Answer
$$L = e - {e^{ - 1}}$$
Work Step by Step
$$\eqalign{
& x = {e^t},\,\,\,\,y = {e^{ - t}},\,\,\,\,z = \sqrt 2 t,\,\,\,\,\,\,\,\,0 \leqslant t \leqslant 1 \cr
& \cr
& {\text{Calculate the derivatives with respect to }}t{\text{ for }}x\left( t \right),\,\,\,y\left( t \right){\text{ and z}}\left( t \right) \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{e^t}} \right] = {e^t} \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{e^{ - t}}} \right] = - {e^{ - t}} \cr
& \frac{{dz}}{{dt}} = \frac{d}{{dt}}\left[ {\sqrt 2 t} \right] = \sqrt 2 \cr
& \cr
& {\text{Find the arc length using }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} } dt \cr
& {\text{for the interval }}0 \leqslant t \leqslant 1,\,\,{\text{ }}a = 0{\text{ and }}b = 1.{\text{ Then}}{\text{,}} \cr
& L = \int_0^1 {\sqrt {{{\left( {{e^t}} \right)}^2} + {{\left( { - {e^{ - t}}} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} } dt \cr
& L = \int_0^1 {\sqrt {{e^{2t}} + {e^{ - 2t}} + 2} } dt \cr
& {\text{factoring}} \cr
& L = \int_0^1 {\sqrt {{{\left( {{e^t} + {e^{ - t}}} \right)}^2}} } dt \cr
& L = \int_0^1 {\left( {{e^t} + {e^{ - t}}} \right)} dt \cr
& \cr
& {\text{Integrate and evaluate}} \cr
& L = \left[ {{e^t} - {e^{ - t}}} \right]_0^1 \cr
& L = \left[ {{e^1} - {e^{ - 1}}} \right] - \left[ {{e^0} - {e^{ - 0}}} \right] \cr
& L = e - {e^{ - 1}} \cr} $$
