Fundamental confusions of calculus

Discussion in 'Physics & Math' started by arfa brane, Feb 11, 2012.

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  1. Pete It's not rocket surgery Registered Senior Member

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    There's one other major sticking point between Tach and I that I'd really like confirmed:
    I've been led to believe by the aforementioned video, and also by [post=2901729]temur[/post], that the total derivative dx/dt is not meaningful for a function like x = f(y, t) without some expression (even a hypothetical one) relating y and t.

    Have I been misled?
     
    Last edited: Feb 13, 2012
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    You did make a mistake. The partial derivative expression will have the sin term in there too. If x is independent of theta then in this case the partial and total derivatives will coincide.

     
    Last edited: Feb 13, 2012
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  5. przyk squishy Valued Senior Member

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    Its presence. Because it wasn't there in your example. You never mentioned \(u\) or \(v\) or how \(f\) was supposed to depend on them.

    It's neither a variable nor a function. \(\theta\) is a variable. \(\sin \,:\, \mathbb{R} \to [-1,\, 1]\) is a function. \(\sin(\theta)\) is an expression given in terms of a function and a variable, that was part of the larger expression \(3 \theta \,+\, \sin(\theta)^{2} \,+\, \log(x)\) that you used to define \(f\).

    No, and there's no standard definition or convention that makes it clear.

    No, you just made that condition up, and it isn't even well defined. As Guest pointed out, any part of that expression could be considered a function. The mapping \(\theta \mapsto 3\theta\) is a function for instance. So is \(\theta \mapsto 3\theta \,+\, \sin(\theta)^{2}\), for that matter.

    For the record, here's how Mathematical Analysis (second edition) by Tom Apostol defines partial derivation, on page 115:
    This makes the partial derivatives completely independent of how you chose to write the definition of the function f. It doesn't matter if you gave the definition of f in terms of other functions.

    This is also false. For example, for the functions
    \( f \,:\, \mathbb{R}^{n} \to \mathbb{R} \,:\, x \mapsto f(x) \,, \\ g \,:\, \mathbb{R}^{m} \to \mathbb{R}^{n} \,:\, y \mapsto g(y) \,, \)​
    the chain rule for the partial derivatives of \(h = f \circ g\) is
    \( \frac{\partial h}{\partial y^{k}} \,=\, \sum_{j=1}^{n} \frac{\partial f}{\partial x^{j}} \, \frac{\partial g^{j}}{\partial y^{k}} \,. \)​
    The chain rule does not automatically mean you're calculating a total derivative. For the example above that's only the special case where \(m = 1\).
     
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  7. Tach Banned Banned

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    The textbook says clearly that it will not, because f is a function of \(\theta, u,v\).

    The term in sin comes in only in the total derivative through \(u\).
     
  8. Pete It's not rocket surgery Registered Senior Member

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    D'oh!

    \(\sin 2\theta = 2\sin\theta\cos\theta\)

    Problem solved.

    Anyone want to take a break from Tach and look at the rest of post 37 for me?
     
  9. Tach Banned Banned

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    They are both present, in the post that shows both the theory and the example. See post 18.
     
  10. Tach Banned Banned

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    This is an excellent video, it tells you exactly what I have been telling you. Using the exact terminology in the video:

    \(\frac{dT}{dt}\) is the total derivative

    \(\frac{\partial T}{\partial t}\) is the partial derivative

    The relationship is:

    \(\frac{dT}{dt}=\frac{\partial T}{\partial t} + (\frac{\partial T}{\partial x} \frac{dx}{dt}+\frac{\partial T}{\partial y} \frac{dy}{dt})\)

    Now, you can replace \(t\) with \(\theta\), \(x\) with \(u\) and you will be getting what I have been telling you all along. Pay special attention to the definition of the partial derivative.
     
  11. Tach Banned Banned

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    Now, post 18 tells you the same exact thing, with a twist. If y is not a function of t then:

    \(\frac{dT}{dt}=\frac{\partial T}{\partial t} + \frac{\partial T}{\partial x} \frac{dx}{dt}\)

    This is the part that you seem to have trouble with because you are stuck on the idea that the above is somewhat connected to your claim that "\(\frac{dy}{dt}\) does not make sense".
     
  12. Pete It's not rocket surgery Registered Senior Member

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    No, it's a different situation.
    The function he's describing is T = f(x, y, t), where x = g(t) and y=h(t).
    Note that all the function parameters other than t are dependent on t.
    There is no variable independent of t.

    Note what he says at 4:25
    In other words, dT/dt is not meaningful unless all the function parameters are either t or depend on t.
    It only becomes meaningful when:
     
  13. Tach Banned Banned

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    I told you that THIS is exactly your problem. Once you manage to get past this hangup, I think that you will understand the rest. There is NOTHING stopping you to calculate the total derivative wrt t if y is not a function of t, you are misinterpreting his words. What he's telling you is that T is a :

    -direct function of t

    -indirect function of t (through x and y) . It is not mandatory that BOTH x and y depend on t.

    Case and point:

    \(T=T(t,x,y,z)\)

    but the bug moves only in x and y, so

    \(\frac{dT}{dt}=\frac{\partial T}{\partial t} + (\frac{\partial T}{\partial x} \frac{dx}{dt}+\frac{\partial T}{\partial y} \frac{dy}{dt})\)

    There is no term in z, THOUGH T depends on z.
     
