To find the unit tangent vector $\mathbf{T}(t)$ for the vector function $\mathbf{r}(t) = 4t \mathbf{i} + 8t^2 \mathbf{j} + 2t \mathbf{k}$, follow these steps:

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

The derivative $\mathbf{r}'(t)$ is the velocity vector, and it is obtained by differentiating $\mathbf{r}(t)$ component-wise:

$\mathbf{r}'(t) = \frac{d}{dt} \left( 4t \mathbf{i} + 8t^2 \mathbf{j} + 2t \mathbf{k} \right)$

$\mathbf{r}'(t) = 4 \mathbf{i} + 16t \mathbf{j} + 2 \mathbf{k}$

Step 2: Find the magnitude $|\mathbf{r}'(t)|$

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

$|\mathbf{r}'(t)| = \sqrt{(4)^2 + (16t)^2 + (2)^2}$

$|\mathbf{r}'(t)| = \sqrt{16 + 256t^2 + 4}$

$|\mathbf{r}'(t)| = \sqrt{20 + 256t^2}$

Step 3: Compute the unit tangent vector $\mathbf{T}(t)$

The unit tangent vector $\mathbf{T}(t)$ is given by:

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

Substitute $\mathbf{r}'(t) = 4 \mathbf{i} + 16t \mathbf{j} + 2 \mathbf{k}$ and $|\mathbf{r}'(t)| = \sqrt{20 + 256t^2}$:

$\mathbf{T}(t) = \frac{4 \mathbf{i} + 16t \mathbf{j} + 2 \mathbf{k}}{\sqrt{20 + 256t^2}}$

This is the unit tangent vector.

Would you like more details, or do you have any questions?

Here are 5 related questions to explore further:

1. How do you find the unit normal vector $\mathbf{N}(t)$?
2. What is the formula for the curvature $\kappa(t)$ of a curve?
3. How do you compute the torsion of a space curve?
4. What are the differences between tangent and normal vectors?
5. How does the concept of arc length relate to unit tangent vectors?

Tip: Always simplify your vector expressions as much as possible to avoid computational errors in later steps.