91Ó°ÊÓ

In mathematics, a vector field is a function that assigns a vector to every point in a particular space. Vector fields are usually visualized as arrows with varying direction and magnitude, representing their values at different points. In a two-dimensional plane, for example, a vector field can be denoted as \( \mathbf{F}(x, y) = M(x, y)\mathbf{i} + N(x, y)\mathbf{j} \), where \( M(x, y) \) and \( N(x, y) \) are the components of the vector field in the \( x \) and \( y \) directions, respectively.

• Each vector in the field depicts how a small particle would move at that point.
• These fields appear ubiquitously in physics for modeling speed, forces, velocity, and gravitational fields.

In the problem, the vector field defined by \( \mathbf{F}(x, y) \) has components \( M(x, y) = 2xe^{-y} \) and \( N(x, y) = 2y - x^2e^{-y} \). Understanding these components is crucial for analyzing the movement within the field and later checking if the field is conservative.

A conservative vector field is one where the line integral from one point to another is path-independent; only initial and final points matter. This property significantly simplifies computations. It is rooted in potential energy concepts, where the work done by a force field is the same regardless of the path taken.

• For a field to be conservative, there must exist a potential function \( f \) such that \( abla f = \mathbf{F} \).
• Mathematically, this involves checking that the partial derivatives of the components align: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

In our exercise, we found \( \frac{\partial M}{\partial y} = -2xe^{-y} \) and \( \frac{\partial N}{\partial x} = -2xe^{-y} \). Since these conditions match, the vector field in question is conservative, allowing us to compute integrals more efficiently.

A potential function is a scalar function whose gradient yields the given vector field. Finding this function helps in evaluating line integrals over conservative fields by reducing path-dependent calculations to simple evaluations. Here are the steps to determine a potential function:

• Start by solving \( \frac{\partial f}{\partial x} = M(x, y) \). Integrate \( M \) with respect to \( x \) to receive part of the potential function, plus an undetermined function of \( y \).
• To find this unknown function of \( y \), ensure that the differentiation with respect to \( y \) aligns with \( N(x,y) \).
• Finally, solve for any constants.

For the vector field in the problem, integrating \( 2xe^{-y} \) with respect to \( x \) gives \( x^2e^{-y} \), with the function \( g(y) \) determining \( y \'s \) effect. Finding that \( g'(y) = 2y \) and thus \( g(y) = y^2 + C \), the potential function is \( f(x, y) = x^2e^{-y} + y^2 \). This function allows us to compute a line integral between two points simply by subtracting their potential values.

Path independence is a key property of conservative fields, denoting that the integral of a field over a curve depends only on the endpoints, not the particular path taken. This concept simplifies many calculations within vector calculus, making complex path computations unnecessary in certain scenarios.

• To leverage path independence, ensure that the vector field is conservative.
• Calculate the potential function, if applicable, and perform simple endpoint evaluation.

Applying this property in the given exercise, we compute the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) solely based on the values \( f(2,1) \) and \( f(1,0) \). Thus, despite the infinite paths possible between \( (1,0) \) and \( (2,1) \), the integral resolves to \( 4e^{-1} \) without needing any specific trajectories.