91Ó°ÊÓ

To understand line integrals, it's crucial to start with the parametrization of curves. When given a curve, it can often be described using a vector function, commonly denoted as \( \mathbf{r}(t) \). This function describes the curve in terms of a parameter \( t \), simplifying the description of paths in space.
For example, in our problem, the curve \( C \) is parameterized by \( \mathbf{r}(t) = 2t \mathbf{i} + 3t \mathbf{j} - t^2 \mathbf{k} \). This shows:\

• \( x(t) = 2t \)
• \( y(t) = 3t \)
• \( z(t) = -t^2 \)

This means as \( t \) varies from \(-1\) to \(1\), the curve sweeps out a path in three-dimensional space. Parametrization allows for easy computation of derivatives, integrals, and intersections which are essential for vector calculus.

The dot product is a fundamental operation in vector analysis. It's the product of two vectors and serves to find the projection of one vector onto another. The formula for the dot product between two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is \( \mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z \).
In our problem, we compute the dot product of the vector field \( \mathbf{F}(t) = 2t \mathbf{i} + t^2 \mathbf{j} + 3t \mathbf{k} \) and the derivative \( \frac{d \mathbf{r}}{dt} = 2 \mathbf{i} + 3 \mathbf{j} - 2t \mathbf{k} \). The resulting expression \( \mathbf{F}(t) \cdot \frac{d \mathbf{r}}{dt} = 4t - 3t^2 \) is crucial for evaluating the line integral. The value of this dot product simplifies the process of integration over the parameter \( t \). This operation helps determine the contribution of \( \mathbf{F} \) along the path \( C \).

Vector fields assign a vector to every point in space and are often used to model physical phenomena such as gravitational fields, magnetic fields, and fluid flow. In this exercise, the vector field is given as \( \mathbf{F}(x, y, z) = x \mathbf{i} - z \mathbf{j} + y \mathbf{k} \). This means at each point \( (x, y, z) \), there is a vector based on the coordinates.
Substituting our parametric equations, \( x = 2t \), \( y = 3t \), and \( z = -t^2 \) into \( \mathbf{F} \) provides us with the vector field along the curve \( C \).

• The substitution results in \( \mathbf{F}(t) = 2t \mathbf{i} + t^2 \mathbf{j} + 3t \mathbf{k} \).

This calculated vector field is used to compute the line integral, illustrating how \( \mathbf{F} \) interacts with the path defined by \( C \). Understanding vector fields aids in visualizing and solving complex integrals over paths.

Integration over intervals evaluates the area under a curve between two points, building a bridge between straight algebraic lines and curved geometries. For a line integral, we integrate a scalar product over the interval defined by the parameter \( t \).
In this problem, the goal is to integrate \( \mathbf{F}(t) \cdot \frac{d \mathbf{r}}{dt} = 4t - 3t^2 \) over \(-1 \leq t \leq 1 \).

• Split the integral into two parts: \( \int_{-1}^{1} 4t \, dt \) and \( \int_{-1}^{1} -3t^2 \, dt \).
• Evaluate these integrals to get \( 0 \) and \(-2 \), resulting in \( 0 - 2 = -2 \).

Integration here formalizes how we accumulate the effects of the vector field along the curve. This process reveals the total effect of \( \mathbf{F} \) along the path from \( t = -1 \) to \( t = 1 \).