9.9 Independence of Path

INTRODUCTION

In this section we refer to a piecewise-smooth curve C between an initial point A and a terminal point B as a path of integration or simply a path. We begin the discussion by considering line integrals in 2-space. Recall, in (10) of Section 9.8 we saw that if F(x, y) = P(x, y)i + Q(x, y)j is a vector field in 2-space and a path C is defined by the vector function then a line integral can be written as

where dr = dxi + dyj. Generally the value of a line integral depends on the path of integration. Stated another way, if C1 and C2 are two different paths between the same points A and B, then we expect that . However, there is an important exception. As we see in this section, line integrals that involve a certain kind of vector field F depend not on the path C but only on the endpoints A = (f(a), g(a)) and B = (f(b), g(b)) of the path.

Note: To avoid needless repetition we assume throughout that the component functions of the vector field F are continuous in some region of 2- or 3-space, its component functions have continuous first partial derivatives in the region, and the path C of integration lies entirely in the region.

EXAMPLE 1 Path Independence

The integral has the same value on each path C between (0, 0) and (1, 1) shown in FIGURE 9.9.1. You may recall from Problems 11–14 in Exercises 9.8 that on these paths

You are also urged to verify on the curves and between (0, 0) and (1, 1). The relevance of all this is to suggest that the integral does not depend on the path joining these two points. We continue this discussion in Examples 2 and 3.

4 graphs. Graph a. An upward sloping curve is graphed on an x y coordinate plane. The curve, labeled y = x^2, begins at a marked point at the origin, goes up and to the right with increasing steepness, and ends at the marked point (1, 1). Graph b. An upward sloping line is graphed on an x y coordinate plane. The line, labeled y = x, begins at a marked point at the origin, goes up and to the right, and ends at the marked point (1, 1). Graph c. A function is graphed on an x y coordinate plane. The function begins at a marked point at the origin, goes vertically up until the point (0, 1), then goes horizontally to the right, and ends at the marked point (1, 1). Graph d. A function is graphed on an x y coordinate plane. The function begins at a marked point at the origin, goes horizontally to the right until the point (1, 0), then goes vertically up, and ends at the marked point (1, 1).

FIGURE 9.9.1 Line integral in Example 1 is the same on four paths

≡

Conservative Vector Fields

Before proceeding, we need to introduce a special kind of vector field F called a conservative field.

DEFINITION 9.9.1 Conservative Vector Field

A vector function F in 2- or 3-space is said to be conservative if F can be written as the gradient of a scalar function ϕ. The function ϕ is called a potential function for F.

In other words, F is conservative if there exists a function ϕ such that A conservative vector field is also called a gradient vector field.

EXAMPLE 2 Conservative Vector Field

The integral in Example 1 can be interpreted as a line integral of a vector field F along a path C. If F(x, y) = yi + xj and dr = dxi + dyj, then Now consider the function The gradient of the scalar function ϕ is

Because we conclude that F(x, y) = yi + xj is a conservative vector field and that φ is a potential function for F. ≡

Of course, not every vector field is a conservative field although many vector fields encountered in physics are conservative. For present purposes, the importance of conservative vector fields will become evident as we continue our study of line integrals.

Path Independence

If the value of a line integral is the same for every path in a region connecting the initial point A and terminal point B, then the integral is said to be independent of the path. In other words, a line integral of F along C is independent of the path if for any two paths C1 and C2 between A and B.

A Fundamental Theorem

The next theorem establishes an important relationship between the value of a line integral and the fact that its path of integration C lies within a conservative vector field F. In addition, it provides a means of evaluating these line integrals in a manner that is akin to the Fundamental Theorem of Calculus:

(1)

where f(x) is an antiderivative of In the next theorem, known as the Fundamental Theorem for Line Integrals, the gradient of a scalar function ϕ plays the part of the derivative in (1).

THEOREM 9.9.1 Fundamental Theorem

Suppose C is a path in an open region R of the xy-plane and is defined by , If F(x, y) = P(x, y)i + Q(x, y)j is a conservative vector field in R and ϕ is a potential function for F, then

, (2)

where and

PROOF:

We will prove the theorem for a smooth path C. Because ϕ is a potential function for F we have

Then using we can write the line integral of F along the path C as

In view of the Chain Rule (see (8) in Section 9.4),

,

it follows that

≡

For piecewise-smooth curves, the foregoing proof must be modified by considering each smooth arc of the curve C.

