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Determine whether or not \newlineFF is a conservative vector field. If it is, find a function \newlineff such that \newlineF=\(\newlineabla f\). (If the vector field is not conservative, enter DNE.)\newlinef(x,y)=(x,y)=(yex+sin(y))i+(ex+xcos(y))jf(x,y)=\square(x,y)=(ye^{x}+\sin(y))i+(e^{x}+x \cos(y))j

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Q. Determine whether or not \newlineFF is a conservative vector field. If it is, find a function \newlineff such that \newlineF=\(\newlineabla f\). (If the vector field is not conservative, enter DNE.)\newlinef(x,y)=(x,y)=(yex+sin(y))i+(ex+xcos(y))jf(x,y)=\square(x,y)=(ye^{x}+\sin(y))i+(e^{x}+x \cos(y))j
  1. Check for Conservative: To check if FF is conservative, we need to verify if the curl of FF is zero. The vector field FF is given by F=(yex+sin(y))i+(ex+xcos(y))jF = (y e^x + \sin(y))\mathbf{i} + (e^x + x \cos(y))\mathbf{j}. Calculate the partial derivatives: /y\partial/\partial y of (yex+sin(y))=ex+cos(y)(y e^x + \sin(y)) = e^x + \cos(y), /x\partial/\partial x of (ex+xcos(y))=exxsin(y)(e^x + x \cos(y)) = e^x - x \sin(y).
  2. Calculate Partial Derivatives: Compare the partial derivatives: y\frac{\partial}{\partial y} of the ii-component = ex+cos(y)e^x + \cos(y), x\frac{\partial}{\partial x} of the jj-component = exxsin(y)e^x - x \sin(y). Since these are not equal, the curl of FF is not zero.
  3. Compare Partial Derivatives: Since the curl of F\mathbf{F} is not zero, F\mathbf{F} is not a conservative vector field.

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