Question

i need the answer to be on a MatLab window

1. Consider the following equation, which represents the concentration (c, in mg/ml) of a drug in the bloodstream over time (t, in seconds). Assume we are interested in a concentration of c2 mg/ml C3te-0.4t A. Estimate the times at which the concentration is 2 mg/ml using a graphical method Be sure to show your plot(s). Hint: There are 2 real solutions B. Use MATLAB to apply the secant method (e.g. secant.m) to find each of the solutions/roots with a tolerance of 1 x 10-3 on the absolute relative approximate error C. Calculate the first two iterations of the secant method by hand for starting values of t 0 and t2 seconds. D. Either by hand or by writing a MATLAB function, compute 3 iterations of the bisection method to find the solution/root within the starting interval of [4.5, 5.5] seconds. For each iteration, show the 3 relevant values of t and the corresponding values of f (t) Use the symbolic solve function to find a root of the equation. Express the result as a double (double-precision floating point value) Use the function fzero with suitable starting values to find both of the solutions/roots E. F.

Answer 1

A)

PLEASE REFER BELOW CODE

close all
clear all
clc

%given expression
%A)
t = 0:0.1:2; %taking range of t from 0 to 2 with increment of 0.1
c = 3 . t . exp(-0.4 * t); %given expression

plot(t,c);
title('t vs c');
xlabel('time(t)');
ylabel('concentration(c)');

index = find((c-2) <= 0.00);
fprintf('Number of times concentration is 2 mg/ml = %d\n', length(index));

PLEASE REFER BELOW OUTPUT

Number of times concentration is 2 mg/ml = 10
>>

B) PLEASE REFER BELOW CODE:

close all
clear all
clc

%2) Secant method

%a=input('Input expression : ','s');
%f=inline(a);
f = @(t) ( 3 t exp(-0.4 * t))
%at t = 0;
t(1)= 3 0 exp(-0.4 * 0);

%at t = 2;
t(2)=3 2 exp(-0.4 * 2);

n = 10^-3;
iteration=0;

for i=3:1000
t(i) = t(i-1) - (f(t(i-1)))*((t(i-1) - t(i-2)) ./(f(t(i-1)) - f(t(i-2))));
iteration=iteration+1;
if abs((t(i)-t(i-1))./t(i)) < n
root=t(i)
iteration=iteration
break
end
end

f =

function_handle with value:

@(t)(3*t*exp(-0.4*t))

>>

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