In a cylinder compute the normal curvature at $p_0$ in an arbitrary direction by three different methods:
1.) Using the equation: $II_p((\cos\theta)v_1+(\sin\theta)v_2)=k_1\cos^2\theta+k_2\sin^2\theta$.
2.) By parametrizing a curve $\gamma$ in an arbitrary direction and differentiating the restriction of $N$ to $\gamma$.
3.) By parametrizing a curve $\gamma$ in an arbitrary direction and computing the normal component of $\gamma''(0)$.
A cylinder given as $C(r)=\{(r\cos\theta,r\sin\theta,z)\in R^3 \mid \theta\in[0,2\pi),z∈R\}$ and $p_0=(r\cos\theta,r\sin\theta,z_0)\in C_r$.
So I believe im looking for the normal curvature at $p_0$ in the direction of $(\cos\theta)v_1+(\sin\theta)v_2$, and I need to start by finding the principal curvatures $k_1$ and $k_2$.
The normal curvatures at $p_0$ occur in the “around” direction $v_1 =(− \sin θ_0, \cos θ_0, 0)$ and the “upward” direction $v_2 = (0, 0, 1)$, so we will express the Weingarten map with respect to this basis.
The curve $\gamma_1(t)=(r \cos(θ_0 + t/r), r \sin(θ_0 + t/r), z_0) $ in $ C(r)$ satisfies $γ_1(0) = p_0$ and $γ_1(0) = v_1$. The restriction of $N$ to $γ_1$ is $N_1(t) = N(γ1(t)) = (\cos(θ_0 + t/r),\sin(θ_0 + t/r), 0)$, so $W_{p_0} (v1) = −N'_1(0) = \frac{-1}{r}v_1.$
Similarly $W_{p_0} (v_2) = −N'_2(0) = (0, 0, 0) = 0 · v_2$.
Then $k_1=-\frac{-1}{r}$ and $k_2=0$.