Consider the following vector function.

**r**(*t*) = 6*t*, 

1

2

*t*2, *t*2

(a) Find the unit tangent vector **T**(*t*).


**T**(*t*)

 = 

(6√36+5t2​,t√36+5t2​,2t√36+5t2​)
 
 


(b) Use this formula to find the curvature: |T'(t)|/|r'(t)|

Let's break down the problem into two parts: finding the unit tangent vector \(\mathbf{T}(t)\), and then using that to calculate the curvature.

### Given Vector Function:

\[
\mathbf{r}(t) = (6t, \frac{1}{2}t^2, t^2)
\]

### Part (a): Find the Unit Tangent Vector \(\mathbf{T}(t)\)

1. **Find the derivative of \(\mathbf{r}(t)\):**

We need to differentiate \(\mathbf{r}(t)\) with respect to \(t\):

\[
\mathbf{r}'(t) = \left( \frac{d}{dt}(6t), \frac{d}{dt}\left(\frac{1}{2}t^2\right), \frac{d}{dt}(t^2) \right) = (6, t, 2t)
\]

2. **Find the magnitude of \(\mathbf{r}'(t)\):**

To get the magnitude of \(\mathbf{r}'(t)\), we use the formula:

\[
|\mathbf{r}'(t)| = \sqrt{6^2 + t^2 + (2t)^2} = \sqrt{36 + t^2 + 4t^2} = \sqrt{36 + 5t^2}
\]

3. **Find the unit tangent vector \(\mathbf{T}(t)\):**

The unit tangent vector is found by normalizing \(\mathbf{r}'(t)\):

\[
\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \left( \frac{6}{\sqrt{36 + 5t^2}}, \frac{t}{\sqrt{36 + 5t^2}}, \frac{2t}{\sqrt{36 + 5t^2}} \right)
\]

So, the unit tangent vector is:

\[
\mathbf{T}(t) = \left( \frac{6}{\sqrt{36 + 5t^2}}, \frac{t}{\sqrt{36 + 5t^2}}, \frac{2t}{\sqrt{36 + 5t^2}} \right)
\]

### Part (b): Use the formula to find the curvature \(\kappa(t)\)

The curvature \(\kappa(t)\) is given by:

\[
\kappa(t) = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}
\]

1. **Find \(\mathbf{T}'(t)\):**

To find \(\mathbf{T}'(t)\), differentiate each component of \(\mathbf{T}(t)\) with respect to \(t\).

- First component:

\[
\frac{d}{dt} \left( \frac{6}{\sqrt{36 + 5t^2}} \right) = 6 \cdot \left( \frac{-5t}{(36 + 5t^2)^{3/2}} \right) = \frac{-30t}{(36 + 5t^2)^{3/2}}
\]

- Second component:

\[
\frac{d}{dt} \left( \frac{t}{\sqrt{36 + 5t^2}} \right) = \frac{\sqrt{36 + 5t^2} - t \cdot \frac{5t}{\sqrt{36 + 5t^2}}}{(36 + 5t^2)} = \frac{36}{(36 + 5t^2)^{3/2}}
\]

- Third component:

\[
\frac{d}{dt} \left( \frac{2t}{\sqrt{36 + 5t^2}} \right) = \frac{2\sqrt{36 + 5t^2} - 2t \cdot \frac{5t}{\sqrt{36 + 5t^2}}}{(36 + 5t^2)} = \frac{72}{(36 + 5t^2)^{3/2}}
\]

2. **Find \(|\mathbf{T}'(t)|\):**

Now, we calculate the magnitude of \(\mathbf{T}'(t)\):

\[
|\mathbf{T}'(t)| = \sqrt{\left( \frac{-30t}{(36 + 5t^2)^{3/2}} \right)^2 + \left( \frac{36}{(36 + 5t^2)^{3/2}} \right)^2 + \left( \frac{72}{(36 + 5t^2)^{3/2}} \right)^2}
\]

Simplify the expression:

\[
|\mathbf{T}'(t)| = \frac{\sqrt{900t^2 + 1296 + 5184}}{(36 + 5t^2)^{3/2}} = \frac{\sqrt{1296 + 900t^2}}{(36 + 5t^2)^{3/2}}
\]

3. **Curvature:**

Finally, we plug in \(|\mathbf{T}'(t)|\) and \(|\mathbf{r}'(t)|\) into the formula for curvature:

\[
\kappa(t) = \frac{\frac{\sqrt{1296 + 900t^2}}{(36 + 5t^2)^{3/2}}}{\sqrt{36 + 5t^2}} = \frac{\sqrt{1296 + 900t^2}}{(36 + 5t^2)^2}
\]

This gives the curvature \(\kappa(t)\).

---

Would you like more details on the differentiation steps or simplifications? Here are five related questions to explore further:

1. How is the unit normal vector \(\mathbf{N}(t)\) related to the tangent vector \(\mathbf{T}(t)\)?
2. Can you find the torsion of the curve \(\mathbf{r}(t)\)?
3. What is the geometric interpretation of the curvature \(\kappa(t)\)?
4. How would the problem change if the vector function were in three-dimensional space?
5. What are some applications of curvature in physics and engineering?

**Tip:** Remember that curvature gives you the "bending" of a curve at a point, while the unit tangent vector indicates the direction of the curve's motion.

Consider the following vector function. r(t) = 6t, 1/2t^2, t^2 (a) Find the unit tangent vector T(t). (b) Use this formula to find the curvature: |T'(t)|/|r'(t)|.

Learn how to compute the unit tangent vector T(t) and curvature κ(t) for the vector function r(t) = (6t, 1/2t^2, t^2). This problem covers key concepts in vector calculus, including differentiation, unit tangent vectors, and curvature calculations. Suitable for undergraduate calculus students.

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