# Prove that the planar curve obtained by projecting $\alpha$ into its osculating plane at $P$ has the same curvature at $P$ as $\alpha$.

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!

• 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. – Sarah T. Jan 22 at 1:36
• Incidentally, this is recognizably a problem from Ted Shifrin's differential geometry notes. I've edited to add a reference. – Sarah T. Jan 22 at 1:41
• Okay cool. But since the parameterization is local, shouldn't we be able to ignore the higher order terms? – user624065 Jan 22 at 1:41
• Yes, higher order terms can be ignored. – Sarah T. Jan 22 at 1:44
• 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 $1+3ks+3(ks)^2+(ks)^3$ – user624065 Jan 22 at 21:28