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# User:Michael B. Porter/Teaching/Curvature problem

Find the curvature of the curve ${\displaystyle {\textbf {r}}(t)=t^{3}{\textbf {i}}+t^{2}{\textbf {k}}}$.

There are two formulas we learned for curvature:

${\displaystyle \kappa ={\frac {|{\textbf {r}}'\times {{\textbf {r}}''}|}{|{\textbf {r}}'|^{3}}}}$

${\displaystyle \kappa ={\frac {|{\textbf {T}}'|}{|{\textbf {r}}'|}}}$

There was a lot more calculation involved in using the second formula. Here is the solution using the first formula:

${\displaystyle {\textbf {r}}'=3t^{2}{\textbf {i}}+2t{\textbf {k}}}$

${\displaystyle {\textbf {r}}''=6t{\textbf {i}}+2{\textbf {k}}}$

${\displaystyle {\textbf {r}}'\times {\textbf {r}}''={\begin{vmatrix}{\textbf {i}}&{\textbf {j}}&{\textbf {k}}\\3t^{2}&0&2t\\6t&0&2\end{vmatrix}}=0{\textbf {i}}-(6t^{2}-12t^{2}){\textbf {j}}+0{\textbf {k}}=6t^{2}{\textbf {j}}}$

${\displaystyle |{\textbf {r}}'\times {\textbf {r}}''|=6t^{2}}$

${\displaystyle |{\textbf {r}}'|={\sqrt {9t^{4}+4t^{2}}}}$

so

${\displaystyle \kappa ={\frac {6t^{2}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}}$

And here is the solution using the second formula:

${\displaystyle \mathbf {T} ={\frac {{\textbf {r}}'}{|{\textbf {r}}'|}}={\frac {3t^{2}}{\sqrt {9t^{4}+4t^{2}}}}{\textbf {i}}+{\frac {2t}{\sqrt {9t^{4}+4t^{2}}}}{\textbf {k}}}$

${\displaystyle \mathbf {T} '={\frac {d}{dt}}{\frac {3t^{2}}{\sqrt {9t^{4}+4t^{2}}}}{\textbf {i}}+{\frac {d}{dt}}{\frac {2t}{\sqrt {9t^{4}+4t^{2}}}}{\textbf {k}}}$

The derivatives are a little tricky:

${\displaystyle {\frac {d}{dt}}{\frac {3t^{2}}{\sqrt {9t^{4}+4t^{2}}}}={\frac {6t{\sqrt {9t^{4}+4t^{2}}}-3t^{2}({\frac {1}{2}}(9t^{4}+4t^{2})^{-{\frac {1}{2}}})(36t^{3}+8t)}{9t^{4}+4t^{2}}}={\frac {6t(9t^{4}+4t^{2})-3t^{2}({\frac {1}{2}})(36t^{3}+8t)}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}}$

${\displaystyle ={\frac {54t^{5}+24t^{3}-54t^{5}-12t^{3}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}={\frac {12t^{3}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}}$

${\displaystyle {\frac {d}{dt}}{\frac {2t}{\sqrt {9t^{4}+4t^{2}}}}={\frac {2{\sqrt {9t^{4}+4t^{2}}}-2t({\frac {1}{2}}(9t^{4}+4t^{2})^{-{\frac {1}{2}}})(36t^{3}+8t)}{9t^{4}+4t^{2}}}={\frac {2(9t^{4}+4t^{2})-2t({\frac {1}{2}})(36t^{3}+8t)}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}}$

${\displaystyle ={\frac {18t^{4}+8t^{2}-36t^{4}-8t^{2}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}={\frac {-18t^{4}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}}$

so

${\displaystyle \mathbf {T} '={\frac {12t^{3}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}{\textbf {i}}-{\frac {18t^{4}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}{\textbf {k}}}$

${\displaystyle |\mathbf {T} '|={\frac {\sqrt {(12t^{3})^{2}+(18t^{4})^{2}}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}={\frac {6t^{2}{\sqrt {(2t)^{2}+(3t^{2})^{2}}}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}={\frac {6t^{2}{\sqrt {9t^{4}+4t^{2}}}}{(9t^{4}+4t^{2})^{\frac {3}{2}}}}={\frac {6t^{2}}{9t^{4}+4t^{2}}}}$

and finally,

[itex]\kappa=\frac{|\textbf{T}'|}{|\textbf{r}'|}=\frac{\frac{6t^2}{9t^4+4t^2}}{\sqrt{9t^4+4t^2}}=\frac{6t^2}{(9t^4+4t^2)^\frac{3}{2}}