To find the unit tangent vector $\mathbf{T}(t)$ at $t = 1$ for the vector function $\mathbf{r}(t) = 4t \, \hat{i} + 2t^2 \, \hat{j} + 4t \, \hat{k}$, follow these steps:

1. Compute the derivative $\mathbf{r}'(t)$ (velocity vector).

Take the derivative of each component of $\mathbf{r}(t)$:

$\mathbf{r}'(t) = \frac{d}{dt}(4t \, \hat{i} + 2t^2 \, \hat{j} + 4t \, \hat{k})$

$\mathbf{r}'(t) = 4 \, \hat{i} + 4t \, \hat{j} + 4 \, \hat{k}$

2. Evaluate $\mathbf{r}'(t)$ at $t = 1$:

$\mathbf{r}'(1) = 4 \, \hat{i} + 4(1) \, \hat{j} + 4 \, \hat{k}$ $\mathbf{r}'(1) = 4 \, \hat{i} + 4 \, \hat{j} + 4 \, \hat{k}$ So, the velocity vector at $t = 1$ is: $\mathbf{r}'(1) = \langle 4, 4, 4 \rangle$

3. Compute the magnitude of $\mathbf{r}'(1)$:

$|\mathbf{r}'(1)| = \sqrt{4^2 + 4^2 + 4^2} = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3}$

4. Compute the unit tangent vector $\mathbf{T}(t)$:

The unit tangent vector is given by:

$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$

At $t = 1$, this becomes:

$\mathbf{T}(1) = \frac{\langle 4, 4, 4 \rangle}{4\sqrt{3}} = \left\langle \frac{4}{4\sqrt{3}}, \frac{4}{4\sqrt{3}}, \frac{4}{4\sqrt{3}} \right\rangle = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle$

Thus, the unit tangent vector at $t = 1$ is:

$\mathbf{T}(1) = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle$

Would you like more details or further explanations?

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Here are 5 related questions for further exploration:

1. How do you find the curvature $\kappa(t)$ of a vector function $\mathbf{r}(t)$?
2. What is the relationship between the unit tangent vector $\mathbf{T}(t)$ and the unit normal vector $\mathbf{N}(t)$?
3. How can you compute the arc length of a curve given by $\mathbf{r}(t)$?
4. What is the significance of the binormal vector $\mathbf{B}(t)$ in the Frenet-Serret frame?
5. How do you find the acceleration vector from the position vector $\mathbf{r}(t)$?

Tip: The unit tangent vector is useful for understanding the direction of motion along a curve, and it is always tangent to the curve at the point of interest.