Path-Independent Vector Fields and the Fundamental Theorem of Calculus for Line Integrals

Section 4.4 Path-Independent Vector Fields and the Fundamental Theorem of Calculus for Line Integrals

Motivating Questions

What characteristic of a vector field $F$ will make $\int_{C} F \cdot d r$ have the same value for every oriented curve from a point $P$ to a point $Q ?$
What special properties do gradient vector fields have?
Given a gradient vector field $F,$ how can we efficiently find a potential function $f$ so that $F = \nabla f ?$

In Activity 4.3.3, Example 4.3.5, and Activity 4.3.4, we have encountered situations where

C_{1}

and

C_{2}

are different oriented curves from a point

P

to a point

Q

and

\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r .

In this section, we explore vector fields which have the property that for all points

P

and

Q,

C_{1}

and

C_{2}

are oriented paths from

P

Q,

then

\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r .

🔗

Preview Activity 4.4.1.

🔗

In Activity 4.3.3, we considered the vector field

F (x, y) = ⟨ y^{2}, 2 x y + 3 ⟩

and two different oriented curves from

(1, 0)

(- 1, 0) .

We found that the value of the line integral of

F

was the same along those two oriented curves.

🔗

(a)

🔗

Verify that

F (x, y) = ⟨ y^{2}, 2 x y + 3 ⟩

is a gradient vector field by showing that

F = \nabla f

for the function

f (x, y) = x y^{2} + 3 y .

Answer.

Calculate

\frac{\partial f}{\partial x}

and

\frac{\partial f}{\partial y} .

Solution.

Since

\frac{\partial f}{\partial x} = y^{2}

and

\frac{\partial f}{\partial y} = 2 x y + 3,

F = \nabla f .

🔗

(b)

🔗

Calculate the change in the output of the scalar function

f

over the curves

C_{1}

and

C_{2} .

In other words, what is the difference in the output of

f

at the start of the curve and the end of the curve. How does this value compare to the value of the line integral

\int_{C_{1}} F \cdot d r

you found in Activity 4.3.3?

Answer.

0

Solution.

Algebraically,

f (- 1, 0) - f (1, 0) = 0 - 0 = 0 .

There is no net change in the output of

f .

The line integral is also

0 .

🔗

(c)

🔗

Let

C_{3}

be the line segment from

(1, 1)

(3, 4) .

Calculate

\int_{C_{3}} F \cdot d r

as well as

f (3, 4) - f (1, 1) .

What do you notice?

Answer.

56

Solution.

The line segment

C_{3}

is given by

r (t) = ⟨ 1 + 2 t, 1 + 3 t ⟩

0 \leq t \leq 1 .

Thus,

\int_{C_{3}} F \cdot d r = \int_{0}^{1} ⟨ (1 + 3 t)^{2}, 2 (1 + 2 t) (1 + 3 t) ⟩ \cdot ⟨ 2, 3 ⟩ d t = 56

Note here that

f (3, 4) - f (1, 1) = 60 - 4 = 56

which is the same as the result of the line integral.

🔗

(d)

🔗

Earlier in the chapter, we’ve used Clairaut’s Theorem to argue that a vector field in

R^{2}

is not a gradient vector field when

\partial F_{1} / \partial y \neq \partial F_{2} / \partial x .

Earlier in this preview activity, you verified that a given vector field was the gradient of a particular function of two variables. Clairaut’s Theorem holds for functions of three variables. However, in that case there are six mixed partials to calculate, and thus it can be rather tedious.The remaining parts of this preview activity suggest a process for determining if a vector field in

R^{3}

is a gradient vector field as well as finding a potential function for the vector field.

🔗

Let

G (x, y, z) = ⟨ 3 e^{y^{2}} + z \sin (x), 6 x y e^{y^{2}} - z, 3 z^{2} - y - \cos (x) ⟩

and

H (x, y, z) = ⟨ 3 x^{2} y, x^{3} + 2 y z^{3}, x z + 3 y^{2} z^{2} ⟩ .

🔗

G

and

H

are to be gradient vector fields, then there are functions

g

and

h

for which

G = \nabla g

and

H = \nabla h .

What would this tell us about the partial derivatives

g_{x},

g_{y},

g_{z},

h_{x},

h_{y},

and

h_{z} ?

Answer.

