91Ó°ÊÓ

In calculus, a position vector function, often denoted as \( \mathbf{r}(t) \), describes a path or curve in space. This vector assigns a unique position to each parameter \( t \), effectively tracing out the path of a moving point in three-dimensional space. For the given problem, the position vector is expressed as:

• \( \mathbf{r}(t) = (t \cos t) \mathbf{i} + (t \sin t) \mathbf{j} + \left(\frac{2 \sqrt{2}}{3} t^{3/2}\right) \mathbf{k} \)

Here, each component of the vector corresponds to a dimension in space:

• The \( \mathbf{i} \) component involves both \( t \) and \( \cos t \), signifying how it affects movements in the horizontal plane.
• The \( \mathbf{j} \) component uses \( \sin t \), influencing vertical movements in the plane.
• The \( \mathbf{k} \) component is a function of \( t^{3/2} \), adding a third dimension typically associated with depth or altitude.

Understanding the position vector is fundamental as it influences subsequent calculations, particularly derivatives and arc lengths.

Finding the derivative of a position vector is crucial for understanding the behavior of the curve it represents. The derivative, denoted as \( \mathbf{r}'(t) \), indicates the velocity of a point on the curve, showing how the position changes with time.For our exercise:

• The derivative is calculated as:
• \( \mathbf{r}'(t) = (\cos t - t \sin t) \mathbf{i} + (\sin t + t \cos t) \mathbf{j} + (\sqrt{2} t^{1/2}) \mathbf{k} \).

Each component of \( \mathbf{r}'(t) \) highlights the instantaneous rate of change:

• \( \cos t - t \sin t \) involves the horizontal velocity.
• \( \sin t + t \cos t \) represents vertical velocity.
• \( \sqrt{2} t^{1/2} \) describes changes in the third dimension, shifting the point in depth.

Differentiating accurately sets the stage for calculating the tangent vector, a core step in arc length determination.

The magnitude of a vector function helps to understand the length or size of the vector at any point. For the vector's derivative \( \mathbf{r}'(t) \), its magnitude represents the speed or rate at which the point moves along the curve.To find the magnitude of \( \mathbf{r}'(t) \), we calculate:

• \( |\mathbf{r}'(t)| = \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2 + (\sqrt{2} t^{1/2})^2} \).

After simplifying, the expression becomes:

• \( \sqrt{1 + t^2 + 2t} \).

This length is vital, as it forms the foundation for calculating the unit tangent vector and the curve's arc length.

Determining the arc length of a curve provides a measure of the 'distance' traveled along that curve. This is essential for applications involving trajectory planning and path analysis. The arc length \( L \) of a curve given by \( \mathbf{r}(t) \) from \( t = 0 \) to \( t = \pi \) is calculated as:

• \( L = \int_{0}^{\pi} |\mathbf{r}'(t)| \, dt = \int_{0}^{\pi} \sqrt{1 + t^2 + 2t} \, dt \).

This integral represents the cumulative sum of infinitesimal distances along the curve, where \( |\mathbf{r}'(t)| \) is the derivative's magnitude. For many curves, such integrals might not yield simple expressions and may require numerical methods for evaluation. In this exercise, the integral numerically evaluates to:

• \( L \approx 8.0039 \).

This provides a quantifiable measurement of the segment of the curve between these bounds, rounding out the physics of motion described by the path.