what does it mean to be conservative math

sixteen.three: Conservative Vector Fields

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• Professor (Mathematics) at Massachusetts Institute of Technology (Strang) & University of Wisconsin-Stevens Indicate (Herman)
• Source: OpenStax CNX

Learning Objectives

• Draw unproblematic and closed curves; define continued and but connected regions.
• Explain how to discover a potential function for a conservative vector field.
• Use the Fundamental Theorem for Line Integrals to evaluate a line integral in a vector field.
• Explicate how to test a vector field to determine whether it is conservative.

In this section, we continue the study of conservative vector fields. We examine the Key Theorem for Line Integrals, which is a useful generalization of the Fundamental Theorem of Calculus to line integrals of conservative vector fields. We besides detect show how to examination whether a given vector field is conservative, and determine how to build a potential function for a vector field known to be conservative.

Curves and Regions

Before standing our report of bourgeois vector fields, we need some geometric definitions. The theorems in the subsequent sections all rely on integrating over certain kinds of curves and regions, so we develop the definitions of those curves and regions here. We first define two special kinds of curves: closed curves and simple curves. As nosotros have learned, a closed curve is one that begins and ends at the aforementioned bespeak. A simple curve is one that does not cantankerous itself. A curve that is both closed and elementary is a simple closed curve (Figure \(\PageIndex{one}\)).

Effigy \(\PageIndex{1}\). Types of curves that are simple or non elementary and airtight or not airtight.

DEFINITION: Closed Curves

Bend \(C\) is a closed curve if at that place is a parameterization \(\vecs r(t)\), \(a≤t≤b\) of \(C\) such that the parameterization traverses the curve exactly one time and \(\vecs r(a)=\vecs r(b)\). Curve \(C\) is a unproblematic curve if \(C\) does not cross itself. That is, \(C\) is simple if in that location exists a parameterization \(\vecs r(t)\), \(a≤t≤b\) of \(C\) such that \(\vecs r\) is one-to-1 over \((a,b)\). It is possible for \(\vecs r(a)=\vecs r(b)\), significant that the simple curve is also airtight.

Example \(\PageIndex{1}\): Determining Whether a Bend Is Unproblematic and Closed

Is the curve with parameterization \(\vecs{r}(t)=\left\langle\cos t,\frac{\sin(2t)}{2}\right\rangle\), \(0≤t≤two\pi\) a elementary closed curve?

Solution

Note that \(\vecs{r}(0)=⟨1,0⟩=\vecs r(two\pi)\); therefore, the curve is closed. The bend is non simple, however. To come across this, note that \(\vecs{r}\left(\frac{\pi}{ii}\correct)=⟨0,0⟩=\vecs{r}\left(\frac{three\pi}{2}\correct)\), and therefore the curve crosses itself at the origin (Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{2}\) . A bend that is closed just non simple.

Exercise \(\PageIndex{1}\)

Is the curve given by parameterization \(\vecs{r}(t)=⟨2\cos t,3\sin t⟩\), \(0≤t≤6\pi\), a simple closed bend?

Hint

Sketch the curve.

Respond

Yes

Many of the theorems in this chapter chronicle an integral over a region to an integral over the boundary of the region, where the region's boundary is a uncomplicated closed curve or a union of elementary closed curves. To develop these theorems, we demand two geometric definitions for regions: that of a connected region and that of a simply connected region. A continued region is ane in which at that place is a path in the region that connects any two points that prevarication inside that region. A simply connected region is a connected region that does non accept any holes in it. These 2 notions, along with the notion of a unproblematic closed curve, permit us to country several generalizations of the Primal Theorem of Calculus later on in the chapter. These two definitions are valid for regions in any number of dimensions, but we are but concerned with regions in ii or three dimensions.

DEFINITION: connected regions

A region D is a connected region if, for whatever 2 points \(P_1\) and \(P_2\), there is a path from \(P_1\) to \(P_2\) with a trace contained entirely inside D. A region D is a simply connected region if D is continued for any simple closed curve C that lies inside D, and curve C can exist shrunk continuously to a bespeak while staying entirely inside D. In two dimensions, a region is only continued if it is connected and has no holes.

All simply connected regions are connected, simply not all connected regions are simply connected (Figure \(\PageIndex{three}\)).

Figure \(\PageIndex{3}\): Non all connected regions are merely connected. (a) Only continued regions take no holes. (b) Connected regions that are not just continued may accept holes simply you lot tin nevertheless find a path in the region between whatsoever 2 points. (c) A region that is non connected has some points that cannot be connected by a path in the region.

Exercise \(\PageIndex{2}\)

Is the region in the below image continued? Is the region simply continued?

Hint

Consider the definitions.

Reply

The region in the figure is connected. The region in the figure is not simply connected.

Fundamental Theorem for Line Integrals

Now that we understand some basic curves and regions, let'due south generalize the Central Theorem of Calculus to line integrals. Recall that the Cardinal Theorem of Calculus says that if a function \(f\) has an antiderivative \(F\), then the integral of \(f\) from \(a\) to \(b\) depends only on the values of \(F\) at \(a\) and at \(b\)—that is,

\[\int_a^bf(x)\,dx=F(b)−F(a).\]

If we think of the slope as a derivative, so the same theorem holds for vector line integrals. We show how this works using a motivational instance.

Instance \(\PageIndex{two}\): Evaluating a Line Integral and the Antiderivatives of the Endpoints

Let \(\vecs{F}(ten,y)=⟨2x,4y⟩\). Calculate \(\displaystyle \int_C \vecs{F} \cdot d\vecs{r}\), where C is the line segment from \((0,0)\) to \((2,ii)\) (Figure \(\PageIndex{4}\)).

Solution

We use the method from the previous section to calculate \(\int_C \vecs{F} \cdot d\vecs{r}\). Bend C can be parameterized by \(\vecs{r}(t)=⟨2t,2t⟩\), \(0≤t≤1\). Then, \(\vecs{F}(\vecs r(t))=⟨4t,8t⟩\) and \(\vecs r′(t)=⟨two,2⟩\), which implies that

\[\begin{marshal*} \int_C \vecs{F}·d\vecs{r} &=\int_0^1⟨4t,8t⟩·⟨2,ii⟩dt \\[4pt] &=\int_0^ane(8t+16t)dt=\int_0^i 24tdt\\[4pt] &={\big[12t^two\big]}_0^1=12. \end{marshal*}\]

Effigy \(\PageIndex{4}\): The value of line integral \(\int_C \vecs{F} \cdot d\vecs{r}\) depends only on the value of the potential part of \(\vecs{F}\) at the endpoints of the curve.

