ckfeister
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Final question part (ii) don't know what I'm trying to find...

https://imgur.com/a/xeKtw
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mupsman2312
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(Original post by ckfeister)
Final question part (ii) don't know what I'm trying to find...

https://imgur.com/a/xeKtw
The arc length formula does not involve 2pi - I think that you're getting confused with the surface-area of revolution formula.

Your limits should be pi/2 and theta (lower and upper limits respectively) - remember that, at (0, 1), the angle of the line segment from the origin to this point to the initial line is pi/2.

This is what I would do (don't take my word for it - I haven't worked through the problem yet!):

Spoiler:
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Set \displaystyle \int_{pi/2}^{theta} (sqrt{(tanh^2(theta)})d(theta) = s

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ckfeister
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(Original post by mupsman2312)
The arc length formula does not involve 2pi - I think that you're getting confused with the surface-area of revolution formula.

Your limits should be pi/2 and theta (lower and upper limits respectively) - remember that, at (0, 1), the angle of the line segment from the origin to this point to the initial line is pi/2.

This is what I would do (don't take my word for it - I haven't worked through the problem yet!):

Spoiler:
Show


Set \displaystyle \int_{pi/2}^{theta} (sqrt{(tanh^2(theta)})d(theta) = s

I get this and I know this part is wrong


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ckfeister
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It didn't work the thetea and pi/2 method.
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simon0
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Okay, a few things:

1)  \displaystyle \int^{b}_{a} f(x) \, dx = F(b) - F(a) , where a < b.

2) The formula for s should have  \theta as the variable.

3) Derive the function s. You are also told that  y = \textrm{sech} (\theta) .
You should be able to find the function y in terms of s.

Is that okay. :-)
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simon0
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Also to add, regarding your working (you attached to in your first post), the main method is fine, except for the multiple  2 \pi .

So your formula for s is:

 \displaystyle s = \int^{ t= \theta }_{t=0} \sqrt{\mathrm{tanh} ^{2} (t)} \, dt .

By tweaking your working using my recomendations (in my previous post in this thread), you should be able obtain the answer.
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ckfeister
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(Original post by simon0)
Okay, a few things:

1)  \displaystyle \int^{b}_{a} f(x) \, dx = F(b) - F(a) , where a < b.

2) The formula for s should have  \theta as the variable.

3) Derive the function s. You are also told that  y = \textrm{sech} (\theta) .
You should be able to find the function y in terms of s.

Is that okay. :-)
Well thats the first for (1) never done that before and thx
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simon0
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Glad to hear!

Did you obtain the answer?
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akaash13
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I think s is the arclength itself but the limits of the integral should be 0 and theta (so you'll have to use t or some other parameter in the integral). Also you substituted in the limits the wrong way round
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ckfeister
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(Original post by simon0)
Glad to hear!

Did you obtain the answer?
I'd be doing it later on as I've just done from 8am-7:30pm studying and I'm tired but I think I get the idea it is

 s = ln(cosh(\theta))



 e^s = cosh(\theta)

\frac{1}{e^s}

\frac{1}{cosh(/theta)}

 e^-^s = y Am I correct?
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simon0
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(Original post by ckfeister)
I'd be doing it later on as I've just done from 8am-7:30pm studying and I'm tired but I think I get the idea it is

 s = ln(cosh(\theta))



 e^s = cosh(\theta)

\frac{1}{e^s}

\frac{1}{cosh(/theta)}

 e^-^s = y Am I correct?
Generally yes.
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