Question #283402

Determine whether the following vector fields are conservative on R 2 . If the vector field is conservative then find a potential function. F =<ex cos y, −ex sin y>.


1
Expert's answer
2021-12-29T13:19:38-0500
F=<excosy,exsiny>\vec F =<e^x \cos y, −e^x \sin y>

Py=exsiny,Qx=exsiny\dfrac{\partial P}{\partial y}=-e^x\sin y, \dfrac{\partial Q}{\partial x}=-e^x\sin y

Py=exsiny=Qx\dfrac{\partial P}{\partial y}=-e^x\sin y=\dfrac{\partial Q}{\partial x}

F\vec F satisfies the condition Py=Qx.P_y=Q_x. Moreover, it is defined on all of R2,R^2, hence it is conservative.

Let us find a potential function f(x,y)f(x, y) for F.\vec F . We want


fx=P=excosyf_x=P=e^x\cos y

fy=Q=exsinyf_y=Q=-e^x\sin y

f=Pdx=excosydx=excosy+g(y)f=\int Pdx=\int e^x\cos ydx=e^x\cos y+g(y)

fy=exsiny+g(y)f_y=-e^x\sin y+g'(y)

Then


exsiny+g(y)=exsiny-e^x\sin y+g'(y)=-e^x\sin y

=>g(y)=0=>g(y)=C=>g'(y)=0=>g(y)=C

The potential function is


f(x,y)=excosy+Cf(x, y)=e^x\cos y+C

Choosing the constant С=0,С = 0, we obtain the potential function


f(x,y)=excosyf(x, y)=e^x\cos y


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