Stationary Points of an implicitly defined function?

Mr.Brown

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 :)

cristo

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

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

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

quick example:
g(x,y,z)=x+y-z
g(x,y,z)=0
grad(g)=i+j-k

Dick

And the surface g(x,y,z)=0 would be the plane z=(x+y).

cristo

Yes, I thought I was missing something. I don't quite know what I was doing in my last post!

Mr.Brown

yea that´s exactly what i meant. How is that puzzle solved ? :)

Dick

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

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

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.