To compute the length of the curve $\mathbf{r}(t) = \langle 2t, \ln t, t^2 \rangle$ over the interval $1 \leq t \leq 8$, we can use the formula for the arc length of a parametric curve:

$L = \int_{a}^{b} \|\mathbf{r}'(t)\| \, dt$

where $\mathbf{r}'(t)$ is the derivative of the vector-valued function $\mathbf{r}(t)$, and $\|\mathbf{r}'(t)\|$ is the magnitude of the derivative.

Step 1: Compute $\mathbf{r}'(t)$

Given $\mathbf{r}(t) = \langle 2t, \ln t, t^2 \rangle$, we differentiate each component with respect to $t$:

$\mathbf{r}'(t) = \left\langle \frac{d}{dt}(2t), \frac{d}{dt}(\ln t), \frac{d}{dt}(t^2) \right\rangle = \langle 2, \frac{1}{t}, 2t \rangle$

Step 2: Compute the magnitude $\|\mathbf{r}'(t)\|$

The magnitude of $\mathbf{r}'(t)$ is:

$\|\mathbf{r}'(t)\| = \sqrt{(2)^2 + \left( \frac{1}{t} \right)^2 + (2t)^2}$ $\|\mathbf{r}'(t)\| = \sqrt{4 + \frac{1}{t^2} + 4t^2}$

Step 3: Set up the integral for the arc length

The arc length $L$ is:

$L = \int_{1}^{8} \sqrt{4 + \frac{1}{t^2} + 4t^2} \, dt$

Step 4: Solve the integral

This is a non-trivial integral, so let's evaluate it numerically.

Let me compute that.The length of the curve $\mathbf{r}(t) = \langle 2t, \ln t, t^2 \rangle$ over the interval $1 \leq t \leq 8$ is approximately $65.08$.

Would you like further details or have any questions about this solution?

Here are some related questions you might find helpful:

1. How do you compute the derivative of a parametric curve in general?
2. What is the geometric significance of arc length?
3. Can arc length be computed without finding the explicit derivative of the vector function?
4. How can you numerically approximate integrals if they are too complex to solve analytically?
5. What applications in physics use the concept of arc length?

Tip: When computing arc lengths, be mindful of the interval over which you're integrating, as different intervals can dramatically change the result.