91Ó°ÊÓ

The unit tangent vector helps us understand the direction in which a curve is heading at any given point. It can be thought of as a "directional arrow" for the curve. To find it, we start by taking the derivative of the position vector \( \mathbf{r}(t) \) to get \( \mathbf{r}'(t) \). This tells us the velocity or instant rate of change of the curve. However, \( \mathbf{r}'(t) \) may not have a length of one, which is necessary for a unit vector.
To make it a unit vector, we divide \( \mathbf{r}'(t) \) by its magnitude, \( \| \mathbf{r}'(t) \| \). This results in the unit tangent vector, symbolized as \( \mathbf{T}(t) \). It is defined by:

• \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \)

In our specific problem, \( \| \mathbf{r}'(t) \| \) was calculated to be 5, leading to the unit tangent vector \( \mathbf{T}(t) = \frac{(3 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j} + 4 \mathbf{k}}{5} \). This vector gives us insight into how the curve is progressing through space.

The principal normal vector, \( \mathbf{N}(t) \), provides information on the curve's "bending" direction. While the unit tangent vector points in the direction of motion, the principal normal vector points towards the center of the curve's osculating circle, which can be thought of as a best-fit circle that describes the curve's bending.
To find \( \mathbf{N}(t) \), we first differentiate the unit tangent vector \( \mathbf{T}(t) \), resulting in \( \mathbf{T}'(t) \). This differentiated vector isn't automatically a unit vector. We then normalize \( \mathbf{T}'(t) \) to get \( \mathbf{N}(t) \). The calculation is:

• \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|} \)

For our exercise, this led to \( \mathbf{N}(t) = -\cos t \mathbf{i} - \sin t \mathbf{j} \). This vector lies perpendicular to \( \mathbf{T}(t) \) and plays a crucial role in understanding the curve's concavity.

Torsion, denoted \( \tau \), measures how a space curve twists out of the plane of curvature. While curvature gives us insight into how sharply a curve bends, torsion tells us about its "3D twistage." If a curve lies completely in a plane, its torsion will be zero.
To calculate torsion, we use the formula:

• \( \tau = \frac{-(\mathbf{B}(t) \cdot \mathbf{r}''(t))}{\| \mathbf{r}'(t) \|^2} \)

Where \( \mathbf{B}(t) \) is the binormal vector, another vector essential in the TNB (tangent-normal-binormal) frame. This formula derives torsion from the relationship between the binormal vector and the second derivative of the position vector.
For the exercise at hand, torsion was found to be \( \frac{12}{25} \). This value shows that the given curve has a 3D twist, offering a deeper view into the path's spatial dynamics.

Curvature is an important concept that quantifies how much a curve bends at any point along its path. A larger curvature indicates a tighter turn, while smaller curvature suggests a gentler bend. It is denoted by \( \kappa \) (kappa).
Calculating curvature involves the derivative of the unit tangent vector. Specifically, it is the magnitude of \( \mathbf{T}'(t) \) divided by the magnitude of \( \mathbf{r}'(t) \):

• \( \kappa = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t) \|} \)

This formula gives insight into the rate of change of the tangent vector's direction. A tight curve results in a larger \( \kappa \), indicating significant directional changes.
Understanding curvature alongside torsion provides a comprehensive view of a curve's geometry, both in terms of bending and how it spreads out into three dimensions.