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### Hyperbolic Functions

Package of problems

## Problem

Please be aware that this resource is still in a somewhat draft form.

The shape a heavy rope or chain hangs in is called a catenary. What is the equation of a catenary, and how could we work it out?

There are multiple ways to solve this problem; we’ll explore two of them.

### Setting up the problem: Tension in a rope

(If you have not learnt about tension, you can just use the result from this section in the following sections.)

This diagram shows a rope hanging. We have drawn axes so that the rope hangs in the $(x,y)$-plane with the $y$-axis vertically upwards and the lowest point of the rope on the $y$-axis.

Here, $T_0$ is the tension at the lowest point of the rope, while $T$ is the tension at a point $P$ which is at a distance of $s$ along the rope from the lowest point. At $P$, the rope makes an angle of $\psi$ with the horizontal.

The rope has weight $w$ per unit length.

Prove that $\tan\psi=\Bigl(\dfrac{w}{T_0}\Bigr)s$.

From now on, we’ll make our life easier by assuming that $w/T_0=1$, so that the equation becomes $\tan\psi=s$.

This is a legitimate thing to do since we have not specified a unit for $w$ (which is constant for any particular rope) and $T_0$ is a constant for any particular situation, so we could choose a unit for measuring $w$ which made $w/T_0=1$.

If you want to be more general, you could still simplify things by writing $w/T_0=1/a$, so that the equation becomes $\tan\psi=s/a$. But the remainder of this task is tricky enough without this additional complication, so we will stick to the case $a=1$.

If the rope has the equation $y=f(x)$, prove that $f'(x)=s$.

### An iterative approach

We’ll start by assuming a very simple (wrong) formula for the rope length $s$, use that to work out a first approximation for $f(x)$, and improve it step-by-step.

We will also assume that the rope has a $y$-intercept of $1$.

#### Step 1

Assume that $s=x$, that is, assume the weight is evenly distributed horizontally. (This is the case for ropes holding up bridges.)

What is $f_1(x)$ in this case, where $f'_1(x)=s$?

#### Step 2

Now assume that the weight is distributed uniformly along the shape given by the answer to Step 1, that is to say, $s$ is now the arc length of $y=f_1(x)$.

What is $f_2(x)$ approximately in this case, where $f'_2(x)=s$? Work out an answer as a polynomial, valid for small values of $x$, including one term more than in $f_1(x)$.

(We’re looking for an approximation, so do not try to use any complicated integration techniques here – the most you should have to integrate is a polynomial. See the suggestions if you want one.)

#### Step 3

Now assume that the weight is distributed uniformly along the shape given by the answer to Step 2, that is to say, $s$ is now the arc length of $y=f_2(x)$.

What is $f_3(x)$ approximately in this case, where $f'_3(x)=s$? Work out an answer as a polynomial, valid for small values of $x$, including one term more than in $f_2(x)$.

#### In the limit

1. Can you predict $f_4(x)$? What about $f_5(x)$? And how will the sequence of polynomials $f_1(x)$, $f_2(x)$, $f_3(x)$, … continue further?

2. What is the power series, $f(x)$, that this sequence is tending to? What do we call this function?

3. Can you prove that this limit function $f(x)$ satisfies the equation $f'(x)=s$, where $s$ is the arc length of $y=f(x)$?

4. In the above procedure, what would Step $n$ have been? Can you prove that Step $n$ would have given the function $f_n(x)$ which you predicted?

(See the suggestions for help on this step if you need it.)

5. Why do you think we assumed that the curve has a $y$-intercept of $1$?

### A direct approach

Can you solve the equation $f'(x)=s$ directly?

More generally, can you solve the equation $f'(x)=\dfrac{s}{a}$, where $f(0)=a$?