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1 3.2 Accumulation 36 © Furgolle/Image_Point FR/Corbis 13.2 Accumulation Getting Started In prior sections, we calculated the average rate of change and estimated the instantaneous rate of change (i.e. the derivative) of a function. There are many situations where we know more about the rate at which a function is changing than we do about the value of the function itself. For example, the rate at which the temperature of a cooling object changes, the rate at which a rumor spreads through a population, or the rate at which a person learns a new task. This raises a question: If we know the rate of change at which something has occurred, how do we determine the accumulated amount. In other words, if we know the rate of change of an event how can we determine the total? One example of this is when a nurse cares for a patient who is taking medication intravenously. Suppose a nurse is directed by an obstetrician to give a woman oxytocin (a medication generally given before or after the delivery of a baby). If the nurse monitors his patient every thirty minutes and records the information on Table 13.5, how can he estimate how much medication the patient has received over the two hour period? Learning Objectives:
- beginning with the instantaneous rate of change, work “backwards” to estimate accumulated total
- calculate the area under a curve using approximation techniques
- express the contextual meaning of the area bounded by a curve in a given situation
1 3.2 Accumulation 37 Table 13. t Time (minutes) f Flow rate of oxytocin ( ml min
The techniques used to answer a question such as this and the concepts behind it are very important and foundational to the understanding of Calculus. In this and the subsequent section we investigate both. We begin our study with a seemingly unrelated reflection on the concept of area. However, possessing a deep understanding of area and knowing the process of how to calculate it is essential. The Concept of Area Area is one topic that students of mathematics generally study early on and revisit frequently during subsequent courses. We think of area in the following way: Let’s look at the concept of area starting at a basic level from which we will build the tools necessary to understand and solve accumulation problems (e.g. amount of oxytocin given to a patient). Your focus throughout the following example should be on analyzing the process behind calculating the areas of each figure. Example 1 is designed to be straightforward and its parts logically coherent allowing the spotlight to be placed on the procedure. Our goal is to draw out and recognize the common process used throughout the example so we are enabled Area is the amount of space enclosed in a two-dimensional figure. Area is typically measured by counting the number of 1 unit by 1 unit squares needed to cover a given 2-dimensional space.
1 3.2 Accumulation 39 3 cm 4 cm 8 cm d. A trapezoid with one base ( b 1 ) 8 cm, the other base ( b 2 ) 4 cm, and the height ( h ) 3 cm. Solution: a. To find the area, we are to determine the number of square units (square centimeters in this case) that are enclosed in Figure 13.10. After dividing the figure into 1 cm units on both sides of the square, we count that there are 16 square centimeters. We can also see that there are 4 rows and 4 columns of square centimeters reflecting the standard area formula for a square ( (^) A W (^) = s s ⋅ = s^2 ). For our purposes it is most important to note that the process we used to arrive at our answer was to take the original figure, divide it into smaller 1 cm x 1 cm units, and add to get the total area. Figure 13. 4 cm 4 cm
40 Chapter 13: Looking Ahead to Calculus b. Using a similar technique as was done in part a., we divide the 3 cm x 4 cm rectangle in Figure 13.11 into 1 cm units over the length and width and count that there are 12 square centimeters. We can also see that there are 3 rows and 4 columns of square centimeters reflecting the standard area formula for a
rectangle ( A X = l w ⋅ ). For our purposes it is most important to note that the
process we used to arrive at our answer was to again take the original figure, divide it into smaller 1 cm x 1 cm units, and add to get the total area. Figure 13. c. Recognizing that the triangle is one-half of the rectangle in part b., we use a similar technique with Figure 3.12 as was done in parts a. and b. We again divide the rectangle with base 4 cm and height 3 cm into 1 cm units over the length and width and count that there are 12 square centimeters. However, since this triangle is one-half of the rectangle we can only count one-half of the 12 square centimeters and get 6 square centimeters. We can also see that there are one-half of the 3 rows and 4 columns of square centimeters in the rectangle reflecting the standard area formula for a triangle (
A X (^) = b h g ). For our purposes it is most important to note that the primary process we followed 3 cm 4 cm
42 Chapter 13: Looking Ahead to Calculus Having revisited the process behind finding the area in some elementary problems, let’s now extend the method further by estimating the area of an irregular-shaped figure in the following example. Example 2 : Calculating Area – Irregular Figures a. Show and explain how the process used in Example 1 can be adapted to estimate the area of
the region bounded by f ( ) x = −0.75 x^2 + 3 x + 4 , the y − axis , the x − axis , and x = 4 as
