Parametric Differentiation

Builds on: aqa-alevel-maths-pure-2 / parametric-equations — this lesson is the A2-level deep dive.

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.

Spec Mapping — AQA 7357 Sections C (Coordinate geometry) and G (Differentiation). This lesson is synoptic — it covers the parametric-differentiation content, linking parametric curve work in Section C with chain-rule differentiation in Section G. Refer to the official AQA specification document for exact wording.

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The Chain Rule for Parametric Differentiation

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

$$dy/dx = (dy/dt) div (dx/dt)$$dy/dx=(dy/dt)div(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:

$$dy/dx = dy/dt imes dt/dx = (dy/dt) / (dx/dt)$$dy/dx=dy/dtimesdt/dx=(dy/dt)/(dx/dt)

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Basic Examples

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

$$\begin{aligned} dx/dt = 2t \\ dy/dt = 3t^{2} \\ dy/dx = (3t^{2}) / (2t) = 3t/2 \end{aligned}$$dx/dt=2tdy/dt=3t2dy/dx=(3t2)/(2t)=3t/2​

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

$$\begin{aligned} dx/d \theta = -3sin \theta \\ dy/d \theta = 3cos \theta \\ dy/dx = (3cos \theta ) / (-3sin \theta ) = -cos \theta / sin \theta = -cot \theta \end{aligned}$$dx/dθ=−3sinθdy/dθ=3cosθdy/dx=(3cosθ)/(−3sinθ)=−cosθ/sinθ=−cotθ​

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

$$\begin{aligned} dx/dt = 2 \\ dy/dt = 2t - 4 \\ dy/dx = (2t - 4) / 2 = t - 2 \end{aligned}$$dx/dt=2dy/dt=2t−4dy/dx=(2t−4)/2=t−2​

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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.

$$\begin{aligned} dx/dt = 3t^{2} \\ dy/dt = 4t \\ dy/dx = 4t / (3t^{2}) = 4 / (3t) \\ At t = 2: dy/dx = 4 / 6 = 2/3 \end{aligned}$$dx/dt=3t2dy/dt=4tdy/dx=4t/(3t2)=4/(3t)Att=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).

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Equations of Tangents

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

1. Find dy/dx at that parameter value (this is the gradient m).
2. Find the coordinates (x, y) at that parameter value.
3. 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

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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

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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.

$$\begin{aligned} dx/dt &= 3t^{2} - 3 \\ dy/dt &= 2t \\ dy/dx &= 2t / (3t^{2} - 3) \\ \text{Stationary points} &: dy/dt = 0 \implies 2t = 0 \implies t = 0 \\ \text{Check } dx/dt \ne 0 \text{ at } t = 0 &: dx/dt = -3 \ne 0 \;\checkmark \\ \text{At } t = 0 &: x = 0,\; y = -4 \end{aligned}$$dx/dtdy/dtdy/dxStationary pointsCheck dx/dt=0 at t=0At t=0​=3t2−3=2t=2t/(3t2−3):dy/dt=0⟹2t=0⟹t=0:dx/dt=−3=0✓:x=0,y=−4​

So the stationary point is at $(0, -4)$(0,−4).

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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).

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Second Derivatives (Extension)

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

$$d^{2}y/dx^{2} = (d/dt(dy/dx)) / (dx/dt)$$d2y/dx2=(d/dt(dy/dx))/(dx/dt)

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

$$\begin{aligned} dy/dx = 3t/2 \\ d/dt(dy/dx) = d/dt(3t/2) = 3/2 \\ dx/dt = 2t \\ d^{2}y/dx^{2} = (3/2) / (2t) = 3/(4t) \end{aligned}$$dy/dx=3t/2d/dt(dy/dx)=d/dt(3t/2)=3/2dx/dt=2td2y/dx2=(3/2)/(2t)=3/(4t)​

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Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

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.

(c) Find the equation of the normal 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)

$$\begin{aligned} dx/dt = 4 \\ dy/dt = -4/t^{2} \\ dy/dx = (-4/t^{2}) / 4 = -1/t^{2} \end{aligned}$$dx/dt=4dy/dt=−4/t2dy/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)

$$\begin{aligned} \text{Normal gradient} &= 4 \\ \text{Normal} &: y - 2 = 4(x - 8) \\ &\;\;\; y = 4x - 30 \end{aligned}$$Normal gradientNormal​=4: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

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Synoptic Links

This topic connects to:

• Parametric equations (intro) (aqa-alevel-maths-coordinate-geometry-depth / parametric-equations-intro) — the parametric curves on which this differentiation technique operates.
• Differentiation techniques (aqa-alevel-maths-pure-2 / differentiation-techniques) — the chain rule underpinning dy/dx = (dy/dt)/(dx/dt).
• Rates of change (aqa-alevel-maths-calculus-applications / rates-of-change) — connected-rates problems use the same parameter-chain pattern with t as time.

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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 $\dot{\mathbf{r}} = (\dot{x}, \dot{y})$r˙=(x˙,y˙​) is the parametric derivative of position). The AQA formula booklet provides $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​ in the Differentiation section, so the formula itself is given — but candidates are still expected to know when and how to apply it.

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Worked example with full mark scheme

Question (8 marks):

A curve $C$C is defined parametrically by $x = 2t^2 - 3$x=2t2−3 and $y = t^3 - 4t$y=t3−4t for $t \in \mathbb{R}$t∈R.

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

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

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

Solution with mark scheme:

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

$\dfrac{dx}{dt} = 4t, \qquad \dfrac{dy}{dt} = 3t^2 - 4$dtdx​=4t,dtdy​=3t2−4

M1 — both parametric derivatives correct. A common slip is to differentiate $y = t^3 - 4t$y=t3−4t as $3t^2 - 4t$3t2−4t, carrying the linear coefficient through incorrectly.

