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Math Help - partial derivative question

  1. #1
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    partial derivative question

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

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

    Answer key states in the first step that:

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

    not sure how they got \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:  \sec^{2} \phi \  \frac{d \phi}{d R} = - \frac{X_{L}}{R^{2}}

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

    so   \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; April 25th 2010 at 10:39 PM.
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  3. #3
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    edit: when you substitute \tan^2 \phi for \frac {X_L}{R} the whole fraction becomes squared, however, in the answer just the R in the denominator is squared, not the 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 \tan^2 \phi for \frac {X_L}{R} the whole fraction becomes squared, however, in the answer just the R in the denominator is squared, not the 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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