$1.2.12.$ A boy inflates a balloon. With a ball radius of $10$ cm, the rate of increase in the radius is $1$ $\frac{mm}{s}$. How much air does the boy exhale every second?
The volume of the ball is defined by the expression $$V = \frac{4}{3} \pi r^3$$ Then, the rate of change of the hugram or the volume of air that the boy exhales every second: $$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$ Given $v = \frac{dr}{dt},$ we obtain: $$\frac{dV}{dt} = 4 \pi r^2 v=126\;sm^3$$
$$\frac{dV}{dt} = 4 \pi r^2 v$$