Why bad partial differentiation?

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dodovoscek
dodovoscek 2013-3-9
Hi, I've a problem with this:
input parameter to my function is degree-it's number
There is a reason why I need m0,phi0 and not l0
m=sym('m',[1,degree]);
m=[m0 m];
l=sym('l',[degree,1]);
phi=sym('phi',[1,degree]);
phi=[phi0 phi] ;
for i=1:length(phi)
phi(i)=strcat(char(phi(i)),'(t)');
end
B=(1/2)*l(1)*m(2)*diff(phi(1),t)*diff(phi(2),t)*cos(phi(2))
phit=phi(2)
subs(diff(subs(B, phit, 'phit'),'phit'),'phit', phit)
I want to derive according to phi1(t), and the result is 0, but I expect (-1/2)*l1*m1*diff(phi0(t),t)*diff(phi1(t),t)*sin(phi1(t))
BUT, when I derive this B=(1/2)*l(1)*m(2)*diff(phi(1),t)*cos(phi(2)), it's all righ,
Any help would be appreciated
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Walter Roberson
Walter Roberson 2013-3-9
Could I ask you to show the value of B before the differentiation ?
dodovoscek
dodovoscek 2013-3-9
B=(1/2)*l1*m1*diff(phi0(t),t)*diff(phi1(t),t)*cos(phi1(t)) and I want to derive it according to phi1(t), but the way I mentioned returns 0,and
result=diff(B,phi(2)) doesnt work at all, just error

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Walter Roberson
Walter Roberson 2013-3-9
The MuPAD Symbolic Toolbox cannot differentiate with respect to a function. (Maple cannot either.)
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Walter Roberson
Walter Roberson 2013-3-10
In that case, you are differentiating with respect to a variable, not a function. Keep in mind that you are passing in 'phit' as a string, not phit by value (not quotation marks), so the symbol formed from 'phit' is going to be divorced from the phi(2) that appears in the expression.
The existence of the factor in your original expression of diff(phi(2),t) makes it especially questionable to differentiate a multiple of that factor by phi(2).
dodovoscek
dodovoscek 2013-3-11
Thanks, now everything is clear and I have found a way haw to avoid it :) thaks for your time
Best Regards
Domino

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