APPLIC ATIONS TO ECONOMICS AND BIOLOGY In this section we consider some applications of integration to economics (consumer sur-

8.4 APPLIC ATIONS TO ECONOMICS AND BIOLOGY In this section we consider some applications of integration to economics (consumer sur-

plus) and biology (blood flow, cardiac output). Others are described in the exercises.

CONSUMER SURPLUS

Recall from Section 4.7 that the demand function p 共x兲 is the price that a company has to charge in order to sell units of a commodity. Usually, selling larger quantities requires x

p=p(x)

lowering prices, so the demand function is a decreasing function. The graph of a typical demand function, called a demand curve, is shown in Figure 1. If X is the amount of the commodity that is currently available, then P 苷 p共X兲 is the current selling price. We divide the interval 关0, X兴 into subintervals, each of length n ⌬x 苷 X兾n , and let

be the right endpoint of the th subinterval, as in Figure 2. If, after the first i x P i ⫺ 1 units were sold, a total of only x i units had been available and the price per unit had been set at p 共x i 兲 dollars, then the additional ⌬x units could have been sold (but no more). The

(X, P)

x *苷x i i

consumers who would have paid p 共x i 兲 dollars placed a high value on the product; they would have paid what it was worth to them. So, in paying only dollars they have saved P

FIGURE 1

an amount of

A typical demand curve

i 兲 ⫺ P兴 ⌬x

SECTION 8.4 APPLICATIONS TO ECONOMICS AND BIOLOGY

Considering similar groups of willing consumers for each of the subintervals and adding the savings, we get the total savings:

i苷1 兺

关p共x i 兲 ⫺ P兴 ⌬x

(This sum corresponds to the area enclosed by the rectangles in Figure 2.) If we let nl⬁ (X, P) ,

this Riemann sum approaches the integral

1 X y 关p共x兲 ⫺ P兴 dx

FIGURE 2

which economists call the consumer surplus for the commodity. The consumer surplus represents the amount of money saved by consumers in pur-

chasing the commodity at price , corresponding to an amount demanded of . Figure 3 P X

p=p(x)

shows the interpretation of the consumer surplus as the area under the demand curve and above the line p苷P .

V EXAMPLE 1 The demand for a product, in dollars, is

consumer surplus

(X, P)

p 苷 1200 ⫺ 0.2x ⫺ 0.0001x 2

P p=P

Find the consumer surplus when the sales level is 500.

SOLUTION Since the number of products sold is

X 苷 500 , the corresponding price is

FIGURE 3

P 苷 1200 ⫺ 共0.2兲共500兲 ⫺ 共0.0001兲共500兲 2 苷 1075

Therefore, from Definition 1, the consumer surplus is

500 共1200 ⫺ 0.2x ⫺ 0.0001x y 2

0 关p共x兲 ⫺ P兴 dx 苷 y

兲 dx

苷 500 y 共125 ⫺ 0.2x ⫺ 0.0001x

2 兲 dx

苷 125x ⫺ 0.1x ⫺

BLOOD FLOW In Example 7 in Section 3.7 we discussed the law of laminar flow:

2 ⫺r 2

4␩l

which gives the velocity of blood that flows along a blood vessel with radius and length v R l at a distance from the central axis, where is the pressure difference between the ends r P of the vessel and is the viscosity of the blood. Now, in order to compute the rate of blood ␩

flow, or flux (volume per unit time), we consider smaller, equally spaced radii r 1 ,r 2 , ....

CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION

The approximate area of the ring (or washer) with inner radius r i ⫺ 1 and outer radius is r i

(See Figure 4.) If ⌬r is small, then the velocity is almost constant throughout this ring and can be approximated by v 共r i 兲 . Thus the volume of blood per unit time that flows across the ring is approximately

共2␲r i ⌬r 兲 v 共r i 兲 苷 2␲r i v 共r i 兲 ⌬r

FIGURE 4

and the total volume of blood that flows across a cross-section per unit time is approxi- mately

兺 2␲r i v 共r i 兲 ⌬r

i苷1

This approximation is illustrated in Figure 5. Notice that the velocity (and hence the vol- ume per unit time) increases toward the center of the blood vessel. The approximation gets better as n increases. When we take the limit we get the exact value of the flux (or dis- charge ), which is the volume of blood that passes a cross-section per unit time:

FIGURE 5

n F 苷 lim l 2␲r i v 共r ⬁ i 兺 兲 ⌬r 苷 i苷1 y 2␲r

v 共r兲 dr

2 2 苷 y 2␲r 共R ⫺r 兲 dr

0 4␩l

␲P r苷R R

y 共R r ⫺r 兲 dr 苷 R 2␩l ⫺ 0 2␩l 冋 2 4 册

2 3 ␲P

r苷0

2␩l 冋 2 4 册 苷 8␩l

␲P R 4 R 4 ␲PR 4

The resulting equation

2 F苷 ␲PR

8␩l

is called Poiseuille’s Law; it shows that the flux is proportional to the fourth power of the radius of the blood vessel.

aorta

C ARDIAC OUTPUT

vein pulmonary arteries

Figure 6 shows the human cardiovascular system. Blood returns from the body through the

pulmonary

pulmonary veins

veins, enters the right atrium of the heart, and is pumped to the lungs through the pul-

arteries

monary arteries for oxygenation. It then flows back into the left atrium through the pulmo- nary veins and then out to the rest of the body through the aorta. The cardiac output of the heart is the volume of blood pumped by the heart per unit time, that is, the rate of

right

left

flow into the aorta.

atrium

atrium

The dye dilution method is used to measure the cardiac output. Dye is injected into the

right atrium and flows through the heart into the aorta. A probe inserted into the aorta

pulmonary

veins

measures the concentration of the dye leaving the heart at equally spaced times over a time

be the concentration of the dye at time t . If we divide 关0, T 兴 into subintervals of equal length ⌬t , then the amount of dye that flows

interval 关0, T 兴 until the dye has cleared. Let c 共t兲

vein

past the measuring point during the subinterval from t苷t i ⫺ 1 to t苷t i is approximately

FIGURE 6

SECTION 8.4 APPLICATIONS TO ECONOMICS AND BIOLOGY

where is the rate of flow that we are trying to determine. Thus the total amount of dye F is approximately

兺 c 共t i 兲F ⌬t 苷 F 兺 c 共t i 兲 ⌬t

i苷1

i苷1

and, letting nl⬁ , we find that the amount of dye is

A苷F T y

c 共t兲 dt

Thus the cardiac output is given by

3 A F苷 T

y 0 c共t兲 dt

where the amount of dye A is known and the integral can be approximated from the con-

centration readings.

V t EXAMPLE 2 c 共t兲 t c 共t兲

A 5-mg bolus of dye is injected into a right atrium. The concentration of the dye (in milligrams per liter) is measured in the aorta at one-second intervals as

0 0 6 6.1 shown in the chart. Estimate the cardiac output. 1 0.4 7 4.0 SOLUTION

2 2.8 8 2.3 Here A苷5 , ⌬t 苷 1 , and T 苷 10 . We use Simpson’s Rule to approximate the

3 6.5 9 1.1 integral of the concentration:

5 8.9 y c 共t兲 dt ⬇ 1 3 0 关0 ⫹ 4共0.4兲 ⫹ 2共2.8兲 ⫹ 4共6.5兲 ⫹ 2共9.8兲 ⫹ 4共8.9兲

⬇ 41.87 Thus Formula 3 gives the cardiac output to be

F苷 10 ⬇

y 41.87

0 c共t兲 dt