# 16.3: Path Independence, Conservative Fields, and Potential Functions

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For certain vector fields, the amount of work required to move a particle from one point to another is dependent only on its initial and final positions, not on the path it takes. This section will discuss the properties of these vector fields.

Definition: Path Independent and Conservative

Let (mathbf{F}) be a vector field defined on an open region D in space, and suppose that for any two points A and B in D the line integral

[int_{C}^{ }mathbf{F}cdot mathit{d}mathbf{r}]

along a path C from A to B in D is the same over all paths from A to B. Then the integral [int_{C}^{ }mathbf{F}cdot mathit{d}mathbf{r}] is path independent in D and the field F is conservative on D.

## Potential Function

Definition: If F is a vector field defined on D and [mathbf{F}= riangledown f] for some scalar function f on D, then f is called a potential function for F. You can calculate all the line integrals in the domain F over any path between A and B after finding the potential function f

[int_{A}^{B}mathbf{F}cdot mathit{d}mathbf{r}=int_{A}^{B} riangledown fmathit{d}mathbf{r}=mathit{f(B)}-mathit{f(A)}]

This can be related back to the Fundamental Theorem of Calculus, since the gradient can be thought of as similar to the derivative. Another important property of conservative vector fields is that the integral of F around any closed path D is always 0.

## Assumptions on Curves, Vector Fields, and Domains

For computational sake, we have to assume the following properties regarding the curves, surfaces, domains, and vector fields:

1. The curves we consider are piecewise smooth, meaning they are composed of many infinitesimally small, smooth pieces connected end to end.
2. We assume that the domain D is a simply connected open region, meaning that any two points in D can be joined by a smooth curve within the region and that every loop in D can be contracted to a point in D without ever leaving D.

Theorem 1: Fundamental Theorem of Line Integrals

Let C be a smooth curve joining the point A to point B in the plane ore in space and parametrized by (mathbf{r}(t)). Let f be a differentiable function with a continuous gradient vector (mathbf{F}=igtriangledown{f}) on a domain D containing C. Then (int_{C}mathbf{F}cdot dmathbf{r}=f(B)-f(A)).

Proof

Suppose that A and B are two points in region D and that the curve C is given by [mathbf{r}(t)=xmathbf{i}+ymathbf{j}+zmathbf{k}] is a smooth curve in D that joins points A and B. Along C, f is a differentiable function of t and

[egin{align*} dfrac{partial f }{partial t}&=dfrac{partial f }{partial x}dfrac{partial x }{partial t}+dfrac{partial f }{partial y}dfrac{partial y }{partial t}+dfrac{partial f }{partial z}dfrac{partial z }{partial t} &=igtriangledown fcdot left ( dfrac{mathrm{d} x}{mathrm{d} t}mathbf{i}+dfrac{mathrm{d} y}{mathrm{d} t}mathbf{j}dfrac{mathrm{d} z}{mathrm{d} t}mathbf{k} ight ) &=igtriangledown fcdot dfrac{mathrm{d} mathbf{r}}{mathrm{d} t} &=mathbf{F}cdot dfrac{mathrm{d} mathbf{r}}{mathrm{d} t} end{align*}]

Therefore,

[int_{C}mathbf{F}cdot dmathbf{r}=int_{t=a}^{t=b}mathbf{F}cdot dfrac{mathrm{d} mathbf{r}}{mathrm{d} t}dt=int_a^bdfrac{mathrm{d} f}{mathrm{d} t}dt onumber]

*Note:

[mathbf{r}(a)=A, ; mathbf{r}(b)=B onumber]

Which means:

[left. f(g(t),h(t),k(t)) ight|_a^b=f(B)-f(A) onumber]

Thus proving Theorem 1. This shows us that the integral of a gradient field is easy to compute, provided we know the function (f).

(square)

As mentioned earlier, this is very similar to the Fundamental Theorem of Calculus both in theory and importance. Like the FTC, it provides us with a way to evaluate line integrals without limits of Riemann sums.

Theorem 2: Conservative Fields are Gradient Fields

Let (mathbf{F}=Mhat{mathbf{i}}+Nhat{mathbf{j}}+Phat{mathbf{k}}) be a vector field whose components are continuous throughout an open connected region D in space. Then F is conservative if and only it F is a gradient field (igtriangledown f) for a differentiable function f.

Proof

If F is a gradient field, then (mathbf{F}=igtriangledown f) for a differentiable function f. By Theorem 1, we know that [int_Cmathbf{F}cdot dmathbf{r}=f(B)-f(A)] and that the value of the line integral depends only on the two endpoints, not on the path. The line integral is said to be independent and F is a conservative field.

However, suppose F is a conservative vector field and we want to find some function f on D such that (igtriangledown f=mathbf{F}). First, we must pick a point A in the domain D such that (f(A)=0). For any other point B, we must define (f(B)) as equal to [int_Cmathbf{F}cdot dmathbf{r},] where the curve C is any smooth path in D from A to B. Because F is conservative, we know that (f(B)) is not dependant on C and vice versa. In order to show that (igtriangledown f=mathbf{F},) we need to show that

[dfrac{partial f}{partial x}=M, dfrac{partial f}{partial y}=N, dfrac{partial f}{partial z}=P. onumber]

Suppose B has coordinates ((x,y,z)) and a nearby point (B_0=(x_0,y,z).) By definition, then, the value of function f at the nearby point is [int_{C_0}mathbf{F}cdot dmathbf{r},] where (C_0) is any path from A to (B_0.) We can take path C to be the union between path (C_0) and line segment L from B to (B_0). Therefore,

[f(x,y,z)=int_{C_0}mathbf{F}cdot dmathbf{r}+int_Lmathbf{F}cdot dmathbf{r} onumber]

We can differentiate this integral, arriving at:

[dfrac{partial }{partial x}f(x,y,z)=dfrac{partial}{partial x}left ( int_{C_0}mathbf{F}cdot dmathbf{r}+int_Lmathbf{F}cdot dmathbf{r} ight ) onumber]

Only the last term of the above equation is dependent on x, so

[dfrac{partial }{partial x}f(x,y,z)=dfrac{partial }{partial x}int_Lmathbf{F}cdot dmathbf{r} onumber]

