Parametric Differentiation

This lesson covers how to find derivatives (gradients) of curves defined by parametric equations. This is a core A-Level topic that combines parametric equations with differentiation. You will learn how to find dy/dx without first converting to Cartesian form, and how to use this to find equations of tangents and normals to parametric curves.

The Chain Rule for Parametric Differentiation

If x and y are both functions of a parameter t, then by the chain rule:

dy/dx = (dy/dt) ÷ (dx/dt)

provided dx/dt ≠ 0.

This formula avoids the need to eliminate the parameter — you can differentiate directly in terms of t.

Derivation: Since y is a function of t and t is a function of x:

dy/dx = dy/dt × dt/dx = (dy/dt) / (dx/dt)

Basic Examples

Example 1: Find dy/dx for the curve x = t², y = t³.

dx/dt = 2t
dy/dt = 3t²

dy/dx = (3t²) / (2t) = 3t/2

Example 2: Find dy/dx for x = 3cos θ, y = 3sin θ.

dx/dθ = −3sin θ
dy/dθ = 3cos θ

dy/dx = (3cos θ) / (−3sin θ) = −cos θ / sin θ = −cot θ

Example 3: Find dy/dx for x = 2t + 1, y = t² − 4t.

dx/dt = 2
dy/dt = 2t − 4

dy/dx = (2t − 4) / 2 = t − 2

Finding the Gradient at a Specific Point

Evaluate dy/dx at the given value of the parameter.

Example 4: The curve has parametric equations x = t³, y = 2t² − 1. Find the gradient at the point where t = 2.

dx/dt = 3t²
dy/dt = 4t

dy/dx = 4t / (3t²) = 4 / (3t)

At t = 2: dy/dx = 4 / 6 = 2/3

The gradient at t = 2 is 2/3.

The point is x = 8, y = 7, so the point is (8, 7).

Equations of Tangents

To find the equation of the tangent at a given parameter value:

Find dy/dx at that parameter value (this is the gradient m).
Find the coordinates (x, y) at that parameter value.
Use y − y₁ = m(x − x₁).

Example 5: Find the equation of the tangent to the curve x = t², y = 2t at the point where t = 3.

At t = 3: x = 9, y = 6. Point: (9, 6).

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

dy/dx = 2/6 = 1/3

Tangent: y − 6 = (1/3)(x − 9)
         y − 6 = x/3 − 3
         y = x/3 + 3

or equivalently, x − 3y + 9 = 0.

Example 6: Find the equation of the tangent to x = 4cos θ, y = 4sin θ at θ = π/3.

At θ = π/3: x = 4cos(π/3) = 4 × 1/2 = 2
             y = 4sin(π/3) = 4 × √3/2 = 2√3

dx/dθ = −4sin θ = −4sin(π/3) = −4 × √3/2 = −2√3
dy/dθ = 4cos θ = 4cos(π/3) = 4 × 1/2 = 2

dy/dx = 2 / (−2√3) = −1/√3 = −√3/3

Tangent: y − 2√3 = (−√3/3)(x − 2)
         3(y − 2√3) = −√3(x − 2)
         3y − 6√3 = −√3 x + 2√3
         √3 x + 3y = 8√3

Equations of Normals

The normal at a point is perpendicular to the tangent. Its gradient is the negative reciprocal of the tangent gradient.

Example 7: Find the equation of the normal to x = t², y = t³ at t = 1.

At t = 1: x = 1, y = 1.

dy/dx = 3t/(2) = 3/2 (at t = 1)

Wait — let us recalculate properly.
dx/dt = 2t = 2
dy/dt = 3t² = 3

dy/dx = 3/2

Normal gradient = −2/3

Normal: y − 1 = (−2/3)(x − 1)
        3(y − 1) = −2(x − 1)
        3y − 3 = −2x + 2
        2x + 3y − 5 = 0

Stationary Points of Parametric Curves

A stationary point occurs where dy/dx = 0.

Since dy/dx = (dy/dt) / (dx/dt), a stationary point occurs where dy/dt = 0 (provided dx/dt ≠ 0 at that parameter value).

