-2/3$ and the point coordinates evaluated correctly using exact surd values for $\sin(\pi/4)$ and $\cos(\pi/4)$ .

Step 4 — form the tangent equation.

Using $y - y_1 = m(x - x_1)$ :

$y - \sqrt{2} = -\dfrac{2}{3}\left(x - \dfrac{3\sqrt{2}}{2}\right)$

Multiply through by $3$ : $3y - 3\sqrt{2} = -2x + 3\sqrt{2}$ , hence $2x + 3y = 6\sqrt{2}$ .

A1 — final answer in the form $ax + by = c$ with $a = 2$ , $b = 3$ , $c = 6\sqrt{2}$ , all exact.

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A curve $C$ has parametric equations $x = t^2 + 1$ , $y = 2t - 3$ for $t \in \mathbb{R}$ .

(a) Find a Cartesian equation of $C$ , giving your answer in the form $x = f(y)$ . (3)

(b) Find $\dfrac{dy}{dx}$ in terms of $t$ , and determine the value of $t$ at which the tangent to $C$ is parallel to the line $y = x$ . (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — isolating $t$ from the simpler parametric equation: from $y = 2t - 3$ , $t = \dfrac{y + 3}{2}$ .
M1 (AO1.1b) — substituting into $x = t^2 + 1$ .
A1 (AO1.1b) — $x = \left(\dfrac{y + 3}{2}\right)^2 + 1 = \dfrac{(y + 3)^2}{4} + 1$ .

(b)

M1 (AO1.1b) — $\dfrac{dx}{dt} = 2t$ , $\dfrac{dy}{dt} = 2$ , so $\dfrac{dy}{dx} = \dfrac{2}{2t} = \dfrac{1}{t}$ .
M1 (AO2.2a) — setting $\dfrac{1}{t} = 1$ (gradient of $y = x$ ).
A1 (AO1.1b) — $t = 1$ .

Total: 6 marks split AO1 = 5, AO2 = 1. This question rewards algebraic precision (the squaring step in (a) is a classic mark-loser if the bracket is dropped) and reasoning (recognising what "parallel to $y = x$ " means as a gradient condition).

Synoptic links

Connects to:

Section H — Parametric differentiation: the rule $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ is itself an instance of the chain rule. Once $y$ and $x$ are both functions of $t$ , $\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx}$ , with $\dfrac{dt}{dx} = 1/(dx/dt)$ provided $dx/dt \neq 0$ . Points where $dx/dt = 0$ correspond to vertical tangents — a frequent A* discriminator.
Section G — Trigonometric identities: every elimination-by-trig question reduces to an identity. Pythagorean ( $\cos^2 + \sin^2 = 1$ , $\sec^2 - \tan^2 = 1$ , $\cosec^2 - \cot^2 = 1$ ), double-angle ( $\sin 2t = 2\sin t \cos t$ ), and sum-to-product identities all generate parametric-to-Cartesian conversions.
Section C — Conics (implicit and parametric forms): the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ has standard parametrisation $x = a\cos t$ , $y = b\sin t$ . The hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ uses $x = a\sec t$ , $y = b\tan t$ . Recognising these standard forms turns 8-mark conversions into 2-mark inspections.
Paper 3 — Mechanics, projectile motion: a particle launched at speed $u$ at angle $\alpha$ above horizontal has position $x = (u\cos\alpha)t$ , $y = (u\sin\alpha)t - \tfrac{1}{2}gt^2$ . Eliminating $t$ produces the Cartesian trajectory $y = x\tan\alpha - \dfrac{gx^2}{2u^2\cos^2\alpha}$ — a parabola. This is the canonical applied parametric problem, and AQA examines it on Paper 3.
Section I — Integration (areas under parametric curves): the area between a parametric curve and the $x$ -axis is $\int y \dfrac{dx}{dt}\,dt$ over the parameter range. This is an extension of the Cartesian $\int y\,dx$ formula via substitution, and Year 2 questions sometimes pair parametrisation with integration in a single 10-mark item.

Mark-scheme literacy

Parametric questions on AQA 7357 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Differentiating each component, applying $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ , applying trig identities to eliminate $t$ , evaluating at given parameter values
AO2 (reasoning / interpretation)	25–35%	Choosing the right elimination strategy (algebraic vs trigonometric), justifying validity at boundary parameter values, interpreting "parallel"/"perpendicular" gradient conditions, recognising standard conic parametrisations
AO3 (problem-solving)	10–20%	Modelling contexts (projectile range, point-of-closest-approach), parametrising a curve given geometric constraints, identifying parameter values for stationary points or tangent contacts

Examiner-rewarded phrasing: "using the identity $\cos^2 t + \sin^2 t = 1$ "; "since $dx/dt \neq 0$ at this value of $t$ , the chain rule gives…"; "the Cartesian equation valid for the range of $t$ given is…". Phrases that lose marks: writing $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ with the wrong derivative on top (a sign error in $\dfrac{dx}{dt}$ when $x = a\cos t$ is the most common); concluding with a Cartesian equation but failing to state the range of validity (e.g. $-3 \leq x \leq 3$ for the ellipse with $a = 3$ ); leaving a tangent equation in point-gradient form when the question asks for $ax + by = c$ .

A specific AQA pattern to watch: questions sometimes ask for the Cartesian equation "and state the range of values of $x$ (or $y$ ) for which it represents the curve". The implicit Cartesian form $x^2/9 + y^2/4 = 1$ is satisfied by the whole ellipse — but a parametric range $0 \leq t \leq \pi$ traces only the upper half. The implicit form alone undersells the parametrisation; you must add $y \geq 0$ to be precise.

Grade-band model answers

3-mark question

Question: A curve has parametric equations $x = 2t + 1$ , $y = t^2$ . Find a Cartesian equation of the curve.

Grade C response (~180 words):

From $x = 2t + 1$ , rearrange: $t = \dfrac{x - 1}{2}$ .

Substitute into $y = t^2$ :

$y = \left(\dfrac{x - 1}{2}\right)^2 = \dfrac{(x - 1)^2}{4}$ .

So the Cartesian equation is $y = \dfrac{(x - 1)^2}{4}$ .

Examiner commentary: Full marks (3/3). The candidate isolates $t$ from the linear parametrisation (the easier of the two), substitutes cleanly into the quadratic component, and produces a Cartesian form. The answer is left in the natural form $y = f(x)$ which is acceptable in the absence of a specified format. Some candidates lose marks here by expanding the bracket unnecessarily — $\dfrac{x^2 - 2x + 1}{4}$ is mathematically equivalent but no cleaner, and any algebraic slip in the expansion costs the A1.

Grade A response (~220 words):*

Parametric Equations

Parametric Equations

Introduction

Worked Example 1

Converting from Parametric to Cartesian

Method

Worked Example 2

Worked Example 3

Parametric Equations of Common Curves

Differentiation with Parametric Equations

Worked Example 4

Worked Example 5

Finding Stationary Points

Worked Example 6

Areas Under Parametric Curves

Worked Example 7

Second Derivatives

Worked Example 8

Exam Tips

A-Level Deep Dive: Parametric Equations

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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