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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.

Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve: r(t) = (-2t cos t)i + (2t sin t)j + (4√2 / 3)t^(3/2) k for 0 ≤ t ≤ π.

Learn how to find the unit tangent vector for the vector-valued function r(t) = (-2t cos t)i + (2t sin t)j + (4√2 / 3)t^(3/2) k, and calculate the length of the curve from t = 0 to t = π. This solution involves vector calculus, differentiation, and tangent vectors.

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