MatLab codes for the used by forensics to determine the exact time of death.

Question

Evalistus Simon 2022-10-11

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回答： Sam Chak 2022-10-13

Hello:

Please help me write MatLab codes for the solution of the model used by forensics to determine the exact time of death derived from Newton's law of cooling given that T(t)= 70+10e^kt and T(0)=80. I need it for my project and am not so familiar with MATLAB

Those are the only informations i was given

Here I was trying but i do not really have a clue on how to continue

clc; clear ; close all;

%T(t)=70+10e^kt T(0)=80

F = @(t,T(t)) 70+10*exp(k*t);

T_0 = 80;

plot(t,T_0)

Thank you for your help.

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Answer 1

Davide Masiello 2022-10-11

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编辑：Davide Masiello 2022-10-13

在 MATLAB Online 中打开

if you already have an expression for T, why would you solve the ODE?

anyways, the correct syntax is

clear,clc

k = 0.5

k = 0.5000

F = @(t,T) k*(70-T);

tspan = [0 5];

T0 = 80;

[t,T] = ode45(F,tspan,T0);

plot(t,T)

legend('k = 0.5')

You could even solve it analytically

syms t T(t) k

eqn = diff(T,t) == k*(70-T);

sol = dsolve(eqn,T(0)==80)

sol =

%Solution for several values of k

for n = 0.1:0.2:1

solk = subs(sol,k=n);

fplot(solk,[0 5]),hold on

end

legend([repmat('k = ',5,1),num2str((0.1:0.1:0.5)')],'Location','best')

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Evalistus Simon 2022-10-13

Thank you very much.

Davide Masiello 2022-10-13

My pleasure. If the answer was helpful, don't forget to Accept it!

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Answer 2

Sam Chak 2022-10-13

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Hi @Evalistus Simon

The equation derived from Newton's Law of Cooling is given by

The ambient temperature is 70°F. Suppose that the coroner measured the body's temperature when they first arrive at the scene at 8:15 AM. Then, an hour later a second temperature is taken.

If the first temperature is 72.5°F, and the second temperature is 72.0°F, then the coroner can solve for the cooling rate k.

k = - log(2.5/2)
k = -0.2231

Substituting this value back to the equation when the 1st temperature is taken, and solve for

fun = @(t) 70 + 10*exp(-0.2231*t) - 72.5
fun = function_handle with value:
    @(t)70+10*exp(-0.2231*t)-72.5
tsol = fsolve(fun, 10)
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
tsol = 6.2138