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.

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).

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

A curve has parametric equations x = t + 1/t, y = t − 1/t. Find dy/dx when t = 2.

dx/dt = 1 − 1/t², dy/dt = 1 + 1/t².

At t = 2: dx/dt = 1 − 1/4 = 3/4, dy/dt = 1 + 1/4 = 5/4.

dy/dx = (5/4) / (3/4) = 5/3.

Finding Equations of Tangents and Normals

Once you have dy/dx at a particular parameter value, you can find the tangent and normal lines.

Tangent: Use the point-gradient form with the gradient dy/dx and the point (x, y) at the given parameter value.

Normal: The normal gradient is −1/(dy/dx), i.e. the negative reciprocal.

Example

A curve is defined by x = 2t, y = t² − 1. Find the equation of the tangent at t = 3.

Step 1: Find the point. x = 6, y = 8. Point is (6, 8).

Step 2: Find dy/dx. dx/dt = 2, dy/dt = 2t = 6. dy/dx = 6/2 = 3.

Step 3: Tangent equation: y − 8 = 3(x − 6), so y = 3x − 10.

The normal at the same point: gradient = −1/3. y − 8 = (−1/3)(x − 6), so y = −x/3 + 10.

Stationary Points

A stationary point occurs when dy/dx = 0. Since dy/dx = (dy/dt)/(dx/dt), this happens when dy/dt = 0 (provided dx/dt ≠ 0).

Example: Find the stationary point of x = t², y = t³ − 3t.

dy/dt = 3t² − 3 = 0, so t² = 1, t = ±1.

When t = 1: x = 1, y = 1 − 3 = −2. Point (1, −2). When t = −1: x = 1, y = −1 + 3 = 2. Point (1, 2).

Check dx/dt = 2t ≠ 0 at t = ±1 ✓.

Vertical Tangents

A vertical tangent occurs when dx/dt = 0 and dy/dt ≠ 0. At such points, dy/dx is undefined (the gradient is infinite).

Second Derivative

To find d²y/dx², differentiate dy/dx with respect to t and divide by dx/dt:

d²y/dx² = (d/dt(dy/dx)) / (dx/dt)

Example: For x = t², y = t³: dy/dx = 3t/2.

d/dt(3t/2) = 3/2.

d²y/dx² = (3/2) / (2t) = 3/(4t).

Exam Tip: When finding stationary points from parametric equations, set dy/dt = 0 (not dy/dx = 0 directly). Then check dx/dt ≠ 0 at those parameter values. If both dy/dt = 0 and dx/dt = 0, more careful analysis is needed.

Key Terms

Term	Definition
Chain rule	dy/dx = (dy/dt) × (dt/dx)
Tangent	A line touching the curve at a point, with gradient dy/dx
Normal	A line perpendicular to the tangent at the point of tangency
Stationary point	A point where dy/dx = 0

Summary

For parametric curves, dy/dx = (dy/dt) / (dx/dt).
Use this gradient to find equations of tangents and normals.
Stationary points occur when dy/dt = 0 (with dx/dt ≠ 0).
Vertical tangents occur when dx/dt = 0 (with dy/dt ≠ 0).
The second derivative uses d²y/dx² = (d/dt(dy/dx)) / (dx/dt).

A-Level Deep Dive: Parametric Differentiation

Spec mapping

Edexcel 9MA0 Year 2 Pure specification, section 9 — Differentiation, sub-strand on parametric methods covers simple functions and relations defined parametrically, for first derivative only (refer to the official specification document for exact wording). This is examined principally on 9MA0-02 Paper 2 (Pure Mathematics 2), but the topic is deeply synoptic and questions are routinely lifted from the dy/dx step into a tangent/normal demand drawn from section 4 (Coordinate geometry in the (x, y) plane) or a stationary-point analysis from section 9 (Differentiation, applications). Crucially, the parametric differentiation formula $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ is NOT printed in the Edexcel formula booklet — candidates must memorise it, along with its derivation from the chain rule. The same omission applies to the second-derivative form $\dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) \div \dfrac{dx}{dt}$ . Synoptic appearances: section 8 (Parametric equations of curves) supplies the curves themselves and the conversion to Cartesian form; section 4 (Coordinate geometry) supplies the tangent/normal point-slope machinery; 9MA0-03 Mechanics uses the same logic when velocity $\mathbf{v} = (dx/dt, dy/dt)$ and acceleration are computed from parametric position vectors; and section 10 (Integration) revisits the chain rule structure when computing the area under a parametric curve via $\int y \,(dx/dt)\,dt$ .

