Didn't mean to bother you, but I don't know exactly what is going on here, I'm trying to have a grasp on proving the following formula:
$\displaystyle \kappa = \frac{\Vert\dot \gamma \times \ddot \gamma\Vert}{\Vert\dot \gamma\Vert^3}\quad$ where $\kappa$ stands for curvature, and $\gamma$ is a parameterised curve of time $t$.
Here I upload an extract of what I'm attending atm.
In line $4$, the third identity what has exactly been done by author?
Are we allowed to do such things in elemantary calculus? I mean sending $ds$ of $d/ds$ into the numerator of $\displaystyle \frac{d\gamma/dt}{ds/dt}$ and pulling its $dt$ back where $ds$ was, without even touching the denominator?! And the funny thing is after applying this substituting, $d/dt$ became the operator operating only on the numerator, and not the denominator.
Am I missing something?! Another proof of this would also be much obliged ;-)
The curvature is the norm of the second derivative of the curve; however, one uses not the usual derivative with respect to $t$, but the derivative with respect to the unit-speed parameter $s$. This derivative is given by $$ \frac{d}{ds} f(t) = \frac{1}{\| \dot \gamma \|} \frac{df}{dt}\,. $$ The unit-speed parameter is characterized by $$ \frac{ds}{dt} = \| \dot \gamma \|\,,$$ and taken together you obtain the suspicious-looking formula $$ \frac{d}{ds} f(t) = \frac{df/dt}{ds/dt}\,. $$