Curvature of a curve on a surface

Question

I dont know how to solve this problem in Do Carmo's Differential geometry of Curves and Surfaces. Can anyone help me?

Let $C$ be a regular curve on a surface $S$ with Gaussian curvature $K>0$ and principle curvatures $k_1$ and $k_2$. Show that the curvature $k$ of $C$ at $p$ satisfies: $k\ge min\{|k_1|,|k_2|\}$.

Do I have to use the normal curvature to solve the problem?

Answer 1

I think you wanted to ask

Show that the normal curvature $k_n$ of $C$ at $p$ satisfies:

$$k_n \ge min\{|k_1|,|k_2|\}$$

$$k_n \le max\{|k_1|,|k_2|\}$$

Or

$$ |k_1|> k_n >| k_2|$$

This comes out of Euler scalar normal curvature identity by finding max/min values by differentiation.

$$ k_n = k_1 cos^2 \theta + k_2 sin^2 \theta $$ Where $\theta$ is angle curve makes to direction 1.

EDIT1 If at all another curvature is to be brought in, that is geodesic torsion$ \tau_g$ rather than $ k_g$ included in the Mohr Circle.. and makes more clear their magnitudes geometrically.

Answer 2

Yes you have to use the normal curvature. Because of \begin{equation}\kappa^2=\kappa_g^2+\kappa_n^2 \end{equation} it is sufficient to prove $\kappa_n\geq \min \{k_1,k_2\}$, where $\kappa_g, \kappa_n$ denote the geodesic and normal curvature. From now on we may assume that the curve $c:I\to S$ is parametrised by arc length. Therefore we can write (at $t_0$ arbitrary) $c'=\cos(\alpha) w_1 + \sin(\alpha) w_2$ for some angle $\alpha.$ Then \begin{equation}\kappa_n=\langle c'', \nu\circ c \rangle= -\langle c',(d\nu)(c')\rangle=II(c',c') \end{equation} and by using the bilinearity of the second fundamental form we get (depending on the sign convention of the principal curvatures) \begin{equation}II(c',c')=\cos(\alpha)^2 k_1 + \sin(\alpha)^2 k_2 \end{equation} We can choose $\nu$ in such a way that $k_1,k_2>0$. Furthermore we may assume $k_1\geq k_2.$ Putting everything together we obtain \begin{equation}\kappa\geq \kappa_n=II(c',c')=\cos(\alpha)^2 k_1 + \sin(\alpha)^2 k_2\geq \cos(\alpha)^2 k_2 + \sin(\alpha)^2k_2=k_2=\min\{k_1,k_2\}. \end{equation}

Answer 3

The normal curvature, by Euler's formula is $$k_n (\theta) = k_1 \cos^2 \theta + k_2 \sin^2 \theta$$ Also, by definition we have that normal curvature is $k_n = k \cos \theta$. Since curvature $k>0$ so $|k|=k$. We also use the fact $|\cos \theta| \leq 1$.

By hypotesis $K = k_1 \cdot k_2 >0$ and then the principal curvatures must be either both positive of both negative. So $k_1, k_2$ having the same sign means that we will never have to deal with $\pm k_1 \cos^2 \theta \mp k_2 \sin^2 \theta$.

Therefore

$$k = |k| \geq |k \cos \theta| = |k_n| = \pm k_n = \pm(k_1 \cos^2 \theta + k_2 \sin^2 \theta) = \pm k_1 \cos^2 \theta \pm k_2 \sin^2 \theta = |k_1| \cos^2 \theta + |k_2| \sin^2 \theta \geq \min \{|k_1|,|k_2|\} \cos^2 \theta + \min \{|k_1|,|k_2|\} \sin^2 \theta = \min \{|k_1|,|k_2|\}.$$