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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. Gravitational and electric fields are examples of such vector fields. This section will discuss the properties of these vector fields.

Definition: Path Independent and Conservative

Let 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

CFdr

along a path C from A to B in D is the same over all paths from A to B. Then the integral CFdr 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 F=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

BAFdr=BAfdr=f(B)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 r(t). Let f be a differentiable function with a continuous gradient vector F=f on a domain D containing C. Then CFdr=f(B)f(A).

Proof

Suppose that A and B are two points in region D and that the curve C is given by r(t)=xi+yj+zk is a smooth curve in D that joins points A and B. Along C, f is a differentiable function of t and

ft=fxxt+fyyt+fzzt=f(dxdti+dydtjdzdtk)=fdrdt=Fdrdt

Therefore,

CFdr=t=bt=aFdrdtdt=badfdtdt

*Note:

r(a)=A,r(b)=B

Which means:

f(g(t),h(t),k(t))|ba=f(B)f(A)

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

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 F=Mˆi+Nˆj+Pˆk 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 f for a differentiable function f.

Proof

If F is a gradient field, then F=f for a differentiable function f. By Theorem 1, we know that CFdr=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 f=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 CFdr, 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 f=F, we need to show that

fx=M,fy=N,fz=P.

Suppose B has coordinates (x,y,z) and a nearby point B0=(x0,y,z). By definition, then, the value of function f at the nearby point is C0Fdr, where C0 is any path from A to B0. We can take path C to be the union between path C0 and line segment L from B to B0. Therefore,

f(x,y,z)=C0Fdr+LFdr

We can differentiate this integral, arriving at:

xf(x,y,z)=x(C0Fdr+LFdr)

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

xf(x,y,z)=xLFdr

Now, if we parametrize L such that

r(t)=ti+yj+zk

where x0tx Then,

drdt=i Fdrdt=M

and

Lfdr=xx0M(t,y,z)dt]

Substitution gives us

xf(x,y,z)=xxx0M(t,y,z)dt=M(x,y,z)

by the FTC. The partial derivatives

fy=N

and

fz=P

follow similarly, showing that

F=f

In other words, F=f is only true when, for any two point A and B in the region D, CFdr 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:

  • CFdr=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

Fdr

has the same value over any two paths C1 & C2 from A to B.

We reverse the direction of C2 to make the path C2 from B to A.

Together, the two curves C1 & C2 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:

C1FdrC2Fdr=C1Fdr+C2Fdr=CFdr=0

Therefore, the integrals over C1 & C2 must be equal.

Part 2

We want to show that the integral over Fdr is zero for any closed loop C. We pick two points A & B on C and use them to break C into 2 pieces: C1 from A to B and C2 from B back to A.

Therefore:

CFdr=C1Fdr+C2Fdr=BAFdrBAFdr=0

Finding Potentials for Conservative Fields

Component Test for Conservative Fields: Let F=M(x,y,z)ˆi+N(x,y,z)ˆj+P(x,y,z)ˆk 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

Px=Mz

Py=Nz

and

Nx=My.

*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

Mdx+Ndy+Pdz=fxdx+fydy+fzdz=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

Px=Mz

Py=Nz

and

Nx=My

Notice, this is the same as saying the field F=Mˆi+Nˆj+Pˆk is conservative.

Example 16.3.1

Suppose the force field F=f is the gradient of the function f(x,y,z)=1x2+y2+z2. 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:

CFdr=f(0,0,2)f(1,0,0)=14(1)=34

Example 16.3.2

Show that

F=(excosy+yz)ˆi+(xzexsiny)ˆj+(xy+z)ˆk

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=excosy+yz

N=xzexsiny

P=xy+z

and calculate

Px=y=Mz

Py=x=Nz

Nx=exsiny=My.

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.

fx=excosy+yz

fy=xzexsiny

fz=xy+z

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

f(x,y,z)=(ex/cosy+yz)dx=excosy+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.

y(f(x,y,z))=ex/siny+xz+gy=xzexsiny

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)=excosy+xyz+h(z)

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

z(f(x,y,z))=xy+dhdz=xy+z

which means that

dhdz=z

so we can find h(z) by integrating:

h(z)=z22+C.

Therefore,

f(x,y,z)=excosy+xyz+z22+C.

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

Example 16.3.3

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

(2,3,1)(1,1,1)ydx+xdy+4dz

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:

Nx=1=My

Nz=0=Py

and

Nz=0=Py.

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

ydx+xdy+4dz=df

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:

fx=y,fy=x,fz=4

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

The second equation tells us that fy=x+gy=x

Therefore,

gy=0

Hence,

f(x,y,z)=xy+h(z)

The third equation tells us that fz=0+dhdz=4 so h(z)=4z+C

Therefore,

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

By substitution, we find that:

f(2,3,1)f(1,1,1)=2+C(5+C)=3

References

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

Contributors and Attributions

  • Alagu Chidambaram (UCD)
  • Integrated by Justin Marshall.


16.3: Path Independence, Conservative Fields, and Potential Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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