In this lesson, Grant/Khan taught that $K=\frac{|\vec{s}'(t) \times \vec{s}''(t)}{||\vec{s}'(t)||^3}$:
=
If the formula is right, is it generalisable to all dimensions? I applied it to $\vec{s}(t)= (\cos t, \sin t, \frac{t}{5})$ to find the curvature function for the graph (a helix) and got the following:
- first & second derivative of $\vec{s}(t)$:
$\vec{s}'(t)= \begin{bmatrix} -\sin t\\\cos t \\ \frac{1}{5}\end{bmatrix}$ , $\vec{s} ''(t) =\begin{bmatrix} - \cos t\\ -\sin t \\ 0\end{bmatrix}$
- Cross the numerator, and divided it by the denominator:
$\frac{\frac{1}{5}(\sin t - \cos t t)+1}{\frac{26\sqrt{26}}{125}}$, which seem to be the wrong answer.
I've verified that the derivatives, cross product and the magnitude in the denominator is correct. What aim I missing here?
The right answer is simply a constant $K=\frac{25}{26}$ which makes sense. The only difference is that I followed the formula then normalized it at the end. But Grant normalized the numerator first.
for the numerator $\vec{s}'(t) \times \vec{s} ''(t) = \begin{bmatrix} -\sin t\\\cos t \\ \frac{1}{5}\end{bmatrix} \times \begin{bmatrix} - \cos t\\ -\sin t \\ 0\end{bmatrix} = \begin{bmatrix} \frac15 \sin t\\ -\frac15 \cos t \\ 1\end{bmatrix} $ with norm $n=\sqrt{\frac{1}{25}+1}$
for the denominator, norm is $d=\sqrt{\frac{1}{25}+1}$
Finally curvature is $$ k=\frac{n}{d^3}=\frac{1}{\frac{1}{25}+1}=\frac{25}{26}$$