Conservative Vector Fields

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 ${\bf{F}}$ is conservative, then calculating $\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}$ has two steps: first, find a potential function $f$ for ${\bf{F}}$ and, second, calculate $f(p_1)-f(P_0)$ , where $P_1$ is the endpoint of $C$ and $P_0$ is the starting point. To use this theorem for a conservative field ${\bf{F}}$ , we must be able to find a potential function $f$ for ${\bf{F}}$ . Therefore, we must answer the following question: Given a conservative vector field ${\bf{F}}$ , how do we find a function $f$ such that $\nabla{f}={\bf{F}}$ ? 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 ${\bf{F}}(x,y)=\langle2xy^3,3x^2y^2+\cos{(y)}\rangle$ , thereby showing that ${\bf{F}}$ is conservative.

Show Solution

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

$f_x=2xy^3\text{ and }f_y=3x^2y^2+\cos{y}$ .

Integrating the equation $f_x=2xy^3$ with respect to $x$ yields the equation

$f(x,y)=x^2y^2+h(y)$ .

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 $f(x,y)=x^2y^2+h(y)$ can be confirmed by taking the partial derivative with respect to $x$ :

$\frac{\partial{f}}{\partial{x}}=\frac{\partial}{\partial{x}}(x^2y^3)\frac{\partial}{\partial{x}}(h(y))=2xy^3+0=2xy^3$ .

Since $f$ is a potential function for ${\bf{F}}$ ,

$f_y=3x^2y^2+\cos{(y)}$ ,

and therefore

$3x^2y^2+h^\prime(y)=3x^2y^2+\cos{(y)}$ .

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

To verify that $f$ is a potential function, note that $\nabla{f}=\langle2xy^3,3x^2y^2+\cos{y}\rangle={\bf{F}}$ .

try it

Find a potential function for ${\bf{F}}(x,y)=\langle{e}^xy^3+y,3e^xy^2+x\rangle$ .

Show Solution

$f(x,y)=e^xy^3+xy$ .

The logic of the previous example extends to finding the potential function for any conservative vector field in $\mathbb{R}^2$ . Thus, we have the following problem-solving strategy for finding potential functions:

Problem solving strategy: finding a potential function for a conservative vector field ${\bf{F}}(X,Y)=<{P}(X,Y),Q(X,Y)>$

Integrate $P$ with respect to $x$ . This results in a function of the form $g(x, y)+h(y)$ , where $h(y)$ is unknown.
Take the partial derivative of $g(x, y)+h(y)$ with respect to $y$ , which results in the function $g_y(x, y)+h'(y)$ .
Use the equation $g_y(x, y)+h'(y)=Q(x ,y)$ to find $h'(y)$ .
Integrate $h'(y)$ to find $h(y)$ .
Any function of the form $f(x,y)=g(x,y)+h(y)+C$ , where $C$ is a constant, is a potential function for ${\bf{F}}$ .

We can adapt this strategy to find potential functions for vector fields in $\mathbb{R}^3$ , as shown in the next example.

Example: finding a potential function in $\mathbb{R}^3$

Find a potential function for ${\bf{F}}(x,y,z)=\langle2xy,x^2+2yz^3,3y^2z^2+2z\rangle$ , thereby showing that ${\bf{F}}$ is conservative.

Show Solution

Suppose that $f$ is a potential function. Then, $\nabla{f}={\bf{F}}$ and therefore $f_x=2xy$ . Integrating this equation with respect to $x$ yields the equation $f(x, y, z)=x^{2}y+g(y, z)$ 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,

$x^{2}+2yz^{3}=f_y=x^{2}+g_y$ .

Therefore,

$g_y=2yz^{3}$ .

Integrating this function with respect to $y$ yields

$g(y, z)=y^{2}z^{3}+h(z)$

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

$f(z,y,z)=x^2+y+g(y,z)=x^2y+y^2z^3+h(z)$ .

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

$3y^2z^2+2z=g_z=3y^2z^2+h^\prime(z)$ .

