I am reading an article which has a fixed $S^1 \times I$ where $S^1$ is the unit circle and $I \subset \mathbb{R}$ with coordinates $(\theta,t)$. Fix a Riemannian metric $$ds^2 = dt^2 + f(t)^2d\theta^2.$$It says that the Gauss curvature depends only on $t$ and is given by $$K(t) = -\frac{f''(t)}{f(t)}.$$
I can't reach this expression. I thought about using the curvature operator $$R : \mathfrak{X}(S^1 \times I) \times \mathfrak{X}(S^1 \times I) \times \Gamma(T\mathbb{R}^3\mid_{S^1 \times I}) \rightarrow \Gamma(T\mathbb{R}^3\mid_{S^1 \times I})$$ given by $$R(X,Y) = \nabla_X \nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}.$$And the problem would reduce to finding the coordinate expression for the connection. Then I would use the formula of Gaussian curvature: $$K = \langle R(e_1,e_2)e_2,e_1\rangle.$$ But this means that we only need the Christoffel symbols to compute. Indeed, I computed and got $$\Gamma_{11}^1 = 0, \Gamma_{12}^1 = \Gamma_{21}^1 = f^3(t)f'(t), \ \Gamma_{22}^1 = 0,$$and $$\Gamma_{11}^2 = f(t)f'(t), \ \Gamma_{12}^2 = \Gamma_{21}^2 = \Gamma_{22}^2 = 0.$$But then, using the formula for the Gaussian curvature we see that $$R(\partial_\theta,\partial_t)\partial_t = 0.$$And this would imply that the Gaussian curvature is zero and it shouldn't according with the article. (Notice that it is known that the Gaussian curvature is zero in the cylinder...)
Thank you in advance.
Edit: I found this post which express the Gausian curvature for a warped product space in terms of the function associated. I see that this answers my question by making $u = t$ and $v = \theta$.
New question: How did the person that answered derive the $\omega_1$ and $\omega_2$?