Stationary Points of an implicitly defined function?

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Homework Help Overview

The discussion revolves around the concept of stationary points in the context of an implicitly defined function g(x,y,z) = 0. Participants are exploring whether such a function can have stationary points and how to interpret the gradient in this scenario.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the relationship between the gradient of g and the existence of stationary points, particularly when g is defined as zero. There is a discussion about the implications of grad(g) being zero and how it relates to the definition of stationary points.

Discussion Status

There is an ongoing examination of the apparent contradiction between the statements regarding stationary points and the gradient. Some participants have provided examples to illustrate their points, while others are seeking clarification on how to resolve the confusion surrounding the definitions and interpretations involved.

Contextual Notes

Participants are considering the implications of defining a surface through g(x,y,z)=0 and how this affects the interpretation of stationary points in relation to implicit functions. There is an acknowledgment that g=0 does not imply g is zero everywhere, which raises further questions about the nature of the function and its derivatives.

Mr.Brown
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Hi
i just got a short question about definition if i got an implicitly defined function g(x,y,z) = 0 and then be asked whether g hast stationary points.
How to answer that intuitively i´d say no g = 0 = constant hence no stationary points but if i do grad(g) = ( 0,0,0) i get stationary points.
So what´s the answer for this ?
And if it´s the grad(g) thing how to interpret that kind of stationary point geometricaly.
Thanks and bye :)
 
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I don't understand what you're doing here. [tex]\nabla g=\frac{\partial g}{\partial x}\bold{i}+\frac{\partial g}{\partial y}\bold{j}+\frac{\partial g}{\partial z}\bold{k}[/tex]. But this is identically zero, if g=0 (unless I'm missing something obvious).
 
He is seeing a contradiction between 2 statements:
if grad(g) = ( 0,0,0) then there is stationary points
and if g(x,y,z) = 0 then there is no stationary points.
since g(x,y,z) = 0 then grad(g) = ( 0,0,0)
I think that what he means.
 
If it is a surface defined by g(x,y,z)=0 (a level surface), there is no need that grad(g)=0. What would be zero is grad(g).t for t a tangent vector to the surface.
 
quick example:
g(x,y,z)=x+y-z
g(x,y,z)=0
grad(g)=i+j-k
 
ziad1985 said:
quick example:
g(x,y,z)=x+y-z
g(x,y,z)=0
grad(g)=i+j-k

And the surface g(x,y,z)=0 would be the plane z=(x+y).
 
Yes, I thought I was missing something. I don't quite know what I was doing in my last post!
 
ziad1985 said:
He is seeing a contradiction between 2 statements:
if grad(g) = ( 0,0,0) then there is stationary points
and if g(x,y,z) = 0 then there is no stationary points.
since g(x,y,z) = 0 then grad(g) = ( 0,0,0)
I think that what he means.

yea that´s exactly what i meant. How is that puzzle solved ? :)
 
Look at ziad1985's example. g(x,y,z)=0 in implicit function definition is not meant to mean g is zero everywhere.
 
  • #10
yea that´s what i understand in a sense it defines a z(x,y) or x(z,y) and so on so when i ask what stationary points does g have i mean what stationary points does z(x,y) have ?
 
  • #11
If you want stationary points of z(x,y), then find an expression for z(x,y), take partial derivatives wrt x and y and set them both equal to zero.
 

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