# Determine whether or not F is a conservative vector field. If it is, find a func

Determine whether or not F is a conservative vector field. If it is, find a function f such that F=delf. $$\displaystyle{F}{\left({x},{y}\right)}={\left({3}{x}^{{2}}-{2}{y}^{{2}}\right)}{i}+{\left({4}{x}{y}+{3}\right)}{j}$$

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davonliefI
The vector field $$\displaystyle{F}=\Pi+{Q}{j}$$ is convervative if and only if
$$\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}={\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}$$
The given vector field is $$\displaystyle{F}{\left({x},{y}\right)}={\left({3}{x}^{{2}}-{2}{y}^{{2}}\right)}{i}+{\left({4}{x}{y}+{3}\right)}{j}$$
Here $$\displaystyle{P}={3}{x}^{{2}}-{2}{y}^{{2}}$$, Therefore $$\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}=-{4}{y}$$
And $$\displaystyle{Q}={4}{x}{y}+{3}$$, Therefore $$\displaystyle{\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}={4}{y}$$
$$\displaystyle{\frac{{{d}{P}}}{{{\left.{d}{y}\right.}}}}\ne{\frac{{{d}{Q}}}{{{\left.{d}{x}\right.}}}}$$
Therefore F is NOT conservative.