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

$f(x,y,z)=x^2y+y^2z^3+z^2$ .

To verify that $f$ is a potential function, note that $\nabla{f}=\langle2xy,x^2+2yz^3,3y^2z^2+2z\rangle$ .

try it

Find a potential function for ${\bf{F}}(x,y,z)=\langle12x^2,\cos{y}\cos{z},1-\sin{y}\sin{z}\rangle$ .

Show Solution

$f(x,y,z)=4x^3+\sin{y}\cos{z}+z$

You can view the transcript for “CP 6.29” here (opens in new window).

We can apply the process of finding a potential function to a . Recall that, if an object has unit mass and is located at the origin, then the gravitational force in $\mathbb{R}^2$ that the object exerts on another object of unit mass at the point $(x, y)$ is given by vector field

$\large{{\bf{F}}(x,y)=-G\left\langle\frac{x}{(x^2+y^2)^{3/2}},\frac{y}{(x^2+y^2)^{3/2}}\right\rangle}$ ,

where $G$ is the universal gravitational constant. In the next example, we build a potential function for ${\bf{F}}$ , thus confirming what we already know: that gravity is conservative.

Example: finding a potential function

Find a potential function $f$ for ${\bf{F}}(x,y)=-G\left\langle\frac{x}{(x^2+y^2)^{3/2}},\frac{y}{(x^2+y^2)^{3/2}}\right\rangle$ .

Show Solution

Suppose that $f$ is a potential function. Then, $\nabla{f}={\bf{F}}$ and therefore

$\large{f_x=\frac{-Gx}{(x^2+y^2)^{3/2}}}$ .

To integrate this function with respect to $x$ , we can use $u$ -substitution. If $u=x^{2}+y^{2}$ then $\frac{du}2=x \ dx$ , so

$\begin{aligned} \displaystyle\int\frac{-Gx}{(x^2+y^2)^{3/2}}&=\displaystyle\int\frac{-G}{2u^{3/2}} \ du \\ &=\frac{G}{\sqrt{u}}+h(y) \\ &=\frac{G}{\sqrt{x^2+y^2}}+h(y) \end{aligned}$

for some function $h(y)$ . Therefore,

$\large{f(x,y)=\frac{G}{\sqrt{(x^2+y^2)}}+h(y)}$ .

Since $f$ is a potential function for ${\bf{F}}$ ,

$\large{f_y=\frac{-Gy}{(x^2+y^2)^{3/2}}}$ .

Since $f(x,y)=\frac{G}{\sqrt{x^2+y^2}}+h(y)$ , $f_y$ also equals $\frac{-Gy}{(x^2+y^2)^{3/2}}+h^\prime(y)$ .

Therefore,

$\large{\frac{-Gy}{(x^2+y^2)^{3/2}}+h^\prime(y)=\frac{-Gy}{(x^2+y^2)^{3/2}}}$ ,

which implies that $h^\prime(y)=0$ . Thus, we can take $h(y)$ to be any constant; in particular, we can let $h(y)=0$ . The function

$f(x,y)=\frac{G}{\sqrt{x^2+y^2}}$

is a potential function for the gravitational field ${\bf{F}}$ . To confirm that $f$ is a potential function, note that

$\begin{aligned} \nabla{f}&=\left\langle-\frac12\frac{G}{(x^2+y^2)^{3/2}}(2x),-\frac12\frac{G}{(x^2+y^2)^{3/2}}(2y)\right\rangle \\ &=\left\langle\frac{-Gx}{(x^2+y^2)^{3/2}},\frac{-Gy}{(x^2+y^2)^{3/2}}\right\rangle \\ &={\bf{F}} \end{aligned}$ .

try it

Find a potential function $f$ for the three-dimensional gravitational force ${\bf{F}}(x,y,z)=\left\langle\frac{-Gx}{(x^2+y^2+z^2)^{3/2}},\frac{-Gy}{(x^2+y^2+z^2)^{3/2}},\frac{-Gz}{(x^2+y^2+z^2)^{3/2}}\right\rangle$ .

