How to find the potential funtion for a force

In summary: Fy = -4axy/ (x^2 + y^2)^3In summary, the homework statement is to find the potential function and the work done in moving from (0, 1) to (1, 0) along the curve.
  • #1
vande060
186
0

Homework Statement



F1 = (a x)/(x^2 + y^2)^2 i + (a y)/(x^2 + y^2)^2 j

determine the potential function (if there is one) and the work done in moving from (0, 1) to (1, 0) along the curve

x2 + y2 = 1 (let a = 5.0).

Homework Equations


F1 = (a x)/(x^2 + y^2)^2 i + (a y)/(x^2 + y^2)^2 j

x2 + y2 = 1 (let a = 5.0)

The Attempt at a Solution



All right, so from lecture i gathered that for this potential function to be determined this force must be conservative and must pass the component test. Already at this point i begin to get confused, so does this mean that i just have to make sure that each components derivative is the same? DO i differentiate the i component with respect to x and the j component with respect to y like this:

x' = 5(y^2 - 3x^2)/(x^2 + y^2)^3

y' = 5(x^2 - 3y^2)/(y^2 + x^2)^3

am i right this far, I am am so unsure about this any help would be appreciated
 
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  • #2
If you know vector calculus, you would say the field F is conservative if [itex]\nabla\times \mathbf{F}=0[/itex]. For the two-dimensional case, this reduces to

[tex]\frac{\partial}{\partial y}F_x-\frac{\partial}{\partial x}F_y = 0[/tex]

where F=Fxi+Fyj.

There are a few problems with your work so far. First off, you're a bit too sloppy with your notation. For example, x' means the derivative of x with respect to something. You're not differentiating x; you're differentiating the x-component of F1. And because you have two variables, x and y, you need to say which one you're differentiating with respect to. Second, your derivatives are wrong. I'm not sure how you came up with what you got.

But overall, your approach is correct. :smile:
 
  • #3
Okay. I used mathematica for that one, but ill try to run through it myself(i am a bit rusty)

5x/(x^2 + y^2)^2

differentiate with i respect to x, or df/dy

= [5*(x^2 +y^2)^2 - 5x*2*(x^2 + y^2)*(2x + 2y) ]/ (x^2 + y^2)^4

= 5*[(x^2 + y^2) - 2x*(2x+2y)] / (x^2 + y^2)^3

= 5*[x^2 + y^2 - 4x^2 -4xy] / (x^2 + y^2)^3

=5*[ -2x^2 + y^2 - 4xy] / (x^2 + y^2)^3

is that much right?

and similarly for differentiating with respect to y

5y/(x^2 + y^2)^2

= [5*(x^2 +y^2)^2 - 5y*2*(x^2 + y^2)*(2x + 2y) ]/ (x^2 + y^2)^4

= 5*[(x^2 + y^2) - 2y*(2x+2y)] / (x^2 + y^2)^3

= 5*[x^2 + y^2 - 4y^2 -4xy] / (x^2 + y^2)^3

=5*[ -2y^2 + x^2 - 4xy] / (x^2 + y^2)^3

is it right? as it stands [tex]
\frac{\partial}{\partial y}F_x-\frac{\partial}{\partial x}F_y = 0
[/tex]

wont be equal to zero as it should, which would mean this is not a conservative force
 
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  • #4
You differentiated Fx with respect to x. You want to differentiate it with respect to y. Same problem with Fy. That actually simplifies the problem as no longer have to use the quotient rule because the numerators are constant.

This is what I got for the derivatives:

[tex]-\frac{4axy}{(x^2+y^2)^3}[/tex]
 
  • #5
vela said:
You differentiated Fx with respect to x. You want to differentiate it with respect to y. Same problem with Fy. That actually simplifies the problem as no longer have to use the quotient rule because the numerators are constant.

