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The unit tangent vector, often represented as \( \mathbf{T}(t) \), plays a critical role in understanding the direction of a space curve. It gives the direction in which the curve is heading at any point. To find the unit tangent vector, one must first determine the derivative of the position vector, \( \mathbf{r}(t) \). This derivative, \( \mathbf{r}'(t) \), represents the velocity of the curve. However, to turn this velocity vector into a unit tangent vector, you need to normalize it.

Normalization means adjusting the length of the vector to one unit without changing its direction. To achieve this, divide \( \mathbf{r}'(t) \) by its magnitude \( \| \mathbf{r}'(t) \| \). In our example, once you compute \( \mathbf{r}'(t) \), you find its magnitude is \( e^t \sqrt{2} \). This leads to the unit tangent vector:
\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{1}{\sqrt{2}} ((\cos t - \sin t) \mathbf{i} + (\sin t + \cos t) \mathbf{j}) \]

This provides a standardized way to describe the direction of motion along the space curve.

The unit normal vector, denoted \( \mathbf{N}(t) \), is a vector that is perpendicular to the tangent vector at any given point on the curve. It shows how the direction of the curve changes and helps in studying the behavior of curves through their bending.

To find the unit normal vector, begin with the derivative of the unit tangent vector \( \mathbf{T}'(t) \). This derivative gives insight into how \( \mathbf{T}(t) \) changes, highlighting the curvature of the curve. From our exercise, the differentiation yields:
\[ \mathbf{T}'(t) = \frac{1}{\sqrt{2}} ( -\sin t - \cos t ) \mathbf{i} + \frac{1}{\sqrt{2}} ( \cos t - \sin t ) \mathbf{j} \]

Next, find the magnitude \( \| \mathbf{T}'(t) \| \), which turns out to be 1 in this case. Normalize \( \mathbf{T}'(t) \) by dividing it by its magnitude to get \( \mathbf{N}(t) \):
\[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|} = \frac{-1}{\sqrt{2}} ( (\sin t + \cos t) \mathbf{i} + (\cos t - \sin t) \mathbf{j} ) \]

The unit normal vector is crucial while discussing concepts like curvature as it indicates how the direction of the curve changes.

Curvature \( \kappa(t) \) measures how sharply a curve bends. It quantifies the deviation of the curve from being a straight line. High curvature means the curve bends sharply, while low curvature indicates a gentler turn.

In mathematical terms, curvature is found by taking the magnitude of the derivative of the tangent vector, \( \| \mathbf{T}'(t) \| \), and dividing it by the magnitude of the derivative of the position vector, \( \| \mathbf{r}'(t) \| \):
\[ \kappa(t) = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t) \|} \]

This formula gives a consistent way to compute curvature. For our specific space curve:
\[ \kappa(t) = \frac{1}{e^t \sqrt{2}} \]

This result implies that as \( t \) increases, the curvature decreases, indicating the curve becomes less sharp or more straightened.
Understanding curvature is vital not just in mathematics, but in fields like physics and engineering, where it helps in analyzing paths, trajectories, and the shape of objects and structures.