Now, if we parametrize (L) such that

[mathbf{r}(t)=tmathbf{i}+ymathbf{j}+zmathbf{k} onumber]

where (x_0leq t leq x) Then,

[dfrac{mathrm{d} r}{mathrm{d} t}=mathbf{i}] [mathbf{F}cdot dfrac{mathrm{d} r}{mathrm{d} t}=M onumber]

and

[int_Lmathbf{f}cdot dmathbf{r}=int_{x_0}^xM(t,y,z)dt] onumber]

Substitution gives us

[dfrac{partial }{partial x}f(x,y,z)=dfrac{partial }{partial x}int_{x_0}^xM(t,y,z)dt=M(x,y,z) onumber]

by the FTC. The partial derivatives

[dfrac{partial f}{partial y}=N onumber]

and

[dfrac{partial f}{partial z}=P onumber]

[mathbf{F}=igtriangledown f onumber]

(square)

In other words, (mathbf{F}=igtriangledown f) is only true when, for any two point A and B in the region D, (int_Cmathbf{F}cdot dmathbf{r}) is independent of the path C that joins the two points in D.

Theorem 3: Looper Property of Conservative Fields

The following statements are equivalent:

• (oint_{C}mathbf{F}cdot dmathbf{r}=0) around every loop (closed curve C) in D.
• The field F is conservative on D.

Proof

Part 1

We want to show that for any two points A and B in D, the ingtegral of

[mathbf{F}cdot dmathbf{r} onumber]

has the same value over any two paths (C_1) & (C_2) from A to B.

We reverse the direction of (C_2) to make the path (-C_2) from B to A.

Together, the two curves (C_1) & (-C_2) make a closed loop, which we will call C.

If you recall from earlier in this section, the integral over a closed loop for a conservative field is always 0:

[egin{align*} int_{C_1}mathbf{F}cdot dmathbf{r}-int_{C_2}mathbf{F}cdot dmathbf{r}&=int_{C_1}mathbf{F}cdot dmathbf{r}+int_{-C_2}mathbf{F}cdot dmathbf{r} &=int_Cmathbf{F}cdot dmathbf{r} &=0 end{align*}]

Therefore, the integrals over (C_1) & (C_2) must be equal.

Part 2

We want to show that the integral over (mathbf{F}cdot dmathbf{r}) is zero for any closed loop C. We pick two points A & B on C and use them to break C into 2 pieces: (C_1) from A to B and (C_2) from B back to A.

Therefore:

[egin{align*} oint _Cmathbf{F}cdot dmathbf{r}&=int_{C_1}mathbf{F}cdot dmathbf{r}+int_{C_2}mathbf{F}cdot dmathbf{r} &=int_A^Bmathbf{F}cdot dmathbf{r}-int_A^Bmathbf{F}cdot dmathbf{r} &=0 end{align*}]

(square)

## Finding Potentials for Conservative Fields

Component Test for Conservative Fields: Let (mathbf{F}=M(x,y,z)hat{ extbf{i}} + N(x,y,z) hat{ extbf{j}}+ P(x,y,z) hat{ extbf{k}} ) be a field on a connected and simply connected domain whose component functions have continuous first partial derivatives. Then, F is conservative if and only if

[dfrac{partial P }{partial x}=dfrac{partial M}{partial z}]

[dfrac{partial P }{partial y}=dfrac{partial N}{partial z}]

and

[dfrac{partial N }{partial x}=dfrac{partial M}{partial y}.]

*Note: See Example 2

Definition: Exact Differential Forms

Any expression

[M(x,y,z)dx+N(x,y,z)dy+P(x,y,z)dz]

is a differential form. A differential form is exact on a domain D in space if

[M,dx+N,dy+P,dz=dfrac{partial f}{partial x}dx+dfrac{partial f}{partial y}dy+dfrac{partial f}{partial z}dz=df]

for some scalar function f throughout D.

Component Test for Exactness of (Mdx+Ndy+Pdz): The differential form (Mdx+Ndy+Pdz) is exact on a connected and simply connected domain if and only if

[dfrac{partial P }{partial x}=dfrac{partial M}{partial z}]

[dfrac{partial P }{partial y}=dfrac{partial N}{partial z}]

and

[dfrac{partial N }{partial x}=dfrac{partial M}{partial y}]

Notice, this is the same as saying the field (mathbf{F}=Mhat{mathbf{i}}+Nhat{mathbf{j}}+Phat{mathbf{k}}) is conservative.

Example (PageIndex{1})

Suppose the force field (mathbf{F}=igtriangledown f) is the gradient of the function (f(x,y,z)=-dfrac{1}{x^2+y^2+z^2}). Find the work done by F in moving an object along a smooth curve C joining ((1,0,0)) to ((0,0,2)) that does not pass through the origin.

Solution

Since we know that this is a conservative field, we can apply Theorem 1, which shows that regardless of the curve C, the work done by F will be as follows:

[egin{align(} int_{C}mathbf{F}cdot dmathbf{r}&=f(0,0,2)-f(1,0,0) &=-dfrac{1}{4}-(-1) &=dfrac{3}{4} end{align(}]

Example (PageIndex{2})

Show that

[mathbf{F}=(e^x cos y+yz)hat{mathbf{i}}+(xz-e^xsin y)hat{mathbf{j}}+(xy+z)hat{mathbf{k}} onumber]

is conservative over its natural domain and find a potential function for it.

Solution

The natural domain of F is all of space, which is connected and simply connected. Let's define the following:

[M=e^xcos y+yz onumber]

[N=xz-e^xsin y onumber]

[P=xy+z onumber]

and calculate

[dfrac{partial P }{partial x}=y=dfrac{partial M}{partial z} onumber]

[dfrac{partial P }{partial y}=x=dfrac{partial N}{partial z} onumber]

[dfrac{partial N }{partial x}=-e^xsin y=dfrac{partial M}{partial y}. onumber]

Because the partial derivatives are continuous, F is conservative. Now that we know there exists a function f where the gradient is equal to F, let's find f.

[dfrac{partial f }{partial x}=e^xcos y+yz onumber]

[dfrac{partial f }{partial y}=xz-e^xsin y onumber]

[dfrac{partial f }{partial z}=xy+z onumber ]

If we integrate the first of the three equations with respect to x, we find that

[f(x,y,z)=int(e^x/cos y+yz)dx=e^xcos y+xyz+g(y,z)]

where (g(y,z)) is a constant dependent on (y) and (z) variables. We then calculate the partial derivative with respect to (y) from this equation and match it with the equation of above.

