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 16.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises 16.3 - Page 968: 34


$\nabla g=\nabla F=F(proved)$

Work Step by Step

We have $ F=\nabla F $ and $\int_{0,0,0}^{x,y,z} F \cdot \space dr=\int_{0,0,0}^{x,y,z} \nabla \space F \space dr $ $\implies dr=f(x,y,z)-f(0,0,0)$ Since, $\dfrac{\partial g}{\partial x}=\dfrac{\partial f}{\partial x}-0,\dfrac{\partial f}{\partial y}=0$ This implies that $\nabla g=\nabla F=F(proved)$
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