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 967: 27


$f(x,y) =\dfrac{x^2}{y} -\dfrac{1}{y}+C$

Work Step by Step

We have:$M=\dfrac{2x}{y}; N=\dfrac{1-x^2}{y^2}; P=0$ and $\dfrac{\partial N}{\partial z}=\dfrac{\partial P}{\partial y}=0\\ \dfrac{\partial M}{\partial z}=\dfrac{\partial P}{\partial x}=0 \\ \dfrac{\partial N}{\partial x}=\dfrac{\partial M}{\partial y}=\dfrac{-2x}{y^2}$ This implies that $F$ is conservative, so $F =\nabla f$ Thus, $ \dfrac{\partial f}{\partial x}=\dfrac{2x}{y} $ $\implies f =\dfrac{x^2}{y} +k (y) ...(1)$ $\dfrac{\partial f}{\partial y}=\dfrac{1-x^2}{y}=\dfrac{-x^2}{y}+\dfrac{dk}{dy}$ $\implies \dfrac{ dk}{dy}=\dfrac{1}{y^2} \\k(y) =\dfrac{-1}{y} +C ...(2)$ From equations (1) and (2), we have: $f(x,y) =\dfrac{x^2}{y} -\dfrac{1}{y}+C$
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