Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Every 12 minutes between 4:00 P.M. and 6:00 P.M., the rate (in gallons per minute) at which water flowed out of a town's water tank was measured. The results are shown in the table below. How much water was used in this 2-hour span? \begin{tabular}{lcccccc} \hline Time & \(4: 00\) & \(4: 12\) & \(4: 24\) & \(4: 36\) & \(4: 48\) & \(5: 00\) \\ Flow (gal/min) & 65 & 71 & 68 & 78 & 105 & 111 \\ \hline Time & \(5: 12\) & \(5: 24\) & \(5: 36\) & \(5: 48\) & \(6: 00\) & \\ Flow \((\mathrm{gal} / \mathrm{min})\) & 108 & 144 & 160 & 152 & 148 & \\ \hline \end{tabular}

Step-by-Step Solution

Answer

13,744 gallons

Step 1: Understand the Problem

The problem requires calculating the total amount of water used over a 2-hour period from 4:00 P.M. to 6:00 P.M. by considering the provided flow rates every 12 minutes.

Step 2: Identify the Time Intervals

Note that the flow rate is measured every 12 minutes. Therefore, from 4:00 P.M. to 6:00 P.M., there are 10 intervals of 12 minutes each. Each interval represents 12 minutes of water flow at the given rate.

Step 3: Calculate the Water Used in Each Interval

To find the amount of water used in each interval, multiply the flow rate by the duration (12 minutes). For example, at 4:00 P.M., the rate is 65 gal/min, so the water used from 4:00 to 4:12 is: \[ \text{Water } = 65 \, \text{gal/min} \times 12 \, \text{min} = 780 \, \text{gallons} \]

Step 4: Repeat for All Intervals

Repeat this calculation for each interval using the corresponding flow rates: - 4:12 to 4:24: \(71 \times 12 = 852\) gallons- 4:24 to 4:36: \(68 \times 12 = 816\) gallons- 4:36 to 4:48: \(78 \times 12 = 936\) gallons- 4:48 to 5:00: \(105 \times 12 = 1260\) gallons- 5:00 to 5:12: \(111 \times 12 = 1332\) gallons- 5:12 to 5:24: \(108 \times 12 = 1296\) gallons- 5:24 to 5:36: \(144 \times 12 = 1728\) gallons- 5:36 to 5:48: \(160 \times 12 = 1920\) gallons- 5:48 to 6:00: \(152 \times 12 = 1824\) gallons

Step 5: Sum the Water Used in All Intervals

Add up all the gallons from each interval to find the total water used:\[780 + 852 + 816 + 936 + 1260 + 1332 + 1296 + 1728 + 1920 + 1824 = 13744 \, \text{gallons}\]

Step 6: Conclude the Calculation

The total amount of water used between 4:00 P.M. and 6:00 P.M is 13,744 gallons.

Key Concepts

Riemann Sum

Riemann Sum is a method for approximating the total area under a curve on a graph, which in many cases directly relates to calculating the total quantity over a certain range. In the case of the exercise, we want to figure out how much water flowed out based on given rates and timings.
The idea here is to use each interval's water flow rate to estimate the water used. Each measured rate (like 65 gal/min) represents the height of a rectangle in a graph, with the base being the time interval. For this exercise, with consistent 12-minute intervals, each flow rate forms a rectangle that represents water volume. The sum of the areas of these rectangles gives an approximation of the volume of water used over the entire period.

• The Riemann Sum simplifies estimating volumes when exact measures aren't easy to calculate directly.
• By using consistent intervals, you streamline the calculations, often making these methods achievable even manually.

Water Flow Rate

Water Flow Rate in this problem is how fast water moves out of the tank, measured in gallons per minute (gal/min). It varies with time, defining how much water leaves the system during specific intervals.
As the water flow rate fluctuates throughout the day, measuring these rates at intervals helps in understanding the system's dynamics. This understanding allows for predicting and calculating total water usage efficiently.

• Flow rate provides insights into how demand changes over time in a system, guiding management and optimization strategies.
• The exercise assumes uniform flow within each 12-minute period, streamlining calculations.

Definite Integral

The Definite Integral is a concept in calculus that sums up an infinite number of infinitely small data points in a certain range. It helps calculate the total quantity represented by the area under a curve, such as determining total water outflow over time.
In practical terms, the definite integral sums these small bits, just like calculating Riemann sum rectangles. In this exercise, the total water used is derived by discreetly integrating the water flow rate over given intervals.

• This concept makes a powerful tool because it accurately sums complex, often fluctuating data into usable total figures.
• Learning to transition from Riemann sums to definite integrals is a valuable skill that furthers one's understanding of calculus and its applications.

Volume Calculation

Volume Calculation in this context involves determining how much water flowed out of the tank over a period by adding up individual volumes calculated for each interval.
Each interval's flow rate, when multiplied by the time span, gives the water volume for that duration. Summing all calculated volumes gives the total volume for the complete time period.

• The process highlighted in the exercise is a simple but effective approach to solving real-world volume flow problems.
• Ensuring that all data (time and rates) are correctly applied and considered helps in achieving accurate results.