Notice that \(\vecs{F}=\vecs \nabla f\), where \(f(x,y)=10^2+2y^ii\). If we recall of the slope as a derivative, then \(f\) is an "antiderivative" of \(\vecs{F}\). In the case of single-variable integrals, the integral of derivative \(g′(x)\) is \(thousand(b)−g(a)\), where a is the start point of the interval of integration and b is the endpoint. If vector line integrals work like single-variable integrals, then nosotros would look integral \(\vecs{F}\) to be \(f(P_1)−f(P_0)\), where \(P_1\) is the endpoint of the curve of integration and \(P_0\) is the start point. Notice that this is the case for this example:

\[\int_C \vecs{F} \cdot d\vecs{r}=\int_C \vecs \nabla f \cdot d\vecs{r}=12 \nonumber\]

and

\[f(2,ii)−f(0,0)=four+viii−0=12. \nonumber\]

In other words, the integral of a "derivative" tin be calculated by evaluating an "antiderivative" at the endpoints of the curve and subtracting, just equally for unmarried-variable integrals.

The following theorem says that, under certain weather, what happened in the previous case holds for whatsoever gradient field. The same theorem holds for vector line integrals, which nosotros call the Fundamental Theorem for Line Integrals.

Theorem: THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS

Let C be a piecewise smooth bend with parameterization \(\vecs r(t)\), \(a≤t≤b\). Allow \(f\) be a function of two or iii variables with first-society partial derivatives that exist and are continuous on C. And so,

\[\int_C \vecs \nabla f \cdot d\vecs{r}=f(\vecs r(b))−f(\vecs r(a)). \label{FunTheLine}\]

Proof

First,

\[\int_C \vecs \nabla f \cdot d \vecs{r}=\int_a^b \vecs \nabla f( \vecs r(t)) \cdot \vecs r′(t)\,dt. \nonumber \]

Past the chain rule,

\[\dfrac{d}{dt}(f( \vecs r(t))= \vecs \nabla f( \vecs r(t)) \cdot \vecs r′(t) \nonumber\]

Therefore, past the Fundamental Theorem of Calculus,

\[\begin{marshal*} \int_C \vecs \nabla f \cdot d \vecs{r} &=\int_a^b \vecs \nabla f( \vecs r(t)) \cdot \vecs r′(t)dt \\[4pt] &=\int_a^b\dfrac{d}{dt}(f( \vecs r(t))dt \\[4pt] &={\big[f( \vecs r(t))\big]}_{t=a}^{t=b}\\[4pt] &=f( \vecs r(b))−f( \vecs r(a)). \stop{align*}\]

\(\square\)

We know that if \(\vecs{F}\) is a conservative vector field, there is a potential function \(f\) such that \( \vecs \nabla f= \vecs F\). Therefore

\[\int_C \vecs F·d\vecs r=\int_C\vecs \nabla f·d\vecs{r}=f(\vecs r(b))−f(\vecs r(a)).\]

In other words, just as with the Fundamental Theorem of Calculus, computing the line integral \(\int_C \vecs F·d\vecs{r}\), where \(\vecs{F}\) is conservative, is a two-step process:

1. Detect a potential function ("antiderivative") \(f\) for \(\vecs{F}\) and
2. Compute the value of \(f\) at the endpoints of \(C\) and calculate their difference \(f(\vecs r(b))−f(\vecs r(a))\).

Keep in mind, nonetheless, at that place is 1 major divergence between the Primal Theorem of Calculus and the Fundamental Theorem for Line Integrals:
A function of i variable that is continuous must have an antiderivative. Even so, a vector field, even if it is continuous, does non demand to have a potential function.

Example \(\PageIndex{3}\): Applying the Central Theorem

Calculate integral \(\int_C \vecs{F} \cdot d\vecs{r}\), where \(\vecs{F}(x,y,z)=⟨2x\ln y,\dfrac{ten^two}{y}+z^2,2yz⟩\) and \(C\) is a curve with parameterization \(\vecs{r}(t)=⟨t^2,t,t⟩\), \(ane≤t≤e\)

1. without using the Central Theorem of Line Integrals and
2. using the Fundamental Theorem of Line Integrals.

Solution

i. First, permit's calculate the integral without the Fundamental Theorem for Line Integrals and instead use the method we learned in the previous department:

\[\begin{align*} \int_C \vecs{F} \cdot dr &=\int_1^due east\vecs F(\vecs r(t)) \cdot \vecs r′(t)\,dt\\[4pt] &=\int_1^e⟨2t^ii\ln t,\dfrac{t^iv}{t}+t^2,2t^2⟩ \cdot ⟨2t,1,i⟩\,dt\\[4pt] &=\int_1^e(4t^3\ln t+t^three+3t^2)\,dt \\[4pt] &=\int_1^e 4t^3\ln t \,dt+\int_1^due east(t^3+3t^2)\,dt \\[4pt] &=\int_1^e 4t^3\ln t\,dt+{\Big[\dfrac{t^4}{4}+t^3\Large]}_1^e \\[4pt] &=\int_1^e 4t^3\ln t\,dt+\dfrac{e^4}{4}+due east^3 −\dfrac{1}{4} −one \\[4pt] &= \int_1^east 4t^iii\ln t\,dt+\dfrac{e^four}{4}+due east^3 −\dfrac{five}{4}\end{align*}\]

Integral \(\displaystyle \int_1^eastward t^iii\ln t\,dt\) requires integration by parts. Permit \(u=\ln t\) and \(dv=t^3\). Then \(u=\ln t\), \(dv=t^3\)

and

\[du=\dfrac{1}{t}\,dt, \;\;v=\dfrac{t^4}{4}.\nonumber\]

Therefore,

\[\brainstorm{align*} \int_1^e t^iii\ln t\,dt &={\Big[\dfrac{t^4}{4}\ln t\Big]}_1^eastward−\dfrac{ane}{4}\int_1^e t^3\,dt \\[4pt] &=\dfrac{e^4}{4}−\dfrac{ane}{4}\left(\dfrac{e^4}{four}−\dfrac{i}{iv}\right). \terminate{marshal*}\]

Thus,

\[\begin{align*} \int_C \vecs F \cdot d\vecs{r} &= 4\int_1^e t^3\ln t\, dt\quad +\quad \dfrac{due east^iv}{4}+eastward^3 − \dfrac{5}{4} \\[4pt] &=4\left(\dfrac{due east^4}{4}−\dfrac{1}{4}\left(\dfrac{e^4}{4}−\dfrac{1}{4}\right)\right)+\dfrac{e^4}{4}+eastward^3−\dfrac{5}{4}\\[4pt] &=e^iv−\dfrac{due east^4}{4}+\dfrac{1}{4}+\dfrac{e^4}{4}+e^three−\dfrac{v}{4} \\[4pt] &=e^4+east^three−one. \end{align*}\]

two. Given that \(f(ten,y,z)=x^2\ln y+yz^2\) is a potential function for \(\vecs F\), let's use the Key Theorem for Line Integrals to summate the integral. Note that

\[\begin{align*} \int_C \vecs F \cdot d\vecs{r} &=\int_C \vecs \nabla f \cdot d\vecs{r} \\[4pt] &=f(\vecs r(east))−f(\vecs r(1)) \\[4pt] &=f(e^2,e,eastward)−f(1,1,1)\\[4pt] &=e^4+e^3−1. \end{align*}\]

This calculation is much more straightforward than the calculation we did in (a). As long equally we accept a potential function, calculating a line integral using the Key Theorem for Line Integrals is much easier than calculating without the theorem.