This is what I got for the derivatives:

[tex]-\frac{4axy}{(x^2+y^2)^3}[/tex]

i have never differentiated with respect to anything before, so i am sorry but don't understand how you got that answer and why the numerator is constant.

if you factor out the constant ax, and then differentiate 1/(x^2 +y)^2 i understand that
it becomes -2ax/(x^2 + y^2)^3, but i don't get how you get from this to the final answer of

[tex]-\frac{4axy}{(x^2+y^2)^3}[/tex]
 
  • #6
vande060 said:
i have never differentiated with respect to anything before.
You have always differentiated with respect to some variable. Usually it was x or t, and you were only working with one variable, so it was understood which variable you were differentiating with respect to.
so i am sorry but don't understand how you got that answer and why the numerator is constant.
When you take a partial derivative with respect to one variable, all other variables are held constant, so with Fx, because you're differentiating with respect to y and y does not appear in the numerator, the numerator is essentially a constant.
if you factor out the constant ax, and then differentiate 1/(x^2 +y)^2 i understand that it becomes -2ax/(x^2 + y^2)^3, but i don't get how you get from this to the final answer of

[tex]-\frac{4axy}{(x^2+y^2)^3}[/tex]
Did you forget to apply the chain rule?
 
  • #7
okay i have never done a partial derivative, but i understand how it happens now thanks to you and my calculus book :) . I see how much more simple it is that complete derivatives.

ax * -2/(x^2 + y^2)^3 * (0 + 2y)

= -4axy/ (x^2 + y^2)^3

and..

ay * -2/(x^2 + y^2)^3 *(0 + 2x)

= -4axy/ (x^2 + y^2)^3
so it is a constant force after all

[-4axy/ (x^2 + y^2)^3] - [-4axy/ (x^2 + y^2)^3] = 0my book now begins to talk about parametrization and integrating the original components, but I am not sure what they mean by such .

i guess it means to eliminate one of the variables by substitution, but no examples are given in my book. do you understand the next logical step?
 
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  • #8
Looks good, but it's a conservative force, not a constant force.
 
  • #9
vela said:
Looks good, but it's a conservative force, not a constant force.

okay :redface:
 
  • #10
okay so i know that work = integral from a to b fxdx and a conservative force only matters between the beginning and end points of displacement, but i still have multiple variables in my components, integrating will be hard. since i know that the path is x^2 + y^2 = 1, can i somehow preform a substitution to eliminate one variable?
 
  • #11
Yup, that's generally what you do. Use the equation of the path so eliminate one variable from the integrals. Note that the combination x2+y2 appears in the denominator of the force and, on the path, that combination is equal to 1.

You could, alternatively, write x and y in terms of a third parameter and rewrite the integrals in terms of the parameter. For example, in this case, the path is part of the unit circle, so you could write x=cos θ and y=sin θ and integrate with respect to θ.
 
  • #12
okay bear with me. Thanks for your help so far i think i am starting to (slowly) understand.

so i can make y^2 = 1 - x^2 and x^2 = 1 - y^2

the the original components become

ax/(x^2 + y^2)^2 ---> substituting y^2 = 1- x^2 ---> ax/1

ay/(x^2 + y^2)^2 ---> substituting x^2 = 1- y^2 ---> ay/1

and i can now integrate from 0 to 1 for fx and from 1 to 0 for fy right? is it right that they are two separate integrals for each coordinate?

so it would be integral from 0 to 1 (ax^2)/2

and integral from 1 to 0 (ay^2)/2
 
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  • #13
Yup. The work along some contour C is given by

[tex]W=\int_C \mathbf{F}\cdot d\mathbf{r} = \int_C (F_x dx + F_y dy)[/tex]

So you have an x integral and a y integral. The total work is their sum.
 
