6.3 Conservative Vector Fields - Calculus Volume 3 | OpenStax (2023)

Proof

By Equation 6.9,

$${\displaystyle {\int }_C\text{∇}f.d\text{r}={\displaystyle {\int }_a^b\text{∇}f\left(\text{r}\left(t\right)\right).{\text{r}}^{\prime }\left(t\right)dt.}}$$

By the chain rule,

$$\frac{d}{dt}(f\left(\text{r}\left(t\right)\right)=\text{∇}f\left(\text{r}\left(t\right)\right).{\text{r}}^{\prime }\left(t\right).$$

Therefore, by the Fundamental Theorem of Calculus,

$$\begin{array}{cc}\hfill {\displaystyle {\int }_C\nabla }f.d\text{r}& ={\displaystyle {\int }_a^b\text{∇}f\left(\text{r}\left(t\right)\right).{\text{r}}^{\prime }\left(t\right)dt}\hfill \\ & ={\displaystyle {\int }_a^b\frac{d}{dt}(f\left(\text{r}\left(t\right)\right)dt}\hfill \\ & ={\left[f\left(\text{r}\left(t\right)\right)\right]}_{t=a}^{t=b}\hfill \\ & =f\left(\text{r}\left(b\right)\right)-f\left(\text{r}\left(a\right)\right).\hfill \end{array}$$

□

We know that if F is a conservative vector field, there are potential functions $f$ such that $\text{∇}f=\text{F}.$ Therefore ${\displaystyle {\int }_C\text{F}·d\text{r}}={\displaystyle {\int }_C\text{∇}f·d\text{r}}=f\left(\text{r}(b)\right)-f\left(\text{r}(a)\right).$ In other words, just as with the Fundamental Theorem of Calculus, computing the line integral ${\displaystyle {\int }_C\text{F}·d\text{r}},$ where F is conservative, is a two-step process: (1) find a potential function (“antiderivative”) $f$ for F and (2) compute the value of $f$ at the endpoints of C and calculate their difference $f\left(\text{r}\left(b\right)\right)-f\left(\text{r}\left(a\right)\right).$ Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals. A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.

Example 6.29 illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.

The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if F is conservative and C is a closed curve, then the circulation of F along C is zero—that is, ${\displaystyle {\int }_C\text{F}·d\text{r}}=0.$ To see why this is true, let $f$ be a potential function for F. Since C is a closed curve, the terminal point r(b) of C is the same as the initial point r(a) of C—that is, $\text{r}\left(a\right)=\text{r}\left(b\right).$ Therefore, by the Fundamental Theorem for Line Integrals,

$$\begin{array}{cc}\hfill {\displaystyle {\oint }_C\text{F}·d\text{r}}& ={\displaystyle {\oint }_C\text{∇}f·d\text{r}}\hfill \\ & =f\left(\text{r}(b)\right)-f\left(\text{r}(a)\right)\hfill \\ & =f\left(\text{r}(b)\right)-f\left(\text{r}(b)\right)\hfill \\ & =0.\hfill \end{array}$$

Recall that the reason a conservative vector field F is called “conservative” is because such vector fields model forces in which energy is conserved. We have shown gravity to be an example of such a force. If we think of vector field F in integral ${\displaystyle {\oint }_C\text{F}·d\text{r}}$ as a gravitational field, then the equation ${\displaystyle {\oint }_C\text{F}·d\text{r}}=0$ follows. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.

The second important consequence of the Fundamental Theorem for Line Integrals is that line integrals of conservative vector fields are independent of path—meaning, they depend only on the endpoints of the given curve, and do not depend on the path between the endpoints.

The second consequence is stated formally in the following theorem.

Proof

Let D denote the domain of F and let $C_1$ and $C_2$ be two paths in D with the same initial and terminal points (Figure 6.29). Call the initial point $P_1$ and the terminal point $P_2.$ Since F is conservative, there is a potential function $f$ for F. By the Fundamental Theorem for Line Integrals,

$${\displaystyle {\int }_{C_1}\text{F}·d\text{r}}=f(P_2)-f(P_1)={\displaystyle {\int }_{C_2}\text{F}·d\text{r}}.$$

Therefore, ${\displaystyle {\int }_{C_1}\text{F}·d\text{r}}={\displaystyle {\int }_{C_2}\text{F}·d\text{r}}$ and F is independent of path.

□

Figure 6.29 The vector field is conservative, and therefore independent of path.

To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides it is too difficult for him. Therefore he returns to camp and takes the non-steep path to the top. All three hikers are traveling along paths in a gravitational field. Since gravity is a force in which energy is conserved, the gravitational field is conservative. By independence of path, the total amount of work done by gravity on each of the hikers is the same because they all started in the same place and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total amount of energy each one expended is not the same, but the net energy expended against gravity is the same for all three hikers.

We have shown that if F is conservative, then F is independent of path. It turns out that if the domain of F is open and connected, then the converse is also true. That is, if F is independent of path and the domain of F is open and connected, then F is conservative. Therefore, the set of conservative vector fields on open and connected domains is precisely the set of vector fields independent of path.