Theorem 9.9.1 shows that if F is a conservative vector field in an open region in 2- or 3-space, then depends only on the initial and terminal points A and B of the path C, and not on C itself. In other words, line integrals of conservative vector fields are independent of the path. Such integrals are often written

. (3)

EXAMPLE 3 Example 1 Revisited Again

Evaluate where C is a path with initial point (0, 0) and terminal point (1, 1).

SOLUTION

The path C shown in FIGURE 9.9.2 represents any piecewise-smooth curve with initial and terminal points (0, 0) and (1, 1). We have just seen that F = yi+ xj is a conservative vector field defined at each point of the xy-plane and that is a potential function for F. Thus, in view of (2) of Theorem 9.9.1 and (3), we can write

≡

A graph. A piecewise-smooth curve is graphed on an x y coordinate plane. The curve labeled C, begins at the marked origin (0, 0), goes up and to the right initially, changes direction and continues going up to the left, changes direction once again and goes up to the right, reaches a high point, goes down to the right, and ends at the marked point (1, 1).

FIGURE 9.9.2 Piecewise-smooth curve in Example 3

In using the Fundamental Theorem of Calculus (1), any antiderivative of can be used, such as f(x)+ K, where K is a constant. Similarly, a potential function for the vector field in Example 2 is where K is a constant. We may disregard this constant when using (2) of Theorem 9.9.1 since

Some Terminology

We say that a region (in the plane or in space) is connected if every pair of points A and B in the region can be joined by a piecewise-smooth curve that lies entirely in the region. A region R in the plane is simply connected if it is connected and every simple closed curve C lying entirely within the region can be shrunk, or contracted, to a point without leaving R. The last condition means that if C is any simple closed curve lying entirely in R, then the region in the interior of C also lies entirely in R. Roughly put, a simply connected region has no holes in it. The region R in FIGURE 9.9.3(a) is a simply connected region. In Figure 9.9.3(b) the region R shown is not connected, or disconnected, since A and B cannot be joined by a piecewise-smooth curve C that lies in R. The region in Figure 9.9.3(c) is connected but not simply connected because it has three holes in it. The representative curve C in the figure surrounds one of the holes, and so cannot be shrunk to a point without leaving the region. This last region is said to be multiply connected. Lastly, a region R is said to be open if it contains no boundary points.

3 visual representations. Visual representation a. Caption. Connected region R. Figure. A region labeled R contains a piecewise-smooth curve and 2 marked points. The curve labeled C, begins at a point labeled A, and ends at a point labeled B. Visual representation b. Caption. R is not connected. Figure. 2 disconnected regions. The region on the left contains a point labeled A, and the region on the right contains a point labeled B. Visual representation c. Caption. Multiply connected region R. Figure. A region labeled R contains 3 holes and a closed curve. The closed curve, labeled C, goes around one of the holes in the region.

FIGURE 9.9.3 Regions in the plane

In an open connected region R, the notions of path independence and a conservative vector field are equivalent. This means: If F is conservative in R, then is independent of the path C, and conversely, if is independent of the path, then F is conservative.

We state this formally in the next theorem.

THEOREM 9.9.2 Equivalent Concepts

In an open connected region R, is independent of the path C if and only if the vector field F is conservative in R.

PROOF:

If F is conservative in R, then we have already seen that is independent of the path C as a consequence of Theorem 9.9.1.

For convenience we prove the converse for a region R in the plane. Assume that is independent of the path in R and that (x0, y0) and (x, y) are arbitrary points in the region R. Let the function ϕ(x, y) be defined as

where C is an arbitrary path in R from (x0, y0) to (x, y) and F = Pi + Qj. See FIGURE 9.9.4(a). Now choose a point (x1, y), x1 ≠ x, so that the line segment from (x1, y) to (x, y) is in R. See Figure 9.9.4(b). Then by path independence we can write

Now,

since the first integral does not depend on x. But on the line segment between (x1, y) and (x, y), y is constant so that dy = 0. Hence, By the derivative form of the Fundamental Theorem of Calculus we then have

Likewise we can show that Hence, from

we conclude that F is conservative. ≡

2 visual representations. Visual representation a. A region labeled R contains a piecewise-smooth curve. The curve labeled C, begins at the marked point (x subscript 0, y subscript 0), and ends at the marked point (x y). Visual representation b. A region labeled R contains a curve. The curve labeled C, consists of a piecewise-smooth curve and a horizontal line segment. The piecewise-smooth curve begins at the marked point (x subscript 0, y subscript 0), and ends at the marked point (x subscript 1, y). The line begins at the marked point (x subscript 1, y), and ends at the marked point (x y).

