finding a potential function for conservative vector fields - math insight

8/2/2019 Finding a Potential Function for Conservative Vector Fields - Math Insight

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The process of finding a potential function of a conservative vector field is a multi-step procedure that

involves both integration and differentiation, while paying close attention to the variables you are

integrating or differentiating with respect to. For this reason, given a vector field , we recommend that

you first determine that that is indeed conservative before beginning this procedure. That way you

know a potential function exists so the procedure should work out in the end.

In this page, we focus on finding a potential function of a two-dimensional conservative vector field. We

address three-dimensional fields in another page.

We introduce the procedure for finding a potential function via an example. Let's use the vector field

The first step is to check if is conservative. Since

we conclude that the scalar curl of is zero, as

Next, we observe that is defined on all of , so there are no tricks to worry about. The vector fieldis indeed conservative.

Since is conservative, we know there exists some potential function so that . As a first step

toward finding , we observe that the condition means that

Finding a potential function for conservative

vector fields

F

F

F(

x,

y) = (

yc o s

x+ , s i n

x+ 2 x y 2

y) .

y

2

F

F

2

x

F

1

y

= ( s i n x

+ 2 x y 2y

) = c o s x

+ 2y

x

= (y

c o sx

+ ) = c o s x

+ 2y

,

y

y

2

F

= 0 .

F

2

x

F

1

y

F R

2

F

F f

f=

F

f f = F

( , ) = ( , ) = ( y c o s x + , s i n x + 2 x y 2 y ) .

f

x

f

y

F

1

F

2

y

2


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This vector equation is two scalar equations, one for each component. We need to find a function

that satisfies the two conditions

and

Let's take these conditions one by one and see if we can find an that satisfies both of them. (We

know this is possible since is conservative. If were path-dependent, the procedure that follows

would hit a snag somewhere.)

Let's start with condition . We can take the equation

and treat as though it were a number. In other words, we pretend that the equation is

for some number . We can integrate the equation with respect to and obtain that

But, then we have to remember that really was the variable so that

But actually, that's not right yet either. Since we were viewing as a constant, the integration constant

could be a function of and it wouldn't make a difference. Thepartial derivative of any function of

with respect to is zero. We can replace with any function of , say , and condition will be

satisfied. A new expression for the potential function is

If you are still skeptical, try taking the partial derivative with respect to of defined by equation

. Since does not depend on , we can conclude that . Indeed, condition is

satisfied for the of equation .

Now, we need to satisfy condition . We can take the of equation (so we know

f(

x,

y)

(x

,y

) =y

c o sx

+

f

x

y

2

( 1 )

(x

,y

) = s i n x

+ 2 x y 2y

.

f

y

( 2 )

f(

x,

y)

F F

( 1 )

(x

,y

) =y

c o sx

+ ,

f

x

y

2

y

(x

) =a

c o sx

+

df

dx

a

2

a x

f(

x) =

as i n

x+

x+

C.

a

2

a y

f(

x,

y) =

ys i n

x+

x+

C.

y

2

y

C y y

x C y g (y

) ( 1 )

f(

x,

y) =

ys i n

x+

x+

g(

y) .

y

2

( 3 )

x f (x

,y

)

( 3 ) g ( y )x g

( y ) = 0

x

( 1 )

f(

x,

y) ( 3 )

( 2 )f

(x

,y

) ( 3 )


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that condition will be satisfied) and take its partial derivative with respect to , obtaining

Comparing this to condition , we are in luck. We can easily make this satisfy condition

as long as

If the vector field had been path-dependent, we would have found it impossible to satisfy both

condition and condition . We would have run into trouble at this point, as we would have

found that would have to be a function of as well as . Since is a function of alone, our

calculation verifies that is conservative.

If we let

for some constant , then

and we have satisfied both conditions.

Combining this definition of with equation , we conclude that the function

is a potential function for You can verify that indeed

With this in hand, calculating the integral

is simple, no matter what path is. We can apply the gradient theorem to conclude that the integral is

simply , where is the beginning point and is the ending point of . (For this reason,

if is a closed curve, the integral is zero.)

( 1 )y

(x

,y

) = ( y s i n x + x + g ( y ) ) = s i n x + 2 y x + ( y ) .

f

y

y

y

2

d g

dy

( 2 )f

(x

,y

)

( 2 )

(y

) = 2 y

.

dg

dy

F

( 1 ) ( 2 )

d g

d y

x y

d g

d y

y

F

g(

y) = +

ky

2

k

(x

,y

) = s i n x

+ 2 y x 2y

,

f

y

g(

y) ( 3 )

f(

x,

y) =

ys i n

x+

x +

ky

2

y

2

F.

f

= (y

c o sx

+ , s i n x

+ 2 x y 2y

) =F

(x

,y

) .y

2

F

d s

C

C

f(

q)

f(

p) p q

C

C


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We might like to give a problem such as find

where is the curve given by the following graph.

The answer is simply

(The constant is always guaranteed to cancel, so you could just set .)

If the curve is complicated, one hopes that is conservative. It's alwas a good idea to check if is

conservative before computing its line integral

You might save yourself a lot of work.

Nykamp DQ, Finding a potential function for conservative vector fields. FromMah Inigh.

http://mathinsight.org/conservative_vector_field_find_potential

Keywords: conservative, gradient theorem, path independent, potential function

F d s

C

C

F

d s

C

=f

(

/ 2 , 1 ) f

(

, 2 )

= s i n

/ 2 + 1 + k

( 2 s i n (

) 4

4 +k

)

2

= s i n

/ 2 + + 3 = + 2

9

2

9

2

k k = 0

CF F

F d s .

C

Cite this as


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finding a potential function for conservative vector fields - math insight

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