Answer: Find Force for Potential Energy Function V

[mbrmbrg]

[SOLVED] given potential: find F

Homework Statement

Find the force for the following potential energy function:
$$V=ce^{-(\alpha x+\beta y+\gamma z)}$$

Homework Equations

$$\mathbf{F}=-\nabla V$$

$$F_x=- \frac{\partial V}{\partial x}$$
$$F_y=- \frac{\partial V}{\partial y}$$
$$F_z=- \frac{\partial V}{\partial z}$$

The Attempt at a Solution

By the chain rule:

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

Additionally,
$$\mathbf{F}=F_x+F_y+F_z$$

So add it up and factor out the common factors and get
$$\mathbf{F}=-(\alpha x+\beta y+\gamma z)ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})$$

And that would be my final answer, except that the back of the book says that
$$\mathbf{F}=ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})$$

Can you help me find my error?

Thanks!

 

[mbrmbrg]

[SOLVED] given potential find F

Homework Statement

Find the force for the following potential energy function:
$$V=ce^{-(\alpha x+\beta y+\gamma z)}$$

Homework Equations

$$\mathbf{F}=-\nabla V$$

$$F_x=- \frac{\partial V}{\partial x}$$

$$F_y=- \frac{\partial V}{\partial y}$$

$$F_z=- \frac{\partial V}{\partial z}$$

The Attempt at a Solution

By the chain rule:

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

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

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

Additionally,
$$\mathbf{F}=F_x+F_y+F_z$$

So add it up and factor out the common factors and get
$$\mathbf{F}=-(\alpha x+\beta y+\gamma z)ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})$$

And that would be my final answer, except that the back of the book says that
$$\mathbf{F}=ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})$$

Can you help me find my error?

Thanks!

 

[daschaich]

Check the derivatives

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

etc. Recall

$$\frac{\partial}{\partial x}e^{f(x)} = e^{f(x)}\frac{\partial f}{\partial x}.$$

 

[cepheid]

You used the chain rule incorrectly.

$$\textrm{Let}\ \ \ u(x,y,z) = \alpha x+\beta y+\gamma z$$

Then

$$\frac{\partial }{\partial x}(ce^{-u}) = \frac{d}{du}(ce^{-u}) \frac{\partial u}{\partial x}$$

$$= -\alpha ce^{-(\alpha x+\beta y+\gamma z)}$$

likewise for the partial derivatives wrt y and z.

 

[hunter151]

As I recall, chain rule goes like "Derivative of the inner function multiplied by the derivative of the outer function evaluated at the inner function".

 

[mbrmbrg]

Right... it's the easy math that bites me in the butt.
Thanks!