\[ U(x, y) = a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \]

This question aims to find an expression for the **Force f** which is expressed in terms of the **unit vectors**Â **i^** and **j^**.

The concepts needed for this question include **potential energy function, conservative forces,** and **unit vectors. Potential Energy Function** is a function that is defined as the **position** of the **object** only for the **conservative forces** like **gravity. Conservative forces** are those forces that do not depend on the **path** but only on the **initial** and **final positions** of the object.

## Expert Answer

The given **potential energy function** is given as:

\[ U(x, y) = a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \]

The **conservative force** of **motion** in **two dimensions** is the **negative partial derivative** of its potential energy function multiplied by its respective **unit vector.** The formula for **conservative force** in terms of its potential energy function is given as:

\[ \overrightarrow{F} = – \Big( \dfrac { dU }{ dx } \hat{i} + \dfrac { dU }{ dy } \hat{j} \Big) \]

Substituting the value of **U** in the above equation to get the expression for **Force f**.

\[ \overrightarrow{F} = – \Big( \dfrac { d }{ dx } a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \hat{i} + \dfrac { d }{ dy } a \Big( \dfrac{1} {x^2} + \dfrac{1} {y^2} \Big) \hat{j} \Big) \]

\[ \overrightarrow{F} = – \Big( a \dfrac { d }{ dx } \Big( \dfrac{1} {x^2} \Big) \hat{i} + a \dfrac { d }{ dy } \Big( \dfrac{1} {y^2} \Big) \hat{j} \Big) \]

\[ \overrightarrow{F} = 2a \dfrac{ 1 }{ x^3 } \hat{i} + 2a \dfrac{ 1 }{ y^3 } \hat{j} \]

\[ \overrightarrow{F} = 2a \Big( \dfrac{ 1 }{ x^3 } \hat{i} + \dfrac{ 1 }{ y^3 } \hat{j} \Big) \]

## Numerical Result

The **expression** for the **force** $\overrightarrow {f}$ is expressed in terms of the **unit vectors** $\hat{i}$ and $\hat{j}$ is calculated to be:

\[ \overrightarrow{F} = \Big( \dfrac{ 2a }{ x^3 } \hat{i} + \dfrac{ 2a }{ y^3 } \hat{j} \Big) \]

## Example

**Potential energy function** is given for an object moving in **XY-plane.** Derive an expression for the **force** **f** expressed in terms of the **unit vectors** $\hat{i}$ and $\hat{j}.

\[ U(x, y) = \big( 3x^2 + y^2 \big) \]

We can derive an expression for **force** by taking the **negative** of the **partial derivative** of the **potential energy function** and multiplying it by respective **unit vectors.** The formula is given as:

\[ \overrightarrow{F} = – \Big( \dfrac { dU }{ dx } \hat {i} + \dfrac { dU }{ dy } \hat {j} \Big) \]

\[ \overrightarrow{F} = – \Big( \dfrac { d }{ dx } \big( 3x^2 + y^2 \big) \hat {i} + \dfrac { d }{ dy } \big( 3x^2 + y^2 \big) \hat {j} \Big) \]

\[ \overrightarrow{F} = – \big( 6x \hat {i} + 2y \hat {j} \big) \]

\[ \overrightarrow{F} = – 6x \hat {i}\ -\ 2y \hat {j} \]

The expression of **force** **f** is calculated to be $- 6x \hat {i}\ -\ 2y \hat {j}$