Question

A balloon, which always remains spherical has a variable radius. find the rate at which its volume is increasing with the radius when the later is 10 cm.

Answer 1

Answer:

$the \: volume \: of \: a \: sphere \: (v) \: with \: radius \: (r) \: is \: given \: by \\ v = \frac{4}{3}\pi {r}^{2} \\ the \: rate \: of \: change \: of \: volume \: (v) \: with \: respect \: to \: its \: radius \: (r) \: is \: given \: by \\ \frac{dv}{dr} = \frac{d}{dr} \:( \frac{4}{3}\pi {r}^{2}) = \frac{4}{3}\pi(3 {r}^{2}) = 4\pi {r}^{2} \\ therefore \: when \: radius = 10cm \\ \frac{dv}{dr} = 4\pi ({10}^{2}) = 400\pi \\ hence \: the \: volume \: of \: the \: baloon \: is \: increasing \: at \: the \: rate \: of \\ 400\pi {cm}^{3}$