Mupad partial time derivative

2020 10 15
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更新时间：2020 10 19
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Hi, I've been working with mupad for some time, but I'm really surprised by the recent situation:
I've been doing some time derivative, but when the expression includes higher devirative of the same variable, it suddenly equals zero, idk why.
Any ideas how to fix this?
this is what i mean:
code:
A:=q(t)
AA:=diff(A|[q(t)=s],s)|[s=q(t)]
B:=(q(t)*diff(q(t),t))
BB:=diff(B|[q(t)=s],s)|[s=q(t)]
I'd expect diff(q(t),t)) instead of 0.

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Walter Roberson
Walter Roberson 2020-10-16

1 个投票

You substitute s for q(t), so B transforms into s*diff(s,t) . s is constant with respect to t so diff(s,t) is 0 and s*0 is 0 (provided that s is finite) so s*diff(s,t) is going to collapse to 0. Then you take the derivative of that 0 with respect to s, which is going to be 0.

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Witold Sure
Witold Sure 2020-10-16
Thank you for your explanation Walter.
Can you say how the code supposed to be composed?
Walter Roberson
Walter Roberson 2020-10-16
If you were to take q(t)*diff(q(t),t) and somehow differentiate with respect to function q(t), then I would expect that by the chain rule,
diff(q(t),t)*diff(q(t),t) + q(t)*diff(q(t),t,t)
which would be
diff(q(t),t)^2 + q(t)*diff(q(t),t,t)
not diff(q(t),t) ?
Witold Sure
Witold Sure 2020-10-17
This is true when you differentiate with respect to time.
When you differentiate expression q(t)*diff(q(t),t) with respect to q(t) the resoult is 1*diff(q(t),t).
https://www.wolframalpha.com/input/?i=d%5Bq*%28q%27%29%5D%2F%28dq%29
You helped me a lot, so far, Walter. I know what's the problem now!
I even have an idea how to get rid of this problem. Maybe you can help me out further.
When differentiating ' q(t)*diff(q(t),t) ' with respect to q(t), ' diff(q(t),t) ' sould be treated like any other constant, that's why it equals 1*diff(q(t),t).
B:=(q(t)*diff(q(t),t))
BB:=diff(B|[q(t)=s],s)|[s=q(t)]
To obtain such a resoult, I need to not only assign q(t)=s in B, but for the moment of differentiation, diff(q(t),t) should be replaced by some constant, for instance 'c'.
The resoult will be 'c'.
Then afterwards, I will assign diff(q(t),t) for 'c', again.
Do you know how to assign q(t)=s and diff(q(t),t)=c simultaneously?
Witold Sure
Witold Sure 2020-10-17
OK, I've happened to found an answer.
B:=(q(t)*diff(q(t),t))
BB:=diff(B|diff(q(t),t)=c|[q(t)=s],s)|[s=q(t)]|[c=diff(q(t),t)]
Be well!
Walter Roberson
Walter Roberson 2020-10-19
By the way, as of R2020b, the MATLAB interface to diff() can differentiate with respect to a function derivative.
Witold Sure
Witold Sure 2020-10-19
Good to know, It must make those operations way easier!

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