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Thread: partial derivative question

  1. #1
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    partial derivative question

    seems pretty simple but the $\displaystyle tan(\phi)$ on the left side threw me off

    $\displaystyle tan(\phi) = \frac {X_L}{R}$ ; find $\displaystyle \frac {\partial \phi}{\partial R}$

    Answer key states in the first step that:

    $\displaystyle \frac {\partial \phi}{\partial R}= \frac {1}{(\frac {X_L}{R^2}) + 1} * \frac {-X_L}{R^2}$

    not sure how they got $\displaystyle \frac {1}{(\frac {X_L}{R^2}) + 1} $
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  2. #2
    Super Member Random Variable's Avatar
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    EDIT: $\displaystyle \sec^{2} \phi \ \frac{d \phi}{d R} = - \frac{X_{L}}{R^{2}}$

    but $\displaystyle \sec^{2} \phi = 1+\tan^{2}\phi $

    so $\displaystyle \frac{d \phi}{d R} = - \frac{1}{1+\tan^{2} \phi}\frac{X_{L}}{R^{2}} = - \frac{1}{1+(\frac{X_{L}}{R})^{2}} \frac{X_{L}}{R^{2}} $
    Last edited by Random Variable; Apr 25th 2010 at 09:39 PM.
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  3. #3
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    edit: when you substitute $\displaystyle \tan^2 \phi$ for $\displaystyle \frac {X_L}{R}$ the whole fraction becomes squared, however, in the answer just the $\displaystyle R$ in the denominator is squared, not the $\displaystyle X_L$
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by dben View Post
    edit: when you substitute $\displaystyle \tan^2 \phi$ for $\displaystyle \frac {X_L}{R}$ the whole fraction becomes squared, however, in the answer just the $\displaystyle R$ in the denominator is squared, not the $\displaystyle X_L$
    It has to be a misprint.
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  5. #5
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    yea, i put it down for the day and looked at it again tonight and realized it is a misprint. The final answer that is printed only works if the whole fraction is squared...Thanks!

    dave
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