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When calculating a line integral around a closed curve like a circle, parameterization is essential. It is the process of expressing the coordinates of a curve as functions of a single parameter, typically denoted as \( t \). For a circle centered at the origin with a radius of 2, this can be done using trigonometric functions:

\[ x = 2 \cos(t) \text{ and } y = 2 \sin(t), \quad t \in [0, 2\pi] \]

This parameterization allows us to traverse the circle in a counter-clockwise direction, covering every point on the circle exactly once. Each value of \( t \) corresponds to a unique point on the circle, where \( t = 0 \) and \( t = 2\pi \) both correspond to the initial point, completing a full loop.

In multivariable calculus, a vector field assigns a vector to every point in space. For the given problem, the vector field \( \mathbf{G}(x, y) \) is defined as \( (y+x) \mathbf{i}+(y-x) \mathbf{j} \). This means each point \((x, y)\) on the plane has a corresponding vector which can be visualized as an arrow pointing in a certain direction.

By parameterizing the curve, we can express this vector field in terms of the parameter \( t \). Substituting our parameterized \( x \) and \( y \) values into \( \mathbf{G}(x, y) \) results in the vector field being expressed as:

\[ \mathbf{G}(t) = (2 \sin(t) + 2 \cos(t))\mathbf{i} + (2 \sin(t) - 2 \cos(t))\mathbf{j} \]

This reformulation makes it easier to compute how the vector field interacts with the curve.

The dot product is a mathematical operation that takes two vectors and returns a scalar. It is vital in line integrals to measure how much a vector field contributes along a curve. To compute it, you multiply corresponding components of the vectors and add the results.

For the dot product of the vectors from the problem, \( \mathbf{G} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) is a small segment of the curve, we have:

\[(2\sin(t) + 2\cos(t))(-2\sin(t)) + (2\sin(t) - 2\cos(t))(2\cos(t)) \]

This simplifies (after some calculations) to:

\[-4\sin(t)^2 - 4\cos(t)^2 \]

Using the Pythagorean identity where \( \sin^2(t) + \cos^2(t) = 1 \), the expression further reduces to \( -4 \cdot 1 = -4 \). This result represents the interaction of the vector field with the curve at each point t.

A differential element is an infinitesimally small piece of a curve, often expressed as \( d\mathbf{r} \) in vector form. In line integrals, it helps represent the curve's path, which the vector field interacts with.

For our parameterized circle, the differential changes in \( x \) and \( y \) with respect to \( t \) are given by:

• \( dx = -2\sin(t) \, dt \)
• \( dy = 2\cos(t) \, dt \)

Thus, \( d\mathbf{r} \), the differential vector element along the curve, can be expressed as:

\[ d\mathbf{r} = (-2\sin(t) \mathbf{i} + 2\cos(t) \mathbf{j}) \, dt \]

This expression reflects the instantaneous direction and magnitude of the infinitesimal path the curve takes at any point \( t \). Understanding \( d\mathbf{r} \) is crucial for setting up and solving the line integral, as it indicates how the curve's path interacts with the vector field.