Worked example with full mark scheme

Question (8 marks):

A curve $C$ has parametric equations $x = t^2 - 1, \qquad y = t^3 - 3t, \qquad t \in \mathbb{R}.$

(a) Find $\dfrac{dy}{dx}$ in terms of $t$ , simplifying your answer. (2)

(b) Find the coordinates of the stationary points of $C$ and determine their nature. (4)

(c) Find the equation of the tangent to $C$ at the point where $t = 2$ , giving your answer in the form $y = mx + c$ . (2)

Solution with mark scheme:

(a) Step 1 — differentiate each parametric equation with respect to $t$ .

$\dfrac{dx}{dt} = 2t, \qquad \dfrac{dy}{dt} = 3t^2 - 3 = 3(t^2 - 1).$

M1 — both derivatives correct. The middle slip here is differentiating $t^3 - 3t$ as $3t^2 - 3t$ (treating $3t$ as if its derivative were $3t$ rather than $3$ ).

Step 2 — apply the parametric formula.

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3(t^2 - 1)}{2t} = \dfrac{3(t - 1)(t + 1)}{2t}.$

A1 — correct quotient, simplified or factorised. The factorisation is not required for the mark, but it makes part (b) immediate.

(b) Step 3 — locate stationary points by setting the numerator to zero.

A stationary point on a parametric curve occurs when $\dfrac{dy}{dx} = 0$ , i.e. when $\dfrac{dy}{dt} = 0$ and $\dfrac{dx}{dt} \neq 0$ .

$3(t^2 - 1) = 0 \implies t = \pm 1$ . Check $dx/dt = 2t$ : at $t = 1$ , $dx/dt = 2 \neq 0$ ✓. At $t = -1$ , $dx/dt = -2 \neq 0$ ✓.

M1 — setting $dy/dt = 0$ and obtaining both $t$ values.

Step 4 — convert parameter values to coordinates.

At $t = 1$ : $x = 1 - 1 = 0$ , $y = 1 - 3 = -2$ . Point $(0, -2)$ .

At $t = -1$ : $x = 1 - 1 = 0$ , $y = -1 + 3 = 2$ . Point $(0, 2)$ .

A1 — both coordinate pairs correct.

Step 5 — classify nature using the second derivative.

Differentiate $\dfrac{dy}{dx} = \dfrac{3(t^2 - 1)}{2t} = \dfrac{3t}{2} - \dfrac{3}{2t}$ with respect to $t$ :

$\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = \dfrac{3}{2} + \dfrac{3}{2t^2}.$

Then $\dfrac{d^2y}{dx^2} = \dfrac{(3/2) + (3/(2t^2))}{2t} = \dfrac{3t^2 + 3}{4t^3}$ .

At $t = 1$ : $\dfrac{d^2y}{dx^2} = \dfrac{6}{4} = \dfrac{3}{2} > 0$ — minimum at $(0, -2)$ .

At $t = -1$ : $\dfrac{d^2y}{dx^2} = \dfrac{6}{-4} = -\dfrac{3}{2} < 0$ — maximum at $(0, 2)$ .

M1 A1 — correct second-derivative method and correct classification of both points.

Point: $x = 4 - 1 = 3$ , $y = 8 - 6 = 2$ , so $(3, 2)$ .

Gradient: $\dfrac{dy}{dx}\bigg|_{t=2} = \dfrac{3(4 - 1)}{2 \cdot 2} = \dfrac{9}{4}$ .

