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Notation of a Parametric Derivative

  1. May 19, 2010 #1

    Char. Limit

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    Gold Member

    Now first, as I'm sure you all know, for any two parametric equations x(t) and y(t), the slope of the curve thus generated is equal to the quotient (dy/dt)/(dx/dt). Also, we know that the concavity of the function (or change of slope, as I'm not sure if concavity is actually a correct term) is equal to the quotient (dy'/dt)/(dx/dt), where y'=dy/dx=(dy/dt)/(dx/dt). Now to my question... Is it allowed to represent dy'/dt as... I'll switch to latex now to make sure it comes out right...

    [tex]\frac{d^2 y}{dx dt}[/tex]

    ...this?
     
  2. jcsd
  3. May 19, 2010 #2

    Mark44

    Staff: Mentor

    I'll take a stab at what I think you're trying to say.

    You have a curve defined by the parametric equation x = x(t) and y = y(t). Assuming that these functions are differentiable, we can talk about
    [tex]\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

    Then
    [tex]\frac{d^2y}{dx^2} = \frac{d}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

    [tex]= \frac{d}{dt} \left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right) \frac{dx}{dt}[/tex]
     
  4. May 19, 2010 #3
    Actually, I think, by the chain rule, it should be

    [tex]\frac{d}{dx} \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{d}{dt} \left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right) \frac{dt}{dx}[/tex]
     
  5. May 19, 2010 #4

    Mark44

    Staff: Mentor

    You are correct, Unit.
     
  6. May 19, 2010 #5

    Char. Limit

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    Gold Member

    Actually, I'm just wondering if d/dt(dy/dx) can be represented by the notation in the OP in laTEX. It seems like an odd notation, with mixed total derivatives or something.
     
  7. May 19, 2010 #6

    Mark44

    Staff: Mentor

    AFAIK, no. For partials, yes, but I don't recall ever seeing ordinary derivatives stacked up like that.
     
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