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Parametric derivatives

  1. Jan 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Given a pair of parametric equations,
    ##x=f(t)## and ##y=g(t)## ,
    The first derivative is given by
    ##\frac{dy}{dx}=\frac{g'(t)}{f'(t)}##
    and the second derivative is actually
    ##\frac{d}{dt}(\frac{dy}{dx})##
    But why we cannot find the second derivative of a parametric equation by doing this?
    ##\frac{d^2 y}{dx^2}=\frac{g''(t}{f''(t)}##

    2. Relevant equations

    Parametric derivatives

    3. The attempt at a solution
    I know the parametric derivatives can be done by treating
    ##\frac{dy}{dt}## and ##\frac{dx}{dt}## as a fraction, but why can't we do that to second derivatives?
     
  2. jcsd
  3. Jan 6, 2014 #2

    vanhees71

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    Start with deriving the correct equation
    [tex]\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{g'(t)}{f'(t)}.[/tex]
    Then you should get an idea, how to get the 2nd derivative.
     
  4. Jan 6, 2014 #3

    tiny-tim

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    hi sooyong94! :smile:
    you mean ##\frac{d}{dx}\left(\frac{dy}{dx}\right)## :wink:
    no, that's dv/du where u(t) = f'(t), v(t) = g'(t)
     
  5. Jan 6, 2014 #4
    I don't get it though... I known I can use quotient rule to find the second derivative...
     
  6. Jan 6, 2014 #5

    Curious3141

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    Because you're not really treating them as fractions. You're using Chain Rule. World of difference.
     
  7. Jan 6, 2014 #6

    tiny-tim

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    yes, chain rule: dy/dx = dy/dt*dt/dx = dy/dt*(1/(dx/dt)) = g'(t)/f'(t) …

    nothing to do with quotients :wink:
     
  8. Jan 6, 2014 #7
    I got it right now.... So it's about chain rule, not treating them as fractions.
     
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