- #1

mbrmbrg

- 496

- 2

**[SOLVED] given potential: find F**

## Homework Statement

Find the force for the following potential energy function:

[tex]V=ce^{-(\alpha x+\beta y+\gamma z)}[/tex]

## Homework Equations

[tex]\mathbf{F}=-\nabla V[/tex]

[tex]F_x=- \frac{\partial V}{\partial x}[/tex]

[tex]F_y=- \frac{\partial V}{\partial y}[/tex]

[tex]F_z=- \frac{\partial V}{\partial z}[/tex]

## The Attempt at a Solution

By the chain rule:

[tex]F_x=-\frac{\partial V}{\partial x}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\alpha)][/tex]

[tex]F_y=-\frac{\partial V}{\partial y}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\beta)][/tex]

[tex]F_z=-\frac{\partial V}{\partial z}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\gamma)][/tex]

Additionally,

[tex]\mathbf{F}=F_x+F_y+F_z[/tex]

So add it up and factor out the common factors and get

[tex]\mathbf{F}=-(\alpha x+\beta y+\gamma z)ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})[/tex]

And that would be my final answer, except that the back of the book says that

[tex]\mathbf{F}=ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})[/tex]

Can you help me find my error?

Thanks!