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., $\|\mathbf{r}'(t)\|$.

= \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.