4. [0/1 Points ]DETAILSMY NOTESSCALCET916.3.007.Determine whether or not F is a coliservative vector fleld. If it is, find a function f such that \mathrm{F}=\(\newlineabla f \). (If the vector field is not conservative, enter DNE.)F(x,y)=(yex+sin(y))i+(ex+xcos(y))jf(x,y)=
Q. 4. [0/1 Points ]DETAILSMY NOTESSCALCET916.3.007.Determine whether or not F is a coliservative vector fleld. If it is, find a function f such that \mathrm{F}=\(\newlineabla f \). (If the vector field is not conservative, enter DNE.)F(x,y)=(yex+sin(y))i+(ex+xcos(y))jf(x,y)=
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(x,y)=(yex+sin(y))i+(ex+xcos(y))j. Calculate the partial derivatives: ∂y∂ of the first component: ∂y∂(yex+sin(y))=ex+cos(y). ∂x∂ of the second component: ∂x∂(ex+xcos(y))=ex+cos(y).
Verify Conservative Field: Since the partial derivatives are equal, the curl of F is zero, indicating that F is a conservative vector field.
Find Potential Function: Now, find a potential function f such that F = \(\newlineabla f\). Start by integrating the first component of F with respect to x:f(x,y)=∫(yex+sin(y))dx=yex+xsin(y)+g(y), where g(y) is a function of y only.
Differentiate with Respect to y: Next, differentiate this potential function with respect to y to match the second component of F:∂y∂f=ex+xcos(y)+g′(y).Since ∂y∂f should equal the second component of F, which is ex+xcos(y), we set g′(y)=0.
Integrate g′(y)=0: Integrate g′(y)=0 to find g(y):g(y)=C, where C is a constant.
Substitute g(y)=C: Substitute g(y)=C into the potential function:f(x,y)=yex+xsin(y)+C.Since the constant C does not affect the gradient, we can set C=0 for simplicity.
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