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FURTHER APPLICATIONS OF INTEGRATION 9 FURTHER APPLICATIONS OF INTEGRATION 9.4 Applications to Economics and Biology In this section, we will learn about: Some applications of integration to economics and biology. CONSUMER SURPLUS DEMAND CURVE Recall from Section 4.7 that the demand The graph of a typical demand function, function p(x) is the price a company has to called a demand curve, is shown. charge in order to sell x units of a commodity. Usually, selling larger quantities requires lowering prices. If X is the amount of the commodity that is currently available, then P = p(X) is the current selling price. So, the demand function is a decreasing function. CONSUMER SURPLUS CONSUMER SURPLUS We divide the interval [0, X] into n Suppose, after the first xi-1 units were sold, subintervals, each of length ∆x = X/n. a total of only xi units had been available and the price per unit had been set at p(xi) dollars. Then, we let xi* = xi be the right endpoint of the i th subinterval. Then, the additional ∆x units could have been sold (but no more). 1 CONSUMER SURPLUS CONSUMER SURPLUS The consumers who would have Thus, in paying only P dollars, they have paid p(xi) dollars placed a high value saved an amount of: on the product. (savings per unit)(number of units) They would have paid what it was worth to them. = [p(xi) - P] ∆x CONSUMER SURPLUS CONSUMER SURPLUS Taking similar groups of willing consumers This sum corresponds to the area for each of the subintervals and adding enclosed by the rectangles. the savings, we get the total savings: n ∑ [ p ( x ) − P ] ∆x i i =1 CONSUMER SURPLUS Definition 1 CONSUMER SURPLUS If we let n → ∞, this Riemann sum The consumer surplus represents the amount approaches the integral of money saved by consumers in purchasing ∫ x 0 the commodity at price P, corresponding to [ p( x) − P]dx an amount demanded of X. Economists call this the consumer surplus for the commodity. 2 CONSUMER SURPLUS CONSUMER SURPLUS The figure shows the interpretation of The demand for a product, in dollars, is the consumer surplus as the area under Example 1 p = 1200 − 0.2 x − 0.0001x 2 the demand curve and above the line p = P. Find the consumer surplus when the sales level is 500. CONSUMER SURPLUS CONSUMER SURPLUS Example 1 The number of products sold is X = 500. So, from Definition 1, the consumer surplus is: ∫ 500 0 So, the corresponding price is: P = 1200 − (0.2)(500) − (0.0001)(500) = 1075 Example 1 2 [ p ( x ) − P ] dx = ∫ 500 = ∫ 500 0 0 (1200 − 0.2 x − 0.0001 x 2 − 1075) dx (125 − 0.2 x − 0.0001x 2 ) dx 500 x3 = 125 x − 0.1 x 2 − (0.0001) 3 0 = (125)(500) − (0.1)(500) 2 − (0.0001)(500) 2 3 = $33, 333.33 BLOOD FLOW LAW OF LAMINAR FLOW In Example 7 in Section 3.7, v(r ) = we discussed the law of laminar flow: P (R2 − r 2 ) 4η l This gives the velocity v of blood that flows v (r ) = P (R2 − r 2 ) 4η l along a blood vessel with radius R and length l at a distance r from the central axis, where P is the pressure difference between the ends of the vessel. η is the viscosity of the blood. 3 BLOOD FLOW BLOOD FLOW Now, in order to compute the rate of The approximate area of the ring (or blood flow, or flux (volume per unit time), washer) with inner radius ri-1 and outer we consider smaller, equally spaced radii radius ri is: 2π ri ∆r r1, r2, … where ∆r = ri − ri −1 BLOOD FLOW BLOOD FLOW If ∆r is small, then the velocity is almost Therefore, the volume of blood per constant throughout this ring and can be unit time that flows across the ring is approximated by v(ri). approximately (2π ri ∆r )v(ri ) = 2π ri v(ri ) ∆r BLOOD FLOW BLOOD FLOW The total volume of blood that flows The approximation is illustrated here. across a cross-section per unit time is approximately n ∑ 2π r v(r ) ∆r i Notice that the velocity (and hence the volume per unit time) increases toward the center of the blood vessel. The approximation gets better as n increases. i i =1 4 FLUX BLOOD FLOW When we take the limit, we get the exact F = lim ∑ 2π ri v( ri ) ∆r n value of the flux (or discharge). n →∞ i =1 R = ∫ 2π rv (r ) dr 0 R This is the volume of blood that passes a cross-section per unit time. = ∫ 2π r 0 P ( R 2 − r 2 ) dr 4η l = πP R 2 ( R r − r 3 ) dr 2η l ∫0 = π P 2 r2 r4 π P R 4 R 4 π PR 4 R − = − = 2η l 2 4 r =0 2ηl 2 4 8η l r =R POISEUILLE’S LAW Equation 2 The resulting equation F= π PR 8η l CARDIAC OUTPUT The figure shows the human 4 cardiovascular system. is called Poiseuille’s Law. It shows that the flux is proportional to the fourth power of the radius of the blood vessel. CARDIAC OUTPUT CARDIAC OUTPUT Blood returns from the body through the veins, Then, it flows back into the left atrium enters the right atrium of the heart, and is (through the pulmonary veins) and then out pumped to the lungs to the rest of the body through the pulmonary (through the aorta). arteries for oxygenation. 5 CARDIAC OUTPUT DYE DILUTION METHOD The cardiac output of the heart is the volume The dye dilution method of blood pumped by the heart per unit time, that is, the rate of flow is used to measure the cardiac into the aorta. output. DYE DILUTION METHOD DYE DILUTION METHOD Dye is injected into the right atrium A probe inserted into the aorta measures and flows through the heart into the concentration of the dye leaving the heart the aorta. at equally spaced times over a time interval [0, T ] until the dye has cleared. DYE DILUTION METHOD DYE DILUTION METHOD Let c(t) be the concentration of the dye Thus, the total amount of dye is at time t. approximately n n ∑ c(t ) F ∆t = F ∑ c(t ) ∆t Let’s divide [0, T ] into subintervals of equal length ∆t. i i i =1 Then, the amount of dye that flows past the measuring point during the subinterval from t = ti-1 to t = ti is approximately (concentration)(volume) = c(ti)(F∆t) i =1 Letting n → ∞, we find that the amount of dye is: T A = F ∫ c(t ) dt 0 where F is the rate of flow we are trying to determine. 6 DYE DILUTION METHOD Formula 3 Thus, the cardiac output is given by F= ∫ 0 Example 2 atrium. The concentration of the dye (in A T CARDIAC OUTPUT A 5-mg bolus of dye is injected into a right milligrams per liter) is measured in the aorta c(t ) dt at one-second intervals, as shown. where the amount of dye A is known Estimate the cardiac output. and the integral can be approximated from the concentration readings. CARDIAC OUTPUT Example 2 Here, A = 5, ∆t = 1, and T = 10. CARDIAC OUTPUT Example 2 We use Simpson’s Rule to approximate the integral of the concentration: ∫ 10 0 c(t ) dt ≈ 13 [0 + 4(0.4) + 2(2.8) + 4(6.5) + 2(9.8) + 4(8.9) + 2(6.1) + 4(4.0) + 2(2.3) + 4(1.1) + 0] ≈ 41.87 CARDIAC OUTPUT Example 2 Thus, Formula 3 gives the cardiac output to be: F= A ∫ 10 0 ≈ c(t )dt 5 ≈ 0.12L / s = 7.2L / min 41.87 7