Download Applications to Economics and Biology

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).
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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.
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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.
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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
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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
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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.
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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.
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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
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