# Angle between two curves on a surface

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).
• Well, tell me Euler's formula and what it says if $\theta$ corresponds to a direction of normal curvature $0$. Commented Feb 7, 2016 at 17:55
• 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). Commented Feb 7, 2016 at 18:18
• Not quite right. You should have $\cos^2\theta$ and $\sin^2\theta$. Does that help? Commented Feb 7, 2016 at 18:53
• I am guessing that now it is time to use some trigonometric identities + the fact that $\alpa = \theta_1 - \theta_2$ . Commented Feb 7, 2016 at 19:15