The radius of a sphere is increasing at a rate \(3cm s^{-1}\). Find the rate of increase in the surface area, when the radius is 2cm.
- \(8\pi cm^{2}s^{-1}\)
- \(16\pi cm^{2}s^{-1}\)
- \(24\pi cm^{2}s^{-1}\)
-
\(48\pi cm^{2}s^{-1}\)
The correct answer is: D
Explanation
Surface area of sphere, \( A = 4\pi r^{2}\)
\(\frac{\mathrm d A}{\mathrm d r} = 8\pi r\)
The rate of change of radius with time \(\frac{\mathrm d r}{\mathrm d t} = 3cm s^{-1}\)
\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)
= \(8\pi \times 2cm \times 3cm s^{-1} = 48\pi cm^{2}s^{-1}\)