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Math Help - How do derivate in mathematica 7.0

  1. #1
    MHF Contributor arbolis's Avatar
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    How do derivate in mathematica 7.0

    I've been able to derivate some function but not the following one:
    Code:
    Dirac := (m dx[t])/(2 Sqrt[dx[t]^2 + dy[t]^2 + dz[t]^2])
    .
    I want to derivate the Dirac's function with respect to t. I tried
    Code:
    D[Dirac[dx, dy, dz], t]
    but then I get
    Code:
    ((m 
    \!\(\*SuperscriptBox["dx", "\[Prime]",
    MultilineFunction->None]\)[t])/(
       2 Sqrt[dx[t]^2 + dy[t]^2 + dz[t]^2]) - (m dx[t] (2 dx[t] 
    \!\(\*SuperscriptBox["dx", "\[Prime]",
    MultilineFunction->None]\)[t] + 2 dy[t] 
    \!\(\*SuperscriptBox["dy", "\[Prime]",
    MultilineFunction->None]\)[t] + 2 dz[t] 
    \!\(\*SuperscriptBox["dz", "\[Prime]",
    MultilineFunction->None]\)[t]))/(
       4 (dx[t]^2 + dy[t]^2 + dz[t]^2)^(3/2)))[dx, dy, dz]
    which isn't what I'm looking for, I believe (I don't even know what this output mean). Notice that dx, dy and dz aren't the differentials x, y and z. Rather, they're functions which depends on t.
    Any idea?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by arbolis View Post
    I've been able to derivate some function but not the following one:
    Code:
    Dirac := (m dx[t])/(2 Sqrt[dx[t]^2 + dy[t]^2 + dz[t]^2])
    .
    I want to derivate the Dirac's function with respect to t. I tried
    Code:
    D[Dirac[dx, dy, dz], t]
    but then I get
    Code:
    ((m 
    \!\(\*SuperscriptBox["dx", "\[Prime]",
    MultilineFunction->None]\)[t])/(
       2 Sqrt[dx[t]^2 + dy[t]^2 + dz[t]^2]) - (m dx[t] (2 dx[t] 
    \!\(\*SuperscriptBox["dx", "\[Prime]",
    MultilineFunction->None]\)[t] + 2 dy[t] 
    \!\(\*SuperscriptBox["dy", "\[Prime]",
    MultilineFunction->None]\)[t] + 2 dz[t] 
    \!\(\*SuperscriptBox["dz", "\[Prime]",
    MultilineFunction->None]\)[t]))/(
       4 (dx[t]^2 + dy[t]^2 + dz[t]^2)^(3/2)))[dx, dy, dz]
    which isn't what I'm looking for, I believe (I don't even know what this output mean). Notice that dx, dy and dz aren't the differentials x, y and z. Rather, they're functions which depends on t.

    Any idea?
    They are not even functions, at best and with some caveats they are functionals.

    CB
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    They are not even functions, at best and with some caveats they are functionals.

    CB
    Ok I see. It makes sense since what I call the Dirac's function arises from Euler-Lagrange equation.
    Written differently, I want to derivate the following expression with respect to time: \frac{m \dot x}{2 \sqrt {\dot x^2 + \dot y^2 + \dot z^2}} where the dot represent the derivative with respect to time.
    The variables are \dot x, \dot y and \dot z all depends on time. Do you think Mathematica can do it?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by arbolis View Post
    Ok I see. It makes sense since what I call the Dirac's function arises from Euler-Lagrange equation.
    Written differently, I want to derivate the following expression with respect to time: \frac{m \dot x}{2 \sqrt {\dot x^2 + \dot y^2 + \dot z^2}} where the dot represent the derivative with respect to time.
    The variables are \dot x, \dot y and \dot z all depends on time. Do you think Mathematica can do it?
    I'm pretty sure Mathematica can, but can't tell you how, I gave up my Mathematica licence a number of years ago now.

    CB
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  5. #5
    MHF Contributor arbolis's Avatar
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    Ok no problem. I'll be waiting now for any further help from anyone.
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  6. #6
    Senior Member roninpro's Avatar
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    Try this:

    Code:
    D[m x'[t]/(2 Sqrt[x'[t]^2 + y'[t]^2 + z'[t]^2]), t]
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