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: 63



Work Step by Step

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