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:
4 minutes to 8 minutes:
0 minutes to 24 minutes:
complete the statement to explain the meaning of the rate of change in this situation.
the amount of water remaining in the pool is at a rate of
what do the values above tell you about ( v(t) )?
( v(t) ) is

Explanation:

Step1: Recall the formula for average rate of change

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

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

From the graph, when \(t = 0\), \(V(0)=45\) (assuming each grid - square is \(1\) unit for \(t\) and \(1\) unit for \(V(t)\) in the vertical direction, and by visual inspection). 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 \(t = 24\), \(V(24)=0\)

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

Step5: Interpret the rate of change

The formula \(\frac{\Delta V}{\Delta t}\) represents the change in the amount of water (\(\Delta V\)) over the change in time (\(\Delta t\)). A negative rate of change means the amount of water is decreasing.

Step6: Analyze the function \(V(t)\)

Since the average rate of change (which is similar to the slope of the secant line) is constant over the intervals (the first two intervals have the same rate of change, and the overall rate of change is also a constant - like behavior, and the graph is a straight line), \(V(t)\) is a linear function.

Answer:

• \(0\) minutes to \(4\) minutes: \(-1.25\)
• \(4\) minutes to \(8\) minutes: \(-1.25\)
• \(0\) minutes to \(24\) minutes: \(-1.875\)
• The amount of water remaining in the pool is \(\text{decreasing}\) at a rate of \(\text{gallons per minute}\)
• \(V(t)\) is \(\text{a linear function}\)