When are the fibres totally geodesic? Let $I \subseteq \mathbb{R}$ be an open interval and $(M,g_0)$ a Riemannian manifold. Given a smooth function $f : I \to (0, \infty)$, consider $I \times_f M$ with the Riemannian metric
$$ g((r,u), (s,v)) = rs + f(t)^2 g_0(u,v), \quad (r,u), (s,v) \in T_{t} I \oplus  T_p M $$
For general $f$, it is known that the leaves $I \times \{p\}$ are totally geodesic, while the fibres $M_t :=\{t\} \times M$ are only totally umbilical.
My question is:

What conditions on $f$ ensure that each $M_t$ is also totally geodesic? (e.g. $f$ affine,..) 

 A: If one has a smooth family of compact minimal submanifolds $\Sigma_t$ in a Riemannian manifold, then by the first variation formula, 
$$ \frac{d}{dt} Vol (\Sigma_t) = \int_{\Sigma _t } V_t \cdot H_t d\mu_t = 0,$$
here $V_t = \partial_t \Sigma_t$ is the variational vector fields and $H_t$ is the mean curvature vector, which is zero by definition. Thus the volume is unchanged and using the expression of the metric, it implies that $f$ is constant. 
The above does not work if $M$ is non-compact. In general, you can just calculate: Let $(x_1, \cdots, x_n)$ be a local coordinates of $(M, g_0)$ with Christoffel symbols $\Gamma_{ij}^k$. Then in the product manifold with the extra coordinates $t$, the new Christoffel symbols are 
\begin{align}
\overline\Gamma_{ij}^t &= -\frac 12g^{tt} \partial_t g_{ij} = -\frac{f'}{2} (g_0)_{ij}.
\end{align} 
Thus 
$$ \overline\nabla _{\frac{\partial}{\partial x_i}} \frac{\partial}{\partial x_j} = \overline \Gamma_{ij}^k \frac{\partial}{\partial x_k } + \overline\Gamma_{ij}^t \frac{\partial}{\partial t}$$
which implies 
$$A_{ij} = g\left( \overline\nabla _{\frac{\partial}{\partial x_i}} \frac{\partial}{\partial x_j}, \frac{\partial}{\partial t}\right) = -\frac{f'}{2} (g_0)_{ij}.$$
and $H = -\frac{nf'}{2f}$. This implies that $A= 0$ if and only if $H = 0$ if and only if $f' = 0$. Thus if each $M_t$ is totally geodesic, than $f' = 0$ for all $t$ and thus $f$ is constant. 
