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Big differential

  1. Oct 13, 2009 #1
    Big differential, just want to make sure I am doing correct thing here.

    a is the only changing dimension, r and l are constants


    I want to differentiate [tex]\frac{\delta\Theta}{\delta\ a}[/tex]

    So what I did was using

    [tex]\Theta=arccos a[/tex]

    [tex] cos\Theta=a[/tex] differentiate this

    [tex] -sin\Theta \frac{\delta\Theta}{\delta\ a}=1[/tex]

    therefore with trig identities and a rearrange:

    [tex]\frac{\delta\Theta}{\delta\ a}=\frac{-1}{\sqrt{1-a^{2}}}[/tex]

    where [tex]a = \left(\frac{r^{2}+\left(r+l-a\right)^{2}-l^{2}}{2r\left(r+l-a\right)}\right)[/tex]

    So the differential of theta with respect to a is

    [tex]\frac{\delta\Theta}{\delta\ a}=\frac{-a^{'}}{\sqrt{1-a^{2}}}[/tex]

    This involves the quotient rule, and I end up the expression below: I took out a factor of 4 top and bottom of the differential of a, hence the 3/2 coefficient

    [tex]\frac{\delta\Theta}{\delta\ a}= \frac{ra^{2}-\left(2r^{2}-rl\right)a + \left(2r^{3}+3lr^{2}+l^{2}r\right)}{ra^{3}-\frac{3}{2}r^{2}a^{2}+\left(2r^{3}+3lr^{2}+rl^{2}\right)a-\left(r^{4}-2r^{3}l-r^{2}l^{2}\right)}
    \sqrt{1-\left(\frac{a^{2}-2a\left(r-l\right)+\left(2r^{2}+2rl\right)}{-2ra + 2r^{2} +2rl}\right)[/tex]

    Last edited: Oct 13, 2009
  2. jcsd
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