Instance \(\PageIndex{three}\) illustrates a nice feature of the Central Theorem of Line Integrals: information technology allows u.s. to summate more easily many vector line integrals. Every bit long as we have a potential function, computing the line integral is just a matter of evaluating the potential function at the endpoints and subtracting.

Exercise \(\PageIndex{3}\)

Given that \(f(x,y)={(x−one)}^2y+{(y+ane)}^2x\) is a potential function for \(\vecs F(x,y)=⟨2xy−2y+{(y+ane)}^2,{(10−1)}^2+2yx+2x⟩\), calculate integral \(\int_C \vecs F·d\vecs r\), where \(C\) is the lower half of the unit of measurement circumvolve oriented counterclockwise.

Hint

The Fundamental Theorem for Line Intervals says this integral depends but on the value of \(f\) at the endpoints of \(C\).

Answer

ii

The Fundamental Theorem for Line Integrals has 2 of import consequences. The beginning consequence is that if \(\vecs{F}\) is conservative and \(C\) is a closed curve, and so the circulation of \(\vecs{F}\) along \(C\) is cipher—that is, \(\int_C \vecs F·d\vecs r=0\). To run across why this is true, let \(f\) be a potential office for \(\vecs{F}\). Since \(C\) is a closed bend, the last indicate \(\vecs r(b)\) of \(C\) is the same equally the initial \(\vecs r(a)\) of \(C\)—that is, \(\vecs r(a)=\vecs r(b)\). Therefore, by the Cardinal Theorem for Line Integrals,

\[\brainstorm{marshal} \oint_C \vecs F·d\vecs r &=\oint_C \vecs \nabla f·d\vecs r\\[4pt] &=f(\vecs r(b))−f(\vecs r(a)) \\[4pt] &=f(\vecs r(b))−f(\vecs r(b)) \\[4pt] &=0. \end{align}\]

Recall that the reason a conservative vector field \(\vecs{F}\) is called "conservative" is because such vector fields model forces in which free energy is conserved. We have shown gravity to be an instance of such a force. If we think of vector field \(\vecs{F}\) in integral \(\oint_C \vecs F·d\vecs r\) as a gravitational field, then the equation \(\oint_C \vecs{F}·d\vecs{r}=0\) follows. If a particle travels along a path that starts and ends at the same identify, and so the piece of work done by gravity on the particle is naught.

The 2nd important outcome of the Primal Theorem for Line Integrals (Equation \ref{FunTheLine}) is that line integrals of conservative vector fields are contained of path—meaning, they depend only on the endpoints of the given curve, and exercise not depend on the path between the endpoints.

DEFINITION: Path Independence

Let \(\vecs{F}\) be a vector field with domain \(D\); it is independent of path (or path contained) if

\[\int_{C_1} \vecs{F}·d\vecs{r}=\int_{C_2} \vecs{F}·d\vecs{r}\]

for whatever paths \(C_1\) and \(C_2\) in \(D\)with the same initial and last points.

The second event is stated formally in the following theorem.

Theorem: CONSERVATIVE FIELDS

If \(\vecs{F}\) is a conservative vector field, then \(\vecs{F}\) is independent of path.

Proof

Let \(D\) announce the domain of \(\vecs{F}\) and let \(C_1\) and \(C_2\) be two paths in \(D\) with the same initial and terminal points (Effigy \(\PageIndex{5}\)). Call the initial point \(P_1\) and the terminal point \(P_2\). Since \(\vecs{F}\) is conservative, there is a potential office \(f\) for \(\vecs{F}\). Past the Fundamental Theorem for Line Integrals,

\[\int_{C_1} \vecs{F}·d\vecs{r}=f(P_2)−f(P_1)=\int_{C_2} \vecs{F}·d\vecs{r}. \nonumber\]

Therefore, \(\int_{C_1}\vecs F·d\vecs r=\int_{C_2}\vecs F·d\vecs r\) and \(\vecs{F}\) is contained of path.

\(\foursquare\)

To visualize what independence of path means, imagine three hikers climbing from base army camp to the pinnacle of a mountain. Hiker i takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is non steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides information technology is too hard for him. Therefore he returns to campsite 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 full amount of work done by gravity on each of the hikers is the same because they all started in the same place and concluded in the aforementioned place. The work done past the hikers includes other factors such as friction and musculus move, so the full amount of free energy each 1 expended is not the aforementioned, simply the net energy expended confronting gravity is the aforementioned for all iii hikers.

Figure \(\PageIndex{4}\): The vector field is bourgeois, and therefore independent of path.

We have shown that if \(\vecs{F}\) is conservative, then \(\vecs{F}\) is contained of path. It turns out that if the domain of \(\vecs{F}\) is open up and connected, then the converse is likewise true. That is, if \(\vecs{F}\) is contained of path and the domain of \(\vecs{F}\) is open and connected, so \(\vecs{F}\) is conservative. Therefore, the fix of conservative vector fields on open and connected domains is precisely the ready of vector fields independent of path.

Theorem: THE PATH INDEPENDENCE TEST FOR CONSERVATIVE FIELDS

If \(\vecs{F}\) is a continuous vector field that is contained of path and the domain \(D\) of \(\vecs{F}\) is open and connected, and so \(\vecs{F}\) is conservative.

