Lecture Note
Advanced Engineering Mathematics, 7th edition
by
Peter V. O’Neil,
Math 301 (Methods of Applied Mathematics)
Chapter 12 (Vector Integral Calculus)
12.4 Independence of Path and Potential Theory
Dr T. A. Apalara
IMPORTANT!! These instructor notes are made available for convenience to
reduce the amount of note-taking in class and also in case it is not possible to
attend class due to circumstances beyond control (e.g. illness). These notes are
neither intended as a replacement for class attendance and participation nor as a
substitute for using the textbook. The class presentation may differ from these
notes. Moreover, some other problems not in these notes may be discussed in class.
Math 301
Week #3
12.4 Independence of Path and Potential Theory
Dr T. A. Apalara
§§ 12.4 Independence of Path and Potential Theory
Objective: In this chapter, you will know, insha Allaah:
☛ How to show that a vector field F is conservative.
☛ How to find a potential for a conservative field.
In physics, a conservative force field is one that is derivable from a potential. We will use the
same terminology.
Definition − Conservative Vector Field: Let D be a set of points in the plane. A vector
field F(x, y) is conservative on D if for some real-valued φ(x, y) (scalar field), we have
F = ∇φ =
∂φ
∂φ
∂φ
i+
j+
k
∂x
∂y
∂z
for all (x, y) in D.
Remark 1. ☛ We call φ a potential function, or potential, for F.
☛ If φ is a potential function for F, then so is φ + c for any constant c, because ∇(φ + c) = ∇φ.
For this reason we often speak of a potential function for F, rather than the potential function.
☛ Recall that, if F(x, y)Z= f (x, y)i+g(x, y)j+h(x, y)k and R(t)
Z = x(t)i+y(t)j+z(t)k is a position
F · dR is another way of writing
function for C, then
that is
C
f (x, y)dx + g(x, y)dy + h(x, y)dz,
C
Z
Z
F · dR =
C
f (x, y)dx + g(x, y)dy + h(x, y)dz.
C
Z
F · dR throughout this section because we want
We will make frequent use of the notation
C
to examine the effect on this integral when F has a potential function, and for this we will use
vector notation.
THEOREM 12.2: Let F be conservative in a region D (of the plane or of 3-space). Let C be
a path from P0 to P1 in D. Then
Z
F · dR = φ (P1 ) − φ (P0 ) .
C
Proof. Suppose C has differentiable coordinate functions x = x(t), y = y(t), z = z(t) for a ≤ t ≤ b,
where P0 = (x(a), y(a), z(a)) and P1 = (x(b), y(b), z(b)) . Since F is conservative, we have F = ∇φ.
1
Math 301
Week #3
12.4 Independence of Path and Potential Theory
Dr T. A. Apalara
So
Z
Z
F · dR =
C
C
Z b
∂φ
∂φ
∂φ
dx +
dy +
dz =
∂x
∂y
∂z
Z
b
a
Z b
∂φ dx ∂φ dx ∂φ dx
+
+
∂x dt
∂y dt
∂z dt
dt
d
φ (x(t), y(t), z(t)) dt =
d (φ (x(t), y(t), z(t)))
dt
a
a
=φ (x(b), y(b), z(b)) − φ (x(a), y(a), z(a)) = φ (P1 ) − φ (P0 ) .
=
Remark 2.
☛ The above result states that
Z
F · dR = φ(terminal point of C) − φ(initial point of C).
C
Z
F · dR depends only on the endpoints of C. In other words, it shows
Hence, the value of
C
that the line integral of a conservative vector field over a path is the difference in values of a
potential function at end points of the path.
☛ This is the line integral version of the fundamental theorem of calculus, and applies to line
integrals of conservative vector fields.
☛ Another consequence of F being conservative is that the actual path itself does not influence
the outcome, only the end points of the path. If weZchose a different path K between the
same end points, we would obtain the same result for
F · dR. This suggests the concept of
K
independence of path of a line integral.
Z
Definition − Independence of Path:
F · dR is independent of path on a set D of points
C
in the plane if for any points P0 and P1 in D, the line integral has the same value over any paths
in D having initial point P0 and terminal point P1 . in D.
Theorem: Let φ and its first partial
Z derivatives be continuous for all (x, y, z) in a set D of points
in the space. Let F = ∇φ. Then
F · dR is independent of path in D. Further, if C is a simple
C
I
closed path in D, then
F · dR = 0.
