Find 
T hat(t), N hat(t), and B hat(t)
 for 
r(t) = 

2 cos(t) + 2t sin(t), 2 sin(t) − 2t cos(t), 4

.
 (Your instructors prefer angle bracket notation < > for vectors.)

We are tasked with finding the unit tangent vector \(\hat{T}(t)\), the unit normal vector \(\hat{N}(t)\), and the binormal vector \(\hat{B}(t)\) for the vector function:

\[
\mathbf{r}(t) = \langle 2 \cos(t) + 2t \sin(t), 2 \sin(t) - 2t \cos(t), 4 \rangle
\]

### Step 1: Find the Tangent Vector \(\mathbf{T}(t)\)
To find \(\mathbf{T}(t)\), we first need the derivative of \(\mathbf{r}(t)\), i.e., \(\mathbf{r}'(t)\).

\[
\mathbf{r}'(t) = \frac{d}{dt} \langle 2 \cos(t) + 2t \sin(t), 2 \sin(t) - 2t \cos(t), 4 \rangle
\]

#### Derivatives of each component:
1. \( \frac{d}{dt}[2 \cos(t) + 2t \sin(t)] = -2 \sin(t) + 2 \sin(t) + 2t \cos(t) = 2t \cos(t) \)
2. \( \frac{d}{dt}[2 \sin(t) - 2t \cos(t)] = 2 \cos(t) - 2(\cos(t) - t \sin(t)) = -2t \sin(t) \)
3. \( \frac{d}{dt}[4] = 0 \)

Thus, 
\[
\mathbf{r}'(t) = \langle 2t \cos(t), -2t \sin(t), 0 \rangle
\]

### Step 2: Find the Magnitude of \(\mathbf{r}'(t)\)
Next, we compute the magnitude of \(\mathbf{r}'(t)\), i.e., \(\|\mathbf{r}'(t)\|\).

\[
\|\mathbf{r}'(t)\| = \sqrt{(2t \cos(t))^2 + (-2t \sin(t))^2 + 0^2} 
= \sqrt{4t^2 \cos^2(t) + 4t^2 \sin^2(t)} 
= \sqrt{4t^2(\cos^2(t) + \sin^2(t))} 
= \sqrt{4t^2} = 2|t|
\]

### Step 3: Find the Unit Tangent Vector \(\hat{T}(t)\)
The unit tangent vector \(\hat{T}(t)\) is the normalized derivative of \(\mathbf{r}(t)\):

\[
\hat{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{1}{2|t|} \langle 2t \cos(t), -2t \sin(t), 0 \rangle = \langle \cos(t), -\sin(t), 0 \rangle
\]

### Step 4: Find the Unit Normal Vector \(\hat{N}(t)\)
To find the unit normal vector \(\hat{N}(t)\), we first differentiate \(\hat{T}(t)\):

\[
\hat{T}'(t) = \frac{d}{dt} \langle \cos(t), -\sin(t), 0 \rangle = \langle -\sin(t), -\cos(t), 0 \rangle
\]

Next, find the magnitude of \(\hat{T}'(t)\):

\[
\|\hat{T}'(t)\| = \sqrt{(-\sin(t))^2 + (-\cos(t))^2 + 0^2} = \sqrt{\sin^2(t) + \cos^2(t)} = 1
\]

Thus, the unit normal vector is:

\[
\hat{N}(t) = \frac{\hat{T}'(t)}{\|\hat{T}'(t)\|} = \langle -\sin(t), -\cos(t), 0 \rangle
\]

### Step 5: Find the Binormal Vector \(\hat{B}(t)\)
The binormal vector \(\hat{B}(t)\) is the cross product of \(\hat{T}(t)\) and \(\hat{N}(t)\):

\[
\hat{B}(t) = \hat{T}(t) \times \hat{N}(t)
\]

Now, compute the cross product:

\[
\hat{T}(t) = \langle \cos(t), -\sin(t), 0 \rangle, \quad \hat{N}(t) = \langle -\sin(t), -\cos(t), 0 \rangle
\]

\[
\hat{B}(t) = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\cos(t) & -\sin(t) & 0 \\
-\sin(t) & -\cos(t) & 0
\end{vmatrix} = \langle 0, 0, -\cos(t)(-\cos(t)) - (-\sin(t))(-\sin(t)) \rangle = \langle 0, 0, 1 \rangle
\]

Thus, the binormal vector is:

\[
\hat{B}(t) = \langle 0, 0, 1 \rangle
\]

### Final Results:
1. \(\hat{T}(t) = \langle \cos(t), -\sin(t), 0 \rangle\)
2. \(\hat{N}(t) = \langle -\sin(t), -\cos(t), 0 \rangle\)
3. \(\hat{B}(t) = \langle 0, 0, 1 \rangle\)

Would you like further clarification on any step, or a breakdown of any concept?

---

Here are five related questions:
1. How is the unit tangent vector \(\hat{T}(t)\) derived from a vector function?
2. What is the significance of the binormal vector \(\hat{B}(t)\) in the context of a curve?
3. How do you compute the curvature of a curve given \(\hat{T}(t)\) and \(\hat{N}(t)\)?
4. Can you explain how the Frenet-Serret formulas relate to \(\hat{T}(t)\), \(\hat{N}(t)\), and \(\hat{B}(t)\)?
5. How would the vectors change if the third component of \(\mathbf{r}(t)\) were not constant?

**Tip:** The binormal vector \(\hat{B}(t)\) always points perpendicular to both the tangent and normal vectors, defining the "twist" of the curve in space.

Find T hat(t), N hat(t), and B hat(t) for r(t) = 2 cos(t) + 2t sin(t), 2 sin(t) − 2t cos(t), 4. (Your instructors prefer angle bracket notation < > for vectors.)

This problem involves finding the unit tangent vector T̂(t), unit normal vector N̂(t), and binormal vector B̂(t) for the parametric curve r(t) = 2 cos(t) + 2t sin(t), 2 sin(t) − 2t cos(t), 4. The solution applies vector calculus concepts and Frenet-Serret formulas, suitable for students in advanced calculus courses.

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