Chapter 10 - Parametric And Polar Curves; Conic Sections - 10.1 Parametric Equations; Tangent Lines And Arc Length For Parametric Curves - Exercises Set 10.1 - Page 704: 65

$$L = \frac{{5\sqrt 5 - 8}}{3}$$

$$\eqalign{ & x = {t^2},{\text{ }}y = \frac{1}{3}{t^3}{\text{ on the interval }}\left( {0 \leqslant t \leqslant 1} \right) \cr & {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{t^2}} \right] = 2t \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{1}{3}{t^3}} \right] = {t^2} \cr & {\text{Use the Arc Length Formula For Parametric Curves}} \cr & L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} \cr & L = \int_0^1 {\sqrt {{{\left( {2t} \right)}^2} + {{\left( {{t^2}} \right)}^2}} dt} \cr & L = \int_0^1 {\sqrt {{t^2}\left( {4 + {t^2}} \right)} dt} \cr & L = \int_0^1 {t\sqrt {4 + {t^2}} dt} \cr & {\text{Integrating}} \cr & L = \frac{1}{2}\int_0^1 {2t\sqrt {4 + {t^2}} dt} \cr & L = \frac{1}{2}\left[ {\frac{{{{\left( {4 + {t^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^1 \cr & L = \frac{1}{3}\left[ {{{\left( {4 + {t^2}} \right)}^{3/2}}} \right]_0^1 \cr & L = \frac{1}{3}\left[ {{{\left( {4 + {1^2}} \right)}^{3/2}} - {{\left( {4 + {0^2}} \right)}^{3/2}}} \right] \cr & L = \frac{1}{3}\left[ {{5^{3/2}} - 8} \right] \cr & L = \frac{{5\sqrt 5 - 8}}{3} \cr} $$