    Last edited: Feb 13, 2012
  14. Pete It's not rocket surgery Registered Senior Member

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    Is anyone other than Tach able to help me out with the questions in post 41 and post 37 (only after the second quote)?

    I feel they've been lost in the noise.
     
  15. przyk squishy Valued Senior Member

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    Post 18 is here. This is how you define your example:
    You do not say here that \(f\) has parameters \(u\) and \(v\). The reader is left to infer that from earlier in the same post. You also do not state how \(f\) depends on \(u\) and \(v\). The reader is left to guess that too.

    Incidentally during the five years I spent studying physics at university, including calculus intensive courses like thermodynamics and analytical mechanics, I never saw anyone use the notation
    to indicate a function's parameter dependence.
     
  16. Tach Banned Banned

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    You need to understand that Pete and I have we been going over this for about 50 posts PRIOR to post 18.



    Isn't this clear enough?

    This is your problem. So, it all boils down that you did not read the post. Paradoxically, you seemed to have understood it first time.
     
  17. Pete It's not rocket surgery Registered Senior Member

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    I can see why. I've used that notation myself (don't know if I made it up, or copied it from Tach), and it introduces confusion. I did it in post 37 in this thread. Perhaps that's why no one's answered it. Should I go back and fix it? (edit - done)

    What do you do if you have used f(...), g(...), h(...), and you want to describe another function? Just keep climbing through the alphabet, risking ambiguities?
     
    Last edited: Feb 13, 2012
  18. Pete It's not rocket surgery Registered Senior Member

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    No.
    Like przyk said, you haven't described u as a function of theta, or f as a function of u.

    Perhaps you meant this, which I believe would be correct:
    \(\begin{align} u &= sin^2\theta \\ f &= 3 \theta + u + ln(x) \\ \frac{\partial f}{\partial \theta} &= 3 \end{align}\)​
    or this (also correct):
    \(\begin{align} u &= sin\theta \\ f &= 3 \theta + u^2 + ln(x) \\ \frac{\partial f}{\partial \theta} &= 3 \end{align}\)​

    ...but note that this is also correct:
    \(\begin{align} u &= 3\theta \\ f &= u + sin^2\theta + ln(x) \\ \frac{\partial f}{\partial \theta} &= \sin2\theta \end{align}\)​

    And, as you've been told several times, this is also correct:
    \(\begin{align} u &= 0 \\ f &= u + 3\theta + sin^2\theta + ln(x) \\ \frac{\partial f}{\partial \theta} &= 3 + \sin2\theta \end{align}\)​
     
  19. AlphaNumeric Fully ionized Registered Senior Member

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    Tach, I explicitly worked through the definition of the partial derivative for the example you'd given. It came out to include the sin term. I explicitly asked you to address this. I also explicitly asked you to address how your logic completely disagrees with all of Lagrangian and Hamiltonian mechanics. I also gave an explicit example for that.

    You are not actually engaging in a discussion here, you're just pointing at the book and saying "Look, look!". Yes, we've looked at it. And several of us have explained to you its context and application, ie the implicit/explicit parametrisation. Some of us use Hamiltonian mechanics a lot. And the distinction between \(\frac{dL}{dt}\) and \(\frac{\partial L}{\partial t}\) where \(L = L(q_{i},\dot{q}_{j},t)\) and \(q_{i} = q_{i}(t)\) is a very important one. Another example, which przyk brought up, is statistical mechanics, where there's partial and total derivatives everywhere. This isn't some result no one here has any experience with, it's something used a lot.

    When you just point at a book and say "Look, look, there's an equation!" you're giving the impression your experience with this stuff isn't on a working level. Of course that isn't a crime but it does call into question whether or not you understand the application of the result. No one is denying the equation, the problem is your understanding of it.

    Please address the explicit examples I gave, namely the definition of the partial derivative and the role of the partial derivative in Hamiltonian mechanics. If I have to ask a third time I'm giving you a warning for trolling because avoiding addressing the definition of a mathematical construct is trolling.
     
  20. przyk squishy Valued Senior Member

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    Absolutely not.

    No. Your use of non standard definitions and conventions, and the problems that causes, are your own fault and nobody else's.
     
  21. Pete It's not rocket surgery Registered Senior Member

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    Try Mathematica:
    "In Mathematica, D(f,x) gives a partial derivative, with all other variables assumed independent of x. Dt(f,x) gives a total derivative, in which all variables are assumed to depend on x."

    ...so you're defining the bug's position in z over time, specifically z=0, so dz/dt = 0

    If you don't know how the bug moves (or not) in z, then how can you determine dT/dt?
     
  22. Tach Banned Banned

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    This is not what the example says, besides it is irrelevant to solving your misunderstandings.

    This is not what the example says, besides it is irrelevant to solving your misunderstandings.



    You know, if you spent less time on your pettiness relative to notation and more time in trying to solve your lack of understanding as to why the total derivative does not depend on \(v\), you could have used your time and a much more productive manner.
     
  23. Tach Banned Banned

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    Nope, I am simply defining T as a function of z, which is very natural, if you stop and think about it.

    The bug moves only in x,y as explained. Do you really want to understand this or do you want to be right no matter what?
     
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