MATLAB ENGINEERING PROBLEM This problem must be solved using matlab, not by hand

Question

MATLAB ENGINEERING PROBLEM

This problem must be solved using matlab, not by hand. Please post a screenshot of your code.

"2.34. You arrive at your lab at 8 A.M. and add an indeterminate quantity of bacterial cells to a flask. At 11 A.M you measure the number of cells using a spectrophotometer (the absorbance of light is directly related to the number ofcells and determinefromaprevious calibration that theflaskcontains 3850cells, andat 5 P.M. the cell count has reached 36,530. (a) Fiteach of the following formulas to thetwo givendata points (thatis,determinethevaluesofthetwo constants in each formula): linear growth, C Co kt; exponential growth, C Coe'r;power-law growth. C ktb. In these expressions, Co is the initial cell concentration and kand b are constants. (b) Select the most reasonable of the three formulas and justify your selection. (c) Estimate the initial number of cells present at 8 A.M. (t 30). State any assumptions you make. (d) The culture needs to be split into two equal parts once the number of cells reaches 2 million. Estimate the time at which you would have to come back to perform this task. State any assumptions you make. If this is a routine operation that you must perform often, what does your result suggest about the scheduling of the experiment?

Explanation / Answer

Kindly copy the below in MATLAB script and run.

Thanks

clc
clear all
format long

%%

t=0 ; % 8 am
t1=3 ; % 11 am
t2=9 ; % 5 pm

c1=3850;
c2=36250; % c1 and c2 are number of cells at time t1 and t2, resp.

c=[c1;c2];
t=[t1;t2];

% for C=C0+kt
linear= polyfit(t,c,1); %
linear_t0=polyval(linear,0);

% for C=C0 exp(kt)

exp=fit(t,c,'exp1')
exp_t0= feval(exp,0);

% for C=k t^b

power=fit(t,c,'power1');
%power_t0 = feval(power,0);

fit=[linear_t0 exp_t0 ] ;

% comments
%As can be observed from the results that we have the following output
%for : C=C0+kt C0=-12350.000 k= 5400.0
% C=C0 exp(kt) C0 = 1255 k = 0.3737
% C=k t^b k = 408.9 k = 2.041

%(b) We select exponetial model because for linear we get negative C at
%t=0 which is not possible. For power fit, at t=0, it is not possible to
% evaluate the data.

% Hence we select the exponential data

% (c) at t=0, at 8 am , we have 1254 number of cells. (exp_t0)

% (d) time taken to reach 2 million can be calculated by:

t3=1/0.3737*(log(2e5/1255));

% this comes to be nearly 13.5 hours. Hence one needs to come back at 9.30
% PM to change the perforrm the task
% Once the cells have been split into 2 each one million then time taken
% after that to reach 2 million is
t4=1/0.3737*(log(2e5/1e5));
% which is approximately 1.9 hours. This has to be repeacted again and
% again

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