Find the least/greatest distance from point to sphere

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SUMMARY

The discussion focuses on finding the greatest and least distances from the point P(2,1,-2) to the sphere defined by the equation x² + y² + z² = 1 using Lagrange Multipliers. Participants emphasize minimizing the squared distance, d² = (x - 2)² + (y - 1)² + (z + 2)², to simplify calculations. The constraint is that the point (x,y,z) must lie on the surface of the sphere. Additionally, it is noted that the maximum and minimum distance points lie along the line connecting the point P and the center of the sphere (0,0,0).

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Homework Statement


Find the greatest and least distances from the point P(2,1,-2) to the sphere x^2 + y^2 + z^2 = 1 using Lagrange Multipliers.

(The question had an = before the z^2 which was a typo and I think it should of been a + so just saying in case I made a wrong correction.)


Homework Equations


Partial differentiation.
Lagrange Multiplier equation.


The Attempt at a Solution


I tried visualizing this geometrically and I see that it is a sphere centered at (0,0,0) with a radius of 1 and I need to find the shortest line from the point to the surface of the sphere so I was thinking of taking a vector <2,1,-2> and dotting it with the gradient of f(x,y,z) and finding x, y and z values such that the dot product is 0. I'm not entirely sure if this approach is good (please, tell me if it is or isn't since I'm curious) but I have to do it using Lagrange Multipliers so I'm left thinking: "What's the constraint explicitly?" I'm having a bit of trouble understanding what the greatest distance is as well. Is it the line from the point to an edge of the surface area of the sphere facing the point (which is a circle) such that we maximize the distance by making the line joining the point to the sphere be the hypotenuse of a triangle?

A push in the right direction or any help whatsoever would be greatly appreciated!
Thanks in advance!
 
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Do it the Lagrange multiplier way and then look back at your 'vector way'. Pick a point (x,y,z). The thing you want to minimize is the distance from P(2,1,-2) to (x,y,z). The constraint is that (x,y,z) has to lie on your surface.
 
Dick said:
Do it the Lagrange multiplier way and then look back at your 'vector way'. Pick a point (x,y,z). The thing you want to minimize is the distance from P(2,1,-2) to (x,y,z). The constraint is that (x,y,z) has to lie on your surface.

Equivalently, you want to minimize (or maximize) the distance^2, which will be a lot easier to work with.

RGV
 
I took into consideration what you both said but I'm still having trouble to actually start. Could you guys give me the first line of what I am supposed to do so that I can see it and continue myself?
 
s3a said:
I took into consideration what you both said but I'm still having trouble to actually start. Could you guys give me the first line of what I am supposed to do so that I can see it and continue myself?

Do you know the formula for the distance between (x,y,z) and (2,1,-2)?

RGV
 
It's been a long time since I used it but yes I do :)

d = sqrt[(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2]
d^2 = (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2
 
It's been a long time since I used it but yes I do :)

d = sqrt[(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2]
d^2 = (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2
 
So if the point we're interested in is (2,1,-2) you want to minimize (x-2)2+(y-1)2+(z+2)2
 
s3a said:
It's been a long time since I used it but yes I do :)

d = sqrt[(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2]
d^2 = (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2

Ok, so continue. You want to extremize d^2.
 
  • #10
By the way, although the problem asks you to use Lagrange multipliers, as a check you can use the fact that the max and min distance points lie on the line through the point (2, 1, -2) and (0, 0, 0), the center of the given sphere. Put x= 2t, y= t, z= -2t into the equation of the sphere to find those points.
 
  • #11
Alright, so am I on the right track now?
 

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  • #12
s3a said:
Alright, so am I on the right track now?

Yes, you've got the setup right.
 

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