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Posted by vswede on Sep-15-2005 17:35:

math help

well i need some help with my math worksheet and i am so lost. taking ap calc sucks...

took some time drawing that so pleeeeasee help me
thanks

Posted by UWM on Sep-15-2005 17:40:

Here's a Taylor series I had to derive for my last problem set:

P = (RT/V1) [([2]V1+ ..

Blah fuck it it's way too long to write out.

Posted by DJ Mikey Mike on Sep-15-2005 17:45:

If you really need help with something as straight forward as that then perhaps you've chosen the wrong subject to study.

Posted by igottaknow on Sep-15-2005 17:48:

cant help but here's something to make you laugh

edit: whats the world coming to when you cant even steal a crummy b/w mspaint gif?

Posted by jonze on Sep-15-2005 17:56:

quote:
Originally posted by igottaknow
cant help but here's something to make you laugh

try hosting the pic yourself...its pretty funny though.

Posted by vswede on Sep-15-2005 18:01:

quote:
Originally posted by DJ Mikey Mike
If you really need help with something as straight forward as that then perhaps you've chosen the wrong subject to study.

OY! your like 4 years older than me
just say how u do it dont be all pessimistic

Posted by Mane on Sep-15-2005 18:25:

ok if i coun good the resul should be 3/5

Posted by RenderedDream on Sep-15-2005 18:29:

just use the cauchy's theorem
derive upper and down parts separately, eventually you'll have a real limit

Posted by Neuro'99 on Sep-15-2005 18:29:

quote:
Originally posted by Mane
ok if i coun good the resul should be 3/5

Yeah I got the same, its a few years since I did my calculus course however :P

Use L'Hopital's rule if you know what that is heh

Posted by UWM on Sep-15-2005 18:31:

3x/(2x^2+5x) = 3x/2x^2 + 3x/5x = 3/x + 3/5

lim x-->0 (3/2x + 3/5)

Ding.

Posted by Orbax on Sep-15-2005 18:35:

Posted by UWM on Sep-15-2005 18:38:

Your shitty stick figure comics always crack me up.

Posted by Mebot on Sep-15-2005 18:38:

haha

Posted by colonelcrisp on Sep-15-2005 18:39:

now bear in mind i havent taken calc for over 4 years but from what i remember, using l'hopital's rule, (take derrivative of top and bottom)
your limit approaching zero is aprox 3/5 (three fifths)

Posted by TheDarkOne on Sep-15-2005 19:01:

ye, it's 3/5. I'll explain to you tomorrow at school.

Posted by RenderedDream on Sep-15-2005 22:38:

quote:
Originally posted by colonelcrisp
now bear in mind i havent taken calc for over 4 years but from what i remember, using l'hopital's rule, (take derrivative of top and bottom)
your limit approaching zero is aprox 3/5 (three fifths)

it's cauchy's =P
i think...

Posted by noikeee on Sep-15-2005 22:40:

quote:
Originally posted by RenderedDream
it's cauchy's =P
i think...

wtf, you're still alive and posting?

Posted by Cloudburst on Sep-15-2005 23:32:

Heh, calculus... Just sold my book to a younger student...

Posted by OrZonE on Sep-16-2005 01:49:

quote:
Originally posted by RenderedDream
it's cauchy's =P
i think...

L'Hospital

Posted by _Nut_ on Sep-16-2005 02:04:

quote:
Originally posted by UWM
3x/(2x^2+5x) = 3x/2x^2 + 3x/5x = 3/x + 3/5

lim x-->0 (3/2x + 3/5)

Ding.

Your epiphany makes perfect sense now. Or was the the Taylor series derivation you had earlier?

Posted by Minhaj on Sep-16-2005 02:28:

2+2 is four

Posted by Vlad on Sep-16-2005 02:44:

Posted by sandstorm03 on Sep-16-2005 04:14:

d (tan(3x))
/dt = 3/(cos3x)^2

d(2x^2 +5x)
/dt = 4x + 5

3/(cas(3x)^2)+4x+5) x-> 0 cos(3x)^2+4x->0

3/5

Posted by Yan on Sep-16-2005 04:19:

quote:
Originally posted by sandstorm03
d (tan(3x))
/dt = 3/(cos3x)^2

d(2x^2 +5x)
/dt = 4x + 5

3/(cas(3x)^2)+4x+5) x-> 0 cos(3x)^2+4x->0

3/5

Errr... Wha?

Posted by RenderedDream on Sep-16-2005 12:13:

quote:
Originally posted by paranoik0
wtf, you're still alive and posting?

voltei agr das ferias =D
2 meses na suica a trabalhar..