calc word problem

[sonniebeth22]

i'm having a massive amount of trouble w/ this problem.....any help is appreciated!

Find the length of a flexible cable suspended between horizontal supports that are 800 ft apart, if the sag in the cable is prescribed to be 40 ft. (answers to be 1/10th of a ft)

 

[tkhunny]

Small wonder you may be strugglinh with this. Many have strggled so.

Check this out and see if it leads anywhere: http://mathworld.wolfram.com/Catenary.html You'll at least know what you're working with.

 

[Matt]

Hi Sonnie,

Please let us know if you need more help. In tkhunny's link you can find an explicit formula for the sagging cable, which you can integrate to find the arc length of the cable (or, you could simply use the formula for the arc length the website provides, but most likely the intention of the problem is to have you integrate something). If you still need help, then please feel free to let us know.

 

[sonniebeth22]

Thanks for responding!! I really appreciate it....but the trouble lies in the formula...i know i have to use:
y=a*cosh(x/a)
i have no idea how to get "a" if a is the height and all i know is the sag is 40 ft.

thanks so much for you help!

 

[Matt]

Hi sonniebeth,

It is not possible to determine an exact value for a, because in order to do so, you would have to solve a transcendental equation. But it is possible to approximate a. Let's do this.

In general, it always helps to draw a picture. Here's an image of a typical graph of f(x)=a*cosh(x/a), superimposed with the distance between the two supports and the sag labeled appropriately:

I hope it is clear from the diagram that f(400)=40+a. We can use this single piece of information to find out what a is supposed to be:

. . . . 40 + a = f(400) = a*cosh(400/a)

Thus, we can approximate the value of a by finding a numerical solution to the equation

. . . . a*cosh(400/a) - 40 - a = 0

You could try using Newton's method to solve this equation. You might remember learning about Newton's method in calculus 1.