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A function describing a tilted tube

  1. Jun 7, 2013 #1
    Hi

    I have a function, which is cylindrical symmetric given by
    [tex]
    f(x, y, z) = \exp(-x^2-z^2)
    [/tex]
    For a given [itex]y[/itex], the function [itex]\exp(-x^2-z^2) = c[/itex] traces out a circle (where c is a constant). A contourplot of [itex]f(x, 0, z)[/itex] is attached.

    However, this is for [itex]y=0[/itex] (currently, I get the same plot for an arbitrary value of y). I am interested in constructing a function, which is identical to [itex]f[/itex], but where the center of the above circle increases linearly with y. In other words, at [itex]y=y'[/itex] I want my function to have the same contour plot as attached, but its center should be at [itex]y=y'[/itex].

    Is it possible to construct such a function? I guess this is merely a tube, which is tilted.
     

    Attached Files:

  2. jcsd
  3. Jun 7, 2013 #2

    tiny-tim

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    Hi Niles! :smile:

    Essentially, this is a function of r, where in the simple case r is the distance from the y-axis,

    ie r2 = x2 + z2.

    In the slanted case, r is the distance (parallel to the x-z-plane) from (ky',y',0),

    so r2 = (x - ky')2 + z2. :wink:
     
  4. Jun 9, 2013 #3
    Thanks tiny-tim, that is a good explanation.
     
  5. Jun 9, 2013 #4
    If I want to rotate this function by an angle α around the y-axis (for a given slope k), then I need to invoke the rotation matrix. So we now have
    [tex]
    f(x, y, z) = \exp(-(x-ky)^2)\exp(-z^2)
    [/tex]
    and in order to rotate it I would use
    [tex]
    f(x\cos(\alpha) - z\sin(\alpha), y, x\sin(\alpha) + z\cos(\alpha))
    [/tex]
    Is this correct?
     
    Last edited: Jun 9, 2013
  6. Jun 9, 2013 #5

    tiny-tim

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    Hi Niles! :smile:

    Yes, that looks ok.
     
  7. Jun 9, 2013 #6
    Thanks. However, I think I am not 100% correct - simply because we're not rotating the function along the [itex]y[/itex]-axis (which is what I suggested above), but along the line [itex]ky=x[/itex]. I'll have to work on this a little more.... I can let you know how it turns out.

    By the way, can I slant [itex]f[/itex] independenty in both [itex]x[/itex] and [itex]z[/itex]? So if I want a gradient in both of these dimensions, I am allowed to make the trivial extension
    [tex]
    f(x, y, z) = \exp(-(x-k'y)^2)\exp(-(z-k''z)^2)
    [/tex]
    I believe so, because the two dimensions are independent of eachother.
     
  8. Jun 9, 2013 #7

    tiny-tim

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    let's see …

    that makes r the distance from (x,y,z) to (k'y,y,k''z),

    so yes, that would be from the axis x/k' = z'k'' = y, but slanted so as to be parallel to the x-z plane :smile:
     
  9. Jun 10, 2013 #8
    Thanks, that's also what I thought.. I'm interested in seeing how to generalize this result to ellipses. I'll best create a new thread in order to keep things organized, but I will refer to this thread.
     
    Last edited: Jun 10, 2013
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