91Ó°ÊÓ

An irrotational field is a vector field where the curl is zero. This means there is no rotation at any point in the field. Understanding this concept is crucial for determining if a vector field can be expressed as a gradient of a potential function. If a vector field is irrotational, it can be considered conservative.
In more formal terms, if you have a vector field \( \vec{v} \), and you compute its curl using the formula \( abla \times \vec{v} \), the result should be zero for the field to be irrotational. In our specific case, to check whether the vector field \( \vec{v} = (y, x, 0) \) is irrotational, the calculation resulted in\[ abla \times \vec{v} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \y & x & 0\end{vmatrix} = 0. \]This zero result shows no rotation, thus confirming the irrotational nature of \( \vec{v} \). This property leads us to the conclusion that \( \vec{v} \) can indeed be associated with a potential function.

A gradient field is a special type of vector field that can be expressed as the gradient of some scalar function. If a vector field \( \vec{v} \) is irrotational, it is also a gradient field. For a gradient field, there must exist a scalar function \( f(x, y, z) \) such that the field is the gradient of \( f \), written as \( abla f = \vec{v} \).
In the exercise, for the vector field \( \vec{v} = (y, x, 0) \), we know it is irrotational, hence there exists a potential function \( f \) that \( \vec{v} \) can be derived from. The partial derivatives needed to form \( \vec{v} \) are derived from \( f \) as follows:

• \( \frac{\partial f}{\partial x} = y \)
• \( \frac{\partial f}{\partial y} = x \)
• \( \frac{\partial f}{\partial z} = 0 \)

By integrating these partial derivatives, we find the potential function:
First, integrate \( \frac{\partial f}{\partial x} = y \) with respect to \( x \), which gives \( f(x, y, z) = xy + g(y, z) \), a general form. Then, integrating \( \frac{\partial f}{\partial y} = x \) with respect to \( y \), we refine that to \( f(x, y, z) = xy + h(x, z) \). Because these forms are compatible, we determine \( f(x, y, z) = xy + C \), where \( C \) is a constant.

The concept of a total differential relates to finding how a small change in a function, due to changes in its variables, can be expressed. For a function \( f(x, y, z) \), the total differential \( \mathrm{d}f \) is given by:\[ \mathrm{d}f = \frac{\partial f}{\partial x} \mathrm{d}x + \frac{\partial f}{\partial y} \mathrm{d}y + \frac{\partial f}{\partial z} \mathrm{d}z \]This gives a linear approximation of how \( f \) changes with its inputs. In the problem, we verify if \( y \mathrm{d}x + x \mathrm{d}y \) is the total differential of the function \( f(x, y, z) = xy + C \).
The partial derivatives from the potential function give us:

• \( \frac{\partial f}{\partial x} = y \)
• \( \frac{\partial f}{\partial y} = x \)
• \( \frac{\partial f}{\partial z} = 0 \)

Inserting these values into the total differential formula results in:\[ \mathrm{d}f = y \mathrm{d}x + x \mathrm{d}y + 0 \mathrm{d}z \]This expression perfectly matches \( y \mathrm{d}x + x \mathrm{d}y \), confirming that this is indeed the total differential of our function \( f(x, y, z) = xy + C \). This means any small change in \( f \) can be predicted by this formula as \( x \), \( y \), and \( z \) undergo minor adjustments.