University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 13 - Section 13.1 - Functions of Several Variables - Exercises - Page 683: 63



Work Step by Step

Remember that the level curve for $f(x,y,z)=\sqrt {x^2+y^2+z^2}$ has the form of $c=\sqrt {x^2+y^2+z^2}$ ...(1) Here, we have $x=1,y=-1,z=\sqrt 2$ Then $c=\sqrt {(1)2+(-1)^2+(\sqrt 2)^2} \implies c=2 $ Thus, equation (1), becomes: $2=\sqrt {x^2+y^2+z^2}$ Hence, $4=x^2+y^2+z^2$
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