Chapter 11 - Vectors and Vector-Valued Functions - 11.9 Curvature and Normal Vectors - 11.9 Exercises - Page 852: 16

$${\bf{T}}\left( t \right) = \left\langle {\cos t, - \sin t} \right\rangle {\text{ and }}\kappa \left( t \right) = \cos t$$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {t,\ln \cos t} \right\rangle \cr & {\text{Calculate }}{\bf{v}}\left( t \right) \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left\langle {t,\ln \cos t} \right\rangle \cr & {\bf{v}}\left( t \right) = \left\langle {1,\frac{{ - \sin t}}{{\cos t}}} \right\rangle \cr & {\bf{v}}\left( t \right) = \left\langle {1, - \tan t} \right\rangle \cr & {\text{Find the unit tangent vector }}{\bf{T}} \cr & {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} \cr & {\bf{T}}\left( t \right) = \frac{{\left\langle {1, - \tan t} \right\rangle }}{{\left| {\left\langle {1, - \tan t} \right\rangle } \right|}} = \frac{{\left\langle {1, - \tan t} \right\rangle }}{{\sqrt {1 + {{\tan }^2}t} }} \cr & {\bf{T}}\left( t \right) = \frac{{\left\langle {1, - \tan t} \right\rangle }}{{\sqrt {{{\sec }^2}t} }} \cr & {\bf{T}}\left( t \right) = \frac{{\left\langle {1, - \tan t} \right\rangle }}{{\sec t}} \cr & {\bf{T}}\left( t \right) = \left\langle {\cos t, - \sin t} \right\rangle \cr & \cr & {\text{Therefore}}{\text{, the curvature is}} \cr & \kappa \left( t \right) = \frac{1}{{\left| {\bf{v}} \right|}}\left| {\frac{{d{\bf{T}}}}{{dt}}} \right| \cr & \frac{{d{\bf{T}}}}{{dt}} = \frac{1}{{\sec t}}\left| {\frac{d}{{dt}}\left\langle {\cos t, - \sin t} \right\rangle } \right| \cr & \frac{{d{\bf{T}}}}{{dt}} = \cos t\left| {\left\langle { - \sin t, - \cos t} \right\rangle } \right| \cr & \kappa \left( t \right) = \cos t\sqrt {{{\sin }^2}t + {{\cos }^2}t} \cr & \kappa \left( t \right) = \cos t \cr & \cr & {\bf{T}}\left( t \right) = \left\langle {\cos t, - \sin t} \right\rangle {\text{ and }}\kappa \left( t \right) = \cos t \cr} $$