Finding Specific Extrema when grad(F)=constant & Lagrange Gives y=-z/2

  • Thread starter The Head
  • Start date
  • #1
125
0

Homework Statement:

Consider a particle that moves in x^2+2y^2+2z^2<=4 and is subject to the force field f(x,y,z)= (1,-2z,-2y). If f=grad(F), find the values in this region that maximize f.

Relevant Equations:

grad(F)=0 to find extrema
Lagrange Mulitpliers Technique
I found that f= x -2yz. To maximize f, I can first inspect the solutions to grad(F)=0. z=y=0 pops out, but I'm not sure what to do with the x-component equaling 1. Do we just include (x,0,0) as a solution? I think the problem wants specifics though, based on what I've seen previously from problem sets by this instructor.

Using Lagrange Multipliers, we get:
1=2xλ
-2z=4yλ
-2y=4zλ

Working through this, I find that z(1-4λ)=0. If z=0, then y=0. If λ=1/4, then y=-z/2 and x=2. So it appears we have (x,0,0) (as before) and (2, -z/2, z).

I'm not sure how to work through these to get the specifics here. Am I missing something, or is this a departure from this instructor's typical problems? Also, I don't think I need to parameterize and look for points inside the boundary here because of the Laplacian being equal to zero (I read that somewhere, but could be mistaken).
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,254
... f=grad(F), find the values in this region that maximize f.

grad(F)=0 to find extrema
Don't you think that is a little contradictory ? Why should f=0 be a maximum ?

Or is your notation so sloppy it is beyond deciphering ? How is a maximum of a vector defined ?
 
  • #3
125
0
Don't you think that is a little contradictory ? Why should f=0 be a maximum ?

Or is your notation so sloppy it is beyond deciphering ? How is a maximum of a vector defined ?
Oops, grad(f)=0. All the work is correct consistent, I just must have held the shift button after hitting the parenthesis. In any case, the the derivatives are set equal to zero, which is I set (1,-2z,-2y)=0, then also did the Lagrange stuff.
 
  • #5
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,254
I set (1,-2z,-2y)=0
That is ##\vec f = \vec 0## ? Searching for a "maximum" of the vector f ?

Please render the complete and litteral problem statement, exactly as given to you.
 
  • #6
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,254
I just must have held the shift button after hitting the parenthesis
Several times, it seems to me. Aren't you looking for a maximum of F instead of f ?
 
  • #7
125
0
Several times, it seems to me. Aren't you looking for a maximum of F instead of f ?
Attached is the question, and I'm working on part c. I'm maximizing f, and thus looking at F, because it's the gradient. My work was showing setting (1,-2z,-2y)=0, but I don't know what 1=0 implies for the x-component. Also, with Lagrange I don't get anything that creates definitive values, but I believe I need them.

Also, not sure why I'm being criticized repeatedly for a typo-- it happens :smile:

Apologies about the confusion.
 

Attachments

Last edited:
  • #8
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,254
I'm not trying to burn you with criticism, I am trying to help.

You wrote
is subject to the force field f(x,y,z)= (1,-2z,-2y). If f=grad(F)
but meant to write is subject to the force field ##\ \vec F(x,y,z)= (1,-2z,-2y).\ ## If ##\ \vec F=\vec\nabla (f) ##...

You found, correctly, ##f = x-2yz##.

What did you do to show that ##\vec F ## is conservative ?

In part (b) you found the ##W## that @etotheipi mentions, and finding extrema for that amounts to finding extrema for ##f##.

Lagrange multipliers work for constraints that can be written as equalities. Make an educated guess and search on ##x^2+2y^2+2z^2\ {\bf =} \ 4\;##. Check later.

(afraid to add insult to injury :rolleyes:):
... To maximize f, I can first inspect the solutions to grad(F)=0. z=y=0 pops out, but I'm not sure what to do with the x-component equaling 1. Do we just include (x,0,0) as a solution?
I don't follow at all. ##\vec \nabla f = (1,0,0)## not zero anywhere.

Anyway, you have the constraint (plus the guess) and want to tackle
Using Lagrange Multipliers, we get:
1=2xλ
-2z=4yλ
-2y=4zλ
But that is four unknowns and three equations. You need a fourth equation: the constraint !

Working through this, I find that z(1-4λ)=0. If z=0, then y=0. If λ=1/4, then y=-z/2 and x=2. So it appears we have (x,0,0) (as before) and (2, -z/2, z).
I get ##\ z(1-4\lambda^2) = 0\ ## (or did I just bump into another typo :wink: )

Keep going !

## \ ##
 
  • Like
Likes The Head and etotheipi
  • #9
125
0
I'm not trying to burn you with criticism, I am trying to help.

You wrote
but meant to write is subject to the force field ##\ \vec F(x,y,z)= (1,-2z,-2y).\ ## If ##\ \vec F=\vec\nabla (f) ##...

You found, correctly, ##f = x-2yz##.

What did you do to show that ##\vec F ## is conservative ?

In part (b) you found the ##W## that @etotheipi mentions, and finding extrema for that amounts to finding extrema for ##f##.

Lagrange multipliers work for constraints that can be written as equalities. Make an educated guess and search on ##x^2+2y^2+2z^2\ {\bf =} \ 4\;##. Check later.

(afraid to add insult to injury :rolleyes:):
I don't follow at all. ##\vec \nabla f = (1,0,0)## not zero anywhere.

Anyway, you have the constraint (plus the guess) and want to tackle

But that is four unknowns and three equations. You need a fourth equation: the constraint !

I get ##\ z(1-4\lambda^2) = 0\ ## (or did I just bump into another typo :wink: )

Keep going !

## \ ##
Haha, yeah, I definitely made a mistake there. λ=1/2 or -1/2. Thanks for the hint on adding the constraint, that makes sense and helps pull it together conceptually.

OK so, I have a few paths: λ=1/2 implies y=-z and x=1, so I get (1, -z, z), whereas λ=-1/2 implies y=z and x= -1, so I get (-1, z, z). We still have the y=z=0 path from before, so I get (x, 0, 0).

Using the constraint for (1, -z, z) we see that z=sqrt(3)/2 or z=-sqrt(3)/2. I get the same thing for (-1, z, z). Then for (x, 0, 0), I get x=2 and x=-2.

So in total, I think my critical points are
(1, -sqrt(3)/2, sqrt(3)/2), (1, sqrt(3)/2, -sqrt(3)/2), (-1, sqrt(3)/2, sqrt(3)/2), (-1, -sqrt(3)/2, -sqrt(3)/2), (2, 0, 0), (-2, 0, 0). I can plug those into W to find which are absolute global max and mins. And then that's basically it I believe.

And to be clear, the reason I can ignore the points inside the boundary is because the laplacian of f = 0, and if it weren't, I'd need to parameterize the f?
 
Last edited:
  • #10
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,254
λ=1/2 implies y=-z and x=1, so I get (1, -z, z)
Plus ##x^2+2y^2+2z^2\ {\bf =} \ 4\;## gets you one single point as answer for this ##\lambda##. Check by sketching the ##yz## plane for x = 1.
 
  • #11
BvU
Science Advisor
Homework Helper
2019 Award
13,527
3,254
All ok now ?
 

Related Threads on Finding Specific Extrema when grad(F)=constant & Lagrange Gives y=-z/2

Replies
5
Views
3K
Replies
4
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
5
Views
3K
Replies
8
Views
6K
  • Last Post
Replies
2
Views
3K
Top