Determine whether or not F is a conservative vector field. If it is, find a function f such that F=del f. F(x,y)=e^xsinyi+ e^xsinyj

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tnie54 · Accepted Answer

Step 1  Given that  F(x,y)=(y2−2x)i+2xyj  we need to  determine whether or not F is conservative vector field  Recall that if F(x,y)=P(x,y)i+Q(x,y)j, than in order to check the conservative vector field we need to chack whether  ∂P∂y=∂Q∂x  in the domian of F.Here P(x,y)=y2−2xandQ(x,y)=2xy.  Step 2  Observe that  ∂P∂y=2y  ∂Q∂x=2y.  Since on the domain of F, which is R2,farc{∂P}{∂y}=∂Q∂x. Note that R2 is simply connected and open and hence using Theorem 6, F is a concervative vector field and hence there exists a function f such that  F=△downf.  Our next goal is to determine the f. From the above relation, we have  F=triand≤downf⇒(y2−2x)i+2xyj=fx(x,y)i+fy(x,y)j  ⇒fx(x,y)=y2−2x,andfy(x,y)=2xy.

Determine whether or not F is a conservative vector field.

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