Related rate of volume of a sphere

Start with the volume formula $V = \frac{4}{3}\pi r^3$, where both $V$ and $r$ are functions of time $t$. Differentiate both sides implicitly with respect to $t$: $\dfrac{dV}{dt} = \dfrac{4}{3}\pi \cdot 3r^2 \cdot \dfrac{dr}{dt} = 4\pi r^2 \dfrac{dr}{dt}$. This equation relates the rate of change of volume to the rate of change of radius.

Now substitute the known values $r = 3$ cm and $\dfrac{dr}{dt} = 0.02$ cm/s: $\dfrac{dV}{dt} = 4\pi (3)^2 (0.02) = 4\pi(9)(0.02) = 0.72\pi$. Thus the volume is increasing at a rate of $0.72\pi \approx 2.26$ cm$^3$/s at that instant.

The answer is increase. Since dV/dt equals four pi r squared times dr/dt, and dr/dt is held constant, the growth rate of volume depends on r squared. As r increases, r squared increases, so dV/dt grows larger. This matches the pattern seen when comparing the rate at r equals three to what would happen at a larger radius.