How to find the potential funtion for a force

  • Thread starter vande060
  • Start date
1. The problem statement, all variables and given/known data

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).



2. Relevant 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)



3. 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, im am so unsure about this any help would be appreciated
 
Last edited:

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
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:
 
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
 
Last edited:

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
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]
 
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 dont 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 dont get how you get from this to the final answer of

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

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
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 dont 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 dont 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?
 
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] = 0


my book now begins to talk about parametrization and integrating the original components, but im 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?
 
Last edited:

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
Looks good, but it's a conservative force, not a constant force.
 
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?
 

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
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.
 
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
 
Last edited:

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
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.
 
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
 

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
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.
 
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 dont 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?
 
Last edited:

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
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 dont 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.
 
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
 
Last edited:

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,311
1,037
Yup, looks good! (Off by a sign perhaps?)

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

Want to reply to this thread?

"How to find the potential funtion for a force" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top