# Stationary Points of an implicitly defined function?

by Mr.Brown
Tags: defined, function, implicitly, points, stationary
 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 :)
 Mentor P: 8,262 I don't understand what you're doing here. $$\nabla g=\frac{\partial g}{\partial x}\bold{i}+\frac{\partial g}{\partial y}\bold{j}+\frac{\partial g}{\partial z}\bold{k}$$. But this is identically zero, if g=0 (unless I'm missing something obvious).
 P: 239 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.
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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.
 P: 239 quick example: g(x,y,z)=x+y-z g(x,y,z)=0 grad(g)=i+j-k
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