MATLAB ENGINEERING PROBLEM This problem must be solved using

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?

Solution

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