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:

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

1. 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.
2. The family of parabolas $f_\alpha(t) = \langle t, \alpha t^2\rangle$.
3. 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 $g$.

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.

• Thanks! Is it possible to require the arc length between any pair of values $a$ and $b$ to be exactly $|b-a|$? I imagine it's not generally possible, but perhaps for some of these curves? – user326210 Oct 22 '16 at 4:03
• Yes, it's possible in general; just reparametrize by arclength. I forgot you could do that for some reason. – ಠ_ಠ Oct 22 '16 at 4:13
• Oh, right! Then for the example curve families, will it be possible to satisfy both the second and third constraints simultaneously? That is, can you simultaneously parameterize $g$ with respect to arc length and so that for each $\alpha$, $g(\alpha)$ is on the graph of $f_\alpha$ ? – user326210 Oct 22 '16 at 5:20
• Yep that's guaranteed by the way I've defined g above – ಠ_ಠ Oct 22 '16 at 9:45

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.