Finding Equilibrium Points of V(x,y) = x^2+y^2 - z

[Somefantastik]

$$V(x,y) = x^{2}+y^{2} - z;$$

I need to find the equilibrium points of this guy. I can do this in my sleep for two dimensions but I'm not getting anything useful with this particular function.

What would the graph look like for V(x,y) = c, c is arbitrary constant. It's a cone, I know, but I'm not sure how to graph it out.

Any suggestions?

If anybody cares, this is what the plot looks like in matlab. Looks like the EP is (0,0,0).

 

[adriank]

What is z? Is it a constant, or did you mean V(x, y, z)?

 

[Somefantastik]

I meant V(x,y,z)

 

[Office_Shredder]

What's your definition of equilibrium point? I've only heard it used in the context of differential equations.

With respect to graphing, notice that for each fixed z=constant plane, you just the the equation for a circle centered at the origin with radius sqrt(c+z), so you expect it to look like a cone, but with a parabolic shape

 

[Somefantastik]

Yes, this is part of a dynamical systems problem. It's really frustrating when I get stuck on the Calculus parts.

Here, the equilibrium point (or critical point) is defined as V(x,y,z) = 0.

 

[NoMoreExams]

I'm not sure what the problem is, you have a function of 3 variables, the set of equil. points will be $$z = x^2 + y^2$$

 

[HallsofIvy]

If V is a potential energy function, then the equilibrium points are at the minimum (stable equilibrium) and maximum (unstable equilibrium) values of V(x,y,z). The minimum and maximum values of a function of three variables occur where the gradient,
$$\nabla V= 2x\vec{i}+ 2y\vec{j}- \vec{k}= \vec{0}$$
Since the $\vec{k}$ component is never 0, there are NO equilibrium points.

 

[Somefantastik]

That's what I thought too. I don't know how the phase portrait would look with no EQ PTS. I guess the solutions just pass through [somewhere].