[dfrac{partial }{partial y}(f(x,y,z))=-e^x/sin y+xz+dfrac{partial g}{partial y}=xz-e^xsin y onumber]

This means that the partial derivative of (g) with respect to (y) is 0, thus eliminating (y) from (g) entirely and leaving it as a function of (z) alone.

[f(x,y,z)=e^xcos y+xyz+h(z) onumber]

We then repeat the process with the partial derivative with respect to (z).

[dfrac{partial }{partial z}(f(x,y,z))=xy+dfrac{mathrm{d} h}{mathrm{d} z}=xy+z onumber]

which means that

[dfrac{mathrm{d} h}{mathrm{d} z}= z onumber]

so we can find (h(z)) by integrating:

[h(z)=dfrac{z^2}{2}+C. onumber]

Therefore,

[f(x,y,z)=e^xcos y+xyz+dfrac{z^2}{2}+C. onumber]

We still have infinitely many potential functions for F-one at each value of C.

Example (PageIndex{3})

Show that ( ydx+xdy+4dz) is exact and evaluate the integral

[int_{(1,1,1)}^{(2,3,-1)}ydx+xdy+4dz onumber]

over any path from ((1,1,1)) to ((2,3,-1)).

Solution

We let (M=y), (N=x), and (P=4). Apply the Test for Exactness:

[dfrac{partial N}{partial x}=1=dfrac{partial M}{partial y} onumber]

[dfrac{partial N}{partial z}=0=dfrac{partial P}{partial y} onumber]

and

[dfrac{partial N}{partial z}=0=dfrac{partial P}{partial y}. onumber]

This proves that (ydx+xdy+4dz) is exact, so

[ydx+xdy+4dz=df onumber]

for some function f, and the integral's value is (f(2,3,-1)-f(1,1,1)).

We find fo up to a constant by integrating the following equations:

[dfrac{partial f}{partial x}=y, dfrac{partial f}{partial y}=x, dfrac{partial f}{partial z}=4 onumber]

From the first equation, we get that (f(x,y,z)=xy+g(y,z))

The second equation tells us that (dfrac{partial f}{partial y}=x+dfrac{partial g}{partial y}=x)

Therefore,

[dfrac{partial g}{partial y}=0 onumber]

Hence,

[f(x,y,z)=xy+h(z) onumber]

The third equation tells us that (dfrac{partial f}{partial z}=0+dfrac{d h}{d z}=4) so (h(z)=4z+C)

Therefore,

[f(x,y,z)=xy+4z+C onumber]

By substitution, we find that:

[f(2,3,-1)-f(1,1,1)=2+C-(5+C)=-3 onumber]

## References

1. Weir, Maurice D., Joel Hass, and George B. Thomas. Thomas' Calculus: Early Transcendentals. Boston: Addison-Wesley, 2010. Print.

## Math Insight

The process of finding a potential function of a conservative vector field is a multi-step procedure that involves both integration and differentiation, while paying close attention to the variables you are integrating or differentiating with respect to. For this reason, given a vector field $dlvf$, we recommend that you first determine that that $dlvf$ is indeed conservative before beginning this procedure. That way you know a potential function exists so the procedure should work out in the end.

In this page, we focus on finding a potential function of a two-dimensional conservative vector field. We address three-dimensional fields in another page.

We introduce the procedure for finding a potential function via an example. Let's use the vector field egin dlvf(x,y) = (y cos x+y^2, sin x+2xy-2y). end

The first step is to check if $dlvf$ is conservative. Since egin pdiff &= pdiff<>(sin x+2xy-2y) = cos x+2y pdiff &= pdiff<>(y cos x+y^2) = cos x+2y, end we conclude that the scalar curl of $dlvf$ is zero, as egin pdiff - pdiff = 0. end Next, we observe that $dlvf$ is defined on all of $R^2$, so there are no tricks to worry about. The vector field $dlvf$ is indeed conservative.

Since $dlvf$ is conservative, we know there exists some potential function $f$ so that $abla f = dlvf$. As a first step toward finding $f$, we observe that the condition $abla f = dlvf$ means that egin left(pdiff,pdiff ight) &= (dlvfc_1, dlvfc_2) &= (y cos x+y^2, sin x+2xy-2y). end This vector equation is two scalar equations, one for each component. We need to find a function $f(x,y)$ that satisfies the two conditions egin pdiff(x,y) = y cos x+y^2 label end and egin pdiff(x,y) = sin x+2xy -2y. label end Let's take these conditions one by one and see if we can find an $f(x,y)$ that satisfies both of them. (We know this is possible since $dlvf$ is conservative. If $dlvf$ were path-dependent, the procedure that follows would hit a snag somewhere.)

Let's start with condition eqref. We can take the equation egin pdiff(x,y) = y cos x+y^2, end and treat $y$ as though it were a number. In other words, we pretend that the equation is egin diff(x) = a cos x + a^2 end for some number $a$. We can integrate the equation with respect to $x$ and obtain that egin f(x)= a sin x + a^2x +C. end But, then we have to remember that $a$ really was the variable $y$ so that egin f(x,y) = y sin x + y^2x +C. end But actually, that's not right yet either. Since we were viewing $y$ as a constant, the integration &ldquoconstant&rdquo $C$ could be a function of $y$ and it wouldn't make a difference. The partial derivative of any function of $y$ with respect to $x$ is zero. We can replace $C$ with any function of $y$, say $g(y)$, and condition eqref will be satisfied. A new expression for the potential function is egin f(x,y) = y sin x + y^2x +g(y). label end If you are still skeptical, try taking the partial derivative with respect to $x$ of $f(x,y)$ defined by equation eqref. Since $g(y)$ does not depend on $x$, we can conclude that $displaystyle pdiff<> g(y) = 0$. Indeed, condition eqref is satisfied for the $f(x,y)$ of equation eqref.

