Math 4600: Homework 1
Gregory Handy
[1.1] MATLAB primer. (not graded)
Let me know if you have any questions!
[1.2] Use the following parameter values for the model of the periodic arterial pulse that we considered in class: Rs = 17.5 mmHg/(liter/min), Csa = 0.00175 liters/mmHg, ∆V0 = 0.07 liters,
T = (1/80) min. Answer the following questions:
(a) Find the systolic pressure, the diastolic pressure, the pulse pressure;
(b) Make a graph of the systemic arterial pressure as a function of time, include a few periods of
heart beat;
(c) If a person has all parameters exactly the same as above, except his heart pumps 1.5 times as
much blood per beat, what blood pressure would that person have?
For this question, and those in 3 and 4, you should use the following MATLAB function,
function [ psa ] = psa_function(R_s, C_sa, deltaV_0, T, t )
psa = deltaV_0/C_sa .* exp(-t/(R_s*C_sa)).*(1-exp(-T/(R_s*C_sa)))^(-1);
end
(a) The following MATLAB script can be use
R_s = 17.5;
C_sa = 0.00175;
deltaV_0 = 0.07;
T = 1/80;
psa_0 = psa_function(R_s, C_sa, deltaV_0, T, 0)
psa_T =
% systolic 119.3568
psa_function(R_s, C_sa, deltaV_0, T, T) % diastolic 79.35678
pulse_pressure = psa_0-psa_T
% pulse 40
(b) Using the following code to produce the necessary graph:
t = [0:T/100:T];
psa_period = psa_function(R_s, C_sa, deltaV_0, T, t);
% loops through the different periods
for i = 1:4
plot(t+T*(i-1), psa_period)
hold on
end
set(gca,’fontsize’,16)
xlabel(’Time (min)’)
ylabel(’Psa(t)’)
1
120
115
110
Psa(t)
105
100
95
90
85
80
75
0
0.01
0.02
0.03
Time (min)
0.04
0.05
Figure 1: Systemic arterial pressure as a function of time.
(c) If his heart pumps 1.5 times as much blood per beat, we need to change multiply ∆V0 by 1.5, meaning
∆V0 = 0.07 ∗ 1.5 = 0.1050. Rerunning the above code, we find
deltaV_0=0.07*1.5;
psa_0 = psa_function(R_s, C_sa, deltaV_0, T, 0)
psa_T = psa_function(R_s, C_sa, deltaV_0, T, T)
pulse_pressure = psa_0-psa_T
% 179.0352
% 119.0352
% 60
[1.3] During exercise the arterioles in an excercising muscle dilate, and the systemic resistance falls.
Let us assume that it falls to half its value.
(a) First, make an unrealistic assumption that the circulatory system does not compensate for this
change, i.e. that all other parameters remain the same. Plot the new time course of systemic
arterial pressure. What would happen to the systolic, diastolic and pulse pressure?
(b) In reality, the fall in systemic resistance is compensated in multiple ways, most noticeably by
the increase of heart rate, i.e. by decrease in T . Assume that T changes proportionally to Rs .
Plot the new time course of systemic arterial pressure. What are the changes in the systolic,
diastolic and pulse pressure?
(a) We start by assuming Rs = 17.5/2. Use the following code to find the new pressures and to plot the
graph
R_s = 17.5/2;
C_sa = 0.00175;
deltaV_0 = 0.07;
T = 1/80;
psa_0 = psa_function(R_s, C_sa, deltaV_0, T, 0)
psa_T = psa_function(R_s, C_sa, deltaV_0, T, T)
pulse_pressure = psa_0-psa_T
% 40
% 71.6913
% 31.6913
t = [0:T/100:T];
psa_period =
psa_function(R_s, C_sa, deltaV_0, T, t);
plot(t, psa_period)
hold on
plot(t+T, psa_period)
plot(t+2*T, psa_period)
set(gca,’fontsize’,16)
2
xlabel(’Time (min)’)
ylabel(’Psa(t)’)
axis([0 3*T 30 75])
75
70
65
Psa(t)
60
55
50
45
40
35
30
0
0.01
0.02
Time (min)
0.03
Figure 2: Systemic arterial pressure as a function of time.
(b) We now assume that T compensates by also decreasing by a factor of 2, meaning T = (1/80)/2. This
is captured by the following code
T = (1/80)/2;
psa_0 = psa_function(R_s, C_sa, deltaV_0, T, 0)
psa_T = psa_function(R_s, C_sa, deltaV_0, T, T)
pulse_pressure = psa_0-psa_T
% 40
% 119.3568
% 79.3568
t = [0:T/100:T];
psa_period =
psa_function(R_s, C_sa, deltaV_0, T, t);
plot(t, psa_period)
hold on
plot(t+T, psa_period)
plot(t+2*T, psa_period)
set(gca,’fontsize’,16)
xlabel(’Time (min)’)
ylabel(’Psa(t)’)
axis([0 3*T 75 125])
120
Psa(t)
110
100
90
80
0
0.005
0.01
Time (min)
0.015
Figure 3: Systemic arterial pressure as a function of time.
3
[1.4] Suppose there is a patient, whose heart does not respond properly to exercise, and the heart
rate remains constant.
(a) Average value of a periodic function can be found by avearging over one period. Find a formula
for average Psa as a function of Rs , ∆V0 , and T . Suppose that the patient performs exercise
(Rs falls). How much the stroke volume needs to change to compensate for the fall in Rs to
keep the mean pressure unchanged?
(b) If during exercise the stroke volume changes enough to keep the mean pressure unchanged,
with these new parameter values plot the new time course of systemic arterial pressure. What
are the changes in the systolic, diastolic and pulse pressure?
(a) To find the average value, we find the following integral
Ave Psa =
=
=
=
=
−1
∆V0
−T
−t
1 − exp
exp
dt
Csa
Rs Csa
Rs Csa
0
−1 Z T
−T
∆V0
−t
1 − exp
dt
exp
T Csa
Rs Csa
Rs Csa
0
−1 T
∆V0
−T
−t
1 − exp
−Rs Csa exp
T Csa
Rs Csa
Rs Csa 0
−1 ∆V0 Rs
−T
−T
1 − exp
1 − exp
T
Rs Csa
Rs Csa
∆V0 Rs
.
T
1
T
Z
T
Thus, if RS were to fall by a factor of α, then ∆V0 would need to increase by a factor of α in order for
the mean pressure to remain unchanged.
(b) The following code can be used to find the new pressures and the graph
R_s = 17.5/2;
deltaV_0 = 0.07*2;
C_sa = 0.00175;
T = 1/80;
psa_0 = psa_function(R_s, C_sa, deltaV_0, T, 0) % 143.3827
psa_T = psa_function(R_s, C_sa, deltaV_0, T, T) % 63.3827
pulse_pressure = psa_0-psa_T
% 80
t = [0:T/100:T];
psa_period =
psa_function(R_s, C_sa, deltaV_0, T, t);
plot(t, psa_period)
hold on
plot(t+T, psa_period)
plot(t+2*T, psa_period)
set(gca,’fontsize’,16)
xlabel(’Time (min)’)
ylabel(’Psa(t)’)
axis([0 3*T 60 145])
4
140
130
Psa(t)
120
110
100
90
80
70
60
0
0.01
0.02
Time (min)
0.03
Figure 4: Systemic arterial pressure as a function of time.
[1.5] Suppose that the stroke volume and the heartbeat period randomly change from beat to beat
by as much as 20%. Plot the resulting pulse time course for the first four beats and describe how
the pulse characteristics change between the shorter and the longer beats.
This questions requires a slight modification to our Psa function defined above (one that takes in an initial
condition, since it is not always going to be the same value):
function [ psa ] = psa_function_v2(R_s, C_sa, deltaV_0, t_j, psa_j_minus, t )
psa_initial = psa_j_minus + deltaV_0/C_sa;
psa = psa_initial *exp(-(t-t_j)/(R_s*C_sa));
end
The following code will use both of our Psa functions, though it could be cleaned up to only use one
R_s = 17.5;
C_sa = 0.00175;
deltaV_0 = 0.07;
T = 1/80;
t = [0:T/100:T];
% Plots the first, non-randomized period
psa_period = psa_function(R_s, C_sa, deltaV_0, T, t);
plot(t, psa_period)
hold on
for i = 1:3
t_j = t(end);
psa_j_minus = psa_period(end);
% allow for a random change in deltaV_0 and T (up to 20%)
deltaV_0 = deltaV_0*(.8 + (1.2-.8)*rand(1));
T = T*(.8 + (1.2-.8)*rand(1));
t = [t_j:T/100:T+t_j];
psa_period = psa_function_v2(R_s, C_sa, deltaV_0, t_j, psa_j_minus,t);
plot(t, psa_period)
end
set(gca,’fontsize’,16)
xlabel(’Time (min)’)
5
140
130
Psa(t)
120
110
100
90
80
70
0
0.01
0.02
0.03
0.04
Time (min)
0.05
0.06
Figure 5: Systemic arterial pressure as a function of time. Note: a different graph will be produced each
time that you run the code!
6