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Conservative Vector
Fields
In the previous section we saw that if we knew that the vector field F was conservative then
∫ F id r
was independent of path
...
In this section we want to look at two questions
...
For
higher dimensional vector fields we’ll need to wait until the final section in this chapter to answer
this question
...
Theorem
Let F = P i + Q j be a vector field on an open and simply-connected region D
...
Let’s take a look at a couple of examples
...
(a) F ( x, y ) = ( x 2 − yx ) i + ( y 2 − xy ) j [Solution]
(
) (
)
(b) F ( x, y ) = 2xe xy + x 2 ye xy i + x 3e xy + 2 y j
[Solution]
Solution
Okay, there really isn’t too much to these
...
(
) (
)
(a) F ( x, y ) = x − yx i + y − xy j
2
2
In this case here is P and Q and the appropriate partial derivatives
...
[Return to Problems]
(
) (
)
(b) F ( x, y ) = 2xe xy + x 2 ye xy i + x 3e xy + 2 y j
Here is P and Q as well as the appropriate derivatives
...
[Return to Problems]
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
...
First, let’s assume that the vector field is conservative and so we know that a potential
function, f ( x, y ) exists
...
f ( x, y ) = ∫ P ( x, y ) dx
or
f ( x, y ) = ∫ Q ( x, y ) dy
We saw this kind of integral briefly at the end of the section on iterated integrals in the previous
chapter
...
Example 2 Determine if the following vector fields are conservative and find a potential
function for the vector field if it is conservative
...
∂P
= 8 x3 y 3
∂y
∂Q
3 3
= 8x y
∂x
P = 2x y + x
3
4
Q = 2x y + y
4
3
So, the vector field is conservative
...
From the first fact
above we know that,
∂f
= 2 x3 y 4 + x
∂x
∂f
= 2 x4 y3 + y
∂y
From these we can see that
f ( x, y ) = ∫ 2x3 y 4 + x dx
or
f ( x, y ) = ∫ 2x 4 y 3 + y dy
We can use either of these to get the process started
...
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
...
Here is the first integral
...
We now need to now determine h ( y )
...
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
...
So, if we
differentiate our function with respect to y we know what it should be
...
∂f
= 2 x 4 y 3 + h′ ( y ) = 2 x 4 y 3 + y = Q
∂y
From this we can see that,
h′ ( y ) = y
Notice that since h′ ( y ) 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
...
1 2
h ( y ) = ∫ h′ ( y ) dy = ∫ y dy = y + c
2
So, putting this all together we can see that a potential function for the vector field is,
1 4 4 1 2 1 2
f ( x, y ) = x y + x + y + c
2
2
2
Note that we can always check our work by verifying that ∇f = F
...
[Return to Problems]
(
) (
)
(b) F ( x, y ) = 2xe xy + x 2 ye xy i + x 3e xy + 2 y j
Okay, this one will go a lot faster since we don’t need to go through as much explanation
...
Let’s start with the following,
∂f
= 2 xe xy + x 2 ye xy
∂x
∂f
= x3e xy + 2 y
∂y
This means that we can do either of the following integrals,
f ( x, y ) = ∫ 2xe xy + x 2 ye xy dx
or
f ( x, y ) = ∫ x3e xy + 2 y dy
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
...
So, from the second integral we get,
f ( x, y ) = x 2e xy + y 2 + h ( x )
Notice that this time the “constant of integration” will be a function of x
...
Sometimes this will happen
and sometimes it won’t
...
f ( x, y ) = x 2e xy + y 2 + c
[Return to Problems]
Now, as noted above we don’t have a way (yet) of determining if a three-dimensional vector field
is conservative or not
...
In this case we will use the fact that,
∂f
∂f
∂f
∇f = i +
j + k = Pi +Q j + Rk = F
∂x
∂y
∂z
Let’s take a quick look at an example
...
∂f
= 2 xy 3 z 4
∂x
∂f
= 3x 2 y 2 z 4
∂y
∂f
= 4 x2 y3 z3
∂z
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
...
f ( x, y, z ) = ∫ 2xy 3 z 4 dx = x 2 y 3 z 4 + g ( y, z )
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
...
Doing this gives,
∂f
= 3x 2 y 2 z 4 + g y ( y, z ) = 3x 2 y 2 z 4 = Q
∂y
Of course we’ll need to take the partial derivative of the constant of integration since it is a
function of two variables
...
This means that we now know the potential function must be in the following form
...
∂f
= 4 x 2 y 3 z 3 + h′ ( z ) = 4 x 2 y 3 z 3 = R
∂z
So,
h′ ( z ) = 0
⇒
h( z) = c
The potential function for this vector field is then,
f ( x, y , z ) = x 2 y 3 z 4 + c
Note that to keep the work to a minimum we used a fairly simple potential function for this
example
...
However, we should be careful to remember that this usually won’t be the case
and often this process is required
...
Each
would have gotten us the same result
...
Example 4 Find a potential function for the vector field,
F = ( 2x cos ( y ) − 2z 3 ) i + ( 3 + 2 ye z − x 2 sin ( y ) ) j + ( y 2e z − 6xz 2 ) k
Solution
Here are the equalities for this vector field
...
f ( x, y, z ) = ∫ y 2e z − 6xz 2 dz = y 2e z − 2xz 3 + g ( x, y )
The “constant of integration” for this integration will be a function of both x and y
...
Doing this gives,
∂f
= −2 z 3 + g x ( x, y ) = 2 x cos ( y ) − 2 z 3 = P
∂x
So, it looks like we’ve now got the following,
g x ( x, y ) = 2 x cos ( y )
⇒
The potential function for this problem is then,
g ( x, y ) = x 2 cos ( y ) + h ( y )
f ( x, y, z ) = y 2e z − 2 xz 3 + x 2 cos ( y ) + h ( y )
To finish this out all we need to do is differentiate with respect to y and set the result equal to Q
...
We need to work one final example in this section
...
⎝ 2 ⎠
Solution
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
...
∫ F id r = ∫ ∇f id r = f ( r (1) ) − f ( r ( 0 ) )
C
where,
So, the integral is,
C
r (1) = −2,1
r ( 0 ) = −1, 0
∫ F id r = f ( −2,1) − f ( −1, 0 )
C
⎛ 21 ⎞ ⎛ 1
⎞
= ⎜ + c⎟ −⎜ + c⎟
⎝ 2
⎠ ⎝2
⎠
= 10