Let $\alpha(s)$ be a regular curve with $\kappa\neq0$ at P, where $\kappa$ is the curvature. Prove that the planar curve obtained by projecting $\alpha(s)$ into its osculating plane at $P$ has the same curvature at $P$ as $\alpha(s)$. (This is problem 1.2.14 in Ted Shifrin's differential geometry notes, ().)
The local projection onto the osculating plane is given by $(s,\frac{s^2\kappa}{2})$. However, i'm not sure if this equation holds only for when curves are parametrized by arc length.
Anyway, lets say that this projection is the correct projection for $\alpha(s)$ even though it's not stated to be parameterized by arc length.
The curvature is given by $\frac{|\alpha'(s) \times \alpha''(s)|}{\alpha'(s)}$.
Let $\gamma=(s,\frac{s^2\kappa}{2})$.
$\rightarrow$ $\gamma'=(1,s\kappa+\frac{s^2\kappa'}{2})$
$\rightarrow$ $\gamma''=(0,\kappa+s\kappa'+s\kappa'+\frac{s^2\kappa''}{2})$
$\rightarrow$ $|\gamma'|=(1^2+(sk)^2+(\frac{s^2\kappa'}{2})^2+s^3\kappa\kappa')^{1/2}$
So, if I apply the equation for curvature to $\gamma$, in theory I should get $\kappa$. Is this reasoning correct? Thanks!
$\endgroup$ 5 Reset to default