Each of the partial derivatives of

g

and

h

would have to be as follows:

g_{x} = 3 e^{y^{2}} + z \sin (x),

g_{y} =, 6 x y e^{y^{2}} - z,

g_{z} = 3 z^{2} - y - \cos (x),

h_{x} = x^{2} y,

h_{y} = x^{3} + 2 y z^{3},

and

h_{z} = x z + 3 y^{2} z^{2} .

Solution.

Each of the partial derivatives of

g

and

h

would have to be as follows:

g_{x} = 3 e^{y^{2}} + z \sin (x),

g_{y} =, 6 x y e^{y^{2}} - z,

g_{z} = 3 z^{2} - y - \cos (x),

h_{x} = x^{2} y,

h_{y} = x^{3} + 2 y z^{3},

and

h_{z} = x z + 3 y^{2} z^{2} .

🔗

(e)

🔗

Let

g_{1} (x, y, z) = 3 e^{y^{2}} + x y z - z \sin (x) .

Calculate

\partial g_{1} / \partial x .

Could

g_{1}

be a potential function for the vector field

G ?

Answer.

\frac{\partial g_{1}}{\partial x} = 3 e^{y^{2}} + y z - z \cos (x) .

No.

Solution.

The partial of

g_{1}

with respect to

x

would be

3 e^{y^{2}} + y z - z \cos (x) .

Since the partial of

g_{1}

with respect to

x

is not the same as the

x

-component of

G,

g_{1}

cannot be a potential function for

G .

🔗

(f)

🔗

Find a function

g

so that

\partial g / \partial x = 3 e^{y^{2}} + z \sin (x) .

Find a function

h

so that

\partial h / \partial x = 3 x^{2} y .

Answer.

g (x, y, z) = 3 x e^{y^{2}} - z \cos (x)

and

h (x, y, z) = x^{3} y

Solution.

g (x, y, z) = 3 x e^{y^{2}} - z \cos (x)

and

h (x, y, z) = x^{3} y

🔗

(g)

🔗

When finding the most general anti-derivative for a function of one variable, we add a constant of integration (usually denoted by

C

) to capture the fact that any constant will become

0

through differentiation.

🔗

(i)

🔗

When taking the partial derivative with respect to

x

of a function of

x,

y,

and

z,

what variables can appear in terms that become

0

in the partial derivative because they are treated as constants?

Answer.

y

and

z

Solution.

Any terms that do not contain an

x

-variable will be treated as a constant, so

y

and

z

are allowed.

🔗

(ii)

🔗

What does this tell you should be added to

g

and

h

in the previous part to make them the most general possible functions with the desired partial derivatives with respect to

x ?

Answer.

g (x, y, z) = 3 x e^{y^{2}} - z \cos (x) + C_{1} (y, z)

and

h (x, y, z) = x^{3} y + C_{2} (y, z)

Solution.

The most general form of

g

and

h

are

g (x, y, z) = 3 x e^{y^{2}} - z \cos (x) + C_{1} (y, z)

and

h (x, y, z) = x^{3} y + C_{2} (y, z),

where

C_{1}

and

C_{2}

are functions.

🔗

(h)

🔗

Now calculate

\partial g / \partial y

and

\partial h / \partial y .

Explain why this tells you that we must have

g (x, y, z) = 3 x e^{y^{2}} - z \cos (x) - y z + m_{1} (z)

🔗

and

h (x, y, z) = x^{3} y + y^{2} z^{3} + m_{2} (z)

🔗

for some functions

m_{1}

and

m_{2}

depending only on

z .

Answer.

\frac{\partial g}{\partial y} = 6 x y e^{y^{2}} + \frac{\partial C_{1}}{\partial y},

\frac{\partial h}{\partial y} = x^{3} + \frac{\partial C_{2}}{\partial y} .

Solution.

We find that

\frac{\partial g}{\partial y} = 6 x y e^{y^{2}} + \frac{\partial C_{1}}{\partial y}

and

\frac{\partial h}{\partial y} = x^{3} + \frac{\partial C_{2}}{\partial y} .

This means that we need to add terms to

g

and

h

y

and

z

such that the partial with respect to

y

will correspond to the second component of

G

and

H .

🔗

(i)

🔗

Calculate

\partial g / \partial z

and

\partial h / \partial z

for the functions in the part above. Notice that

m_{1}

and

m_{2}

are functions of

z

alone, so taking a partial derivative with respect to

z

is the same as taking an ordinary derivative, and thus you may use the notation

m_{1}^{'} (z)

and

m_{2}^{'} (z) .

