Arc Length

[jrs7892]

Hi. I'm having trouble completing this homework problem.. any help would be greatly appreciated!

Find the length of the parabolic segment r =2/(1-cos?) , ?/2 ? ? ? ? .

I know I need to calculate r[sup:29elfnaf]2[/sup:29elfnaf], and (dr/d?)[sup:29elfnaf]2[/sup:29elfnaf], but once I do that I'm at a loss for how to do the integration.

r[sup:29elfnaf]2[/sup:29elfnaf]= 4/((1-cos?)[sup:29elfnaf]2[/sup:29elfnaf]); (dr/d?)[sup:29elfnaf]2[/sup:29elfnaf]= 4sin[sup:29elfnaf]2[/sup:29elfnaf]?/(1-cos?)[sup:29elfnaf]4[/sup:29elfnaf]).

Thanks in advance for your assistance!!

 

[jrs7892]

I guess I forgot to mention that the formula for arc length we're using is (forgive me.. I don't know how to use the integration symbol on here):

Integration from ? to ?/2, ? r[sup:xyrn38za]2[/sup:xyrn38za] + (dr/d?)2.. this is where I'm stuck now. I have the stuff under the radical, just don't know how to integrate it properly.

Thanks again!

 

[Deleted member 4993]

I guess I forgot to mention that the formula for arc length we're using is (forgive me.. I don't know how to use the integration symbol on here):

Integration from ? to ?/2, ? r[sup:2z27wddy]2[/sup:2z27wddy] + (dr/d?)2.. this is where I'm stuck now. I have the stuff under the radical, just don't know how to integrate it properly.

Thanks again!

My previous post was incorrect - I deleted it.

 

[BigGlenntheHeavy]

$\displaystyle r \ = \ \frac{2}{1-cos(\theta)}, \ \frac{\pi}{2} \ \le \ \theta \ \le \ \pi, \ find \ arc \ length.$

$\displaystyle Hence, \ s \ = \ \int_{\alpha}^{\beta}\sqrt{[f(\theta)]^2+[f'(\theta)]^2}d\theta.$

$\displaystyle Now, \ since \ f(\theta) \ = \ r, \ [f(\theta)]^2 \ = \ \frac{4}{[1-cos(\theta)]^2}, \ and$

$\displaystyle f'(\theta) \ = \ \frac{-2sin(\theta)}{[1-cos(\theta)]^2}, \ [f'(\theta)]^2 \ = \ \frac{4sin^2(\theta)}{[1-cos(\theta)]^4}$

$\displaystyle Ergo, \ we \ have \ s_{arc \ length} \ = \ \int_{\pi/2}^{\pi}\sqrt{\frac{4}{[1-cos(\theta)]^2}+\frac{4sin^2(\theta)}{[1-cos(\theta)]^4}}d\theta$

$\displaystyle Now, \ can \ you \ take \ it \ from \ here?, \ have \ fun.$

 

[jrs7892]

Actually, THIS is the point I'm stuck at. I included the stuff from under the square root in my first post.. What I need help with is the integration.

Thanks!

 

[galactus]

This is rough to integrate.

It whittles down to $\displaystyle 2^{\frac{3}{2}}\int_{\frac{\pi}{2}}^{\pi}\sqrt{\frac{1}{(1-cos{\theta})^{3}}}d{\theta}$

which is a booger. I even tried changing it to rectangular, but it still turned out to be difficult to integrate.

Use some sort of tech to get the solution. Afterall, setting up the problem is what takes the human brain. Let some sort of software do the grunt work.

For a check, I ran it through my calculator in an indefinite form, and it returned nothing. That means it has no antiderivative that can be obtained by elementary means.

You can use some sort of numerical method, such as Simpson's Rule to approximate an answer.

If someone sees a slick way of going about it, I would like to see it.

 

[jrs7892]

I agree that it's a booger, but this has to done by hand. We haven't used Simpson's rule or any tech methods so all I have available is Trig substitution, integration by parts etc. Will have to see if the tutoring center at school can help.

 

[Deleted member 4993]

This is rough to integrate.

It whittles down to $\displaystyle 2^{\frac{3}{2}}\int_{\frac{\pi}{2}}^{\pi}\sqrt{\frac{1}{(1-cos{\theta})^{3}}}d{\theta}$

which is a booger. I even tried changing it to rectangular, but it still turned out to be difficult to integrate.

Use some sort of tech to get the solution. Afterall, setting up the problem is what takes the human brain. Let some sort of software do the grunt work.

For a check, I ran it through my calculator in an indefinite form, and it returned nothing. That means it has no antiderivative that can be obtained by elementary means.

You can use some sort of numerical method, such as Simpson's Rule to approximate an answer.

If someone sees a slick way of going about it, I would like to see it.

Actually it simplifies a little if you substitute:

$\displaystyle 2\phi \ = \ \theta$

then you'll be left with:

$\displaystyle \int_{\pi}^{2\pi}cosec^3(\phi) d\phi$

That one is a fairly standard integration.

