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Please help me with parametric derivatives

  1. Sep 4, 2012 #1
    I'm given that x=cos3θ and that y=sin3θ

    if (d2y/dx2)=[(dy/dθ)/(dx/dθ)]/[dx/dθ] is right, wouldn´t the second derivative of the parametric be:

    1/3c3θ ??


    I got this by using dy/dθ=3sin2θ,

    and dx/dθ=-3cos2θsinθ




    any idea what's wrong? or is it right?
     
  2. jcsd
  3. Sep 5, 2012 #2
    I'm pretty sure this isn't correct. you need to wrap a (d/dθ) around the first term ([(dy/dθ)/(dx/dθ)]), which gives you
    ((d/dθ)[(dy/dθ)/(dx/dθ)])/[dx/dθ]
    =((d/dθ)[(dy/dθ)/(dx/dθ)])/[1/(dx/dθ)]
    =[(dx/dθ)*(d^2y/dθ^2)-(d^2x/dθ^2)(dy/dθ)]/[(d^3x/dθ^3)]
    or (x' y'' - y' x'')/y'''
    where ' means derivative wrt θ
     
  4. Sep 5, 2012 #3
    [tex]\frac {d^2y} {dx^2} = \frac {d \frac {dy}{dx} } {dx}
    = \frac { d\frac {dy}{dx} } {d\theta} \frac {d\theta} {dx}
    = \frac { d (\frac {dy}{d\theta} \frac {d\theta} {dx})} {d\theta} \frac {d\theta} {dx}
    = \frac { d [\frac {dy}{d\theta} / \frac {dx} {d\theta}]} {d\theta} / \frac {dx} {d\theta}
    = \frac { d \frac {y'} {x'}} {d\theta} / x'
    = \frac {y''x' - y'x''}{x'^3}
    [/tex]
     
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