To solve the problem of finding the unit tangent vector $\mathbf{T}(t)$ at the given point $t = 4$, we proceed as follows:

1. Compute the derivative of $\mathbf{r}(t)$, denoted as $\mathbf{r}'(t)$, which gives the tangent vector.

The function $\mathbf{r}(t)$ is defined as: $\mathbf{r}(t) = \left( t^2 - 4t, 1 + 5t, \frac{t^3}{3} + \frac{t^2}{2} \right)$

a) First component: $t^2 - 4t$

$\frac{d}{dt}(t^2 - 4t) = 2t - 4$

b) Second component: $1 + 5t$

$\frac{d}{dt}(1 + 5t) = 5$

c) Third component: $\frac{t^3}{3} + \frac{t^2}{2}$

$\frac{d}{dt} \left( \frac{t^3}{3} + \frac{t^2}{2} \right) = t^2 + t$

Thus, the derivative is: $\mathbf{r}'(t) = \left( 2t - 4, 5, t^2 + t \right)$

2. Evaluate $\mathbf{r}'(t)$ at $t = 4$

Substitute $t = 4$ into $\mathbf{r}'(t)$: $\mathbf{r}'(4) = \left( 2(4) - 4, 5, 4^2 + 4 \right) = (8 - 4, 5, 16 + 4) = (4, 5, 20)$

3. Compute the magnitude of $\mathbf{r}'(t)$ at $t = 4$

The magnitude of the vector $\mathbf{r}'(4) = (4, 5, 20)$ is: $\|\mathbf{r}'(4)\| = \sqrt{4^2 + 5^2 + 20^2} = \sqrt{16 + 25 + 400} = \sqrt{441} = 21$

4. Find the unit tangent vector $\mathbf{T}(t)$ at $t = 4$

The unit tangent vector is given by: $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}$ At $t = 4$: $\mathbf{T}(4) = \frac{1}{21}(4, 5, 20) = \left( \frac{4}{21}, \frac{5}{21}, \frac{20}{21} \right)$

Thus, the unit tangent vector at $t = 4$ is: $\mathbf{T}(4) = \left( \frac{4}{21}, \frac{5}{21}, \frac{20}{21} \right)$

Let me know if you'd like more details or if you have any questions!

Here are some related questions you might want to explore:

1. How do you find the tangent vector for a given vector function at any point?
2. What is the significance of a unit tangent vector?
3. How does one compute curvature using a tangent vector?
4. What is the difference between a tangent vector and a normal vector?
5. How is the magnitude of a vector related to its unit vector?

Tip: The unit tangent vector always points in the direction of motion but with a magnitude of 1.