Thomas' Calculus 13th Edition

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

Chapter 14: Partial Derivatives - Section 14.1 - Functions of Several Variables - Exercises 14.1 - Page 789: 64



Work Step by Step

Here, the level curve for $f(x,y,z)=\dfrac{x-y+z}{2x+y-z}$ has the form of $c=\dfrac{x-y+z}{2x+y-z}$ As we are given that $x=1,y=0,z=-2$ Then $c=\dfrac{1-0+(-2)}{2+0-(-2)}$ or, $ c=\dfrac{1-0-2}{2+0+2}=-\dfrac{1}{4} $ Now, $c=\dfrac{x-y+z}{2x+y-z} \implies -(\dfrac{1}{4}) =\dfrac{x-y+z}{2x+y-z}$ or, $6x-3y+3z=0$ Hence, $2x-y+z=0$
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