Tangent: $y - 2 = \dfrac{9}{4}(x - 3) \implies y = \dfrac{9}{4}x - \dfrac{27}{4} + 2 = \dfrac{9}{4}x - \dfrac{19}{4}$ .

M1 for substitution into the gradient and point formulae; A1 for the correct tangent in the requested form.

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

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): A curve $C$ is defined by the parametric equations $x = 1 + 2\sin\theta$ , $y = 4 + \cos 2\theta$ , for $0 \leq \theta \leq \pi$ .

(a) Show that $\dfrac{dy}{dx} = -2\sin\theta$ . (3)

(b) Hence find the equation of the normal to $C$ at the point where $\theta = \pi/6$ , giving your answer in the form $ax + by + c = 0$ with integer $a$ , $b$ , $c$ . (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — differentiating $x$ and $y$ to obtain $dx/d\theta = 2\cos\theta$ and $dy/d\theta = -2\sin 2\theta$ .
M1 (AO1.1b) — applying the double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ inside $dy/d\theta$ to obtain $dy/d\theta = -4\sin\theta\cos\theta$ .
A1 (AO2.1) — forming the quotient and cancelling $\cos\theta$ to reach the printed answer $dy/dx = -2\sin\theta$ , with a brief justification that $\cos\theta \neq 0$ on the relevant arc.

(b)

M1 (AO1.1b) — substituting $\theta = \pi/6$ : gradient of curve = $-2\sin(\pi/6) = -1$ , so normal gradient = $1$ (negative reciprocal); coordinates $(1 + 2 \cdot 1/2, 4 + \cos(\pi/3)) = (2, 9/2)$ .
M1 (AO1.1b) — forming $y - 9/2 = 1 \cdot (x - 2)$ and rearranging.
A1 (AO2.5) — final form $2x - 2y + 5 = 0$ with integer coefficients (or any positive-leading-coefficient equivalent).

Total: 6 marks split AO1 = 4, AO2 = 2. This is the typical Paper 2 parametric question — heavy AO1 procedural fluency with two AO2 marks reserved for the trigonometric simplification in (a) and the integer-form presentation in (b).

Synoptic links

Connects to:

Section 9 — Chain rule: the parametric formula $dy/dx = (dy/dt)/(dx/dt)$ is not a separate rule but the chain rule $dy/dx = (dy/dt)(dt/dx)$ rewritten with $dt/dx = 1/(dx/dt)$ . Examiners reward candidates who write the chain-rule line first, then derive the quotient — it shows the technique is understood, not memorised.
Section 9 — Implicit differentiation: parametric and implicit differentiation are two solutions to the same problem (curves without explicit $y = f(x)$ ). For a curve like $x^2 + y^2 = 1$ , you can either parametrise $x = \cos t$ , $y = \sin t$ and use parametric differentiation, or differentiate implicitly to get $2x + 2y(dy/dx) = 0$ . Both routes give $dy/dx = -x/y$ .
Section 7 — Trigonometric differentiation: when the parametrisation is trigonometric ( $x = a\cos\theta$ , $y = b\sin\theta$ for an ellipse), every parametric question becomes a trigonometric differentiation question, with $d/d\theta(\sin\theta) = \cos\theta$ and $d/d\theta(\cos\theta) = -\sin\theta$ as the building blocks. Identities like $\sin 2\theta = 2\sin\theta\cos\theta$ then collapse the quotient.
9MA0-03 Mechanics — velocity and acceleration: if a particle's position is given by $\mathbf{r}(t) = (x(t), y(t))$ , its velocity vector is $\mathbf{v} = (dx/dt, dy/dt)$ and the direction of motion has gradient $dy/dx = (dy/dt)/(dx/dt)$ — the same parametric formula. Direction-of-motion questions in Mechanics 2 use this identity directly.
Section 10 — Integration, parametric area: the area under a parametric curve from $t = a$ to $t = b$ is $\int_a^b y(t)\,(dx/dt)\,dt$ , which is exactly the chain rule run in reverse. Confidence with the parametric derivative formula transfers immediately to the parametric area formula.