Example 8: Find the stationary points of the curve x = t³ − 3t, y = t² − 4.

dx/dt = 3t² − 3
dy/dt = 2t

dy/dx = 2t / (3t² − 3)

For stationary points: dy/dt = 0  ⟹  2t = 0  ⟹  t = 0

Check dx/dt ≠ 0 at t = 0: dx/dt = −3 ≠ 0  ✓

At t = 0: x = 0, y = −4.

Stationary point at (0, −4).

Vertical Tangents

When dx/dt = 0 (and dy/dt ≠ 0), the gradient dy/dx is undefined, and the tangent is vertical.

Example 9: For x = t³ − 3t, y = t² − 4, find where the tangent is vertical.

dx/dt = 3t² − 3 = 0
3(t² − 1) = 0
t = ±1

At t = 1: x = 1 − 3 = −2, y = 1 − 4 = −3. Point: (−2, −3).
At t = −1: x = −1 + 3 = 2, y = 1 − 4 = −3. Point: (2, −3).

At both points, the tangent is vertical (x = −2 and x = 2 respectively).

Second Derivatives (Extension)

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

d²y/dx² = (d/dt(dy/dx)) / (dx/dt)

Example 10: For x = t², y = t³, find d²y/dx².

dy/dx = 3t/2

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

dx/dt = 2t

d²y/dx² = (3/2) / (2t) = 3/(4t)

Worked Examination-Style Problem

Problem: A curve C is defined by x = 4t, y = 4/t, where t > 0.

(a) Find dy/dx in terms of t.

(b) Find the equation of the tangent at the point where t = 2.

(d) The tangent and normal at t = 2 meet the x-axis at P and Q respectively. Find the distance PQ.

Solution:

(a)

dx/dt = 4
dy/dt = −4/t²

dy/dx = (−4/t²) / 4 = −1/t²

(b) At t = 2: x = 8, y = 2. Point: (8, 2).

dy/dx = −1/4

Tangent: y − 2 = −(1/4)(x − 8)
         4y − 8 = −x + 8
         x + 4y − 16 = 0
         or y = −x/4 + 4

(c)

Normal gradient = 4

Normal: y − 2 = 4(x − 8)
        y = 4x − 30

(d)

Tangent meets x-axis (y = 0): x + 0 − 16 = 0  ⟹  x = 16. So P = (16, 0).
Normal meets x-axis (y = 0): 0 = 4x − 30  ⟹  x = 15/2. So Q = (15/2, 0).

PQ = |16 − 15/2| = |32/2 − 15/2| = 17/2 = 8.5

Summary

Parametric differentiation: dy/dx = (dy/dt) / (dx/dt).
Tangent: find dy/dx at the parameter value, then use y − y₁ = m(x − x₁).
Normal: gradient is the negative reciprocal of the tangent gradient.
Stationary points: dy/dt = 0 (with dx/dt ≠ 0).
Vertical tangents: dx/dt = 0 (with dy/dt ≠ 0).
Second derivative: d²y/dx² = (d/dt(dy/dx)) / (dx/dt).

Exam Tip: When finding tangents and normals to parametric curves, always state the point (x, y) explicitly before writing the equation. Show the separate calculations of dx/dt and dy/dt clearly — the examiner awards marks for each stage. If the question gives a specific value of t, evaluate dy/dx numerically before forming the line equation.

A-Level Deep Dive: Parametric Differentiation

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section G: Differentiation (Year 2 content) covers simple functions and relations defined parametrically, for first derivative only (refer to the official specification document for exact wording). This sub-strand sits within the wider Year 2 differentiation topic, immediately after implicit differentiation and before differentiating inverse trigonometric functions. It is examined on Papers 1 and 2 (both Pure papers may include parametric calculus) and connects directly to Section H — Integration (parametric integrals for area under a parametrically-defined curve), Section J — Vectors (where curves traced by position vectors are parametric in the time variable) and Paper 3 — Mechanics (where velocity $\dot{\mathbf{r}} = (\dot{x}, \dot{y})$ is the parametric derivative of position). The AQA formula booklet provides $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ in the Differentiation section, so the formula itself is given — but candidates are still expected to know when and how to apply it.