Proof

We prove the theorem for vector fields in \(ℝ^2\). The proof for vector fields in \(ℝ^3\) is like. To show that \(\vecs F=⟨P,Q⟩\) is conservative, we must notice a potential function \(f\) for \(\vecs{F}\). To that end, permit \(X\) be a fixed point in \(D\). For any point \((x,y)\) in \(D\), permit \(C\) be a path from \(X\) to \((10,y)\). Define \(f(x,y)\) past \(f(x,y)=\int_C \vecs F·d\vecs r\). (Annotation that this definition of \(f\) makes sense but because \(\vecs{F}\) is independent of path. If \(\vecs{F}\) was not independent of path, then it might be possible to find another path \(C′\) from \(X\) to \((x,y)\) such that \(\int_C \vecs F·d\vecs r≠\int_C \vecs F·d\vecs r\), and in such a case \(f(x,y)\) would not exist a function.) We want to bear witness that \(f\) has the property \(\vecs \nabla f=\vecs F\).

Since domain \(D\) is open, it is possible to find a disk centered at \((x,y)\) such that the disk is contained entirely inside \(D\). Let \((a,y)\) with \(a<x\) be a signal in that disk. Let \(C\) be a path from \(X\) to \((x,y)\) that consists of two pieces: \(C_1\) and \(C_2\). The offset piece, \(C_1\), is any path from \(C\) to \((a,y)\) that stays inside \(D\); \(C_2\) is the horizontal line segment from \((a,y)\) to \((ten,y)\) (Effigy \(\PageIndex{6}\)). Then

\[f(x,y)=\int_{C_1} \vecs F·d\vecs r+\int_{C_2}\vecs F \cdot d\vecs r.\nonumber\]

The showtime integral does not depend on \(10\), then

\[f_x(x,y)=\dfrac{∂}{∂x}\int_{C_2} \vecs F \cdot d\vecs r. \nonumber\]

If we parameterize \(C_2\) by \(\vecs r(t)=⟨t,y⟩\), \(a≤t≤x\), and so

\[\begin{align*} f_x(x,y) &=\dfrac{∂}{∂ten}\int_{C_2} \vecs F \cdot d\vecs r \\[4pt] &=\dfrac{∂}{∂10}\int_a^x \vecs F(\vecs r(t)) \cdot \vecs r′(t)\,dt \\[4pt] &=\dfrac{∂}{∂x}\int_a^10 \vecs F(\vecs r(t)) \cdot \dfrac{d}{dt}(⟨t,y⟩)\,dt \\[4pt] &=\dfrac{∂}{∂x}\int_a^x \vecs F(\vecs r(t)) \cdot ⟨1,0⟩\,dt \\[4pt] &=\dfrac{∂}{∂x}\int_a^x P(t,y)\,dt.\\[4pt] \end{align*}\]

By the Fundamental Theorem of Calculus (part ane),

\[f_x(x,y)=\dfrac{∂}{∂10}\int_a^10 P(t,y)\,dt=P(10,y).\nonumber\]

Figure \(\PageIndex{6}\). Here, \(C_1\) is whatever path from \(C\) to \((a,y)\) that stays inside \(D\), and \(C_2\) is the horizontal line segment from \((a,y)\) to \((10,y)\).

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

Therefore \(\vecs \nabla f=\vecs F\) and \(\vecs{F}\) is conservative.

\(\foursquare\)

We accept spent a lot of time discussing and proving the theorems above, just nosotros can summarize them simply: a vector field \(\vecs F\) on an open up and continued domain is conservative if and but if it is contained of path. This is important to know considering conservative vector fields are extremely of import in applications, and these theorems give usa a dissimilar way of viewing what information technology means to be bourgeois using path independence.

Case \(\PageIndex{four}\): Showing That a Vector Field Is Not Conservative

Apply path independence to show that vector field \(\vecs F(10,y)=⟨x^2y,y+5⟩\) is not conservative.

Solution

We tin can indicate that \(\vecs{F}\) is not conservative by showing that \(\vecs{F}\) is non path contained. Nosotros practise so by giving ii different paths, \(C_1\) and \(C_2\), that both get-go at \((0,0)\) and end at \((1,i)\), and still \(\int_{C_1} \vecs F \cdot d\vecs r≠\int_{C_2} \vecs F \cdot d\vecs r\).

Allow \(C_1\) be the curve with parameterization \(\vecs r_1(t)=⟨t,\,t⟩\), \(0≤t≤one\) and let \(C_2\) exist the curve with parameterization \(\vecs r_2(t)=⟨t,\,t^2⟩\), \(0≤t≤ane\) (Figure \(\PageIndex{7}\).). Then

\[\begin{marshal*} \int_{C_1} \vecs{F}·d\vecs r &=\int_0^1 \vecs F(\vecs r_1(t))·\vecs r_1′(t)\,dt \\[4pt] &=\int_0^ane⟨t^3,t+5⟩·⟨1,i⟩\,dt=\int_0^1(t^3+t+5)\,dt\\[4pt] &={\Big[\dfrac{t^4}{four}+\dfrac{t^two}{ii}+5t\Large]}_0^ane=\dfrac{23}{four} \end{marshal*}\]

and

\[\brainstorm{align*} \int_{C_2}\vecs F·d\vecs r &=\int_0^1 \vecs F(\vecs r_2(t))·\vecs r_2′(t)\,dt \\[4pt] &=\int_0^1⟨t^4,t^2+5⟩·⟨1,2t⟩\,dt=\int_0^i(t^four+2t^iii+10t)\,dt \\[4pt] &={\Big[\dfrac{t^5}{5}+\dfrac{t^4}{2}+5t^2\Big]}_0^ane=\dfrac{57}{10}. \terminate{marshal*}\]

Since \(\int_{C_1} \vecs F \cdot d\vecs r≠\int_{C_2} \vecs F \cdot d\vecs r\), the value of a line integral of \(\vecs{F}\) depends on the path between ii given points. Therefore, \(\vecs{F}\) is non independent of path, and \(\vecs{F}\) is non conservative.

Effigy \(\PageIndex{7}\): Curves \(C_1\) and \(C_2\) are both oriented from left to right.

Exercise \(\PageIndex{four}\)

Show that \(\vecs{F}(ten,y)=⟨xy,\,x^2y^2⟩\) is not path independent past considering the line segment from \((0,0)\) to \((0,2)\) and the piece of the graph of \(y=\dfrac{x^two}{ii}\) that goes from \((0,0)\) to \((0,two)\).

Hint

Calculate the respective line integrals.

Answer

If \(C_1\) and \(C_2\) represent the 2 curves, and so \[\int_{C_1} \vecs F \cdot d\vecs r≠\int_{C_2} \vecs F \cdot d\vecs r. \nonumber\]

Conservative Vector Fields and Potential Functions

Every bit we take learned, the Primal Theorem for Line Integrals says that if \(\vecs{F}\) is conservative, so computing \(\int_C \vecs F·d\vecs r\) has ii steps: beginning, find a potential function \(f\) for \(\vecs{F}\) and, 2nd, summate \(f(P_1)−f(P_0)\), where \(P_1\) is the endpoint of \(C\) and \(P_0\) is the starting point. To apply this theorem for a conservative field \(\vecs{F}\), we must exist able to find a potential function \(f\) for \(\vecs{F}\). Therefore, nosotros must answer the post-obit question: Given a conservative vector field \(\vecs{F}\), how practise we find a function \(f\) such that \(\vecs \nabla f=\vecs{F}\)? Before giving a general method for finding a potential function, let'southward motivate the method with an example.

