91Ó°ÊÓ

A vector field is a function that assigns a vector to every point in a space. Imagine it as a field of arrows, where each arrow has both direction and magnitude, representing the vector at that point. For example, the vector field given in the exercise is \(\mathbf{F} = \frac{2x}{y}\mathbf{i} + \left(\frac{1-x^2}{y^2}\right)\mathbf{j}\). Each pair \((x, y)\) in the plane has a corresponding vector, influenced by these expressions.
Vector fields are crucial in physics and engineering as they can represent various entities like the velocity of a fluid across space, magnetic fields, or gravitational fields. In this context, finding a potential function for a vector field means identifying a scalar function whose gradient matches the original vector field.

The gradient of a scalar function is a vector field that points in the direction of the greatest rate of increase of the function. It is denoted by \(abla f\), surprising many students because it turns a function into something new—a field of vectors.
In mathematical terms, for a function \(f(x, y)\), the gradient \(abla f\) is given by:

• \(\frac{\partial f}{\partial x}\) for the \(x\)-component
• \(\frac{\partial f}{\partial y}\) for the \(y\)-component

This is why the exercise seeks \(f(x, y)\) for which the gradient equals \(\mathbf{F}\). If such \(f(x, y)\) is found, it's known as a potential function, indicating that the vector field is conservative or can be expressed as the gradient of some scalar field.

Partial derivatives are a way of exploring how functions change with respect to individual variables, while keeping others constant. In essence, it’s as if you’re slightly tweaking one of the knobs on a multivariable function to see what happens. For example, when calculating \(\frac{\partial f}{\partial x}\) in a two-variable function \(f(x, y)\), you treat \(y\) as a constant.
Partial derivatives are the building blocks for gradients. In the exercise, they are crucial for the reconstruction process of the potential function. Knowing \(\frac{\partial f}{\partial x} = \frac{2x}{y}\) and \(\frac{\partial f}{\partial y} = \frac{1-x^2}{y^2}\) helped solve for \(f(x, y)\) by integrating these derivatives. This highlights their importance in connecting vector fields to scalar potential functions.

Integration is the reverse process of differentiation. It's a way to accumulate quantities, finding a function given its derivative. In the context of finding a potential function, integration helps work backwards from the gradient to retrieve the original function \(f(x, y)\).
The solution required integrating the partial derivatives with respect to their variables. The exercise first integrates \(\frac{2x}{y}\) with respect to \(x\) to find a partial expression of \(f(x, y)\). This introduces an arbitrary function \(g(y)\), since integration with respect to \(x\) means that terms involving only \(y\) remain unknown post-integration.
Thus, another integration step (this time solving for \(g(y)\)) completes the process, revealing the complete form of the potential function. This methodical integration not only uncovers the potential function but also ensures that all components of the gradient align correctly with the given vector field, ensuring consistency and correctness.