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Critical points of doughnut

  • #1
798
1

Homework Statement


A perfectly smooth two-dimensional surface that has the shape of a doughnut is suspended in the (x, y, z)-Cartesian space. Let the doughnut be positioned so that an observer facing the plane y=0 will see through the hole. For example, the doughnut is standing up vertically, rather than lying flat. Next, define f(x, y, z)=z.

1. Consider the restrictions of f to the doughnut surface. How many critical points are there?
2. How many of those critical points are of the minimum or maximum type?

(It is suffice to sketch the surface and indicate which of the points are max/min.)

Essentially in 2 space, the shape will merely be two circles with different radii, and the domain will include all points on and between the boundaries of these circles. Also, since an observer sees the hole facing y=0, the circles will be in the xz-plane.

What I don't really understand is the statement "f(x, y, z)=z". Are they restricting the domain to points on the z-axis? If so, there will be 4 critical points, 2 relative max, and two relative min.
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
I think the observer is given only to define the orientation of the doughnut?

I would just consider it as a 2D surface embedded in 3D space. Now consider any point on the surface, we can assign it a value, in this case =z. This is the restriction of f(x,y,z) to the surface. if you paramteriesed teh surface you coudl plot f vs the parameters.

Your last statement appears correct with that interpretation
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,808
933

Homework Statement


A perfectly smooth two-dimensional surface that has the shape of a doughnut is suspended in the (x, y, z)-Cartesian space. Let the doughnut be positioned so that an observer facing the plane y=0 will see through the hole. For example, the doughnut is standing up vertically, rather than lying flat.
If I understand this correctly, the torus is given by the parametric equations
[itex]x= rsin(\theta)[/itex] , [itex]y= (R+ rcos(\theta))cos(\phi)[/itex], [itex]z= (R+ rcos(\theta))sin(\phi)[/itex].

Next, define f(x, y, z)=z.

1. Consider the restrictions of f to the doughnut surface. How many critical points are there?
2. How many of those critical points are of the minimum or maximum type?

(It is suffice to sketch the surface and indicate which of the points are max/min.)

Essentially in 2 space, the shape will merely be two circles with different radii, and the domain will include all points on and between the boundaries of these circles. Also, since an observer sees the hole facing y=0, the circles will be in the xz-plane.
No, that would be the projection of the torus to the yz-plane.

What I don't really understand is the statement "f(x, y, z)=z". Are they restricting the domain to points on the z-axis? If so, there will be 4 critical points, 2 relative max, and two relative min.[/QUOTE]
No, they are restricting the function f(x, y, z)= z to the surface of the torus.

So f(x,y,z)= z, restricted to the torus, is [itex]f(\theta, \phi)= (R+ rcos(\theta))sin(\phi)[/itex]
The critical points will be where the derivatives, with respect to [itex]\theta[/itex] and [itex]\phi[/itex] are 0.
 
  • #4
798
1
Thank you for the response. What I did was I graphed a "doughnut" in the xz-plane. The critical points are at (x, z1)=(0, a), (x, z2)=(0, b), (x, z3)=(0, c), (x, z4)=(0, d)
where
d<c<b<a
 
  • #5
798
1
@HallsofIvy
Are the angles measured with respect to the z axis in the x-direction, and then the z-axis in the y-direction?
 

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