# Arc length question

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#1
Final question part (ii) don't know what I'm trying to find...

https://imgur.com/a/xeKtw
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1 year ago
#2
(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

1
#3
(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

I get this and I know this part is wrong

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#4
It didn't work the thetea and pi/2 method.
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1 year ago
#5
Okay, a few things:

1) where a < b.

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

3) Derive the function s. You are also told that .
You should be able to find the function y in terms of s.

Is that okay. :-)
2
1 year ago
#6
Also to add, regarding your working (you attached to in your first post), the main method is fine, except for the multiple .

So your formula for s is:

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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#7
(Original post by simon0)
Okay, a few things:

1) where a < b.

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

3) Derive the function s. You are also told that .
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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1 year ago
#8

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1 year ago
#9
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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#10
(Original post by simon0)

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

\frac{1}{e^s}

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

Am I correct?
0
1 year ago
#11
(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

\frac{1}{e^s}

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

Am I correct?
Generally yes.
0
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