C
Z
Remark 3. The independence of path states that, when F = ∇φ, the value of
F · dR depends
C
only on the values of φ(x, y, z) at the end points of the path, and not where the path goes in
between. For the last conclusion of the theorem, if C is a closed path in D, then the initial and
terminal points coincide, hence the difference between the values of φ at the terminal and initial
points is zero .
Z
F · dR is independent of path in D.
Theorem: If F is conservative in D, then
C
2
Math 301
Week #3
12.4 Independence of Path and Potential Theory
Dr T. A. Apalara
Z
F · dR is independent of path in D if and only if
Theorem:
C
I
F · dR = 0
C
for every closed path in D.
Z
F · dR over paths in some region
Remark 4. Thus far, we have the following implications for
C
D:
Z
F · dR.
☛ Conservative F =⇒ independence of path of
C
☛ Independence of path in D ⇐⇒ integrals over all closed paths in D are zero.
Theorem (Test for a Convervative Field in the Plane): Let F(x, y) = f (x, y)i + g(x, y)j be
a vector field and D a simply connected domain. Suppose f, g, ∂f /∂y, and ∂g/∂x are continuous
on D. Then, F is conservative on D if and only if
∂f
∂g
=
∂y
∂x
for all (x, y) in D.
In 3−space there is a similar test for a vector field to be conservative, with adjustments to accommodate the extra dimension. A set S of points in R3 is a domain if it satisfies the following two
conditions:
☛ If P is a point in S, then there is a sphere about P that encloses only points of S.
☛ Between any two points of S there is a path lying entirely in S.
Furthermore, S is simply connected if every simple closed path in S is the boundary of a surface
in S. With this notion of simple connectivity in 3−space, we can state a three-dimensional version
of test of conservative field in 3−space.
Theorem (Test for a Conservative Field in the 3−Space): Let F(x, y, z) = f (x, y, z)i +
g(x, y, z)j + h(x, y, z)k be a vector field over a simply connected domain S in R3 . Then, F is
conservative on S if and only if
∇ × F = O.
Remark 5. This test can also be used for a plane (2−space) by putting zero in the k component,
that is F(x, y) = f (x, y)i + g(x, y)j + 0k. Now compute
i
j
∂
∂
∇×F=
∂x
∂y
f (x, y) g(x, y)
So the condition ∇ × F = O =⇒
k
∂
∂z
0
=
∂g
∂f
−
∂x ∂y
k.
∂g
∂f
=
if the vector field is in the plane.
∂x
∂y
3
Math 301
Week #3
12.4 Independence of Path and Potential Theory
Dr T. A. Apalara
Example 1. Given the force F = 3x2 yz 2 i + x3 z 2 + ez j + 2x3 yz + yez k.
(a) Show that F is conservative.
(b) Find a potential function for F.
(c) Find the work done by F along any path from (0, 0, 0) to (−1, 3, −2).
Solution
(a) f (x, y, z) = 3x2 yz 2 , g(x, y, z) = x3 z 2 + ez , and h(x, y, z) = 2x3 yz + yez
i
j
∂
∂
∇×F=
∂x
∂y
3x2 yz 2 x3 z 2 + ez
∂
=
(2x3 yz + yez ) −
∂y
∂ 3 2
(x z + ez ) −
+
∂x
k
∂
∂z
2x3 yz + yez
∂ 3 2
∂
∂
(x z + ez ) i +
(3x2 yz 2 ) −
(2x3 yz + yez ) j
∂z
∂z
∂x
∂
(3x2 yz 2 ) k = O
∂y
So F is conservative.
(b) A potential function φ must satisfy
∂φ
=3x2 yz 2 ,
∂x
∂φ
=x3 z 2 + ez ,
∂y
∂φ
=2x3 yz + yez .
∂z
(1)
(2)
(3)
Consider (1), we have
Z
φ=
3x2 yz 2 dx = x3 yz 2 + ψ(y, z)
∂ψ(y, z)
∂φ
=x3 z 2 +
= x3 z 2 + ez =⇒ ψ(y, z) = yez + ϕ(z)
∂y
∂y
∴ φ =x3 yz 2 + yez + ϕ(z)
∂φ
=2x3 yz + yez + ϕ′ (z) = 2x3 yz + yez =⇒ ψ(z) = k.
∂z
We can take the constant k = 0, so a potential function is φ = x3 yz 2 + yez .