Step 2 — apply the parametric chain rule.

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2 - 4}{4t}$dxdy​=dx/dtdy/dt​=4t3t2−4​

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

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

$3t^2 - 4 = 0 \implies t^2 = \dfrac{4}{3} \implies t = \pm \dfrac{2}{\sqrt{3}}$3t2−4=0⟹t2=34​⟹t=±3​2​

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 $t$t back into the original parametric equations.

For $t = \dfrac{2}{\sqrt{3}}$t=3​2​: $x = 2 \cdot \dfrac{4}{3} - 3 = \dfrac{8}{3} - 3 = -\dfrac{1}{3}$x=2⋅34​−3=38​−3=−31​, and $y = \left(\dfrac{2}{\sqrt{3}}\right)^3 - 4 \cdot \dfrac{2}{\sqrt{3}} = \dfrac{8}{3\sqrt{3}} - \dfrac{8}{\sqrt{3}} = \dfrac{8 - 24}{3\sqrt{3}} = -\dfrac{16}{3\sqrt{3}} = -\dfrac{16\sqrt{3}}{9}$y=(3​2​)3−4⋅3​2​=33​8​−3​8​=33​8−24​=−33​16​=−9163​​.

For $t = -\dfrac{2}{\sqrt{3}}$t=−3​2​: $x = -\dfrac{1}{3}$x=−31​ (same, since $x$x depends on $t^2$t2), and $y = +\dfrac{16\sqrt{3}}{9}$y=+9163​​.

M1 — substituting at least one root back into both $x$x and $y$y. A1 — both stationary points correct: $\left(-\tfrac{1}{3},\, -\tfrac{16\sqrt{3}}{9}\right)$(−31​,−9163​​) and $\left(-\tfrac{1}{3},\, \tfrac{16\sqrt{3}}{9}\right)$(−31​,9163​​).

(c) Step 1 — find the point on the curve at $t = 2$t=2.

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

B1 — point of tangency correct.

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

$\dfrac{dy}{dx}\bigg|_{t=2} = \dfrac{3 \cdot 4 - 4}{4 \cdot 2} = \dfrac{8}{8} = 1$dxdy​​t=2​=4⋅23⋅4−4​=88​=1

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

Step 3 — write the tangent.

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

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

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

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A curve has parametric equations $x = \cos 2\theta$x=cos2θ and $y = \sin\theta$y=sinθ for $0 \leq \theta \leq \pi$0≤θ≤π.

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

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

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — differentiating: $\dfrac{dx}{d\theta} = -2\sin 2\theta$dθdx​=−2sin2θ and $\dfrac{dy}{d\theta} = \cos\theta$dθdy​=cosθ.
• M1 (AO1.1b) — forming the quotient $\dfrac{dy}{dx} = \dfrac{\cos\theta}{-2\sin 2\theta}$dxdy​=−2sin2θcosθ​.
• A1 (AO2.5) — simplifying using $\sin 2\theta = 2\sin\theta\cos\theta$sin2θ=2sinθcosθ: $\dfrac{\cos\theta}{-4\sin\theta\cos\theta} = -\dfrac{1}{4\sin\theta}$−4sinθcosθcosθ​=−4sinθ1​.

(b)

• M1 (AO1.1b) — recognising vertical tangent corresponds to $\dfrac{dx}{d\theta} = 0$dθdx​=0, i.e. $\sin 2\theta = 0$sin2θ=0.
• A1 (AO1.1b) — solving on $0 \leq \theta \leq \pi$0≤θ≤π gives $\theta = 0,\, \tfrac{\pi}{2},\, \pi$θ=0,2π​,π, with care needed at $\theta = \tfrac{\pi}{2}$θ=2π​ where $\dfrac{dy}{d\theta} = 0$dθdy​=0 also (a singular point) and at the endpoints.
• A1 (AO2.5) — coordinates $(\cos\pi, \sin(\pi/2)) = (-1, 1)$(cosπ,sin(π/2))=(−1,1) at $\theta = \tfrac{\pi}{2}$θ=2π​, 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.

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Synoptic links

Connects to:

1. Chain rule (Section G, earlier in Year 2): the formula $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​ is the chain rule $\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx}$dxdy​=dtdy​⋅dxdt​ rewritten using $\dfrac{dt}{dx} = 1 \big/ \dfrac{dx}{dt}$dxdt​=1/dtdx​. Recognising this connection is what allows the parametric formula to be remembered (or re-derived) under exam pressure.

2. Coordinate geometry — parametric curves (Section F): before differentiating, students learn to plot and recognise parametric curves (circles $x = r\cos t,\, y = r\sin t$x=rcost,y=rsint; ellipses $x = a\cos t,\, y = b\sin t$x=acost,y=bsint). The same parametrisations reappear here, now to be differentiated. The classical "ladder sliding down a wall" related-rates problem is naturally parametric.

3. Tangent and normal equations (Section F): the gradient $\dfrac{dy}{dx}$dxdy​ at a point is the slope of the tangent; the normal has slope $-1 / (dy/dx)$−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.

4. Parametric integration / area under a curve (Section H, Year 2): the area under a parametric curve from $t = a$t=a to $t = b$t=b is $\int_a^b y \dfrac{dx}{dt}\, dt$∫ab​ydtdx​dt. The same $dx/dt$dx/dt that appears in the denominator of parametric differentiation appears as a multiplicand here — same toolkit, different role.