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 788: 62



Work Step by Step

Here, the level curve for $f(x,y,z)=\ln (x^2+y+z^2)$ has the form of $c=\ln (x^2+y+z^2)$ Also, $x=-1;y=2;z=1$ Then $c=\ln [(-1)^2+(2)+(1)^2]$ or,$ c= \ln 4 $ Now, $c=\ln (x^2+y+z^2) \implies \ln 4=\ln (x^2+y+z^2)$ or, $e^{(\ln 4)}=e^{[\ln (x^2+y+z^2)]}$ Hence, $(x^2+y+z^2)=4$
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