Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.1 Vector Fields - Exercises - Page 919: 38

Answer

By inspection, a potential function for ${\bf{F}}$ is $f = \frac{1}{2}{x^2}$. We prove that ${\bf{G}} = \left( {y,0} \right)$ is not conservative.

Work Step by Step

1. We have the vector field: ${\bf{F}} = \left( {x,0} \right)$. Let $f$ be a potential function for ${\bf{F}}$. Then, $\nabla f = {\bf{F}} = \left( {x,0} \right)$. Since $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right)$, so $\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right) = \left( {x,0} \right)$ $\frac{{\partial f}}{{\partial x}} = x$, ${\ \ \ \ \ }$ $\frac{{\partial f}}{{\partial y}} = 0$ So, $f = \frac{1}{2}{x^2}$. 2. We have the vector field: ${\bf{G}} = \left( {y,0} \right)$. Write ${\bf{G}} = \left( {{G_1},{G_2}} \right) = \left( {y,0} \right)$. Evaluate: $\frac{{\partial {G_1}}}{{\partial y}} = 1$, ${\ \ \ \ \ }$ $\frac{{\partial {G_2}}}{{\partial x}} = 0$ Since $\frac{{\partial {G_1}}}{{\partial y}} \ne \frac{{\partial {G_2}}}{{\partial x}}$, by Theorem 1, ${\bf{G}} = \left( {y,0} \right)$ is not conservative.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.