Now, we need to satisfy condition eqref. We can take the $f(x,y)$ of equation eqref (so we know that condition eqref will be satisfied) and take its partial derivative with respect to $y$, obtaining egin pdiff(x,y) &= pdiff<> left( y sin x + y^2x +g(y) ight) &= sin x + 2yx + diff(y). end Comparing this to condition eqref, we are in luck. We can easily make this $f(x,y)$ satisfy condition eqref as long as egin diff(y)=-2y. end If the vector field $dlvf$ had been path-dependent, we would have found it impossible to satisfy both condition eqref and condition eqref. We would have run into trouble at this point, as we would have found that $diff$ would have to be a function of $x$ as well as $y$. Since $diff$ is a function of $y$ alone, our calculation verifies that $dlvf$ is conservative.

If we let egin g(y) = -y^2 +k end for some constant $k$, then egin pdiff(x,y) = sin x + 2yx -2y, end and we have satisfied both conditions.

Combining this definition of $g(y)$ with equation eqref, we conclude that the function egin f(x,y) = ysin x + y^2x -y^2 +k end is a potential function for $dlvf.$ You can verify that indeed egin abla f = (ycos x + y^2, sin x + 2xy -2y) = dlvf(x,y). end

With this in hand, calculating the integral egin dlint end is simple, no matter what path $dlc$ is. We can apply the gradient theorem to conclude that the integral is simply $f(vc)-f(vc )$, where $vc$ is the beginning point and $vc$ is the ending point of $dlc$. (For this reason, if $dlc$ is a closed curve, the integral is zero.)

We might like to give a problem such as find egin dlint end where $dlc$ is the curve given by the following graph.

The answer is simply egin dlint &= f(pi/2,-1) - f(-pi,2) &=-sin pi/2 + frac<2>-1 + k - (2 sin (-pi) - 4pi -4 + k) &=- sin pi/2 + frac<9pi> <2>+3= frac<9pi> <2>+2 end (The constant $k$ is always guaranteed to cancel, so you could just set $k=0$.)

If the curve $dlc$ is complicated, one hopes that $dlvf$ is conservative. It's always a good idea to check if $dlvf$ is conservative before computing its line integral egin dlint. end You might save yourself a lot of work.

1.1 Functions and Their Graphs

1.2 Combining Functions Shifting and Scaling Graphs

1.3 Trigonometric Functions

1.4 Graphing with Software

1.6 Inverse Functions and Logarithms

2.1 Rates of Change and Tangents to Curves

2.2 Limit of a Function and Limit Laws

2.3 The Precise Definition of a Limit

2.6 Limits Involving Infinity Asymptotes of Graphs

3.1 Tangents and the Derivative at a Point

3.2 The Derivative as a Function

3.4 The Derivative as a Rate of Change

3.5 Derivatives of Trigonometric Functions

3.7 Implicit Differentiation

3.8 Derivatives of Inverse Functions and Logarithms

3.9 Inverse Trigonometric Functions

3.11 Linearization and Differentials

4 Applications of Derivatives

4.1 Extreme Values of Functions

4.2 The Mean Value Theorem

4.3 Monotonic Functions and the First Derivative Test

4.4 Concavity and Curve Sketching

4.5 Indeterminate Forms and L’Hôpital’s Rule

5.1 Area and Estimating with Finite Sums

5.2 Sigma Notation and Limits of Finite Sums

5.4 The Fundamental Theorem of Calculus

5.5 Indefinite Integrals and the Substitution Method

5.6 Definite Integral Substitutions and the Area Between Curves

6 Applications of Definite Integrals

6.1 Volumes Using Cross-Sections

6.2 Volumes Using Cylindrical Shells

6.4 Areas of Surfaces of Revolution

6.6 Moments and Centers of Mass

7 Integrals and Transcendental Functions

7.1 The Logarithm Defined as an Integral

7.2 Exponential Change and Separable Differential Equations

7.4 Relative Rates of Growth

8 Techniques of Integration

8.1 Using Basic Integration Formulas

8.3 Trigonometric Integrals

8.4 Trigonometric Substitutions

8.5 Integration of Rational Functions by Partial Fractions

8.6 Integral Tables and Computer Algebra Systems

9 First-Order Differential Equations

9.1 Solutions, Slope Fields, and Euler’s Method

9.2 First-Order Linear Equations

9.4 Graphical Solutions of Autonomous Equations

9.5 Systems of Equations and Phase Planes

10 Infinite Sequences and Series

10.5 Absolute Convergence The Ratio and Root Tests

10.6 Alternating Series and Conditional Convergence

10.8 Taylor and Maclaurin Series

10.9 Convergence of Taylor Series

10.10 The Binomial Series and Applications of Taylor Series

11 Parametric Equations and Polar Coordinates

11.1 Parametrizations of Plane Curves

11.2 Calculus with Parametric Curves

11.4 Graphing Polar Coordinate Equations

11.5 Areas and Lengths in Polar Coordinates

11.7 Conics in Polar Coordinates

12 Vectors and the Geometry of Space

12.1 Three-Dimensional Coordinate Systems

12.5 Lines and Planes in Space

13 Vector-Valued Functions and Motion in Space

13.1 Curves in Space and Their Tangents

13.2 Integrals of Vector Functions Projectile Motion

13.4 Curvature and Normal Vectors of a Curve

13.5 Tangential and Normal Components of Acceleration

13.6 Velocity and Acceleration in Polar Coordinates

14.1 Functions of Several Variables

14.2 Limits and Continuity in Higher Dimensions

14.5 Directional Derivatives and Gradient Vectors

14.6 Tangent Planes and Differentials

14.7 Extreme Values and Saddle Points

14.9 Taylor’s Formula for Two Variables

14.10 Partial Derivatives with Constrained Variables

15.1 Double and Iterated Integrals over Rectangles

15.2 Double Integrals over General Regions

15.3 Area by Double Integration

15.4 Double Integrals in Polar Form

15.5 Triple Integrals in Rectangular Coordinates
15.6 Moments and Centers of Mass

## 16.3: Path Independence, Conservative Fields, and Potential Functions

In the previous section we saw that if we knew that the vector field (vec F) was conservative then (intlimits_<>) was independent of path. This in turn means that we can easily evaluate this line integral provided we can find a potential function for (vec F).

In this section we want to look at two questions. First, given a vector field (vec F) is there any way of determining if it is a conservative vector field? Secondly, if we know that (vec F) is a conservative vector field how do we go about finding a potential function for the vector field?

The first question is easy to answer at this point if we have a two-dimensional vector field. For higher dimensional vector fields we’ll need to wait until the final section in this chapter to answer this question. With that being said let’s see how we do it for two-dimensional vector fields.

