3-dimensional potential energy problem

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Homework Help Overview

The problem involves calculating the force on a particle based on its potential energy expressed as a function of its position in the x-y plane. The potential energy is given by a specific polynomial equation, and the task includes finding equilibrium points.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find the gradient of the potential energy and its implications for calculating force. There are questions about taking vector derivatives and simplifying the potential energy function. Some participants consider separating the potential into components and question the assumptions regarding constants during differentiation.

Discussion Status

The discussion is active, with participants offering guidance on the use of partial derivatives to find force components. There is recognition of the need to consider constraints related to the particle's position in the x-y plane, and some participants express uncertainty about their understanding of the mathematical concepts involved.

Contextual Notes

Participants mention the lack of specific information regarding derivatives with respect to time or position, and there is a focus on the implications of assuming certain variables are constant during differentiation.

winhog
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I have a problem on my homework that says the potential energy of a particle is given by its position in the x-y plane according to

P.E. = x^3 + 8x^2 + 34yz

and I have to calculate the force on the particle at point (x,y,z), and all equilibrium points.

dU = 3x^2 dx + 16x dx + 34y dz + 34z dy

and F = -dU/dr with r being the vector position...

-F = 3x^2 dx/dr + 16x dx/dr + 34y dz/dr + 34z dy/dr

but that can't be the answer can it? I'm not given any dx/dr's or anything like that...I have a feeling there's some simple math thing I'm missing.

Any hints anyone can think of?
 
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You need to find the gradient of the potential - it's a vector quantity!
 
Hmmm...I think my problem might be I've never learned how to take a vector derivative and this might be above my head mathematically :confused:

Maybe I can separate the potential energy into x and y components...but i really don't know how.

I guess since the particle is on the x-y plane, dz = 0...so my equation can be simplified to

F = (-3x^2 - 16x) dx/dr - 34z dy/dr .

Can I change dx/dr and dy/dr into (x-direction) and (y-direction) in terms of the force? Then use pythagorean theorem to find the total force? Am I just rambling?
 
It's really straightforward:

The x-component of the force is the partial derivative of the potential with respect to x. For the y-component use the partial wrt y and for the z-component use the partial wrt z. Voila!
 
Tide said:
It's really straightforward:

The x-component of the force is the partial derivative of the potential with respect to x. For the y-component use the partial wrt y and for the z-component use the partial wrt z. Voila!

One caveat: the force is MINUS the gradient of the potential, [itex]\vec F = - \vec \nabla V[/itex]

or, to be more specific,

[itex]F_x = - {\partial V(x,y,z) \over \partial x}[/itex]

[itex]F_y = - {\partial V(x,y,z) \over \partial y}[/itex]

[itex]F_z = - {\partial V(x,y,z) \over \partial z}[/itex]

If the particle is indeed constrained to the xy plane, then the z component of the force will be canceled by a normal force. In the x and y component, one must then set the value of z corresponding to the z plane one is in (z=0 or some other value).

Pat
 
nrqed,

Thanks - I meant to type "proportional to" but somehow it didn't come out!
 
If I take the derivative with respect to, say, x, can I assume y and z are constants? If so, I messed up on a pretty simple problem :blushing:
 
That's essentially what you do when you take partial derivatives which applies here.
 
Thanks for the help guys!
 

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