0606/22 May/Jun 2025, Question 6: Connected Rates of Change and Small Changes
Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 6 marks
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The question, in short
Given a sphere's volume grows at a constant 2π cm³ s⁻¹, find how fast its surface area changes when the volume is 36π cm³.
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Solution · 6 marks
- 1(4/3)πr³ = 36π, so r = 3
- 2dV/dr = 4πr² = 36π and dS/dr = 8πr = 24π
- 3dr/dt = (dV/dt)/(dV/dr) = 2π/(36π) = 1/18
- 4dS/dt = dS/dr × dr/dt = 24π × 1/18
- 5dS/dt = 4π/3 ≈ 4.19 cm² s⁻¹
Answer4π/3 cm² s⁻¹ (4.19)
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