Math Is Fun Forum

There have been times I have been a bit selfish (very few occasions, and more often, unknowingly), but it is past now. I tried to be a good man, a good member of the society.

I feel a person ought to become better, as there is no point in brooding at a later date.

mathland wrote:

ganesh wrote:

I am single. A person can be unmarried, married, happily married, or separated. It is one's decision.

In my case, I am happy to be single and independent.

I am happy to be single as well. Less heartaches to deal with.

Thanks. But COVID-19 has been as nightmare for many nations, developing and developed. As a matter of fact, it has been a pandemic.

We should fight together, against a common foe, the deadly virus.

mathland wrote:

ganesh wrote:

A tough situation.

Very tough. Impossible to deal with. Crushing blow.

Hope things get better.

Ok. Here we go. Before I get on to the catenary itself, there's a couple of topics I need as preliminaries.

Arc Length Formula.

https://i.imgur.com/NLPKqX6.gif

This works for any curve, not just a catenary.  Δs is the length along the curve.

The diagram shows that Δx, Δy and Δs almost make a right angled triangle.  Not quite, because the 'hypotenuse' isn't a straight line.  However, as Δx tends to zero, the curve gets ever closer to straight and so in the limit we can assume pythagoras theorem.

Divide by Δx^2

Let Δx tend to zero:

Hyperbolic Trig Functions.

Sinh is pronounced sinch

I leave the following as exercises for the interested reader:

The Catenary

By observation of hanging ropes etc. it is clear that there is a minimum point.  It is usual to put the y axis through that point.
The curve looks a bit like a parabola, but it isn't quite that. Here's a rough diagram of the situation.  If you have done any mechanics questions with 'light inextensible strings' you may be used to the tension being constant along the string.  The catenary calculation does not have weight free strings.  The assumption is that the string has a constant weight along its length.  This means that, if the weight is w per unit length (let's say feet) the 1 foot weighs w, 2 weighs 2w, and s feet weighs ws.

The following applies to ropes, strings, chains and wires.  As the original post was for a wire I shall use that from now on.

The tension varies from point to point.  O is the lowest point and P some point elsewhere along the wire.  There are three forces acting on OP.  The horizontal tension T0 at O; the tension T at P, acting at angle alpha, and the weight of OP, ws, acting vertically downwards.

https://i.imgur.com/5AXJcsX.gif

The rope is in equilibrium, meaning the forces are balanced so that it isn't moved left or right, up or down.  So the vertical forces must balance and the horizontal forces must separately balance.

The tension, T, varies, but can be eliminated from the equations by dividing the first by the second.  And tan alpha is the gradient of the line and so may be written dy/dx.

T0 and w are both constants so it makes sense to simplify by replacing with a single constant. For reasons which will become clear later, w/T0 = 1/a rather than the other way up.  Don't get hung up on this; it's just a constant.

So we have a differential equation:

Now you might want to jump in with an easy integral, s/a, but wait!  The variable is 's' and the variable we are wanting to integrate with respect to is x.  Can't be done!  You have to have both variables the same.

That's why we need the arc length formula. We can change the 'dx' into a 'ds' using this. 

We have

and

If either is solved alone I'll have a equation not y in terms of x, but y (or x) in terms of s.  Not what we want.  So both need to be solved and then I have to see if s can be eliminated.  Don't worry; I've sneaked a look ahead and it can

The y one is directly integrable:

If we choose that minimum point to be (0,a) then C = 0.

The other is harder.  It can be done using substitution.  This technique sometimes works if you can find a good substitution.  Experience helps.  In this case we've got a ready sub that we can try.  You have to change all instants of the old variable to the new one.  At the end you have to convert back to the old variable.

Try s = a sinh(u/a) where u is the new variable.  ds/du = cosh(u/a). 

Therefore

This constant is also zero as sinh(0) = 0

So x = a arc sinh (s/a)   =>    s = a . sinh(x/a)

I can sub that into the y equation thus:

I'll take a break at this point and come back to the actual question later.

Bob

(10, 24000) is not a point.  You can use (0,24000) or (10, 2000)

Bob

Let's look at the horizontal case first as it's easy to see.

On the line y = a, take any two points, (b,a) and (c,a)

The gradient is the difference in y over the difference in x so we have

m = (a-a)/c-b) = 0/(c-b) = 0

For a vertical line, say, x = d, choose two points, (d,e) and (d,f)

m = (f-e)/(d-d) = (f-e) / 0

What do you get when you divide a number by zero?

Another way of answering this is to ask how many zeros must we add together to reach a total of (f-e).

It's just not a question we can answer. 

You could ask what is the limit of (f-e)/Δx as Δx tends to zero?

It tends to ∞.

In studying ∞, mathematicians have realised it's best not to regard ∞ as a number.  It doesn't obey the usual rules of numbers:

such as ∞ + ∞ = ∞

That's why the word 'undefined' is used for this gradient.

For a moment let's suspend the 'not a number' idea and play with the equation of the line anyway.

Suppose I write m = ∞ and make the equation y = ∞x + c

We know that (d,3) lies on the line so y = ∞d + c = 3

So c = 3 - ∞d.  How are you going to work that out?

Bob