1. **Problem:** Find the length of the curve given by $$\mathbf{r}(t) = \sqrt{2}t\mathbf{i} + e^t\mathbf{j} + e^{-t}\mathbf{k}$$ for $$0 \leq t \leq 8$$. 2. **Formula for curve length:** The length $$L$$ of a curve $$\mathbf{r}(t)$$ from $$t=a$$ to $$t=b$$ is given by: $$ L = \int_a^b \|\mathbf{r}'(t)\| \, dt $$ where $$\mathbf{r}'(t)$$ is the derivative of $$\mathbf{r}(t)$$. 3. **Find the derivative:** $$ \mathbf{r}'(t) = \frac{d}{dt}(\sqrt{2}t)\mathbf{i} + \frac{d}{dt}(e^t)\mathbf{j} + \frac{d}{dt}(e^{-t})\mathbf{k} = \sqrt{2}\mathbf{i} + e^t\mathbf{j} - e^{-t}\mathbf{k} $$ 4. **Find the magnitude of $$\mathbf{r}'(t)$$:** $$ \|\mathbf{r}'(t)\| = \sqrt{(\sqrt{2})^2 + (e^t)^2 + (-e^{-t})^2} = \sqrt{2 + e^{2t} + e^{-2t}} $$ 5. **Simplify the expression inside the square root:** Recall that $$e^{2t} + e^{-2t} = 2\cosh(2t)$$, so: $$ \|\mathbf{r}'(t)\| = \sqrt{2 + 2\cosh(2t)} = \sqrt{2(1 + \cosh(2t))} $$ 6. **Use the identity:** $$1 + \cosh(2t) = 2\cosh^2(t)$$, so: $$ \|\mathbf{r}'(t)\| = \sqrt{2 \cdot 2 \cosh^2(t)} = \sqrt{4 \cosh^2(t)} = 2\cosh(t) $$ 7. **Set up the integral for length:** $$ L = \int_0^8 2\cosh(t) \, dt $$ 8. **Integrate:** $$ \int 2\cosh(t) \, dt = 2\sinh(t) + C $$ 9. **Evaluate definite integral:** $$ L = 2\sinh(8) - 2\sinh(0) = 2\sinh(8) - 0 = 2\sinh(8) $$ **Final answer:** $$ \boxed{L = 2\sinh(8)} $$ This is the length of the curve from $$t=0$$ to $$t=8$$.