Symbolic Substitution(subs) is not working , why?

4 vues (au cours des 30 derniers jours)
Mirroyal Ismayilov
Mirroyal Ismayilov le 27 Juil 2018
Commenté : Walter Roberson le 1 Août 2018
T(t) =
(25*diff(r1(t), t)^2)/12 + (25*diff(r2(t), t)^2)/8 + (25*cos(r1(t))^2*diff(r1(t), t)^2)/4 + (25*sin(r1(t))^2*diff(r1(t), t)^2)/4 + (3*(5*cos(r1(t))*diff(r1(t), t) + (5*cos(r2(t))*diff(r2(t), t))/2)^2)/2 + (3*(5*sin(r1(t))*diff(r1(t), t) + (5*sin(r2(t))*diff(r2(t), t))/2)^2)/2
T is gained from somewhere which is long story but it works perfectly fine. My code is following:
syms T r1(t) k1
subs(T,r1,k1);
P1=diff(T,k1);
Because I can't differentiate in terms of symbolic function, I wanted to replace r1 with k1, but it doesn't replace and subs function gives back these:
ans(t) =
(25*diff(r2(t), t)^2)/8 + (75*cos(r2(t))^2*diff(r2(t), t)^2)/8 + (75*sin(r2(t))^2*diff(r2(t), t)^2)/8
Somehow r1(t)'s are disappeared but couldn't been replaced with k1. Because there is not k1 in substituted function,differentiation(diff) gives back just 0(zero) which is normal.
Can someone explain me the situation please?

Réponse acceptée

Walter Roberson
Walter Roberson le 27 Juil 2018
No, the substitution is working. But you have diff(r1(t),t) and when you replace r1(t) with k1, that becomes diff(k1, t) and since k1 is independent of t, the diff() becomes 0.
  2 commentaires
Mirroyal Ismayilov
Mirroyal Ismayilov le 27 Juil 2018
Modifié(e) : Mirroyal Ismayilov le 31 Juil 2018
Mr Walter, Although subs is working, still I can't get answer as wanted since whenever I substitute symbolic function with symbolic variables, it automatically makes things zero. I had to use subs because I don't still understand why it is not possible to derive the symbolic function in terms of symbolic function. Do you have any suggestion?
Walter Roberson
Walter Roberson le 1 Août 2018
Perhaps you want to use functionalDerivative()
But what I recommend is that you look at the first example in odeFunction() to see how to convert equations with derivatives into systems of equations.

Connectez-vous pour commenter.

Plus de réponses (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by