Answer.

\frac{\partial g}{\partial z} = - \cos (x) - y + \frac{\partial m_{1}}{\partial z}

and

\frac{\partial h}{\partial z} = 3 y^{2} z^{2} + \frac{\partial m_{2}}{\partial z}

Solution.

\frac{\partial g}{\partial z} = - \cos (x) - y + \frac{\partial m_{1}}{\partial z}

and

\frac{\partial h}{\partial z} = 3 y^{2} z^{2} + \frac{\partial m_{2}}{\partial z}

🔗

(j)

🔗

Explain why

G

is a gradient vector field but

H

is not a gradient vector field. Find a potential function for

G .

Answer.

g (x, y, z) = 3 x e^{y^{2}} - z \cos (x) - y z + z^{3}

Solution.

g (x, y, z) = 3 x e^{y^{2}} - z \cos (x) - y z + z^{3},

then

\nabla g = G .

There is no

m_{2} (z)

that will have a partial with respect to

z

x z

since

m_{2} (z)

cannot have a dependence on

x .

🔗

Subsection 4.4.1 Path-Independent Vector Fields

🔗

We say that a vector field

F

defined on a domain

D

is path-independent if

\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r

whenever

C_{1}

and

C_{2}

are oriented paths in

D

such that both curves start at the same point

P

and end at point

Q .

🔗

In Activity 4.3.3 and Example 4.3.5, we encountered situations where we had evidence that a vector field was path-independent. However, since the definition of path-independence requires that the value of the line integral be the same for every pair of points in the domain and every possible path from one point to the other, it doesn’t appear that verifying a vector field is path-independent is an easy task.

🔗

Fortunately, one familiar class of vector fields can be shown to be path-independent. Let

f : R^{3} \to R

be a function for which

\nabla f

is continuous on a domain

D .

Suppose that

P

and

Q

are points in

D

and let

C

be a smooth oriented path from

P

Q .

Let’s take a look at

\int_{C} \nabla f \cdot d r

by fixing an arbitrary parametrization

r (t)

C,

a \leq t \leq b .

Since

\nabla f (r (t)) = ⟨ f_{x} (r (t)), f_{y} (r (t)), f_{z} (r (t)) ⟩,

we know that

\int_{C} \nabla f \cdot d r = \int_{a}^{b} \nabla f (r (t)) \cdot r^{'} (t) d t = \int_{a}^{b} ⟨ f_{x} (r (t)), f_{y} (r (t)), f_{z} (r (t)) ⟩ \cdot r^{'} (t) d t

🔗

r (t) = ⟨ x (t), y (t), z (t) ⟩,

then the integrand above is

⟨ f_{x} (r (t)), f_{y} (r (t)), f_{z} (r (t)) ⟩ \cdot ⟨ x^{'} (t), y^{'} (t), z^{'} (t) ⟩ = f_{x} (x (t), y (t), z (t)) x^{'} (t) + f_{y} (x (t), y (t), z (t)) y^{'} (t) + f_{z} (x (t), y (t), z (t)) z^{'} (t) .

🔗

Notice that this is exactly what the chain rule tells us

\frac{d}{d t} f (r (t))

is equal to. Therefore,

\int_{C} \nabla f \cdot d r = \int_{a}^{b} \frac{d}{d t} f (r (t)) d t = f (r (b)) - f (r (a)) = f (Q) - f (P) .

🔗

In other words, gradient vector fields are path-independent vector fields, and we can evaluate line integrals of gradient vector fields by using a potential function. (Technically the argument above assumed that

C

was smooth, but we can replace

C

by a piecewise smooth curve by splitting the line integral up into the sum of finitely many line integrals along smooth curves.)

🔗

This result is so important that it is frequently called the Fundamental Theorem of Calculus for Line Integrals, because of its similarity to the Fundamental Theorem of Calculus, which can be written as

\int_{a}^{b} f^{'} (x) d x = f (b) - f (a) .

🔗

Theorem 4.4.1. Fundamental Theorem of Calculus for Line Integrals.

🔗

Let

f

be a function for which

\nabla f

is continuous on a domain

D .

P

and

Q

are points in

D

and

C

is a piece-wise smooth oriented path from

P

Q

D,

then

\int_{C} \nabla f \cdot d r = f (Q) - f (P) .

