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Arc length of a curve (help)


mb12777

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Okay, so I've got a bunch of questions to do with hyperbolic functions, arc lengths and surface areas of revolution. I'm stuck on the final question, it's an integration that I have no idea how to even start...

I know the arc length as an integral, but I'm having trouble actually integrating it... My y' was 2√(a2 + x2), which is definitely correct as the first part of the question was to show that. So in this next part, a = 1, so...

L = (integral) [ √(5 + 4x2) ]

The limits are 1 and 0, but I don't even know how to integrate this indefinitely, never mind definitely... So... Help?

Edit: I've used Wolfram Alpha's integral calculator tool to find out what this integrates to, but still have no idea how to get there...

Further edit: The value of L is supposed to be 1/8(12 + 5ln5)

Edited by mb12777
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Try substituting x = b sinh(t) with a fixed parameter b. For a good choice of b, everything is going to turn into cosh.

Hidden hint, only use if needed:

Use 1 + sinh(t)2 = cosh(t)2 to simplify the nasty square root.

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That's just a number. If you want a better hint from computer algebra, put in a variable as upper limit, or let it compute the indefinite integral. edit... wait, why is there no "a" in the result you quoted? There should be.

And, well, what do you know about sinh and cosh?

Edited by Z-Man
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There isn't an 'a', and it's a show that question so I know that's what it should come out as.

I know their exponential forms, a few identities, and differentiation of their inverse functions / integration to their inverse functions

Edit: hold on, I think some of the first part might be useful here, I'll put it in my first post

Yet another edit: Never mind, there shouldn't be an a there, I've been really dumb... In the first part I differentiated some function sinh-1(x/a), but in this part a=1.... *facepalm self*

Edits for days: right, I've chaged the first post and I'll try doing it without the 'a' complicating things now...

Edited by mb12777
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A couple of identities between sinh and cosh? Jolly good. Choose b so that the expression under the square root becomes one side of one of them.

And I suppose you also know the derivatives and integrals of sinh and cosh themselves.

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Yep, I'm just getting on with it now, I've used a substitution of x=(2/√5)sinh(u), and I'm hoping it'll integrate to an inverse hyperbolic function, which is where the ln will come from?

Edit: One of my new limits is sinh-1(2/√5), I hope I'm doing this right, haha...

I like edits: Successful integration! :D Now it's just plugging the values of my limits in and tidying up, thanks for the help

Edited by mb12777
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Sorry to keep bothering you, one last question!

So I've got 5/4 [0.5sinh(2u) + u], that's with a lower limit of 0 and an upper limit of sinh-1(2/√5). I get that sinh(sinh-1(2/√5)) would be 2/√5, but it's sinh(2u) instead of sinh(u). Now, when I put this all into my calculator for 2u, I got an answer of exactly 2.4.... Spooky, right?

So, is there some identity to do with the inverse hyperbolic functions that I'm missing out on here, or is it just pure coincidence?

Edit: wait, never mind, I'll try and derive something from the logarithmic form of sinh-1(x)

No more edits: Done, thanks again for the help

Edited by mb12777
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