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The concept of the unit tangent vector, represented as \( \mathbf{T}(t) \), is fundamental in understanding the direction of a curve at any given point. It is derived by normalizing the velocity vector. This process involves scaling the velocity vector such that its magnitude becomes 1, thereby making it a "unit" vector.To find the unit tangent vector, first compute the velocity vector \( \mathbf{v}(t) \) by differentiating the position vector. Then, calculate the magnitude of this velocity vector.The unit tangent vector is given by:\[\mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}\]This formula ensures that the resulting vector has a length of 1. It maintains the direction of \( \mathbf{v}(t) \) while normalizing its magnitude.In the provided example, the unit tangent vector at \( t = 0 \) is computed as:\[\mathbf{T}(0) = \left\langle \frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{6} \right\rangle\]This vector points in the same direction as the velocity vector at \( t = 0 \), indicating the direction of the curve while remaining of unit length.

The velocity vector is a vector tangent to the path of a moving object at any instant. It provides rich information about the motion, including direction and rate of change. For a position vector function \( \mathbf{r}(t) \),the velocity vector \( \mathbf{v}(t) \) can be found by differentiating:\[\mathbf{v}(t) = \mathbf{r}'(t)\]This computation involves taking the derivative of each component of \( \mathbf{r}(t) \).In the given exercise, the position vector is \( \mathbf{r}(t) = \langle 2e^t, e^t \cos t, e^t \sin t \rangle \),and its derivative provides the velocity vector:\[\mathbf{v}(t) = \langle 2e^t, e^t(\cos t - \sin t), e^t(\sin t + \cos t) \rangle\]This vector dictates the trajectory's direction and speed at every point \( t \). It illustrates how our object travels through space, with its components expressing changes along specific axes.At \( t = 0 \), the velocity vector computes to \( \langle 2, 1, 1 \rangle \),indicating the object's movement direction at that specific moment in time.

The magnitude of a velocity vector is crucial as it depicts the speed of an object at a specific point. For a vector \( \mathbf{v} = \langle a, b, c \rangle \), the magnitude is computed by:\[\|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2}\]This measurement is scalar, meaning it represents size or length without direction, providing a pure numerical value reflective of speed.From the example, the velocity vector at \( t = 0 \) is \( \langle 2, 1, 1 \rangle \). The magnitude at this point is:\[\|\mathbf{v}(0)\| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}\]This result signifies the object's speed magnitude at \( t = 0 \).The significance of understanding the magnitude lies in its utility for normalizing vectors and calculating arc lengths, which helps describe curves' geometric properties.