91Ó°ÊÓ

When dealing with motion in three-dimensional space, a Position Vector like \( \mathbf{r}(t) \) helps us describe the path of a point. The Position Vector is expressed in terms of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) along the x, y, and z axes, respectively.

In our example, the Position Vector is given by \( \mathbf{r}(t) = (-3 \sin t) \mathbf{i} + (-3 \sin t) \mathbf{j} + (1 \cos t) \mathbf{k} \). Each component gives us the location of the point along its respective axis as a function of \( t \), typically representing time.
This vector describes a curve in space by showing how the position of the point changes as \( t \) changes. Understanding the Position Vector is crucial because it forms the basis for further calculations such as velocity, acceleration, and curvature.

The First Derivative of a Position Vector is essentially its velocity vector. It shows how the position is changing with respect to time. By differentiating the Position Vector \( \mathbf{r}(t) \) with respect to \( t \), we get the First Derivative, \( \mathbf{r'}(t) \).

For our example:\[ \mathbf{r'}(t) = (-3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} - (1 \sin t) \mathbf{k} \]
This derivative is computed by differentiating each component of \( \mathbf{r}(t) \) independently. Differentiating \(-3 \sin t\) with respect to \( t \) gives \(-3 \cos t\), and similarly for the other components.

• Each component now represents the instantaneous rate of change of that specific coordinate.
• This vector is vital as it helps determine other properties of the curve, like acceleration and curvature.

The Cross Product of two vectors results in a third vector that is orthogonal to both of the original vectors. It provides insight into the orientation and twist of a curve in space.

In the case of curvature, we compute the Cross Product of the First Derivative \( \mathbf{r'}(t) \) and the Second Derivative \( \mathbf{r''}(t) \):
The formula can be written using a determinant expansion involving the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).
This can be shown as:

• Place vectors \( \mathbf{r'}(t) \) and \( \mathbf{r''}(t) \) as rows in a 3x3 matrix with the unit vectors as the first row.
• Compute the determinant to find the Cross Product vector.

In our case, the calculation involved results in:
\( \mathbf{r'}(t) \times \mathbf{r''}(t) = (3\cos^2 t - 9\sin^2 t)\mathbf{i} + (3\sin^2 t - 9\cos^2 t)\mathbf{j} + 9\sin t\cos t(\sin t + \cos t)\mathbf{k} \).
The resulting vector is used to calculate the curvature of the vector.

Once we have the Cross Product vector from the previous step, the next task in curvature calculation is to find the Magnitude. The Magnitude of a vector represents its length and is integral to determining the curvature accurately.

To calculate the Magnitude of a vector, say \( \mathbf{A} \), with components \( (a, b, c) \):

• Use the formula \( \|\mathbf{A}\| = \sqrt{a^2 + b^2 + c^2} \).

This process is applied to both the First Derivative \( \mathbf{r'}(t) \) and the Cross Product \( \mathbf{r'}(t) \times \mathbf{r''}(t) \).
For example:
\( \| \mathbf{r'}(t) \| = 3 \) is straightforward since it simplifies due to trigonometric identities.
Finally, the Magnitude of \( \mathbf{r'}(t) \times \mathbf{r''}(t) \) is slightly more complex, involving several squared terms, but follows the same principle. This Magnitude is crucial for finding the curvature using the formula:
\( \kappa(t) = \frac{ \| \mathbf{r'}(t) \times \mathbf{r''}(t) \| }{\| \mathbf{r'}(t) \| ^3} \).
Understanding how to calculate these Magnitudes is essential to analyzing the curve's behavior.