91Ó°ÊÓ

To find a potential function, we need to identify a scalar function \( f(x, y) \) from a conservative vector field \( \vec{F} \). A conservative vector field allows us to express it as the gradient of some scalar function. For \( \vec{F} = (2x + y, 2y + x) \), we check if there is a function such that its gradient \( abla f = \vec{F} \):- \( \frac{\partial f}{\partial x} = 2x + y \)- \( \frac{\partial f}{\partial y} = 2y + x \)To find \( f(x, y) \), integrate each component.

Integration Process

Start by integrating \( \frac{\partial f}{\partial x} \) concerning \( x \),yielding: \( f(x, y) = x^2 + xy + g(y) \). Here, \( g(y) \) is an arbitrary function of \( y \). Next, differentiate this result concerning \( y \), and equate to \( \frac{\partial f}{\partial y} \) to solve for \( g(y) \). Integrating gives \( g(y) = y^2 + C \). Substitute \( g(y) \) back into \( f(x, y) \), and the potential function becomes \( f(x, y) = x^2 + xy + y^2 + C \).

The fundamental theorem of line integrals connects potential functions with line integrals, providing a simplified evaluation method for the work done by a conservative vector field. If a vector field is conservative, i.e., \( \vec{F} = abla f \), then the line integral of \( \vec{F} \) over a curve \( C \) that starts at point \( A \) and ends at \( B \) equals the difference in the potential function values at these points.- \( \int_{C} \vec{F} \cdot d \vec{r} = f(B) - f(A) \)This theorem shows the path-independence property, which states that the integral's value depends only on endpoints.

Application

For the vector field \( \vec{F} = (2x+y, 2y+x) \), using this theorem means calculating \( f(0, 0) - f(0, 0) = 0 \), showing that the computed integral over curve \( C \) is zero.

The curl of a vector field gives us a measure of the field's rotation at a point. For a two-dimensional field, the curl is a scalar quantity.To calculate the curl in \( \vec{R}^2 \), for \( \vec{F} = (M, N) \), use the formula:- \( \text{curl} \, \vec{F} = \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \)For our vector field \( \vec{F} = (2x+y, 2y+x) \),this results in:- \( \text{curl} \, \vec{F} = \frac{\partial}{\partial x}(2y + x) - \frac{\partial}{\partial y}(2x + y) = 1 - 1 = 0 \)

Interpretation

A curl of zero indicates that \( \vec{F} \) is conservative, confirming our calculation that \( \vec{F} \) is a gradient field and supporting using potential functions.

A line integral, particularly in the context of vector fields, calculates the work done by a vector field along a specified path or curve.For vector field \( \vec{F} \) and curve \( C \) parametrized by \( \vec{r}(t) \), the line integral is:- \( \int_{C} \vec{F} \cdot d \vec{r} = \int \vec{F}(\vec{r}(t)) \cdot \frac{d \vec{r}}{dt} dt \)

Direct Calculation

For \( \vec{F} = (2x + y, 2y + x) \) with \( \vec{r}(t) = \langle t^2 - t, t^3 - t \rangle \),calculate \( \vec{F}(\vec{r}(t)) \) and \( \frac{d \vec{r}}{dt} \), leading to the integral of their dot product over \( t \) from 0 to 1.The detailed work involves substitution and integration,resulting in the overall work done by the vector field along \( C \).

Parametrization of curves turns geometric paths into equations depending on a parameter, usually \( t \).This transformation is key in calculus and helps calculate integrals like the line integral.For a curve \( C \) represented by \( \vec{r}(t) \), the vector function \( \vec{r}(t) = \langle x(t), y(t) \rangle \) describes the curve's path by parameter \( t \).

Understanding Parametrization

Using the given \( \vec{r}(t) = \langle t^2 - t, t^3 - t \rangle \),the line integral over \( C \) can be evaluated by transforming the problem into an integral with respect to \( t \). The curve's form dictates how the vector field interacts with the path, and correct parametrization is essential for accurate calculations.Parametrization allows a simple, manageable form for complex curves, facilitating line integral evaluations.