Learning Objectives
- Explain how to find a potential function for a conservative vector field.
- Use the Fundamental Theorem for Line Integrals to evaluate a line integral in a vector field.
- Explain how to test a vector field to determine whether it is conservative.
Conservative Vector Fields and Potential Functions
As we have learned, the Fundamental Theorem for Line Integrals says that if [latex]{\bf{F}}[/latex] is conservative, then calculating [latex]\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}[/latex] has two steps: first, find a potential function [latex]f[/latex] for [latex]{\bf{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 for a conservative field [latex]{\bf{F}}[/latex], we must be able to find a potential function [latex]f[/latex] for [latex]{\bf{F}}[/latex]. Therefore, we must answer the following question: Given a conservative vector field [latex]{\bf{F}}[/latex], how do we find a function [latex]f[/latex] such that [latex]\nabla{f}={\bf{F}}[/latex]? Before giving a general method for finding a potential function, let’s motivate the method with an example.
Example: finding a potential function
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.
try it
Find a potential function for [latex]{\bf{F}}(x,y)=\langle{e}^xy^3+y,3e^xy^2+x\rangle[/latex].
The logic of the previous example extends to finding the potential function for any conservative vector field in [latex]\mathbb{R}^2[/latex]. Thus, we have the following problem-solving strategy for finding potential functions:
Problem solving strategy: finding a potential function for a conservative vector field [latex]{\bf{F}}(X,Y)=<{P}(X,Y),Q(X,Y)>[/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], which 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].
- 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]{\bf{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.
Example: finding a potential function in [latex]\mathbb{R}^3[/latex]
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.
try it
Find a potential function for [latex]{\bf{F}}(x,y,z)=\langle12x^2,\cos{y}\cos{z},1-\sin{y}\sin{z}\rangle[/latex].
We can apply the process of finding a potential function 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
[latex]\large{{\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],
where [latex]G[/latex] is the universal gravitational constant. In the next example, we build a potential function for [latex]{\bf{F}}[/latex], thus confirming what we already know: that gravity is conservative.
Example: finding a potential function
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].
try it
Find a potential function [latex]f[/latex] for the three-dimensional gravitational force [latex]{\bf{F}}(x,y,z)=\left\langle\frac{-Gx}{(x^2+y^2+z^2)^{3/2}},\frac{-Gy}{(x^2+y^2+z^2)^{3/2}},\frac{-Gz}{(x^2+y^2+z^2)^{3/2}}\right\rangle[/latex].
Testing a Vector Field
Until now, we have worked with vector fields that we know are conservative, but if we are not told that a vector field is conservative, we need to be able to test whether it is conservative. Recall that, if [latex]{\bf{F}}[/latex] is conservative, then [latex]{\bf{F}}[/latex] has the cross-partial property (see The Cross-Partial Property of Conservative Vector Fields Theorem). That is, if [latex]{\bf{F}}=\langle{P},Q,R\rangle[/latex] is conservative, then [latex]P_y=Q_x[/latex], [latex]P_z=R_x[/latex], and [latex]Q_z=R_y[/latex], So, if [latex]{\bf{F}}[/latex] has the cross-partial property, then is [latex]{\bf{F}}[/latex] conservative? If the domain of [latex]{\bf{F}}[/latex] is open and simply connected, then the answer is yes.
theorem: the cross-partial test for conservative fields
If [latex]{\bf{F}}=\langle{P},Q,R\rangle[/latex] is a vector field on an open, simply connected region [latex]D[/latex] and [latex]P_y=Q_x[/latex], [latex]P_z=R_x[/latex], and [latex]Q_z=R_y[/latex] throughout [latex]D[/latex], then [latex]{\bf{F}}[/latex] is conservative.
Although a proof of this theorem is beyond the scope of the text, we can discover its power with some examples. Later, we see why it is necessary for the region to be simply connected.
Combining this theorem with the cross-partial property, we can determine whether a given vector field is conservative:
theorem: the cross-partial property of conservative fields
Let [latex]{\bf{F}}=\langle{P},Q,R\rangle[/latex] be a vector field on an open, simply connected region [latex]D[/latex]. Then [latex]P_y=Q_x[/latex], [latex]P_z=R_x[/latex], and [latex]Q_z=R_y[/latex] throughout [latex]D[/latex] if and only if [latex]{\bf{F}}[/latex] is conservative.
The version of this theorem in [latex]\mathbb{R}^2[/latex] is also true. If [latex]{\bf{F}}=\langle{P},Q\rangle[/latex] is a vector field on an open, simply connected domain in [latex]\mathbb{R}^2[/latex], then [latex]{\bf{F}}[/latex] is conservative if and only if [latex]P_y=Q_x[/latex].
Example: determining whether a vector field is conservative
Determine whether vector field [latex]{\bf{F}}(x,y,z)=\langle{x}y^2z,x^2yz,z^2\rangle[/latex] is conservative.
Example: determining whether a vector field is conservative
Determine vector field [latex]{\bf{F}}(x,y)=\left\langle{x}\ln{(y)},\frac{x^2}{2y}\right\rangle[/latex] is conservative.
try it
Determine whether [latex]{\bf{F}}(x,y)=\langle\sin{x}\cos{y},\cos{x}\sin{y}\rangle[/latex] is conservative.
