Question

Question not in drive.

Short answer

The graph of given vector function \({\bf{r}}(t)\) and its corresponding curvature function \(\kappa (t)\) is attached inside.

Step-by-step solution

Step 1: Given information

The curvature of the curve given by the vector function \({\bf{r}}\) is

\(\kappa (t) = \frac{{\left| {{r^1}(t) \times {r^{''}}(t)} \right|}}{{{{\left| {{r^'}(t)} \right|}^3}}}\)

Given that

\(r(t) = t{e^t},{e^{ - t}},\sqrt 2 t\)

Therefore,

\({r^'}(t) = {e^t} + t{e^t}, - {e^{ - t}},\;\;\;\sqrt 2 \)

And

\({r^{''}}(t) = 2{e^t} + t{e^t},{e^{ - t}},0\)

\(\begin{aligned}{l}\left| {{r^'}(t)} \right| = \sqrt {{{\left( {{e^t} + t{e^t}} \right)}^2} + {{\left( { - {e^{ - t}}} \right)}^2} + {{(\sqrt 2 )}^2}} \\{\left| {{r^'}(t)} \right|^3} = {e^{2t}}{(t + 1)^2} + {e^{ - 2t}} + {2^{\frac{3}{2}}}\end{aligned}\)

\({r^'}(t) \times {r^{''}}(t) = \left| {\begin{aligned}{*{20}{c}}i&j&k\\{{e^t} + t{e^t}}&{ - {e^{ - t}}}&{\sqrt 2 }\\{2{e^t} + t{e^t}}&{{e^{ - t}}}&0\end{aligned}} \right|\)

\( = {\bf{i}}\left| {\begin{aligned}{*{20}{c}}{ - {{\rm{e}}^{ - t}}}&{\sqrt 2 }\\{{{\rm{e}}^{ - t}}}&0\end{aligned}} \right| - {\bf{j}}\left| {\begin{aligned}{*{20}{c}}{{{\rm{e}}^t} + t{{\rm{e}}^t}}&{\sqrt 2 }\\{2{{\rm{e}}^t} + t{{\rm{e}}^t}}&0\end{aligned}} \right| + {\bf{k}}\left| {\begin{aligned}{*{20}{c}}{{{\rm{e}}^t} + t{{\rm{e}}^t}}&{ - {{\rm{e}}^{ - t}}}\\{2{{\rm{e}}^t} + t{{\rm{e}}^t}}&{{{\rm{e}}^{ - t}}}\end{aligned}} \right|\)

\( = - \sqrt 2 {e^{ - t}}i + \sqrt 2 \left( {2{e^t} + t{e^t}} \right)j + (3 + 2t)k\)

\(\left| {{r^'}(t) \times {r^{''}}(t)} \right| = \sqrt {2{e^{ - 2t}} + 2{e^{2t}}{{(2 + t)}^2} + {{(3 + 2t)}^2}} \)

Step 2: Final proof

\(\kappa (t) = \frac{{\left| {{r^'}(t) \times {r^{''}}(t)} \right|}}{{{{\left| {{r^'}(t)} \right|}^3}}} = \frac{{\sqrt {2{e^{ - 2t}} + 2{e^{2t}}{{(2 + t)}^2} + {{(3 + 2t)}^2}} }}{{{{\left( {{e^{2t}}{{(t + 1)}^2} + {e^{ - 2t}} + 2} \right)}^{\frac{3}{2}}}}}\)

Graph for the function \(y = \kappa (t)\) is shown below :

Graph for the vector function is shown below :

\(r(t) = t{e^t},{e^{ - t}},\sqrt 2 t\)

The graph of given vector function \({\bf{r}}(t)\) and its corresponding curvature function \(\kappa (t)\) is attached inside.