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0606/21 May/Jun 2025, Question 9: Connected Rates of Change and Small Changes

Cambridge IGCSE Additional Mathematics 0606 · Paper 21 (Calculator) · Calculator · 12 marks

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The question, in short

Sector AOB, radius 15 and angle 6π/5, is rolled into a cone with A and B meeting; work out the cone's surface area, top circle, radius and height, then water-flow rates.

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

Part (a) · 2 marks

Calculate the curved surface area of the cone formed from the sector.

  1. 1Curved surface area = area of the sector
  2. 2= (1/2) × 15² × 6π/5
  3. 3= 135π ≈ 424

Answer135π cm² (424.1 to 424.2)

Part (b) · 2 marks

Calculate the circumference of the cone's horizontal circular top.

  1. 1Circumference of the top = arc length AB
  2. 2= 15 × 6π/5 = 18π ≈ 56.5

Answer18π cm (56.5 to 56.6)

Part (c) · 2 marks

Use the circumference to obtain the top radius and the cone's perpendicular height.

  1. 12πR = 18π, so R = 9
  2. 2Slant height = 15, so height² = 15² − 9² = 144
  3. 3Height = 12

Answerradius = 9, height = 12

Part (di) · 1 mark

For water depth h and water-surface radius r, express r as a formula involving h.

  1. 1Similar triangles: r/h = 9/12
  2. 2r = (3/4)h

Answerr = (3/4)h

Part (dii) · 5 marks

Given water enters at 27 cm³ per second, find how fast the depth grows when h = 4.

  1. 1V = (1/3)π((3/4)h)² × h = (3π/16)h³
  2. 2dV/dh = (9π/16)h²
  3. 3dh/dt = dV/dt ÷ dV/dh
  4. 4At h = 4: dV/dh = 9π
  5. 5dh/dt = 27/(9π) = 3/π ≈ 0.955

Answer0.955 cm/s

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