Worked example with full mark scheme

Question (8 marks):

A curve $C$ is defined parametrically by $x = 2t^2 - 3$ and $y = t^3 - 4t$ for $t \in \mathbb{R}$ .

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

(b) Find the coordinates of the stationary points of $C$ . (3)

(c) Find the equation of the tangent to $C$ at the point where $t = 2$ , giving your answer in the form $y = mx + c$ . (3)

Solution with mark scheme:

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

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

M1 — both parametric derivatives correct. A common slip is to differentiate $y = t^3 - 4t$ as $3t^2 - 4t$ , carrying the linear coefficient through incorrectly.

Step 2 — apply the parametric chain rule.

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2 - 4}{4t}$

A1 — correct expression. Note this is valid only where $dx/dt \neq 0$ , i.e. $t \neq 0$ .

(b) Step 1 — stationary points occur where $dy/dx = 0$ , which (since the denominator is non-zero there) means $dy/dt = 0$ .

$3t^2 - 4 = 0 \implies t^2 = \dfrac{4}{3} \implies t = \pm \dfrac{2}{\sqrt{3}}$

M1 — setting the numerator to zero and solving. A frequent error is to set the denominator to zero, confusing stationary points with vertical tangents.

Step 2 — compute coordinates by substituting each $t$ back into the original parametric equations.

For $t = \dfrac{2}{\sqrt{3}}$ : $x = 2 \cdot \dfrac{4}{3} - 3 = \dfrac{8}{3} - 3 = -\dfrac{1}{3}$ , and $y = \left(\dfrac{2}{\sqrt{3}}\right)^3 - 4 \cdot \dfrac{2}{\sqrt{3}} = \dfrac{8}{3\sqrt{3}} - \dfrac{8}{\sqrt{3}} = \dfrac{8 - 24}{3\sqrt{3}} = -\dfrac{16}{3\sqrt{3}} = -\dfrac{16\sqrt{3}}{9}$ .

For $t = -\dfrac{2}{\sqrt{3}}$ : $x = -\dfrac{1}{3}$ (same, since $x$ depends on $t^2$ ), and $y = +\dfrac{16\sqrt{3}}{9}$ .

M1 — substituting at least one root back into both $x$ and $y$ . A1 — both stationary points correct: $\left(-\tfrac{1}{3},\, -\tfrac{16\sqrt{3}}{9}\right)$ and $\left(-\tfrac{1}{3},\, \tfrac{16\sqrt{3}}{9}\right)$ .

$x = 2 \cdot 4 - 3 = 5$ , $y = 8 - 8 = 0$ . So the point is $(5, 0)$ .

B1 — point of tangency correct.

Step 2 — find the gradient at $t = 2$ .

$\dfrac{dy}{dx}\bigg|_{t=2} = \dfrac{3 \cdot 4 - 4}{4 \cdot 2} = \dfrac{8}{8} = 1$

M1 — substituting $t = 2$ into the gradient expression.

Step 3 — write the tangent.

$y - 0 = 1(x - 5)$ , i.e. $y = x - 5$ .

A1 — correct equation in the requested form, $m = 1$ and $c = -5$ .

Total: 8 marks (M3 A3 B1, plus an embedded M1 in (b)).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A curve has parametric equations $x = \cos 2\theta$ and $y = \sin\theta$ for $0 \leq \theta \leq \pi$ .

(a) Find $\dfrac{dy}{dx}$ in terms of $\theta$ , simplifying your answer. (3)

(b) Find the values of $\theta$ at which the tangent to the curve is vertical, and state the corresponding $(x, y)$ coordinates. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — differentiating: $\dfrac{dx}{d\theta} = -2\sin 2\theta$ and $\dfrac{dy}{d\theta} = \cos\theta$ .
M1 (AO1.1b) — forming the quotient $\dfrac{dy}{dx} = \dfrac{\cos\theta}{-2\sin 2\theta}$ .
A1 (AO2.5) — simplifying using $\sin 2\theta = 2\sin\theta\cos\theta$ : $\dfrac{\cos\theta}{-4\sin\theta\cos\theta} = -\dfrac{1}{4\sin\theta}$ .

