We are tasked with finding the unit tangent vector to the given vector-valued function:

$\mathbf{r}(t) = (-2t \cos t) \mathbf{i} + (2t \sin t) \mathbf{j} + \left( \frac{4\sqrt{2}}{3} t^{3/2} \right) \mathbf{k}, \quad 0 \leq t \leq \pi.$

Steps to solve:

1. Compute the derivative of $\mathbf{r}(t)$ to find the tangent vector $\mathbf{r}'(t)$.

2. Normalize the tangent vector $\mathbf{r}'(t)$ to obtain the unit tangent vector $\mathbf{T}(t)$.

Step 1: Compute $\mathbf{r}'(t)$

The derivative of $\mathbf{r}(t)$ with respect to $t$ is calculated component-wise.

• For the i-component: $\frac{d}{dt} \left( -2t \cos t \right) = -2 \cos t + 2t \sin t.$

• For the j-component: $\frac{d}{dt} \left( 2t \sin t \right) = 2 \sin t + 2t \cos t.$

• For the k-component: $\frac{d}{dt} \left( \frac{4\sqrt{2}}{3} t^{3/2} \right) = \frac{4\sqrt{2}}{2} \cdot t^{1/2} = \frac{2\sqrt{2}}{3} t^{1/2}.$

Thus, the tangent vector $\mathbf{r}'(t)$ is: $\mathbf{r}'(t) = \left( -2 \cos t + 2t \sin t \right) \mathbf{i} + \left( 2 \sin t + 2t \cos t \right) \mathbf{j} + \frac{2\sqrt{2}}{3} t^{1/2} \mathbf{k}.$

Step 2: Normalize the tangent vector

To obtain the unit tangent vector, we need to compute the magnitude of $\mathbf{r}'(t)$:

$|\mathbf{r}'(t)| = \sqrt{\left( -2 \cos t + 2t \sin t \right)^2 + \left( 2 \sin t + 2t \cos t \right)^2 + \left( \frac{2\sqrt{2}}{3} t^{1/2} \right)^2}.$

Once we have the magnitude, the unit tangent vector is given by:

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

Let me compute the exact expression for the magnitude and simplify the unit tangent vector.The derivative of the vector function $\mathbf{r}(t)$, $\mathbf{r}'(t)$, is:

$\mathbf{r}'(t) = \left( 2t \sin t - 2 \cos t \right) \mathbf{i} + \left( 2t \cos t + 2 \sin t \right) \mathbf{j} + 2\sqrt{2} t^{1/2} \mathbf{k}.$

The magnitude of the tangent vector $\mathbf{r}'(t)$ is:

$|\mathbf{r}'(t)| = 2\sqrt{2} \cdot \sqrt{0.5 t^2 + t + 0.5}.$

Step 3: Unit Tangent Vector

The unit tangent vector is found by dividing each component of $\mathbf{r}'(t)$ by the magnitude:

$\mathbf{T}(t) = \frac{1}{2\sqrt{2} \cdot \sqrt{0.5 t^2 + t + 0.5}} \left[ \left( 2t \sin t - 2 \cos t \right) \mathbf{i} + \left( 2t \cos t + 2 \sin t \right) \mathbf{j} + 2\sqrt{2} t^{1/2} \mathbf{k} \right].$

Would you like me to simplify the expression further or compute the length of the curve portion over the given interval?

Here are some related questions:

1. How is the arc length of a curve calculated using integrals?
2. What is the significance of a unit tangent vector in vector calculus?
3. How can you interpret the components of the tangent vector geometrically?
4. How does the tangent vector change as $t$ varies from 0 to $\pi$?
5. What are other types of vectors associated with curves, like normal and binormal vectors?

Tip: The unit tangent vector always has a magnitude of 1, ensuring it only indicates direction without affecting the vector's length.