# I want to take time derivative of the function

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Haseeb Hashim on 29 Mar 2022
Edited: Walter Roberson on 15 Aug 2022
syms theta(t)
theta2 = 0:0.001:2*pi;
r_x = 11.6;
r_y = 4.3;
r2 = 3.5;
omega2 = 17; % Given in RPM
omega2 = omega2*2*pi/60; % In rad/s
link3_linear = sqrt((r_x - r2*cos(theta)).^2 + (r_y - r2*sin(theta).^2));
Now I have this code but how can we get the numerical answer by substituting for theta dot which comes as a result of the derivative with time 3 we can use the subs function but what are we substituting for or what would be the theta dot interpretation in MATLAB
I hope you understand my question
I just want the velocity of link3_linear by taking it sderivative and then substituting the angular velocity theta dot in the resulting derivative

Torsten on 29 Mar 2022
Edited: Torsten on 29 Mar 2022
link3_linear = sqrt((r_x - r2*cos(theta)).^2 + (r_y - r2*sin(theta).^2))
d_theta_dt = diff(theta,t)
##### 1 CommentShowHide None
Haseeb Hashim on 29 Mar 2022
So MATLAB stores theta_dot in form of diff(theta,t) as theta is a function of time
Understood. Wonderful @Torsten thanks man I was stuck here for a long time, Now I can complete my assignment on time thankss again man

R2017a

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