FIGURE 9.9.4 Region R in the proof of Theorem 9.9.2

Integrals Around Closed Paths

Recall from Section 9.8 that a path, or curve, C is said to be closed when its initial point A is the same as the terminal point B. If C is a parametric curve defined by a vector function then C is closed when A = B; that is, r(a) = r(b). The next theorem is an immediate consequence of Theorem 9.9.1.

THEOREM 9.9.3 Equivalent Concepts

In an open connected region R, is independent of the path if and only if for every closed path C in R.

PROOF:

First, we show that if is independent of the path, then for every closed path C in R. To see this let us suppose A and B are any two points on C and that C = C1 ∪ C2, where C1 is a path from A to B and C2 is a path from B to A. See FIGURE 9.9.5(a). Then

(4)

where –C2 is now a path from A to B. Because of path independence Thus (4) implies that

2 visual representations. Visual representation a. Caption. C = C subscript 1 union C subscript 2. Figure. A closed path consists of 2 curves. The first curve, labeled C subscript 1, begins at a point labeled A, and ends at a point labeled B. The second curve, labeled C subscript 2, begins at the point labeled B, and ends at the point labeled A. Visual representation b. Caption. C = C subscript 1 union (negative C subscript 2). Figure. A closed path consists of 2 curves. The first curve, labeled C subscript 1, begins at a point labeled A, and ends at a point labeled B. The second curve, labeled C subscript 2, also begins at the point labeled A, and ends at the point labeled B.

FIGURE 9.9.5 Paths in the proof of Theorem 9.9.3

Next, we prove the converse that if for every closed path C in R, then is independent of the path. Let C1 and C2 represent any two paths from A to B and so C = C1 ∪ (–C2) is a closed path. See Figure 9.9.5(b). It follows from or

that Hence is independent of the path. ≡

Suppose F is a conservative vector field defined over an open connected region and C is a closed path lying entirely in the region. When the results of the preceding theorems are put together we conclude that:

. (5)

The symbol ⇔ in (5) is read “equivalent to” or “if and only if.”

Test for a Conservative Field

The implications in (5) show that if the line integral is not path independent, then the vector field is not conservative. But there is an easier way of determining whether F is conservative. The following theorem is a test for a conservative vector field that uses the partial derivatives of the component functions of F = Pi + Qj.

THEOREM 9.9.4 Test for a Conservative Field

Suppose F(x, y) = P(x, y)i + Q(x, y)j is a conservative vector field in an open region R, and that P and Q are continuous and have continuous first partial derivatives in R. Then

(6)

for all (x, y) in R. Conversely, if the equality (6) holds for all (x, y) in a simply connected region R, then F = Pi + Qj is conservative in R.

PARTIAL PROOF:

We prove the first half of the theorem. Assume that the component functions of the conservative vector field F = Pi + Qj are continuous and have continuous first partial derivatives in an open region R. Since F is conservative there exists a potential function ϕ such that

Thus and Now

From continuity of the partial derivatives we have as was to be shown. ≡

EXAMPLE 4 Using Theorem 9.9.4

The conservative vector field F(x, y) = yi+ xj in Example 2 is continuous and has component functions that have continuous first partial derivatives throughout the open region R consisting of the entire xy-plane. With the identifications P = y and Q = x it follows from (6) of Theorem 9.9.4,

≡

EXAMPLE 5 Using Theorem 9.9.4

Determine whether the vector field is conservative.

SOLUTION

With and we find

Because for all points in the plane, it follows from Theorem 9.9.4 that F is not conservative. ≡

EXAMPLE 6 Using Theorem 9.9.4

Determine whether the vector field is conservative.

