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Critical points of doughnut

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data
    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.
     
  2. jcsd
  3. Nov 16, 2011 #2

    lanedance

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    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
     
  4. Nov 16, 2011 #3

    HallsofIvy

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    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].

    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.
     
  5. Nov 16, 2011 #4
    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
     
  6. Nov 16, 2011 #5
    @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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