Mark-scheme literacy

Parametric differentiation questions on 9MA0-02 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Differentiating each parametric equation, forming the quotient $dy/dx$ , substituting parameter values
AO2 (reasoning / interpretation)	25–35%	Recognising stationary points come from $dy/dt = 0$ (not $dy/dx = 0$ algebraically), justifying $dx/dt \neq 0$ , simplifying with trigonometric identities
AO3 (problem-solving)	0–10%	Multi-step problems combining parametric differentiation with mechanics or with curve sketching; rare on standalone parametric questions

Examiner-rewarded phrasing: "by the chain rule, $dy/dx = (dy/dt)/(dx/dt)$ "; "stationary points occur where $dy/dt = 0$ provided $dx/dt \neq 0$ "; "checking $dx/dt \neq 0$ at $t = 1$ "; "using the second-derivative test, $d^2y/dx^2 = \frac{d}{dt}(dy/dx) \div (dx/dt)$ ". Phrases that lose marks: writing $dy/dx = (dx/dt)/(dy/dt)$ (quotient inverted); setting $dy/dx = 0$ algebraically and dividing by $dx/dt$ as if it were a constant; classifying a stationary point by the sign of $dy/dt$ (which is by construction zero) rather than by the second derivative or a sign-change argument; quoting a second-derivative formula as $d^2y/dx^2 = (d^2y/dt^2)/(d^2x/dt^2)$ , which is a very common and entirely incorrect shortcut.

A specific Edexcel pattern to watch: questions asking for "the equation of the tangent in the form $y = mx + c$ " demand the explicit slope-intercept form; leaving the answer as $y - 2 = (9/4)(x - 3)$ is mathematically correct but loses the final A1.

Grade-band model answers

3-mark question

Question: A curve has parametric equations $x = 3t^2$ , $y = t^3 - 4t$ . Find the value of $\dfrac{dy}{dx}$ at the point where $t = 2$ .

Grade C response (~210 words):

I differentiate each one:

$dx/dt = 6t$ , $dy/dt = 3t^2 - 4$ .

So $dy/dx = (3t^2 - 4)/(6t)$ .

At $t = 2$ : $dy/dx = (3 \cdot 4 - 4)/(6 \cdot 2) = 8/12 = 2/3$ .

So the gradient is $2/3$ .

Examiner commentary: Full marks (3/3). The candidate executes the procedure cleanly: differentiates both parametric equations correctly, forms the quotient in the right order $(dy/dt)/(dx/dt)$ , and substitutes $t = 2$ into the resulting expression rather than into the original equations and then differentiating. The simplification $8/12 = 2/3$ is correct. What's missing — and what would matter on a longer question — is any named invocation of the chain rule. On a 6+ mark question the absence of "by the chain rule" or "since $dx/dt \neq 0$ " can lose AO2 marks. For a 3-mark procedural question this is fine.

Grade A response (~280 words):*

By the chain rule, for a curve given parametrically as $x = x(t)$ , $y = y(t)$ , we have

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \quad \text{provided } \dfrac{dx}{dt} \neq 0.$

Differentiating each parametric equation with respect to $t$ :

$\dfrac{dx}{dt} = 6t, \qquad \dfrac{dy}{dt} = 3t^2 - 4.$

Substituting into the parametric formula:

$\dfrac{dy}{dx} = \dfrac{3t^2 - 4}{6t}.$

Evaluating at $t = 2$ (and noting that $dx/dt = 12 \neq 0$ at this point, so the gradient is well-defined):

$\dfrac{dy}{dx}\bigg|_{t=2} = \dfrac{3 \cdot 4 - 4}{6 \cdot 2} = \dfrac{8}{12} = \dfrac{2}{3}.$

Parametric Differentiation

Parametric Differentiation

The Chain Rule for Parametric Differentiation

Worked Examples

Example 1: Simple Polynomial

Example 2: Trigonometric Parametrisation

Example 3: Finding the Gradient at a Point

Finding Equations of Tangents and Normals

Example

Stationary Points

Vertical Tangents

Second Derivative

Key Terms

Summary

A-Level Deep Dive: Parametric Differentiation

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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