Answer
\[{\mathbf{B}} = \frac{{\left\langle {t\sin t - \cos t,\sin t + t\cos t, - {t^2}} \right\rangle }}{{\sqrt {{t^4} + {t^2} + 1} }}{\text{ and }}\tau = - \frac{{{t^2} }}{{{{\left( {\sqrt {{t^4} + {t^2} + 1} } \right)}^2}}}\]
Work Step by Step
\[\begin{gathered} {\mathbf{r}}\left( t \right) = \left\langle {\sin t - t\cos t,\cos t + t\sin t,t} \right\rangle \hfill \\ {\text{Calculate }}{\mathbf{v}}\left( t \right){\text{, }}{\mathbf{a}}\left( t \right){\text{ and }}{\mathbf{a}}'\left( t \right) \hfill \\ {\mathbf{v}}\left( t \right) = {\mathbf{r}}'\left( t \right) \hfill \\ {\mathbf{v}}\left( t \right) = \left\langle {t\sin t,t\cos t,1} \right\rangle \hfill \\ {\mathbf{a}}\left( t \right) = {\mathbf{v}}'\left( t \right) \hfill \\ {\mathbf{a}}\left( t \right) = \left\langle {\sin t + t\cos t,\cos t - t\sin t,0} \right\rangle \hfill \\ \hfill \\ {\mathbf{a}}'\left( t \right) = \left\langle {\cos t + \cos t - t\sin t, - \sin t - \sin t - t\cos t} \right\rangle \hfill \\ {\mathbf{a}}'\left( t \right) = \left\langle {2\cos t - t\sin t, - 2\sin t - t\cos t,0} \right\rangle \hfill \\ \hfill \\ {\text{Calculate }}{\mathbf{v}} \times {\mathbf{a}}{\text{ and }}\left| {{\mathbf{v}} \times {\mathbf{a}}} \right| \hfill \\ {\mathbf{v}} \times {\mathbf{a}} = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ {t\sin t}&{t\cos t}&1 \\ {\sin t + t\cos t}&{\cos t - t\sin t}&0 \end{array}} \right| \hfill \\ {\mathbf{v}} \times {\mathbf{a}} = \left( {t\sin t - \cos t} \right){\mathbf{i}} + \left( {\sin t + t\cos t} \right){\mathbf{j}} - {t^2}\left( {{{\sin }^2}t + {{\cos }^2}t} \right){\mathbf{k}} \hfill \\ {\mathbf{v}} \times {\mathbf{a}} = \left( {t\sin t - \cos t} \right){\mathbf{i}} + \left( {\sin t + t\cos t} \right){\mathbf{j}} - {t^2}{\mathbf{k}} \hfill \\ \left| {{\mathbf{v}} \times {\mathbf{a}}} \right| = \sqrt {{t^2}{{\sin }^2}t + {{\cos }^2}t + {{\sin }^2}t + {t^2}{{\cos }^2}t + {t^4}} \hfill \\ \left| {{\mathbf{v}} \times {\mathbf{a}}} \right| = \sqrt {{t^2}{{\sin }^2}t + 1 + {t^2}{{\cos }^2}t + {t^4}} \hfill \\ \left| {{\mathbf{v}} \times {\mathbf{a}}} \right| = \sqrt {{t^4} + {t^2} + 1} \hfill \\ {\text{Calculate the unit binormal vector:}}\,{\text{ }}{\mathbf{B}} = \frac{{{\mathbf{v}} \times {\mathbf{a}}}}{{\left| {{\mathbf{v}} \times {\mathbf{a}}} \right|}} \hfill \\ {\mathbf{B}} = \frac{{\left\langle {t\sin t - \cos t,\sin t + t\cos t, - {t^2}} \right\rangle }}{{\sqrt {{t^4} + {t^2} + 1} }} \hfill \\ \hfill \\ {\text{Calculate }}\tau \hfill \\ \tau = \frac{{\left( {{\mathbf{v}} \times {\mathbf{a}}} \right) \cdot {\mathbf{a}}'}}{{{{\left| {{\mathbf{v}} \times {\mathbf{a}}} \right|}^2}}} \hfill \\ \tau = \frac{{\left\langle {t\sin t - \cos t,\sin t + t\cos t, - {t^2}} \right\rangle \cdot \left\langle {2\cos t - t\sin t, - 2\sin t - t\cos t,0} \right\rangle }}{{{{\left( {\sqrt {{t^4} + {t^2} + 1} } \right)}^2}}} \hfill \\ \tau = - \frac{{{t^2}}}{{{{\left( {\sqrt {{t^4} + {t^2} + 1} } \right)}^2}}} \hfill \\ \end{gathered} \]
