The length of a curve given by a vector function $\mathbf{r}(t)$ from $t = a$ to $t = b$ is calculated using the formula:

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

Given the vector function:

$\mathbf{r}(t) = \sqrt{2} \, t \, \mathbf{i} + e^t \, \mathbf{j} + e^{-t} \, \mathbf{k}$

1. Step 1: Differentiate the components of $\mathbf{r}(t)$

   The derivative $\mathbf{r}'(t)$ is found by differentiating each component with respect to $t$.

   $\mathbf{r}'(t) = \frac{d}{dt} \left( \sqrt{2} t \right) \mathbf{i} + \frac{d}{dt} \left( e^t \right) \mathbf{j} + \frac{d}{dt} \left( e^{-t} \right) \mathbf{k}$

   $\mathbf{r}'(t) = \sqrt{2} \mathbf{i} + e^t \mathbf{j} - e^{-t} \mathbf{k}$

2. Step 2: Find the magnitude of $\mathbf{r}'(t)$

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

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

   Simplifying:

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

3. Step 3: Set up the integral for the length

   The length of the curve is given by:

   $L = \int_0^5 \sqrt{2 + e^{2t} + e^{-2t}} \, dt$

This integral may not have a simple closed-form solution, so it can be evaluated numerically.

I'll compute this integral for you.The length of the curve $\mathbf{r}(t) = \sqrt{2} \, t \, \mathbf{i} + e^t \, \mathbf{j} + e^{-t} \, \mathbf{k}$ from $t = 0$ to $t = 5$ is approximately 148.41 units.

Would you like further details or have any questions?

Here are 5 related questions to deepen understanding:

1. How do you interpret the physical meaning of the length of a curve in 3D space?
2. What is the significance of the parameter $t$ in the curve equation?
3. How would you find the length if the limits of $t$ were different?
4. What numerical methods are used for evaluating integrals that don't have closed-form solutions?
5. How do you compute arc length for curves in 2D versus 3D?

Tip: When dealing with vector functions, always calculate the derivative of each component separately before finding the magnitude of the vector.