91Ó°ÊÓ

A vector field is called conservative if it has a corresponding potential function. To determine if a vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \) is conservative, we check if there is a scalar function \( f \) such that \( abla f = \mathbf{F} \). This means that the gradient of the potential function \( f \) equals the vector field \( \mathbf{F} \).
To further validate conservativeness, the condition \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \) must hold. This condition ensures that the vector field is path-independent and thus conservative. For the given vector field \( \mathbf{F}(x, y) = \frac{1}{x+y} \mathbf{i} + \frac{1}{x+y} \mathbf{j} \), when we calculate partial derivatives \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \), both result in \( -\frac{1}{(x+y)^2} \). Because these derivatives are equal, \( \mathbf{F} \) is confirmed to be conservative. This means any line integral involving \( \mathbf{F} \) depends only on the endpoints of the curve, not the path taken.

A potential function \( f(x, y) \) is a scalar function that acts as the 'antidote' to a conservative vector field. When a vector field \( \mathbf{F} \) is conservative, it means we can express it as the gradient of some potential function \( f \). In the given exercise, our task is to find \( f(x, y) \) such that \( abla f = \mathbf{F} \).
To find this function, we solve \( \frac{\partial f}{\partial x} = \frac{1}{x+y} \) and \( \frac{\partial f}{\partial y} = \frac{1}{x+y} \). Integrating with respect to one variable and treating the other as constant helps us discover \( f \). In this case, integrating \( \frac{1}{x+y} \) with respect to \( x \) gives \( f(x, y) = \ln|x+y| + g(y) \), where \( g(y) \) is a function of \( y \).
Another condition from \( \frac{\partial f}{\partial y} \) confirms \( g(y) \) must be a constant because differentiating \( \ln|x+y| \) with respect to \( y \) already matches \( \frac{1}{x+y} \). Thus, the potential function is \( f(x, y) = \ln|x+y| \). This function simplifies evaluating the line integral over any path within the conservative field.

The Fundamental Theorem for Line Integrals creates a simple way to evaluate line integrals when dealing with conservative vector fields. It states that the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) is simply the difference in values of the potential function \( f \) at the endpoints of the curve. This means, for a conservative field with a known potential \( f(x, y) \), the work done moving from point \( \mathbf{r}(a) \) to \( \mathbf{r}(b) \) is \( f(\mathbf{r}(b)) - f(\mathbf{r}(a)) \).
In our exercise, the path \( C \) is from \( (1, 0) \) to \( (0, 1) \). The potential function \( f(x, y) = \ln|x+y| \) gives us the following evaluations: \( f(0, 1) = \ln|0+1| = 0 \) and \( f(1, 0) = \ln|1+0| = 0 \). Subsequently, the difference \( f(0, 1) - f(1, 0) = 0 \) implies zero work done by the field \( \mathbf{F} \) over the specified path. This result shows that the path's influence is nullified under conservative conditions.