partial derivative question

**dben** · April 25th 2010, 09:16 PM

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 $\frac {1}{(\frac {X_L}{R^2}) + 1}$

**Random Variable** · April 25th 2010, 09:26 PM

EDIT: $\sec^{2} \phi \ \frac{d \phi}{d R} = - \frac{X_{L}}{R^{2}}$

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

so $\frac{d \phi}{d R} = - \frac{1}{1+\tan^{2} \phi}\frac{X_{L}}{R^{2}} = - \frac{1}{1+(\frac{X_{L}}{R})^{2}} \frac{X_{L}}{R^{2}}$

**dben** · April 25th 2010, 09:58 PM

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 $X_L$

**Random Variable** · April 26th 2010, 09:58 PM

Originally Posted by dben

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.

**dben** · April 26th 2010, 10:37 PM

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

Thread: partial derivative question

LinkBack

Thread Tools

Display