We are given the vector function $\mathbf{r}(t) = \langle 3\sin(t), 8t, 3\cos(t) \rangle$, and we need to find:

(a) The unit tangent vector $\mathbf{T}(t)$. (b) The unit normal vector $\mathbf{N}(t)$.

Part (a) - Finding the Unit Tangent Vector $\mathbf{T}(t)$

The unit tangent vector is given by: $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$ where $\mathbf{r}'(t)$ is the derivative of $\mathbf{r}(t)$ with respect to $t$, and $|\mathbf{r}'(t)|$ is its magnitude.

1. Find $\mathbf{r}'(t)$: $\mathbf{r}'(t) = \frac{d}{dt} \langle 3\sin(t), 8t, 3\cos(t) \rangle = \langle 3\cos(t), 8, -3\sin(t) \rangle$

2. Find $|\mathbf{r}'(t)|$ (the magnitude of $\mathbf{r}'(t)$): $|\mathbf{r}'(t)| = \sqrt{(3\cos(t))^2 + (8)^2 + (-3\sin(t))^2} = \sqrt{9\cos^2(t) + 64 + 9\sin^2(t)}$ Since $\cos^2(t) + \sin^2(t) = 1$, we have: $|\mathbf{r}'(t)| = \sqrt{9 + 64} = \sqrt{73}$

3. Find $\mathbf{T}(t)$: Now, the unit tangent vector is: $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \frac{1}{\sqrt{73}} \langle 3\cos(t), 8, -3\sin(t) \rangle$ Thus, the unit tangent vector is: $\mathbf{T}(t) = \left\langle \frac{3\cos(t)}{\sqrt{73}}, \frac{8}{\sqrt{73}}, \frac{-3\sin(t)}{\sqrt{73}} \right\rangle$

Part (b) - Finding the Unit Normal Vector $\mathbf{N}(t)$

The unit normal vector $\mathbf{N}(t)$ is given by: $\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}$ where $\mathbf{T}'(t)$ is the derivative of $\mathbf{T}(t)$, and $|\mathbf{T}'(t)|$ is its magnitude.

1. Find $\mathbf{T}'(t)$: Differentiate $\mathbf{T}(t)$ with respect to $t$: $\mathbf{T}'(t) = \frac{d}{dt} \left\langle \frac{3\cos(t)}{\sqrt{73}}, \frac{8}{\sqrt{73}}, \frac{-3\sin(t)}{\sqrt{73}} \right\rangle = \left\langle \frac{-3\sin(t)}{\sqrt{73}}, 0, \frac{-3\cos(t)}{\sqrt{73}} \right\rangle$

2. Find $|\mathbf{T}'(t)|$: $|\mathbf{T}'(t)| = \sqrt{\left( \frac{-3\sin(t)}{\sqrt{73}} \right)^2 + 0^2 + \left( \frac{-3\cos(t)}{\sqrt{73}} \right)^2}$ $|\mathbf{T}'(t)| = \sqrt{\frac{9\sin^2(t)}{73} + \frac{9\cos^2(t)}{73}} = \sqrt{\frac{9(\sin^2(t) + \cos^2(t))}{73}} = \sqrt{\frac{9}{73}} = \frac{3}{\sqrt{73}}$

3. Find $\mathbf{N}(t)$: Now, the unit normal vector is: $\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} = \frac{1}{\frac{3}{\sqrt{73}}} \left\langle \frac{-3\sin(t)}{\sqrt{73}}, 0, \frac{-3\cos(t)}{\sqrt{73}} \right\rangle$ $\mathbf{N}(t) = \left\langle -\sin(t), 0, -\cos(t) \right\rangle$

Thus, the unit normal vector is: $\mathbf{N}(t) = \langle -\sin(t), 0, -\cos(t) \rangle$

Summary:

• $\mathbf{T}(t) = \left\langle \frac{3\cos(t)}{\sqrt{73}}, \frac{8}{\sqrt{73}}, \frac{-3\sin(t)}{\sqrt{73}} \right\rangle$
• $\mathbf{N}(t) = \langle -\sin(t), 0, -\cos(t) \rangle$

Would you like more details or clarification on any part? Here are some related questions you might consider:

1. How do we find the binormal vector $\mathbf{B}(t)$?
2. What is the geometric interpretation of the tangent and normal vectors?
3. How do we find the curvature of the curve $\mathbf{r}(t)$?
4. What is the relationship between the tangent and normal vectors?
5. How can we generalize this approach to find vectors for any parametric curve?

Tip: The unit tangent vector always points in the direction of motion, while the unit normal vector points towards the center of curvature of the curve.