Finding the minimum distance between two curves

[Charles Link]

TL;DR

The problem is to compute the minimum distance between the two curves y=4-x^2 and y=(x-3)^2 in two dimensions.

The other day a friend of mine gave me the above problem which I found rather interesting. I was able to get a numerical solution. I'll post how I solved it later in the thread. Others may want to try it and see what they come up with.

 

[anuttarasammyak]

Square of the distance is
$$L^2=(y_1-y_2)^2+(x_1-x_2)^2=(4-x_1^2-(x_2-3)^2)^2+(x_1-x_2)^2$$
The condition of minimum is
$$\frac{\partial L^2}{\partial x_1}=\frac{\partial L^2}{\partial x_2}=0$$
Intersection of these two curves would give an answer.

 

[Charles Link]

@anuttarasammyak That's one part of what I did to solve it. Those two partial derivative expressions are really too clumsy to solve exactly. I used them though by getting an approximate $x$ for where the two $x$ values occur by first solving for the $x$ where the vertical distance is a minimum. That turns out to be at $x=3/2$.

I then let $x_1=3/2+ \Delta_1$ and $x_2=3/2+ \Delta_2$ and solved for $\Delta_ 1$ and $\Delta_2$ to first order using the two partial derivative expressions.

Others may want to try the calculation and/or have alternative ways of solving it, so I'm leaving just a summary of the solution rather than showing all of the details.