91Ó°ÊÓ

Parametric equations are a way to define a curve by expressing the coordinates of the points on the curve as functions of a single variable, usually denoted by \( t \). This approach is particularly useful for representing complex curves and can simplify the evaluation of integrals along these curves.
In the given exercise, the curve \( C \) is defined by the parametric equation \( \mathbf{r}(t) = 2 \mathbf{i} + t \mathbf{j} + (2 - 2t) \mathbf{k} \). Here:

• \( x \) is a constant 2.
• \( y \) is represented as \( t \), meaning it changes linearly with \( t \).
• \( z \) is \( 2 - 2t \), which decreases linearly as \( t \) increases.

Without parametric equations, expressing these relationships would be more cumbersome. They allow us to easily substitute values for \( x \), \( y \), and \( z \) into the integrand. By translating the curve into a function of \( t \), we prepare it for further mathematical manipulation, such as integration.

Vector calculus is a branch of mathematics that concerns the differentiation and integration of vector fields. It is crucial for analyzing curves, surfaces, and their behaviors in multi-dimensional space.
In this problem, the parametric representation of the curve allows us to calculate the line integral of a scalar field along a vector field. The given curve \( r(t) \) is composed of its components \( 2 \mathbf{i} + t \mathbf{j} + (2 - 2t) \mathbf{k} \).
To compute the line integral \( \int_{C}(x y+y+z) \, ds \), we need to express the scalar field in terms of parametric components. Then, we find the corresponding differential element \( ds \) to consider the length contribution of the curve:

• \( ds = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \)
• In our example, this simplifies to \( \sqrt{5} \, dt \) since \( \frac{dy}{dt} = 1 \) and \( \frac{dz}{dt} = -2 \).

Overall, vector calculus tools allow us to navigate multi-dimensional space mathematically and to understand the described curvature's intrinsic properties.

A definite integral gives the accumulation of quantities along a defined interval, with specific limits. In the context of line integrals, it often helps evaluate accumulated values along a path or curve.
In the given task, we are evaluating \( \int_{0}^{1} (t + 2) \sqrt{5} \, dt \), which sums the contributions of the function \( (t + 2) \) weighted by the line element \( ds = \sqrt{5} \, dt \) over the interval from \( t = 0 \) to \( t = 1 \).
Calculating a definite integral involves finding the antiderivative of the function within the integrand and evaluating it at the upper and lower bounds of the interval:

• The antiderivative of \( t + 2 \) is \( \frac{t^2}{2} + 2t \).
• Evaluated from 0 to 1, this yields \( \frac{1^2}{2} + 2 \times 1 - (\frac{0^2}{2} + 2 \times 0) = \frac{5}{2} \).

Multiplying by the constant \( \sqrt{5} \), we find the final value of the line integral: \( \frac{5\sqrt{5}}{2} \). Each step, from finding the integrand to evaluating the final expression, is guided by the rules of definite integration.