Question

Exercise: Bacteria Growth and Interpretation You can see that doing this with a for loop can be easier than repeatedly typing the same commands for every a value. Using a for loop also makes the desired values or parameters easier to change Consider various populations of bacteria whose populations are modelled by the function P(t) 100 e a.Sat, where P(t) is the population, t is the time in hours, and a is some positive constant. Your job is to create a plot of how different populations behave for different values of a. Use a for loop to go through values of a in the interval [1,2] in steps of 0.1 (a 1, a 1.1, a 1.2 a 2), and plot all of them on the same figure (11 different functions). Hint: Before the for loop, use a figure command to create a new figure window and hold on to keep all plots on the same figure Use your figure to answer the following questions 1. Using your understanding of exponential functions, identify the general order or pattern of which function (a value) corresponds to which line 2, which population grows the fastest as t →oo? At what value of t does it becomes the biggest population? (Similar to previous labs, you may need to use the axis function and the zoom tool in the figure window to see this. You may also want to adjust the domain of t values you are using) 3. Choose three of the populations and read the figure to estimate the doubling time (amount of time takes for the population to double) for each. Does this match what you can find analytically using the function expression? Can you find a relationship between the doubling time and a value? 4. In a paragraph or two, please discuss similarities/differences in doing matrix math by-hand versus via Matlab. When is one more appropriate than another? If so, when? What are some important considerations an engineer should think about when using "for" loops?
MATLAB

Thank you!

Answer 1

clc;
close all;
clear all;

t1=1:0.1:24;
j=1;
for a=1:0.1:2 % a increments with 0.1 step size
i=1;
for t=1:0.1:24 % t increments with 0.1 step size
p(i)=(1000/a)*exp(0.5*a*t);
i=i+1;
end
b(j)=plot(t1,p);
j=j+1;
hold on
end
hold off
xlabel('time');
ylabel('p(t)');
title('bacterial growth');
%plot marks, just to label plots for different values of a
M1 = "a=1";
M2 = "a=1.1";
M3 = "a=1.2";
M4 = "a=1.3";
M5 = "a=1.4";
M6 = "a=1.5";
M7 = "a=1.6";
M8 = "a=1.7";
M9 = "a=1.8";
M10 = "a=1.9";
M11 = "a=2";

legend([b(1),b(2),b(3),b(4),b(5),b(6),b(7),b(8),b(9),b(10),b(11)],[M1,M2,M3,M4,M5,M6,M7,M8,M9,M10,M11]);

bacterial growth x 10 14 12 a1.3 a=1.4 10 a=1.7 8 a:1.9 a-2 6 4 2 0 15 20 25 10 time

1) from the above figure, it was derivable that with increasing value of a the rate of growth of bacteria increased.

2)from the fig(2), it can be realized as t-->infinity(t=1000) the maximum growth rate was observed for a=2. we believe that for further increasing a, the growth rate increases.

bacterial growth 17879710307 a=1 a=1.1 a=1.2 ㅡㅡ a=1.3 a=1.4 11.4399 a=1.6 a=1.7 a=1.8 a=1.9 a-2 0 100 200 300 400 500 600 700 800 900 1000 time

fig(2)

4) Matrix math by hand approach works for smaller time sets making derivation of conclusions being more based on human understanding while Matlab based approach enables a broader view of data for higher time set consideration.

For for loop, Matlab usually employs quite a slow approach. One aspect of Matlab, being the vectorized approach provides faster conclusion.