The radius of an air bubble is increasing at the rate of 0.5 cm/sec. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Explanation

`V=frac{4}{3} pi r^{3}`



The rate of change of volume `(V)` with respect to time `(t)` is given by, ` frac{d V}{d t} =frac{4}{3} pi frac{d}{d r}(r^{3}) cdot frac{d r}{d t} =frac{4}{3} pi(3 r^{2}) frac{d r}{d t} =4 pi r^{2} frac{d r}{d t} `


It is given that `frac{d r}{d t}=frac{1}{2} {cm} / {s}`.


Therefore, when `r=1 {cm}_{text {, }}`



`frac{d V}{d t}=4 pi(1)^{2}(frac{1}{2})=2 pi {cm}^{3} / {s}`