SOLUTION

With and we find

The components of F are continuous and have continuous partial derivatives. Thus (6) holds throughout the xy-plane, which is a simply connected region. From the converse in Theorem 9.9.4 we can conclude that F is conservative. ≡

We have one more important question to answer in this section:

If F is a conservative vector field, how does one find a potential function φ for F? (7)

In the next example we give the answer to the question posed in (7).

EXAMPLE 7 Integral That Is Path Independent

(a) Show that where is independent of the path C between (–1, 0) and (3, 4).

(b) Find a potential function φ for F.

(c) Evaluate

SOLUTION

(a) Identifying and yields

The vector field F is conservative because (6) holds throughout the xy-plane and as a consequence the integral is independent of the path between any two points A and B in the plane.

(b) Because F is conservative there is a potential function φ such that

(8)

In partial integration with respect to x, the variable y is treated as a constant. Similarly, in partial integration with respect to y we treat x as a constant.

Employing partial integration on the first expression in (8) gives

(9)

where g(y) is the “constant” of integration. Now we take the partial derivative of (9) with respect to y and equate it to the second expression in (8):

From the last equality we find Integrating again gives where C is a constant. Thus

(10)

(c) We can now use Theorem 9.9.2 and the potential function (10) (without the constant) to evaluate the line integral:

. ≡

Note: Since the integral in Example 7 was shown to be independent of the path in part (a), we can evaluate it without finding a potential function. We can integrate along any convenient curve C connecting the given points. In particular, the line y = x + 1 is such a curve. Using x as a parameter then gives

Conservative Vector Fields in 3-Space

For a three-dimensional conservative vector field

and a piecewise-smooth space curve the basic form of (2) is the same:

. (11)

If C is a space curve, a line integral is independent of the path whenever the three-dimensional vector field

is conservative. The three-dimensional analogue of Theorem 9.9.4 goes like this: If F is conservative and P, Q, and R are continuous and have continuous first partial derivatives in some open region of 3-space, then

. (12)

Conversely, if (12) holds throughout an appropriate region of 3-space, then F is conservative.

The necessity of (12) can be seen from (5) of Section 9.7. If F is conservative then F = and curl that is,

Setting the three components of the vector curl F equal to 0 yields (12).

EXAMPLE 8 Integral That Is Path Independent

(a) Show that the line integral

is independent of the path C between (1, 1, 1) to (2, 1, 4).

(b) Evaluate .

SOLUTION

(a) With the identifications

we see that the equalities

hold throughout 3-space. From (12) we conclude that F is conservative and therefore the integral is independent of the path.

(b) The path C shown in FIGURE 9.9.6 represents any path with initial and terminal points (1, 1, 1) and (2, 1, 4). To evaluate the integral we again illustrate how to find a potential function for F using partial integration.

A graph. A curve is graphed on a three dimensional x y z coordinate system. The curve labeled C, begins at the marked point (1, 1, 1), goes down, reaches a low point, goes up, and ends at the marked point (2, 1, 4).

FIGURE 9.9.6 Representative path C in Example 8

First we know that

Integrating the first of these three equations with respect to x gives

The derivative of this last expression with respect to y must then be equal to Q:

Hence,

Consequently,

The partial derivative of this last expression with respect to z must now be equal to the function R:

From this we get and Disregarding K, we can write

(13)

Finally, from (11) and the potential function (13) we obtain

≡

Conservation of Energy

In a conservative force field F, the work done by the force on a particle moving from position A to position B is the same for all paths between these points. Moreover, the work done by the force along a closed path is zero. See Problem 29 in Exercises 9.9. For this reason, such a force field is also said to be conservative. In a conservative field F the law of conservation of mechanical energy holds:

For a particle moving along a path in a conservative field,

kinetic energy + potential energy = constant.

See Problem 31 in Exercises 9.9.

A frictional force such as air resistance is nonconservative. Nonconservative forces are dissipative in that their action reduces kinetic energy without a corresponding increase in potential energy. Stated in another way, if the work done depends on the path, then F is nonconservative.

9.9 Exercises Answers to selected odd-numbered problems begin on page ANS-25.

In Problems 1–10, show that the given line integral is independent of the path. Evaluate in two ways: (a) Find a potential function φ and then use Theorem 9.9.1, and (b) Use any convenient path between the endpoints of the path.

