\(V = \frac{2}{3} \pi r^{3}\) Given: \(\frac{\mathrm d V}{\mathrm d t} = 18\pi m^{3} s^{-1}\) \(\frac{\mathrm d V}{\mathrm d t} = \frac{\mathrm d V}{\mathrm d r} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d V}{\mathrm d r} = 2\pi r^{2}\) \(18\pi = 2\pi r^{2} \times \frac{\mathrm d r}{\mathrm d t}\) \(\frac{\mathrm d r}{\mathrm d t} = \frac{18\pi}{2\pi r^{2}} = \frac{9}{r^{2}}\) The rate of change of the radius when r = 6m, \(\frac{\mathrm d r}{\mathrm d t} = \frac{9}{6^{2}} = \frac{1}{4}\) = \(0.25 ms^{-1}\)