shown in Figure 1 3.14. Figure 13. b. Show another technique using accumulated areas to estimate the total enclosed area. c. What could be done differently to come up with a more accurate estimate for the area of the bounded region? Explain using words and diagrams. Solution: a. The process that we followed in Example 1 was to divide the original figures into smaller regions, determine the area of each, and then find the sum. Using this
1 3.2 Accumulation 43 approach, the area of Figure 13.15 in this example can be estimated by dividing it into a semicircle (Region 1) and a square (Region 2), find the area of each, and then calculate the sum. Note that this area is not the exact area but an estimate because the “top” of the figure is not a semi-circle so using this technique we have an underestimate for the area of Region 1. Figure 13. Area of Region 1 (semicircle): 2 semicircle
A = π r 2 2 2 semicircle
(2 units) (4 units ) 2 units 2 2
A = π = π = π
Area of Region 2 (square): A = s^2 W 2 2 A W = (4 units) =16 units
1 3.2 Accumulation 45 c. To get a more accurate estimate than what we got in b. for the enclosed region,
we need to reduce the amount of area included above the function f ( ) x. One
technique would be to increase the number of subdivisions as shown in Figure 13.17, find the area of each rectangle, and determine the sum to find the total area. Figure 13. It is important to note that we could get increasingly more accurate estimates if we continue to make additional subdivisions. We can see that by using this technique we get an underestimate because we are not including in our answer portions that are in the enclosed region. We now need to observe how the area of an enclosed region can represent various things in different contexts. Let’s look at an example that exemplifies this.
46 Chapter 13: Looking Ahead to Calculus Example 3 : Contextual Meaning of Area In January of 1970, Gilbert, Arizona, one of the fastest growing cities in America, had a population of 1,971 people. Table 13.6 shows the average growth of the community over each 10 year span until the year 2000 measured each year in January. Table 13. 6 d Decade g Average Population Growth ( people year
Recalling that 1,971 people lived in Gilbert in 1970, how many people lived there in January 2000? To estimate this, we need to determine how many more people took up residence in Gilbert. Over the first 10 year span (1970 to 1980) there were an additional 372 people year which means there were 372 people year x 10 years = 3,720 more people during that decade. We calculate the population growth for each decade and show this graphically in Figure 13.18 as the area of rectangles.
48 Chapter 13: Looking Ahead to Calculus Through these examples we have seen the key concept behind addressing the nursing problem we posed at the outset of this lesson. Let’s now revisit that scenario and see how we can employ what we have learned to estimate the total amount of medication given to the patient using what is known as one of the methods of approximation. Example 4: Area Approximation Recall that a nurse cares for a patient who is taking medication intravenously. The nurse has been directed by the patient’s obstetrician to give a woman oxytocin. If the nurse monitors his patient every thirty minutes and records the information on the following table, estimate how much medication the patient has received over the two-hour period using techniques discussed in the first three examples of this section. Use 3 0 minute time intervals Peer into the Past: The co- Discovery of Calculus – Integral Calculus As with Differential Calculus, Sir Isaac Newton and Wilhelm Leibniz were also instrumental in the development of Integral Calculus. They both made significant progress with the contemporary problems of their day in analytical geometry – drawing tangents to curves (differentiation) and defining areas bounded by curves (integration). Newton discovered that derivatives and integrals were inverses of each other and described differentiation as the “method of fluxions” and integration as “inverse method of fluxions”. With integration both Newton and Leibniz developed techniques for approximating the area of a region bounded by a curve. Leibniz developed the notation that is currently used to notate the integral – an elongated “s”. An interesting historical note is that the two men independently came up with their theories. In England, Newton had essentially put in place his methods of fluxions in 1666 but didn’t make his work public until
- Meanwhile in Paris in 1675 Leibniz evolved his ideas of differential calculus and published his first paper in 1684. Some claimed that Newton was the originator of these ideas but others proclaimed it was Leibniz. Even after their deaths, the controversy raged. The verdict over time, however, has been that both men independently invented their methods and are considered the “Fathers of Calculus”. (Source: www.newton.cam.ac
1 3.2 Accumulation 49 and give a justification for the estimate being an overestimate or underestimate. Table 13. 7 t Time (minutes) f Flow rate of medication ( ml min
Solution: If we take the rate in milliliters per minute at which the medication is given to a patient and multiply by the number of minutes the drug is administered, we get the accumulated total of drug given in milliliters ( ml min ml min ⋅ = ). This is represented graphically by the area of an enclosed region. The technique we have seen in the prior examples for determining area of an irregular figure is to divide the figure into smaller regions, calculate the area of each sub-region, and total the individual areas to get the accumulative amount. This is what we will do here. Let’s begin by plotting the data into a scatter plot in Figure 13.19.
