Answer
$$L = \sqrt 3 $$
Work Step by Step
$$\eqalign{
& x = \frac{1}{2}t,\,\,\,\,y = \frac{1}{3}{\left( {1 - t} \right)^{3/2}},\,\,\,\,z = \frac{1}{3}{\left( {1 + t} \right)^{3/2}},\,\,\,\,\,\,\, - 1 \leqslant t \leqslant 1 \cr
& \cr
& {\text{Calculate the derivatives with respect to }}t{\text{ for }}x\left( t \right),\,\,\,y\left( t \right){\text{ and z}}\left( t \right) \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{1}{2}t} \right] = \frac{1}{2} \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{1}{3}{{\left( {1 - t} \right)}^{3/2}}} \right] = \frac{1}{3}\left( {\frac{3}{2}} \right){\left( {1 - t} \right)^{1/2}} = \frac{1}{2}{\left( {1 - t} \right)^{1/2}} \cr
& \frac{{dz}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{1}{3}{{\left( {1 + t} \right)}^{3/2}}} \right] = \frac{1}{3}\left( {\frac{3}{2}} \right){\left( {1 + t} \right)^{1/2}} = \frac{1}{2}{\left( {1 + t} \right)^{1/2}} \cr
& \cr
& {\text{Find the arc length using }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} } dt \cr
& {\text{for the interval }} - 1 \leqslant t \leqslant 1,\,\,{\text{ }}a = - 1{\text{ and }}b = 1.{\text{ Then}}{\text{,}} \cr
& L = \int_{ - 1}^1 {\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left[ {\frac{1}{2}{{\left( {1 - t} \right)}^{1/2}}} \right]}^2} + {{\left[ {\frac{1}{2}{{\left( {1 + t} \right)}^{1/2}}} \right]}^2}} } dt \cr
& L = \int_{ - 1}^1 {\sqrt {\frac{1}{4} + \frac{1}{4}\left( {1 - t} \right) + \frac{1}{4}\left( {1 + t} \right)} } dt \cr
& L = \int_{ - 1}^1 {\sqrt {\frac{1}{4} + \frac{1}{4} + \frac{1}{4}} } dt \cr
& L = \int_{ - 1}^1 {\sqrt {\frac{3}{4}} } dt \cr
& \cr
& {\text{Integrate and evaluate}} \cr
& L = \frac{{\sqrt 3 }}{2}\left[ t \right]_{ - 1}^1 \cr
& L = \frac{{\sqrt 3 }}{2}\left( {1 - \left( { - 1} \right)} \right) \cr
& L = \sqrt 3 \cr} $$
