If the volume of a hemisphere is increasing at a steady rate of 18π m\(^{3}\) s\(^{-1}\), at what rate is its radius changing when its is 6m?
\(V = \frac{2}{3} \pi r^{3}\)
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,
\(\frac{\mathrm d r}{\mathrm d t} = \frac{9}{6^{2}} = \frac{1}{4}\)
= \(0.25 ms^{-1}\)
There is an explanation video available below.
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