Consider the curve 

**r**(*t*) = 3 sin(*t*), 8*t*, 3 cos(*t*).

(a)

Find the unit tangent vector 

**T**(*t*).

**T**(*t*) = 
 
(b)

Find the unit normal vector 

**N**(*t*).

**N**(*t*) = 
 

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.

Consider the curve r(t) = 3 sin(t), 8t, 3 cos(t). (a) Find the unit tangent vector T(t). (b) Find the unit normal vector N(t).

This problem explores how to find the unit tangent vector T(t) and the unit normal vector N(t) for the parametric curve r(t) = <3 sin(t), 8t, 3 cos(t)>. The solution involves vector calculus concepts and detailed step-by-step derivations for university-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation