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1 BME 210 Spring 2011 Name: Solutions _______ EXAMINATION #1 (1 hour, 15 minutes) 1. (20 points) A dye-dilution procedure was performed in a patient in which 0.02 grams of dye were injected into the superior vena cava and the dye concentrations (C(t)) given in the table below were measured in the aorta at several times. t – time (sec) C(t) – dye concentration (mg/L) 5 0 10 0 15 0 20 0 25 30 30 10 35 4 40 1 Using these data, estimate the area under the dye concentration-time curve by (a) trapezoidal approximation, and (b) Simpson’s method. Include the units for your result. In each case, use the area under the curve to calculate the cardiac output for this patient, given in liters per minute. a) Trapezoidal approximation T ( f ; P ) = h 2 ( f ( x 1 ) + f ( x n )) + h f ( x i ) i = 2 n − 1 ∑ T ( f ; p ) = 5 2 (0 + 1) + 5 ⋅ (0 + 0 + 0 + 30 + 10 + 4) = 222.5 mg ⋅ sec L ˙ Q ≅ D T ( f ; P ) = 20 mg 222.5 mg ⋅ sec L = .08988 L sec ⋅ 60 sec min = 5.39 L min

2 b) Simpson’s method We have an even number of points, so we have to use trapezoidal integrations for the first or last point. Using trapezoidal integration for the first point gives: S ( f ; P ) = h 3 ( f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) + 4 f ( x 5 ) + 2 f ( x 6 ) + 4 f ( x 7 ) + f ( x 8 )) + h 2 ( f ( x 1 ) + f ( x 2 )) S ( f ; p ) = 5 3 (0 + 4 ⋅ 0 + 2 ⋅ 0 + 4 ⋅ 30 + 2 ⋅ 10 + 4 ⋅ 4 + 1) + 5 2 (0 + 0) = 261.67 mg ⋅ sec L ˙ Q ≅ D S ( f ; P ) = 20 mg 261.67 mg ⋅ sec L = .0764 L sec ⋅ 60 sec min = 4.59 L min Alternatively, using trapezoidal approximation for the last point: S ( f ; p ) = h 3 ( f ( x 1 ) + 4 ⋅ f ( x 2 ) + 2 ⋅ f ( x 3 ) + 4 ⋅ f ( x 4 ) + 2 ⋅ f ( x 5 ) + 4 ⋅ f ( x 6 ) + f ( x 7 )) + h 2 ( f ( x 7 ) + f ( x 8 )) S ( f ; p ) = 5 3 (0 + 4 ⋅ 0 + 2 ⋅ 0 + 4 ⋅ 0 + 2 ⋅ 30 + 4 ⋅ 10 + 4) + 5 2 (4 + 1) = 185.83 mg ⋅ sec L ˙ Q ≅ D S ( f ; P ) = 20 mg 185.83 mg ⋅ sec L = 0.108 L sec ⋅ 60 sec min = 6.46 L min For this example, the first solution gives a better approximation.

3 2. (10 points) The curve shown below was recorded following a dye dilution experiment performed to measure blood flow through the pancreas. Flow is assumed to be constant (F=0.1 L/min) during the experiment (from 0 to 10min). If total dye injected at time 0 is 20mg (D=20 mg) find C0. F = D C ( t ) dt 0 10 ∫ C ( t ) dt 0 10 ∫ = C ( t ) dt + C ( t ) dt 4 10 ∫ 2 4 ∫ = (4 − 2) C 0 + (10 − 4) C 0 2 = 5 C 0 F = D 5 C 0 C 0 = D 5 F = 20 mg 5min ⋅ 0.1 L min = 40 mg L

4 3. (15 points) The following program computes cardiac output from dye dilution data. To do so, it performs a numerical integration of dye concentration vector y by rectangular approximation and Simpson’s method. The definitions of M, s1, and s2 needed in the final AUC equations (indicated in the circles) have been deleted from the program. What do M, s1, and s2 have to be so that the program will function correctly? Write in your answer in the rectangles below. You may use more than one line if you need to, but you may NOT use any loops . % Program to calculate Cardiac Output using dye dilution data clc; clear; data=importdata( 'dye.dat' ); D=input( 'Please enter the amount of ICG dye: ' ); %Numerical integration x=data(:,1); y=data(:,2); %time and dye vectors from data file n=length(x); %number of data points h=x(2)-x(1); %sample interval %Rectangular approximation M =sum(y(1:n-1) ; r_auc= h* M ; %AUC from rectangular approximation %Simpson's method s1=y(2:2:n-1); s2=y(3:2:n-2); s_auc=h/3*(y(1)+4*sum( s1 )+2*sum( s2 )+y(n)); %AUC from Simpson’s method %Calculate cardiac output and transfer its value into (L/min) r_co=D/r_auc*60; s_co=D/s_auc*60; disp( 'The area under the dye concentration curve for' ) disp( 'Rectangular and Simpson method are: ' ) disp([ num2str(r_auc), ' ' ,num2str(s_auc), '(mg-sec/L) respectively' ]) disp( ' ' ) disp( 'Cardic Output of rectangular and Simpson method are: ' ) disp([num2str(r_co), ' ' , num2str(s_co), ' (L/min) respectively' ]);

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