🔗

Activity 4.4.2.

🔗

Calculate each of the following line integrals.

🔗

(a)

🔗

\int_{C} \nabla f \cdot d r

f (x, y) = 3 x y^{2} - \sin (x) + e^{y}

and

C

is the top half of the unit circle oriented from

(- 1, 0)

(1, 0) .

Answer.

0

Solution.

Because

f (- 1, 0) = - \sin (- 1) + 1

and

f (1, 0) = s i n (1) + 1,

then

\int_{C} \nabla f \cdot d r = (\sin (1) + 1) - (- \sin (- 1) + 1) = 0 .

🔗

(b)

🔗

\int_{C} \nabla g \cdot d r

g (x, y, z) = x z^{2} - 5 y^{3} \cos (z) + 6

and

C

is the portion of the helix

r (t) = ⟨ 5 \cos (t), 5 \sin (t), 3 t ⟩

from

(5, 0, 0)

(0, 5, 9 π / 2) .

Answer.

0

Solution.

Because

g (5, 0, 0) = 0 - 0 + 6

and

g (0, 5, \frac{9 π}{2}) = 0 - 0 + 6,

then

\int_{C} \nabla g \cdot d r = (6) - (6) = 0 .

🔗

(c)

🔗

\int_{C} \nabla h \cdot d r

h (x, y, z) = 3 y^{2} e^{y^{3}} - 5 x \sin (x^{3} z) + z^{2}

and

C

is the curve consisting of the line segment from

(0, 0, 0)

(1, 1, 1),

followed by the line segment from

(1, 1, 1)

(- 1, 3, - 2),

followed by the line segment from

(- 1, 3, - 2)

(0, 0, 10) .

Answer.

100

Solution.

Because

h (0, 0, 0) = 0

and

h (0, 0, 10) = 0 - 0 + 100,

then

\int_{C} \nabla h \cdot d r = (100) - (0) = 100 .

🔗

By following the methodology laid out in Preview Activity 4.4.1 to show that a given vector field is a gradient vector field, the Fundamental Theorem of Calculus for Line Integrals makes evaluating a large number of line integrals simpler now.

🔗

Activity 4.4.3.

🔗

Calculate each of the following line integrals.

🔗

(a)

🔗

\int_{C} F \cdot d r

F (x, y) = ⟨ 2 x, 2 y ⟩

and

C

is the line segment from

(1, 2)

(- 1, 0) .

Hint.

Find

f (x, y)

such that

\nabla f = F .

Answer.

4

Solution.

We can choose

f (x, y) = x^{2} + y^{2} + C

and note that

\nabla f = F .

so by the Fundamental Theorem of Calculus for Line Integrals,

\int_{C} F \cdot d r = f (1, 2) - f (- 1, 0) = 5 - 1 = 4 .

🔗

(b)

🔗

\int_{C} G \cdot d r

G (x, y) = ⟨ 4 x^{3} - 12 y \cos (x y), 9 y^{2} - 12 x \cos (x y) ⟩

and

C

is the portion of the unit circle from

(0, - 1)

(0, 1) .

Answer.

6

Solution.

We can choose

g (x, y) = x^{4} - 12 \sin (x y) + 3 y^{3} + C

and note that

\nabla g = G .

so by the Fundamental Theorem of Calculus for Line Integrals,

\int_{C} G \cdot d r = g (0, 1) - g (0, - 1) = 3 - (- 3) = 6 .

🔗

(c)

🔗

\int_{C} H \cdot d r

H (x, y, z) = ⟨ H_{1}, H_{2}, H_{3} ⟩

with

\begin{aligned} H_{1} (x, y, z) & = e^{z^{2}} + 2 x y^{3} z + \cos (x) - y^{3} \sin (x) \\ H_{2} (x, y, z) & = 2 y e^{y^{2}} + 3 x^{2} y^{2} z + 3 y^{2} z^{2} + 3 y^{2} \cos (x) \\ H_{3} (x, y, z) & = x^{2} y^{3} + 2 x z e^{z^{2}} + 2 y^{3} z - 4 z^{3} \end{aligned}

🔗

and

C

is the curve consisting of the line segment from

(1, 1, 1)

(3, 0, 3),

followed by the line segment from

(3, 0, 3)

(1, 5, - 1),

followed by the line segment from

(1, 5, - 1)

(0, 0, 0) .

