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: 69

$$L = \frac{{\sqrt {10} }}{{2{e^2}}}\left( {{e^4} - 1} \right)$$

$$\eqalign{ & x = {e^{2t}}\left( {\sin t + \cos t} \right),{\text{ }}y = {e^{2t}}\left( {\sin t - \cos t} \right){\text{ on }}\left( { - 1 \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[ {{e^{2t}}\left( {\sin t + \cos t} \right)} \right] \cr & \frac{{dx}}{{dt}} = 2{e^{2t}}\left( {\sin t + \cos t} \right) + {e^{2t}}\left( {\cos t - \sin t} \right) \cr & \frac{{dx}}{{dt}} = \left( {2\sin t + 2\cos t + \cos t - \sin t} \right){e^{2t}} \cr & \frac{{dx}}{{dt}} = \left( {\sin t + 3\cos t} \right){e^{2t}} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{e^{2t}}\left( {\sin t - \cos t} \right)} \right] \cr & \frac{{dy}}{{dt}} = 2{e^{2t}}\left( {\sin t - \cos t} \right) + {e^{2t}}\left( {\cos t + \sin t} \right) \cr & \frac{{dy}}{{dt}} = \left( {2\sin t - 2\cos t + \cos t + \sin t} \right){e^{2t}} \cr & \frac{{dy}}{{dt}} = \left( {3\sin t - \cos t} \right){e^{2t}} \cr & \cr & {\left( {\frac{{dx}}{{dt}}} \right)^2} = {\left[ {\left( {\sin t + 3\cos t} \right){e^{2t}}} \right]^2} \cr & {\left( {\frac{{dx}}{{dt}}} \right)^2} = {e^{4t}}\left( {{{\sin }^2}t + 6\sin t\cos t + 9{{\cos }^2}t} \right) \cr & {\left( {\frac{{dy}}{{dt}}} \right)^2} = {\left[ {\left( {3\sin t - \cos t} \right){e^{2t}}} \right]^2} \cr & {\left( {\frac{{dy}}{{dt}}} \right)^2} = {e^{4t}}\left( {9{{\sin }^2}t - 6\sin t\cos t + {{\cos }^2}t} \right) \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}^1 {\sqrt {{e^{4t}}\left( {{{\sin }^2}t + 6\sin t\cos t + 9 - 6\sin t\cos t + {{\cos }^2}t} \right)} dt} \cr & L = \int_{ - 1}^1 {\sqrt {{e^{4t}}\left( {10 + 6\sin t\cos t - 6\sin t\cos t} \right)} dt} \cr & L = \int_{ - 1}^1 {\sqrt {10{e^{4t}}} dt} \cr & L = \sqrt {10} \int_{ - 1}^1 {{e^{2t}}dt} \cr & {\text{Integrating}} \cr & L = \frac{{\sqrt {10} }}{2}\left[ {{e^{2t}}} \right]_{ - 1}^1 \cr & {\text{Integrating}} \cr & L = \frac{{\sqrt {10} }}{2}\left[ {{e^2} - {e^{ - 2}}} \right] \cr & L = \frac{{\sqrt {10} }}{2}\left( {{e^2} - \frac{1}{{{e^2}}}} \right) \cr & L = \frac{{\sqrt {10} }}{{2{e^2}}}\left( {{e^4} - 1} \right) \cr} $$