Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Practice Exercises - Page 864: 24


$-(\dfrac{1}{2r^2l}) \sqrt{\dfrac{T}{\pi \omega}}, -(\dfrac{1}{2rl^2}) \sqrt{\dfrac{T}{\pi \omega}}, \dfrac{1}{4rl\sqrt{T\pi \omega}}, -\dfrac{\sqrt T}{(4rl \omega)\sqrt{r \omega}}$

Work Step by Step

$\dfrac{\partial f}{\partial r}=f_r=-(\dfrac{1}{2r^2l}) \sqrt{\dfrac{T}{\pi \omega}}$; $\dfrac{\partial f}{\partial l}==f_l=-(\dfrac{1}{2rl^2})\sqrt{\dfrac{T}{\pi \omega}}$; $\dfrac{\partial f}{\partial T}=f_T=\dfrac{1}{4rl\sqrt{T\pi \omega}}$; Thus, $\dfrac{\partial f}{\partial \omega}=f_{\omega}=\dfrac{1}{2rl}\sqrt {\dfrac{T}{\omega}}(-\dfrac{1}{2 \omega ^{-3/2}})=-\dfrac{\sqrt T}{4rl \omega\sqrt{r \omega}}$
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