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.

 

[anuttarasammyak]

continued from #2:
With the help of Wolfram I get one real solution. https://www.wolframalpha.com/input?i=Solve+(4-x^2-(y-3)^2)(-2x)+(x-y)=0+and++(4-x^2-(y-3)^2)2(y-3))+(x-y)=0.+

where $x=x_1,y=x_2$ Let us confirm it.
It is easily observed that the two curve formula coincide when $-x=y-3=-a$ thus a should satisfy
$$4a^3-6a-3=0$$
which derives all the solutions. Using real $a=2^{-1/3}+2^{-2/3}=1.42366...$.
The minimum distance is
$$L=\sqrt{4a^4-4a^2-12a+25}=\sqrt{-6a^2-9a+25}=\sqrt{19-12*2^{-1/3}-15*2^{-2/3}}=0.16182....$$
when $x_1=a, x_2=3-a$

I hope I commit no more careless mistakes.

 

[Charles Link]

@anuttarasammyak We are going to need to look over your two partial derivative expressions carefully, because your solution is incorrect and not even in the right ballpark. I got, doing it all by hand with no computer that $(x_1,y_1)=(1.43,1.96)$, and $(x_2,y_2)=(1.57,2.04)$ for a minimum distance of $L=0.16$. These are approximate coordinates, and only accurate to the two decimals.

Edit: and I looked over your partial derivative expressions=they look like they may be correct. The values for the real solution might also be correct, but I don't have a computer that can compute them. It looks like you did get $x_1=1.42366$ and $x_2=1.57634$ which is close enough to what I got that they are most likely correct and even exact in the form Wolfram has them. I think your error may be when you finished up.

Try computing $L$ again, but use $(x_1,y_1)$ and $(x_2,y_2)$. I don't know where your expression for $L$ using $a$ came from. Note: $L=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

With that one correction, I think everything else you did may be correct. :)

and I see you have now edited your post and made the correction. Very good=excellent. :)

 

[anuttarasammyak]

@anuttarasammyak We are going to need to look over your two partial derivative expressions carefully, because your solution is incorrect and not even in the right ballpark.

I am sorry for my careless mistakes on L. I corrected #4. It seems in good accord with your result.

 

[Charles Link]

I am sorry for my careless mistakes on L. I corrected #4. It seems in good accord with your result.

@anuttarasammyak Thank you very much for your solution above. It is excellent. <3 <3

 

[Charles Link]

@anuttarasammyak Perhaps you also noticed this already, but I find it interesting that with your exact solution, the slopes of the curves are precisely parallel, as they should be, at $x_1$ and $x_2$, with slopes of $-2 x_1 =-2a$ and $2(x_2-3)=-2a$.