# Answer: Find Force for Potential Energy Function V

• mbrmbrg
In summary, the chain rule is used to find the force for a potential energy function. The potential energy function is given by V=ce^{-(\alpha x+\beta y+\gamma z)} and the force is found to be -\alpha ce^{-(\alpha x+\beta y+\gamma z)}
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:

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

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

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!

[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:

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

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

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

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

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!

Check the derivatives

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

etc. Recall

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

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.

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".

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

## 1. What is the formula for finding force from a potential energy function?

The formula for finding force from a potential energy function is F = -dV/dx, where F is the force, V is the potential energy function, and x is the position.

## 2. How is potential energy related to force?

Potential energy and force are directly related. The force acting on an object is equal to the negative gradient of the potential energy function at a given point. This means that as the potential energy increases, the force acting on the object also increases.

## 3. Can you explain the concept of a conservative force?

A conservative force is a type of force that does not dissipate energy, meaning that the work done by the force is independent of the path taken. This means that the total energy of the system is conserved, and it is possible to define a potential energy function for the force.

## 4. How can I use the potential energy function to find the force acting on an object?

To find the force acting on an object from a potential energy function, you can use the formula F = -dV/dx. This means that you need to take the derivative of the potential energy function with respect to the position variable, and then multiply by -1.

## 5. Is it possible to have multiple potential energy functions for the same force?

Yes, it is possible to have multiple potential energy functions for the same force. However, all of these potential energy functions will have the same gradient, meaning that they will all produce the same force. This is because the force is determined by the change in potential energy, not the absolute value of the potential energy itself.

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