AQA A-Level Mathematics Paper 1, June 2025: Question 13

9 marks · Medium difficulty · Multi-step Problem

Differentiate parametric equations involving quadratic and exponential terms to find dy/dx, determine the equation of a tangent at t = 0, and find the justified Cartesian equation in the form y = f(x).

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Question

Question text

13 A curve C has parametric equations

x = 4(4t + 1)2

y = e–4t

– 1 t

for ≤ ≤ 0

13 (a) Find in terms of t

[3 marks]

13 (b) Find an equation of the tangent to C at the point where t = 0

[3 marks]

13 (c) Find a Cartesian equation for C in the form y = f (x)

Fully justify your answer.

[3 marks]

(16)

Mark scheme

Show the mark scheme

Q Marking instructions AO Marks Typical solution

13(a) Obtains 32 4(t +1) dx

= 32 4(t +1)

Can be unsimplified 1.1b B1 dt

Obtains −4e−4t 1.1b B1 dy

= −4e−4t

dy dx dt

Uses their their to

dt dt −4t −4t

dy dy −4e −e

obtain an expression for = =

dx 3.1a B1F dx 32 4(t +1) 8 4(t +1)

Can be unsimplified

ISW

ACF

Subtotal 3

13(b) Substitutes t = 0 to obtain

dy t = 0

values for x, y and their 3.1a M1

dx x = 4

PI by correct values y = 1

Obtains at least two of x = 4 ,

dy 1 dy 1

y =1or = − 1.1a M1 = −

dx 8 dx 8

Obtains

1 y −1= − (x − 4)

y −1= − (x − 4) 8

8 1.1b A1

ACF – A-LEVEL MATHEMATICS – 7357/1 –

ISW

Subtotal 3

13(c) Eliminates t correctly to form a x

Cartesian equation. May use 4t +1=

positive or negative square root 2

or both. The positive square root is used

3.1a M1 1

Condone a sign error when since − t 0

rearranging either parametric 4

equation. x − 2

4t =

x x 2

1− 1+ x−2

Obtains y = e 2 or y = e 2 −

y = e 2

1 x

or y = e 2 1.1a M1

May be unsimplified.

Gives a valid explanation for

selecting the appropriate square

root

16 1

for example − t or y e 2.4 R1

AND

1− x

Obtains y = e 2 OE

Subtotal 3

Question 13 Total 9

How to answer it

Parametric Curves: Differentiation, Tangents & Cartesian Conversion

Summary of Assessment

What This Question Tests

This question assesses key Pure Year 2 skills in coordinate geometry and calculus:

Parametric Differentiation: Applying the chain rule to individual parametric equations and calculating dy/dx = (dy/dt) ÷ (dx/dt) .
Chain Rule on Composite Expressions: Differentiating power brackets (ax + b)ⁿ and exponential terms e^(kt) .
Tangents to Curves: Evaluating coordinates and gradient at a specific parameter value ( t = 0 ) and writing straight line equations using y - y₁ = m(x - x₁) .
Eliminating the Parameter & Mathematical Reasoning: Inverting parametric relationships involving squares and square roots, and strictly justifying the choice of sign ( ± ) based on the domain restriction -¼ ≤ t ≤ 0 .

Part (a) • 3 Marks

Finding dy/dx in Terms of the Parameter t

Differentiating x = 4(4t + 1)² and y = e⁻⁴ᵗ

📐 Step-by-Step Calculation

Differentiate x with respect to t (Chain Rule):
x = 4(4t + 1)²
dx/dt = 4 × 2(4t + 1)¹ × 4 = 32(4t + 1)
Differentiate y with respect to t:
y = e⁻⁴ᵗ
dy/dt = -4e⁻⁴ᵗ
Combine using the parametric derivative rule:
dy/dx = (dy/dt) / (dx/dt) = -4e⁻⁴ᵗ / [32(4t + 1)] = -e⁻⁴ᵗ / [8(4t + 1)]

✅ Final Answer

dy/dx = -e⁻⁴ᵗ / [8(4t + 1)]
(Unsimplified equivalents like -4e⁻⁴ᵗ / [32(4t + 1)] are also awarded full marks, with ISW applied).

❌ Common Errors

Forgetting the inner derivative: Differentiating 4(4t + 1)² to get 8(4t + 1) instead of multiplying by the inner derivative of 4t (which is 4).
Inverting the quotient: Writing dx/dt ÷ dy/dt instead of dy/dt ÷ dx/dt .
Sign slip on exponential: Missing the negative sign when differentiating e⁻⁴ᵗ .

