Arc Length

Firstly, the curve $\textbf{r}(t)$ doesn't pass through the origin, but we can proceed and pretend that $t=0$ corresponds to the origin point, since that's what the correct answer assumes. In reality the arc-length calculated is between the point (0,1,1) and the second one given. Also notice that the second point given corresponds to $t=1$ . We know that the arc-length of a parametric curve is given by

$L = \int_a^b |\textbf{r}'(t)|\,dt = \int_a^b \sqrt{\Big(\frac{dx}{dt} \Big)^2 + \Big( \frac{dy}{dt}\Big)^2 + \Big(\frac{dz}{dt} \Big)^2} \, dt$

where $\textbf{r}'(t) = \langle \sqrt{2}, e^t, -e^{-t}\rangle$ . Plugging this into the above equation with $a = t_1 = 0$ and $b=t_2=1$ gives

$L = \int_0^1 \sqrt{e^{2t} + 2 + \frac{1}{e^{2t}} }\,dt$

and notice that $e^{2t} + 2 + \frac{1}{e^{2t}} = (e^t + \frac{1}{e^t})^2$ to give

$L = \int_0^1 \sqrt{(e^t + \frac{1}{e^t})^2}\,dt = \int_0^1 e^t + \frac{1}{e^t} \, dt = e^t \big|_0^1 - e^{-t}\big|_0^1 = \boxed{e-\frac{1}{e}}$