Question

The potential energy of a certain particle is given by \(U=10 x^{2}+35 z^{3}\). Find the force vector exerted on the particle.

Short answer

Answer: The force vector exerted on the particle is \(F = \langle -20x, -105z^2 \rangle\).

Step-by-step solution

Calculate the partial derivative of U with respect to x

To find the partial derivative of U with respect to x, we differentiate U with respect to x, treating z as a constant. \[\frac{\partial U}{\partial x} = 20x\]

Calculate the partial derivative of U with respect to z

To find the partial derivative of U with respect to z, we differentiate U with respect to z, treating x as a constant. \[\frac{\partial U}{\partial z} = 105z^2\]

Represent the partial derivatives as a force vector

We will represent the force vector (F) as \[F = \langle F_x, F_z \rangle\]

Calculate the components of the force vector

To calculate the components of the force vector, we apply the negative gradient of the potential energy function. So, we will have \[F_x = -\frac{\partial U}{\partial x} = -20x\] and \[F_z = -\frac{\partial U}{\partial z} = -105z^2\]

Write the force vector

Now that we have the components of the force vector, we can write it as: \[F = \langle -20x, -105z^2 \rangle\] Therefore, the force vector exerted on the particle is \(F = \langle -20x, -105z^2 \rangle\).

Key concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as we tweak one variable at a time while keeping others constant. In mathematics, this is particularly useful for functions of multiple variables.
For instance, if a potential energy function depends on variables like \(x\) and \(z\), you can find out how the function changes with respect to \(x\) while keeping \(z\) fixed by computing the partial derivative \(\frac{\partial U}{\partial x}\). Similarly, you can determine its change with respect to \(z\) using \(\frac{\partial U}{\partial z}\).

• To find the partial derivative with respect to \(x\), \(\frac{\partial U}{\partial x}\), consider \(z\) as a constant.
• To find the partial derivative with respect to \(z\), \(\frac{\partial U}{\partial z}\), hold \(x\) constant.

This technique is fundamental when analyzing how each variable independently influences the function’s behavior.

Potential Energy Function

A potential energy function represents the potential energy (energy stored due to position) of a system or particle as a function of its position or coordinates.
For instance, in the potential energy function given by \(U = 10x^2 + 35z^3\), the potential energy is dependent on the particle's position denoted by \(x\) and \(z\).

• The term \(10x^2\) describes how potential energy varies with \(x\).
• The term \(35z^3\) explains how it changes with \(z\).

Understanding the potential energy function is crucial because it reveals how the stored energy in a system changes with position, which in turn can help us determine the force exerted on the system.

Gradient of a Function

The gradient of a function is a vector that indicates the direction of the greatest rate of increase of the function and its magnitude represents the rate of increase. For scalar fields like potential energy, the gradient is essential in determining forces.
The negative gradient of a potential energy function gives us the corresponding force field exerted by the potential. Mathematically, the force \(\mathbf{F}\) is computed as \(\mathbf{F} = -abla U\).
In our example with \(U = 10x^2 + 35z^3\), the gradient vector \(abla U\) is composed of its partial derivatives \(\frac{\partial U}{\partial x}\) and \(\frac{\partial U}{\partial z}\).

• Hence, \(abla U = \langle 20x, 105z^2 \rangle\).
• The force vector is then \(\mathbf{F} = \langle -20x, -105z^2 \rangle\).

The negative sign indicates the direction of force is opposite to the increase in potential energy, a principle rooted in physics.