Inspired by this question and these lecture notes which belong to it, where derivatives of specific points is prescribed to be some known vectors.
Let us consider the $${\bf c}'(0) = {\bf v_0}\\{\bf c}'(1) = {\bf v_1}$$
equations, and say that we don't know $\bf v_0$ or $\bf v_1$, but that we want to punish certain orientations.
$${\bf T_0} {\bf c}'(0) - {\bf d_0}={\bf 0}\\{\bf T_1} {\bf c}'(1) - {\bf d_1}={\bf 0}$$ Or maybe (which would be a softer version) to do least squares fit : $$\|{\bf c}(0)-{\bf p_0}\|_2+\|{\bf c}(1)-{\bf p_1}\|_2+\|{\bf T_0} {\bf c}'(0) - {\bf d_0}\|_2+\|{\bf T_1}{\bf c}'(1) - {\bf d_1}\|_2$$
Where $\bf T_0,T_1$ are two known tensors and $\bf d_0, d_1$ are two known vectors.
Just to show the power of the approach, we could imagine $$\begin{cases}{\bf T_0 = I - \hat v_0}{\bf \hat v_0}^T\\{\bf T_1 = I - \hat v_1}{\bf \hat v_1}^T\\\bf d_0=0\\\bf d_1=0\end{cases}$$
Which would allow any vectors in the same orientation as $\bf v_0$ and $\bf v_1$, respectively, but with unknown magnitudes.
We will also realize that the special case $$\begin{cases}\bf T_0= I\\\bf T_1 = I\\\bf d_0=v_0\\ \bf d_1=v_1\end{cases}$$ would take us back to the original formulation.
How would solving this problem differ from solving the original formulation?
Or rephrased: Is there some framework of solving the original question which would be rich enough to allow us to add cost terms, like the one above
This approach does not work. There is at least one particular reason for this.
Consider ${\bf c}'(1) = [\epsilon_1,\epsilon_2]^T$
Our optimization can pick any $\epsilon_1,\epsilon_2$ arbitrarily close to 0, and we can get any orientation of the tangent vector.
What we would like to express in this case is something like $$\frac{{\bf c}'(1)}{|{\bf c}'(1)|}$$ And then multiply with tensors. But this is very non-linear. It will not fit into our linear algebra framework.
I would think we could do better to do an approach with integral of polar coordinates or some such thing.