Rate of Change of a Volume Function

Question

The volume of water in a pool, in gallons, is modeled by $$V(t)=8t^2-32t+4$$, where $$t$$ is the time in hours. The table shows the rate of change of the volume for several values of $$t$$. All of the following statements are true EXCEPT:

| t (hours) | $$dV/dt$$ (gallons/hour) |
|-----------|-------------------------|
| 1         | -16                     |
| 2         | 0                       |
| 3         | 16                      |

Accepted Answer

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

Answer

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

Answer

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

Answer

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

Rate of Change of a Volume Function

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