Proof

We prove the theorem for vector fields in ${ℝ}^2.$ The proof for vector fields in ${ℝ}^3$ is similar. To show that $\text{F}=\langle P,Q\rangle$ is conservative, we must find a potential function $f$ for F. To that end, let X be a fixed point in D. For any point $\left(x,y\right)$ in D, let C be a path from X to $\left(x,y\right).$ Define $f$$\left(x,y\right)$ by $f(x,y)={\displaystyle {\int }_C\text{F}·d\text{r}.}$ (Note that this definition of $f$ makes sense only because F is independent of path. If F was not independent of path, then it might be possible to find another path $C^{\prime }$ from X to $\left(x,y\right)$ such that ${\displaystyle {\int }_C\text{F}·d\text{r}}\ne {\displaystyle {\int }_C\text{F}·d\text{r}},$ and in such a case $f$$\left(x,y\right)$ would not be a function.) We want to show that $f$ has the property $\text{∇}f=\text{F}.$

Since domain D is open, it is possible to find a disk centered at $\left(x,y\right)$ such that the disk is contained entirely inside D. Let $\left(a,y\right)$ with $a<x$ be a point in that disk. Let C be a path from X to $\left(x,y\right)$ that consists of two pieces: $C_1$ and $C_2.$ The first piece, $C_1,$ is any path from X to $\left(a,y\right)$ that stays inside D; $C_2$ is the horizontal line segment from $\left(a,y\right)$ to $\left(x,y\right)$ (Figure 6.30). Then

$$f(x,y)={\displaystyle {\int }_{C_1}\text{F}·d\text{r}}+{\displaystyle {\int }_{C_2}\text{F}·d\text{r}.}$$

The first integral does not depend on x, so

$$f_x=\frac{\partial }{\partial x}{\displaystyle {\int }_{C_2}\text{F}.d\text{r}.}$$

If we parameterize $C_2$ by $\text{r}\left(t\right)=\langle t,y\rangle ,a\le t\le x,$ then

$$\begin{array}{cc}\hfill f_x& =\frac{\partial }{\partial x}{\displaystyle {\int }_{C_2}\text{F}}.d\text{r}\hfill \\ & =\frac{\partial }{\partial x}{\displaystyle {\int }_a^x\text{F}}\left(\text{r}\left(t\right)\right).{\text{r}}^{\prime }\left(t\right)dt\hfill \\ & =\frac{\partial }{\partial x}{\displaystyle {\int }_a^x\text{F}}\left(\text{r}\left(t\right)\right).\frac{d}{dt}\left(\langle t,y\rangle \right)dt\hfill \\ & =\frac{\partial }{\partial x}{\displaystyle {\int }_a^x\text{F}}\left(r\left(t\right)\right).\langle 1,0\rangle dt\hfill \\ & =\frac{\partial }{\partial x}{\displaystyle {\int }_a^{x}P}\left(t,y\right)dt.\hfill \end{array}$$

By the Fundamental Theorem of Calculus (part 1),

$$f_x=\frac{\partial }{\partial x}{\displaystyle {\int }_a^{x}P}\left(t,y\right)dt=P\left(x,y\right).$$

Figure 6.30 Here, $C_1$ is any path from X to $\left(a,y\right)$ that stays inside D, and $C_2$ is the horizontal line segment from $\left(a,y\right)$ to $\left(x,y\right).$

A similar argument using a vertical line segment rather than a horizontal line segment shows that $f_y=Q\left(x,y\right).$

Therefore $\text{∇}f=\text{F}$ and F is conservative.

□

We have spent a lot of time discussing and proving Path Independence of Conservative Fields and The Path Independence Test for Conservative Fields, but we can summarize them simply: a vector field F on an open and connected domain is conservative if and only if it is independent of path. This is important to know because conservative vector fields are extremely important in applications, and these theorems give us a different way of viewing what it means to be conservative using path independence.

Example 6.30

Showing That a Vector Field Is Not Conservative

Use path independence to show that vector field $\text{F}(x,y)=\langle x^{2}y,y+5\rangle$ is not conservative.

See Also

16.3: Conservative Vector FieldsCalculus III - Conservative Vector FieldsHow to determine if a vector field is conservative

Solution

We can indicate that F is not conservative by showing that F is not path independent. We do so by giving two different paths, $C_1$ and $C_2,$ that both start at $(0,0)$ and end at $(1,1),$ and yet ${\displaystyle {\int }_{C_1}\text{F}}.dr\ne {\displaystyle {\int }_{C_2}\text{F}.d\text{r}.}$

Let $C_1$ be the curve with parameterization $r_1(t)=\langle t,t\rangle ,0\le t\le 1$ and let $C_2$ be the curve with parameterization $r_2(t)=\langle t,t^2\rangle ,0\le t\le 1$ (Figure 6.31). Then

$$\begin{array}{cc}\hfill {\displaystyle {\int }_{C_1}\text{F}·dr}& ={\displaystyle {\int }_0^1\text{F}\left(r_1(t)\right)·r_1{}^{\prime }(t)dt}\hfill \\ & ={\displaystyle {\int }_0^1\langle t^3,t+5\rangle ·\langle 1,1\rangle dt}={\displaystyle {\int }_0^1\left(t^3+t+5\right)dt}\hfill \\ & ={\left[\frac{t^4}{4}+\frac{t^2}{2}+5t\right]}_0^1=\frac{23}{4}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\displaystyle {\int }_{C_2}\text{F}·dr}& ={\displaystyle {\int }_0^1\text{F}\left(r_2(t)\right)·r_2\prime (t)dt}\hfill \\ & ={\displaystyle {\int }_0^1\langle t^4,t^2+5\rangle ·\langle 1,2t\rangle dt}={\displaystyle {\int }_0^1\left(t^4+2t^3+10t\right)dt}\hfill \\ & ={\left[\frac{t^5}{5}+\frac{t^4}{2}+5t^2\right]}_0^1=\frac{57}{10}.\hfill \end{array}$$

Since ${\displaystyle {\int }_{C_1}\text{F}.d\text{r}}\ne {\displaystyle {\int }_{C_2}\text{F}.d\text{r},}$ the value of a line integral of F depends on the path between two given points. Therefore, F is not independent of path, and F is not conservative.

Figure 6.31 Curves $C_1$ and $C_2$ are both oriented from left to right.