Thomas' Calculus 13th Edition

$${\bf{T}} = \frac{1}{{\sqrt {1 + {t^2}} }}\left( {{\bf{i}} - t{\bf{j}}} \right),\,\,\,\,\,\,\,\,\,{\bf{N}} = - \frac{t}{{\sqrt {1 + {t^2}} }}{\bf{i}} - \frac{1}{{\sqrt {1 + {t^2}} }}{\bf{j}}\,,\,\,\,\,\,\,\,\,\,\kappa = \frac{1}{{2{{\left( {1 + {t^2}} \right)}^{3/2}}}}$$
\eqalign{ & {\bf{r}}\left( t \right) = \left( {2t + 3} \right){\bf{i}} + \left( {5 - {t^2}} \right){\bf{j}} \cr & \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( {2t + 3} \right){\bf{i}} + \left( {5 - {t^2}} \right){\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = 2{\bf{i}} - 2t{\bf{j}} \cr & \cr & {\text{We calculate }}{\bf{T}}{\text{ from the velocity vector}} \cr & {\bf{v}}\left( t \right) = 2{\bf{i}} - 2t{\bf{j}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 2t} \right)}^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {4 + 4{t^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {4\left( {1 + {t^2}} \right)} \cr & \left| {{\bf{v}}\left( t \right)} \right| = 2\sqrt {1 + {t^2}} \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{{2{\bf{i}} - 2t{\bf{j}}}}{{2\sqrt {1 + {t^2}} }} \cr & {\bf{T}} = \frac{{2{\bf{i}}}}{{2\sqrt {1 + {t^2}} }} - \frac{{2t{\bf{j}}}}{{2\sqrt {1 + {t^2}} }} \cr & {\bf{T}} = \frac{1}{{\sqrt {1 + {t^2}} }}\left( {{\bf{i}} - t{\bf{j}}} \right) \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 {1 + {t^2}} }}\left( {{\bf{i}} - t{\bf{j}}} \right)} \right) \cr & \frac{{d{\bf{T}}}}{{dt}} = \frac{d}{{dt}}\left( {{{\left( {1 + {t^2}} \right)}^{ - 1/2}}{\bf{i}} - t{{\left( {1 + {t^2}} \right)}^{ - 1/2}}{\bf{j}}} \right) \cr & \frac{{d{\bf{T}}}}{{dt}} = - \frac{1}{2}{\left( {1 + {t^2}} \right)^{ - 3/2}}\left( {2t} \right){\bf{i}} - \left( {\frac{1}{2}{{\left( {1 + {t^2}} \right)}^{ - 3/2}}\left( {2t} \right)t + {{\left( {1 + {t^2}} \right)}^{ - 1/2}}} \right){\bf{j}} \cr & \frac{{d{\bf{T}}}}{{dt}} = - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{i}} - \left( {\frac{{{t^2}}}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}} + \frac{1}{{\left( {1 + {t^2}} \right)}}} \right){\bf{j}} \cr & \frac{{d{\bf{T}}}}{{dt}} = - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{i}} - \frac{1}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{j}} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \sqrt {{{\left( { - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}} \right)}^2} + {{\left( {\frac{1}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}} \right)}^2}} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \sqrt {\frac{{{t^2}}}{{{{\left( {1 + {t^2}} \right)}^3}}} + \frac{1}{{{{\left( {1 + {t^2}} \right)}^3}}}} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \sqrt {\frac{{{t^2} + 1}}{{{{\left( {1 + {t^2}} \right)}^3}}}} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \sqrt {\frac{1}{{{{\left( {1 + {t^2}} \right)}^2}}}} \cr & \left| {\frac{{d{\bf{T}}}}{{dt}}} \right| = \frac{1}{{1 + {t^2}}} \cr & {\bf{N}} = \frac{{d{\bf{T}}/dt}}{{\left| {d{\bf{T}}/dt} \right|}} = \frac{{ - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{i}} - \frac{1}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}{\bf{j}}}}{{1/\left( {1 + {t^2}} \right)}} \cr & {\bf{N}} = - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}\left( {1 + {t^2}} \right){\bf{i}} - \frac{1}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}\left( {1 + {t^2}} \right){\bf{j}} \cr & {\bf{N}} = - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{1/2}}}}{\bf{i}} - \frac{1}{{{{\left( {1 + {t^2}} \right)}^{1/2}}}}{\bf{j}} \cr & {\bf{N}} = - \frac{t}{{\sqrt {1 + {t^2}} }}{\bf{i}} - \frac{1}{{\sqrt {1 + {t^2}} }}{\bf{j}} \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}{{2\sqrt {1 + {t^2}} }}\left( {\frac{1}{{1 + {t^2}}}} \right) \cr & \kappa = \frac{1}{{2{{\left( {1 + {t^2}} \right)}^{3/2}}}} \cr}