# Potential and conservative

1. Jul 10, 2011

### athrun200

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
The answer should be $-(x^{2}-1)cos^{2}\theta$
But I get a constant -1.
Why?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jul 11, 2011

### lanedance

can you explain what you are attempting to do? I would approach as follows

say phi exists then you know
$$F_x = \frac{\partial\phi}{\partial x}$$
$$F_y = \frac{\partial\phi}{\partial y}$$

so integrating the first gives
$$\phi = \int F_x dx= \int 2x cos^2y dx =?$$

3. Jul 11, 2011

### athrun200

Sorry that I skip some steps and forget to give explanation.

This is my approach.

This yields $$x^2 cos^2y+C$$

But how to find the constant?

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4. Jul 11, 2011

### lanedance

not quite as you are only integrating w.r.t. to x, so c could be a function of y
$$\phi = \int F_x dx= \int 2x cos^2y dx = x^2cos^2y + c(y)$$

now also do it from the other direction
$$\phi = \int F_y dy= ?$$

5. Jul 11, 2011

### athrun200

$(x^2+1)\frac{cos2y}{2}+C(x)$

it can then simplfied as
$(x^2+1)\frac{2cos^2y-1}{2}+C(x)$

6. Jul 11, 2011

### lanedance

$$\phi = \int F_x dx = \int F_y dy$$

also I would use a different constant function for the 2nd integral (say c(y) for first and d(x) for the second)

and from there you should be able to deduce withther it is possible to to solve c(x) & d(y)

7. Jul 11, 2011

### athrun200

Last edited by a moderator: Apr 26, 2017
8. Jul 11, 2011

### athrun200

It's my first time to solve the problem like this.
So I stuck in here

I would like to ask, what's wrong with my method above?
I have solved a lot of problems by using that method.(I learn it from mathematical methods in the physical sciences Chapter 6 sec8)

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9. Jul 11, 2011

### lanedance

Last edited by a moderator: Apr 26, 2017
10. Jul 11, 2011

### lanedance

stick to whatever you're comfortable with... took me awhile to figure out what you're doing as there's a few jumps (also the big pics are difficult to read on a laptop screen), but i think you set phi=0 at the origin and do a path interval over F to find the form of phi if it exists, in essence i think they're the same thing though yours has the advantage of setting one value to zero in the integral

Last edited: Jul 11, 2011
11. Jul 11, 2011

### lanedance

also note the potential is only unique upto a scalar additive constant ie (phi) or (phi +1) give the same force field

12. Jul 11, 2011

### lanedance

this is how i would approach it using my method (i re-named c & d as f & g to be clear they are functions)
$$\phi(x,y) = \int F_x dx = \int 2x cos^2(y) = x^2cos^2(y) + f(y)$$
$$\phi(x,y) = \int F_y dx = \int -(x^2+1) sin(2y) dy = (x^2+1) \frac{1}{2}cos(2y) +g(x) = (x^2+1) (cos^2(y)-\frac{1}{2})+g(x)$$

in the last line we may as well group all the x only terms in a function, so let
$$h(x) = +\frac{1}{2}(x^2-1)+g(x)$$

equating and re-arranging a little gives
$$x^2cos^2(y)+ cos^2(y)+h(x) = x^2cos^2(y) + f(y)$$

$$h(x) = 0$$
$$f(y) = cos^2(y)$$

so we get
$$\phi(x,y) = (x^2+1)cos^2(y)+ c$$

where c is any constant

13. Jul 11, 2011

### lanedance

after all that... and re-reading the posts i think it was only post 11 you needed!

you can only ever determine phi upto a additive scalar constant, as the constant will always be sent to zero in the differentiation to get the field