Download Honors Project 14: Measuring Cardiac Output

Math 163, Fall 2001
Due Date:
Name(s):
Honors Project 14: Measuring Cardiac Output
Objective
To illustrate how integration is applied to the measurement of cardiac output. (Note: The text [1] covers
some of the same material starting on p. 601.)
Background Required
Basic concepts of integration.
Narrative
Cardiac output is, by definition, the volume of blood pumped by the heart in a specific period of time. The
cardiac output of an individual at rest is, for example, on the order of 5 liters per minute. The measurement
of cardiac output is important since it is an indicator of certain diseases of the blood, heart, and vascular
system (the body’s system of arteries, arterioles, capillaries, and veins).
Cardiac output is measured by a technique known as dye dilution: A known amount D of dye is injected
into a main vein near the heart. Blood carrying the dye circulates through the right ventricle of the heart,
to the lungs, through the left auricle and left ventricle of the heart, and then into the arterial system; it
then returns to the left auricle and left ventricle of the heart. The concentration of the dye passing through
an artery or vein is monitored at specific time intervals, and from these concentrations cardiac output is
computed. In this project we discuss how it is computed.
In any small volume dV of blood, the amount dD of dye is the product of the concentration c of the dye
and the volume dV . (We are assuming a “small” volume of blood is a volume for which the concentration is
nearly constant.) Thus dD = c dV. If we monitor the concentration of dye along an artery leading from the
left ventricle, then the volume dV of blood is the rate R at which blood is being pumped by the heart —
the cardiac output through the given artery — times the length dt of time over which the dye concentration
is measured. Thus dV = R dt, and we have
dD = c dV = c(t)R dt,
where we now write c = c(t) to reflect the fact that the concentration of dye in volumes of blood that pass
the monitoring point depends on time. Integrating both sides of this equation from the time t = T0 the dye
is injected until the time t = T1 at which all blood has circulated once through the body, and assuming that
all blood passes through the monitoring point, we have
T1
D=R
c(t) dt.
t=T0
It follows that
R = T1
D
t=T0
c(t) dt
.
For later reference, we note that T1 − T0 is typically on the order of 21 seconds.
One interesting aspect of this computation is that while the amount D of dye injected into the bloodstream
can be controlled, and is hence known, we have no way of knowing the total volume of blood in an individual,
and our computation of R avoids this issue.
Another interesting aspect of this computation is that in practice, we might expect to see real concentration
vs. time monitoring data as illustrated in Fig. 1. This data displays several interesting features. First it
is discrete, not continuous since it will inevitably take some small amount dt = ∆t of time to measure
1
concentration. (Typically dt = ∆t is on the order of 1 sec.) Second, it has peaks and valleys, and the peaks
decrease in height and the valleys (at least initially) increase in height as time increases. This is damped
periodicity: the periodicity arises from the continual recirculation of the dye through the bloodstream and
past the point of monitoring; the damping arises from the dissipation of the dye into the total volume of
blood over time. (As time goes on, mixing of the dye in the bloodstream occurs. Indeed, as some dye is
completes one pass through the bloodstream it might well overtake dye that is in the process of starting an
earlier pass through the bloodstream.)
c
t
T0 = 0
T1
Fig. 1: Typical Concentration Levels.
The significance of this later observation (see Fig. 2) is that while we may know T0 , we must guess T1
and the concentration levels near T1 . This is one reason our measurement of cardiac output can only be an
approximation. (There are other reasons, too. Since there is no way to eliminate all such reasons, it is not
practical to try to refine the method of dye dilution beyond a certain point.)
Finally, note (again see Fig. 1) that “eventually” (if monitored long enough) we should see almost flat
concentration levels: as dye is eliminated from the bloodstream its concentration should eventually reach 0.
c
t
T1
T0 = 0
Fig. 2: Concentration Levels Near T1 .
As the following example illustrates, since measurements are discrete we use Riemann sum approximations
T1
c(t) dt ≈
c(ti )∆t
t=T0
to compute cardiac output. (Later, in MATH 164, we will see how we can improve on Riemann sum
approximations vis the Trapezoidal Rule and Simpson’s Rule.)
Exercise 1: After 5 mg of dye is injected, we obtain the concentration levels illustrated below
(t is in secs and c(t) is in mg/liter)
t
0
1
2
3
4
5
c(t)
0.00
0.20
0.77
1.63
2.69
3.81
Fig. 3:
t
c(t)
t
c(t)
6 4.84 12 4.74
7 5.67 13 3.76
8 6.19 14 2.75
9 6.35 15 1.82
10 6.13 16 1.10
11 5.57 17 0.65
A Set of Concentration
2
t
c(t)
18 0.54
19 0.75
20 1.25
21 1.97
22 2.81
23 3.68
Levels.
To compute the cardiac output we begin by observing that c has a relative minimum near 18 sec.
By plotting concentration levels near t = 18 sec, and using the fact that T1 is generally around 21
sec, we “revise” the data set to try to take into account the fact that near T1 , some dye making
its second pass by the monitoring point gets comingled with dye making its first pass through
the monitoring point. (This is an imperfect process which is done simply by trying to make a
good guess.)
8
t
0
1
2
3
4
5
c(t)
0.00
0.20
0.77
1.63
2.69
3.81
t
6
7
8
9
10
11
c(t)
4.84
5.67
6.19
6.35
6.13
5.57
t
12
13
14
15
16
17
c(t)
4.74
3.76
2.75
1.82
1.10
0.60
t
18
19
20
21
22
c(t)
0.30
0.15
0.05
0.00
6
4
2
0
0
10
20
30
Fig. 4: “Revised” Concentration Levels.
T1
Since the sum of the “revised” concentration levels is 59.11, t=T
c(t) dt ≈ 59.11, so R ≈
0
5/59.11 = 0.0846 liters per second, or 5.08 liters per minute.)
Tasks
1. Try a different “revision” of the original data for the example considered above, and see how it affects
your answer.
2. Compute the cardiac output for the hypothetical concentrations illustrated graphically in each part of
Fig. 5.
c
c0
c
c0
c
c0
t
T1
T0 = 0
t
t
T1/2
T0 = 0
(a)
T0 = 0
T1
(b)
T1/3
2T1/3
T1
(c)
Fig. 5: Hypothetical Concentration Levels.
3. Assuming that a 4.8 mg dose of dye is injected and that we obtain the concentration levels illustrated
below (t is in secs and c(t) is in mg/liter), compute the cardiac output.
t
0
1
2
3
4
5
c(t)
0.00
0.20
0.76
1.62
2.66
3.77
t
6
7
8
9
10
11
c(t)
4.80
5.64
6.17
6.34
6.15
5.61
t
12
13
14
15
16
17
3
c(t)
4.81
3.84
2.83
1.90
1.16
0.70
t
18
19
20
21
22
23
c(t)
0.55
0.72
1.19
1.88
2.70
3.56
References
[1] Calculus, 4th Ed., James Stewart, Brooks/Cole Pub. Co., 1999.
[2] Measuring Cardiac Output, Brindell Horelick and Sinan Koont, UMAP Module 71, Birkhauser-Boston
Inc., 1979.
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