# Finding the maximum time of ode45

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Collin Parker el 30 de Abr. de 2021
Comentada: Collin Parker el 30 de Abr. de 2021
I have a question about my code. I am supposed to find the time at which ymax occurs. I have no idea how to set this up. Thank you.
% Physical constants
m = 45; % kg
l = 10; % m
d = 3.3; % m
rg = 4; % m
I = m*rg^2; % kg.m^2
% Derive the equations of motion and solve them below using ode45
Dz = @(t,z)[z(2);((10*sin(0.1*t)*cos(z(1)*d)-(k/100*(z(1))))/(I+(m*d^2)))];
[t,z] = ode45(Dz,(0:0.1:330),[0;0]);
% Compute the maximum deflection magnitude and the first time of occurence
y = tan(z(:,1)).*l;
ymax = max(abs(y))
tmax =
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VBBV el 30 de Abr. de 2021
Editada: VBBV el 30 de Abr. de 2021
Use the index of the matrix y where the max selection occurs

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Benjamin Großmann el 30 de Abr. de 2021
The max function is able to output the index of the max value:
[ymax, idx_max] = max(abs(y))
tmax = t(idx_max)
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Collin Parker el 30 de Abr. de 2021
Thank you!

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