Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.4 - Curvature and Normal Vectors of a Curve - Exercises 13.4 - Page 765: 16

$${\bf{T}} = \frac{1}{{\sqrt 2 }}\tanh t{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}} + \frac{1}{{\sqrt 2 }}{\text{sech}}t{\bf{k}},\,\,\,\,\,\,{\bf{N}} = {\text{sech}}t{\bf{i}} + \tanh t{\bf{k}},\,\,\,\,\,\,\,\kappa = \frac{1}{2}{\text{sec}}{{\text{h}}^2}t$$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {\cosh t} \right){\bf{i}} - \left( {\sinh t} \right){\bf{j}} + t{\bf{k}} \cr & {\text{Calculate }}{\bf{v}}\left( t \right).{\text{ Use }}{\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left( {\cosh t} \right){\bf{i}} - \left( {\sinh t} \right){\bf{j}} + t{\bf{k}}} \right] \cr & {\bf{v}}\left( t \right) = \sinh t{\bf{i}} - \cosh t{\bf{j}} + {\bf{k}} \cr & \cr & {\text{We calculate }}{\bf{T}}{\text{ from the velocity vector}} \cr & {\bf{v}}\left( t \right) = \sinh t{\bf{i}} - \cosh t{\bf{j}} + {\bf{k}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( {\sinh t} \right)}^2} + {{\left( { - \cosh t} \right)}^2} + {{\left( 1 \right)}^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\sinh }^2}t + {{\cosh }^2}t + 1} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\cos }^2}t - 1 + {{\cosh }^2}t + 1} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt 2 \cosh t \cr & \cr & {\text{use }}{\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} \cr & {\bf{T}} = \frac{{\sinh t{\bf{i}} - \cosh t{\bf{j}} + {\bf{k}}}}{{\sqrt 2 \cosh t}} \cr & {\bf{T}} = \frac{1}{{\sqrt 2 }}\tanh t{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}} + \frac{1}{{\sqrt 2 }}{\text{sech}}t{\bf{k}} \cr & \cr & {\text{Calculate }}{\bf{N}}\left( t \right){\text{ using the equation }}{\bf{N}} = \frac{{d{\bf{T}}/dt}}{{\left| {d{\bf{T}}/dt} \right|}}{\text{ }}\left( {{\text{see page 764}}} \right).{\text{ Then}} \cr & \frac{{d{\bf{T}}}}{{dt}} = \frac{d}{{dt}}\left( {\frac{1}{{\sqrt 2 }}\tanh t{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}} + \frac{1}{{\sqrt 2 }}{\text{sech}}t{\bf{k}}} \right) \cr & \frac{{d{\bf{T}}}}{{dt}} = \frac{1}{{\sqrt 2 }}{\text{sec}}{{\text{h}}^2}t{\bf{i}} + \frac{1}{{\sqrt 2 }}\sec {\text{h}}t\tanh t{\bf{k}} \cr & \cr & and \cr & \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \sqrt {{{\left( {\frac{1}{{\sqrt 2 }}{\text{sec}}{{\text{h}}^2}t} \right)}^2} + {{\left( {\frac{1}{{\sqrt 2 }}\sec {\text{h}}t\tanh t} \right)}^2}} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \sqrt {\frac{1}{2}{\text{sec}}{{\text{h}}^4}t + \frac{1}{2}{\text{sec}}{{\text{h}}^2}t{{\tanh }^2}t} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \sqrt {\frac{1}{2}{\text{sec}}{{\text{h}}^2}t\left( {{\text{sec}}{{\text{h}}^4}t + {{\tanh }^2}t} \right)} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \frac{{{\text{sech}}t}}{{\sqrt 2 }} \cr & \cr & {\bf{N}} = \frac{{d{\bf{T}}/dt}}{{\left| {d{\bf{T}}/dt} \right|}} = \frac{{\frac{1}{{\sqrt 2 }}{\text{sec}}{{\text{h}}^2}t{\bf{i}} + \frac{1}{{\sqrt 2 }}\sec {\text{h}}t\tanh t{\bf{k}}}}{{\frac{{{\text{sech}}t}}{{\sqrt 2 }}}} \cr & {\bf{N}} = \frac{{d{\bf{T}}/dt}}{{\left| {d{\bf{T}}/dt} \right|}} = {\text{sech}}t{\bf{i}} + \tanh t{\bf{k}} \cr & \cr & {\text{Calculate }}\kappa {\text{ using the equation }}\kappa = \frac{1}{{\left| {\bf{v}} \right|}}\left| {\frac{{d{\bf{T}}}}{{dt}}} \right|{\text{ }}\left( {{\text{see page 764}}} \right).{\text{ Then}} \cr & \kappa = \frac{1}{{\left| {\bf{v}} \right|}}\left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \frac{1}{{\sqrt 2 \cosh t}}\left( {\frac{{{\text{sech}}t}}{{\sqrt 2 }}} \right) \cr & \kappa = \frac{1}{{\left| {\bf{v}} \right|}}\left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \frac{1}{2}{\text{sec}}{{\text{h}}^2}t \cr} $$