Rate of Change of a Volume Function

For $V(t)=8*t^2-32*t+4$, application of the power rule yields $V'(t)=16*t-32$.

The derivative of the volume function represents the rate at which the pool’s water volume changes over time.

Differentiating $V(t)=8*t^2-32*t+4$ gives $V'(t)=16*t+32$, indicating the rate of volume change increases linearly.

At $t=2$, substituting into the derivative yields $V'(2)=16*2-32=0$, correctly showing no instantaneous change in volume.