Charles Link's latest activity

@Gavran Very good=Excellent. :) You have an interesting approach where you translated the first curve by ## (a,b) ## until it made...

Let ## (x_1,y_1) ## be a point on the curve ## f(x)=4-x^2 ##, which is the closest point on the curve ## f(x) ## to the curve ##...

Just a couple additional comments: I thought I did pretty well by getting a numerical solution by writing out the partial derivative...

@pasmith I think it should read ##-2(x_1+x_2-3)=0 ##. Otherwise, yes, very good=excellent. :) and I think I see where he got the...

Yes, sometime I'm going to need to try Vieta's substitution on this cubic and see if I can get the same answers that Wolfram got. :)...

If $$L = (y_1(x_1) - y_2(x_2))^2 + (x_1 - x_2)^2$$ then the partial derivatives are $$\begin{split} \frac{\partial L}{\partial x_1} &=...

For the calculation with the slopes mentioned in post 8, we have ## m=\frac{2-(4-a^2)}{3/2-a}=\frac{(3-a-3)^2-2}{3-a-3/2}## and these...

After transformation of ##X_2=3-x_2## the two partial differential equations are $$-2(4-x_1^2 -X_2^2)x_1+(x_1+X_2-3)=0$$ $$-2(4-x_1^2...

@anuttarasammyak Perhaps you also noticed this already, but I find it interesting that with your exact solution, the slopes of the...

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

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

continued from #2: With the help of Wolfram I get one real solution...

@anuttarasammyak We are going to need to look over your two partial derivative expressions carefully, because your solution is...

@anuttarasammyak That's one part of what I did to solve it. Those two partial derivative expressions are really too clumsy to solve...

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...