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Tangent to Catenary

  1. Jan 18, 2014 #1
    Hi all, this question stems from a homework question but is not the homework question itself, more a discussion on something I found, hence why I have put it here.

    The question involved using variational calculus to minimise the surface area of a soap bubble to find the shape it would take. The restrictions were that [itex]r=a[/itex] for [itex]z=\pm b[/itex] and I found the radius of the bubble as a function of height z to be:

    [tex]r(z) = c\cosh(z/c)[/tex]

    which is a catenary as expected. The constant c is constrained by the boundary conditions s.t:

    [tex]a/c = \cosh(b/c)[/tex]

    which the question points out only has solutions for the ratio [itex]b/a<m_c[/itex] where [itex]m_c[/itex] is some critical value. The question does not ask us to find [itex]m_c[/itex] but I wished to do so.

    I plotted a selection of functions of the form [itex]a/c = \cosh(b/c)[/itex] for various [itex]c[/itex] and observed that the tangents appeared to form a straight line through the origin (which I have also added in red to help see it). This is consistent with what the question points out, but I can't seem to show why this is the case analytically (ie. that each catenary touches the same line through the origin). This line, incidentally, has a gradient of 1.5089 which I found using NR.

    Could anyone point out why the catenaries form in this way, because I can't show (analytically) that they do.

    Cheers,
    James
     

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  3. Jan 19, 2014 #2
  4. Jan 19, 2014 #3
    Thanks for pointing that out, at least now I know what it is called and that it seems to be a general result (for some envelope). But is it possible to show that, in this case, the envelope is a straight line with gradient [itex]m_c[/itex] as above.

    I find the following two restrictions:

    [tex]a=c\cosh(b/c)[/tex]
    [tex]c/b = \tanh(b/c)[/tex]

    But no way of showing that this leads to the same gradient for each choice of c.
     
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