Arc Length of y=ln((e^(x)+1)/(e^(x)-1))

  • #1
Arc length of y=ln((e^(x)+1)/(e^(x)-1)) on [a,b]

Using L=[tex]\int\sqrt{1+(y')^2}dx[/tex] on [a,b]

I am having difficulties differentiating y and plugging the results back in to get a useful integral. So far I have y'=2e^(x)/(e^(2x)-1)
 

Answers and Replies

  • #2
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I get y' = -2ex/(e2x - 1). IOW, same as what you got except for the sign in the numerator.

Use that to get 1 + (y')2, which simplifies enough so that you can actually get the square root.
 
  • #3
Ah, thanks a ton!
 
  • #4
Nope, still got nothing. I still can't figure it out.
 
  • #6
Thanks, I figured it out a little while afterwards and lost my internet connection.
 
  • #7
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Just want to bump this question, its on my assignment and I'm stuck at the sqr root part.

So far, I got (like posted already)

int{ sqrt(1+((-2e^x)/(e^(2x)-1))^2dx)

unfortunately I don't know where to go from here. I tried multiplying out to get for the 1+y'^2:

1+(4-4e^x+e^2x)/(e^4x-2e^2x+1)

but I can't help but think I messed up in there. my brain is feeling like mush at this point, but the assignments due tomorrow and I've just been spinning tires for the last hour :(
 
  • #8
Char. Limit
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I don't think that you squared y' correctly...
 
  • #9
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does (4e^2x)/(e^4x-2e^2x+1) sound about right? it simplifies to 4/(e^4(x)-1)
 
  • #10
Char. Limit
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That sounds a lot better. Now just add [itex]\frac{e^{4x}-1}{e^{4x}-1}[/itex] to it and go from there.
 
  • #11
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thanks for your help so far, but I'm still stuck at the next step

after replacing the 1 with e^4x-1/e^4x-1 and simplifying, how do I take the sqrt of (e^(4x)+3/e^4(x)-1)?
 
  • #12
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does (4e^2x)/(e^4x-2e^2x+1) sound about right? it simplifies to 4/(e^4(x)-1)

The first part is right for (y')2, but it doesn't simplify to your second expression. How did you get e4x - 1? Also, it's better to leave the denominator unexpanded.

See if you can finish this.

1 + (y')2 = 1 + 4e2x/(e2x - 1)2 = ?

Replace 1 with (e2x - 1)2/(e2x - 1)2 so you can get a common denominator.
 
  • #13
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That sounds a lot better. Now just add [itex]\frac{e^{4x}-1}{e^{4x}-1}[/itex] to it and go from there.
You don't want to do this.
 
  • #14
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I simplified by cancelling the e^(2x) from the top/bottom, but now that I think about it you can't do that with addition/subtraction, correct?

ok, so I redid it and now have sqrt(1+4e2x). after substituting then adding, the (e2x-1)2 cancels out on both top and bottom leaving ones right? so then the root would be something around 2ex, but I get stuck at the same place as my last post: how do I deal with the one? I know I can't separate it into sqrt(1)+sqrt(4e2x)...

sqrt(1+4e2x)
 
  • #15
Char. Limit
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Sorry mark, but I went with the idea that he was correct (wrong in hindsight) and produced the best expression for 1. Now he'd need a different expression.
 
  • #16
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I simplified by cancelling the e^(2x) from the top/bottom, but now that I think about it you can't do that with addition/subtraction, correct?
You can cancel only when an expression appears as a factor in numerator and denominator. You can't cancel expressions that make up a sum or difference. For example, this is wrong: (4 + 1)/(4 + 2) = 1/2.
ok, so I redid it and now have sqrt(1+4e2x).
That's wrong, too. It seems to me that your algebra skills are weak. Unless you take steps to bring them up, algebra errors will continue to plague you, and keep you from completing problems in calculus.

You should have
1 + (y')2 = 1 + 4e2x/(e2x - 1)2
= (e2x - 1)2/(e2x - 1)2 + 4e2x/(e2x - 1)2

If you expand the e2x - 1)2 expression in the numerator (but not the same expression in the denominator), and then add all the terms in the numerator, you should get something that is a perfect square. This makes it easy to take the square root, which is your next step.

after substituting then adding, the (e2x-1)2 cancels out on both top and bottom leaving ones right? so then the root would be something around 2ex, but I get stuck at the same place as my last post: how do I deal with the one? I know I can't separate it into sqrt(1)+sqrt(4e2x)...

sqrt(1+4e2x)
 
  • #17
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thanks, I think I got it. my brain just wasn't functioning at all last night and I kept making the same mistakes. ended up with the integral of 2 to 3 being (e2x+1)/e2x-1)dx, and then solution being ln(ex-e-x) from 2 to 3. but you are absolutely right, I took four years off between high school and uni and forgot all the basic rules. when I graduated I got 1500 in checks for achieving the top grades in pure math 30 and math 31, but I now know my teachers were just jokes. I never even heard of the unit circle until calc 1 last semester!
 
  • #18
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Hi, I'm new here and I have this same problem that I'm trying to tackle at the moment (sorry for the necroposting).

I'm currently stuck at sqrt((e^2x + 1)^2 / (e^2x + 1)(e^2x - 1))

This is what I've thought of so far:
Cancel out e^2x - 1 from numerator and denominator to give:
sqrt((e^2x + 1)/(e^2x - 1))...then i get stuck there.

I closest thing for me to get unstuck from above is to turn e^2x+1 and e^2x -1 to something that's a squared notation, but I can't get anywhere..
 
  • #19
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Hi, I'm new here and I have this same problem that I'm trying to tackle at the moment (sorry for the necroposting).

I'm currently stuck at sqrt((e^2x + 1)^2 / (e^2x + 1)(e^2x - 1))

This is what I've thought of so far:
Cancel out e^2x - 1 from numerator and denominator to give:
sqrt((e^2x + 1)/(e^2x - 1))...then i get stuck there.
There is no factor of e^(2x) - 1 in the numerator, so how can you cancel it from numerator and denominator?
I closest thing for me to get unstuck from above is to turn e^2x+1 and e^2x -1 to something that's a squared notation, but I can't get anywhere..
 
  • #20
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Oh I meant (e^2x + 1) from

sqrt ((e^2x + 1)^2) / ((e^2x + 1)(e^2x - 1))
 
  • #21
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It might help to go back and read through this entire thread, ignoring the misleading advice from one poster.
 
  • #22
Char. Limit
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I already apologized for that...
 
  • #23
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I already apologized for that...
I wasn't trying to cast blame. I just wanted to warn this poster about a wrong turn.
 

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