1 3.2 Accumulation 51 We now superimpose rectangles of width 3 0 units in Figure 13.21, which in this case represent the 30 minute time intervals the nurse checked on the patient, upon the shaded area. Figure 13. 21 Using Table 13.8, we now find the area of the four rectangles and total them to arrive at estimation for the shaded area representing the total amount of medication given. Table 13. 8 Area of rectangle 1 + Area of rectangle 2 + Area of rectangle 3 + Area of rectangle 4 t Time (minutes) f Flow rate of medication ( ml min
ml (0 min, 0.5 ) min ml (90 min, 0.7 ) min ml (60 min, 0.8 ) min ml (30 min, 1 ) min
52 Chapter 13: Looking Ahead to Calculus
= l w 1 1 + l w 2 2 +l w 3 3 + l w 4 4
ml ml ml ml (0.5 )(30 min) (1.0 )(30 min) (0.8 )(30 min) (0.7 )(30 min) min min min min
= (15 ml) + (30 ml) + (24 ml) +(21 ml)
= 90 ml
It appears that our estimate of 90 ml of oxytocin is an underestimate of the actual amount given because we can see in the diagram that there is a large portion of the enclosed area in the first 30 minute interval that is not included in our solution. However, there are smaller portions area regions in the last three 30 minute intervals that are overestimates. Therefore, the approximation of 90 ml may be relatively accurate. (Note: It is important to recall that we saw in the prior example that we would get a more accurate estimate by increasing the number of subintervals but we are told in this example to use 30 minute intervals.) Summary In this section, you learned that if the rate of change at which something occurs is known, you can determine the accumulated amount. This was done by estimating the area of an enclosed region. If the region is irregularly shaped you estimated the area by dividing it into smaller regions and adding to get the total area. In the following exercises you will see that this technique is a powerful tool that empowers you to solve many types of accumulation problems.
54 Chapter 13: Looking Ahead to Calculus and the x − axis. Use the indicated number of intervals n.
- 1 ≤ x ≤ 6 and n = 5.
- 0 ≤ x ≤ 2 and n = 2.
- 0 ≤ x ≤ 3 and n = 6.
- 2 ≤ x ≤ 10 and n = 8. In Exercises 13 – 16 give the unit of measure for what the area under the curve described in each would represent.
- The dependent variable is the rate at which money is spent on advertising in millions of dollars per year and the independent variable is the year.
1 3.2 Accumulated Change 55 © Stuart Westmorland/Corbis
- The dependent variable is the rate at which the number of dishwashers is sold per month and the independent variable is the month.
- The dependent variable is the rate at which the concentration of a drug in the bloodstream of a patient is changing in milligrams per hour and the independent variable is the hour.
- The dependent variable is the rate at which the level of production at a computer chip manufacturer is made in thousands of chips per hour and the independent variable is the hour. For each rate of change function in Exercises 17 – 22, determine what the units would be for the area under the curve of the graph for each function. [RW/M] 17. Rate of new Medicare Enrollees
m t ( ) = −0.00944 t + 0.663million
Medicare enrollees per year, m , since 1980 ( t = 0 ). (Source: Statistical Abstract of the United States, 2006, Table 132). [RW/M] 18. Rate of Cassette Tape Sales
0.5277 1.
t t t e c t e e
cassette tape sales in millions of dollars per year, c , since 1990 ( t = 0 ).(Source: Statistical Abstract of the United States, 2006, Table 1003). [RW/M] 19. Rate of Pineapple Consumption
