Answer
$$L = \frac{2}{{27}}\left( {37\sqrt {37} - 10\sqrt {10} } \right)$$
Work Step by Step
$$\eqalign{
& x = \sqrt t - 2,{\text{ }}y = 2{t^{3/4}}{\text{ on the interval }}\left( {1 \leqslant t \leqslant 16} \right) \cr
& {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sqrt t - 2} \right] = \frac{1}{{2\sqrt t }} \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {2{t^{3/4}}} \right] = \frac{3}{2}{t^{ - 1/4}} \cr
& {\text{Use the Arc Length Formula For Parametric Curves}} \cr
& L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} \cr
& L = \int_1^{16} {\sqrt {{{\left( {\frac{1}{{2\sqrt t }}} \right)}^2} + {{\left( {\frac{3}{2}{t^{ - 1/4}}} \right)}^2}} dt} \cr
& L = \int_1^{16} {\sqrt {{{\left( {\frac{1}{{2\sqrt t }}} \right)}^2} + {{\left( {\frac{3}{{2{t^{1/4}}}}} \right)}^2}} dt} \cr
& L = \int_1^{16} {\sqrt {{{\left( {\frac{1}{{2\sqrt t }}} \right)}^2} + \frac{9}{{4{t^{1/2}}}}} dt} \cr
& {\text{Let }}u = \sqrt t ,{\text{ }}{u^2} = t,{\text{ 2}}udu = dt \cr
& t = 1 \Rightarrow u = 1 \cr
& t = 16 \Rightarrow u = 4 \cr
& L = \int_1^4 {\sqrt {{{\left( {\frac{1}{{2u}}} \right)}^2} + \frac{9}{{4u}}} \left( {2u} \right)du} \cr
& L = 2\int_1^4 {u\sqrt {\frac{1}{{4{u^2}}} + \frac{9}{{4u}}} du} \cr
& L = 2\int_1^4 {u\sqrt {\frac{{1 + 9u}}{{4{u^2}}}} du} \cr
& L = 2\int_1^4 {\frac{u}{{2u}}\sqrt {1 + 9u} du} \cr
& L = \int_1^4 {\sqrt {1 + 9u} du} \cr
& {\text{Integrating}} \cr
& L = \frac{1}{9}\int_1^4 {\sqrt {1 + 9u} du} \cr
& L = \frac{1}{9}\left[ {\frac{{{{\left( {1 + 9u} \right)}^{3/2}}}}{{3/2}}} \right]_0^1 \cr
& L = \frac{2}{{27}}\left[ {{{\left( {1 + 9u} \right)}^{3/2}}} \right]_0^1 \cr
& L = \frac{2}{{27}}\left[ {{{\left( {1 + 36} \right)}^{3/2}} - {{\left( {1 + 9} \right)}^{3/2}}} \right] \cr
& L = \frac{2}{{27}}\left[ {{{37}^{3/2}} - {{10}^{3/2}}} \right] \cr
& L = \frac{2}{{27}}\left( {37\sqrt {37} - 10\sqrt {10} } \right) \cr} $$
