Partial derivatives - proving force is conservative

[Oblio]

I'm trying to show that the force F= k [x, 2y, 3z] (where k is a constant)

is conservative.

If I take the cross product of:

\nabla x F, that equals \frac{\partial}{\partial y} F_{z} - \frac{\partial}{\partial z} F_{y}

= \frac{\partial}{\partial y} k3z - \frac{\partial}{\partial z} k2y


isn't everything in the derivatives constants? ... or is that the point. lol

Thanks a lot

 

[Hootenanny]

You've only worked out the x component of the cross product, what about the rest?

 

[Oblio]

You've only worked out the x component of the cross product, what about the rest?

I wasn't sure if I had to, the example in my book only did F(x).

Is that much correct though?
If you bring all the constants out, can you be left with nothing in the derivative?

k3z \frac{\partial}{\partial y} - k2y \frac{\partial}{\partial z}

 

[Hootenanny]

I wasn't sure if I had to, the example in my book only did F(x).

Is that much correct though?
If you bring all the constants out, can you be left with nothing in the derivative?

k3z \frac{\partial}{\partial y} - k2y \frac{\partial}{\partial z}

Yup, looks good to me, now all you need to do is continue on with working out the rest of the cross products, do you know how to do that?

 

[Oblio]

(Didn't mean to double post, sorry. I usually go in the introductory physics and I realized after I was in the advanced; incorrectly I thought)

 

[Hootenanny]

(Didn't mean to double post, sorry. I usually go in the introductory physics and I realized after I was in the advanced; incorrectly I thought)

No worries, just don't do it again, we have big sticks you know...

 

[Oblio]

Yup, looks good to me, now all you need to do is continue on with working out the rest of the cross products, do you know how to do that?

In the same way I guess... no?

I didn't know I could take everything of a derivative either. cool!

 

[Oblio]

No worries, just don't do it again, we have big sticks you know...

Not answering my questions might be worse then a good ol' fashion beating :)

 

[Oblio]

I wasn't sure if I had to, the example in my book only did F(x).

Is that much correct though?
If you bring all the constants out, can you be left with nothing in the derivative?

k3z \frac{\partial}{\partial y} - k2y \frac{\partial}{\partial z}

Doesn't that leave me with not 0 ? (which I want?)

k3z-k2y ?

 

[CompuChip]

Doesn't look good to me , you better not be left with nothing in the derivative.
But you're almost there -- I agree with all steps except the one where you take everything in front of the derivative. You can indeed take the numbers and k's out. Then you just need to think about dz/dy and dy/dz (hint: y does not depend on x and z - trivially).

 

[Hootenanny]

Okay, so for the full definition;

If we have some vector function such that

\underline{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{j}

Then;

curl(\underline{F}) = \nabla\times\underline{F} = \left|\begin{array}<br /> \hat{i} &amp; \hat{j} &amp; \hat{i} \\<br /> \partial/\partial x &amp; \partial/\partial y &amp; \partial/\partial z \\<br /> F_x &amp; F_y &amp; F_z \\<br /> \end{array}\right|

 

[Oblio]

In the same way I guess... no?

I didn't know I could take everything of a derivative either. cool!

Doesn't look good to me , you better not be left with nothing in the derivative.
But you're almost there -- I agree with all steps except the one where you take everything in front of the derivative. You can indeed take the numbers and k's out. Then you just need to think about dz/dy and dy/dz (hint: y does not depend on x and z - trivially).

I know \frac{\partial y}{\partial z} would be 0 since they're unrelated, but again, the example in my book does indeed take z out of a simliar situation.

 

[Hootenanny]

One can take everything in front of the derivative since they are independent of the variable we are taking the derivative wrt.

 

[Oblio]

That leaves me non zero though, that's not ok is it?

 

[Hootenanny]

That leaves me non zero though, that's not ok is it?

You must have made an error, the curl of that vector field is definitely zero. If you show me your working, I'll point out where you've gone wrong.

 

[Oblio]

You must have made an error, the curl of that vector field is definitely zero. If you show me your working, I'll point out where you've gone wrong.

Well just factoring out all those constants,

k3z-k2y \neq 0 right?

 

[Hootenanny]

Well just factoring out all those constants,

k3z-k2y \neq 0 right?

Yes, but you don't just have k3z-k2y do you? What else do you have?

 

[Oblio]

Your saying the y and z components will make it cancel?

I thought this was supposed to (looking at my example again :P )

 

[Hootenanny]

Okay, let's look at this again;

\frac{\partial}{\partial y} k3z - \frac{\partial}{\partial z} k2y = 0 - 0 = 0

Since you are taking the derivatives wrt different variables than the respective components both the derivative are equal to zero. Or we can rewrite it like this

k3z \frac{\partial}{\partial y} 1- k2y\frac{\partial}{\partial z}1 = 0 - 0 = 0

Make sense?

 

[Oblio]

im getting Fy to cancel out, but Fx and Fz adding up to 2Fx and -2k3z

 

[Oblio]

(i posted that last one before seeing YOUR last one)

 

[Oblio]

Okay, let's look at this again;

\frac{\partial}{\partial y} k3z - \frac{\partial}{\partial z} k2y = 0 - 0 = 0

Since you are taking the derivatives wrt different variables than the respective components both the derivative are equal to zero. Or we can rewrite it like this

k3z \frac{\partial}{\partial y} 1- k2y\frac{\partial}{\partial z}1 = 0 - 0 = 0

Make sense?

So, its zero since their derivatives are zero?

 

[Hootenanny]

So, its zero since their derivatives are zero?

Yup, essentially your multiplying some constants by zero.

 

[Oblio]

I thought you were saying something different before, that's why i was confused over something simple:P

 

[Oblio]

I'm supposed to substitute to verify, using F = -\nabla U

hows this look?

U(x) = -\int Fx dx = - \int kx = -kxtF = -\nabla U, so...=- \frac{\partial}{\partial x} (-kxt) = kt = F (x)

I don't know how to work limits into latex but i guess its technically xo -->x

?

 

[Oblio]

I'm only not sure because Fx dx in the integral are Fx and dx prime...

 

[Oblio]

from what I'm reading I think I might need to add a path of the particle?

yes?
I'm confused because the answer seems like its right lol.

 

[CompuChip]

--- Post #22 ---

Don't get confused, what you had there was correct.
Now what are the equations for the y and z components of the curl?
What do you get when working them out (show the calculation)?