Find curve passing through family of curves I have a family of smooth curves $f_\alpha : \mathbb{R}\rightarrow \mathbb{R}^2$, for $\alpha\in [0,1]$. 
The family of curves is also smooth if you fix $x$ and vary $\alpha$. That is, for each fixed $x$, the curve $\alpha \mapsto f_\alpha(x)$ is also smooth.
I'm looking for a curve $g : [0,1] \rightarrow \mathbb{R}^2$ with the following properties:


*

*$g$ is smooth.

*$g$ intersects every curve in the family $\{f_\alpha : \alpha \in [0,1]\}$. In particular, $g(\alpha)$ should be a point on the graph $f_\alpha$.

*$g$ is "steady-going". I'm still looking for a suitable definition, but I mean something like the arc length of $g$ between any pair of values $a$ and $b$ is equal to $|b-a|$.


I have been struggling to find a calculus-based solution for some special cases. Evidently, this problem amounts to finding a smooth function $t(\alpha)$ that picks a specific "time" along each curve in the family so that we can define
$$g(\alpha) = f_{\alpha}(t(\alpha)).$$
The special cases I care about are, for example:


*

*The family of circles $f_\alpha(t) = \langle \alpha\cos(t), \alpha \sin(t)\rangle$, for which there are apparently several possible $g$ in the radial direction.

*The family of parabolas $f_\alpha(t) = \langle t, \alpha t^2\rangle$.

*The family $f_\alpha(t) = \langle t, (t-1)\cdot\log(\alpha) + \frac{1}{2}\log{(1+t)}\rangle $


This last example is the one that motivates this post in the first place. If I can find a solution for this concrete example, it will be sufficient — but knowledge about alternate or more general cases would be extremely helpful as well.

My partial solution is to take advantage of the fact that for fixed $t\in \mathbb{R}$, $f_\alpha$ is smooth as a function of $\alpha$. Then instead we can consider the two-parameter family:
$F:\mathbb{R}^2 \rightarrow \mathbb{R}^2,\quad F(\alpha, t) = f_\alpha(t)$.
Then we can define a vector field on each point of the graph of $F$, which is the partial derivative of $F$ with respect to $\alpha$. The curves traced out by this vector field might be useful as candidates for $g$.
 A: Looks like you want your family of curves to be a smooth homotopy $f: [0,1] \times \mathbb{R} \to \mathbb{R}^2$. Then fixing the first argument at $\alpha \in [0,1]$ gives you a smooth curve $f_\alpha: \mathbb{R} \to \mathbb{R}^2$. Fixing the second argument at $x_0 \in \mathbb{R}^2$ gives you a smooth curve $g_{x_0}: [0, 1] \to \mathbb{R}^2$ which passes through each point $f_\alpha(x_0)$. 
For your last requirement, just reparametrize your curve $g_{x_0}$ by arclength to get the desired result.
A: Your idea of choosing $g(\alpha) = f_\alpha (t(\alpha))$ for some increasing function $\alpha$ is a good one. Since we want to enforce a constant-speed condition, we calculate the speed of this curve using the multivariable chain rule:
$$ |g'(\alpha)| = \left|F_\alpha(\alpha,t(\alpha)) + F_x(\alpha,t(\alpha)) t'(\alpha)\right|.$$
The condition that this is equal to a constant $C$ then implies (by squaring and expanding) 
$$ |F_\alpha(\alpha, t(\alpha))|^2 + 2 F_\alpha(\alpha,t(\alpha))F_x(\alpha,t(\alpha))t'(\alpha) + F_x(\alpha,t(\alpha))^2 t'(\alpha)^2 = C^2,$$
which is a fully nonlinear first-order ODE for $t(\alpha)$. I would see what this equation looks like with your particular choice of $F$ - for general $F$ you won't get an explicit solution but you might get lucky.