  1. x2 dx + y2 dy
  2. 2xy dx + x2 dy
  3. (x + 2y) dx + (2x – y) dy
  4. cos x cos y dx + (1 – sin x sin y) dy
  5. on any path not crossing the x-axis
  6. on any path not through the origin
  7. (2y2x – 3) dx + (2yx2 + 4) dy
  8. (5x + 4y) dx + (4x – 8y3) dy
  9. (y3 + 3x2y) dx + (x3 + 3y2x + 1) dy
  10. (2x – y sin xy – 5y4) dx – (20xy3 + x sin xy) dy

In Problems 11–16, determine whether the given vector field is a conservative field. If so, find a potential function ϕ for F.

  1. F(x, y) = (4x3y3 + 3)i + (3x4y2 + 1)j
  2. F(x, y) = 2xy3i + 3y2(x2 + 1)j
  3. F(x, y) = y2 cos xy2i – 2xy sin xy2j
  4. F(x, y) = (x2 + y2 + 1)−2(xi + yj)
  5. F(x, y) = (x3 + y)i + (x + y3)j
  6. F(x, y) = 2e2yi + xe2yj

In Problems 17 and 18, find the work done by the force F(x, y) = (2x + e–y)i + (4y – xe–y)j along the indicated curve.

  1. A graph. A curve is graphed on an x y coordinate plane. The curve labeled y = x^4, begins at the marked origin (0, 0), goes up and to the right with increasing steepness, and ends at the marked point (1, 1).

    FIGURE 9.9.7 Curve for Problem 17

  2. A graph. A parabola is graphed on an x y coordinate plane. The curve labeled y = (x^2) over 4 + (y^2) over 9 = 1, begins on the x axis at the marked point (negative 2, 0), goes up and to the right with decreasing steepness, reaches a high point on the y axis at the point (0, 3), goes down to the right symmetrically, and ends on the x axis at the marked point (2, 0).

    FIGURE 9.9.8 Curve for Problem 18

In Problems 19–24, show that the given integral is independent of the path. Evaluate.

  1. yz dx + xz dy + xy dz
  2. 2x dx + 3y2 dy + 4z3 dz
  3. (2x sin y + e3z) dx + x2 cos y dy + (3xe3z + 5) dz
  4. (2x + 1) dx + 3y2 dy + dz
  5. e2z dx + 3y2 dy + 2xe2z dz
  6. 2xz dx + 2yz dy + (x2 + y2) dz

In Problems 25 and 26, evaluate ∫C F · dr.

  1. F(x, y, z) = (y – yz sin x)i + (x + z cos x)j + y cos xk;
    r(t) = 2ti + (1 + cos t)2j + 4 sin3tk, 0 ≤ t ≤ π/2
  2. F(x, y, z) = (2 – ez)i + (2y – 1)j + (2 – xez)k;
    r(t) = ti + t2j + t3k, (–1, 1, –1) to (2, 4, 8)
  3. The inverse square law of gravitational attraction between two masses m1 and m2 is given by F = –Gm1m2r/r3, where r = xi + yj + zk. Show that F is conservative. Find a potential function for F.
  4. Find the work done by the force F(x, y, z) = 8xy3zi + 12x2y2zj + 4x2y3k acting along the helix r(t) = 2 cos ti + 2 sin tj + tk from (2, 0, 0) to (1, , π/3). From (2, 0, 0) to (0, 2, π/2). [Hint: Show that F is conservative.]
  5. If F is a conservative force field, show that the work done along any simple closed path is zero.
  6. A particle in the plane is attracted to the origin with a force F = rnr, where n is a positive integer and r = xi + yj is the position vector of the particle. Show that F is conservative. Find the work done in moving the particle between (x1, y1) and (x2, y2).
  7. Suppose F is a conservative force field with potential function φ. In physics the function p = –ϕ is called potential energy. Since F = −∇p, Newton’s second law becomes

    By integrating m · + ∇p · = 0 with respect to t, derive the law of conservation of mechanical energy: mv2 + p = constant. [Hint: See Problem 39 in Exercises 9.8.]

  8. Suppose that C is a smooth curve between points A (at t = a) and B (at t = b) and that p is potential energy, defined in Problem 31. If F is a conservative force field and K = mv2 is kinetic energy, show that p(B) + K(B) = p(A) + K(A).