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Etermine whether or not f is a conservative vector field. if it is, find a function f such that f = ∇f. (if the vector field is not conservative, enter dne.) f(x, y) = ex cos(y)i + ex sin(y)j

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A vector field
\mathbf f(x,y) is conservative *if and only if* we can find a scalar function
f(x,y) for which
\\abla f=\mathbf f, which means we're looking for a function
f(x,y) such that


(\partial f)/(\partial x)=e^x\cos y

(\partial f)/(\partial y)=e^x\sin y

Integrating both sides of the first PDE with respect to
x, we get


f(x,y)=e^x\cos y+g(y)

We're assuming the "constant" of integration is some expression independent of
x. However, upon differentiating with respect to
y, we have


(\partial f)/(\partial y)=-e^(\sin y)+(\mathrm dg)/(\mathrm dy)=e^x\sin y

\implies(\mathrm dg)/(\mathrm dy)=2e^x\sin y

which means
g is *not* a function of
y alone, contradicting the assumption of the contrary. So our desired
f(x,y) does not exist, and therefore
\mathbf f(x,y) is *not* conservative.
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