91Ó°ÊÓ

A vector function is a mathematical expression where each element is a function of a single variable, usually time or another continuous parameter. This concept is incredibly useful in describing paths or curves in three-dimensional space. In the given example, the vector function \( \mathbf{r}(t)=t \mathbf{i}-\mathbf{j}+t^{2} \mathbf{k} \) describes a curve.

Each component of this vector function depends on \( t \), the parameter, and represents a unique dimension in space:

• \( t \) relates to the \( x \)-axis
• \(-1\) is constant for the \( y \)-axis
• \( t^2 \) corresponds to the \( z \)-axis

This makes the process of solving integrals over curves more intuitive as each variable can be related separately to the parameter \( t \). Understanding the way vector functions are constructed and used allows us to navigate complex three-dimensional problems by breaking them down into simpler one-variable functions. Indeed, this considerably simplifies processes like parameterization and integration.

Parameterization is the process of defining a curve by a single parameter instead of equations. This is crucial for evaluating line integrals since it relates the point in space directly to the parameter \( t \).

In our problem, the vector function \( \mathbf{r}(t)=t \mathbf{i}-\mathbf{j}+t^{2} \mathbf{k} \) serves as the parameterization for the curve \( C \). Since we parameterized:

• \( x(t) = t \)
• \( y(t) = -1 \)
• \( z(t) = t^2 \)

Consequently, we handle operations over the curve conveniently by manipulating \( t \) only. This process translates higher-dimensional problems into manageable one-dimensional tasks.

Parameterization helps compute derivatives and integrals easier: the derivatives such as \( dx \), \( dy \), and \( dz \) are calculated with respect to \( t \). Such transformations greatly simplify the work needed to find line integrals, which would otherwise require dealing with a multidimensional equation directly.

Definite integrals find the value over a specified range, and in the context of vector calculus, they can compute quantities like work done along a curve. When evaluating a line integral over curve \( C \), these integrals consider both the path taken and the vector field impacting the movement.

Here, for instance, we convert the integral \( \int_{C}(x+y-z)dx \) into \( \int_{0}^{1} (t - 1 - t^2) dt \) after parameterization is applied. This converts a line integral into a regular definite integral over the interval \([0, 1]\).

Key steps involved include:

• Substituting the parameterized expressions for \( x, y, \) and \( z \)
• Replacing \( dx \), \( dy \), or \( dz \) with their respective derivatives in terms of \( t \)
• Solving the resulting definite integral using standard integration techniques

This clean transformation emphasizes how important parameterization is for handling multidimensional integrals effectively through single-variable calculus tools. Ultimately, through correct setup and computation, we conclude that the evaluated value for all parts resulted in the same simple result due to the constraints on \( dy \) and the symmetric nature of the problem.