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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.

Find the unit tangent vector T(t) at the point with the given value of the parameter t, where r(t) = (t^2 - 4t, 1 + 5t, (1/3)t^3 + (1/2)t^2) and t = 4.

Learn how to find the unit tangent vector of a vector function r(t) = (t^2 - 4t, 1 + 5t, (1/3)t^3 + (1/2)t^2) at t = 4. This solution includes step-by-step differentiation, magnitude calculation, and unit vector determination, suitable for Grades 11-12 students.

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