Show Solution

$f(x,y,z)=\frac{G}{\sqrt{x^2+y^2+z^2}}$ .

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 ${\bf{F}}$ is conservative, then ${\bf{F}}$ has the cross-partial property (see The Cross-Partial Property of Conservative Vector Fields Theorem). That is, if ${\bf{F}}=\langle{P},Q,R\rangle$ is conservative, then $P_y=Q_x$ , $P_z=R_x$ , and $Q_z=R_y$ , So, if ${\bf{F}}$ has the cross-partial property, then is ${\bf{F}}$ conservative? If the domain of ${\bf{F}}$ is open and simply connected, then the answer is yes.

theorem: the cross-partial test for conservative fields

If ${\bf{F}}=\langle{P},Q,R\rangle$ is a vector field on an open, simply connected region $D$ and $P_y=Q_x$ , $P_z=R_x$ , and $Q_z=R_y$ throughout $D$ , then ${\bf{F}}$ 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 ${\bf{F}}=\langle{P},Q,R\rangle$ be a vector field on an open, simply connected region $D$ . Then $P_y=Q_x$ , $P_z=R_x$ , and $Q_z=R_y$ throughout $D$ if and only if ${\bf{F}}$ is conservative.

The version of this theorem in $\mathbb{R}^2$ is also true. If ${\bf{F}}=\langle{P},Q\rangle$ is a vector field on an open, simply connected domain in $\mathbb{R}^2$ , then ${\bf{F}}$ is conservative if and only if $P_y=Q_x$ .

Example: determining whether a vector field is conservative

Determine whether vector field ${\bf{F}}(x,y,z)=\langle{x}y^2z,x^2yz,z^2\rangle$ is conservative.

Show Solution

Note that the domain of ${\bf{F}}$ is all of $\mathbb{R}^2$ and $\mathbb{R}^3$ is simply connected. Therefore, we can use Cross-Partial Property of Conservative Fields Theorem to determine whether ${\bf{F}}$ is conservative. Let

$P(x,y,z)=xy^2z,\text{ }Q(x,y,z)=x^2yz,\text{ and }R(x,y,z)=z^2$ .

Since $Q_z=x^{2}y$ and $R_y=0$ , the vector field is not conservative.

Example: determining whether a vector field is conservative

Determine vector field ${\bf{F}}(x,y)=\left\langle{x}\ln{(y)},\frac{x^2}{2y}\right\rangle$ is conservative.

Show Solution

Note that the domain of ${\bf{F}}$ is the part of $\mathbb{R}^2$ in which $y>0$ . Thus, the domain of ${\bf{F}}$ is part of a plane above the x-axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use Cross-Partial Property of Conservative Fields Theorem to determine whether ${\bf{F}}$ is conservative. Let

$P(x,y)=x\ln{(y)}\text{ and }Q(x,y)=\frac{x^2}{2y}$ .

Then $P_y=\frac{x}y=Q_x$ and thus ${\bf{F}}$ is conservative.

try it

Determine whether ${\bf{F}}(x,y)=\langle\sin{x}\cos{y},\cos{x}\sin{y}\rangle$ is conservative.

Show Solution

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:

$\large{{\bf{F}}(x,y)=\frac{y}{x^2+y^2}{\bf{i}}+\frac{-x}{x^2+y^2}{\bf{j}}}$ .

This vector field satisfies the cross-partial property, since

$\large{\frac{\partial}{\partial{y}}\left(\frac{y}{x^2+y^2}\right)=\frac{(x^2+y^2)-y(2y)}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}}$

and

$\large{\frac{\partial}{\partial{x}}\left(\frac{-x}{x^2+y^2}\right)=\frac{-(x^2+y^2)+x(2x)}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}}$ .

