University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Practice Exercises - Page 749: 24

Answer

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

Work Step by Step

$\dfrac{\partial f}{\partial r}=-\dfrac{1}{2r^2l} \sqrt{\dfrac{T}{\pi \omega}}$ and $\dfrac{\partial f}{\partial l}=-\dfrac{1}{2rl^2} \sqrt{\dfrac{T}{\pi \omega}}$ and $\dfrac{\partial f}{\partial T}=\dfrac{1}{4rl\sqrt{T\pi \omega}}$ and $\dfrac{\partial f}{\partial \omega}=\dfrac{1}{2rl}\sqrt {\dfrac{T}{\pi}}(-1/2 \omega ^{-3/2})=-\dfrac{\sqrt T}{4rl \omega\sqrt{\pi \omega}}$ Hence, $-\dfrac{1}{2r^2l} \sqrt{\dfrac{T}{\pi \omega}}, -\dfrac{1}{2rl^2} \sqrt{\dfrac{T}{\pi \omega}}, \dfrac{1}{4rl\sqrt{T\pi \omega}}, -\dfrac{\sqrt T}{4rl \omega\sqrt{\pi \omega}}$
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