The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Arc length question

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

https://imgur.com/a/xeKtw
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

Reply 2
Avatar for ckfeister
ckfeister
OP
19
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


I get this and I know this part is wrong


20180415_181805-compressed.jpg.jpeg
Reply 3
Avatar for ckfeister
ckfeister
OP
19
It didn't work the thetea and pi/2 method.
(edited 6 years ago)
Reply 4
Avatar for simon0
simon0
13
Okay, a few things:

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

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

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

Is that okay. :-)
(edited 6 years ago)
Reply 5
Avatar for simon0
simon0
13
Also to add, regarding your working (you attached to in your first post), the main method is fine, except for the multiple 2π 2 \pi .

So your formula for s is:

s=∫t=0t=θtanh2(t) dt. \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.
(edited 6 years ago)
Reply 6
Avatar for ckfeister
ckfeister
OP
19
Original post by simon0
Okay, a few things:

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

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

3) Derive the function s. You are also told that y=sech(θ) 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
Reply 7
Avatar for simon0
simon0
13
Glad to hear!

Did you obtain the answer?
Reply 8
Avatar for akaash13
akaash13
6
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
Reply 9
Avatar for ckfeister
ckfeister
OP
19
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(θ))[br][br]es=cosh(θ)[br] s = ln(cosh(\theta))[br][br] e^s = cosh(\theta) [br]

\frac{1}{e^s}

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

Unparseable latex formula:

e^-^s = y

Am I correct?
(edited 6 years ago)
Reply 10
Avatar for simon0
simon0
13
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(θ))[br][br]es=cosh(θ)[br] s = ln(cosh(\theta))[br][br] e^s = cosh(\theta) [br]

\frac{1}{e^s}

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

Unparseable latex formula:

e^-^s = y

Am I correct?


Generally yes.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you