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Derivation of higher deratives of parametric equations

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    so dy/dx = y dot / x dot

    and d^2y/dx^2 = d/dx(y dot/x dot)

    can someone please show me the steps to get

    d^2y/dx^2 = (x dot y dot dot - y dot x dot dot)/ (x dot)^ 2

    I've been trying to get to it for the last half an hour and failing

  2. jcsd
  3. Apr 6, 2010 #2


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    so you have
    [tex]\frac{dy}{dx} = \frac{y'}}{x'}} = y'(x')^{-1} [/tex]

    now try differntiating using the product & chain rule
    (i used dashes instead of dots, but same thing)
  4. Apr 6, 2010 #3
    sorry but I don't see where you got x' = y'(x')^-1 from

  5. Apr 6, 2010 #4


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    The product rule and chain rule need to be used here. Use the product rule first then use the chainrule. Example [itex]\frac{d}{dx} \frac{1}{y}=\frac{d}{dy}\left(\frac{1}{y}\right)\frac{dy}{dx}[/itex].

    The first step is using the product rule on:

    [tex]\frac{d}{dx} \frac{\dot{y}}{\dot{x}}[/tex].

    Can you post the results?
    Last edited: Apr 6, 2010
  6. Apr 6, 2010 #5
    That isn't what he meant: he was writing out the equation and it started on a new line.

    What he meant was:

    [tex]\frac{dy}{dx} = \frac{y'}{x'} = y' \cdot (x')^{-1}[/tex]

    and if you ask me you needn't do that step and head straight on to the quotient rule: [tex]\frac{vdu - udv}{v^{2}}[/tex] or [tex]\frac{vu'-uv'}{v^{2}}[/tex] where [tex]y' = u \ ; \ x' = v[/tex]

    Just to expand on this:

    [tex]\frac{du}{dx}=y''[/tex] or [tex]u'=y''[/tex]

    [tex]\frac{dv}{dx}=x''[/tex] or [tex]v'=x''[/tex]

    both give you the same result(duh) and there is not extra calculation needed.
    Last edited: Apr 6, 2010
  7. Apr 6, 2010 #6
    I hope this has solved your problems.=]
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