91Ó°ÊÓ

When dealing with vector fields, a potential function is a scalar function, often denoted as \( f(x, y) \) or \( f(x, y, z) \), such that the vector field is the gradient of this function. This means if you have a vector field \( \mathbf{F} \), and you can find a function \( f \) where \( \mathbf{F} = abla f \), then \( f \) is the potential function of \( \mathbf{F} \).

Here, in our exercise, the vector field \( \mathbf{F}(x, y) = x y^2 \mathbf{i} + x^2 y \mathbf{j} \) is determined to be conservative. To find the potential function \( f(x, y) \), start by considering the relationships \( f_x = xy^2 \) and \( f_y = x^2y \).
- **Step 1**: Integrate \( f_x = xy^2 \) with respect to \( x \): - This gives \( f(x, y) = \frac{1}{2}x^2y^2 + g(y) \), where \( g(y) \) is a function of \( y \).
- **Step 2**: Differentiate the result with respect to \( y \): - Obtain \( f_y = x^2y + g'(y) \) which should equal \( x^2y \). Thus, \( g'(y) = 0 \).

The function \( g(y) \) must be a constant, suggesting our potential function is:\[f(x, y) = \frac{1}{2}x^2y^2 + C\]This function encapsulates the information of the vector field \( \mathbf{F} \).

In vector calculus, a line integral is a type of integral where you integrate a function over a curve. Imagine you are summing up forces, energies, or other quantities along a path in a field.

For conservative vector fields, like \( \mathbf{F} \), the line integral of \( \mathbf{F} \) over a curve \( C \) only depends on the values of the potential function at the endpoints of \( C \). This drastically simplifies the work. Instead of integrating over every point in the path, you simply subtract the potential function's value at the start point from its value at the endpoint:

• If you have \( C \) as given by \( \mathbf{r}(t) \) and endpoints \( \mathbf{r}(0) = (x_0, y_0) \) and \( \mathbf{r}(1) = (x_1, y_1) \), the integral is:
  \( \int_C \mathbf{F} \cdot d\mathbf{r} = f(x_1, y_1) - f(x_0, y_0) \)

In our exercise, the line integral is evaluated by calculating the potential function \( f \) at these points:
- **Step 1**: Evaluate \( f(x,y) = \frac{1}{2}x^2y^2 \) at \( (2,1) \), giving \( f(2,1) = 2 \)
- **Step 2**: Evaluate at \( (0,1) \), giving \( f(0,1) = 0 \).

Therefore, the value of the line integral is \( 2 - 0 = 2 \). This efficient calculation stems from the fact that \( \mathbf{F} \) is conservative.

A gradient field, or a gradient vector field, is one where every vector can be expressed as the gradient of a particular scalar function. In simpler terms, if you have a scalar function \( f \), its gradient \( abla f \) is the vector field that tells you the direction and rate of fastest increase of \( f \).

In our exercise, the vector field \( \mathbf{F}(x, y) = x y^2 \mathbf{i} + x^2 y \mathbf{j} \) was shown to be the gradient field of the potential function \( f(x, y) = \frac{1}{2}x^2y^2 \). Here is how this works:

• The gradient of a function \( f(x, y) \) is calculated as:
  \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).


• For \( f(x, y) = \frac{1}{2}x^2y^2 \), we have:
  \( \frac{\partial f}{\partial x} = xy^2 \) and \( \frac{\partial f}{\partial y} = x^2y \).

This perfectly matches the components of \( \mathbf{F} \), confirming it as a gradient vector field.

Additionally, gradient fields have the useful property of being conservative, meaning the path taken between two points does not affect the work done, only the start and end points do. This is why computing line integrals in these fields is greatly simplified.