Question

PLEASE HELP I HAVE BEEN STUCK ON THIS FOR OVER AN HOUR!!!!!!!
Telephone wire hangs between two poles at x=-b and x=b. It takes the shape of a catenary with equation y=c+acosh(x/a). From this I found that the length of the wire is 2asinh(x/a).
Suppose two telephone poles are 50 ft apart and the length of the wire between the poles is 51 ft. If the lowest point of the wire must be 20 ft above the ground how high up on each pole should the wire be attached?

Answer 1

Hi!

Answer 2

c: thinking.

Answer 3

So how did you find that the length of the wire is 2asinh(x/a)? Integrating with respect to arclength?

Answer 4

I took the integral of sqrt(1+(sinh(x/a))^2)

Answer 5

not a fan of hyperbolic functions :{

Answer 6

Me either :)

Answer 7

\[\large a \sinh\left(\frac{x}{a}\right)|_{x=-b}^{b}\]
Ok cool I'm coming up with the same thing. I think you want to evaluate it at the different b points, then set it equal to 51.. maybe we can do something with that.

Answer 8

\[\large 2a \sinh\left(\frac{b}{a}\right)=51\]

Answer 9

Hmmmmm

Answer 10

b=25 since poles are 50 ft apart

Answer 11

Hmm does that allow us to solve for \(a\) somehow? :\ hmm

Answer 12

the lowest point occurs when x=0 so ...

20 = c+a*cosh(0)

Answer 13

so do we solve for a or c first?

Answer 14

so 20=c+a since cosh(0)=1

Answer 15

will the difference in height be y(25)-y(0)=h
and then to find H=20+h but to do this we still need to find a I think. :/

Answer 16

yeah ... i cheated
http://www.wolframalpha.com/input/?i=2x*sinh%2825%2Fx%29+%3D+51

Answer 17

ya thats all i could think to do also c: lol
if he needs to show work, and understand the problem though :O Hmmmm lol

Answer 18

so you take the length and set it equal to 51 which is the length given.
2asinh(b/a)=51 and then divide by to to get asinh(b/a)=51/2
what do we use the 51/2 for?

Answer 19

is there an inverse of sinh that we can use to solve for a?

Answer 20

so a would be about 72.3843. what would we do from here?

Answer 21

what would y(25)-y(0) be?