(c) Since F is conservative, then it’s independent of path. Thus, the work done by F is
Z
(−1,3,−2)
F · dR =φ (−1, 3, −2) − φ (0, 0, 0) = x3 yz 2 + yez
C
= −12 + 3e−2 .
(0,0,0)
✍
Remark 6. If C is a closed path then
R
F · dR = 0.
C
4
Math 301
Week #3
12.4 Independence of Path and Potential Theory
Dr T. A. Apalara
Example 2. Given the force F = 2x3 y 4 + x i + 2x4 y 3 + y j. Use potentials to evaluate the
work done by this force when C is any path from point A(−1, 0) to point B(−2, 1).
Solution
Note that F is conservative since
∂f
∂g
=
= 8x3 y 3
∂y
∂x
A potential function φ must satisfy
∂φ
=2x3 y 4 + x,
∂x
∂φ
=2x4 y 3 + y,
∂y
(4)
(5)
Consider (4), we have
1
x2
2x3 y 4 + x dx = x4 y 4 +
+ ψ(y)
2
2
y2
∂φ
=2x4 y 3 + ψ ′ (y) = 2x4 y 3 + y =⇒ ψ(y) =
+ k.
∂y
2
Z
φ=
1 4 4
We can take the constant k = 0, so a potential function is φ =
x y + x2 + y 2 .
2
Thus, the work done by F is
Z
F · dR =φ (−2, 1) − φ (−1, 0) =
C
1 4 4
x y + x2 + y 2
2
(−2,1)
= 10.
(−1,0)
✍
END
More examples for your practice
Example. Determine whether F is conservative in the given region D. If the vector field is
conservative, find a potential.
(i) F(x, y) = yi + ex j
(ii) F(x, y) = (2xy 2 + y)i + (2x2 y + ex y)j
(iii) F(x, y) = 2x cos(2y)i − 2x2 sin(2y)j
(iv) F(x, y) = 2x cos(2y)i − 2x2 sin(2y) + 4y 2 j.
Solution
(i) f (x, y) = y and g(x, y) = ex .
∂g
∂f
= ex and
= 1.
∂x
∂y
Since
∂g
∂f
̸=
, we say F is not conservative. Hence it has no potential.
∂x
∂y
5
Math 301
Week #3
12.4 Independence of Path and Potential Theory
Dr T. A. Apalara
(ii) f (x, y) = 2xy 2 + y and g(x, y) = 2x2 y + ex y.
∂f
∂g
= 4xy + ex y and
= 4xy + 1.
∂x
∂y
Since
∂g
∂f
̸=
, we say F is not conservative. Hence it has no potential.
∂x
∂y
(iii) f (x, y) = 2x cos(2y) and g(x, y) = −2x2 sin(2y)
∂g
∂f
= −4x sin(2y) and
= −4x sin(2y).
∂x
∂y
∂g
∂f
=
, we say F is conservative. Next, we find a potential function. A potential
∂x
∂y
∂φ
∂φ
function φ must satisfy
= 2x cos(2y),
= −2x2 sin(2y). So
∂x
∂y
Z
φ = 2x cos(2y) dx = x2 cos(2y) + ψ(y)
Since
∂φ
= − 2x2 sin(2y) + ϕ′ (y) = −2x2 sin(2y) =⇒ ϕ(y) = k.
∂y
=⇒
We can take the constant k = 0, so a potential function is φ = x2 cos(2y).
(iv) f (x, y) = 2x cos(2y) and g(x, y) = −2x2 sin(2y) − 4y 2
∂g
∂f
= −4x sin(2y) and
= −4x sin(2y).
∂x
∂y
∂g
∂f
=
, we say F is conservative. Next, we find a potential function. A potential
∂x
∂y
function φ must satisfy
Since
∂φ
∂φ
= 2x cos(2y) and
= −2x2 sin(2y) − 4y 2 .
∂x
∂y
So
Z
φ=
=⇒
2x cos(2y) dx = x2 cos(2y) + ψ(y)
∂φ
4y 3
= − 2x2 sin(2y) + ϕ′ (y) = −2x2 sin(2y) − 4y 2 =⇒ ϕ(y) = −
+ k.
∂y
3
We can take the constant k = 0, so a potential function is
φ = x2 cos(2y) −
4y 3
.
