Let $\mathcal{M}$ be a surface and $\gamma_1, \gamma_2$ two smooth curves contained in $\mathcal{M}$ in natural parameterization s.t.: $\gamma_1(0)=\gamma_2(0) = p$ , $\ddot{\gamma}_1(0),\ddot{\gamma}_2(0)\in T_p\mathcal{M}$. Let $\alpha\in [0,\pi/2]$ be the angle between the two curves. Given that $\alpha \neq 0$. Let $\lambda_1,\lambda_2$ be the two principal curvatures of $\mathcal{M}$ at $0$, s.t. $0\neq |\lambda_2| \geq |\lambda_1|$.Prove:

$ |\frac{\lambda_1}{\lambda_2}| = \tan({\alpha/2})$

Can Someone please give me a hint. I don't really have a clue how to approach this. I tried using the fernet equations and differentiating $cos({\theta(t)}) = \langle\dot{\gamma_1},\dot{\gamma_2} \rangle$ , but neither one helped me.


HINT: $\gamma_1$ and $\gamma_2$ are asymptotic curves (their tangent vectors are directions of normal curvature $0$). Write down Euler's formula for the normal curvature in direction $\theta$ (relative to the first principal direction, say).

  • $\begingroup$ Hey Ted, I am still having hard time, can you give me an extra hint? $\endgroup$ – UserB95 Feb 7 '16 at 10:28
  • $\begingroup$ Well, tell me Euler's formula and what it says if $\theta$ corresponds to a direction of normal curvature $0$. $\endgroup$ – Ted Shifrin Feb 7 '16 at 17:55
  • $\begingroup$ If I understood correct (since in my course we haven't learned about Euler's formula, so you are welcome to correct me) , euler's formula says that if the angle between $\gamma$ and the principal direction that correspond to $\lambda_1$ is $\theta$ , then we can write $K_{\gamma} = \lambda_1 \cos{\theta} + \lambda_2 \sin{\theta}. $So if the normal curvature is 0 , then $\tan{\theta} = -\frac{\lambda_1}{\lambda_2} $(Assuming that the angle doesn't make $\cos$ vanish). $\endgroup$ – UserB95 Feb 7 '16 at 18:18
  • $\begingroup$ Not quite right. You should have $\cos^2\theta$ and $\sin^2\theta$. Does that help? $\endgroup$ – Ted Shifrin Feb 7 '16 at 18:53
  • $\begingroup$ I am guessing that now it is time to use some trigonometric identities + the fact that $\alpa = \theta_1 - \theta_2$ . $\endgroup$ – UserB95 Feb 7 '16 at 19:15

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