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, (http://alpha.math.uga.edu/~shifrin/ShifrinDiffGeo.pdf).)

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!

  • $\begingroup$ Where you say that the local projection into the osculating plane is given by $\left(s, \frac{\kappa}{2}s^2 \right)$, you should actually have the parameterization $(s + O(s^3), \frac{\kappa_0}{2}s^2 + O(s^3)$, where $\kappa_0$ is the curvature of $s$ at $P$. In particular, $\kappa_0$ is a constant and should not be differentiated. This will make the rest of the computations a lot simpler. $\endgroup$ – Sarah T. Jan 22 at 1:36
  • $\begingroup$ Incidentally, this is recognizably a problem from Ted Shifrin's differential geometry notes. I've edited to add a reference. $\endgroup$ – Sarah T. Jan 22 at 1:41
  • $\begingroup$ Okay cool. But since the parameterization is local, shouldn't we be able to ignore the higher order terms? $\endgroup$ – user624065 Jan 22 at 1:41
  • $\begingroup$ Yes, higher order terms can be ignored. $\endgroup$ – Sarah T. Jan 22 at 1:44
  • $\begingroup$ Okay, if $\gamma(s)=(s,\frac{ks^2}{2})$ then I got that $\gamma'(s) \times \gamma''(s)=\kappa$. However, the denominator $|\gamma'(s)|$^3 is not behaving well... Is there a reason that $\gamma$ is arc length parameterized because $\alpha(s)$ is? Otherwise, the donominator will be $1+3ks+3(ks)^2+(ks)^3$ $\endgroup$ – user624065 Jan 22 at 21:28

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