When using Cross-Partial Property of Conservative Fields Theorem, it is important to remember that a theorem is a tool, and like any tool, it can be applied only under the right conditions. In the case of Cross-Partial Property of Conservative Fields Theorem, the theorem can be applied only if the domain of the vector field is simply connected.
To see what can go wrong when misapplying the theorem, consider the vector field:
[latex]\large{{\bf{F}}(x,y)=\frac{y}{x^2+y^2}{\bf{i}}+\frac{-x}{x^2+y^2}{\bf{j}}}[/latex].
This vector field satisfies the cross-partial property, since
[latex]\large{\frac{\partial}{\partial{y}}\left(\frac{y}{x^2+y^2}\right)=\frac{(x^2+y^2)-y(2y)}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}}[/latex]
and
[latex]\large{\frac{\partial}{\partial{x}}\left(\frac{-x}{x^2+y^2}\right)=\frac{-(x^2+y^2)+x(2x)}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}}[/latex].
Since [latex]{\bf{F}}[/latex] satisfies the cross-partial property, we might be tempted to conclude that [latex]{\bf{F}}[/latex] is conservative. However, [latex]{\bf{F}}[/latex] is not conservative. To see this, let
[latex]\large{{\bf{r}}(t)=\langle\cos{t},\sin{t}\rangle\text{, }0\leq{t}\leq\pi}[/latex]
be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this [latex]C_1[/latex]) and let
[latex]\large{s(t)=\langle\cos{t},-\sin{t}\rangle\text{, }0\leq{t}\leq\pi}[/latex]
be a parameterization of the lower half of a unit circle oriented clockwise (denote this [latex]C_2[/latex]). Notice that [latex]C_1[/latex] and [latex]C_2[/latex] have the same starting point and endpoint. Since [latex]\sin^2t+\cos^2t=1[/latex],
[latex]\large{{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t)=\langle\sin{(t)},-\cos{(t)}\rangle\cdot\langle-\sin{(t)},-=\cos{(t)}\rangle}=-1[/latex]
and
[latex]\large{\begin{aligned} {\bf{F}}(s(t))\cdot{s}^\prime(t)&=\langle-\sin{t},-\cos{t}\rangle\cdot\langle-\sin{t},-\cos{t}\rangle \\ &=\sin^2t+\cos^2t \\ &=1 \end{aligned}}[/latex].
Therefore,
[latex]\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_0^\pi-1 \ dt = -\pi\text{ and }\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_0^\pi1 \ dt=\pi[/latex].
Thus, [latex]C_1[/latex] and [latex]C_2[/latex] have the same starting point and endpoint, but [latex]\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}\ne\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}[/latex]. Therefore, [latex]{\bf{F}}[/latex] is not independent of path and [latex]{\bf{F}}[/latex] is not conservative.
To summarize: [latex]{\bf{F}}[/latex] satisfies the cross-partial property and yet [latex]{\bf{F}}[/latex] is not conservative. What went wrong? Does this contradict Cross-Partial Property of Conservative Fields Theorem? The issue is that the domain of [latex]{\bf{F}}[/latex] is all of [latex]\mathbb{R}^2[/latex] except for the origin. In other words, the domain of [latex]{\bf{F}}[/latex] has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected, Cross-Partial Property of Conservative Fields Theorem does not apply to [latex]{\bf{F}}[/latex].
We close this section by looking at an example of the usefulness of the Fundamental Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always decide whether the Fundamental Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form [latex]\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}[/latex], then our first question should be: Is [latex]{\bf{F}}[/latex] conservative? If the answer is yes, then we should find a potential function and use the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, then the Fundamental Theorem for Line Integrals can’t help us and we have to use other methods, such as using the equation to compute vector line integral.
Example: using the fundamental theorem for line integrals
Calculate line integral [latex]\displaystyle\int_{C}{\bf{F}}\cdot{d}r[/latex], where [latex]{\bf{F}}(x,y,z)=\langle2xe^yz+e^xz,x^2e^yz,x^2e^y+e^x\rangle[/latex] and [latex]C[/latex] is any smooth curve that goes from the origin to [latex](1, 1, 1)[/latex].
try it
Calculate integral [latex]\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}[/latex], where [latex]{\bf{F}}(x,y)=\langle\sin{x}\sin{y},5-\cos{x}\cos{y}\rangle[/latex] and [latex]C[/latex] is a semicircle with starting point [latex](0,\pi)[/latex] and endpoint [latex](0,-\pi)[/latex].
Watch the following video to see the worked solution to the above Try It
Example: work done on a particle
Let [latex]{\bf{F}}(x,y)=\langle2xy^2,2x^2y\rangle[/latex] be a force field. Suppose that a particle begins its motion at the origin and ends its movement at any point in a plane that is not on the x-axis or the y-axis. Furthermore, the particle’s motion can be modeled with a smooth parameterization. Show that [latex]{\bf{F}}[/latex] does positive work on the particle.
try it
Let [latex]{\bf{F}}(x,y)=\langle4x^3y^4,4x^4y^3\rangle[/latex], and suppose that a particle moves from point [latex](4, 4)[/latex] to [latex](1, 1)[/latex] along any smooth curve. Is the work done by [latex]{\bf{F}}[/latex] on the particle positive, negative, or zero?