 

[BigGlenntheHeavy]

$\displaystyle 2\sqrt2\int_{\pi/2}^{\pi}\frac{d\theta}{[1-cos(\theta)]^{3/2}} \ \dot= \ 2.296 \ (trusty \ TI-89).$

 

[BigGlenntheHeavy]

$\displaystyle If \ you \ let \ \theta \ = \ 2\phi, \ then \ you \ get: \ 4\sqrt2\int_{\pi/4}^{\pi/2}\frac{d\phi}{[1-cos(2\phi)]^{3/2}}, \ not \ \int_{\pi}^{2\pi}csc^3(\phi)d\phi$

 

[galactus]

Yep, SK, I should have seen that. I reckon I did not look at it enough. Cool

Duh, I did not even try another sub.

 

[BigGlenntheHeavy]

$\displaystyle galactus, \ show \ me \ if \ you \ let \ \theta \ = \ 2\phi, \ then:$

$\displaystyle 2\sqrt2\int_{\pi/2}^{\pi}\frac{d\theta}{[1-cos(\theta)]^{3/2}} \ = \ \int_{\pi}^{2\pi}csc^3(\phi)d\phi.$

$\displaystyle Another \ thing: \ What \ are \ you \ doing \ sucking \ up \ to \ Subhotosh \ Khan?, \ he \ is \ all \ wet.$

 

[galactus]

Making the sub that SK suggested, by letting $\displaystyle {\theta}=2\phi$, leads to:

$\displaystyle \frac{1}{(1-cos(2\phi))^{3}}$

Which is equal to $\displaystyle \frac{1}{8}csc^{6}(\phi)$

Then, taking the square root gives:

$\displaystyle \frac{\sqrt{2}}{4}csc^{3}(\phi)$

Of course, the $\displaystyle d\theta$ must be incorporated into the integral, but that is the identity.

SK is a very learned fellow. No sucking up. Just giving credit where crdit is due.

 

[BigGlenntheHeavy]

$\displaystyle Hey, \ \int_{\pi}^{2\pi}csc^3(\phi)d\phi \ = \ \ -\infty, \ however \ 2\int_{\pi/4}^{\pi/2}csc^3(\phi)d\phi \ \dot= \ 2.296.$

 

[galactus]

The limits of integration should be $\displaystyle \frac{\pi}{4} \;\ to \;\ \frac{\pi}{2}$

Because $\displaystyle \pi=2\phi\Rightarrow \phi = \frac{\pi}{2}$

$\displaystyle \frac{\pi}{2}=2\phi\Rightarrow \phi=\frac{\pi}{4}$

$\displaystyle 2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}csc^{3}(\phi)d\phi$

gives the correct arc length.

 

[BigGlenntheHeavy]

$\displaystyle OK \ galactus, \ I \ stand \ corrected \ (a \ little) \ on \ this \ one \ as \ Subhotosh \ Khan \ had \ his \ limits$

$\displaystyle of \ integration \ wrong.$

 

[BigGlenntheHeavy]

$\displaystyle OK, \ after \ all \ is \ said \ and \ done, \ we \ have: \ 2\sqrt2\int_{\pi/2}^{\pi}\frac{d\theta}{[1-cos(\theta)]^{3/2}}$

$\displaystyle Now, \ stealing \ Subhotosh \ Khan's \ thunder, \ let \ \theta \ = \ 2\phi.$

$\displaystyle This \ gives \ 4\sqrt2\int_{\pi/4}^{\pi/2}\frac{d\phi}{[1-cos(2\phi)]^{3/2}}$

$\displaystyle Identity: \ cos(2\phi) \ = \ 1-2sin^2(\phi).$

$\displaystyle Hence, \ 4\sqrt2\int_{\pi/4}^{\pi/2}\frac{d\phi}{[1-cos(2\phi)]^{3/2}} \ = \ 2\int_{\pi/4}^{\pi/2}csc^3(\phi)d\phi$

$\displaystyle = \ -csc(\phi)cot(\phi)+ln|csc(\phi)-cot(\phi)|\bigg]_{\pi/4}^{\pi/2} \ \dot= \ 2.296.$

$\displaystyle Note: \ Subhotosh \ Khan, \ I \ under-estimated \ you.$

 

[jrs7892]

Sure wish I would've seen all of this before class today. No one at the tutoring center could come up with the answer.. sigh.

Hopefully this isn't a stupid question, but I see everyone throwing around the ? symbol. Does that stand for something specifically, or is it being used just for substitution as you would use the letter u, etc? I've never seen this before, so it threw me off a bit.

Thanks!

 

[BigGlenntheHeavy]

Hopefully this isn't a stupid question, it is.

$\displaystyle Check \ out \ the \ koine \ alpha-beta.$

 

[galactus]

$\displaystyle {\phi}$ (phi) was just arbitrarily chosen as a substitution. It could have been anything else, such as a u, t, gamma, etc.

The Greek letters are often used for subsitutions and a lot of other things in the math world.

$\displaystyle \alpha \beta \gamma \delta \epsilon \zeta \eta \theta \iota \kappa \lambda \mu \nu \xi o \pi \rho \sigma \tau \upsilon \phi \chi \psi \omega$

For some reason,omicron would not display...go figure.