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Stationary Points of an implicitly defined function?

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Mr.Brown
#1
Mar8-07, 08:01 AM
P: 67
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 iīd say no g = 0 = constant hence no stationary points but if i do grad(g) = ( 0,0,0) i get stationary points.
So whatīs the answer for this ?
And if itīs the grad(g) thing how to interpret that kind of stationary point geometricaly.
Thanks and bye :)
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cristo
#2
Mar8-07, 08:50 AM
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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).
ziad1985
#3
Mar8-07, 11:48 AM
P: 240
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.

Dick
#4
Mar8-07, 11:55 AM
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Stationary Points of an implicitly defined function?

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.
ziad1985
#5
Mar8-07, 12:02 PM
P: 240
quick example:
g(x,y,z)=x+y-z
g(x,y,z)=0
grad(g)=i+j-k
Dick
#6
Mar8-07, 12:05 PM
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Quote Quote by ziad1985 View Post
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).
cristo
#7
Mar8-07, 12:49 PM
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Yes, I thought I was missing something. I don't quite know what I was doing in my last post!
Mr.Brown
#8
Mar8-07, 02:15 PM
P: 67
Quote Quote by ziad1985 View Post
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 thatīs exactly what i meant. How is that puzzle solved ? :)
Dick
#9
Mar8-07, 02:21 PM
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Look at ziad1985's example. g(x,y,z)=0 in implicit function definition is not meant to mean g is zero everywhere.
Mr.Brown
#10
Mar8-07, 02:34 PM
P: 67
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 ?
Dick
#11
Mar8-07, 02:37 PM
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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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