2024 IGCSE Additional Mathematics February/March 0606/22 Q11

5 marks

A cylinder, open at both ends, has base radius \(r\,\mathrm{cm}\) and height \(4r\,\mathrm{cm}\). Its curved surface area is \(S\,\mathrm{cm}^2\).

Given that \(r\) varies with time \(t\), find \(S\) at the instant when \(\dfrac{\mathrm{d}S}{\mathrm{d}t}=6\dfrac{\mathrm{d}r}{\mathrm{d}t}\).[5]

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Answer:\(S=\dfrac9{8\pi}\,\mathrm{cm}^2\), assuming the radius has nonzero instantaneous rate of change.

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