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

In summary, we are using the Lagrange multiplier method to find the point on the curve y = x^2 that is closest to the point (0, b). This involves solving the equations Lx = 2x - 2λx = 0, Ly = 2y - 2b + λ = 0, and Lλ = x^2 - y = 0. We get three critical points: (0, b), (sqrt(b), b), and (-sqrt(b), b), but we must check them to determine which one is the closest point. This involves plugging the points into the distance formula and comparing the results.
  • #1
plexus0208
49
0

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)
Lx = 2x - 2λx = 0
Ly = 2y - 2b + λ = 0
Lλ = x2 - y = 0

Then solve for x and y? Is this right?
 
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  • #2
It's ok so far. Keep going.
 
  • #3
2x = 2λx -> λ = 1
2y = 2b + λ -> y = (2b+1)/2
x^2 = y -> x = ((2b+1)/2)^1/2
 
  • #4
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.
 
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  • #5
So basically, there are three points on the curve that are equally closest to (0,b)?
 
  • #6
plexus0208 said:
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.
 
  • #7
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...
 
  • #8
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.
 
  • #9
Lxx = 2 - 2λ
Lyy = 2
Lxy = 0

if λ = 1, then discriminant = LxxLyy - Lxy2 = 0...
 
  • #10
Why don't you just plug the points you get into the distance formula and figure out which one is closest?
 

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

What is the formula for finding the point on the curve y = x^2 that is closest to the point (0, b)?

The formula for finding the point on the curve y = x^2 that is closest to the point (0, b) is x = ±√b.

How do you determine which value of x to use in the formula for finding the closest point?

To determine which value of x to use in the formula, you must first find the derivative of the curve y = x^2, which is equal to 2x. Then, set the derivative equal to 0 and solve for x. The resulting value of x will be the x-coordinate of the closest point.

Can you use a different point other than (0, b) in the formula for finding the closest point?

Yes, the formula for finding the closest point can be used for any point (a, b). The resulting value of x will be x = ±√(b-a^2).

Is there a faster method for finding the closest point on the curve y = x^2?

Yes, there is a faster method called the Distance Formula, which uses the Pythagorean Theorem to find the shortest distance between two points. This method can be applied to any curve and point, not just y = x^2 and (0, b).

How can you use calculus to find the closest point on a curve?

Calculus can be used to find the closest point on a curve by finding the derivative of the curve and setting it equal to 0. This will give the x-coordinate of the closest point. Then, use the Distance Formula to find the y-coordinate of the closest point. Alternatively, you can use the concept of critical points to find the closest point.

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