1. ## Catenary

Equation of a catenary is
y = a * cosh(x/a) + C

I am told that two telephone poles are 10 m apart and that the poles are 6 m tall. The minimum clearance (distance of the cable to the ground) is 4 m. Question asks how long can the cable be.

So. I know that once I find the equation for the catenary the cable makes, I get the integral and use the arc-length formula. Unfortunately, I don't know how to get the 'a' in the formula.

Here's my attempt so far.
C doesn't really matter. The lowest point at x = 0 is a, since cosh(0) = 1. That means the point at x = -5 and x = 5 is a+2. So I get an equation:
a+2 = a * cosh(5 / a)

But I'm not sure how to solve it.
a * arccosh[(a+2) / a] = 5

That's as far as I got, and I'm rather sure I'm not doing it right. The assignment is focused on linearization and the like (Newton's method, fixed-point iteration, etc.).

2. Since the poles are 6 meters high and the clearance is 4 m, then the sag is 2 m.

We can use Newton's method to find the length.

The sag is given by $S=a\cdot cosh(\frac{b}{a})-a$

b is the distance from the origin to one of the poles. 5 m.

$2=a\cdot cosh(\frac{5}{a})-a$

Let $u=\frac{5}{a}$

Then, $a=\frac{5}{u}$

$2=\frac{5}{u}\left[cosh(u)-1\right]$

$cosh(u)-1=\frac{2u}{5}$

If $f(u)=cosh(u)-\frac{2u}{5}-1$

Then, $u_{n+1}=u_{n}-\frac{cosh(u_{n})-\frac{2u_{n}}{5}-1}{sinh(u_{n})-\frac{2}{5}}$

Try an initial guess for u. Then, once it converges, 'a' can be found by just subbing it into $a=\frac{5}{u}$

The length can be found by plugging 'a' into the length formula:

$L=2a\cdot sinh(\frac{b}{a})$

We can derive the length by:

$y'=sinh(\frac{x}{a}), \;\ 1+(y')^{2}=1+sinh^{2}(\frac{x}{a})=cosh^{2}(\frac{ x}{a})$

$L=2\int_{0}^{b}cosh(\frac{x}{a})dx=\boxed{2a\cdot sinh(\frac{b}{a})}$

### Cosh5a

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