Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 16 - Vector Calculus - 16.3 Exercises - Page 1107: 29

Answer

$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ $\dfrac{\partial P}{\partial z}=\dfrac{\partial R}{\partial x}$ $\dfrac{\partial Q}{\partial z}=\dfrac{\partial R}{\partial y}$

Work Step by Step

Here, we have $P=f_x=\dfrac{\partial f}{\partial x};Q=f_y=\dfrac{\partial f}{\partial y}$ and $R=f_z=\dfrac{\partial f}{\partial z}$ Now, $\dfrac{\partial P}{\partial y}=\dfrac{\partial}{\partial y}(\dfrac{\partial f}{\partial x})=\dfrac{\partial^2 f}{\partial x \partial y}=\dfrac{\partial Q}{\partial x}$ Thus, we have $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ Now, $\dfrac{\partial P}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})=\dfrac{\partial^2 f}{\partial x \partial z}=\dfrac{\partial R}{\partial x}$ Thus, we have $\dfrac{\partial P}{\partial z}=\dfrac{\partial R}{\partial x}$ Now, $\dfrac{\partial Q}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})=\dfrac{\partial^2 f}{\partial z \partial y}=\dfrac{\partial R}{\partial y}$ Thus, we have $\dfrac{\partial Q}{\partial z}=\dfrac{\partial R}{\partial y}$
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