Mark Breakdown:
• B1: Correctly finds dx/dt = 32(4t + 1) (or expanded 128t + 32 ).
• B1: Correctly finds dy/dt = -4e⁻⁴ᵗ .
• B1F: Divides their dy/dt by their dx/dt to form an expression for dy/dx (Follow-through allowed).

Part (b) • 3 Marks

Finding the Equation of the Tangent at t = 0

Evaluating coordinates, gradient, and constructing the linear equation

📐 Step-by-Step Calculation

Find the coordinates (x₁, y₁) at t = 0:
When t = 0 :
x = 4(4(0) + 1)² = 4(1)² = 4
y = e⁻⁴⁽⁰⁾ = e⁰ = 1
So the tangent passes through the point (4, 1).
Find the gradient m at t = 0:
Substitute t = 0 into dy/dx :
m = -e⁰ / [8(4(0) + 1)] = -1 / 8
Use the point-gradient line formula:
y - y₁ = m(x - x₁)
y - 1 = -⅛(x - 4)
Multiplying through gives: 8y - 8 = -x + 4 ⇒ x + 8y - 12 = 0 (or y = -⅛x + 1.5 ).

✅ Acceptable Forms

• y - 1 = -⅛(x - 4)
• y = -⅛x + 1.5 (or y = -⅛x + ³⁄₂ )
• x + 8y - 12 = 0 (or x + 8y = 12 )

🧠 Exam Technique

Any correct form (ACF) is accepted unless the question specifies ax + by + c = 0 . Leaving it in point-slope form y - 1 = -⅛(x - 4) saves time and prevents rearrangement arithmetic blunders!
Remember that e⁰ = 1 , never 0 .

Mark Breakdown:
• M1: Substitutes t = 0 into equations for x , y , and their expression for dy/dx .
• M1: Correctly identifies at least two of: x = 4 , y = 1 , or m = -⅛ .
• A1: Produces a completely correct tangent line equation in any equivalent form.

Part (c) • 3 Marks

Finding and Justifying the Cartesian Equation y = f(x)

Eliminating t and reasoning with the domain restriction -¼ ≤ t ≤ 0

📐 Step-by-Step Calculation

Rearrange x to isolate the squared term:
x = 4(4t + 1)² ⇒ (4t + 1)² = x/4
Take the square root of both sides:
4t + 1 = ±√(x/4) = ±√x / 2
Apply the given domain condition to resolve the sign (Crucial Justification):
The parameter is restricted to: -¼ ≤ t ≤ 0 .
Multiply by 4: -1 ≤ 4t ≤ 0 .
Add 1: 0 ≤ 4t + 1 ≤ 1 .
Since 4t + 1 ≥ 0 , it cannot be negative! Therefore, we must take the positive square root:
4t + 1 = +√x / 2
Express -4t in terms of x:
4t = (√x / 2) - 1
Multiply by -1: -4t = 1 - (√x / 2)
Substitute into y:
y = e⁻⁴ᵗ = e^(1 - √x / 2)

✅ Final Fully Justified Answer

y = e^(1 - √x / 2) (or y = e^( (2 - √x)/2 ) )

Required Justification: Stating that since -¼ ≤ t ≤ 0 , 4t + 1 ≥ 0 (or showing y ≤ e ), hence the positive square root is required.

💡 Key Knowledge: Why the Sign Matters

When solving A² = B , algebraically A = ±√B .
Whenever a question demands "Fully justify your answer" alongside an equation involving squares/roots or restricted domains, examiners are explicitly testing whether you provide mathematical reasoning for choosing the positive or negative branch.

❌ Common Errors & Mark Losses

Omitting the justification (Losing R1): Many candidates reached y = e^(1 - √x/2) simply by assuming square roots are always positive, entirely ignoring the -¼ ≤ t ≤ 0 restriction.
Algebraic slip on the denominator: Writing √(x/4) = √x / 4 instead of √x / 2 .
Sign errors when rearranging for -4t: Getting -4t = -1 - √x/2 or 4t - 1 .

🧠 Examiner Commentary

The mark scheme awards:

M1: Eliminating t to link x and y (even with ± or a minor sign slip).
M1: Reaching y = e^(1 - √x/2) or y = e^(1 ± √x/2) .
R1 (Reasoning): A rigorous argument explaining why the positive branch is selected (e.g., citing t ≥ -¼ ⇒ 4t + 1 ≥ 0 or comparing range/domain).

Topics

Pure Mathematics · C: Coordinate geometry in the (x, y) plane · G: Differentiation

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.