Parametric Differentiation

This lesson covers differentiation of parametric equations as required by the Edexcel A-Level Mathematics specification (9MA0). You need to be able to find dy/dx for curves defined parametrically and use it to find tangents and normals.

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What Are Parametric Equations?

In parametric form, both x and y are expressed in terms of a third variable (the parameter), usually denoted t or θ.

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

Each value of t gives a point (x, y) on the curve.

Example

The parametric equations x = t², y = 2t describe a parabola (equivalent to x = y²/4 or y² = 4x).

t	-2	-1	0	1	2
x = t²	4	1	0	1	4
y = 2t	-4	-2	0	2	4

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The Formula for dy/dx

To find dy/dx for parametric equations, use:

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

provided dx/dt ≠ 0.

This follows from the chain rule: dy/dt = (dy/dx) × (dx/dt), so dy/dx = (dy/dt) / (dx/dt).

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Worked Examples

Worked Example 1

Find dy/dx for x = t², y = 2t.

Solution: dx/dt = 2t, dy/dt = 2

dy/dx = (dy/dt) / (dx/dt) = 2 / (2t) = 1/t

Worked Example 2

Find dy/dx for x = 3t - 1, y = t² + 2t.

Solution: dx/dt = 3, dy/dt = 2t + 2

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

Worked Example 3

Find dy/dx for x = cos t, y = sin t.

Solution: dx/dt = -sin t, dy/dt = cos t

dy/dx = cos t / (-sin t) = -cos t / sin t = -cot t

Alternatively: dy/dx = -1/tan t.

For the unit circle x² + y² = 1, the gradient at angle t is -cos t / sin t, which matches the implicit differentiation result dy/dx = -x/y. ✓

Worked Example 4

Find dy/dx for x = e²ᵗ, y = t³ - t.

Solution: dx/dt = 2e²ᵗ, dy/dt = 3t² - 1

dy/dx = (3t² - 1) / (2e²ᵗ)

Exam Tip: Leave dy/dx in terms of the parameter t. You do not need to convert back to x and y unless the question specifically asks you to.

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Finding Tangents to Parametric Curves

Method

1. Find dy/dx in terms of t.
2. Substitute the given value of t to find the gradient.
3. Find the x and y coordinates at that value of t.
4. Use y - y₁ = m(x - x₁).

Worked Example 5

A curve is defined by x = t² + 1, y = 2t³. Find the equation of the tangent at t = 1.

Solution: dx/dt = 2t, dy/dt = 6t² dy/dx = 6t² / (2t) = 3t

At t = 1:

• dy/dx = 3(1) = 3
• x = 1 + 1 = 2
• y = 2(1) = 2
• Point: (2, 2)

Tangent: y - 2 = 3(x - 2) y = 3x - 4

Worked Example 6

A curve is given by x = 3 cos θ, y = 3 sin θ. Find the equation of the tangent at θ = π/4.

Solution: dx/dθ = -3 sin θ, dy/dθ = 3 cos θ dy/dx = 3 cos θ / (-3 sin θ) = -cos θ / sin θ

At θ = π/4:

• dy/dx = -(√2/2)/(√2/2) = -1
• x = 3 cos(π/4) = 3√2/2
• y = 3 sin(π/4) = 3√2/2

Tangent: y - 3√2/2 = -1(x - 3√2/2) y = -x + 3√2/2 + 3√2/2 y = -x + 3√2

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Finding Normals to Parametric Curves

The normal has gradient -1/m where m = dy/dx.

Worked Example 7

A curve is given by x = t², y = 4t. Find the equation of the normal at the point where t = 2.

Solution: dx/dt = 2t, dy/dt = 4 dy/dx = 4/(2t) = 2/t

At t = 2:

• Tangent gradient: 2/2 = 1
• Normal gradient: -1
• x = 4, y = 8
• Point: (4, 8)

Normal: y - 8 = -1(x - 4) y = -x + 12

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Horizontal and Vertical Tangents

Horizontal tangent (dy/dx = 0): occurs when dy/dt = 0 and dx/dt ≠ 0.

Vertical tangent (dy/dx undefined): occurs when dx/dt = 0 and dy/dt ≠ 0.

Worked Example 8

A curve is defined by x = t³ - 3t, y = t² - 4. Find the values of t where the tangent is horizontal and where it is vertical.

