I want to take time derivative of the function

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));
dr3 = diff(link3_linear,t,1);

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

Answer 1

[Torsten]

link3_linear  = sqrt((r_x - r2*cos(theta)).^2 + (r_y - r2*sin(theta).^2))
d_link3_linear_dt = diff(link3_linear,t)
d_theta_dt = diff(theta,t)
d_link3_linear_dt_subs_d_theta_dt = subs(d_link3_linear_dt,d_theta_dt,1)

[Haseeb Hashim]

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