The volume of a spherical balloon increases by $1cm^3$ every second. What is the rate of growth of the radius when the surface area of the balloon is $100cm^2$
The surface area of a sphere is $4\pi r^2$, and its volume is $\dfrac{4}{3}\pi r^3$.
The answer sheet states that $\dfrac{dV}{dt} =1$, and we need to find $\dfrac{dr}{dt}$, but I don't understand this, can anyone explain?
Best Answer
$\dfrac{dV}{dt}$ is the rate of change of the volume. In particular, $\dfrac{dV}{dt}=1$ cm per second. $\dfrac{dr}{dt}$ is the rate of change of the radius, which is what we want to know.
Now, the surface area of the sphere is actually $4\pi r^2.$ Observe, then, that $$\frac{dV}{dt}=\frac{d}{dt}\left[\frac43\pi r^3\right]=\frac43\pi\frac{d}{dt}\left[r^3\right]=\frac43\pi\cdot 3r^2\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}.$$ Can you take it from there?