Problem on determining if a vector field is conservative
$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
Let $\mathbf{F}(x,y) = 2 x \log y \ \mathbf{i} + x^2 / y \ \mathbf{j}$. Is $\mathbf{F}(x,y)$ conservative?
Solution
Recall that
Determining if a vector field is conservative
$\mathbf{F}(x,y) = u(x,y) \mathbf{i} + v(x,y) \mathbf{j}$ is conservative if and only if
$$\partial_y u(x,y) = \partial_x v(x,y).$$
We identify $u(x,y) = 2 x \log y$ and $v(x,y) = x^2 / y$.
Partial derivatives
Computing the partial derivatives, we get \begin{align}
\partial_y u(x,y) &= 2x / y, \\
\partial_x v(x,y) &= 2x / y.
\end{align}
Because $\partial_y u(x,y) = \partial_x v(x,y)$, $\mathbf{F}(x,y)$ is conservative.
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