I want to take time derivative of the function

Question

Haseeb Hashim 2022-3-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1683514-i-want-to-take-time-derivative-of-the-function

编辑： Walter Roberson 2022-8-15

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

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Answer 1

Torsten 2022-3-29

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https://ww2.mathworks.cn/matlabcentral/answers/1683514-i-want-to-take-time-derivative-of-the-function#answer_929824

编辑：Torsten 2022-3-29

在 MATLAB Online 中打开

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)