Answer
\[{\mathbf{B}} = \frac{{\left\langle {\operatorname{sech} t, - \tanh t,1} \right\rangle }}{{\sqrt 2 }}\]
\[\tau = \frac{1}{2}{\operatorname{sech} ^2}t\]
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = \left\langle {t,\cosh t, - \sinh 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 {1,\sinh t, - \cosh t} \right\rangle \hfill \\
{\mathbf{a}}\left( t \right) = {\mathbf{v}}'\left( t \right) \hfill \\
{\mathbf{a}}\left( t \right) = \left\langle {0,\cosh t, - \sinh t} \right\rangle \hfill \\
{\mathbf{a}}'\left( t \right) = \left\langle {0,\sinh t, - \cosh t} \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}} \\
1&{\sinh t}&{ - \cosh t} \\
0&{\cosh t}&{ - \sinh t}
\end{array}} \right| \hfill \\
{\mathbf{v}} \times {\mathbf{a}} = \left( { - {{\sinh }^2}t + {{\cosh }^2}t} \right){\mathbf{i}} - \sinh t{\mathbf{j}} + \cosh t{\mathbf{k}} \hfill \\
{\mathbf{v}} \times {\mathbf{a}} = {\mathbf{i}} - \sinh t{\mathbf{j}} + \cosh t{\mathbf{k}} \hfill \\
\left| {{\mathbf{v}} \times {\mathbf{a}}} \right| = \sqrt {1 + {{\sinh }^2}t + {{\cosh }^2}t} \hfill \\
\left| {{\mathbf{v}} \times {\mathbf{a}}} \right| = \sqrt {{{\cosh }^2}t + {{\cosh }^2}t} \hfill \\
\left| {{\mathbf{v}} \times {\mathbf{a}}} \right| = \sqrt 2 \cosh t \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 {1, - \sinh t,\cosh t} \right\rangle }}{{\sqrt 2 \cosh t}} \hfill \\
{\mathbf{B}} = \frac{{\left\langle {\operatorname{sech} t, - \tanh t,1} \right\rangle }}{{\sqrt 2 }} \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 {1, - \sinh t,\cosh t} \right\rangle \cdot \left\langle {0,\sinh t, - \cosh t} \right\rangle }}{{{{\left( {\sqrt 2 \cosh t} \right)}^2}}} \hfill \\
\tau = \frac{{0 - {{\sinh }^2}t - {{\cosh }^2}t}}{{2{{\cosh }^2}t}} \hfill \\
\tau = \frac{{1 + {{\cosh }^2}t - {{\cosh }^2}t}}{{2{{\cosh }^2}t}} \hfill \\
\tau = \frac{1}{{2{{\cosh }^2}t}} \hfill \\
\tau = \frac{1}{2}{\operatorname{sech} ^2}t \hfill \\
\end{gathered} \]