#### Theorem

Let (vec F = P,vec i + Q,vec j) be a vector field on an open and simply-connected region (D). Then if (P) and (Q) have continuous first order partial derivatives in (D) and

the vector field (vec F) is conservative.

Let’s take a look at a couple of examples.

1. (vec Fleft( ight) = left( <- yx> ight)vec i + left( <- xy> ight)vec j)
2. (vec Fleft( ight) = left( <2x<<f>^> + y<<f>^>> ight)vec i + left( <<<f>^> + 2y> ight)vec j)

Okay, there really isn’t too much to these. All we do is identify (P) and (Q) then take a couple of derivatives and compare the results.

In this case here is (P) and (Q) and the appropriate partial derivatives.

So, since the two partial derivatives are not the same this vector field is NOT conservative.

Here is (P) and (Q) as well as the appropriate derivatives.

The two partial derivatives are equal and so this is a conservative vector field.

Now that we know how to identify if a two-dimensional vector field is conservative we need to address how to find a potential function for the vector field. This is actually a fairly simple process. First, let’s assume that the vector field is conservative and so we know that a potential function, (fleft( ight)) exists. We can then say that,

Or by setting components equal we have,

By integrating each of these with respect to the appropriate variable we can arrive at the following two equations.

We saw this kind of integral briefly at the end of the section on iterated integrals in the previous chapter.

It is usually best to see how we use these two facts to find a potential function in an example or two.

1. (vec F = left( <2+ x> ight)vec i + left( <2+ y> ight)vec j)
2. (vec Fleft( ight) = left( <2x<<f>^> + y<<f>^>> ight)vec i + left( <<<f>^> + 2y> ight)vec j)

Let’s first identify (P) and (Q) and then check that the vector field is conservative.

So, the vector field is conservative. Now let’s find the potential function. From the first fact above we know that,

From these we can see that

We can use either of these to get the process started. Recall that we are going to have to be careful with the “constant of integration” which ever integral we choose to use. For this example let’s work with the first integral and so that means that we are asking what function did we differentiate with respect to (x) to get the integrand. This means that the “constant of integration” is going to have to be a function of (y) since any function consisting only of (y) and/or constants will differentiate to zero when taking the partial derivative with respect to (x).

Here is the first integral.

where (hleft( y ight)) is the “constant of integration”.

We now need to determine (hleft( y ight)). This is easier than it might at first appear to be. To get to this point we’ve used the fact that we knew (P), but we will also need to use the fact that we know (Q) to complete the problem. Recall that (Q) is really the derivative of (f) with respect to (y). So, if we differentiate our function with respect to (y) we know what it should be.

So, let’s differentiate (f) (including the (hleft( y ight))) with respect to (y) and set it equal to (Q) since that is what the derivative is supposed to be.

From this we can see that,

Notice that since (h'left( y ight)) is a function only of (y) so if there are any (x)’s in the equation at this point we will know that we’ve made a mistake. At this point finding (hleft( y ight)) is simple.

So, putting this all together we can see that a potential function for the vector field is,

Note that we can always check our work by verifying that ( abla f = vec F). Also note that because the (c) can be anything there are an infinite number of possible potential functions, although they will only vary by an additive constant.

Okay, this one will go a lot faster since we don’t need to go through as much explanation. We’ve already verified that this vector field is conservative in the first set of examples so we won’t bother redoing that.

This means that we can do either of the following integrals,

While we can do either of these the first integral would be somewhat unpleasant as we would need to do integration by parts on each portion. On the other hand, the second integral is fairly simple since the second term only involves (y)’s and the first term can be done with the substitution (u = xy). So, from the second integral we get,

Notice that this time the “constant of integration” will be a function of (x). If we differentiate this with respect to (x) and set equal to (P) we get,

So, in this case it looks like,

[h'left( x ight) = 0hspace <0.25in>Rightarrow hspace<0.25in>hleft( x ight) = c]

So, in this case the “constant of integration” really was a constant. Sometimes this will happen and sometimes it won’t.

Here is the potential function for this vector field.

Now, as noted above we don’t have a way (yet) of determining if a three-dimensional vector field is conservative or not. However, if we are given that a three-dimensional vector field is conservative finding a potential function is similar to the above process, although the work will be a little more involved.

In this case we will use the fact that,

Let’s take a quick look at an example.

Okay, we’ll start off with the following equalities.

To get started we can integrate the first one with respect to (x), the second one with respect to (y), or the third one with respect to (z). Let’s integrate the first one with respect to (x).

Note that this time the “constant of integration” will be a function of both (y) and (z) since differentiating anything of that form with respect to (x) will differentiate to zero.

Now, we can differentiate this with respect to (y) and set it equal to (Q). Doing this gives,

Of course we’ll need to take the partial derivative of the constant of integration since it is a function of two variables. It looks like we’ve now got the following,

[left( ight) = 0hspace <0.5in>Rightarrow hspace<0.5in>gleft( ight) = hleft( z ight)]

Since differentiating (gleft( ight)) with respect to (y) gives zero then (gleft( ight)) could at most be a function of (z). This means that we now know the potential function must be in the following form.

To finish this out all we need to do is differentiate with respect to (z) and set the result equal to (R).

[h'left( z ight) = 0hspace <0.25in>Rightarrow hspace<0.25in>,,,hleft( z ight) = c]

The potential function for this vector field is then,

Note that to keep the work to a minimum we used a fairly simple potential function for this example. It might have been possible to guess what the potential function was based simply on the vector field. However, we should be careful to remember that this usually won’t be the case and often this process is required.

Also, there were several other paths that we could have taken to find the potential function. Each would have gotten us the same result.

Let’s work one more slightly (and only slightly) more complicated example.

Here are the equalities for this vector field.

For this example let’s integrate the third one with respect to (z).

The “constant of integration” for this integration will be a function of both (x) and (y).

Now, we can differentiate this with respect to (x) and set it equal to (P). Doing this gives,

So, it looks like we’ve now got the following,

[left( ight) = 2xcos left( y ight)hspace <0.5in>Rightarrow hspace<0.5in>gleft( ight) = cos left( y ight) + hleft( y ight)]

The potential function for this problem is then,

To finish this out all we need to do is differentiate with respect to (y) and set the result equal to (Q).

