\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\pa}{\partial}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\begin{document}
\parindent=0pt

\begin{center}
\textbf{Vector Fields}

\textit{\textbf{Conservative Fields}}
\end{center}

\textbf{Question}

For the following vector field, find whether it is conservative. If
so, find a corresponding potential

$\ds \un{F}(x,y) = \frac{x \un{i} + y \un{j}}{x^2+y^2}$


\textbf{Answer}

\begin{eqnarray*}
F_1 & = & \frac{x}{x^2+y^2}\\
F_2 & = & \frac{y}{x^2+y^2}\\
\Rightarrow \frac{pa F_1}{\pa y} & = & -\frac{2xy}{(x^2+y^2)^2} =
\frac{\pa F_2}{\pa x}.
\end{eqnarray*}
$\Rightarrow \un{F}$ can be conservative.

If $\un{F} = \nabla \phi$
\begin{eqnarray*}
\Rightarrow \frac{\pa \phi}{\pa x} & = & \frac{x}{x^2+y^2}\\
\frac{\pa \phi}{\pa y} & = & \frac{y}{x^2+y^2}\\
\Rightarrow \phi(x,y) & = & \int \frac{x}{x^2+y^2} \,dx\\
& = & \frac{\ln (x^2+y^2)}{2} + C_1(y)\\
\frac{y}{x^2+y^2} & = & \frac{\pa \phi}{\pa y} = \frac{y}{x^2+y^2}
+c_1'(y)\\
\Rightarrow c_1'(y) & = & 0
\end{eqnarray*}
So choose $C_1{y}=0$
$$\Rightarrow \phi(x,y) = \frac{1}{2} \ln (x^2+y^2)$$
is a scalar potential for $\un{F}$, with $\un{F}$ being conservative
everywhere on $\Re^2$ except for the origin. 

\end{document}




















