The problem is asking for the arc length of the curve given by the vector function:

$\mathbf{r}(t) = \langle 3t^2, 2\sqrt{3}t, \ln t \rangle$ over the interval $1 \leq t \leq 6$.

Steps to Solve

1. Formula for Arc Length: The arc length $L$ of a curve defined by a vector function $\mathbf{r}(t)$ from $t = a$ to $t = b$ is given by: $L = \int_a^b \|\mathbf{r}'(t)\| \, dt$ where $\|\mathbf{r}'(t)\|$ is the magnitude of the derivative of $\mathbf{r}(t)$.

2. Find the Derivative: First, compute the derivative of $\mathbf{r}(t)$: $\mathbf{r}'(t) = \left\langle \frac{d}{dt}(3t^2), \frac{d}{dt}(2\sqrt{3}t), \frac{d}{dt}(\ln t) \right\rangle = \langle 6t, 2\sqrt{3}, \frac{1}{t} \rangle$

3. Magnitude of the Derivative: The magnitude of $\mathbf{r}'(t)$ is: $\|\mathbf{r}'(t)\| = \sqrt{(6t)^2 + (2\sqrt{3})^2 + \left(\frac{1}{t}\right)^2}$ Simplifying the terms: $\|\mathbf{r}'(t)\| = \sqrt{36t^2 + 12 + \frac{1}{t^2}}$

4. Set up the Integral: Now, set up the arc length integral: $L = \int_1^6 \sqrt{36t^2 + 12 + \frac{1}{t^2}} \, dt$

This integral is complex and generally would require numerical methods to evaluate. I will compute the result for you.The arc length of the curve $\mathbf{r}(t) = \langle 3t^2, 2\sqrt{3}t, \ln t \rangle$ from $t = 1$ to $t = 6$ is approximately:

$L \approx 106.79$

Let me know if you want more details or have any questions!

Here are 5 related questions you could explore:

1. How is the arc length formula derived from the vector calculus principles?
2. What are some common applications of arc length in physics or engineering?
3. How would you compute the arc length if the curve was in polar coordinates?
4. What changes if one of the vector function components were exponential or trigonometric?
5. How would the computation differ if we integrated over a different interval, say $2 \leq t \leq 5$?

Tip: Understanding how to compute derivatives and magnitudes of vector functions is crucial when dealing with curves in 3D space.