Thomas' Calculus 13th Edition

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

Chapter 16: Integrals and Vector Fields - Section 16.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises 16.2 - Page 955: 3

Answer

$\nabla g(x,y,z)=-[\dfrac{2xi}{(x^2+y^2)}+ \dfrac{2yj}{(x^2+y^2)}]+e^z k$

Work Step by Step

Re-write as: $ g(x,y,z)=e^{z}-\ln (x^2+y^2)$ Thus, the gradient field can be computed as $\nabla g(x,y,z)=e^{z}-\ln (x^2+y^2)=\dfrac{1}{2} [\dfrac{2x}{(x^2+y^2+z^2)}i +\dfrac{2y}{(x^2+y^2+z^2)} j +\dfrac{2z}{(x^2+y^2+z^2)} k ]$ or, $\nabla g(x,y,z)=-\dfrac{2xi}{(x^2+y^2)}+ [-\dfrac{2yj}{(x^2+y^2)}]+e^z k$ or, $\nabla g(x,y,z)=-[\dfrac{2xi}{(x^2+y^2)}+ \dfrac{2yj}{(x^2+y^2)}]+e^z k$
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