Since ${\bf{F}}$ satisfies the cross-partial property, we might be tempted to conclude that ${\bf{F}}$ is conservative. However, ${\bf{F}}$ is not conservative. To see this, let

$\large{{\bf{r}}(t)=\langle\cos{t},\sin{t}\rangle\text{, }0\leq{t}\leq\pi}$

be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this $C_1$ ) and let

$\large{s(t)=\langle\cos{t},-\sin{t}\rangle\text{, }0\leq{t}\leq\pi}$

be a parameterization of the lower half of a unit circle oriented clockwise (denote this $C_2$ ). Notice that $C_1$ and $C_2$ have the same starting point and endpoint. Since $\sin^2t+\cos^2t=1$ ,

$\large{{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t)=\langle\sin{(t)},-\cos{(t)}\rangle\cdot\langle-\sin{(t)},-=\cos{(t)}\rangle}=-1$

and

$\large{\begin{aligned} {\bf{F}}(s(t))\cdot{s}^\prime(t)&=\langle-\sin{t},-\cos{t}\rangle\cdot\langle-\sin{t},-\cos{t}\rangle \\ &=\sin^2t+\cos^2t \\ &=1 \end{aligned}}$ .

Therefore,

$\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_0^\pi-1 \ dt = -\pi\text{ and }\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_0^\pi1 \ dt=\pi$ .

Thus, $C_1$ and $C_2$ have the same starting point and endpoint, but $\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}\ne\displaystyle\int_{C_2}{\bf{F}}\cdot{d}{\bf{r}}$ . Therefore, ${\bf{F}}$ is not independent of path and ${\bf{F}}$ is not conservative.

To summarize: ${\bf{F}}$ satisfies the cross-partial property and yet ${\bf{F}}$ is not conservative. What went wrong? Does this contradict Cross-Partial Property of Conservative Fields Theorem? The issue is that the domain of ${\bf{F}}$ is all of $\mathbb{R}^2$ except for the origin. In other words, the domain of ${\bf{F}}$ 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 ${\bf{F}}$ .

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 $\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}$ , then our first question should be: Is ${\bf{F}}$ 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 $\displaystyle\int_{C}{\bf{F}}\cdot{d}r$ , where ${\bf{F}}(x,y,z)=\langle2xe^yz+e^xz,x^2e^yz,x^2e^y+e^x\rangle$ and $C$ is any smooth curve that goes from the origin to $(1, 1, 1)$ .

Show Solution

Solution

Before trying to compute the integral, we need to determine whether ${\bf{F}}$ is conservative and whether the domain of ${\bf{F}}$ is simply connected. The domain of ${\bf{F}}$ is all of $\mathbb{R}^3$ , which is connected and has no holes. Therefore, the domain of ${\bf{F}}$ is simply connected. Let

$\large{P(x,y,z)=2xe^yz+e^xz,\text{ }Q(x,y,z)=x^2e^yz,\text{and }R(x,y,z)=x^2e^y+e^x}$

so that ${\bf{F}}=\langle{P},Q,R\rangle$ . Since the domain of ${\bf{F}}$ is simply connected, we can check the cross partials to determine whether ${\bf{F}}$ is conservative. Note that

$\begin{aligned} P_y&=2xe^yz=Q_x \\ P_z&=2xe^y+e^x=R_x \\ Q_z&=x^2e^y=R_y \end{aligned}$ .

Therefore, ${\bf{F}}$ is conservative.

To evaluate $\displaystyle\int_{C}{\bf{F}}\cdot{d}r$ using the Fundamental Theorem for Line Integrals, we need to find a potential function $f$ for ${\bf{F}}$ . Let $f$ be a potential function for ${\bf{F}}$ . Then, $\nabla{f}={\bf{F}}$ , and therefore $f_x=2xe^yz+e^xz$ . Integrating this equation with respect to $x$ gives $f(x,y,z)=x^2e^yz+e^xz+h(y,z)$ for some function $h$ . Differentiating this equation with respect to $y$ gives $x^2e^yz+h_y=Q=x^2e^yz$ , which implies that $h_y=0$ . Therefore, $h$ is a function of $z$ only, and $f(x,y,z)=x^2e^yz+e^x+h(z)$ . To find $h$ , note that $f_z=x^2e^y+e^x+h^\prime(z)$ . Therefore, $h^\prime(z)=0$ and we can take $h(z)=0$ . A potential function for ${\bf{F}}$ is $f(x,y,z)=x^2e^yz+e^xz$ .

Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,

$\begin{aligned} \displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_C\nabla{f}\cdot{d}{\bf{r}} \\ &=f(1,1,1)-f(0,0,0) \\ &=2e \end{aligned}$ .

Analysis

Notice that if we hadn’t recognized that ${\bf{F}}$ is conservative, we would have had to parameterize $C$ and use $\displaystyle\int_{C}{\bf{F}}\cdot{\bf{T}}ds=\displaystyle\int_{a}^{b}{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^{\prime}(t)dt$ . Since curve $C$ is unknown, using the Fundamental Theorem for Line Integrals is much simpler.

try it

Calculate integral $\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}$ , where ${\bf{F}}(x,y)=\langle\sin{x}\sin{y},5-\cos{x}\cos{y}\rangle$ and $C$ is a semicircle with starting point $(0,\pi)$ and endpoint $(0,-\pi)$ .

Show Solution

$-10\pi$ .

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 6.32” here (opens in new window).

Example: work done on a particle

Let ${\bf{F}}(x,y)=\langle2xy^2,2x^2y\rangle$ 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 ${\bf{F}}$ does positive work on the particle.

Show Solution

Solution

We show that ${\bf{F}}$ does positive work on the particle by showing that ${\bf{F}}$ is conservative and then by using the Fundamental Theorem for Line Integrals.

To show that ${\bf{F}}$ is conservative, suppose $f(x, y)$ were a potential function for ${\bf{F}}$ . Then, $\nabla{f}={\bf{F}}=\langle2xy^2,2x^2y\rangle$ and therefore $f_x=2xy^{2}$ and $f_y=2x^{2}y$ . Equation $f_x=2xy^{2}$ implies that $f(x, y)=x^{2}y^{2}+h(y)$ . Deriving both sides with respect to $y$ yields $f_y=2x^{2}y+h'(y)$ . Therefore, $h'(y)$ and we can take $h(y)=0$ .

If $f(x, y)=x^{2}y^{2}$ , then note that $\nabla{f}=\langle2xy^2,2x^2y\rangle={\bf{F}}$ , and therefore $f$ is a potential function for ${\bf{F}}$ .

Let $(a, b)$ be the point at which the particle stops is motion, and let $C$ denote the curve that models the particle’s motion. The work done by ${\bf{F}}$ on the particle is $\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}$ . By the Fundamental Theorem for Line Integrals,

$\begin{aligned} \displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_c\nabla{f}\cdot{d}{\bf{r}} \\ &=f(a,b)-f(0,0) \\ &=a^2b^2 \end{aligned}$ .

Since $a\ne0$ and $b\ne0$ , by assumption, $a^2b^2>0$ . Therefore, $\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}>0$ , and ${\bf{F}}$ does positive work on the particle.

Analysis

Notice that this problem would be much more difficult without using the Fundamental Theorem for Line Integrals. To apply the tools we have learned, we would need to give a curve parameterization and use $\displaystyle\int_{C}{\bf{F}}\cdot{\bf{T}}ds=\displaystyle\int_{a}^{b}{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^{\prime}(t)dt$ . Since the path of motion $C$ can be as exotic as we wish (as long as it is smooth), it can be very difficult to parameterize the motion of the particle.

try it

Let ${\bf{F}}(x,y)=\langle4x^3y^4,4x^4y^3\rangle$ , and suppose that a particle moves from point $(4, 4)$ to $(1, 1)$ along any smooth curve. Is the work done by ${\bf{F}}$ on the particle positive, negative, or zero?

Show Solution

Module 6: Vector Calculus