Find the point on the curve y = x^2 that is closest to the point (0, b).

[plexus0208]

Homework Statement

Find the point on the curve y = x² that is closest to the point (0, b).

Homework Equations

Lagrange multipliers

The Attempt at a Solution

L(x,y,λ) = x² + (y-b)² - λ(x²-y)
L_x = 2x - 2λx = 0
L_y = 2y - 2b + λ = 0
L_λ = x² - y = 0

Then solve for x and y? Is this right?

 

[Dick]

It's ok so far. Keep going.

 

[plexus0208]

2x = 2λx -> λ = 1
2y = 2b + λ -> y = (2b+1)/2
x^2 = y -> x = ((2b+1)/2)^1/2

 

[Dick]

You've got a sign error in your y equation. And you've also ignored a possible solution of the first equation. How about x=0? And in the last equation there are two solutions for x.

 

[plexus0208]

So basically, there are three points on the curve that are equally closest to (0,b)?

 

[Dick]

So basically, there are three points on the curve that are equally closest to (0,b)?

Depends on b. The lagrange multiplier method only gives you the critical points. There might be one or three of them depending on the value of b. You then have to check them to figure out which is closest.

 

[plexus0208]

Well, I was given the point (0, b), without any information about b, so I guess I should just put down all three points then...

 

[HallsofIvy]

No, whatever b is you should check to see which points are minimum. You have only shown that they are critical points. It is quite possible that one of the points is a local maximum for the distance.

 

[plexus0208]

L_xx = 2 - 2λ
L_yy = 2
L_xy = 0

if λ = 1, then discriminant = L_xxL_yy - L_xy² = 0...

 

[Dick]

Why don't you just plug the points you get into the distance formula and figure out which one is closest?