tiffany_rhodes
  • tiffany_rhodes
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)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
tiffany_rhodes
  • tiffany_rhodes
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.
tiffany_rhodes
  • tiffany_rhodes
*find the length of the wire.
anonymous
  • anonymous
\(\cosh{x}\) 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 \(a\dfrac{1}{a}\sinh{\dfrac{x}{a}}=\sinh{\dfrac{x}{a}}.\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

tiffany_rhodes
  • tiffany_rhodes
Oh okay, so when you square the derivative would it be sin ^2 (x^2/a^2) ?
tiffany_rhodes
  • tiffany_rhodes
*sin h^2 (x^2/a^a)
primeralph
  • primeralph
You don't square the elements in the sinh. You never do that.
tiffany_rhodes
  • tiffany_rhodes
hmm. so how would I go about using the arc length formula. I need to square the derivative?
primeralph
  • primeralph
the derivative, not the elemental "x". Sinh(x)^2 is not sinh(x^2) as you've done.
tiffany_rhodes
  • tiffany_rhodes
okay so Sinh(x/a)^2 ?
primeralph
  • primeralph
That's a part of it....
tiffany_rhodes
  • tiffany_rhodes
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.
primeralph
  • primeralph
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.
tiffany_rhodes
  • tiffany_rhodes
okay.
primeralph
  • primeralph
Yeah. @sithandwhetever gave some good advice, but may be misleading if you tilt.
dumbcow
  • dumbcow
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 \[= 2a \sinh(b/a)\]
primeralph
  • primeralph
@dumbcow thank you for taking the time.......
tiffany_rhodes
  • tiffany_rhodes
thank you @dumbcow ! :) and everyone else!
dumbcow
  • dumbcow
yw
primeralph
  • primeralph
Can I get a hug? At least?
tiffany_rhodes
  • tiffany_rhodes
lol yes *hug :D
dumbcow
  • dumbcow
haha medals for everyone
primeralph
  • primeralph
This one?
tiffany_rhodes
  • tiffany_rhodes
yes.
primeralph
  • primeralph
Still don't get it?
tiffany_rhodes
  • tiffany_rhodes
so when he said to just plug in the endpoints: 2asinh(b/a) - 2asinh(-b/a) ?
tiffany_rhodes
  • tiffany_rhodes
I get the process, just im unsure on the answer
primeralph
  • primeralph
No that's the final answer.
tiffany_rhodes
  • tiffany_rhodes
hmm my math hw says it's wrong. I have no clue.
primeralph
  • primeralph
Chill.....
tiffany_rhodes
  • tiffany_rhodes
i am chill.
primeralph
  • primeralph
Oh, I didn't mean that in a disrespectful way; I meant hold on
primeralph
  • primeralph
I'm quite sure 2asinh(b/a) is the answer.
primeralph
  • primeralph
What's the answer in the book?
tiffany_rhodes
  • tiffany_rhodes
It's okay. I'm not using a textbook, it's an online site where I enter my answers
tiffany_rhodes
  • tiffany_rhodes
So I have no way to know whether I'm right or not unless it tells me
primeralph
  • primeralph
did you try typing it as 2*a*sinh(b/a)
tiffany_rhodes
  • tiffany_rhodes
yes. It said it was incorrect.
primeralph
  • primeralph
Okay try asinh(b/a) - asinh(-b/a)
tiffany_rhodes
  • tiffany_rhodes
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
primeralph
  • primeralph
I think that's the case.
primeralph
  • primeralph
One more?
tiffany_rhodes
  • tiffany_rhodes
yes, one more problem ;P
tiffany_rhodes
  • tiffany_rhodes
I'll just post it on here
tiffany_rhodes
  • tiffany_rhodes
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)
primeralph
  • primeralph
|dw:1370323126959:dw|
primeralph
  • primeralph
....do you just want the final answer?
tiffany_rhodes
  • tiffany_rhodes
yeah just so I can make sure I get my hw done. I'm going back through all the problems that I had trouble with anyways.
primeralph
  • primeralph
|dw:1370323557443:dw|
primeralph
  • primeralph
if it doesn't take that, simplify it
tiffany_rhodes
  • tiffany_rhodes
It was correct! thanks again
primeralph
  • primeralph
You're welcome.......

Looking for something else?

Not the answer you are looking for? Search for more explanations.