Melody

If the gradient is not increasing and not decreasing then it must be a constant. 

At a point of inflection

\(\frac{d^2y}{dx^2}=0\\~ \\and\\~\\ \frac{dy}{dx}=\int \frac{d^2y}{dx^2}dx=\int 0dx=k \)

I am not sure that I have presented this properly.

Feel free to comment Max.

Sorry guest but that is not correct.

half of 4/5 is 2/5

(2/5)^2 = 4/25

\((x+\frac{2}{5})^2 = x^2+2*\frac{2x}{5}+(\frac{2}{5})^2\)

Thanks Omi    

Yes, I used co-ordinate geometry.

I let the centre point between the two poles be (0,0)

And I let the bottom of one pole be (-b,0) and the other be (b,0)

This shape is a catenary.

The formula for a catenary is   

  \(\boxed{y=c+acosh(\frac{x}{a})\\ where\\ cosh(x) =\frac{e^x+e^{-x}}{2}}\)

The arclength of this section of catenary is L which is 80m

It can be shown that:

\(L=\displaystyle\int_{-b}^{b}\;\sqrt{1+\left(\frac{dy}{dx}\right)^2}\;dx\)

Put all this together and I ended up getting either nonsense or that b/a=0 

It was then that I looked at the question properly and had my Doh! moment.

All suspended ropes, chains etc hang in a catenary. 

There are a lot of good youtube clips available if you want to look at them. 

The mildly embarrassing fact is that I solved it with complicated mathematics.

I got the answer 0 and though I had made a mistake.

I finally looked at it properly and the penny dropped.    Doh !!!

I have no idea LOL

Yes you are right LOL  :)

One question per post!

Try to do them yourself. These are very basic questions.

Surely you could do some of them yourself

 9^5

Please edit your question so that it reads properly.