[h'left( y ight) = 3hspace <0.25in>Rightarrow hspace<0.25in>,,,hleft( y ight) = 3y + c]

The potential function for this vector field is then,

So, a little more complicated than the others and there are again many different paths that we could have taken to get the answer.

We need to work one final example in this section.

Now, we could use the techniques we discussed when we first looked at line integrals of vector fields however that would be particularly unpleasant solution.

Instead, let’s take advantage of the fact that we know from Example 2a above this vector field is conservative and that a potential function for the vector field is,

Using this we know that integral must be independent of path and so all we need to do is use the theorem from the previous section to do the evaluation.

[vec rleft( 1 ight) = leftlangle < - 2,1> ight angle hspace<0.5in>vec rleft( 0 ight) = leftlangle < - 1,0> ight angle ]

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

1. (AB) is the line segment from (Aleft( <0,0> ight)) to (Bleft( <1,1> ight))
2. (AB) is the parabola (y = ) from (Aleft( <0,0> ight)) to (Bleft( <1,1> ight)).

### Example 1.

1. (AB) is the line segment from (Aleft( <0,0> ight)) to (Bleft( <1,1> ight))
2. (AB) is the parabola (y = ) from (Aleft( <0,0> ight)) to (Bleft( <1,1> ight)).

Consider the first case. Obviously, the equation of the line is (y = x.) Then using the formula

If the curve (AB) is parabola (y = ,) we have

that is we have obtained the same answer.

Apply the test (><> ormalsize> = ><> ormalsize>) to determine if the vector field is conservative.

As it can be seen, the vector field (mathbf = left( ight)) is conservative. This explains the result that the line integral is path independent.

## Conservative Vector Fields

Theorem. If is a vector field in the plane, and P and Q have continuous partial derivatives on a region. the following four statements are equivalent:

1. for some function .

2. .

3. If is a closed curve lying in the region --- i.e. a path which starts and ends at the same point --- then

4. ( Path independence) If and are paths in the region which start at the same point and end at the same point, then

To say that these statements are equivalent means that if one of them is true, then all of them are true (and if one of them is false, all of them are false). A field that satisfies any one of these conditions is a conservative field --- or sometimes a gradient field, or sometimes path independent.

Before I show that these statements are equivalent, I'll give a couple of examples.

Example. Show that is a gradient field.

I want a function f such that , i.e.

Integrate the first equation with respect to x:

Since the integral is with respect to x, y is constant, and must be included in the arbitrary constant --- hence . Differentiate with respect to y and set the result equal to above:

I get , so . This time, D is a numerical constant. Since the derivative of a number is 0, and since I just want some potential function (see the previous example), I might as well take . Then , so

Thus, if , then

Definition. A function f such that is called a potential function for .

Example. Let and

Theorem. Suppose is a gradient field, and let for be a path. Then

In other words, to evaluate the integral of a gradient field, just plug the endpoints of the path ( and ) into the potential function (f).

The antiderivative of the derivative of is just , so

Example. Compute , where

To compute this directly, you would need to do the integral

Instead, notice that if , then , which is the field in the integral. Hence, I can compute the integral by plugging the endpoints of the path into f.

(The fact that it comes out to 0, as opposed to a nonzero number, is a coincidence.) The important thing to notice is how easy it was to do the computation!

Now I will go back and prove that the four statements which define a conservative field are equivalent. To prove that they're equivalent, I must show that any one of them follows from any other. I will do it this way:

Proof. First, assume statement 1 is true, so for some f. Since , this means that

But the two second derivatives are equal by equality of mixed partials, so . This is statement 2.

Next, assume statement 2 is true, so . Take a closed curve . I want to show that the integral around is 0. This follows from Green's theorem, which I'll discuss in more detail later. For now, note that the closed curve encloses a region R.

But the double integral on the right is 0, because . Therefore, the integral around a closed curve is 0, and that is statement 3.

Before I do the next step, here is some notation. If is a curve, will denote the same curve traversed in the opposite direction.

If I traverse a curve backward, the line integral flips its sign:

Now assume statement 3 is true: The line integral around a closed curve is 0. Take curves and , both of which start at P and both of which end at Q. I need to show that

Note that is a closed curve. So the integral around is 0:

The second integral is the negative of the integral along :

Finally, move the second integral to the other side:

This is what I wanted to prove, so statement 4 follows from statement 3.

Finally, suppose statement 4 is true: The field is path independent. I want to show that is a gradient field. I need to find a function f such that . To do this, take any path from to and define

By path independence, it does not matter what path I choose. Now I have to show that and .

Consider the path from to made up of segments as shown below:

The horizontal part is for . The velocity vector is , and . So

Therefore, the integral for the horizontal segment is

The vertical part is for . The velocity vector is and . So

Therefore, the integral for the vertical segment is

The line integral along the whole path --- which by definition is f --- is

Remember that I was trying to show that and . Differentiate the last equation with respect to y.

The first integral only involves x, so its derivative with respect to y is 0. For the second integral, apply the Fundamental Theorem of Calculus: The y in the top limit replaces the t in the integrand and I get . So

That is, f has the right y derivative.

If you take a path made up of segments going from to the "other way" --- along the y-axis, then horizontally --- you can show in similar fashion that . This proves that , which is statement 1.

And that finishes the proof that the four statements are equivalent.

The last part of the proof gives a way of constructing a potential function for a gradient field. It's perhaps not the best way for a human being to do this, but would be a reasonable approach if you were doing this on a computer.

Example. Find a potential function for using the line integrals in the proof of the theorem.

I can find f using the formula

Here , so . Likewise, , so . So

By the way, notice that is also a potential function for this field, since and . There are infinitely many potential functions for a gradient field they differ by a numerical constant.

So far, I've discussed vector fields in , 2 dimensions. There are few surprises when you move up to 3 dimensions.

Theorem. Let be a 3-dimensional vector field, and assume its components have continuous partial derivatives. Then the following four statements are equivalent:

1. for some function .

2. .

3. If is a closed curve --- i.e. a path which starts and ends at the same point --- then

4. ( Path independence) If and are paths which start at the same point and end at the same point, then

Once again, a field that satisfies any one of these conditions is a conservative field --- or sometimes a gradient field, or sometimes path independent.