Example \(\PageIndex{5}\): Finding a Potential Function

Notice a potential function for \(\vecs F(ten,y)=⟨2xy^3,3x^2y^2+\cos(y)⟩\), thereby showing that \(\vecs{F}\) is bourgeois.

Solution

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

\[f_x(x,y)=2xy^3 \; \; \text{and} \;\; f_y(ten,y)=3x^2y^2+\cos y. \nonumber\]

Integrating the equation \(f_x(x,y)=2xy^3\) with respect to \(ten\) yields the equation

\[f(10,y)=ten^2y^3+h(y). \nonumber\]

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 \(x\), simply may still be a function of \(y\). The equation \(f(x,y)=x^2y^iii+h(y)\) tin be confirmed by taking the fractional derivative with respect to \(x\):

\[\dfrac{∂f}{∂x}=\dfrac{∂}{∂ten}(x^2y^iii)+\dfrac{∂}{∂x}(h(y))=2xy^three+0=2xy^iii. \nonumber\]

Since \(f\) is a potential office for \(\vecs{F}\),

\[f_y(x,y)=3x^2y^2+\cos(y), \nonumber\]

and therefore

\[3x^2y^ii+g′(y)=3x^2y^2+\cos(y). \nonumber\]

This implies that \(h′(y)=\cos y\), so \(h(y)=\sin y+C\). Therefore, any function of the form \(f(x,y)=x^2y^3+\sin(y)+C\) is a potential part. 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, notation that \(\vecs \nabla f(x,y)=⟨2xy^3,3x^2y^ii+\cos y⟩=\vecs F\).

Exercise \(\PageIndex{v}\)

Find a potential role for \(\vecs{F}(x,y)=⟨due east^xy^3+y,3e^xy^2+x⟩\).

Hint

Follow the steps in Case \(\PageIndex{5}\).

Answer

\(f(x,y)=e^xy^iii+xy\)

The logic of the previous instance extends to finding the potential function for whatever conservative vector field in \(ℝ^2\). Thus, we have the following problem-solving strategy for finding potential functions:

PROBLEM-SOLVING STRATEGY: FINDING A POTENTIAL FUNCTION FOR A Conservative VECTOR FIELD \(\vecs{F}(x,y)=⟨P(10,y),Q(x,y)⟩\)

1. Integrate \(P\) with respect to \(x\). This results in a function of the grade \(g(x,y)+h(y)\), where \(h(y)\) is unknown.
2. Have the partial derivative of \(g(x,y)+h(y)\) with respect to \(y\), which results in the part \(gy(x,y)+h′(y)\).
3. Use the equation \(gy(x,y)+h′(y)=Q(x,y)\) to find \(h′(y)\).
4. Integrate \(h′(y)\) to discover \(h(y)\).
5. Whatsoever function of the class \(f(x,y)=g(x,y)+h(y)+C\), where \(C\) is a constant, is a potential function for \(\vecs{F}\).

We tin can adapt this strategy to detect potential functions for vector fields in \(ℝ^3\), every bit shown in the adjacent example.

Instance \(\PageIndex{6}\): Finding a Potential Function in \(ℝ^3\)

Detect a potential function for \(F(x,y,z)=⟨2xy,x^2+2yz^3,3y^2z^two+2z⟩\), thereby showing that \(\vecs{F}\) is bourgeois.

Solution

Suppose that \(f\) is a potential function. And so, \(\vecs \nabla f= \vecs{F}\) and therefore \(f_x(10,y,z)=2xy\). Integrating this equation with respect to \(x\) yields the equation \(f(x,y,z)=x^2y+g(y,z)\) for some function \(grand\). Find that, in this instance, the constant of integration with respect to \(10\) is a role of \(y\) and \(z\).

Since \(f\) is a potential function,

\[ten^2+2yz^3=f_y(ten,y,z)=x^2+g_y(y,z). \nonumber\]

Therefore,

\[g_y(y,z)=2yz^iii. \nonumber\]

Integrating this function with respect to \(y\) yields

\[grand(y,z)=y^2z^3+h(z) \nonumber\]

for some function \(h(z)\) of \(z\) alone. (Notice that, because we know that \(1000\) is a function of only \(y\) and \(z\), nosotros practice non need to write \(g(y,z)=y^2z^3+h(x,z)\).) Therefore,

\[f(x,y,z)=x^2y+g(y,z)=x^2y+y^2z^iii+h(z). \nonumber\]

To observe \(f\), we now must only observe \(h\). Since \(f\) is a potential role,

\[3y^2z^ii+2z=g_z(y,z)=3y^2z^two+h′(z). \nonumber\]

This implies that \(h′(z)=2z\), so \(h(z)=z^2+C\). Letting \(C=0\) gives the potential office

\[f(x,y,z)=x^2y+y^2z^3+z^two. \nonumber\]

To verify that \(f\) is a potential function, note that \(\vecs \nabla f(10,y,z)=⟨2xy,x^2+2yz^3,3y^2z^2+2z⟩=\vecs F(x,y,z)\).

Practise \(\PageIndex{half-dozen}\)

Observe a potential function for \(\vecs{F}(x,y,z)=⟨12x^2,\cos y\cos z,1−\sin y\sin z⟩\).

Hint

Post-obit Example \(\PageIndex{6}\), begin by integrating with respect to \(x\).

Answer

\(f(x,y,z)=4x^three+\sin y\cos z+z\)

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

\(\vecs F(10,y)=−G\left\langle\dfrac{ten}{ {(x^2+y^2)}^{three/2} },\dfrac{y}{ {(x^2+y^2)}^{3/2} }\right\rangle\),

where \(G\) is the universal gravitational constant. In the side by side example, we build a potential role for \(\vecs{F}\), thus confirming what nosotros already know: that gravity is bourgeois.

Example \(\PageIndex{7}\): Finding a Potential Part

Detect a potential function \(f\) for \(\vecs{F}(x,y)=−Chiliad\left\langle\dfrac{ten}{ {(x^ii+y^ii)}^{3/ii} },\dfrac{y}{ {(x^2+y^2)}^{iii/2} }\right\rangle\).

