Conservative Vector Fields and Potential Functions
The Fundamental Theorem for Line Integrals tells us that if [latex]\mathbf{F}[/latex] is conservative, then calculating [latex]\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}[/latex] has two steps: first, find a potential function [latex]f[/latex] for [latex]\mathbf{F}[/latex], and second, calculate [latex]f(P_1)-f(P_0)[/latex], where [latex]P_1[/latex] is the endpoint of [latex]C[/latex] and [latex]P_0[/latex] is the starting point.
To use this theorem, we need to answer a key question: Given a conservative vector field [latex]\mathbf{F}[/latex], how do we find a function [latex]f[/latex] such that [latex]\nabla f=\mathbf{F}[/latex]? Let’s start with an example to see how this works.
Find a potential function for [latex]{\bf{F}}(x,y)=\langle2xy^3,3x^2y^2+\cos{(y)}\rangle[/latex], thereby showing that [latex]{\bf{F}}[/latex] is conservative.
Suppose that [latex]f(x, y)[/latex] is a potential function for [latex]{\bf{F}}[/latex]. Then, [latex]\nabla{f}={\bf{F}}[/latex], and therefore
[latex]f_x=2xy^3\text{ and }f_y=3x^2y^2+\cos{y}[/latex].
Integrating the equation [latex]f_x=2xy^3[/latex] with respect to [latex]x[/latex] yields the equation
[latex]f(x,y)=x^2y^2+h(y)[/latex].
Notice that since we are integrating a two-variable function with respect to [latex]x[/latex], we must add a constant of integration that is a constant with respect to [latex]x[/latex], but may still be a function of [latex]y[/latex]. The equation [latex]f(x,y)=x^2y^2+h(y)[/latex] can be confirmed by taking the partial derivative with respect to [latex]x[/latex]:
This implies that [latex]h^\prime(y)=\cos{(y)}[/latex], so [latex]h(y)=\sin{y}+C[/latex]. Therefore, any function of the form [latex]f(x,y)=x^2y^3+\sin{(y)}+C[/latex] is a potential function. Taking, in particular, [latex]C=0[/latex] gives the potential function [latex]f(x,y)=x^2y^3+\sin{(y)}[/latex].
To verify that [latex]f[/latex] is a potential function, note that [latex]\nabla{f}=\langle2xy^3,3x^2y^2+\cos{y}\rangle={\bf{F}}[/latex].
The logic of the previous example extends to finding the potential function for any conservative vector field in [latex]\mathbb{R}^2[/latex]. Here’s the strategy:
How to: Find a Potential Function for a Conservative Vector Field [latex]\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle[/latex]
Integrate [latex]P[/latex] with respect to [latex]x[/latex]. This results in a function of the form [latex]g(x, y)+h(y)[/latex], where [latex]h(y)[/latex] is unknown.
Take the partial derivative of [latex]g(x, y)+h(y)[/latex] with respect to [latex]y[/latex]. This results in the function [latex]g_y(x, y)+h'(y)[/latex].
Use the equation [latex]g_y(x, y)+h'(y)=Q(x ,y)[/latex] to find [latex]h'(y)[/latex].
Integrate [latex]h'(y)[/latex] to find [latex]h(y)[/latex].
Write the potential function. Any function of the form [latex]f(x,y)=g(x,y)+h(y)+C[/latex], where [latex]C[/latex] is a constant, is a potential function for [latex]\mathbf{F}[/latex].
We can adapt this strategy to find potential functions for vector fields in [latex]\mathbb{R}^3[/latex], as shown in the next example.
Find a potential function for [latex]{\bf{F}}(x,y,z)=\langle2xy,x^2+2yz^3,3y^2z^2+2z\rangle[/latex], thereby showing that [latex]{\bf{F}}[/latex] is conservative.
Suppose that [latex]f[/latex] is a potential function. Then, [latex]\nabla{f}={\bf{F}}[/latex] and therefore [latex]f_x=2xy[/latex]. Integrating this equation with respect to [latex]x[/latex] yields the equation [latex]f(x, y, z)=x^{2}y+g(y, z)[/latex] for some function [latex]g[/latex]. Notice that, in this case, the constant of integration with respect to [latex]x[/latex] is a function of [latex]y[/latex] and [latex]z[/latex].
Since [latex]f[/latex] is a potential function,
[latex]x^{2}+2yz^{3}=f_y=x^{2}+g_y[/latex].
Therefore,
[latex]g_y=2yz^{3}[/latex].
Integrating this function with respect to [latex]y[/latex] yields
[latex]g(y, z)=y^{2}z^{3}+h(z)[/latex]
for some function [latex]h(z)[/latex] of [latex]z[/latex] alone. (Notice that, because we know that [latex]g[/latex] is a function of only [latex]y[/latex] and [latex]z[/latex], we do not need to write [latex]g(y, z)=y^{2}z^{3}+h(x,z)[/latex].) Therefore,
This implies that [latex]h^\prime(z)=2z[/latex], so [latex]h(z)=z^{2}+C[/latex]. Letting [latex]C=0[/latex] gives the potential function
[latex]f(x,y,z)=x^2y+y^2z^3+z^2[/latex].
To verify that [latex]f[/latex] is a potential function, note that [latex]\nabla{f}=\langle2xy,x^2+2yz^3,3y^2z^2+2z\rangle[/latex].
We can apply this process to a gravitational force. Recall that if an object has unit mass and is located at the origin, then the gravitational force in [latex]\mathbb{R}^2[/latex] that the object exerts on another object of unit mass at the point [latex](x, y)[/latex] is given by vector field
where [latex]G[/latex] is the universal gravitational constant. In the next example, we build a potential function for [latex]\mathbf{F}[/latex], thus confirming what we already know: that gravity is conservative.
Find a potential function [latex]f[/latex] for [latex]{\bf{F}}(x,y)=-G\left\langle\frac{x}{(x^2+y^2)^{3/2}},\frac{y}{(x^2+y^2)^{3/2}}\right\rangle[/latex].
Suppose that [latex]f[/latex] is a potential function. Then, [latex]\nabla{f}={\bf{F}}[/latex] and therefore
To integrate this function with respect to [latex]x[/latex], we can use [latex]u[/latex]-substitution. If [latex]u=x^{2}+y^{2}[/latex] then [latex]\frac{du}2=x \ dx[/latex], so
[latex]\begin{aligned} \displaystyle\int\frac{-Gx}{(x^2+y^2)^{3/2}}&=\displaystyle\int\frac{-G}{2u^{3/2}} \ du \\ &=\frac{G}{\sqrt{u}}+h(y) \\ &=\frac{G}{\sqrt{x^2+y^2}}+h(y) \end{aligned}[/latex]
which implies that [latex]h^\prime(y)=0[/latex]. Thus, we can take [latex]h(y)[/latex] to be any constant; in particular, we can let [latex]h(y)=0[/latex]. The function
[latex]f(x,y)=\frac{G}{\sqrt{x^2+y^2}}[/latex]
is a potential function for the gravitational field [latex]{\bf{F}}[/latex]. To confirm that [latex]f[/latex] is a potential function, note that