  • #14
so how does the potential function play into this, because that is what i was supposed to have found that first, before determining work. is it just the integrals written down? for example, the i component potential function would be integral from a to b (0 to 1)

potential function = [(5(b)^2)/2 - 5(a)^2)/2]

remembering that a = 5
 
  • #15
You want to find a function V(x,y) that satisfies F=-∇V. Do you recall how to solve exact differential equations from calculus? It's the same problem mathematically.
 
  • #16
vela said:
You want to find a function V(x,y) that satisfies F=-∇V. Do you recall how to solve exact differential equations from calculus? It's the same problem mathematically.

Ill admit, i am unfamiliar with solving differential equations. do i need to find x and y values that satisfy ax/(x^2+y^2)^2i + ay/(x^2 + y^2)^2 = df/dxi + df/dyj

is this even on the trail?Also, I don't understand the total work, when i integrate

0 to 1 for fx i get 5(1)^2/2 - 5(0)^2/2 = 2.5

and for fy

1 to 0 5(0)^2/2 - 5(1)^2/2 = -2.5

the sum would then be zero, how can this be, did i make a simple integration error?
 
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  • #17
vande060 said:
Ill admit, i am unfamiliar with solving differential equations. do i need to find x and y values that satisfy ax/(x^2+y^2)^2i + ay/(x^2 + y^2)^2 = df/dxi + df/dyj

is this even on the trail?
No, you're looking for a function V(x,y) where

[tex]\begin{align*}
\frac{\partial V}{\partial x} & = F_x \\
\frac{\partial V}{\partial y} & = F_y \\
\end{align*}
[/tex]

Also, I don't understand the total work, when i integrate

0 to 1 for fx i get 5(1)^2/2 - 5(0)^2/2 = 2.5

and for fy

1 to 0 5(0)^2/2 - 5(1)^2/2 = -2.5

the sum would then be zero, how can this be, did i make a simple integration error?
No, this is the correct answer. Try drawing a sketch of the contour path and the vector field F along the path, and you should see why the work done is 0.
 
  • #18
sorry this has gone so long, but if there is any chance you are still watching this i have since had a new lecture and now believe i have the last step V(x,y) that satisfies F=-∇V.

i think it is 5/2(x^2+y^2) + c

it even works ;) if you give it a test run

w = Vf -Vi

[5/2(1^2 + 0)+ c] - [5/2( 0 + 1^2) + c] = w

0 = w
 
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  • #19
Yup, looks good! (Off by a sign perhaps?)

It's easy enough to check. Just take the gradient and see if you get F.
 
  • #20
vela said:
Yup, looks good! (Off by a sign perhaps?)

It's easy enough to check. Just take the gradient and see if you get F.

thanks for all the help. you really made it click.
 

Related to How to find the potential funtion for a force

1. What is a potential function for a force?

A potential function for a force is a scalar function that describes the relationship between the force acting on an object and the position of that object in space. It is a fundamental concept in classical mechanics and is used to calculate the work done by a force on an object.

2. How do you calculate the potential function for a force?

The potential function for a force can be calculated using the equation U(x,y,z) = -∫F(x,y,z)dx. This means that the potential energy at a given point is equal to the negative integral of the force over the distance traveled in the x, y, and z directions.

3. What is the difference between a conservative and non-conservative force?

A conservative force is one for which the work done is independent of the path taken, only dependent on the starting and ending points. This means that the potential function for a conservative force is path-independent. In contrast, a non-conservative force does not have a potential function and the work done depends on the path taken.

4. Can you find the potential function for any force?

No, not all forces have a potential function. For a force to have a potential function, it must be conservative. This means that the force must not depend on the path taken and must only be a function of the position of the object. Non-conservative forces, such as friction and air resistance, do not have potential functions.

5. How is the potential function for a force related to the concept of energy?

The potential function for a force is closely related to the concept of energy. It represents the potential energy of an object at a specific position in space. The work done by a force can be calculated using the change in potential energy, and the total energy of a system is the sum of its kinetic and potential energy. This relationship is described by the conservation of energy principle.

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