Solution

Suppose that \(f\) is a potential role. Then, \(\vecs \nabla f= \vecs{F}\) and therefore

\[f_x(x,y)=\dfrac{−Gx}{ {(x^2+y^two)}^{3/2} }.\nonumber\]

To integrate this role with respect to \(10\), we tin apply \(u\)-substitution. If \(u=x^ii+y^2\), then \(\dfrac{du}{2}=x\,dx\), and then

\[\brainstorm{align*} \int \dfrac{−Gx}{ {(x^2+y^2)}^{3/2} }\,dx &=\int \dfrac{−G}{2u^{three/2}} \,du \\[4pt] &=\dfrac{G}{\sqrt{u}}+h(y) \\[4pt] &=\dfrac{Chiliad}{\sqrt{ten^ii+y^2}}+h(y) \end{align*}\]

for some role \(h(y)\). Therefore,

\[f(x,y)=\dfrac{G}{ \sqrt{x^two+y^2}}+h(y).\nonumber\]

Since \(f\) is a potential office for \(\vecs{F}\),

\[f_y(x,y)=\dfrac{−Gy}{ {(x^2+y^2)}^{3/two} }\nonumber\].

Since \(f(ten,y)=\dfrac{G}{ \sqrt{x^2+y^2}}+h(y)\), \(f_y(10,y)\) also equals \(\dfrac{−Gy}{ {(ten^2+y^ii)}^{three/2} }+h′(y)\).

Therefore,

\[\dfrac{−Gy}{ {(x^2+y^2)}^{3/2} }+h′(y)=\dfrac{−Gy}{ {(x^2+y^2)}^{three/two} }, \nonumber\]

which implies that \(h′(y)=0\). Thus, we can have \(h(y)\) to exist whatever constant; in particular, we can let \(h(y)=0\). The function

\[f(ten,y)=\dfrac{1000}{ \sqrt{ten^2+y^2} } \nonumber\]

is a potential function for the gravitational field \(\vecs{F}\). To ostend that \(f\) is a potential function, annotation that

\[\brainstorm{align*} \vecs\nabla f(10,y) &=⟨−\dfrac{1}{2} \dfrac{Yard}{ {(ten^2+y^two)}^{iii/2} } (2x),−\dfrac{one}{two} \dfrac{1000}{ {(x^two+y^ii)}^{3/2} }(2y)⟩ \\[4pt] &=⟨\dfrac{−Gx}{ {(x^2+y^ii)}^{3/2} },\dfrac{−Gy}{ {(x^two+y^2)}^{three/2} }⟩\\[4pt] &=\vecs F(x,y). \end{marshal*}\]

Exercise \(\PageIndex{seven}\)

Notice a potential function \(f\) for the three-dimensional gravitational force \(\vecs{F}(x,y,z)=\left\langle\dfrac{−Gx}{ {(x^two+y^2+z^two)}^{3/2} },\dfrac{−Gy}{ {(ten^2+y^2+z^2)}^{3/2} },\dfrac{−Gz}{ {(x^ii+y^2+z^2)}^{3/2} }\right\rangle\).

Hint

Follow the Problem-Solving Strategy.

Answer

\(f(ten,y,z)=\dfrac{G}{\sqrt{x^2+y^two+z^two}}\)

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 bourgeois, we need to exist able to test whether it is bourgeois. Remember that, if \(\vecs{F}\) is conservative, and then \(\vecs{F}\) has the cross-partial property (see The Cross-Partial Holding of Bourgeois Vector Fields). That is, if \(\vecs F=⟨P,Q,R⟩\) is bourgeois, then \(P_y=Q_x\), \(P_z=R_x\), and \(Q_z=R_y\). And so, if \(\vecs{F}\) has the cross-partial property, then is \(\vecs{F}\) conservative? If the domain of \(\vecs{F}\) is open and simply connected, and then the respond is yes.

Theorem: THE CROSS-Fractional Exam FOR Bourgeois FIELDS

If \(\vecs{F}=⟨P,Q,R⟩\) is a vector field on an open, merely connected region \(D\) and \(P_y=Q_x\), \(P_z=R_x\), and \(Q_z=R_y\) throughout \(D\), and then \(\vecs{F}\) is conservative.

Although a proof of this theorem is across the scope of the text, nosotros can notice its ability with some examples. Later, we see why it is necessary for the region to be just connected.

Combining this theorem with the cantankerous-partial property, we can determine whether a given vector field is bourgeois:

Theorem: Cantankerous-Fractional Holding OF Conservative FIELDS

Permit \(\vecs{F}=⟨P,Q,R⟩\) be a vector field on an open, only continued region \(D\). Then \(P_y=Q_x\), \(P_z=R_x\), and \(Q_z=R_y\) throughout \(D\) if and only if \(\vecs{F}\) is conservative.

The version of this theorem in \(ℝ^ii\) is also true. If \(\vecs F(ten,y)=⟨P,Q⟩\) is a vector field on an open, simply continued domain in \(ℝ^2\), then \(\vecs F\) is conservative if and but if \(P_y=Q_x\).

Case \(\PageIndex{eight}\): Determining Whether a Vector Field Is Conservative

Make up one's mind whether vector field \(\vecs F(x,y,z)=⟨xy^2z,10^2yz,z^2⟩\) is conservative.

Solution

Notation that the domain of \(\vecs{F}\) is all of \(ℝ^two\) and \(ℝ^3\) is but connected. Therefore, we can use The Cross-Partial Holding of Conservative Vector Fields to determine whether \(\vecs{F}\) is conservative. Let

\[P(x,y,z)=xy^2z \nonumber\]

\[Q(x,y,z)=10^2yz \nonumber\]

and

\[R(x,y,z)=z^ii.\nonumber\]

Since \(Q_z(x,y,z)=x^2y\) and \(R_y(x,y,z)=0\), the vector field is not conservative.

Example \(\PageIndex{nine}\): Determining Whether a Vector Field Is Conservative

Decide vector field \(\vecs{F}(x,y)=⟨10\ln (y), \,\dfrac{x^2}{2y}⟩\) is bourgeois.

Solution

Note that the domain of \(\vecs{F}\) is the function of \(ℝ^2\) in which \(y>0\). Thus, the domain of \(\vecs{F}\) is role of a airplane above the \(x\)-axis, and this domain is simply connected (there are no holes in this region and this region is continued). Therefore, we can employ The Cross-Fractional Property of Conservative Vector Fields to make up one's mind whether \(\vecs{F}\) is conservative. Permit

\[P(x,y)=x\ln (y) \;\; \text{and} \;\;\ Q(x,y)=\dfrac{x^2}{2y}. \nonumber\]

Then \(P_y(ten,y)=\dfrac{x}{y}=Q_x(x,y)\) and thus \(\vecs{F}\) is conservative.