The only change in moving from 2 dimensions to 3 dimensions is in statement 2. To see how this is related to the for a 2-dimensional field, take a field and regard it as a 3-dimensional field by making the third component 0, so

(Notice that, e.g., , because P does not contain any z's.) The condition is, in this case, the same as .

In fact, the components of should have continuous first partials, except perhaps at finitely many points. Here's the reason I can allow finitely many "bad points" in this case but none in the 2-dimensional case.

Think of a closed curve as a piece of string. In a rough sense, the reason why the integral of a conservative field around a closed curve is 0 is that the string can be "reeled in" to the basepoint without changing the integral.

When the curve has been reeled in to a single point, the integral over the curve is obviously 0.

In 2 dimensions, a curve can "get stuck" on a single bad point as you reel it in.

However, in 3 dimensions, you have enough room to move the curve around finitely many bad points as you reel it in.

Example. Show that is conservative, and find a potential function for .

Therefore, is conservative. A potential function f must satisfy

Integrate the first equation with respect to x:

Since the integral is with respect to x, y and z are constant, and must be included in the arbitrary constant . Now differentiate with respect to y and set the result equal to above:

I find that , so integrating with respect to y,

This time, z is constant and must be included in the arbitrary constant . Plug this back into f it is

Finally, differentiate with respect to z and set the result equal to above:

I find that , so , where E is a numerical constant. As in an earlier example, I may take . This gives , so

Example. Compute , where

You could compute the integral directly --- would you want to? There must be an easier way .

Since , the field is conservative. If I can find a potential function, I can compute the integral by simply plugging the endpoints of the path into the potential function.

A potential function f must satisfy

Integrate the first equation with respect to x:

Since the integral is with respect to x, y and z are constant, and must be included in the arbitrary constant . Now differentiate with respect to y and set the result equal to above:

I find that , so integrating with respect to y,

This time, z is constant and must be included in the arbitrary constant . Plug this back into f it is

Finally, differentiate with respect to z and set the result equal to above:

I find that , so , where E is a numerical constant. As in an earlier example, I may take . This gives , so

The endpoints of the path are

Example. is a path with positive components from a point P on to a point Q on . Compute , where

The path isn't given --- in fact, its endpoints aren't given. Therefore, the path must not matter, i.e. the integral is probably path independent. In fact, , for . Therefore,

Now Q is on , so for this point . Likewise, P is on , so for this point . Hence,

## Conservative Vector Fields

• Misc
• Section 16.3
• Confusing, connections between pieces?
• Intuitively, connected means you can draw a possibly curved line between any 2 points in the set without leaving the set
• Simply connected means any circle in the set can be pulled tight into a point without leaving the set
• Path independence ?
• You saw an example when we integrated work over 2 flights of stairs that reached the same height&mdashthe integrals were the same, independent of the path of the stairs
• Integral of F same over all paths
• Equivalent to field being conservative
• Main new thing here is name &ldquopotential function,&rdquo a label that&rsquos useful for talking about integrals of conservative fields
• f is potential function for F
• This basically just states a useful connection between 2 kinds of vector fields
• That the integral is 0 follows from the fundamental theorem
• But other direction also interesting, i.e., if you find that integrals around closed loops are always 0 then you know your field is conservative
• Unintuitive now, but we may be able to see where this comes from in a week or so
• M(x,y,z) dx + N(x,y,z) dy + P(x,y,z) dz
• Exactness captures the differential form of those line integrals that happen to be of vector fields
• Let f(x,y,z) = x(y-sin(z 2 )) / (z 2 + e xy ), let F = grad f, and let C be the curve x = sint cost, y = t 2 (&pi/2-t), z = &pi, 0 &le t &le &pi/2
• Find the integral of F around C
• 0, by closed loop theorem and because F is conservative&mdashit&rsquos a gradient field
• By fundamental theorem, integral = f(1, 0, &radic(&pi/2) ) - f( 0, 0, 0 )
• = 1 ( 0-sin(&pi/2) ) / ( &pi/2 + e 0 ) - f(0,0,0) = -1 / (&pi/2 + 1) - 0
• Component test says it is
• So also find potential function
• Green&rsquos Theorem

## Math Insight

Many physical force fields (vector fields) that you are familiar with are conservative vector fields. The term comes from the fact that some kind of energy is conserved by these force fields. The important consequence for us, though, is that as you move an object from point $vc$ to point $vc$, the work performed by a conservative force field does not depend on the path taken from point $vc$ to point $vc$. For this reason, we often refer to such vector fields as path-independent vector fields. Path-independent and conservative are just two terms that mean the same thing.

For example, imagine you have to carry a heavy box from your front door to your bedroom upstairs. Because of the gravity (which can be viewed as a force field), you have to do work to carry the box up.

Here we mean the scientific definition of work, which is force times distance. Although it may feel like work to move the box from one room to another on the same floor, the actual work done against gravity is zero.

Next, imagine that you have two stairways in your house: a gently sloping front staircase, and a steep back staircase. Since the gravitational field is a conservative vector field, the work you must do against gravity is exactly the same if you take the front or the back staircase. As long as the box starts in the same position and ends in the same position, the total work is the same. (In fact, if you decided to first carry the box to your neighbor's house, then carry it up and down your backyard tree, and then in your back door before taking it upstairs, it wouldn't make a difference for this scientific definition of work. The net work you performed against gravity would be the same.)

The line integral of a vector field can be viewed as the total work performed by the force field on an object moving along the path. For the above gravity example, we discussed the work you performed against the gravity field, which is exactly opposite the work performed by the gravity field. We'd need to multiply the line integral by $-1$ to get the work you performed against the gravity field, but that's a technical point we don't need to worry much about.

The vector field $dlvf(x,y)=(x,y)$ is a conservative vector field. (You can read how to test for path-independence later. For now, take it on faith.) It is illustrated by the black arrows in the below figure. We want to compute the integral egin dlint end where $dlc$ is a path from the point $vc=(3,-3)$ (shown by the cyan square) to the point $vc=(2,4)$ (shown by the magenta square). Since $dlvf$ is path-independent, we don't need to know anything else about the path $dlc$ to compute the line integral. Later, you'll learn to compute that the value of the integral is 1, as shown by the magenta line on the slider below the figure.

The below applet demonstrate the path-independence of $dlvf$, as one can see that the integrals along three different paths give the same value. The vector field appears to be path-independent, as promised. (You'd have to check all the infinite number of possible paths from all points $vc$ to all points $vc$ to determine that $dlvf$ was really path-independent. Fortunately, you'll learn some simpler methods.)

