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.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option 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 \(\frac{\mathrm d r}{\mathrm d t} = 3cm s^{-1}\)
\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)
= \(8\pi \times 2cm \times 3cm s^{-1} = 48\pi cm^{2}s^{-1}\)
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 \(\frac{\mathrm d r}{\mathrm d t} = 3cm s^{-1}\)
\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)
= \(8\pi \times 2cm \times 3cm s^{-1} = 48\pi cm^{2}s^{-1}\)