Solution: dx/dt = 3t² - 3 = 3(t² - 1) = 3(t - 1)(t + 1) dy/dt = 2t

Horizontal tangent: dy/dt = 0 → 2t = 0 → t = 0 Check: dx/dt at t = 0 is 3(0 - 1) = -3 ≠ 0. ✓

At t = 0: x = 0, y = -4. Horizontal tangent at (0, -4).

Vertical tangent: dx/dt = 0 → 3(t - 1)(t + 1) = 0 → t = 1 or t = -1 Check: dy/dt at t = 1 is 2 ≠ 0. ✓ dy/dt at t = -1 is -2 ≠ 0. ✓

At t = 1: x = 1 - 3 = -2, y = 1 - 4 = -3. Vertical tangent at (-2, -3). At t = -1: x = -1 + 3 = 2, y = 1 - 4 = -3. Vertical tangent at (2, -3).

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Second Derivative for Parametric Equations

To find d²y/dx² for parametric equations:

d²y/dx² = d/dt(dy/dx) / (dx/dt)

That is, differentiate dy/dx with respect to t, then divide by dx/dt.

Worked Example 9

For x = t², y = t³, find d²y/dx².

Solution: dy/dx = 3t²/(2t) = 3t/2

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

d²y/dx² = (3/2) / (2t) = 3/(4t)

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Common Mistakes

Mistake	Correction
dy/dx = dx/dt ÷ dy/dt (upside down)	dy/dx = (dy/dt) / (dx/dt), not the other way round
Forgetting to find x and y values for the tangent	You need the coordinates as well as the gradient
Converting to Cartesian form unnecessarily	Leave dy/dx in terms of t unless the question says otherwise
d²y/dx² = (d²y/dt²) / (d²x/dt²)	WRONG. The correct formula is d²y/dx² = (d/dt(dy/dx)) / (dx/dt)

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Summary

• For parametric equations x = f(t), y = g(t): dy/dx = (dy/dt) / (dx/dt).
• Leave dy/dx in terms of the parameter t.
• For tangents and normals, substitute the given parameter value to find the gradient and coordinates.
• Horizontal tangents: 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).

A-Level Deep Dive: Parametric Differentiation

Spec mapping

Edexcel 9MA0-02 specification section 9 — Differentiation, sub-strand 9.4 (Year 2 Pure) covers simple functions and relations defined parametrically, for first derivative only (refer to the official specification document for exact wording). The headline rule is the chain-rule consequence $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$dxdy​=dx/dtdy/dt​, valid wherever $\dfrac{dx}{dt} \neq 0$dtdx​=0. The technique is examined on 9MA0-02 Paper 2 alongside implicit differentiation and connected rates of change, and feeds directly into section 8 (Integration, parametric area) and section 7 (Coordinate geometry, tangents and normals to parametric curves). The Edexcel formula booklet does not state the parametric differentiation formula — students must reproduce it from the chain rule.

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Worked example with full mark scheme

Question (8 marks): A curve $C$C has parametric equations $x = 2t^2 - 3$x=2t2−3, $y = t^3 - 4t$y=t3−4t for $t \in \mathbb{R}$t∈R.

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

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

(c) Find an equation of the tangent to $C$C at the point where $t = 2$t=2. (2)

Solution with mark scheme:

(a) Step 1 — differentiate each parameter equation.

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

M1 — both derivatives correct. The most common slip is forgetting that $\dfrac{d}{dt}(-3) = 0$dtd​(−3)=0 and writing $\dfrac{dx}{dt} = 4t - 3$dtdx​=4t−3.

Step 2 — apply the parametric formula.

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

M1 — correct quotient $(dy/dt)/(dx/dt)$(dy/dt)/(dx/dt), not the inverse.

A1 — final form $\dfrac{3t^2 - 4}{4t}$4t3t2−4​ (or equivalent split form $\tfrac{3t}{4} - \tfrac{1}{t}$43t​−t1​).

(b) Step 1 — set $dy/dx = 0$dy/dx=0. Stationary points need $\dfrac{dy}{dt} = 0$dtdy​=0 with $\dfrac{dx}{dt} \neq 0$dtdx​=0:

$3t^2 - 4 = 0 \implies t = \pm \dfrac{2}{\sqrt{3}} = \pm \dfrac{2\sqrt{3}}{3}$3t2−4=0⟹t=±3​2​=±323​​

M1 — equating numerator to zero.

Step 2 — verify denominator non-zero. At $t = \pm \tfrac{2}{\sqrt{3}}$t=±3​2​, $4t \neq 0$4t=0, so genuine stationary points.

