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

In summary, the author is trying to find the equilibrium points of a function, but isn't getting anywhere because he doesn't understand the concept of critical points. He has found that for z=constant, the EP is (0,0,0).
  • #1
Somefantastik
230
0
[tex]V(x,y) = x^{2}+y^{2} - z; [/tex]

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).

cone.jpg
 
Last edited:
Physics news on Phys.org
  • #2
What is z? Is it a constant, or did you mean V(x, y, z)?
 
  • #3
I meant V(x,y,z)
 
  • #4
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
 
  • #5
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.
 
  • #6
I'm not sure what the problem is, you have a function of 3 variables, the set of equil. points will be [tex] z = x^2 + y^2 [/tex]
 
  • #7
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,
[tex]\nabla V= 2x\vec{i}+ 2y\vec{j}- \vec{k}= \vec{0}[/tex]
Since the [itex]\vec{k}[/itex] component is never 0, there are NO equilibrium points.
 
  • #8
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].
 

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

What is the equation for finding equilibrium points?

The equation for finding equilibrium points of V(x,y) = x^2+y^2 - z is V(x,y) = 0. This means that at equilibrium points, the potential energy is equal to zero.

How do I find the equilibrium points for this equation?

To find the equilibrium points, you can set V(x,y) = 0 and solve for x and y. This will give you the coordinates of the equilibrium points.

What is the significance of equilibrium points in this equation?

Equilibrium points represent the stable positions of the system where the potential energy is at its minimum. This means that the system is in a state of balance and will not experience any net force.

Are there any other methods for finding equilibrium points?

Yes, there are other methods such as using partial derivatives and setting them equal to zero, or graphing the equation and finding where it intersects the x and y axes.

What are some real-life applications of finding equilibrium points?

Finding equilibrium points is important in many fields such as physics, chemistry, and engineering. It is used to analyze the stability of systems, determine the optimal values for variables, and predict the behavior of complex systems.

Similar threads

Replies
6
Views
2K
Replies
3
Views
1K
Replies
32
Views
3K
  • Calculus
Replies
8
Views
2K
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
291
  • Calculus and Beyond Homework Help
Replies
2
Views
627
Replies
1
Views
1K
Back
Top