Problem 33 - Precalculus Mathematics for Calculus

Question

A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume \(V\) of water remaining in the tank (in gallons) after \(t\) minutes. $$\begin{array}{|c|c|}\hline t \text { (min) } & V \text { (gal) } \\\\\hline 5 & 694 \\\10 & 444 \\\15 & 250 \\\20 &111 \\\25 & 28 \\\30 & 0 \\\\\hline\end{array}$$ (a) Find the average rates at which water flows from the tank (slopes of secant lines) for the time intervals \([10,15]\) and \([15,20]\) (b) The slope of the tangent line at the point \((15,250)\) represents the rate at which water is flowing from the tank after 15 minutes. Estimate this rate by averaging the slopes of the secant lines in part (a).

Step-by-Step Solution

Answer

The estimated rate at 15 minutes is -33.3 gallons per minute.

Step 1: Understand the Problem

There is a tank with 1000 gallons of water, and the table data gives the volume of water remaining at various points in time. We need to find the average rate of water drainage over specified intervals and estimate the rate of drainage at 15 minutes.

Step 2: Calculate Average Rate for Interval [10, 15]

The formula for average rate of change (slope of a secant line) is \( m = \frac{V_2 - V_1}{t_2 - t_1} \). For the interval [10, 15], the volumes are 444 gallons at \( t = 10 \) and 250 gallons at \( t = 15 \). Therefore, \( m = \frac{250 - 444}{15 - 10} = \frac{-194}{5} = -38.8 \) gallons per minute.

Step 3: Calculate Average Rate for Interval [15, 20]

Use the same formula for the interval [15, 20]. The volumes are 250 gallons at \( t = 15 \) and 111 gallons at \( t = 20 \). So, \( m = \frac{111 - 250}{20 - 15} = \frac{-139}{5} = -27.8 \) gallons per minute.

Step 4: Estimate the Rate at 15 Minutes

To estimate the instantaneous rate of change at \( t = 15 \), we average the slopes from the previous steps. Therefore, the estimated rate is \( \frac{-38.8 + (-27.8)}{2} = \frac{-66.6}{2} = -33.3 \) gallons per minute.

Key Concepts

Secant Lines

Secant lines are straight lines that connect two points on the graph of a function. They help us understand how a function behaves over a specific interval. In this exercise, secant lines show how the volume of water changes in the tank over particular intervals of time.
To find the slope of a secant line, which represents the average rate of change, we use the formula:

• \[\text{slope} = \frac{V_2 - V_1}{t_2 - t_1}\]

For the interval from 10 to 15 minutes, the secant line reflects a water drainage from 444 gallons to 250 gallons. This gives us a slope of -38.8 gallons per minute. The negative sign indicates that water is leaving the tank.
Similarly, for the interval from 15 to 20 minutes, the water volume decreases from 250 to 111 gallons, with the slope of the secant line being -27.8 gallons per minute. These secant lines provide a simple average rate of how fast water is draining over these time spans.

Tangent Line

A tangent line touches the curve of a function at exactly one point and gives us the slope at that particular point. This slope is crucial for understanding the instantaneous rate of change at a specific time.
In our exercise, we want to estimate the rate of water drainage at exactly 15 minutes. The problem helps us to find this by using the average slopes of secant lines just before and after 15 minutes. The tangent line represents what the secant lines approximate over small intervals around the point, offering a more precise rate.
The slope of this tangent line at the point \((15, 250)\) represents the instantaneous rate of change at 15 minutes. Instead of calculating this slope directly, we average the slopes of the secant lines for \([10,15]\) and \([15,20]\), providing an estimated instantaneous rate of -33.3 gallons per minute.

Instantaneous Rate of Change

The instantaneous rate of change describes how fast something happens at a specific moment in time. When dealing with the drainage of water from a tank, it tells us the exact rate at which the water volume decreases at a given time point.
For our task, the point of interest is \(t = 15\) minutes. Calculating the slope of the tangent line at this point offers the best estimate for the instantaneous rate. In essence, it is the limit of the average rates found from closer and closer intervals around 15 minutes.
The process involved approximating by averaging the slopes of secant lines at \([10,15]\) and \([15,20]\):

• \(\text{average slope} = \frac{-38.8 + (-27.8)}{2} = -33.3 \) gallons per minute

While this is an estimation, it closely aligns with what would be exactly calculated if more advanced calculus tools were applied.

Average Rate of Change

The average rate of change between two points on a graph gives an overview of how much a value changes over a certain interval. In this exercise, it represents how much water drains from the tank on average over 5-minute intervals.
The formula to find the average rate of change, using secant line slopes, is:

• \[\text{average rate} = \frac{V_2 - V_1}{t_2 - t_1}\]

The slope calculation for the interval \([10,15]\) showed a decrease of 38.8 gallons per minute, indicating faster drainage compared to the interval \([15,20]\), where it decreased at 27.8 gallons per minute.
This average gives us insight into the drainage process over the specified intervals, summing up how much water exits between measured times rather than at any tangent point. It's a broader measure compared to the instantaneous rate, which focuses on a single moment.