91Ó°ÊÓ

A potential function, denoted by \( \phi(x, y) \), is a scalar function whose gradient is equal to a given vector field. For a vector field to be conservative, it must be expressible as the gradient of a potential function. This means that its components are the partial derivatives of \( \phi(x, y) \) with respect to the field’s variables. In this exercise, we have:

• \( \frac{\partial \phi}{\partial x} = h(x)[x \sin y + y \cos y] \)
• \( \frac{\partial \phi}{\partial y} = h(x)[x \cos y - y \sin y] \)

The challenge is to solve these equations to find the potential function. This is possible when the field satisfies certain conditions, specifically when its curl is zero. If we find such a potential function, then the vector field is conservative, implying that the work done by moving along any path in the field depends only on the endpoints, not the path taken. It's like having a hillside where no energy is lost to friction. The energy only depends on the height, regardless of how you climbed it.

The curl of a vector field is a measure of the field's rotation or 'twisting'. For a two-dimensional vector field \( \mathbf{F}(x, y) \), the curl is simplified as just checking a specific partial derivative equation. The field is conservative if its curl is zero, meaning there is no local rotation. Mathematically, for a vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), the requirement is:\[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0\]For our specific field, we calculated:

• \( \frac{\partial}{\partial y}(h(x)[x \sin y + y \cos y]) \)
• \( \frac{\partial}{\partial x}(h(x)[x \cos y - y \sin y]) \)

These components must be equal for the curl to vanish, ensuring that \( \mathbf{F} \) is conservative. Thus, setting these partial derivatives equal to each other and solving them zeroes out the curl. This condition tells us there’s no "swirling" or circulating force field, and instead, it acts like a gradient of potential.

Partial derivatives are used to describe how a function changes as its variables change, holding other variables constant. They are crucial for analyzing multivariable functions. In this exercise, they help determine whether a vector field is conservative. To find out, we derived:

• \( \frac{\partial}{\partial y}[h(x)(x \sin y + y \cos y)] \to h(x)[x \cos y - y \sin y] + h(x)\cos y \)
• \( \frac{\partial}{\partial x}[h(x)(x \cos y - y \sin y)] \to h'(x)[x \cos y - y \sin y] + h(x)\cos y \)

These derivatives describe how each component of the vector field changes with respect to each variable. Setting these derivatives as conditions ensures the external (nonzero) factor \((x \cos y - y \sin y)\) determines the form of \( h(x) \). The dependability on variable changes is why partial derivatives are foundational in calculus, especially in evaluating field properties.