Answer.

1 - (2 e + \sin (1) + \cos (1))

Solution.

We can choose

h (x, y, z) = x e^{z^{2}} + x^{2} y^{3} z + \sin (x) + y^{3} \cos (x) + e^{y^{2}} - z^{4}

and note that

\nabla h = H .

so by the Fundamental Theorem of Calculus for Line Integrals,

\int_{C} H \cdot d r = h (0, 0, 0) - h (1, 1, 1) = 1 - (e + 1 + \sin (1) + \cos (1) + e - 1) = 1 - (2 e + \sin (1) + \cos (1)) .

🔗

Subsection 4.4.2 Line Integrals Along Closed Curves

🔗

Recall that we call an oriented curve

C

closed if it has the same initial and terminal point. A typical example of a closed curve would be a circle (with an orientation of which way to go around), but we could also consider something like the square with vertices

(1, 1),

(- 1, 1),

(- 1, - 1),

and

(1, - 1),

oriented clockwise (or counterclockwise). Recall that we sometimes use the symbol

\oint

for a line integral when the curve is closed and that if

C = C_{1} + C_{2},

then

\int_{C} F \cdot d r = \int_{C_{1}} F \cdot d r + \int_{C_{2}} F \cdot d r .

🔗

Activity 4.4.4.

🔗

Suppose that

F

is a continuous path-independent vector field (in

R^{2}

R^{3}

) on some domain

D .

🔗

(a)

🔗

Let

P

and

Q

be points in

D

and let

C_{1}

and

C_{2}

be oriented curves from

P

Q .

What can you say about

\int_{C_{1}} F \cdot d r

and

\int_{C_{2}} F \cdot d r ?

Answer.

\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r .

Solution.

Because

F

is a path-independent vector field and both

C_{1}

and

C_{2}

start at

P

and end at

Q,

\int_{C_{1}} F \cdot d r = \int_{C_{2}} F \cdot d r .

🔗

(b)

🔗

Let

C = C_{1} - C_{2} .

Explain why

C

is a closed curve.

Answer.

C

starts and ends at

P .

Solution.

The curve

C

is closed because the end of

C_{1}

is the start of

- C_{2}

and the end of

- C_{2}

is the start of

C_{1},

thus

C

starts and ends at

P .

🔗

(c)

🔗

Calculate

\oint_{C} F \cdot d r .

Answer.

0

Solution.

\oint_{C} F \cdot d r = \int_{C_{1}} F \cdot d r + \int_{- C_{2}} F \cdot d r = \int_{C_{1}} F \cdot d r - \int_{- C_{2}} F \cdot d r

and by part 1, this difference is zero.

🔗

(d)

🔗

Write a sentence that summarizes what we can conclude at this point in the activity.

Solution.

The line integral along every closed curve in a path-independent vector field is

0 .

🔗

(e)

🔗

Now let us suppose that

G

is a continuous vector field on a domain

D

for which

\oint_{C} G \cdot d r = 0

for all closed curves

C .

Pick two points

P

and

Q

D .

Let

C_{1}

and

C_{2}

be oriented curves from

P

Q .

What type of curve is

C = C_{1} - C_{2} ?

Answer.

Closed

Solution.

We have that

C

is a closed curve (from

P

P

) because

C_{1}

will go from

P

Q

and

- C_{2}

will go from

Q

back to

P .

🔗

(f)

🔗

What is

\oint_{C} G \cdot d r ?

Why?

Answer.

0

Solution.

We see that

\oint_{C} G \cdot d r = 0

because (as stated above)

G

is a continuous vector field on a domain

D

for which

\oint_{C} G \cdot d r = 0

for all closed curves

C .

🔗

(g)

🔗

What does that tell you about the relationship between

\int_{C_{1}} G \cdot d r

and

\int_{C_{2}} G \cdot d r ?

Answer.

\int_{C_{1}} G \cdot d r = \int_{C_{2}} G \cdot d r

Solution.

Since

\oint_{C} G \cdot d r = 0 = \int_{C_{1}} G \cdot d r + \int_{- C_{2}} G \cdot d r = \int_{C_{1}} G \cdot d r - \int_{C_{2}} G \cdot d r,

\int_{C_{1}} G \cdot d r = \int_{C_{2}} G \cdot d r .

