$\mathbf{The \ Problem \ is :}$ Let $(M^n,g)$ be a complete, non-compact Riemannian manifold with bounded sectional curvature. Let $\{M_t\}_t$ be an equidistant family of hypersurfaces with $M^n=\cup_{t\in \mathbb{R}} M_t.$ Let $L(t)$ denote the Weingarten map of $M_t.$ We want to show that under this assumptions, $|L(t)|$ is bounded.
Suppose that if $n=2$ and $(M^2,g)$ be a surface with $|K_g| \leq K$ with $K \geq 2.$ Let $l(t_0) \geq K$ for some $t_0\in \mathbf{R}$ and $l(t)\in \mathbf{R}$ denote the shape operator of $M_t.$ Then $l^{\prime}(t_0) \leq \frac{1}{2}l(t_0)^2.$ Also show that the differential equation $f^{\prime}=-\frac{1}{2}f^2$ with $f(t_0)=f_0$ has a solution. Show that this contradicts the assumption that $L(t)$ exists for all time $t.$
$\mathbf{My \ Approach :}$ Firstly, the first thing that comes into mind is the Riccati equation which is : $l^{\prime}(t)+l^2(t)+R_{J(t),\gamma^{\prime}(t)}\gamma^{\prime}(t)=0$ for all $t\in \mathbf{R}$ where $\gamma : \mathbf{R}\to (M^2,g)$ is the geodesic with $\gamma(t_0)=p$ and $\gamma^{\prime}(t_0)=v.$ Then the Jacobi field along $\gamma$ with $J(t_0)=v$ with $\|v\|=1$ and $J^{\prime}(t_0)=w$ with $\|w\|=1$ can be given by $J(t)=\operatorname{exp}_{tv}{tw}.$ Then $K_t:= \langle R_{J(t),\gamma^{\prime}(t)}\gamma^{\prime}(t),J(t) \rangle=K(J(t),\gamma^{\prime}(t))$ and so $K_t \leq K$ for all $t$ but I can't approach further. I am quite sure that I am missing something which will be able to provide a solution but I still can't find it out. Thanks in advance for any help. Also I found several well-known books on Riemannian Geometry but couldn't find such type of problem. It would be so helpful to get a reference for these kind of problems .