In my engineering class we were learning about the hanging cable under its own weight.

If you have coordinate system centered with the hanging cable, the equation of the curve is,

$\displaystyle y=c\cosh (x/c)$

Where, $\displaystyle c$ is the konstant to be determine.

If you know the span $\displaystyle L$ and it sag $\displaystyle h$.

We need to find $\displaystyle c$.

We know that when $\displaystyle x=L/2$ then we have, $\displaystyle y=c+h$

Thus,

$\displaystyle c+h=c\cosh (L/2c)$

Thus,

$\displaystyle 1+h/c=\cosh (L/2c)$

The way the professor taugh us to solve this is to guess through values until it works.

That approach is too primitive for me.

I told him that it might be possible to use the Lambert function to solve it, he slaped me across my face.

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So I have been thinking,

Let $\displaystyle x=1/c$.

Then,

$\displaystyle 1+hx=\frac{e^{.5Lx}+e^{-.5Lx}}{2}$

Thus,

$\displaystyle 2+2hx=e^{.5Lx}+e^{-.5Lx}$

It seems I have been around engineers too much these days, and I learned this trick from them.

$\displaystyle e^{-.5Lx}\approx 0$

(I cannot believe that I did that).

Thus,

$\displaystyle 2+2hx=e^{.5Lx}$

Thus, the problem redues to solving,

$\displaystyle a+bx=e^x$

Which I think is possible using a proper variable substition, right

Thus, we would have an approximate solution through the Lambert function for the solution of konstant $\displaystyle c$.

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I had another approach, more mathematical, is to create error bounds on the solution.

Since,

$\displaystyle 0<e^{-.5Lx}<1$

We solve the equation as if,

$\displaystyle e^{-.5Lx}=0$

And then,

$\displaystyle e^{-.5Lx}=1$

Those would create an interval on which the solution exists.

Which is good because it leads to,

$\displaystyle a+bx=e^x$

Anyone, has any other ideas, how to solve this equation.