Question

water in a play pool
the graph of ( v(t) ) below shows the amount of water in a play pool as time elapses.
determine the average rates of change over each of the following time intervals.
0 minutes to 4 minutes:
-1.25 × gallons per minute
4 minutes to 8 minutes:
-1.25 × gallons per minute
0 minutes to 24 minutes:
1.875 × gallons per minute
complete the statement to explain the meaning of the rate of change in this situation.
the amount of water remaining in the pool is decreasing at a rate of -1.25 × gallons per minute.
what do the values above tell you about ( v(t) )?

Explanation:

Step1: Recall the formula for average rate of change

The formula for the average rate of change of a function \(y = V(t)\) over the interval \([a,b]\) is \(\frac{V(b)-V(a)}{b - a}\).

Step2: Calculate the average rate of change from \(t = 0\) to \(t=4\)

From the graph, when \(t = 0\), \(V(0)=45\) (assuming the \(y\) - intercept is \(45\) as the line starts at the top - left). When \(t = 4\), \(V(4)=40\).

$$ \frac{V(4)-V(0)}{4 - 0}=\frac{40 - 45}{4}=\frac{- 5}{4}=-1.25 $$

Step3: Calculate the average rate of change from \(t = 4\) to \(t = 8\)

When \(t = 4\), \(V(4)=40\) and when \(t = 8\), \(V(8)=35\)

$$ \frac{V(8)-V(4)}{8 - 4}=\frac{35 - 40}{4}=\frac{-5}{4}=-1.25 $$

Step4: Calculate the average rate of change from \(t = 0\) to \(t = 24\)

When \(t = 0\), \(V(0)=45\) and when \(t = 24\), \(V(24)=0\)

$$ \frac{V(24)-V(0)}{24 - 0}=\frac{0 - 45}{24}=\frac{-45}{24}=-1.875 $$

Answer:

• \(0\) minutes to \(4\) minutes: \(-1.25\) gallons per minute
• \(4\) minutes to \(8\) minutes: \(-1.25\) gallons per minute
• \(0\) minutes to \(24\) minutes: \(-1.875\) gallons per minute

The negative values of the average rate of change indicate that the function \(V(t)\) (the amount of water in the pool) is a decreasing function of time \(t\). The constant rate of change (\(-1.25\) for the intervals \(0 - 4\) and \(4 - 8\)) suggests that the function \(V(t)\) is linear (a straight - line graph) over these sub - intervals. The overall rate of change from \(0\) to \(24\) minutes (\(-1.875\)) is also consistent with the linear nature of the function (since for a linear function \(y=mx + b\), the average rate of change over any interval is equal to the slope \(m\)).