Chapter 12 - Vector-Valued Functions - 12.5 Curvature - Exercises Set 12.5 - Page 879: 14

$$\rho = \frac{3}{{\sqrt 2 }}$$

$$\eqalign{ & {\bf{r}}\left( t \right) = {e^t}{\bf{i}} + {e^{ - t}}{\bf{j}} + t{\bf{k}};\,\,\,\,\,\,\,t = 0 \cr & {\text{Calculate the derivatives }}{\bf{r}}'\left( t \right){\text{ and }}{\bf{r}}{\text{''}}\left( t \right){\text{ and evaluate at }}t = 0 \cr & {\bf{r}}'\left( t \right) = {e^t}{\bf{i}} - {e^{ - t}}{\bf{j}} + \left( 1 \right){\bf{k}} \cr & {\bf{r}}'\left( 0 \right) = {e^0}{\bf{i}} - {e^{ - 0}}{\bf{j}} + {\bf{k}} \cr & {\bf{r}}'\left( 0 \right) = {\bf{i}} - {\bf{j}} + {\bf{k}} \cr & and \cr & {\bf{r}}''\left( t \right) = {e^t}{\bf{i}} + {e^{ - t}}{\bf{j}} \cr & {\bf{r}}''\left( 0 \right) = {e^0}{\bf{i}} + {e^{ - 0}}{\bf{j}} \cr & {\bf{r}}''\left( 0 \right) = {\bf{i}} + {\bf{j}} \cr} $$ \[\begin{gathered} {\text{Calculate the cross product }}{\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) \hfill \\ {\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = \left| {\begin{array}{*{20}{c}} {\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\ 1&{ - 1}&1 \\ 1&1&0 \end{array}} \right| \hfill \\ {\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = \left| {\begin{array}{*{20}{c}} { - 1}&1 \\ 1&0 \end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}} 1&1 \\ 1&0 \end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 1&1 \end{array}} \right|{\mathbf{k}} \hfill \\ {\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = - {\mathbf{i}} - \left( { - 1} \right){\mathbf{j}} + \left( {1 + 1} \right){\mathbf{k}} \hfill \\ {\mathbf{r}}'\left( 0 \right) \times {\mathbf{r}}''\left( 0 \right) = - {\mathbf{i}} + {\mathbf{j}} + 2{\mathbf{k}} \hfill \\ \end{gathered} \] $$\eqalign{ & {\text{Use the formula }}\left( 3 \right)\,\,\,\kappa \left( t \right) = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}}.{\text{ Thus}}{\text{, at }}t = 0 \cr & \kappa \left( 0 \right) = \frac{{\left\| { - {\bf{i}} + {\bf{j}} + 2{\bf{k}}} \right\|}}{{{{\left\| {{\bf{i}} - {\bf{j}} + {\bf{k}}} \right\|}^3}}} \cr & \kappa \left( 0 \right) = \frac{{\sqrt {1 + 1 + 4} }}{{{{\left( {\sqrt {1 + 1 + 1} } \right)}^3}}} \cr & \kappa \left( 0 \right) = \frac{{\sqrt 6 }}{{{{\left( {\sqrt 3 } \right)}^3}}} \cr & \kappa \left( 0 \right) = \frac{{\sqrt 2 \sqrt 3 }}{{{{\left( 3 \right)}^{3/2}}}} \cr & \kappa \left( 0 \right) = \frac{{\sqrt 2 }}{3} \cr & \cr & {\text{Calculate the radius of curvature }}\rho {\text{ at }}t = 0 \cr & \rho = \frac{1}{{\kappa \left( 0 \right)}} = \frac{1}{{\sqrt 2 /3}} \cr & \rho = \frac{3}{{\sqrt 2 }} \cr} $$