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: 70

$$L = \ln 3$$

$$\eqalign{ & x = 2{\sin ^{ - 1}}t,{\text{ }}y = \ln \left( {1 - {t^2}} \right){\text{ on }}\left( {0 \leqslant t \leqslant \frac{1}{2}} \right) \cr & {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {2{{\sin }^{ - 1}}t} \right] = 2\left( {\frac{1}{{\sqrt {1 - {t^2}} }}} \right) \cr & \frac{{dx}}{{dt}} = \frac{2}{{\sqrt {1 - {t^2}} }} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\ln \left( {1 - {t^2}} \right)} \right] \cr & \frac{{dy}}{{dt}} = - \frac{{2t}}{{1 - {t^2}}} \cr & \cr & {\left( {\frac{{dx}}{{dt}}} \right)^2} = {\left[ {\frac{2}{{\sqrt {1 - {t^2}} }}} \right]^2} = \frac{4}{{1 - {t^2}}} \cr & {\left( {\frac{{dy}}{{dt}}} \right)^2} = {\left[ { - \frac{{2t}}{{1 - {t^2}}}} \right]^2} = \frac{{4{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}} \cr & {\left( {\frac{{dx}}{{dt}}} \right)^2} + {\left( {\frac{{dy}}{{dt}}} \right)^2} = \frac{4}{{1 - {t^2}}} + \frac{{4{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}} \cr & {\left( {\frac{{dx}}{{dt}}} \right)^2} + {\left( {\frac{{dy}}{{dt}}} \right)^2} = \frac{{4\left( {1 - {t^2}} \right) + 4{t^2}}}{{{{\left( {1 - {t^2}} \right)}^2}}} \cr & {\left( {\frac{{dx}}{{dt}}} \right)^2} + {\left( {\frac{{dy}}{{dt}}} \right)^2} = \frac{4}{{{{\left( {1 - {t^2}} \right)}^2}}} \cr & \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^{1/2} {\sqrt {\frac{4}{{{{\left( {1 - {t^2}} \right)}^2}}}} dt} \cr & L = \int_0^{1/2} {\frac{2}{{1 - {t^2}}}dt} \cr & L = 2\int_0^{1/2} {\frac{1}{{1 - {t^2}}}dt} \cr & {\text{Integrating by the formula }}\int {\frac{{du}}{{{a^2} - {u^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{u + a}}{{u - a}}} \right| + C} \cr & L = 2\left[ {\frac{1}{2}\ln \left| {\frac{{t + 1}}{{t - 1}}} \right|} \right]_0^{1/2} \cr & L = \left[ {\ln \left| {\frac{{t + 1}}{{t - 1}}} \right|} \right]_0^{1/2} \cr & {\text{Integrating}} \cr & L = \ln \left| {\frac{{1/2 + 1}}{{1/2 - 1}}} \right| - \ln \left| {\frac{{0 + 1}}{{0 - 1}}} \right| \cr & L = \ln \left| {\frac{{3/2}}{{ - 1/2}}} \right| \cr & L = \ln 3 \cr} $$