## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 13 - Section 13.1 - Functions of Several Variables - Exercises - Page 683: 62

#### Answer

$(x^2+y+z^2)=4$

#### Work Step by Step

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

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