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The derivative of a vector function is an essential concept in understanding how a vector changes along a curve. When you have a vector function like \( \mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle \), you need to determine how each of its components changes with respect to the variable \( t \). Differentiation is done separately for each component of the vector. So, you differentiate \( 2 \sin t \), \( 5t \), and \( 2 \cos t \) with respect to \( t \), resulting in their respective derivatives. - \( \frac{d}{dt}(2 \sin t) = 2 \cos t \) shows how the first component oscillates due to the sine function.- \( \frac{d}{dt}(5t) = 5 \) indicates linear growth along the y-axis.- \( \frac{d}{dt}(2 \cos t) = -2 \sin t \) reflects the sinusoidal nature due to the cosine function.Together, these create the derivative vector \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \), which represents the tangent's direction and magnitude at each point \( t \) on the curve. This vector serves as a foundational element in calculating a unit tangent vector.

Calculating the magnitude of a vector provides the "length" or "size" of that vector in space. To find the magnitude of the derivative vector \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \), we use the formula for the magnitude in three-dimensional space: \[ ||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2} \]Substituting in the components of our vector:\[ ||\mathbf{r}'(t)|| = \sqrt{(2 \cos t)^2 + 5^2 + (-2 \sin t)^2} \]This simplifies to:\[ \sqrt{4 \cos^2 t + 25 + 4 \sin^2 t} \]Using the Pythagorean identity \( \cos^2 t + \sin^2 t = 1 \), it further reduces to:\[ \sqrt{4 \cdot 1 + 25} = \sqrt{29} \]Thus, the magnitude of \( \mathbf{r}'(t) \) is a constant \( \sqrt{29} \), which indicates that the derivative vector's length doesn't change as \( t \) varies. This constant magnitude is crucial when normalizing the vector to produce a unit tangent vector.

Normalization is the process of converting a vector into a unit vector, which has a magnitude of one but points in the same direction as the original vector. To normalize a vector, you divide each of its components by the vector's magnitude.For the previously derived vector \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \) with magnitude \( \sqrt{29} \), normalization involves:\[ \frac{\langle 2 \cos t, 5, -2 \sin t \rangle}{\sqrt{29}} \]Breaking it down component-wise, we get:

• \( \frac{2 \cos t}{\sqrt{29}} \)
• \( \frac{5}{\sqrt{29}} \)
• \( \frac{-2 \sin t}{\sqrt{29}} \)

The resulting vector \( \mathbf{T}(t) = \left \langle \frac{2 \cos t}{\sqrt{29}}, \frac{5}{\sqrt{29}}, \frac{-2 \sin t}{\sqrt{29}} \right \rangle \) is the unit tangent vector. It maintains the direction of the original derivative vector but has a magnitude of one, making it easier to use for calculations regarding the direction of a curve in space.