Is F=-kx^2+ax^2+bx^4 Conservative?

[anikmartin]

Hello everyone, I have a question about conservative forces.
I am given a function F = -kx^2 + ax^2 + bx^4, where a, b, and k are constants. I am asked to determine if this force is conservative or not, I don't know know how to prove this mathematically or theoretically. Please help!

In this particular example the force is a spring force, spring forces aren't always conservative are they?
Thanks

 

[rhia]

You can use the fact that Work down by a conservative force in a closed path is zero.
I guess spring forces are always conservative if the spring is ideal.

 

[Doc Al]

I am asked to determine if this force is conservative or not, I don't know know how to prove this mathematically or theoretically.

There are several ways to mathematically test that a force is conservative. Here are a few:

(1) You can check that the line integral is zero along any closed path (equivalent to what rhia suggested).
(2) Or you can see if you can find a potential function whose -gradient equals the force function (hint: integrate).
(3) Or you can see if the curl of the force is zero.​

The last two checks should be easy.

 

[rhia]

Hi Doc AI,
Can you please explain the physical significance of the last two checks?
I don't understand things if I can't relate them to physical world.:(
Thanks a lot!
rhia

 

[anikmartin]

Okay I though that if the force was conservative, then the gradient could be used, I didn't know that it proves the force is conservative.

A correction in the force equation

F = -kx + ax^3 + bx^4

After integrating a found the potential equation to be
U = (kx^2)/2 - (ax^4)/4 - (bx^5)/5

using F = -dU/dx.

I found the gradient of U to be

grad(U) = (kx - ax^3 - bx^4)i

Is there a reason for the sign difference? I just learned gradients, in my Calc III class that I am taking right now, two weeks ago so I am still not comfortable with all of its properties.

Okay what if I have a force function of the type:

F = -kx^2y +(y^2)(z^2) + xyz ( I made this one up)

So took three integrals, with respect to x, y, and z seperately, to get the potential function in each direction. Then I found the gradient of the potential function to be

grad(U) = (kyx^2 - (y^2)(z^2) - xyz)i + (k(x^2)y - (y^2)(x^2) - xyz)j + (k(x^2)y - (y^2)(z^2) - xyz)k

To find if the gradient equals the force function, do I find the absolute value of grad(U) , to make it a scalar function? I am not sure how to find if these two are equivalent.

Okay I took a guess to convert the force function into a vector function. First I found the partial derivatives of F with respect to x, y, and z. Then I took the integral with respect to x, y, and z. I found

F = -(k(x^2)y - xyz)i - (k(x^2)y + (y^2)(z^2) + xyz)j + ((z^2)(y^2) +xyz)k

When I took the integral with respect to each I didn't add the Integral constant, so maybe that could account for the difference. According to the scalar function F, when x = O, F = (y^2)(z^2), when z = 0, F = -k(x^2)y . .so these could be the missing constants in my vector function F??
Thanks for all your help.

 

[MiGUi]

I think the simplest way, is the curl.

If \vec{F} = M(x,y,z)\vec{i} + N(x,y,z)\vec{j} + Q(x,y,z)\vec{k}, then

\vec{\nabla} \times \vec{F} = \left| \begin{array}{ccc}<br /> \vec{i} &amp; \vec{j} &amp; \vec{k} \\<br /> \partial/\partial x &amp; \partial/\partial y &amp; \partial/\partial z \\<br /> M &amp; N &amp; Q \end{array} \right| = 0<br />

The reason for this, is the Stokes theorem, which says:

Let be S a surface oriented with a normal vector \vec{n}, limited by a closed simple curve C. If \vec{F} is a vectorial field and its partial derivatives are continuous in that region, then:

\oint_{C} \vec{F}d\vec{r} = \iint_{S} (\vec{\nabla} \times \vec{F})\vec{n}dS

--

If the curl is equal to zero, then the right term is zero as well, and the definition of conservative field is that the left integral is equal to zero.