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 $\kappa(t)$.

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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.