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)=(ye^(x)+sin(y))i+(e^(x)+x cos(y))j

Check for Conservative: To check if F is conservative, we need to verify if the curl of F is zero. The vector field F is given by F = (y e^x + sin(y))i + (e^x + x cos(y))j. Calculate the partial derivatives: ∂/∂y of (y e^x + sin(y)) = e^x + cos(y), ∂/∂x of (e^x + x cos(y)) = e^x - x sin(y).Calculate Partial Derivatives: Compare the partial derivatives: ∂/∂y of the i-component = e^x + cos(y), ∂/∂x of the j-component = e^x - x sin(y). Since these are not equal, the curl of F is not zero.Compare Partial Derivatives: Since the curl of F is not zero, F is not a conservative vector field.

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

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Write equations of cosine functions using properties

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Q. Determine whether or not

\newline

F

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

\newline

f

such that

\newline

F= f

. (If the vector field is not conservative, enter DNE.)

\newline

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

Check for Conservative: To check if $F$ is conservative, we need to verify if the curl of $F$ is zero. The vector field $F$ is given by $F = (y e^x + \sin(y))\mathbf{i} + (e^x + x \cos(y))\mathbf{j}$ . Calculate the partial derivatives: $\partial/\partial y$ of $(y e^x + \sin(y)) = e^x + \cos(y)$ , $\partial/\partial x$ of $(e^x + x \cos(y)) = e^x - x \sin(y)$ .
Calculate Partial Derivatives: Compare the partial derivatives: $\frac{\partial}{\partial y}$ of the $i$ -component = $e^x + \cos(y)$ , $\frac{\partial}{\partial x}$ of the $j$ -component = $e^x - x \sin(y)$ . Since these are not equal, the curl of $F$ is not zero.
Compare Partial Derivatives: Since the curl of $\mathbf{F}$ is not zero, $\mathbf{F}$ is not a conservative vector field.