Chapter 11 - Vectors and Vector-Valued Functions - 11.9 Curvature and Normal Vectors - 11.9 Exercises - Page 852: 42

$${\bf{B}} = \frac{{\left\langle {5\cos t, - 5\sin t, - 2} \right\rangle }}{{\sqrt {29} }}{\text{ and }}\tau = - \frac{5}{{58}}$$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {4\sin t,4\cos t,10t} \right\rangle \cr & {\text{Calculate }}{\bf{v}}\left( t \right){\text{, }}{\bf{a}}\left( t \right){\text{ and }}{\bf{a}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \left\langle {4\cos t, - 4\sin t,10} \right\rangle \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \left\langle { - 4\sin t, - 4\cos t,0} \right\rangle \cr & {\bf{a}}'\left( t \right) = \left\langle { - 4\cos t,4\sin t,0} \right\rangle \cr & \cr & {\text{Calculate }}{\bf{v}} \times {\bf{a}}{\text{ and }}\left| {{\bf{v}} \times {\bf{a}}} \right| \cr} $$ \[{\mathbf{v}} \times {\mathbf{a}} = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ {4\cos t}&{ - 4\sin t}&{10} \\ { - 4\sin t}&{ - 4\cos t}&0 \end{array}} \right|\] $$\eqalign{ & {\bf{v}} \times {\bf{a}} = 40\cos t{\bf{i}} - \left( {40\sin t} \right){\bf{j}} + \left( { - 16{{\cos }^2}t - 16{{\sin }^2}t} \right){\bf{k}} \cr & {\bf{v}} \times {\bf{a}} = 40\cos t{\bf{i}} - 40\sin t{\bf{j}} - 16{\bf{k}} \cr & \left| {{\bf{v}} \times {\bf{a}}} \right| = \sqrt {{{40}^2}{{\cos }^2}t + {{40}^2}{{\sin }^2}t + {{16}^2}} \cr & \left| {{\bf{v}} \times {\bf{a}}} \right| = \sqrt {{{40}^2} + {{16}^2}} = 8\sqrt {29} \cr & \cr & {\text{Calculate the unit binormal vector:}}\,{\text{ }}{\bf{B}} = \frac{{{\bf{v}} \times {\bf{a}}}}{{\left| {{\bf{v}} \times {\bf{a}}} \right|}} \cr & {\bf{B}} = \frac{{{\bf{v}} \times {\bf{a}}}}{{\left| {{\bf{v}} \times {\bf{a}}} \right|}} = \frac{{40\cos t{\bf{i}} - 40\sin t{\bf{j}} - 16{\bf{k}}}}{{8\sqrt {29} }} \cr & {\bf{B}} = \frac{{\left\langle {5\cos t, - 5\sin t, - 2} \right\rangle }}{{\sqrt {29} }} \cr & \cr & {\text{Calculate }}\tau \cr & \tau = \frac{{\left( {{\bf{v}} \times {\bf{a}}} \right) \cdot {\bf{a}}'}}{{{{\left| {{\bf{v}} \times {\bf{a}}} \right|}^2}}} \cr & \tau = \frac{{\left\langle {40\cos t, - 40\sin t, - 16} \right\rangle \cdot \left\langle { - 4\cos t,4\sin t,0} \right\rangle }}{{{{\left( {8\sqrt {29} } \right)}^2}}} \cr & \tau = \frac{{ - 160{{\cos }^2}t - 160{{\sin }^2}t + 0}}{{1856}} \cr & \tau = \frac{{ - 160}}{{1856}} \cr & \tau = - \frac{5}{{58}} \cr} $$