Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.3 - The Fundamental Theorem for Line Integrals - 16.3 Exercise - Page 1095: 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

Consider $P=f_x=\dfrac{\partial f}{\partial x};Q=f_y=\dfrac{\partial f}{\partial y}$ and $R=f_z=\dfrac{\partial f}{\partial z}$ a) $\dfrac{\partial P}{\partial y}=\dfrac{\partial}{\partial y}(\dfrac{\partial f}{\partial x})=\dfrac{\partial^2 f}{\partial x \partial y}=\dfrac{\partial Q}{\partial x}$ b) $\dfrac{\partial P}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})=\dfrac{\partial^2 f}{\partial x \partial z}=\dfrac{\partial R}{\partial x}$ c) $\dfrac{\partial Q}{\partial z}=\dfrac{\partial}{\partial z}(\dfrac{\partial f}{\partial y})=\dfrac{\partial^2 f}{\partial z \partial y}=\dfrac{\partial R}{\partial y}$ Hence, $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ $\dfrac{\partial P}{\partial z}=\dfrac{\partial R}{\partial x}$ $\dfrac{\partial Q}{\partial z}=\dfrac{\partial R}{\partial y}$
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