Question

An object moving in the \(xy\)-plane is acted on by a conservative force described by the potential-energy function \(U(x, y) = \alpha[(1/x^2) + (1/y^2)]\), where a is a positive constant. Derive an expression for the force expressed in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\).

Short answer

\( \vec{F} = \left( \frac{2\alpha}{x^3} \right) \hat{\imath} + \left( \frac{2\alpha}{y^3} \right) \hat{\jmath} \).

Step-by-step solution

Understand the Relationship between Force and Potential Energy

The force acting on an object is related to the potential energy function by the negative gradient of the potential energy. This means the force can be expressed as \( \vec{F} = -abla U(x, y) \).

Calculate the Partial Derivative with Respect to x

Find the partial derivative of the potential energy function \( U(x, y) = \alpha \left( \frac{1}{x^2} + \frac{1}{y^2} \right) \) with respect to \( x \). \[ \frac{\partial U}{\partial x} = \alpha \left( \frac{d}{dx} \left( \frac{1}{x^2} \right) \right) = \alpha \left( -\frac{2}{x^3} \right) = -\frac{2\alpha}{x^3}. \]

Calculate the Partial Derivative with Respect to y

Find the partial derivative of the potential energy function with respect to \( y \). \[ \frac{\partial U}{\partial y} = \alpha \left( \frac{d}{dy} \left( \frac{1}{y^2} \right) \right) = \alpha \left( -\frac{2}{y^3} \right) = -\frac{2\alpha}{y^3}. \]

Construct the Force Vector

The force vector in terms of unit vectors \( \hat{\imath} \) and \( \hat{\jmath} \) is constructed using the negative of the partial derivatives. Thus, the force \( \vec{F} \) is: \[ \vec{F} = - \left( \frac{\partial U}{\partial x} \right) \hat{\imath} - \left( \frac{\partial U}{\partial y} \right) \hat{\jmath} = \left( \frac{2\alpha}{x^3} \right) \hat{\imath} + \left( \frac{2\alpha}{y^3} \right) \hat{\jmath}. \]

Key concepts

Potential Energy

Potential energy is an essential concept in physics. It represents the stored energy of an object due to its position relative to other objects. In our example, the potential energy function is given as \(U(x, y) = \alpha \left[\frac{1}{x^2} + \frac{1}{y^2}\right]\). Here, \(\alpha\) is a constant, and the terms \(\frac{1}{x^2}\) and \(\frac{1}{y^2}\) describe how the potential energy changes with position in the \(x\)- and \(y\)-directions.
Understanding potential energy is crucial because it helps us determine the work done by or against forces. In conservative systems, like the one described in the exercise, potential energy can be converted into kinetic energy and vice versa. This interchange makes analyzing problems straightforward, without needing to account for dissipative forces like friction.

Gradient

The gradient is a vector that points in the direction of the greatest rate of increase of a function. In two dimensions, such as the \(xy\)-plane, the gradient of the potential energy \(U(x, y)\) is given by\[abla U = \left( \frac{\partial U}{\partial x} \right) \hat{\imath} + \left( \frac{\partial U}{\partial y} \right) \hat{\jmath}\]This vector provides valuable information about how the potential energy changes at each point in space.
The relationship between force and potential energy involves the negative of this gradient. For conservative forces, the force is given by \(\vec{F} = -abla U\). This means that the force points in the direction of the steepest decrease of the potential energy, effectively moving objects from high potential energy areas towards lower potential energy regions.

Partial Derivatives

Partial derivatives are a mathematical tool used to analyze functions with multiple variables. They represent the rate of change of a function concerning one variable while holding the others constant.
In this exercise, we compute the partial derivatives of the potential energy function with respect to \(x\) and \(y\):

• For \(x\), \(\frac{\partial U}{\partial x} = -\frac{2\alpha}{x^3}\).
• For \(y\), \(\frac{\partial U}{\partial y} = -\frac{2\alpha}{y^3}\).

These derivatives measure how the potential energy changes with small shifts in the \(x\)- or \(y\)-directions. Mastery of partial derivatives is vital because it allows us to understand how systems respond to changes in different directions, thus helping calculate forces as needed in this problem.

Unit Vectors

Unit vectors are indispensable tools for vector representation in physics. They help specify direction without regard to magnitude. In the context of the problem, the force vector is expressed using unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\).
Each unit vector has a magnitude of one and points in a specific direction:

• \(\hat{\imath}\) represents the x-direction.
• \(\hat{\jmath}\) represents the y-direction.

By using these vectors, we can compactly express the force \(\vec{F}\) as \[ \vec{F} = \left(\frac{2\alpha}{x^3}\right) \hat{\imath} + \left(\frac{2\alpha}{y^3}\right) \hat{\jmath}\] as derived in the solution. Using unit vectors in calculations provides a clear, concise way to convey both magnitude and direction, making complex multidimensional computations more manageable.