6.3 Conservative vector fields  (Page 6/16)

Suppose that $f(x,y)$ is a potential function for F . Then, $\text{∇}f=\text{F},$ and therefore

Integrating the equation $f_x=2x{y}^3$ with respect to x yields the equation

Notice that since we are integrating a two-variable function with respect to x , we must add a constant of integration that is a constant with respect to x , but may still be a function of y . The equation $f(x,y)=x^2{y}^3+h(y)$ can be confirmed by taking the partial derivative with respect to x :

Since $f$ is a potential function for F ,

and therefore

This implies that $h\prime (y)=\text{cos}y,$ so $h(y)=\text{sin}y+C.$ Therefore, any function of the form $f\left(x,y\right)=x^2{y}^3+\text{sin}\left(y\right)+C$ is a potential function. Taking, in particular, $C=0$ gives the potential function $f\left(x,y\right)=x^2{y}^3+\text{sin}\left(y\right).$

To verify that $f$ is a potential function, note that $\text{∇}f=⟨2x{y}^3,3x^2{y}^2+\text{cos}y⟩=\text{F}.$

Suppose that $f$ is a potential function. Then, $\text{∇}f=\text{F}$ and therefore $f_x=2xy.$ Integrating this equation with respect to x yields the equation $f\left(x,y,z\right)=x^{2}y+g\left(y,z\right)$ for some function g . Notice that, in this case, the constant of integration with respect to x is a function of y and z .

Since $f$ is a potential function,

Therefore,

Integrating this function with respect to y yields

for some function $h\left(z\right)$ of z alone. (Notice that, because we know that g is a function of only y and z , we do not need to write $g\left(y,z\right)=y^2{z}^3+h\left(x,z\right).)$ Therefore,

To find $f$ , we now must only find h . Since $f$ is a potential function,

This implies that $h^{\prime }\left(z\right)=2z,$ so $h\left(z\right)=z^2+C.$ Letting $C=0$ gives the potential function

To verify that $f$ is a potential function, note that $\text{∇}f=⟨2xy,x^2+2y{z}^3,3y^2{z}^2+2z⟩=\text{F}.$