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4PM1/01R Jun 2023, Question 6: Connected Rates of Change and Small Changes

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 11 marks

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The question, in short

Cone with vertical angle 60° holds oil of height h and volume V; show V = (1/9)πh³, then find exact rate of change of surface area A when h = 24 and oil drains at 4 cm³/s.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 3 marks

Using the 60° vertical angle, express the cone radius in terms of h and derive the oil volume formula V = (1/9)πh³.

  1. 1Half the vertical angle is 30°, so tan 30° = r/h
  2. 2r = h/√3
  3. 3V = (1/3)πr²h = (1/3)π(h²/3)h
  4. 4V = (1/9)πh³

AnswerV = (1/9)πh³ shown

Part (b) · 8 marks

Relate A to h, then use the constant 4 cm³/s outflow to find the exact rate of change of A at h = 24.

  1. 1dV/dt = −4
  2. 2A = πr² = π(h/√3)² = πh²/3, so dA/dh = 2πh/3
  3. 3dV/dh = πh²/3
  4. 4dA/dt = dA/dh × dV/dt ÷ dV/dh = (2πh/3) × (−4) × 3/(πh²) = −8/h
  5. 5When h = 24: dA/dt = −8/24 = −1/3 cm²/s

Answer−1/3 cm²/s

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