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?

  • A 2.50m/s
  • B 2.00 m/s
  • C 0.25 m/s
  • D 0.20 m/s
jamb 2000mathematics

The correct answer is C. 0.25 m/s

\(V = \frac{2}{3} \pi r^{3}\)

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,

\(\frac{\mathrm d r}{\mathrm d t} = \frac{9}{6^{2}} = \frac{1}{4}\)

= \(0.25 ms^{-1}\)

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