## 16.3: Path Independence, Conservative Fields, and Potential Functions

Math 2321 (Multivariable Calculus), Spring 2021

Course Information
Instructor Class Times Office Hours
Evan Dummit
edummit at northeastern dot edu
(Sec 04) MWR, 1:35pm-2:40pm
(Sec 12) MWR, 4:35pm-5:40pm
Online, via Zoom
W 3:00pm-4:15pm
R 12:15pm-1:15pm
Online, via Zoom
I am teaching two sections of Math 2321, corresponding to the two lecture times listed. You are requested to attend your assigned lecture, but in the event you are not able, you are allowed to attend the other one (the lectures will cover the same content at the same pace). All lectures will be recorded and made available for on-demand viewing at any time.
Math 2321 uses a Piazza page for course discussion. Links to all of the live lectures, office hours, problem sessions, and lecture recordings are hosted there.
Teaching Assistant Recitation Time Office Hours
Anupam Kumar
kumar.anupa at northeastern dot edu
M, 2:50pm-4:30pm
Online, via Zoom
TBA
Online, via Zoom
For detailed information about the course, please consult the 2321 Course Syllabus. (Note: any information given in class or on this webpage supersedes the written syllabus.)
All homework assignments are available on the 2321 WeBWorK page.

Lecture Slides
These are the slides used during the lectures. They will usually be posted ahead of the lecture time.
Date Material
Wed, Jan 20th
Thu, Jan 21st
Lecture 1: Welcome + 3D Graphing (Notes 1.1)
Lecture 2: Vectors + Dot Products (Notes 1.2.1-1.2.2)
Mon, Jan 25th
Wed, Jan 27th
Thu, Jan 28th
Lecture 3: Cross Products, Lines and Planes (Notes 1.2.3-1.2.4)
Lecture 4: Lines and Planes, Vector-Valued Functions (Notes 1.2.4-1.3.1)
Lecture 5: Parametric Curves, Motion in 3-Space (Notes 1.3.1-1.3.2)
Mon, Feb 1st
Wed, Feb 3rd
Thu, Feb 4th
Lecture 6: Limits and Partial Derivatives (Notes 2.1)
Lecture 7: Directional Derivatives and Gradients (Notes 2.2.1)
Lecture 8: Tangent Lines and Planes, Linearization (Notes 2.2.2 + 2.4.1)
Mon, Feb 8th
Wed, Feb 10th
Thu, Feb 11th
Lecture 9: The Chain Rule + Implicit Differentiation (Notes 2.3)
Lecture 10: Critical Points, Minima + Maxima (Notes 2.5.1)
Lecture 11: Optimization on a Region (Notes 2.5.2)
Mon, Feb 15th
Wed, Feb 17th
Thu, Feb 18th
No class university holiday.
Lecture 12: Midterm 1 Review, part 1
Lecture 13: Midterm 1 Review, part 2
Mon, Feb 22nd
Wed, Feb 24th
Thu, Feb 25th
Lecture 14: Lagrange Multipliers (Notes 2.6) [Typos fixed]
Lecture 15: Double Integrals (Notes 3.1.1-3.1.2)
Lecture 16: Computing Double Integrals (Notes 3.1.2-3.1.3)
Mon, Mar 1st
Wed, Mar 3rd
Thu, Mar 4th
Lecture 17: Double Integrals in Polar Coordinates (Notes 3.3.2)
Lecture 18: Triple Integrals in Rectangular Coordinates (Notes 3.2)
Lecture 19: Change of Coordinates, Cylindrical Coordinates (Notes 3.3.1 + 3.3.3)
Mon, Mar 8th
Wed, Mar 10th
Thu, Mar 11th
Lecture 20: Triple Integrals in Cylindrical and Spherical (Notes 3.3.3-3.3.4)
Lecture 21: More Cylindrical and Spherical + Areas, Volumes (Notes 3.3.3-3.4.1)
Lecture 22: Areas, Volumes, Masses, and Moments (Notes 3.4.1-3.4.2)
Mon, Mar 15th
Wed, Mar 17th
Thu, Mar 18th
Lecture 23: Line Integrals (Notes 4.1)
Lecture 24: Midterm 2 Review, part 1
Lecture 25: Midterm 2 Review, part 2
Mon, Mar 22nd
Wed, Mar 24th
Thu, Mar 25th
Lecture 26: Parametric Surfaces (Notes 4.2.1)
Classes cancelled ("CARE Day")
Lecture 27: Surface Integrals (Notes 4.2.2)
Mon, Mar 29th
Wed, Mar 31st
Thu, Apr 1st
Lecture 28: Vector Fields, Work, Circulation, and Flux (Notes 4.3.1-4.3.2)
Lecture 29: Flux Across Surfaces (Notes 4.3.3)
Lecture 30: Conservative Fields and Potential Functions (Notes 4.4)
Mon, Apr 5th
Wed, Apr 7th
Thu, Apr 8th
Lecture 31: Green's Theorem (Notes 4.5)
Lecture 32: Midterm 3 Review, part 1
Lecture 33: Midterm 3 Review, part 2
Mon, Apr 12th
Wed, Apr 14th
Thu, Apr 15th
Classes cancelled ("CARE Day")
Lecture 34: Stokes's Theorem and the Divergence Theorem (Notes 4.6)
Lecture 35: Applications of Vector Calculus, part 1 (Notes 4.7.1)
Mon, Apr 19th
Wed, Apr 21st
Lecture 36: Applications of Vector Calculus, part 2 (Notes 4.7.2-4.7.4)
Lecture 37: Final Exam Review, part 1
Tue, Apr 27th Review: Final Exam Review, part 2

Functions of Several Variables and 3-Space
1.2

Vectors, Dot and Cross Products, Lines and Planes
1.3

Limits and Partial Derivatives
2.2

2.3

Local Extreme Points and Optimization
2.6

Double Integrals in Rectangular Coordinates
3.2

Triple Integrals in Rectangular Coordinates
3.3

Alternative Coordinate Systems and Changes of Variable
3.4

Surfaces and Surface Integrals
4.3

Vector Fields, Work, Circulation, Flux
4.4

Conservative Vector Fields, Path-Independence, and Potential Functions
4.5