Exercise \(\PageIndex{8}\)

Determine whether \(\vecs{F}(x,y)=⟨\sin x\cos y,\,\cos x\sin y⟩\) is conservative.

Hint

Use The Cross-Partial Property of Conservative Vector Fields from the previous department.

Answer

It is conservative.

When using The Cross-Partial Belongings of Bourgeois Vector Fields, information technology is of import to remember that a theorem is a tool, and like any tool, it can exist practical only nether the right conditions. In the case of The Cross-Fractional Property of Bourgeois Vector Fields, the theorem tin can be applied just if the domain of the vector field is simply connected.

To see what can get wrong when misapplying the theorem, consider the vector field from Example \(\PageIndex{four}\):

\[\vecs F(x,y)=\dfrac{y}{x^ii+y^2}\,\lid{\mathbf i}+\dfrac{−ten}{x^2+y^2}\,\hat{\mathbf j}.\]

This vector field satisfies the cantankerous-partial property, since

\[\dfrac{∂}{∂y}\left(\dfrac{y}{ten^ii+y^two}\correct)=\dfrac{(x^ii+y^2)−y(2y)}{ {(ten^2+y^2)}^2}=\dfrac{x^2−y^two}{ {(x^ii+y^2)}^ii}\]

and

\[\dfrac{∂}{∂x}\left(\dfrac{−x}{x^2+y^2}\right)=\dfrac{−(ten^ii+y^two)+x(2x)}{ {(10^2+y^2)}^2}=\dfrac{x^2−y^2}{ {(x^two+y^two)}^two}.\]

Since \(\vecs{F}\) satisfies the cross-partial property, nosotros might exist tempted to conclude that \(\vecs{F}\) is bourgeois. Even so, \(\vecs{F}\) is not conservative. To meet this, allow

\[\vecs r(t)=⟨\cos t,\sin t⟩,\;\; 0≤t≤\pi\]

exist a parameterization of the upper one-half of a unit circle oriented counterclockwise (denote this \(C_1\)) and let

\[\vecs s(t)=⟨\cos t,−\sin t⟩,\;\; 0≤t≤\pi\]

exist a parameterization of the lower half of a unit circumvolve oriented clockwise (denote this \(C_2\)). Find that \(C_1\) and \(C_2\) have the same starting point and endpoint. Since \({\sin}^2 t+{\cos}^two t=1\),

\[\vecs F(\vecs r(t)) \cdot \vecs r′(t)=⟨\sin(t),−\cos(t)⟩ \cdot ⟨−\sin(t), \cos(t)⟩=−1\]

and

\[\vecs F(\vecs southward(t))·\vecs s′(t)=⟨−\sin t,−\cos t⟩·⟨−\sin t,−\cos t⟩={\sin}^two t+{\cos}^2t=1.\]

Therefore,

\[\int_{C_1} \vecs F·d\vecs r=\int_0^{\pi}−i\,dt=−\pi\]

and

\[\int_{C_2}\vecs F·d\vecs r=\int_0^{\pi} i\,dt=\pi.\]

Thus, \(C_1\) and \(C_2\) accept the aforementioned starting signal and endpoint, simply \(\int_{C_1} \vecs F·d\vecs r≠\int_{C_2} \vecs F·d\vecs r\). Therefore, \(\vecs{F}\) is not independent of path and \(\vecs{F}\) is not conservative.

To summarize: \(\vecs{F}\) satisfies the cross-fractional property and all the same \(\vecs{F}\) is not conservative. What went wrong? Does this contradict The Cross-Partial Belongings of Conservative Vector Fields? The effect is that the domain of \(\vecs{F}\) is all of \(ℝ^ii\) except for the origin. In other words, the domain of \(\vecs{F}\) has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected, The Cantankerous-Partial Property of Conservative Vector Fields does non apply to \(\vecs{F}\).

We close this section by looking at an case of the usefulness of the Primal Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always determine whether the Central Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form \(\int_C \vecs F·d\vecs r\), then our outset question should be: Is \(\vecs{F}\) conservative? If the answer is yeah, then nosotros should find a potential function and employ the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, and then the Primal Theorem for Line Integrals cannot assist u.s.a. and we have to apply other methods, such as using the method from the previous section (using \(\vecs F(\vecs r(t))\) and \(\vecs r'(t)\)).

Example \(\PageIndex{x}\): Using the Fundamental Theorem for Line Integrals

Calculate line integral \(\int_C \vecs F·d\vecs r\), where \(\vecs F(x,y,z)=⟨2xe^yz+eastward^xz,\,x^2e^yz,\,x^2e^y+e^x⟩\) and \(C\) is any polish curve that goes from the origin to \((1,1,1)\).

Solution

Before trying to compute the integral, we need to decide whether \(\vecs{F}\) is conservative and whether the domain of \(\vecs{F}\) is only connected. The domain of \(\vecs{F}\) is all of \(ℝ^3\), which is connected and has no holes. Therefore, the domain of \(\vecs{F}\) is just continued. Let

\[P(x,y,z)=2xe^yz+e^xz, \;\; Q(x,y,z)=x^2e^yz, \;\; \text{and} \;\; R(x,y,z)=10^2e^y+east^x \nonumber\]

so that \(\vecs{F}(x,y,z)=⟨P,Q,R⟩\). Since the domain of \(\vecs{F}\) is simply connected, we can check the cross partials to determine whether \(\vecs{F}\) is bourgeois. Note that

\[\begin{align*} P_y(x,y,z) &=2xe^yz=Q_x(x,y,z) \\[4pt]P_z(x,y,z) &=2xe^y+eastward^x=R_x(10,y,z) \\[4pt] Q_z(x,y,z) &=10^2e^y=R_y(x,y,z).\end{marshal*}\]

Therefore, \(\vecs{F}\) is conservative.

To evaluate \(\int_C \vecs F·d\vecs r\) using the Fundamental Theorem for Line Integrals, we need to observe a potential function \(f\) for \(\vecs{F}\). Let \(f\) be a potential function for \(\vecs{F}\). Then, \(\vecs \nabla f=\vecs F\), and therefore \(f_x(x,y,z)=2xe^yz+eastward^xz\). Integrating this equation with respect to \(x\) gives \(f(x,y,z)=x^2e^yz+east^xz+h(y,z)\) for some office \(h\). Differentiating this equation with respect to \(y\) gives \(x^2e^yz+h_y(y,z)=Q(x,y,z)=x^2e^yz\), which implies that \(h_y(y,z)=0\). Therefore, \(h\) is a function of \(z\) simply, and \(f(ten,y,z)=x^2e^yz+east^xz+h(z)\). To find \(h\), annotation that \(f_z=x^2e^y+eastward^x+h′(z)=R=x^2e^y+e^x\). Therefore, \(h′(z)=0\) and we can take \(h(z)=0\). A potential function for \(\vecs{F}\) is \(f(x,y,z)=x^2e^yz+east^xz\).

