Recent content by gtabmx

In the case I'm working with, r will be canceled out, obviously except for the r which has come from the product rule when obtaining the conversion of dx and dy. If you would like to see a particular case then I would be more than glad to offer one. (2*y^4 - 9*x^2*y^2-20*x^4)dx - (3*x*y^3)dy = 0

Hi, I was wondering how to go about converting a homogeneous ODE of the form M(x,y)dx+N(x,y)dy=0 (where, by definition of a homogeneous ODE, M(tx,ty)=(t^a)M(x,y) and N(tx,ty)=(t^a)N(x,y) ) to polar coordinates. I wan to do this because using substitution of y/x=u and dy/dx=u+xdu/dx to make the...

SOLVED! I will explain when I get home. Thanks everyone!

yes, but that formula oversimplifies the problem since it ignores the fact that there is no input so the flux rate ot the hole drops as volume drops, and volume drop rate is not consatnt and relies of the past flux rates and how much water was previously lost. Anyway, cyrusabdollahi, don't...

OK honestly, I'm out for tonight, and as close as we are, its too late and I'll do it tomorow, goodnight and thanks so much And sorry I took up your time, I'm sure, trust me, I'm sure your stuff was more important than mine, so thanks, but I just want to tell you, as much as you may think I...

Did you get the picture and my question about what I am solving for? I wake at 6AM everyday... that's in 5hrs, wow

Oh no problem, its just going to take me an hour more, I mean its 1:14AM here but who cares :P

Here's something http://www.yousendit.com/transfer.php?action=download&ufid=852CA7CB14E5C3E2 solve for what, lambda of Volume, i see where you're going but I don't quite understand what you wnat to use that flux rate equation to get?

I'lll show you something...

Well if you look at that formula i had posted as a mess before, I have tons of cross section areas in there, and if i have time maybe I'll take a crack at it. I'm sure cross sections would be in there since area of slot greatly affects flux rates and XSarea of bottle is what defines the amount...

Alright, if its ok with you we will stop here and I will just use this to prove the data's relation to bernouilli's law and that it is in fact quadratic and that will be good enough, because let's face it, this is useless to either of us since its not in my course coverage and its not your lab...

Its ok, but look, where are we going with this, right now we, I, still have have no way of finding lambda unless I do the experiment, so I have no general law.

No unitsssssssss

ohh wait wait wait heres my deal, i made a stupid mistake probably due to the fact that I've been up till 2AM adn woke up at 6 for the past 3 days so I'm sorry also, i get consistent, but NO UNITSyup sextuple checked no units as simplified its m = m * y^2 - m * y + m u even proved it yourself...

You know I have a feeling that's not right look at the final equations, each term has to be in meters after all the steps are carried out now the first temr is Y^2 times gravity times time squared which is m = Y^2 * m*s(-2) * s^2 so then s^(-2) outs s^2 and you have m = Y^2 * m no?