Stationary Points of an implicitly defined function?

In summary, the conversation discusses the concept of stationary points in relation to an implicitly defined function g(x,y,z) = 0. The question is whether g has stationary points, and intuitively it is believed that there are none since g is a constant. However, the contradiction arises when using the gradient of g, where if it equals (0,0,0), there are stationary points. The conversation concludes that g(x,y,z) = 0 does not necessarily mean g is zero everywhere, and to find stationary points of z(x,y), one must take partial derivatives and set them equal to zero.
  • #1
Mr.Brown
67
0
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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  • #2
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).
 
  • #3
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.
 
  • #4
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.
 
  • #5
quick example:
g(x,y,z)=x+y-z
g(x,y,z)=0
grad(g)=i+j-k
 
  • #6
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).
 
  • #7
Yes, I thought I was missing something. I don't quite know what I was doing in my last post!
 
  • #8
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 ? :)
 
  • #9
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.
 

1. What is the definition of a stationary point of an implicitly defined function?

A stationary point of an implicitly defined function is a point where the derivative of the function with respect to one or more variables is equal to zero. This can also be understood as a point where the slope or gradient of the function's graph is flat or horizontal.

2. How do you find the stationary points of an implicitly defined function?

To find the stationary points of an implicitly defined function, you can use the implicit differentiation method. This involves differentiating the function with respect to the variables and setting the resulting equations equal to zero, then solving for the variables. The resulting points will be the stationary points of the function.

3. Why are stationary points important in mathematics?

Stationary points are important in mathematics because they can tell us information about the behavior of a function. For example, a stationary point can indicate where a function reaches its maximum or minimum value, or where it changes from increasing to decreasing (or vice versa).

4. Can a function have more than one stationary point?

Yes, a function can have more than one stationary point. In fact, a function can have many stationary points depending on its complexity. These points can be local (relative) maximum or minimum points, or they can be inflection points where the curvature of the graph changes.

5. How do you determine if a stationary point is a maximum or minimum?

To determine if a stationary point is a maximum or minimum, you can use the second derivative test. This involves taking the second derivative of the function and evaluating it at the stationary point. If the second derivative is positive, the stationary point is a minimum, and if it is negative, the stationary point is a maximum. If the second derivative is zero, the test is inconclusive and another method may need to be used.

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