AQA A-Level Mathematics Paper 3, June 2025: Question 11

10 marks · Medium difficulty · Modelling

Form a differential equation for the melting of a cuboid block of ice, use the chain rule to find dx/dt, and interpret the result in context.

Practise this question

Question

Question 11 illustrates a cuboid block of ice with dimensions labelled 4x, 2x, and x centimetres. Part (a) states the volume V decreases at a rate proportional to its surface area, and when x = 4 the volume decreases at 7 cm³ per minute; it asks to show that dV/dt = -0.4375x². Part (b) asks to find dV/dx in terms of x. Part (c)(i) asks to find dx/dt using parts (a) and (b). Part (c)(ii) asks to interpret the answer to part (c)(i) in context.

Mark scheme

Show the mark scheme Mark scheme for Question 11. Part (a) awards marks for setting surface area equal to 28x², setting dV/dt = -28kx², solving for k = 1/64 using x = 4 and dV/dt = -7, and concluding dV/dt = -0.4375x² (4 marks). Part (b) calculates V = 8x³ and gives dV/dx = 24x² (2 marks). Part (c)(i) uses the chain rule dx/dt = (dx/dV)(dV/dt) to obtain -7/384 or approximately -0.0182 (2 marks). Part (c)(ii) awards marks for explaining that the dimensions/height/width decrease at a constant rate of 7/384 cm per minute (2 marks). Total 10 marks.

How to answer it

Melting Ice Block: Rates of Change & Geometric Modelling

📋 What this question tests

This multi-part problem evaluates your ability to translate a geometric word problem into mathematical equations and apply calculus techniques:

  • Formulating Differential Equations: Translating phrases like "decreases at a rate proportional to" into mathematical notation ( dV/dt = −k × Surface Area ).
  • Geometric Formulae: Finding expressions for the total surface area and volume of a cuboid in terms of a variable parameter x .
  • Boundary Conditions: Solving for the constant of proportionality using given instantaneous values.
  • Chain Rule for Connected Rates: Using dx/dt = (dx/dV) × (dV/dt) .
  • Real-world Contextual Interpretation: Recognising that when a derivative is a negative constant, the dimension is decreasing at a constant rate, and stating the correct units.
Part (a) • 4 Marks

Forming the Differential Equation for dV/dt

Show that dV/dt = −0.4375x²

📐 Step-by-Step Proof

  1. Find Total Surface Area (SA):
    A cuboid has 3 pairs of opposite identical faces:
    SA = 2(4x × 2x + 4x × x + 2x × x)
    SA = 2(8x² + 4x² + 2x²) = 2(14x²) = 28x²
  2. Set up the differential equation:
    Rate of decrease is proportional to SA:
    dV/dt = −k(SA) = −k(28x²)
  3. Use the given condition to find k:
    When x = 4, volume decreases at 7 cm³ min⁻¹ (so dV/dt = −7 ):
    −7 = −k × 28 × (4)²
    −7 = −k × 28 × 16
    −7 = −448k ⇒ k = 7 / 448 = 1 / 64
  4. Substitute k back:
    dV/dt = −(1/64) × 28x² = −(28/64)x²
    Since 28/64 = 7/16 = 0.4375:
    dV/dt = −0.4375x² (As required)

💡 Key Knowledge

  • "Decreases at a rate of 7": Means the rate of change is negative: dV/dt = −7 . Missing this negative sign leads to sign contradictions.
  • Proportionality: dV/dt ∝ SA implies dV/dt = m(SA) , where m is negative.

❌ Common Errors

  • Forgetting that a cuboid has 6 faces (omitting the factor of 2 or missing faces, giving 14x² instead of 28x²).
  • Jumping to −0.4375x² without showing the intermediate fraction 28/64 or the exact calculation for k . Because this is a "Show that" (AG) question, every step must be clearly justified.
Mark Breakdown:
• B1: Correct expression for surface area ( 28x² or unsimplified equivalent).
• M1: Setting up differential equation of the form dV/dt = m(28x²) or dV/dt ∝ x² .
• M1: Correctly substituting x = 4 and dV/dt = −7 to solve for the constant.
• R1: Completing a fully reasoned argument with no errors seen to reach −0.4375x² .
Part (b) • 2 Marks

Differentiating Volume with Respect to x

Find dV/dx in terms of x

📐 Step-by-Step Calculation

  1. Write down the volume V of the cuboid:
    V = length × width × height
    V = (4x) × (2x) × (x)
    V = 8x³
  2. Differentiate V with respect to x:
    dV/dx = d/dx (8x³) = 8 × 3x²
    dV/dx = 24x²

✅ Final Answer

dV/dx = 24x²

Be sure to explicitly write V = 8x³ first so the method mark is secured even if an arithmetic slip occurs later.

Mark Breakdown:
• M1: Stating or forming the correct expression for the volume: V = 8x³ .
• A1: Correctly obtaining 24x² .
Part (c)(i) • 2 Marks

Connected Rates of Change

Find dx/dt using your results from (a) and (b)

🧠 Exam Technique: Chain Rule

Connect the three rates using the derivative chain rule:

dx/dt = (dx/dV) × (dV/dt)

Remember that dx/dV = 1 / (dV/dx) .

📐 Step-by-Step Calculation

  1. Substitute expressions:
    dx/dt = (1 / 24x²) × (−0.4375x²)
  2. Simplify and cancel x²:
    Notice that x² cancels completely:
    dx/dt = −0.4375 / 24 = −(7/16) / 24 = −7/384
  3. Decimal equivalent:
    dx/dt ≈ −0.0182 cm min⁻¹ (accept any value in range [−0.0183, −0.018])

✅ Correct Answer

dx/dt = −7/384  (or ≈ −0.0182 )

Mark Breakdown:
• M1: Using the chain rule to connect dV/dt , their dV/dx , and dx/dt .
• A1: Correct value of −7/384 (or decimal in [−0.0183, −0.018]).
Part (c)(ii) • 2 Marks

Contextual Interpretation

Interpret, in context, your answer to part (c)(i)

✅ Model Answer (Full 2 Marks)

"The height (or x) of the ice block is decreasing at a constant rate of 7/384 cm per minute (approx. 0.018 cm/min)."

  • Component 1: Identifies dimension (height / edge / x) is decreasing at a constant rate [E1].
  • Component 2: States the numerical value with correct units: 7/384 cm per minute [E1F].

❌ Pitfalls Where Students Lose Marks

  • Missing the word "constant": Because dx/dt does not depend on x or t , the rate does not change over time. Omitting "constant" loses the first E mark!
  • Double negatives: Writing "decreasing at a rate of −0.018 cm/min". Either say "decreasing at a rate of 0.018..." OR "changing at a rate of −0.018...".
  • Missing / wrong units: Units must be length per time: cm per minute (not cm³, cm², or seconds).
Mark Breakdown:
• E1: Identifies that a linear dimension (height/width/length/side) is decreasing at a constant rate.
• E1F: States the correct rate: 7/384 (or [0.018, 0.0183]) cm per minute (follow-through on their constant dx/dt).

Topics

Pure Mathematics · G: Differentiation · H: Integration

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.