🔗

(h)

🔗

Explain why this shows that

G

is path-independent.

Answer.

Arbitrary paths with common initial and terminal points were used.

Solution.

The argument above relies only on the fact that the two paths have the same initial and terminal points.

🔗

We summarize the result of Activity 4.4.4 with the theorem below. Although this theorem is not a terribly useful way to show that a vector field is path-independent, it can be a useful way to show that a vector field is not path-independent: find a closed curve around which the circulation is not zero.

🔗

Theorem 4.4.2.

🔗

Let

F

be a continuous vector field on a domain

D .

The following are equivalent conditions on

F .

$F$ is path-independent.
For every closed curve $C$ in $D,$ the circulation of $F$ along $C$ is zero.

🔗

The following activity gives you a chance to reason about path-independence based purely on a graphical representation of a vector field.

🔗

Activity 4.4.5.

🔗

Explain why neither of the vector fields in Figure 4.4.3 is path-independent.

described in detail following the image — (a) $F$

Answer.

For

F,

consider a square with lower-left corner

(- 1, - 1)

and upper-right corner

(1, 1) .

For

G,

consider a circle centered at

(3, 0) .

Solution.

F,

we consider the square with lower-left corner

(- 1, - 1)

and upper-right corner

(1, 1),

oriented clockwise, to be

C .

Notice that

\oint_{C} F \cdot d r > 0

because the line integrals along the horizontal edges of the square are both positive and the line integrals along the vertical sides are zero because

F

is orthogonal to those sides. Because the line integral along this closed curve is not

0,

F

is not path-independent.

G,

consider the circle

C

of radius

2

centered at the point

(3, 0) .

It appears that

G

is tangent to

C,

but we can confidently say that

G \cdot d r > 0

along

C,

which makes

\oint_{C} G \cdot d r,

and thus

G

is not path-independent.

🔗

Subsection 4.4.3 What other vector fields are path-independent?

🔗

Recall that in single variable calculus, the Second Fundamental Theorem of Calculus tells us that given a constant

c

and a continuous function

f,

there is a unique function

A (x)

for which

A (c) = 0

and

A^{'} (x) = f (x) .

In particular,

A (x) = \int_{c}^{x} f (t) d t

is this function. We are about to investigate an analog of this result for path-independent vector fields, but first we require two additional definitions.

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D

is a subset of

R^{2}

R^{3},

we say that

D

is open provided that for every point in

D,

there is a disc (in

R^{2}

) or ball (in

R^{3}

) centered at that point such that every point of the disc/ball is contained in

D .

For example, the set of points

(x, y)

R^{2}

for which

x^{2} + y^{2} < 1

is open, since we can always surround any point in this set by a tiny disc contained in the set. However, if we change the inequality to

x^{2} + y^{2} \leq 1,

then the set is not open, as any point on the circle

x^{2} + y^{2} = 1

cannot be surrounded by a disc contained in the set; any disc surrounding a point on that circle will contain points outside the set, that is with

x^{2} + y^{2} > 1 .

We will also say that a region

D

is path-connected provided that for every pair of points in

D,

there is a path from one to the other contained in

D .

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Activity 4.4.6.

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Let

F = ⟨ F_{1}, F_{2} ⟩

be a continuous, path-independent vector field on an open, connected region

D .

We will assume that

D

is in

R^{2}

and

F

is a two-dimensional vector field, but the ideas below generalize completely to

R^{3} .

We want to define a function

f

D

by using the vector field

F

and line integrals, much like the Second Fundamental Theorem of Calculus allows us to define an antiderivative of a continuous function using a definite integral. To that end, we assign

f (x_{0}, y_{0})

an arbitrary value. (Setting

f (x_{0}, y_{0}) = 0

is probably convenient, but we won’t explicitly tie our hands. Just assume that

f (x_{0}, y_{0})

is defined to be some number.) Now for any other point

(x, y)

D,

define

f (x, y) = f (x_{0}, y_{0}) + \int_{C} F \cdot d r,

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where

C

is any oriented path from

(x_{0}, y_{0})

(x, y) .

Since

D

is connected, such an oriented path must exist. Since

F

is path-independent,

f

is well-defined. If different paths from

(x_{0}, y_{0})

(x, y)

gave different values for the line integral, then we’d not be sure what

f (x, y)

really is.

