Suppose that $N(t)$ is a family of k-dimensional submanifolds in an n-dimensional manifold $M$, with $n > k$. Then $N(t)$ is a path in the space $\mathcal{M}$ of all k-dimensional submanifolds. I want to know, what is "$N'(t)$"?
If $r(t)$ is a curve in a manifold $M$, which is finite-dimensional, I can easily calculate it's derivative $r'(t)$. How would I get the analogous thing for $N(t)$ as in the situation above?
Some idea:
Suppose our family is $L_t = \{ (x,y) \in \mathbb{R}^2 : y = tx \}$ where $t$ is any number. Then maybe we can write:
$$ \frac{\partial}{\partial t} L_t = \langle 0 , x \rangle $$
This vector field doesn't seem to be "normal" to the line though... So I'm not sure if that's correct.
The setting where I know it's defined is the one of parameterized submanifolds: You fix two manifolds $M$ and $N$ and consider the space of embeddings (or, more generally, smooth maps) maps $N\to M$. A curve in this space (or, a a 1-parameter family of submanifolds) is a smooth map $u: N\times [0,1]\to M$ such that for each $t$, $u(\cdot, t)$ is an embedding. Then the time-derivative of such a map is $u'(t)$, i.e. a vector field in $M$ along the submanifold $u(t)\subset M$.
There is also a version for families of unparameterized submanifolds, in which case the derivative $N'(t)$ is a section of of the normal bundle of $N(t)\subset M$. If you know what the normal bundle (and a section) is, I can explain the details.
Edit. As requested, here is an explicit example. Consider the family of parameterized lines $u: (x,t)\mapsto (x,tx)$ in the plane, where $t$ is the time parameter of the family. (Each graph $y=tx$ is a line of the slope $t$.) The time-derivative of this family is $$ \frac{\partial u}{\partial t}(x,t)= (0,x). $$ Therefore, for each fixed $t$, the time derivative is the restriction to the line $y=tx$ of the vertical vector field $V(x,y)=(0,x)$.