Question

1. a pump is filling up a swimming pool. the volume of water in the pool is a function of time since it has been filling. the table below shows volumes of water, in gallons, at various times during filling.

on average, is the water entering the pool faster over the interval ( 0leq tleq36 ) minutes or ( 16leq tleq53 ) minutes? justify your answer with appropriate calculations.

Explanation:

Step1: Calculate the average rate for \(0\leq t\leq36\)

The average rate of change formula is \(\frac{V(b)-V(a)}{b - a}\).
For \(a = 0\), \(b=36\), \(V(0)=0\), \(V(36) = 1872\).

$$ \frac{V(36)-V(0)}{36-0}=\frac{1872 - 0}{36}=\frac{1872}{36}=52 $$

Step2: Calculate the average rate for \(16\leq t\leq53\)

For \(a = 16\), \(b = 53\), \(V(16)=896\), \(V(53)=2561\).

$$ \frac{V(53)-V(16)}{53 - 16}=\frac{2561-896}{37}=\frac{1665}{37}=45 $$

Answer:

The water is entering the pool faster over the interval \(0\leq t\leq36\) minutes. Since the average rate of change over \(0\leq t\leq36\) is \(52\) gallons per minute and over \(16\leq t\leq53\) is \(45\) gallons per minute, and \(52>45\).