(b)

M1 (AO1.1b) — recognising vertical tangent corresponds to $\dfrac{dx}{d\theta} = 0$ , i.e. $\sin 2\theta = 0$ .
A1 (AO1.1b) — solving on $0 \leq \theta \leq \pi$ gives $\theta = 0,\, \tfrac{\pi}{2},\, \pi$ , with care needed at $\theta = \tfrac{\pi}{2}$ where $\dfrac{dy}{d\theta} = 0$ also (a singular point) and at the endpoints.
A1 (AO2.5) — coordinates $(\cos\pi, \sin(\pi/2)) = (-1, 1)$ at $\theta = \tfrac{\pi}{2}$ , with appropriate comment on endpoint behaviour.

Total: 6 marks split AO1 = 4, AO2 = 2. Parametric questions tend to carry slightly more AO2 weight than ordinary differentiation because identifying which derivative condition (numerator zero vs denominator zero) corresponds to which geometric feature is itself reasoning.

Synoptic links

Connects to:

Chain rule (Section G, earlier in Year 2): the formula $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ is the chain rule $\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx}$ rewritten using $\dfrac{dt}{dx} = 1 \big/ \dfrac{dx}{dt}$ . Recognising this connection is what allows the parametric formula to be remembered (or re-derived) under exam pressure.
Coordinate geometry — parametric curves (Section F): before differentiating, students learn to plot and recognise parametric curves (circles $x = r\cos t,\, y = r\sin t$ ; ellipses $x = a\cos t,\, y = b\sin t$ ). The same parametrisations reappear here, now to be differentiated. The classical "ladder sliding down a wall" related-rates problem is naturally parametric.
Tangent and normal equations (Section F): the gradient $\dfrac{dy}{dx}$ at a point is the slope of the tangent; the normal has slope $-1 / (dy/dx)$ . Once the parametric gradient is known, tangent and normal equations follow as in Cartesian work — but at the parametric value, not the Cartesian one.
Parametric integration / area under a curve (Section H, Year 2): the area under a parametric curve from $t = a$ to $t = b$ is $\int_a^b y \dfrac{dx}{dt}\, dt$ . The same $dx/dt$ that appears in the denominator of parametric differentiation appears as a multiplicand here — same toolkit, different role.
Mechanics — velocity and acceleration vectors (Paper 3): for a particle with position $\mathbf{r}(t) = (x(t), y(t))$ , velocity is $\dot{\mathbf{r}} = (\dot{x}, \dot{y})$ and the path's slope at any instant is $\dot{y}/\dot{x}$ — exactly parametric differentiation with $t$ as physical time. Projectile motion, $x = u_x t,\, y = u_y t - \tfrac{1}{2} g t^2$ , is the canonical example.

Mark-scheme literacy

Parametric differentiation questions on AQA 7357 split AO marks more evenly than basic differentiation:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Correctly differentiating each parametric equation, forming the quotient, substituting values
AO2 (reasoning / interpretation)	25–35%	Identifying which condition (numerator / denominator zero) corresponds to which feature; simplifying the quotient using identities; rejecting invalid $t$ values
AO3 (problem-solving)	5–15%	Modelling contexts (mechanics-flavoured questions), or finding intersection/inflection without explicit prompts

Parametric Differentiation

Parametric Differentiation

The Chain Rule for Parametric Differentiation

Basic Examples

Finding the Gradient at a Specific Point

Equations of Tangents

Equations of Normals

Stationary Points of Parametric Curves

Vertical Tangents

Second Derivatives (Extension)

Worked Examination-Style Problem

Summary

A-Level Deep Dive: Parametric Differentiation

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

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