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

\[\brainstorm{align*} \int_C \vecs F·d\vecs r &=\int_C \vecs \nabla f·d\vecs r\\[4pt] &=f(1,1,1)−f(0,0,0)\\[4pt] &=2e. \finish{align*}\]

Analysis

Notice that if we hadn't recognized that \(\vecs{F}\) is bourgeois, we would have had to parameterize \(C\) and employ the method from the previous section. Since curve \(C\) is unknown, using the Central Theorem for Line Integrals is much simpler.

Practice \(\PageIndex{nine}\)

Calculate integral \(\int_C \vecs F·d\vecs r\), where \(\vecs{F}(10,y)=⟨\sin x\sin y, 5−\cos 10\cos y⟩\) and \(C\) is a semicircle with starting point \((0,\pi)\) and endpoint \((0,−\pi)\).

Hint

Utilize the Fundamental Theorem for Line Integrals.

Respond

\(−ten\pi\)

Example \(\PageIndex{11}\): Work Washed on a Particle

Allow \(\vecs F(x,y)=⟨2xy^2,2x^2y⟩\) be a force field. Suppose that a particle begins its motion at the origin and ends its movement at whatever indicate in a plane that is not on the \(x\)-axis or the \(y\)-axis. Furthermore, the particle's motion can exist modeled with a smooth parameterization. Show that \(\vecs{F}\) does positive work on the particle.

Solution

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

To show that \(\vecs{F}\) is conservative, suppose \(f(x,y)\) were a potential office for \(\vecs{F}\). Then, \(\vecs \nabla f(x,y)=\vecs F(x,y)=⟨2xy^two,2x^2y⟩\) and therefore \(f_x(ten,y)=2xy^2\) and \(f_y(x,y)=2x^2y\). The equation \(fx(x,y)=2xy^2\) implies that \(f(x,y)=x^2y^2+h(y)\). Deriving both sides with respect to \(y\) yields \(f_y(x,y)=2x^2y+h′(y)\). Therefore, \(h′(y)=0\) and nosotros can take \(h(y)=0\).

If \(f(x,y)=x^2y^2\), then note that \(\vecs \nabla f(x,y)=⟨2xy^2,2x^2y⟩=\vecs F\), and therefore \(f\) is a potential part for \(\vecs{F}\).

Let \((a,b)\) be the point at which the particle stops is motion, and let \(C\) denote the bend that models the particle's motion. The work washed by \(\vecs{F}\) on the particle is \(\int_C \vecs{F}·d\vecs{r}\). Past the Fundamental Theorem for Line Integrals,

\[\begin{align*} \int_C \vecs F·d\vecs r &=\int_C \nabla f·d\vecs r \\[4pt] &=f(a,b)−f(0,0)\\[4pt] &=a^2b^2. \end{align*}\]

Since \(a≠0\) and \(b≠0\), by supposition, \(a^2b^ii>0\). Therefore, \(\int_C \vecs F·d\vecs r>0\), and \(\vecs{F}\) does positive work on the particle.

Analysis

Notice that this trouble would exist much more difficult without using the Fundamental Theorem for Line Integrals. To utilise the tools nosotros accept learned, nosotros would demand to give a curve parameterization and use the method from the previous section. Since the path of motion \(C\) can be as exotic as nosotros wish (as long as it is shine), it tin can be very difficult to parameterize the motion of the particle.

Exercise \(\PageIndex{10}\)

Let \(\vecs{F}(x,y)=⟨4x^3y^four,4x^4y^iii⟩\), and suppose that a particle moves from betoken \((4,4)\) to \((ane,1)\) forth whatsoever smooth bend. Is the work washed by \(\vecs{F}\) on the particle positive, negative, or zero?

Hint

Use the Fundamental Theorem for Line Integrals.

Respond

Negative

Key Concepts

• The theorems in this section require curves that are airtight, uncomplicated, or both, and regions that are connected or merely connected.
• The line integral of a conservative vector field can be calculated using the Central Theorem for Line Integrals. This theorem is a generalization of the Fundamental Theorem of Calculus in higher dimensions. Using this theorem usually makes the calculation of the line integral easier.
• Conservative fields are independent of path. The line integral of a conservative field depends only on the value of the potential office at the endpoints of the domain curve.
• Given vector field \(\vecs{F}\), we can test whether \(\vecs{F}\) is bourgeois past using the cantankerous-partial holding. If \(\vecs{F}\) has the cantankerous-partial property and the domain is simply continued, then \(\vecs{F}\) is conservative (and thus has a potential function). If \(\vecs{F}\) is conservative, nosotros tin find a potential function past using the Problem-Solving Strategy.
• The circulation of a conservative vector field on a merely connected domain over a closed curve is zero.

Key Equations

• Central Theorem for Line Integrals
  \(\displaystyle \int_C \vecs \nabla f·d\vecs r=f(\vecs r(b))−f(\vecs r(a))\)
• Circulation of a conservative field over bend C that encloses a simply connected region
  \(\displaystyle \oint_C \vecs \nabla f·d\vecs r=0\)

Glossary

closed curve

a curve that begins and ends at the same bespeak

connected region

a region in which any two points can exist continued by a path with a trace independent entirely inside the region

Key Theorem for Line Integrals

the value of line integral \(\displaystyle \int_C\vecs ∇f⋅d\vecs r\) depends only on the value of \(f\) at the endpoints of \(C: \displaystyle \int_C \vecs ∇f⋅d\vecs r=f(\vecs r(b))−f(\vecs r(a))\)

independence of path

a vector field \(\vecs{F}\) has path independence if \(\displaystyle \int_{C_1} \vecs F⋅d\vecs r=\displaystyle \int_{C_2} \vecs F⋅d\vecs r\) for whatever curves \(C_1\) and \(C_2\) in the domain of \(\vecs{F}\) with the same initial points and terminal points

simple bend

a bend that does not cantankerous itself

simply connected region

a region that is continued and has the holding that any closed curve that lies entirely inside the region encompasses points that are entirely within the region

Contributors and Attributions

• Gilbert Strang (MIT) and Edwin "Jed" Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC four.0 license. Download for complimentary at http://cnx.org.

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Source: https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/16%3A_Vector_Calculus/16.3%3A_Conservative_Vector_Fields

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