Question

Help with an arc length question! There is a telephone wire hanging between two poles x= -b and x= b. It takes the shape of a catenary with the equation y= c + acosh(x/a)

Answer 1

I know the arc length formula is the integral of (1+ f'(t)^2)^(1/2) and the endpoints of integration are -b to b. I just have no idea on how to compute the derivative of y= c+ acosh(x/a). I've never dealt with this function before.

Answer 2

*find the length of the wire.

Answer 3

\(\cosh{x}\) is the hyperbolic cosine function, defined by
\[\cosh{x}=\frac{e^x+e^{-x}}{2}\]
It then follows that if \(f(x)=\cosh{x}\), then
\[f'(x)=\frac{e^x-e^{-x}}{2}=\sinh{x},\]
where \(\sinh{x}\) is the hyperbolic sine function, defined by
\[\sinh{x}=\frac{e^x-e^{-x}}{2}.\]
In general, then, the derivative of \(a\cosh\dfrac{x}{a}\) is
\(a\dfrac{1}{a}\sinh{\dfrac{x}{a}}=\sinh{\dfrac{x}{a}}.\)

Answer 4

Oh okay, so when you square the derivative would it be sin ^2 (x^2/a^2) ?

Answer 5

*sin h^2 (x^2/a^a)

Answer 6

You don't square the elements in the sinh.
You never do that.

Answer 7

hmm. so how would I go about using the arc length formula. I need to square the derivative?

Answer 8

the derivative, not the elemental "x".
Sinh(x)^2 is not sinh(x^2) as you've done.

Answer 9

okay so Sinh(x/a)^2 ?

Answer 10

That's a part of it....

Answer 11

yeah, I really don't know where I'm going wrong. I've never had to deal with these functions, so I have absolutely no idea what i'm doing.

Answer 12

Well, flip back to hyperbolics in your textbook.
I don't have the time to talk about them.
Just know that they follow trig identities similar to sin, cos, tan.

Answer 13

okay.

Answer 14

Yeah. @sithandwhetever gave some good advice, but may be misleading if you tilt.

Answer 15

so plugging into arc length :
\[= \int\limits_{-b} ^{b} \sqrt{1+\sinh^{2}(x/a)} dx\]
here are hyperbolic identities
http://en.wikipedia.org/wiki/Hyperbolic_identities#Useful_relations
\[= \int\limits_{-b} ^{b} \sqrt{\cosh^{2} (x/a)} dx\]
\[=\int\limits_{-b}^{b} \cosh(x/a) dx\]
\[=a \sinh(x/a)\]
plug in limits, note symmetry
\[= 2a \sinh(b/a)\]

Answer 16

@dumbcow thank you for taking the time.......

Answer 17

thank you @dumbcow ! :) and everyone else!

Answer 18

yw

Answer 19

Can I get a hug? At least?

Answer 20

lol yes *hug :D

Answer 21

haha
medals for everyone

Answer 22

This one?

Answer 23

yes.

Answer 24

Still don't get it?

Answer 25

so when he said to just plug in the endpoints: 2asinh(b/a) - 2asinh(-b/a) ?

Answer 26

I get the process, just im unsure on the answer

Answer 27

No that's the final answer.

Answer 28

hmm my math hw says it's wrong. I have no clue.

Answer 29

Chill.....

Answer 30

i am chill.

Answer 31

Oh, I didn't mean that in a disrespectful way; I meant hold on

Answer 32

I'm quite sure 2asinh(b/a) is the answer.

Answer 33

What's the answer in the book?

Answer 34

It's okay. I'm not using a textbook, it's an online site where I enter my answers

Answer 35

So I have no way to know whether I'm right or not unless it tells me

Answer 36

did you try typing it as 2*a*sinh(b/a)

Answer 37

yes. It said it was incorrect.

Answer 38

Okay try asinh(b/a) - asinh(-b/a)

Answer 39

It said it was incorrect again. I can just talk to my math teacher about it tomorrow? She may have made a mistake, it's happened before haha

Answer 40

I think that's the case.

Answer 41

One more?

Answer 42

yes, one more problem ;P

Answer 43

I'll just post it on here

Answer 44

Find the arc length function s(x) for the curve (1/3)x^3 +1/(4x) where x>0 with the starting point (1,7/12)