The volume of a sphere is increasing at a constant rate of 2 cm3 ^3\,3s−1^{-1}−1.
The volume of a sphere of radius r r\,r cm is 43πr3\displaystyle \frac{4}{3}\pi r^334πr3 cm3^33, and its surface area is 4πr2 4\pi r^2\,4πr2 cm2^22.
Show that the rate of increase of the radius when r=2r = 2r=2 is aπ\displaystyle \frac{a}{\pi}πa cm s−1^{-1}−1, where a a\,a is a constant to be found.
Find the rate at which the surface area is increasing when r=2r = 2r=2.