91Ó°ÊÓ

In mathematics, especially vector calculus, a vector field is an assignment of a vector to each point in a subset of space. They are crucial in understanding systems modeled by physics, as they can represent things like magnetic and gravitational fields, where each point in space has a field direction and magnitude.
For the exercise, the vector field is given by \( \mathbf{F}(x, y) = x^2 y \mathbf{i} + 4 \mathbf{j} \). This notation tells us that the vector has two components: a horizontal component \( x^2 y \mathbf{i} \), and a vertical component \( 4 \mathbf{j} \). When evaluating line integrals, understanding how the vector field behaves along a particular path, or curve, is essential. In our example, the curve \( C \) is integrated over this vector field.

• The vector field gives context to the direction and strength of the vectors at each point along the curve.
• Such fields are often visualized as arrows on a graph, pointing in the direction of the vector at that point.

The parametrization of a curve involves expressing the coordinates of the curve as functions of a parameter, usually denoted by \( t \). This creates an easy way to navigate along the curve using \( t \) as a kind of timeline that traces the curve's path.
In the current problem, the curve \( C \) is represented parametrically by \( \mathbf{r}(t) = e^t \mathbf{i} + e^{-t} \mathbf{j} \) for \( 0 \leq t \leq 1 \). This means that for any given \( t \), you can find the corresponding point on the curve.

• Parametrization converts complex curves into manageable components for integration.
• You can imagine the parameter \( t \) as a time scale that moves from a starting point \( t=0 \) to an endpoint \( t=1 \).

Using substituting values of \( t \) allows identification of any position on the curve by plugging \( t \) into the parametric equations.

The dot product, also known as the scalar product, is a way to multiply two vectors to obtain a scalar value. It fundamentally measures how much one vector extends in the direction of another vector.
In the given problem, we're calculating the dot product of \( \mathbf{F}(t) \) and \( d\mathbf{r} \). Upon substitution and differentiation, we found \( d\mathbf{r} = e^t \mathbf{i} - e^{-t} \mathbf{j} \).

• The dot product \( \mathbf{F}(t) \cdot d\mathbf{r} \ = (e^t \mathbf{i} + 4 \mathbf{j}) \cdot (e^t \mathbf{i} - e^{-t} \mathbf{j}) \ = e^{2t} - 4e^{-t} \).
• This multiplication applies the general rule: \( \mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y \).

This operation allows us to combine the components neatly, turning the vector equation into a solvable expression during the integral evaluation.

Definite integrals find the total accumulation of quantities, such as area under a curve, from one point to another. In our exercise, definite integrals are used to evaluate the line integral of the vector field along the parametrized curve.
The integral \( \int_{0}^{1} (e^{2t} - 4e^{-t}) \, dt \) breaks into compute-friendly expressions. Solve them separately and then calculate:

• The integrals are \( \int e^{2t} \, dt = \frac{1}{2} e^{2t} \) and \( \int -4e^{-t} \, dt = 4e^{-t} \).
• Apply the bounds to get: \( \left[ \frac{1}{2} e^{2t} \right]_{0}^{1} \) and \( -4 \left[ e^{-t} \right]_{0}^{1} \).

Evaluate these within their limits to find the final result. By following these steps, we transition from complex vector-based functions to a final scalar result. This is fundamental in physics and engineering, where path integrals are essential in describing systems.