For each of the following vector fields
F, decide whether it is conservative or not by computing the appropriate first order partial derivatives. Type in a potentialfunction f (that is, ∇f = F) with f(0,0)=0. If it is not conservative, type N.
A. F(x,y)=(−16x+2y)i+(2x+10y) j f(x,y)= _____
B. F(x,y)=−8yi−7xj f(x,y)=_____
C. F(x,y)=(−8sin y)i+(4y−8xcosy)j f(x,y)=_____

Respuesta :

(A)

[tex]\dfrac{\partial f}{\partial x}=-16x+2y[/tex]

[tex]\implies f(x,y)=-8x^2+2xy+g(y)[/tex]

[tex]\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y[/tex]

[tex]\implies\dfrac{\mathrm dg}{\mathrm dy}=10y[/tex]

[tex]\implies g(y)=5y^2+C[/tex]

[tex]\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}[/tex]

(B)

[tex]\dfrac{\partial f}{\partial x}=-8y[/tex]

[tex]\implies f(x,y)=-8xy+g(y)[/tex]

[tex]\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x[/tex]

[tex]\implies \dfrac{\mathrm dg}{\mathrm dy}=x[/tex]

But we assume [tex]g(y)[/tex] is a function of [tex]y[/tex] alone, so there is not potential function here.

(C)

[tex]\dfrac{\partial f}{\partial x}=-8\sin y[/tex]

[tex]\implies f(x,y)=-8x\sin y+g(x,y)[/tex]

[tex]\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y[/tex]

[tex]\implies\dfrac{\mathrm dg}{\mathrm dy}=4y[/tex]

[tex]\implies g(y)=2y^2+C[/tex]

[tex]\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}[/tex]

For (A) and (C), we have [tex]f(0,0)=0[/tex], which makes [tex]C=0[/tex] for both.

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