Chapter 6 - Applications of Integration - 6.5 Length of Curves - 6.5 Exercises - Page 450: 28

$$L = 1$$

$$\eqalign{ & y = \ln \left( {x - \sqrt {{x^2} - 1} } \right),{\text{ for }}1 \leqslant x \leqslant \sqrt 2 \cr & {\text{Use the formula }}L = \int_c^d {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} dy} {\text{ }}\left( {\bf{1}} \right) \cr & {e^y} = {e^{\ln \left( {x - \sqrt {{x^2} - 1} } \right)}} \cr & {e^y} = x - \sqrt {{x^2} - 1} \cr & {\text{Solve for }}x \cr & \sqrt {{x^2} - 1} = x - {e^y} \cr & {\left( {\sqrt {{x^2} - 1} } \right)^2} = {\left( {x - {e^y}} \right)^2} \cr & {x^2} - 1 = {x^2} - 2x{e^y} + {e^{2y}} \cr & - 1 = - 2x{e^y} + {e^{2y}} \cr & 2x{e^y} = {e^{2y}} + 1 \cr & x = \frac{{{e^{2y}} + 1}}{{2{e^y}}} = \frac{{{e^y}}}{2} + \frac{1}{{2{e^y}}} \cr & x = \frac{1}{2}{e^y} + \frac{1}{2}{e^{ - y}} \cr & {\text{Differentiate to find }}\frac{{dx}}{{dy}} \cr & \frac{{dx}}{{dy}} = \frac{1}{2}{e^y} - \frac{1}{2}{e^{ - y}} \cr & {\text{Find the new interval of integration}} \cr & {\text{For }}x = 1 \to {e^y} = 1 \to y = 0 \cr & {\text{For }}x = \sqrt 2 \to {e^y} = \sqrt 2 - 1 \to y = \ln \left( {\sqrt 2 - 1} \right) \cr & 0 \leqslant x \leqslant \ln \left( {\sqrt 2 - 1} \right) \cr & \cr & {\text{Therefore, susbsituting into }}\left( {\bf{1}} \right) \cr & L = \int_0^{\ln \left( {\sqrt 2 - 1} \right)} {\sqrt {1 + {{\left( {\frac{1}{2}{e^y} - \frac{1}{2}{e^{ - y}}} \right)}^2}} dy} \cr & L = \int_0^{\ln \left( {\sqrt 2 - 1} \right)} {\sqrt {1 + \frac{1}{4}{e^{2y}} - \frac{1}{2}{e^{y - y}} + \frac{1}{4}{e^{ - 2y}}} dy} \cr & L = \int_0^{\ln \left( {\sqrt 2 - 1} \right)} {\sqrt {\frac{1}{4}{e^{2y}} + \frac{1}{2} + \frac{1}{4}{e^{ - 2y}}} dy} \cr & {\text{Factoring}} \cr & L = \int_0^{\ln \left( {\sqrt 2 - 1} \right)} {\sqrt {{{\left( {\frac{{{e^y} + {e^{ - y}}}}{2}} \right)}^2}} dy} \cr & L = \int_0^{\ln \left( {\sqrt 2 - 1} \right)} {\left( {\frac{{{e^y} + {e^{ - y}}}}{2}} \right)dy} \cr & L = \frac{1}{2}\int_0^{\ln \left( {\sqrt 2 - 1} \right)} {\left( {{e^y} + {e^{ - y}}} \right)dy} \cr & {\text{Integrating}} \cr & L = \frac{1}{2}\left[ {{e^y} - {e^{ - y}}} \right]_0^{\ln \left( {\sqrt 2 - 1} \right)} \cr & L = \frac{1}{2}\left[ {{e^{\ln \left( {\sqrt 2 - 1} \right)}} - {e^{ - \left( {\ln \left( {\sqrt 2 - 1} \right)} \right)}}} \right] - \frac{1}{2}\left[ {{e^0} - {e^0}} \right] \cr & L = \frac{1}{2}\left[ {\sqrt 2 - 1 - \frac{1}{{\sqrt 2 - 1}}} \right] - \frac{1}{2}\left[ 0 \right] \cr & L = \frac{1}{2}\left( {\frac{{3 - 2\sqrt 2 - 1}}{{\sqrt 2 - 1}}} \right) \cr & L = \frac{{2 - 2\sqrt 2 }}{{2\left( {\sqrt 2 - 1} \right)}} \cr & L = - 1 \cr & {\text{Taking the absolute value of the result}} \cr & L = 1 \cr} $$