6.3 Conservative vector fields  (Page 5/16)

Proof

We prove the theorem for vector fields in ${ℝ}^2.$ The proof for vector fields in ${ℝ}^3$ is similar. To show that $\text{F}=⟨P,Q⟩$ 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 C to $\left(a,y\right)$ that stays inside D ; $C_2$ is the horizontal line segment from $\left(a,y\right)$ to $\left(x,y\right)$ ( [link] ). Then

The first integral does not depend on x , so

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

By the Fundamental Theorem of Calculus (part 1),

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.

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We have spent a lot of time discussing and proving [link] and [link] , 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.

Showing that a vector field is not conservative

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

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 $r_1(t)=⟨t,t⟩,0\le t\le 1$ and let $C_2$ be the curve with parameterization $r_2(t)=⟨t,t^2⟩,0\le t\le 1$ ( [link] ). Then

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

and

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

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.

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Conservative vector fields and potential functions

As we have learned, the Fundamental Theorem for Line Integrals says that if F is conservative, then calculating ${\displaystyle {\int }_C\text{F}·dr}$ has two steps: first, find a potential function $f$ for 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 F , we must be able to find a potential function $f$ for F . Therefore, we must answer the following question: Given a conservative vector field F , how do we find a function $f$ such that $\text{∇}f=\text{F}?$ Before giving a general method for finding a potential function, let’s motivate the method with an example.