To determine whether the vector field $\mathbf{F}(x, y) = \frac{8y}{x} \mathbf{i} + \frac{8x^2}{y^2} \mathbf{j}$ is conservative, we follow these steps:

Step 1: Check if the vector field is conservative.

A vector field is conservative if the curl of the field is zero (in two dimensions, this simplifies to checking if $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$, where:

• $P(x, y) = \frac{8y}{x}$ (the component of $\mathbf{i}$),
• $Q(x, y) = \frac{8x^2}{y^2}$ (the component of $\mathbf{j}$).

Calculate $\frac{\partial Q}{\partial x}$:

$Q(x, y) = \frac{8x^2}{y^2} \implies \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \left( \frac{8x^2}{y^2} \right) = \frac{16x}{y^2}.$

Calculate $\frac{\partial P}{\partial y}$:

$P(x, y) = \frac{8y}{x} \implies \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} \left( \frac{8y}{x} \right) = \frac{8}{x}.$

Compare $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$:

$\frac{\partial Q}{\partial x} = \frac{16x}{y^2}, \quad \frac{\partial P}{\partial y} = \frac{8}{x}.$

Since $\frac{\partial Q}{\partial x} \neq \frac{\partial P}{\partial y}$, the vector field is not conservative.

Final Answer:

• The vector field is not conservative.
• Therefore, a potential function does not exist.

Do you have further questions or need clarifications? Here are five related questions:

1. How do you compute the curl of a 2D vector field in general?
2. Why does a zero curl imply that a field is conservative?
3. Can a field be conservative in some region but not in others?
4. What are the physical interpretations of conservative vector fields?
5. How would you verify if a 3D vector field is conservative?

Tip: Always ensure the domain of the vector field is connected and excludes singularities when checking for conservativeness.