Answer
\[{\mathbf{B}} = \frac{{\left\langle { - \sin t,\cos t,2} \right\rangle }}{{\sqrt 5 }}{\text{ and }}\tau = - \frac{1}{5}\]
Work Step by Step
\[\begin{gathered} {\mathbf{r}}\left( t \right) = \left\langle {2\cos t,2\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 { - 2\sin t,2\cos t, - 1} \right\rangle \hfill \\ {\mathbf{a}}\left( t \right) = {\mathbf{v}}'\left( t \right) \hfill \\ {\mathbf{a}}\left( t \right) = \left\langle { - 2\cos t, - 2\sin t,0} \right\rangle \hfill \\ {\mathbf{a}}'\left( t \right) = \left\langle {2\sin t, - 2\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}} \\ { - 2\sin t}&{2\cos t}&{ - 1} \\ { - 2\cos t}&{ - 2\sin t}&0 \end{array}} \right| \hfill \\ {\mathbf{v}} \times {\mathbf{a}} = - 2\sin t{\mathbf{i}} + 2\cos t{\mathbf{j}} + \left( {4{{\sin }^2}t + 4{{\cos }^2}t} \right){\mathbf{k}} \hfill \\ {\mathbf{v}} \times {\mathbf{a}} = - 2\sin t{\mathbf{i}} + 2\cos t{\mathbf{j}} + 4{\mathbf{k}} \hfill \\ \left| {{\mathbf{v}} \times {\mathbf{a}}} \right| = \sqrt {4{{\sin }^2}t + 4{{\cos }^2}t + 16} \hfill \\ \left| {{\mathbf{v}} \times {\mathbf{a}}} \right| = \sqrt {20} = 2\sqrt 5 \hfill \\ \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{{{\mathbf{v}} \times {\mathbf{a}}}}{{\left| {{\mathbf{v}} \times {\mathbf{a}}} \right|}} = \frac{{\left\langle { - 2\sin t,2\cos t,4} \right\rangle }}{{2\sqrt 5 }} \hfill \\ {\mathbf{B}} = \frac{{\left\langle { - \sin t,\cos t,2} \right\rangle }}{{\sqrt 5 }} \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 { - 2\sin t,2\cos t,4} \right\rangle \cdot \left\langle {2\sin t, - 2\cos t,0} \right\rangle }}{{{{\left( {2\sqrt 5 } \right)}^2}}} \hfill \\ \tau = \frac{{ - 4{{\sin }^2}t - 4{{\cos }^2}t + 0}}{{20}} = \frac{{ - 4}}{{20}} \hfill \\ \tau = - \frac{1}{5} \hfill \\ \end{gathered} \]
