The arclength $L$ of a curve given by the vector function $\mathbf{r}(t)$ over the interval $[a, b]$ is computed using the following formula:

$L = \int_a^b \left| \mathbf{r}'(t) \right| \, dt$

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

Given the vector function $\mathbf{r}(t) = \left\langle 6\sqrt{2}t, e^{6t}, e^{-6t} \right\rangle$, we first find its derivative:

$\mathbf{r}'(t) = \left\langle \frac{d}{dt}(6\sqrt{2}t), \frac{d}{dt}(e^{6t}), \frac{d}{dt}(e^{-6t}) \right\rangle$

Computing each component:

1. $\frac{d}{dt}(6\sqrt{2}t) = 6\sqrt{2}$
2. $\frac{d}{dt}(e^{6t}) = 6e^{6t}$
3. $\frac{d}{dt}(e^{-6t}) = -6e^{-6t}$

Thus, the derivative is:

$\mathbf{r}'(t) = \left\langle 6\sqrt{2}, 6e^{6t}, -6e^{-6t} \right\rangle$

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

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

$\left| \mathbf{r}'(t) \right| = \sqrt{\left( 6\sqrt{2} \right)^2 + \left( 6e^{6t} \right)^2 + \left( -6e^{-6t} \right)^2}$

Simplifying each term:

1. $\left( 6\sqrt{2} \right)^2 = 72$
2. $\left( 6e^{6t} \right)^2 = 36e^{12t}$
3. $\left( -6e^{-6t} \right)^2 = 36e^{-12t}$

Thus:

$\left| \mathbf{r}'(t) \right| = \sqrt{72 + 36e^{12t} + 36e^{-12t}}$

Using the identity $e^{12t} + e^{-12t} = 2\cosh(12t)$, we get:

$\left| \mathbf{r}'(t) \right| = \sqrt{72 + 72\cosh(12t)} = 6\sqrt{2(1 + \cosh(12t))}$

Step 3: Set up the integral

Now, we integrate the magnitude over the interval $[0, 1]$:

$L = \int_0^1 6\sqrt{2(1 + \cosh(12t))} \, dt$

This integral does not have an elementary antiderivative, so it can be evaluated numerically.

Step 4: Numerical Evaluation

To compute the arclength, I'll now calculate the value of the integral.

Let's do that:The arclength of the curve $\mathbf{r}(t) = \left\langle 6\sqrt{2}t, e^{6t}, e^{-6t} \right\rangle$ over the interval $0 \leq t \leq 1$ is approximately $403.43$.

Would you like more details on any part of the process or have any questions?

Here are 5 related questions you might find interesting:

1. How is the arclength formula derived for vector functions?
2. Can you compute the arclength of a circular arc given in parametric form?
3. How does the hyperbolic cosine function $\cosh(t)$ arise in this problem?
4. What are some alternative methods for numerically solving integrals?
5. How does the arclength differ for curves in 2D versus 3D?

Tip: Always verify whether the integral you are dealing with can be solved analytically or if it requires numerical methods.