# Tangent to Catenary

1. Jan 18, 2014

### jimbobian

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 $r=a$ for $z=\pm b$ and I found the radius of the bubble as a function of height z to be:

$$r(z) = c\cosh(z/c)$$

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

$$a/c = \cosh(b/c)$$

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

I plotted a selection of functions of the form $a/c = \cosh(b/c)$ for various $c$ 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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2. Jan 19, 2014

### Irid

3. Jan 19, 2014

### jimbobian

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 $m_c$ as above.

I find the following two restrictions:

$$a=c\cosh(b/c)$$
$$c/b = \tanh(b/c)$$

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