0606/13 May/Jun 2026, Question 6: Connected Rates of Change and Small Changes
Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 6 marks
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The question, in short
Cone tank with vertical axis and 30° between axis and sloping edge; relate water depth h, surface radius r and volume V, then use constant inflow 50 cm³ s⁻¹ to find how fast h rises when h = 60.
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Part (a) · 2 marks
Use the 30° angle to express r in terms of h, then substitute into the cone volume formula to show V = (1/9)πh³.
- 1tan 30° = r/h, so r = h/√3
- 2V = (1/3)πr²h
- 3V = (1/3)π(h/√3)² × h = (1/3)π(h²/3)h
- 4V = (1/9)πh³
AnswerV = (1/9)πh³ shown
Part (b) · 4 marks
Differentiate V = (1/9)πh³ with respect to t, substitute dV/dt = 50 and h = 60, and solve for dh/dt in terms of π.
- 1dV/dh = (1/3)πh²
- 2dh/dt = dh/dV × dV/dt
- 3dh/dt = 50/((1/3)πh²)
- 4At h = 60: dh/dt = 50/(1200π) = 1/(24π) cm s⁻¹
Answer1/(24π)
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