A Hanging Rope

Dear Uncle Colin,

I’m designing a small cathedral and have an 80-metre long rope I want to hang between two vertical poles. The poles are both 50 metres high, and I want the lowest point on the rope to be 20 metres above the ground. How far apart should I place the poles?

Geometry And Undefined Definite Integration

Hi, GAUDI, and thanks for your message!

A free-hanging rope generally forms a *catenary* curve, and can be modelled with a function of the form $y = a \cosh\br{\frac{x}{a}} + b$, as long as the turning point is placed at $x=0$.

What we need to do is find an expression for the arclength of the curve between two arbitrary limits and solve for where that is equal to 80.

### Let’s get rid of the $b$ first

First of all, we know that when $x=0$, $y=20$ and when $x=\pm K$, then $y= 50$.

This points us at $20 = a + b$ and $50 = a \cosh\br{\frac{K}{a}} + b$, leaving us with $a\br{\cosh\br{\frac{K}{a}} – 1} = 30$.

### So how about that arclength?

The arclength of a curve can be expressed as $L = \int_a^b \sqrt{ 1 + \br{ \dydx }^2} \dx$. In our case, $\dydx = \sinh\br{\frac{x}{a}}$, so the integral becomes $\int_{-K}^K \sqrt{1 + \sinh^2 \br{\frac{x}{a}}} \dx$. Because $1+\sinh^2(z) = \cosh^2(z)$, this becomes $\int{-K}^{K} \cosh\br{\frac{x}{a}} \dx$, or $2a \sinh \br{\frac{K}{a}}$.

We now have $a\br{\cosh\br{\frac{K}{a}} – 1} = 30$ and $a \sinh \br{\frac{K}{a}} = 40$.

Rearranging gives $a \cosh \br{\frac{K}{a}} = 30 + a$ and $a \sinh \br{\frac{K}{a}} = 40$; squaring both gives $a^2 \cosh^2\br{\frac{K}{a}} = 900 + 60a + a^2$ and $a^2 \sinh^2\br{\frac{K}{a}} = 1600$.

Because $\cosh^2(z) – \sinh^2(z) = 1$, we can subtract these to get $a^2 = -700 + 60a + a^2$, or $a = \frac{35}{3}$.

### But it’s $K$ we want!

What’s the simplest equation we’ve got with a $k$ in? I’d say $a \sinh\br{\frac{K}{a}} = 40$; and now we know that $a = \frac{35}{3}$ this becomes $\sinh\br{\frac{3K}{35}} = \frac{24}{7}$.

Now, $\arsinh(z) = \ln \br{z +\sqrt{z^2 + 1}}$, so we now have $\frac{3}{35}K = \ln \br{ \frac{24}{7} + \sqrt{ \br{\frac{24}{7}}^2 + 1} }$.

That last part under the square root is, reassuringly, two legs of a Pythagorean triple, so $ \sqrt{ \br{\frac{24}{7}}^2 + 1} = \frac{25}{7}$ – and neatly, that makes the argument to the logarithm… just 7!^{1}

We end up with $K = \frac{3}{35} \ln(7)$, so the posts must be $\frac{6}{35}\ln(7)$ metres apart.

Hope that helps — good luck with the cathedral. Let me know when it’s finished!

– Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.