91Ó°ÊÓ

In vector calculus, finding the derivative of a vector function allows us to understand how the position changes with respect to time or any other variable. Consider a vector function given as \( \mathbf{r}(t) = \ln t \mathbf{i} + 2t \mathbf{j} + t^2 \mathbf{k} \). To determine the derivative \( \mathbf{r}^{\prime}(t) \), we differentiate each component of \( \mathbf{r}(t) \) individually:

• For \( \ln t \mathbf{i} \), the derivative is \( \frac{1}{t} \mathbf{i} \).
• For \( 2t \mathbf{j} \), the derivative is \( 2 \mathbf{j} \).
• For \( t^2 \mathbf{k} \), the derivative is \( 2t \mathbf{k} \).

Once differentiated, these parts combine to give \( \mathbf{r}^{\prime}(t) = \frac{1}{t} \mathbf{i} + 2 \mathbf{j} + 2t \mathbf{k} \). This derivative vector shows how the original vector \( \mathbf{r}(t) \) changes, delivering insights into velocity if \( t \) is time.
Understanding the derivative can also be crucial in various applications such as physics and engineering, where vectors often describe motion and forces.

The magnitude of a vector, often denoted as \( \| \mathbf{a} \| \), provides a measure of its length or size in multidimensional space. For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude is calculated using the formula:

• \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \)

In the context of the problem, we need to find the magnitude of the derivative vector \( \mathbf{r}^{\prime}(t) = \frac{1}{t} \mathbf{i} + 2 \mathbf{j} + 2t \mathbf{k} \). Using the magnitude formula, you compute:

• \( \left\| \mathbf{r}^{\prime}(t) \right\| = \sqrt{\left(\frac{1}{t}\right)^2 + 2^2 + (2t)^2} \)
• The simplified expression is \( \sqrt{\frac{1}{t^2} + 4 + 4t^2} \).

This magnitude represents the instantaneous rate of change in position and is crucial when calculating properties like speed or arc length. It gives not just the change direction, but a clear quantitative measure of how fast the change is occurring.

Arc length is the total length along a curve defined by a function over a particular interval. When you have a vector function \( \mathbf{r}(t) \), the arc length along this vector from \( t = a \) to \( t = b \) is given by the integral:

• \( s = \int_{a}^{b} \left\| \mathbf{r}^{\prime}(t) \right\| \, dt \)

For our exercise, we calculate the arc length from \( t = 1 \) to \( t = 3 \) using the derivative magnitude found previously:

• \( \int_{1}^{3} \sqrt{\frac{1}{t^2} + 4 + 4t^2} \, dt \)

This integral measures the continuous extension of our vector function \( \mathbf{r}(t) \) precisely across the given interval. Understanding arc length is crucial in fields such as engineering and physics where the path is not a straight line, as it provides a way to calculate the actual distance traversed.