Answer
$$K = \frac{{\sqrt 2 }}{2}$$
Work Step by Step
\[\begin{gathered}
{\mathbf{r}}\left( t \right) = t{\mathbf{i}} + \frac{1}{t}{\mathbf{j}},{\text{ at }}t = 1 \hfill \\
{\text{By Theorem 12}}{\text{.8 }} \hfill \\
{\text{If }}C{\text{ is a smooth curve given by }}{\mathbf{r}}\left( t \right),{\text{ then the curvature }}K{\text{ of}} \hfill \\
C{\text{ at }}t{\text{ is }}K = \frac{{\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\mathbf{r}}'\left( t \right)} \right\|}^3}}} \hfill \\
{\mathbf{r}}\left( t \right) = t{\mathbf{i}} + \frac{1}{t}{\mathbf{j}} \hfill \\
{\mathbf{r}}'\left( t \right) = {\mathbf{i}} - \frac{1}{{{t^2}}}{\mathbf{j}} \hfill \\
{\mathbf{r}}''\left( t \right) = \frac{2}{{{t^3}}}{\mathbf{j}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
1&{ - \frac{1}{{{t^2}}}}&0 \\
0&{\frac{2}{{{t^3}}}}&0
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{ - \frac{1}{{{t^2}}}}&0 \\
{\frac{2}{{{t^3}}}}&0
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
1&0 \\
0&0
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
1&{ - \frac{1}{{{t^2}}}} \\
0&{\frac{2}{{{t^3}}}}
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \frac{2}{{{t^3}}}{\mathbf{k}} \hfill \\
\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\| = \frac{2}{{{t^3}}} \hfill \\
{\text{Therefore,}} \hfill \\
K = \frac{{\left\| {{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\mathbf{r}}'\left( t \right)} \right\|}^3}}} = \frac{{\frac{2}{{{t^3}}}}}{{{{\left\| {{\mathbf{i}} - \frac{1}{{{t^2}}}{\mathbf{j}}} \right\|}^3}}} = \frac{{2/{t^3}}}{{{{\left( {\sqrt {\left( 1 \right) + {{\left( {\frac{1}{{{t^2}}}} \right)}^2}} } \right)}^3}}} \hfill \\
{\text{Evaluate at }}t = 1 \hfill \\
K = \frac{{2/{{\left( 1 \right)}^3}}}{{{{\left( {\sqrt {\left( 1 \right) + {{\left( {\frac{1}{{{{\left( 1 \right)}^2}}}} \right)}^2}} } \right)}^3}}} = \frac{2}{{{{\left( {\sqrt 2 } \right)}^3}}} \hfill \\
K = \frac{{\sqrt 2 }}{2} \hfill \\
\end{gathered} \]
