# Homework Help: Partial derivatives - proving force is conservative

1. Oct 17, 2007

### 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 alot

2. Oct 17, 2007

### Hootenanny

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

3. Oct 17, 2007

### 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?

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

4. Oct 17, 2007

### Hootenanny

Staff Emeritus
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?

5. Oct 17, 2007

### 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)

6. Oct 17, 2007

### Hootenanny

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

7. Oct 17, 2007

### Oblio

In the same way I guess.... no?

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

8. Oct 17, 2007

### Oblio

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

9. Oct 17, 2007

### Oblio

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

k3z-k2y ?

10. Oct 17, 2007

### 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).

11. Oct 17, 2007

### Hootenanny

Staff Emeritus
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;

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

12. Oct 17, 2007

### Oblio

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.

13. Oct 17, 2007

### Hootenanny

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

14. Oct 17, 2007

### Oblio

That leaves me non zero though, thats not ok is it?

15. Oct 17, 2007

### Hootenanny

Staff Emeritus
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.

16. Oct 17, 2007

### Oblio

Well just factoring out all those constants,

k3z-k2y $$\neq$$ 0 right?

17. Oct 17, 2007

### Hootenanny

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

18. Oct 17, 2007

### Oblio

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

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

19. Oct 17, 2007

### Hootenanny

Staff Emeritus
Okay, lets 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?

20. Oct 17, 2007

### Oblio

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