3
✍
6
12.4 Independence of Path and Potential Theory
387
to think of F as a vector field in 3-space. Now compute
i
j
k ∂g ∂ f
∂/∂
x
∂/∂
y
∂/∂z
∇ ×G=
= ∂ x − ∂ y k.
f (x, y) g(x, y)
0 The 3-space condition ∇ × G = O therefore reduces to equation (12.8) if the vector field is in the
plane.
Theorem 12.7 can be proved when Stokes’s theorem is available to us.
PROBLEMS
SECTION 12.4
21. Prove the law of conservation of energy, which states
that the sum of the kinetic and potential energies of an
object acted on by a conservative force is a constant.
Hint: The kinetic energy is (m/2) R (t) 2 , where m
is the mass and R(t) describes the trajectory of the
particle. The potential energy is −ϕ(x, y, z), where
F = ∇ϕ.
In each of Problems 1 through 10, determine whether F
is conservative in the given region D. If D is not defined
explicitly, it is understood to be the entire plane or 3-space.
If the vector field is conservative, find a potential.
1. F = y 3 i + (3x y 2 − 4)j
2. F = (6y + e x y )i + (6x + xe x y )j
22. Complete the proof of Theorem 12.5 by filling in the
details of the following argument. By differentiation,
it has already been shown that, if F has a potential
function, then
∂g ∂ f
=
.
∂x ∂y
To prove the converse, assume equality of these partial derivatives for (x, y) in D. We must produce a
potential function ϕ for F.
First use Green’s theorem to show that C F ·
dR
= 0 for any closed path in D. Thus conclude that
F · dR is independent of path in D. Choose a point
C
P0 = (a, b) in D. Then, for any (x, y), define
(x,y)
F · dR.
ϕ(x, y) =
3. F = 16xi + (2 − y 2 )j
4. F = 2x y cos(x )i + sin(x )j
2y
2x
i+
j D is the plane with
5. F =
x 2 + y2
x 2 + y2
the origin removed.
2
2
6. F = 2xi − 2yj + 2zk
7. F = i − 2j + k
8. F = yz cos(x)i + (z sin(x) + 1)j + y sin(x)k
9. F = (x 2 − 2)i + x yzj − yz 2 k
10. F = e x yz (1 + x yz)i + x 2 zj + x 2 yk
In each of Problems11 through 20, determine a potential
function to evaluate C F · dR for C any path from the first
point to the second.
P0
This is a function because the integral is independent
of path, hence depends only on (x, y). To show that
∂ϕ/∂ x = f (x, y), first show that
11. F = 3x 2 (y 2 − 4y)i + (2x 3 y − 4x 3 )j; (−1, 1), (2, 3)
ϕ(x + x, y) − ϕ(x, y)
(x+x,y)
f (ξ, η) dξ + g(ξ, η) dη.
=
12. F = e x cos(y)i − e x sin(y)j; (0, 0), (2, π/4)
13. F = 2x yi + (x 2 − 1/y)j; (1, 3), (2, 2) (The path cannot
cross the x - axis).
(x,y)
14. F = i + (6y + sin(y))j; (0, 0), (1, 3)
15. F = (3x y − 6y )i + (2x y − 18x y )j; (0, 0), (1, 1)
2
2
3
3
2
16. F = (y cos(x z) − x yz sin(x z))i + x cos(x z)j − x 2 sin(x z)k;
(1, 0, π ), (1, 1, 7)
17. F = i − 9y 2 zj − 3y 3 k; (1, 1, 1), (0, 3, 5)
18. F = −8y 2 i − (16x y + 4z)j − 4yk; (−2, 1, 1), (1, 3, 2)
19. F = 6x 2 e yz i + 2x 3 ze yz j + 2x 3 ye yz k; (0, 0, 0), (1, 2, −1)
20. F = (y − 4x z)i + xj + (3z 2 − 2x 2 )k; (1, 1, 1), (3, 1, 4)
Parametrize the horizontal line segment from (x, y) to
(x + x, y) by ξ = x + tx for 0 ≤ t ≤ 1 to show that
1
f (x + tx, y) dt.
ϕ(x + x, y) − ϕ(x, y) = x
0
Use this to show that
ϕ(x + x, y) − ϕ(x, y)
= f (x + t0 x, y)
x
for some t0 in (0, 1). Now take the limit as x → 0 to
show that ∂ϕ/∂ x = f (x, y). A similar argument shows
that ∂ϕ/∂ y = g(x, y).
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October 14, 2010
14:53
THM/NEIL
Page-387
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