91Ó°ÊÓ

A vector-valued function, such as \( \mathbf{r}(t) = -t \mathbf{i} + 4t \mathbf{j} + 3t \mathbf{k} \), represents a curve in three-dimensional space where each component (\( \mathbf{i}, \mathbf{j}, \mathbf{k} \)) is a function of the parameter \( t \). This allows us to trace a path in three dimensions by varying \( t \) over a given interval. The curve is composed of a series of vectors, each defined by the particular value of \( t \). In this case, the function describes how the vector changes direction and length as \( t \) moves from 0 to 1. This change is what lets us calculate the arc length, the total distance traveled along the path of the vector.

To find the arc length, we first need to compute the derivative of the vector-valued function, denoted as \( \mathbf{r}'(t) \). The derivative can be thought of as the vector of rates of change for each component of the original vector function. For each component function (e.g., \( -t, 4t, 3t \)), we take the standard derivative with respect to \( t \). Thus, the derivative of our function \( \mathbf{r}(t) = -t \mathbf{i} + 4t \mathbf{j} + 3t \mathbf{k} \) is \( \mathbf{r}'(t) = -\mathbf{i} + 4\mathbf{j} + 3\mathbf{k} \).

• \( -t \) differentiates to \( -1 \)
• \( 4t \) differentiates to \( 4 \)
• \( 3t \) differentiates to \( 3 \)

This derivative tells us the instantaneous rate of change or the direction and magnitude of movement at any point \( t \) along the curve.

Once we have the derivative \( \mathbf{r}'(t) \), we need to compute its magnitude, sometimes called its length. The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is given by the formula \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \). This represents the "length" of the vector, providing a scalar quantity. In our example for \( \mathbf{r}'(t) = -\mathbf{i} + 4\mathbf{j} + 3\mathbf{k} \):

• Calculate \( (-1)^2 + 4^2 + 3^2 \)
• Add the results: \( 1 + 16 + 9 = 26 \)
• The magnitude is \( \sqrt{26} \)

This magnitude \( \sqrt{26} \) will be used in the integral to find the arc length.

The arc length of the curve defined by the vector-valued function is computed through an integral that accounts for the length of \( \mathbf{r}'(t) \) over the interval. For our specific problem, this means finding the length from \( t = 0 \) to \( t = 1 \).The formula for the arc length is:\[ L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \]Using the previously calculated magnitude of \( \mathbf{r}'(t) \), the integral setup becomes:\[ L = \int_{0}^{1} \sqrt{26} \, dt \]Since \( \sqrt{26} \) is a constant, the integral simplifies to:\[ L = \sqrt{26} \times [t]_{0}^{1} = \sqrt{26} \times (1 - 0) \]The length of the arc traced by the parameter \( t \) over the interval is \( \sqrt{26} \). This is the total distance traveled along the curve from start to finish.