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 703: 49

$$\sqrt 3 ,4$$

$$\eqalign{ & {\text{Let }}x = \theta + \cos \theta ,{\text{ }}y = 1 + \sin \theta ,{\text{ }}\theta = \pi /6 \cr & {\text{Calculate the derivatives }}\frac{{dy}}{{d\theta }}{\text{ and }}\frac{{dx}}{{d\theta }} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {1 + \sin \theta } \right] = \cos \theta \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\theta + \cos \theta } \right] = 1 - \sin \theta \cr & {\text{From the equation }}\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }}{\text{ we have,}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{\cos \theta }}{{1 - \sin \theta }} \cr & {\text{Evaluate at }}\theta = \frac{\pi }{6} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{6}}} = \frac{{\cos \left( {\pi /6} \right)}}{{1 - \sin \left( {\pi /6} \right)}} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{6}}} = \frac{{\frac{{\sqrt 3 }}{2}}}{{1 - \frac{1}{2}}} = \sqrt 3 \cr & \cr & {\text{Where }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy'/d\theta }}{{dx/d\theta }},{\text{ and }}y' = \frac{{\cos \theta }}{{1 - \sin \theta }} \cr & \frac{{dy'}}{{d\theta }} = \frac{d}{{dt}}\left[ {\frac{{\cos \theta }}{{1 - \sin \theta }}} \right] \cr & {\text{By the quotient rule}} \cr & \frac{{dy'}}{{d\theta }} = \frac{{\left( {1 - \sin \theta } \right)\left( { - \sin \theta } \right) - \cos \theta \left( { - \cos \theta } \right)}}{{{{\left( {1 - \sin \theta } \right)}^2}}} \cr & \frac{{dy'}}{{d\theta }} = \frac{{ - \sin \theta + {{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\left( {1 - \sin \theta } \right)}^2}}} \cr & \frac{{dy'}}{{d\theta }} = \frac{{1 - \sin \theta }}{{{{\left( {1 - \sin \theta } \right)}^2}}} \cr & \frac{{dy'}}{{d\theta }} = \frac{1}{{1 - \sin \theta }} \cr & {\text{Therefore}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy'/d\theta }}{{dx/d\theta }} = \frac{{\frac{1}{{1 - \sin \theta }}}}{{1 - \sin \theta }} = \frac{1}{{{{\left( {1 - \sin \theta } \right)}^2}}} \cr & {\text{Evaluate at }}\theta = \frac{\pi }{6} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{6}}} = \frac{1}{{{{\left( {1 - \sin \left( {\pi /6} \right)} \right)}^2}}} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{6}}} = \frac{1}{{{{\left( {1 - 1/2} \right)}^2}}} = 4 \cr} $$