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To better understand this mysterious function

f

we’ve now defined, let’s start looking at its partial derivatives.

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(a)

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Since

D

is open, there is a disc (perhaps very small) surrounding

(x, y)

that is contained in

D,

so fix a point

(a, b)

in that disc. Since

D

is path-connected, there is a path

C_{1}

from

(x_{0}, y_{0})

(a, b) .

Let

C_{y}

be the line segment from

(a, b)

(a, y)

and let

C_{x}

be the line segment from

(a, y)

(x, y) .

(See Figure 4.4.4.) Rewrite

f (x, y)

as a sum of

f (x_{0}, y_{0})

and line integrals along

C_{1},

C_{y},

and

C_{x} .

Answer.

f (x, y) = f (x_{0}, y_{0}) + \int_{C_{1}} F \cdot d r + \int_{C_{y}} F \cdot d r + \int_{C_{x}} F \cdot d r

Solution.

Because

C = C_{1} + C_{y} + C_{x},

we have that

f (x, y) = f (x_{0}, y_{0}) + \int_{C} F \cdot d r = f (x_{0}, y_{0}) + \int_{C_{1}} F \cdot d r + \int_{C_{y}} F \cdot d r + \int_{C_{x}} F \cdot d r .

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(b)

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Notice that we can parametrize

C_{y}

⟨ a, t ⟩

for

b \leq t \leq y .

Find a similar parametrization for

C_{x} .

Answer.

⟨ t, y ⟩

for

a \leq t \leq x

Solution.

The curve

C_{x}

can be given by

r (t) = ⟨ t, y ⟩

for

a \leq t \leq x,

since only the

x

-coordinate of points on

C_{x}

varies.

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(c)

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Use the parametrization from above to write

\int_{C_{y}} F \cdot d r

and

\int_{C_{x}} F \cdot d r

as single variable integrals in the manner of Section 4.3. Use the fact that

F (x, y) = ⟨ F_{1} (x, y), F_{2} (x, y) ⟩

to express your integrals in terms of

F_{1}

and

F_{2}

without any dot products.

Answer.

\int_{C_{y}} F \cdot d r = \int_{b}^{y} F_{2} (a, t) d t

and

\int_{C_{x}} F \cdot d r = \int_{a}^{x} F_{1} (t, y) d t

Solution.

Using the parametrization of

C_{x}

and

C_{y}

from the previous part, we notice that

r^{'} (t)

has only one nonzero component in each case. Thus, we have

\int_{C_{y}} F \cdot d r = \int_{b}^{y} 0 + F_{2} (a, t) d t

and

\int_{C_{x}} F \cdot d r = \int_{a}^{x} F_{1} (t, y) + 0 d t .

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(d)

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Rewrite your expression for

f (x, y)

using the single variable integrals above (and a line integral along

C_{1}

Answer.

f (x, y) = f (x_{0}, y_{0}) + \int_{C_{1}} F \cdot d r + \int_{b}^{y} F_{2} (a, t) d t + \int_{a}^{x} F_{1} (t, y) d t

Solution.

Combining the previous parts, we have

\begin{aligned} f (x, y) & = f (x_{0}, y_{0}) + \int_{C_{1}} F \cdot d r + \int_{C_{y}} F \cdot d r + \int_{C_{x}} F \cdot d r \\ = f (x_{0}, y_{0}) + \int_{C_{1}} F \cdot d r + \int_{b}^{y} F_{2} (a, t) d t + \int_{a}^{x} F_{1} (t, y) d t . \end{aligned}

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(e)

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Notice that your expression for

f (x, y)

from the previous part only depends on

x

as the upper limit of an single variable integral. Use the Second Fundamental Theorem of Calculus to calculate

f_{x} (x, y) .

Answer.

f_{x} (x, y) = F_{1} (x, y)

Solution.

The first three terms in our expression for

f (x, y)

do not depend on

x,

and thus when taking the partial derivative with respect to

x,

they vanish. By the Second Fundamental Theorem of Calculus (treating

a, y

as constants),

\frac{\partial}{\partial x} \int_{a}^{x} F_{1} (t, y) d t = F_{1} (x, y) .

Therefore,

f_{x} (x, y) = F_{1} (x, y) .

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(f)

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To calculate

f_{y} (x, y),

we continue to consider a path

C_{1}

from

(x_{0}, y_{0})

(a, b),

but now let

L_{x}

be the line segment from

(a, b)

(x, b)

and let

L_{y}

be the line segment from

(x, b)

(y, b) .

