Number of continuous $[0; 1] \to [0; 1]$ functions for given arc length Just out of pure curiosity ...
Suppose I want to connect the two points $(0,0)$ and $(1,1)$ with the graph of some continuous and differentiable function 
$$f : [0; 1] \to [0; 1]$$
and let $s$ be the arc length of that function in $[0; 1]$.
Of course, the function with minimum $s$ that satisfies the above conditions is $f(x) = x$ with $s = \sqrt 2$. So for $s = \sqrt 2$, exactly one matching function can be found.
But what happens to the number of these functions if $s$ increases?
Surely, more functions can be found to match the given arc length - uncountably many more I suppose due to the nature of the real numbers.
But intuitively, I'd think that the number of such functions grows even more the greater $s$ gets, since there is more "space" the graph can use.
So, despite continuum cardinality, are there any means of measuring the number of such functions against $s$ or is it all the same once that minimal way of $f(x) = x$ as been taken?
And would this change if we limited the ways of constructing such functions to e.g. some elementary ones?
 A: There is probably a sophisticated answer to this; let me just mention an unsophisticated one.  Several possible discrete analogues of the question you ask exhibit the desired behavior.  Consider, for example, the problem of finding a path from $(0, 0)$ to $(n, n)$ in the lattice $\mathbb{Z}^2$ where the only allowable steps are between diagonally adjacent lattice points.  Let the length of such a path be the total number of steps.  Then, of course, there is only one path of length $n$, which is the obvious diagonal one.  However, there are ${n+2 \choose 2} + n$ paths of length $n+2$, already a big leap.
The general answer can be computed as follows.  In a path of length $n+2k$ we need to choose $2j$ of the steps to be up-left or down-right (for some $j$ between $0$ and $k$) and we need to choose $n+2k-2j$ of the steps to be up-right or down-left.  Of the first set of steps, half need to be of one type and half need to be the other, and of the second set of steps, $n+k-j$ need to be of one type and $k-j$ need to be the other.  This gives the total number of paths of length $n+2k$ as
$$\sum_{j=0}^{k} {n+2k \choose 2j} {2j \choose j} {n+2k-2j \choose k-j}.$$
The dominant term when $n$ is large compared to $k$ is the term associated to $j = k$, which is ${n+2k \choose 2k}$ and hence grows asymptotically like $\frac{n^{2k}}{(2k)!}$.  So, again when $n$ is large compared to $k$, it is indeed true that the number of allowable paths grows with the length of the path.
A: Here's my take on an intuitive answer:
Since there are $2^{\aleph_0}$ continuous functions from the unit interval to itself, it is clear that there can be at most that many functions.
Now, given $s > \sqrt{2}$ (which as you well observed the minimal distance), we define $t = s - \sqrt{2}$ and for any $x\in (0,1)$ we can construct a function that has the wanted arc length, simply by "pulling" the value of $f(x)$ smoothly in a way that extends the length of the arc by $t$.
(The construction is very clear when you don't request $f$ to be differentiable, as you can just stick a triangle in the proper length, however the idea should be clear)
So you have at least $|(0,1)| = 2^{\aleph_0}$ functions as such, which means that you have exactly that many.
