Question #214761

Show that line integral is independent of path by finding a potential function for F

F(x,y) = (sin x − x sin y) j + (cos y + y cos x) i


1
Expert's answer
2021-07-09T07:00:53-0400

Let P(x,y)=cosy+ycosxP(x, y)=\cos y+y\cos x and Q(x,y)=sinxxsiny.Q(x, y)=\sin x-x\sin y. Then


Py=siny+cosx\dfrac{\partial P}{\partial y}=-\sin y+\cos x

Qx=cosxsiny\dfrac{\partial Q}{\partial x}=\cos x-\sin y

Py=cosxsiny=Qx\dfrac{\partial P}{\partial y}=\cos x-\sin y=\dfrac{\partial Q}{\partial x}

Then F(x,y)F(x, y) is conservative.


f,f=F\exist f , \nabla f=\vec F


fx=P(x,y)=cosy+ycosx\dfrac{\partial f}{\partial x}=P(x, y)=\cos y+y\cos x

fy=Q(x,y)=sinxxsiny\dfrac{\partial f}{\partial y}=Q(x, y)=\sin x-x\sin y

f=(cosy+ycosx)dx+φ(y)f=\int(\cos y+y\cos x)dx+\varphi(y)

=xcosy+ysinx+φ(y)=x\cos y+y\sin x+\varphi(y)

fy=xsiny+sinx+φ(y)=sinxxsiny\dfrac{\partial f}{\partial y}=-x\sin y+\sin x+\varphi'(y)=\sin x-x\sin y

=>φ(y)=0=>φ(y)=C=>\varphi'(y)=0=>\varphi(y)=C

f(x,y)=xcosy+ysinx+Cf(x, y)=x\cos y+y\sin x+C

Therefore the line integral is independent of path.



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