# Determine whether or not F is a conservative vector field.

Determine whether or not F is a conservative vector field. If it is, find a function f such that F=del f. F(x,y)=e^xsinyi+ e^xsinyj

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tnie54
Step 1
Given that
$$\displaystyle{F}{\left({x},{y}\right)}={\left({y}^{{{2}}}-{2}{x}\right)}{i}+{2}{x}{y}{j}$$
we need to
determine whether or not F is conservative vector field
Recall that if F(x,y)=P(x,y)i+Q(x,y)j, than in order to check the conservative vector field we need to chack whether
$$\displaystyle{\frac{{\partial{P}}}{{\partial{y}}}}={\frac{{\partial{Q}}}{{\partial{x}}}}$$
in the domian of F.Here $$\displaystyle{P}{\left({x},{y}\right)}={y}^{{{2}}}-{2}{x}\text{and}{Q}{\left({x},{y}\right)}={2}{x}{y}$$.
Step 2
Observe that
$$\displaystyle{\frac{{\partial{P}}}{{\partial{y}}}}={2}{y}$$
$$\displaystyle{\frac{{\partial{Q}}}{{\partial{x}}}}={2}{y}$$.
Since on the domain of F, which is $$\displaystyle{R}^{{{2}}},{f}{a}{r}{c}{\left\lbrace\partial{P}\right\rbrace}{\left\lbrace\partial{y}\right\rbrace}={\frac{{\partial{Q}}}{{\partial{x}}}}$$. Note that $$\displaystyle{R}^{{{2}}}$$ is simply connected and open and hence using Theorem 6, F is a concervative vector field and hence there exists a function f such that
$$\displaystyle{F}=\triangle{d}{o}{w}{n}{f}$$.
Our next goal is to determine the f. From the above relation, we have
$$\displaystyle{F}={t}{r}{i}{\quad\text{and}\quad}\le{d}{o}{w}{n}{f}\Rightarrow{\left({y}^{{{2}}}-{2}{x}\right)}{i}+{2}{x}{y}{j}={{f}_{{{x}}}{\left({x},{y}\right)}}{i}+{{f}_{{{y}}}{\left({x},{y}\right)}}{j}$$
$$\displaystyle\Rightarrow{{f}_{{{x}}}{\left({x},{y}\right)}}={y}^{{{2}}}-{2}{x},\text{and}{{f}_{{{y}}}{\left({x},{y}\right)}}={2}{x}{y}$$.