To solve the given problem, we need to:

1. Find the unit tangent vector, $\mathbf{T}(t)$.
2. Find the length of the curve over the given interval, $0 \leq t \leq \pi$.

The given vector function is:

$\mathbf{r}(t) = 4\cos(t) \, \mathbf{i} + 4\sin(t) \, \mathbf{j} + \sqrt{2} \, t \, \mathbf{k}.$

Step 1: Find the unit tangent vector $\mathbf{T}(t)$

1.1: Compute $\mathbf{r}'(t)$, the derivative of $\mathbf{r}(t)$

The derivative of $\mathbf{r}(t)$ is calculated component by component:

$\mathbf{r}'(t) = \frac{d}{dt} \big(4\cos(t)\big) \, \mathbf{i} + \frac{d}{dt} \big(4\sin(t)\big) \, \mathbf{j} + \frac{d}{dt} \big(\sqrt{2}t\big) \, \mathbf{k}.$

$\mathbf{r}'(t) = -4\sin(t) \, \mathbf{i} + 4\cos(t) \, \mathbf{j} + \sqrt{2} \, \mathbf{k}.$

1.2: Compute the magnitude of $\mathbf{r}'(t)$, $\|\mathbf{r}'(t)\|$

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

$\|\mathbf{r}'(t)\| = \sqrt{\big(-4\sin(t)\big)^2 + \big(4\cos(t)\big)^2 + \big(\sqrt{2}\big)^2}.$

Simplify the terms:

$\|\mathbf{r}'(t)\| = \sqrt{16\sin^2(t) + 16\cos^2(t) + 2}.$

Using the Pythagorean identity $\sin^2(t) + \cos^2(t) = 1$, we get:

$\|\mathbf{r}'(t)\| = \sqrt{16(1) + 2} = \sqrt{18} = 3\sqrt{2}.$

1.3: Find the unit tangent vector $\mathbf{T}(t)$

The unit tangent vector is:

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

Substitute $\mathbf{r}'(t) = -4\sin(t) \, \mathbf{i} + 4\cos(t) \, \mathbf{j} + \sqrt{2} \, \mathbf{k}$ and $\|\mathbf{r}'(t)\| = 3\sqrt{2}$:

$\mathbf{T}(t) = \frac{-4\sin(t)}{3\sqrt{2}} \, \mathbf{i} + \frac{4\cos(t)}{3\sqrt{2}} \, \mathbf{j} + \frac{\sqrt{2}}{3\sqrt{2}} \, \mathbf{k}.$

Simplify:

$\mathbf{T}(t) = \frac{-2\sin(t)}{3\sqrt{2}} \, \mathbf{i} + \frac{2\cos(t)}{3\sqrt{2}} \, \mathbf{j} + \frac{1}{3} \, \mathbf{k}.$

Step 2: Find the length of the curve

The formula for the arc length is:

$L = \int_a^b \|\mathbf{r}'(t)\| \, dt.$

Here, $\|\mathbf{r}'(t)\| = 3\sqrt{2}$ (calculated earlier), and the interval is $0 \leq t \leq \pi$. Thus:

$L = \int_0^\pi 3\sqrt{2} \, dt.$

Since $3\sqrt{2}$ is a constant:

$L = 3\sqrt{2} \int_0^\pi 1 \, dt.$

$L = 3\sqrt{2} \big[t\big]_0^\pi = 3\sqrt{2} \big(\pi - 0\big) = 3\sqrt{2} \pi.$

Final Answers:

1. The unit tangent vector is:

$\mathbf{T}(t) = \frac{-2\sin(t)}{3\sqrt{2}} \, \mathbf{i} + \frac{2\cos(t)}{3\sqrt{2}} \, \mathbf{j} + \frac{1}{3} \, \mathbf{k}.$

2. The length of the curve is:

$L = 3\sqrt{2} \pi.$

Would you like further clarifications or a breakdown of any step?
Here are five related questions you might explore:

1. How is the arc length formula derived?
2. What is the significance of a unit tangent vector?
3. How would the solution change if the interval were different (e.g., $[0, 2\pi]$)?
4. What is the geometric interpretation of the curve $\mathbf{r}(t)$?
5. How can curvature or acceleration be computed for this curve?

Tip: Always double-check derivatives and magnitudes, as errors can propagate in multiple steps.