Step 3 — substitute back for coordinates. With $t^2 = \tfrac{4}{3}$t2=34​:

$x = 2 \cdot \dfrac{4}{3} - 3 = \dfrac{8}{3} - 3 = -\dfrac{1}{3}$x=2⋅34​−3=38​−3=−31​ $y = t^3 - 4t = t(t^2 - 4) = t\left(\dfrac{4}{3} - 4\right) = -\dfrac{8t}{3}$y=t3−4t=t(t2−4)=t(34​−4)=−38t​

So at $t = +\tfrac{2}{\sqrt{3}}$t=+3​2​: $y = -\tfrac{16}{3\sqrt{3}} = -\tfrac{16\sqrt{3}}{9}$y=−33​16​=−9163​​. At $t = -\tfrac{2}{\sqrt{3}}$t=−3​2​: $y = +\tfrac{16\sqrt{3}}{9}$y=+9163​​.

M1 — substituting parameter values back into both $x$x and $y$y.

A1 — coordinates $\left(-\tfrac{1}{3}, -\tfrac{16\sqrt{3}}{9}\right)$(−31​,−9163​​) and $\left(-\tfrac{1}{3}, \tfrac{16\sqrt{3}}{9}\right)$(−31​,9163​​).

(c) Step 1 — gradient at $t = 2$t=2.

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

Step 2 — point at $t = 2$t=2. $x = 2(4) - 3 = 5$x=2(4)−3=5, $y = 8 - 8 = 0$y=8−8=0.

M1 — both gradient and coordinates correct.

A1 — tangent equation $y - 0 = 1(x - 5)$y−0=1(x−5), i.e. $y = x - 5$y=x−5.

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): A curve has parametric equations $x = \sin 2t$x=sin2t, $y = \cos^2 t$y=cos2t, for $0 < t < \tfrac{\pi}{2}$0<t<2π​.

(a) Show that $\dfrac{dy}{dx} = -\tfrac{1}{2}$dxdy​=−21​. (4)

(b) Hence write down a Cartesian equation of the curve in the form $y = ax + b$y=ax+b. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — $\dfrac{dx}{dt} = 2\cos 2t$dtdx​=2cos2t and $\dfrac{dy}{dt} = -2\cos t \sin t$dtdy​=−2costsint.
• M1 (AO1.1b) — applying the parametric formula $\dfrac{dy}{dx} = \dfrac{-2\cos t \sin t}{2\cos 2t}$dxdy​=2cos2t−2costsint​.
• M1 (AO2.1) — using the double-angle identity $2\sin t \cos t = \sin 2t$2sintcost=sin2t to simplify the numerator.
• A1 (AO2.4) — recognising $\dfrac{-\sin 2t}{2\cos 2t}$2cos2t−sin2t​ is not the printed answer; instead, using $\cos 2t = 1 - 2\sin^2 t$cos2t=1−2sin2t alongside $y = \cos^2 t = \tfrac{1}{2}(1 + \cos 2t)$y=cos2t=21​(1+cos2t) to confirm a constant gradient $-\tfrac{1}{2}$−21​ across the domain. (Equivalently: differentiate the Cartesian form derived in (b) directly.)

(b)

• M1 (AO3.1a) — using $\cos^2 t = \tfrac{1}{2}(1 + \cos 2t)$cos2t=21​(1+cos2t) and $\sin^2 2t + \cos^2 2t = 1$sin22t+cos22t=1 to eliminate $t$t.
• A1 (AO2.5) — Cartesian form $y = -\tfrac{1}{2}x + \tfrac{1}{2}$y=−21​x+21​ (a straight-line segment, consistent with the constant gradient).

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. This is the AO2-heavy parametric pattern Edexcel favours on Paper 2: technical execution counts for less than recognising that the parametric curve secretly traces a straight line.

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Synoptic links

Connects to:

1. Section 5 — Trigonometry (double-angle identities): parametric forms in $(\sin t, \cos t)$(sint,cost) or $(\sin 2t, \cos^2 t)$(sin2t,cos2t) collapse cleanly only after applying $\sin 2t = 2 \sin t \cos t$sin2t=2sintcost or $\cos 2t = 1 - 2\sin^2 t$cos2t=1−2sin2t. Confidence with these identities turns ugly quotients into one-line simplifications.

2. Section 7 — Coordinate geometry (tangents and normals): once $dy/dx$dy/dx is in terms of $t$t, finding the tangent at a parameter value reduces to substituting and using $y - y_0 = m(x - x_0)$y−y0​=m(x−x0​) exactly as for explicit curves.