- #1

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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)

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- Thread starter mitch_jacky
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- #1

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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)

- #2

Mark44

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Use that to get 1 + (y')

- #3

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Ah, thanks a ton!

- #4

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Nope, still got nothing. I still can't figure it out.

- #5

Mark44

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What do you have for 1 + (y')^{2}?

- #6

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Thanks, I figured it out a little while afterwards and lost my internet connection.

- #7

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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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- #11

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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

Mark44

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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')

See if you can finish this.

1 + (y')

Replace 1 with (e

- #13

Mark44

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You don't want to do this.

- #14

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ok, so I redid it and now have sqrt(1+4e

sqrt(1+4e

- #15

Char. Limit

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- #16

Mark44

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You can cancel only when an expression appears as aI 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?

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.ok, so I redid it and now have sqrt(1+4e^{2x}).

You should have

1 + (y')

= (e

If you expand the e

after substituting then adding, the (e^{2x}-1)^{2}cancels out on both top and bottom leaving ones right? so then the root would be something around 2e^{x}, 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(4e^{2x})...

sqrt(1+4e^{2x})

- #17

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- #18

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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

Mark44

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There is no factor of e^(2x) - 1 in the numerator, so how can you cancel it from numerator and denominator?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..

- #20

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Oh I meant (e^2x + 1) from

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

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

- #21

Mark44

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- #22

Char. Limit

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I already apologized for that...

- #23

Mark44

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I wasn't trying to cast blame. I just wanted to warn this poster about a wrong turn.I already apologized for that...

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