To find a smooth planar curve starting at $\vec{r_0}$ stopping at $\vec{r_1}$ with some additional constraints.

Question

To find a smooth planar curve starting at $\vec{r_0}$ stopping at $\vec{r_1}$ whose unit tangents at start and stop are $\hat{v_0}$ and $\hat{v_1}$ and has the minimum length. Let us assume that the curve is $\vec{r(t)}=x(t)\hat{x}+y(t)\hat{y}$ parametrized versus $0\leq t\leq1$ having the length $L$ to be minimized: $$L=\int_{0}^{1} |\vec{r'(t)}|dt=\int_{0}^{1} \sqrt{x'^2(t)+y'^2(t)}dt$$ where $$\frac{x'(0)\hat{x}+y'(0)\hat{y}}{\sqrt{x'^2(0)+y'^2(0)}}=\hat{v_0},$$ $$\frac{x'(1)\hat{x}+y'(1)\hat{y}}{\sqrt{x'^2(1)+y'^2(1)}}=\hat{v_1},$$ $$x(0)\hat{x}+y(0)\hat{y}=\vec{r_0},$$ and $$x(1)\hat{x}+y(1)\hat{y}=\vec{r_1}.$$ Also: $$\hat{v_0}\times(\vec{r_1}-\vec{r_0})\bullet\hat{v_1}\times(\vec{r_1}-\vec{r_0})<0$$ Any idea how to minimize the $L$?

Answer 1

If you draw a picture using very small parts of the curve satisfying the initial and final conditions, it should be obvious that you can connect those parts (after appropriate turns) with a straight line segment, hence the length of the curve can be made arbitrarily close to the distance between the initial and final points.

It follows that there is no minimum length unless each of the two given unit tangent vectors are in the direction of the displacement vector, but that would break the specified dot product condition.