Modify the process you used to find

f_{x} (x, y)

to find

f_{y} (x, y) .

Answer.

f_{y} (x, y) = F_{2} (x, y)

Solution.

Proceeding as above, we find that

\begin{aligned} f (x, y) & = f (x_{0}, y_{0}) + \int_{C_{1}} F \cdot d r + \int_{L_{x}} F \cdot d r + \int_{L_{y}} F \cdot d r \\ = f (x_{0}, y_{0}) + \int_{C_{1}} F \cdot d r + \int_{a}^{x} F_{1} (t, b) d t + \int_{b}^{y} F_{2} (x, t) d t . \end{aligned}

Only the last term of this expression depends on

y,

so the first three terms vanish when calculating

f_{y} (x, y) .

By the Second Fundamental Theorem of Calculus,

\frac{\partial}{\partial y} \int_{b}^{y} F_{2} (x, t) d t = F_{2} (x, y) .

Therefore,

f_{y} (x, y) = F_{2} (x, y) .

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(g)

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What can you conclude about the relationship between

\nabla f

and

F ?

What does this tell you about

F

beyond that it is path-independent and continuous?

Answer.

\nabla f = F;

F

is a gradient vector field

Solution.

Since

f_{x} (x, y) = F_{1} (x, y)

and

f_{y} (x, y) = F_{2} (x, y),

we can conclude that

\nabla f = F .

Therefore, we know that

F

is a gradient vector field.

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We summarize the result of Activity 4.4.6 below.

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Theorem 4.4.5. Path-Independent Vector Fields.

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F

is a path-independent vector field on an open, path-connected domain

D,

then

F

is a gradient vector field on

D .

Furthermore, if

P

is a point in

D

and

f (P)

is fixed, then for a point

Q

D

and an oriented curve

C

from

P

Q

D,

the function

f (Q) = f (P) + \int_{C} F \cdot d r

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is a potential function for

F .

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Subsection 4.4.4 Summary

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Gradient vector fields are path-independent, and if $C$ is an oriented curve from $(x_{1}, y_{1})$ to $(x_{2}, y_{2}),$ then

$\int_{C} \nabla f \cdot d r = f (x_{2}, y_{2}) - f (x_{1}, y_{1})$

with the analogous result holding if $f$ is a function of three variables.
A vector field is path-independent if and only if the circulation around every closed curve in its domain is $0 .$
If a vector field $F$ is path-independent, then there exists a function $f$ such that $\nabla f = F .$ That is, $F$ is a conservative or gradient vector field.

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Exercises 4.4.5 Exercises

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1.

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For each of the following line integrals:

Determine if the Fundamental Theorem of Calculus for Line Integrals applies.
If the Fundamental Theorem of Calculus for Line Integrals applies, then find the potential function and use this to evaluate the line integral
If the Fundamental Theorem of Calculus for Line Integrals does not apply, then describe where the process laid out in Preview Activity 4.4.1 fails.

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(a)

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The line integral of

F = ⟨ y z, x z, y ⟩

along the helix of radius of 3 given by

r (t) ⟨ 3 \sin (t), 3 \cos (t), \frac{4}{π} t ⟩

- π \leq t \leq π .

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(b)

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The line integral of

F = ⟨ \sin (y z), x z \cos (y z) - z \sin (y), x y \cos (y z) + \cos (y) ⟩

along the line segment from

(0, π, 3)

(2, - 1, 2 π) .

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(c)

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The integral

\int_{C} y^{2} d x + 2 x y d y

where

C

is the parabolic path along

y = x^{2}

from

(- 1, 1)

(4, 16) .

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2.

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Let

F = ⟨ F_{1}, F_{2}, F_{3} ⟩,

where

F_{1} = 3 x y^{2} + z

and

F_{3} = y z^{2} + \cos (z) .

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(a)

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Give a component function

F_{2}

such that

F

is a gradient vector field.

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(b)

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Give a component function

F_{2}

such that

F

is not a gradient vector field.

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3.

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This exercise focuses on reasoning graphically about line integrals and vector fields.

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(a)

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Find a closed curve on which the line integral of the vector field pictured below will not be zero. Be sure to clearly define your curve and explain why the line integral over your curve is non-zero.