7 Fig. 4 shows a cone. The angle between the axis and the slant edge is \(30 ^ { \circ }\). Water is poured into the cone at a constant rate of \(2 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the radius of the water surface is \(r \mathrm {~cm}\) and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\).
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\includegraphics[alt={},max width=\textwidth]{431d496a-a606-4b92-9f5c-e12b074a7ba9-4_363_391_1447_887}
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\caption{Fig. 4}
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- Write down the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\).
- Show that \(V = \frac { \sqrt { 3 } } { 3 } \pi r ^ { 3 }\), and find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\).
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[You may assume that the volume of a cone of height \(h\) and radius \(r\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).] - Use the results of parts (i) and (ii) to find the value of \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) when \(r = 2\).