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Determine whether or not
F is a conservative vector field. If it is, find a function
f such that
F=grad f. (If the vector field is not conservative, enter DNE.

f(x,y)=◻(x,y)=e^(x)sin(y)i+e^(x)cos(y)j

Determine whether or not $\mathrm{F}$ is a conservative vector field. If it is, find a function $f$ such that \mathrm{F}=\(\newlineabla f \). (If the vector field is not conservative, enter DNE. $\newline$ $f(x, y)=\square(x, y)=e^{x} \sin (y) \mathbf{i}+e^{x} \cos (y) \mathbf{j}$

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Write equations of cosine functions using properties

Full solution

Q. Determine whether or not

\mathrm{F}

is a conservative vector field. If it is, find a function

f

such that \mathrm{F}=\(\newlineabla f \). (If the vector field is not conservative, enter DNE.

\newline

f(x, y)=\square(x, y)=e^{x} \sin (y) \mathbf{i}+e^{x} \cos (y) \mathbf{j}

Check Curl of F: To check if $F$ is conservative, we need to see if the curl of $F$ is zero. The vector field $F$ is given by $F = e^x \sin(y)i + e^x \cos(y)j$ .
Compute Partial Derivatives: Compute the partial derivatives: $\newline$ $\frac{\partial}{\partial y} (e^x \sin(y)) = e^x \cos(y)$ , $\newline$ $\frac{\partial}{\partial x} (e^x \cos(y)) = e^x \cos(y)$ .
Verify Conservativity: Since the partial derivative of the $i$ component with respect to $y$ equals the partial derivative of the $j$ component with respect to $x$ , the curl of $extbf{F}$ is zero. This means $extbf{F}$ is conservative.
Find Potential Function: Now, find a potential function $f$ such that \mathbf{F} = \(\newlineabla f\). Start by integrating $e^x \sin(y)$ with respect to $x$ to find $f$ : $f(x, y) = \int e^x \sin(y) \, dx = e^x \sin(y) + g(y)$ , where $g(y)$ is a function of $y$ only.
Integrate with Respect to x: Next, differentiate $f(x, y) = e^x \sin(y) + g(y)$ with respect to $y$ to match it with the $j$ component of $F$ : $\frac{\partial}{\partial y} [e^x \sin(y) + g(y)] = e^x \cos(y) + g'(y)$ .
Differentiate with Respect to y: Since the j component of F is $e^x \cos(y)$ , we set $g'(y) = 0$ . This implies $g(y)$ is a constant, which we can take as $0$ without loss of generality.
Set $g'(y) = 0$ : Thus, the potential function $f$ is $f(x, y) = e^x \sin(y)$ .

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