# Hyperbolic Functions

• Mar 19th 2013, 11:05 PM
KaylaKelley
Hyperbolic Functions
When a particular cable hangs between two poles the shape is a catenary, f(x)=acosh(x/a) where a is the distance from the sag of the cable to the ground. Suppose the poles are 100 feet apart. Find the slope where the cable meets the pole if the sag is 30 feet.

So from f(x) = a cosh(x/a) I got that f(0) = a and f(50)=acosh(50/a) with 50 being the distance from where the sag is 30 feet to the pole.

I need to solve for a somehow with acosh(50/a)-a=30... that being the equation that can tell me how far the sag is from the ground... and then i can plug a into f'(x)=sinh(x/a) and f'(50)=sinh(50/a) to get the derivative at the pole.... but I am stuck solving for a.
• Mar 20th 2013, 07:24 AM
ebaines
Re: Hyperbolic Functions
Are you sure that you are interpreting the problem correctly? If the value of 'a is defined as "the distance from the sag of the cable to the gound," and you are told that "the sag is 30 feet to the pole," perhaps this means that a=30?

But assuming that your interpretation is correct, then solving acosh(50/a) -a = 30 probably requires the use of a numerical technique.
• Mar 20th 2013, 08:06 AM
KaylaKelley
Re: Hyperbolic Functions
The sag is 30 feet. As in if a straight line is drawn from one pole to the other, the cable, at its lowest point is 30ft below the straight line.

My teacher provided that interpretation in class. Im not sure what you mean by numerical technique.....
• Mar 20th 2013, 08:24 AM
jeanshermanF
Re: Hyperbolic Functions
It seems to be like a tough question instead of discussing it here why don't you take help from experts. There are people who can help you with these kind of problems. You can log on to www.artifextutors.com and get help from there talk to there live chat experts and get help from there.

Regards
Jean
• Mar 20th 2013, 09:29 AM
ebaines
Re: Hyperbolic Functions
Quote:

Originally Posted by KaylaKelley
Im not sure what you mean by numerical technique.....

Trial and error is a brute force numerical technique. If you have access to a spreadsheet you can set the formula up, try a value for a, note whether it's too big or too small, then try anouther value, etc. To get you started - if you start with a = 46 you should be able to iterate to a value for 'a' that is accurate to 4 decimal points pretty quickly. Then use that value for 'a' in your slope equation.
• Mar 20th 2013, 09:58 AM
KaylaKelley
Re: Hyperbolic Functions
Oh I see. I graphed it in my calculator and was looking for the x intercept but when I took it to my professor its was wrong and my calculator was giving me something different than his. Thank you for the help though.
• Mar 20th 2013, 10:30 AM
ebaines
Re: Hyperbolic Functions
The catenary shoud look like this:

Attachment 27613
• Mar 21st 2013, 07:15 AM
BobP
Re: Hyperbolic Functions
Hi Kayla,
As was mentioned in an earlier post, solving the equation
$\displaystyle 30 + a = a\cosh\left(\frac{50}{a}\right).$
requires the use of a numerical technique.

There are a number of available methods, you could use the trial and error approach mentioned earlier, or maybe the Bisection method, or perhaps the simple fixed point iteration
$\displaystyle a_{n+1}=a_{n}\cosh\left(\frac{50}{a_{n}}\right)-30.$

Starting with $\displaystyle a_{0}=50,$ this produces the sequence

$\displaystyle 47.154, 46.2417, 46.0121, 45.9597, 45.9480, 45.9454, 45.9449, 45.9447, 45.9447.$

Faster convergence would come from the Newton-Raphson method. This produces the iterative formula, (the bottom line of the fraction is the derivative of the top line),

$\displaystyle a_{n+1}= a_{n}- \frac{30+a_{n}-a_{n}\cosh\left(\frac{50}{a_{n}}\right)}{1-\cosh\left(\frac{50}{a_{n}}\right)+\frac{50}{a_{n} }\sinh\left(\frac{50}{a_{n}}\right)}.$

Again, starting with $\displaystyle a_{0}=50,$ this produces the sequence $\displaystyle 45.4977, 45.9392, 45.9447, 45.9447.$

So it would seem that a = 45.9447 (4dp).