West Coast Connection Forum

Lifestyle => Train of Thought => Topic started by: coola on April 28, 2006, 07:07:50 AM

Title: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 07:07:50 AM
i need help in a differentiation question... if you could show working, that'd be superb... cheers.

1.)   (2y - x)^4 + x^2 = y + 3

2.)  2(x^2 + 1)^3 + (y^2 + 1)^2 = 17
Title: Re: is anyone on this board competent in calculus ?
Post by: Kassem on April 28, 2006, 07:15:57 AM
wat does the whole q say
Title: Re: is anyone on this board competent in calculus ?
Post by: Digital Pimpin' on April 28, 2006, 07:25:31 AM
Damn you're gonna have to use implicit differentiation for this. I would help out but it's been years since I did this, and I would need the weekend to school up.

The only hint I can give you is that both answers will be expressed in terms of x and y, not just x as regular differentiation usually gives you.

Good luck.
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 07:36:11 AM
this is the question: find dy/dx by differentiating implicitly. when applicable, express the result in terms of x and y ...

i just goto find dy/dx dont need to solve for x and y on this one...

dammet ! the problem is i never did calculus in high school, i'm learning all this shit myself from the text book... so i have a few gaps in fundamental rules of calculus..  >:(
Title: Re: is anyone on this board competent in calculus ?
Post by: Kassem on April 28, 2006, 08:25:24 AM
1st one   

4(2y-x)^3 - 2x / 8(2y-x)^3 - 1 = dy/dx
Title: Re: is anyone on this board competent in calculus ?
Post by: nibs on April 28, 2006, 08:41:06 AM
i need help in a differentiation question... if you could show working, that'd be superb... cheers.

1.)   (2y - x)^4 + x^2 = y + 3

2.)  2(x^2 + 1)^3 + (y^2 + 1)^2 = 17


1. (1 - 8(2y -x)^3)^-1 * (2x - (4(2y - x)^3)) = dy/dx

2. 12x(x^2 + 1)^2  / (-4y(y^2 + 1)) = dy/dx
Title: Re: is anyone on this board competent in calculus ?
Post by: MidoriHaze on April 28, 2006, 08:42:17 AM
Damn, questions like these were the reason why i stuck to business maths at school


What field is this applicable to, overall ?
Title: Re: is anyone on this board competent in calculus ?
Post by: nibs on April 28, 2006, 08:53:47 AM
steps
1.)   (2y - x)^4 + x^2 = y + 3

d((2y - x)^4) / dx = 4(2y - x)^3((d((2y - x))/dx = 4(2y - x)^3(2y * dy/dx - 1)

8y(2y - x)^3dy/dx - 4(2y - x)^3 + 2x = dy/dx

 - 4(2y - x)^3 + 2x = dy/dx - 8y(2y - x)^3dy/dx

2x - 4(2y - x)^3  = dy/dx - 8y(2y - x)^3dy/dx

2x - 4(2y - x)^3  = dy/dx(1 - 8y(2y - x)^3)

(2x - 4(2y - x)^3) / (1 - 8y(2y - x)^3)   = dy/dx


2.)  2(x^2 + 1)^3 + (y^2 + 1)^2 = 17

12x(x^2 + 1)^2 + 4y(y^2 + 1)dy/dx = 0

12x(x^2 + 1)^2 = -4y(y^2 + 1)dy/dx

12x(x^2 + 1)^2 / -4y(y^2 + 1) = dy/dx

3x(x^2 + 1)^2 / -y(y^2 + 1) = dy/dx


it looks like my answer agrees with za scarab's, so that is good.
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 09:37:10 AM
thanks for that nibs and scarab.. i'm just going over the workings now, you got alot closer than i did...

the solution is:

[4(2y-x)^3 - 2x] / [8(2y-x)^3 - 1]

and nibs got the second solution right... props on that.


the one i'm finding real tricky is this one too:

(3x^2) / ((y^2)+1) + y = 3x + 1
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 09:41:40 AM
i get the second one !!!  ;D

the only area i fucked up is  -- 12x(x^2 + 1)^2 = -4y(y^2 + 1)dy/dx -- i dont know why, but i fucked up here.. stupid mistake dammet. instead of isolating the dy/dx, i multiplied one of the brackets out, thats what i mean i have gaps even in algebra and BIMDAS...
Title: Re: is anyone on this board competent in calculus ?
Post by: Kassem on April 28, 2006, 10:01:41 AM
is that for an assingment or a final. shit i got calculus for engineering final on the 4th and am getting this shit wrong
Title: Re: is anyone on this board competent in calculus ?
Post by: nibs on April 28, 2006, 10:07:32 AM
is that for an assingment or a final. shit i got calculus for engineering final on the 4th and am getting this shit wrong

your answer wasn't "wrong", you simply omitted the parenthesis  to indicate the correct terms.

your answer:
(4(2y-x)^3 - 2x) / (8(2y-x)^3 - 1) = dy/dx

i had a -1 / -1 factor in mine, so it looked different but also the same

Title: Re: is anyone on this board competent in calculus ?
Post by: Kassem on April 28, 2006, 10:11:13 AM
cool i didn't check it
Title: Re: is anyone on this board competent in calculus ?
Post by: Infinite Trapped in 1996 on April 28, 2006, 10:29:19 AM
thanks for that nibs and scarab..

Ask my Muslim brothers nibs and scarab (Kassem Fawzy), they'll take care of you.  The world has the Muslims to thank for Algebra (invented by Al-Jabeer), Arabic numerals, and the number 0.
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 10:36:30 AM
is that for an assingment or a final. shit i got calculus for engineering final on the 4th and am getting this shit wrong

nah man, these are just exercises we have to do each week.. yeah i'm getting alot of it wrong too  ;D

what studies have you done nibs ?
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 10:38:03 AM
thanks for that nibs and scarab..

Ask my Muslim brothers nibs and scarab (Kassem Fawzy), they'll take care of you.  The world has the Muslims to thank for Algebra (invented by Al-Jabeer), Arabic numerals, and the number 0.

yep i appreciate their help... and we have alot of cultures to thank for what the world is today...
Title: Re: is anyone on this board competent in calculus ?
Post by: E. J. Rizo on April 28, 2006, 10:38:59 AM
 ???
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 10:40:32 AM
try this one !!

(3x^2) / ((y^2)+1) + y = 3x + 1
Title: Re: is anyone on this board competent in calculus ?
Post by: nibs on April 28, 2006, 11:01:35 AM
try this one !!

(3x^2) / ((y^2)+1) + y = 3x + 1

3x^2((y^2)+1)^-1 + y = 3x + 1

6x((y^2)+1)^-1 - 3x^2((y^2)+1)^-2(2ydy/dx) + dy/dx = 3

6x((y^2)+1)^-1 - 6x^2(ydy/dx)((y^2)+1)^-2 + dy/dx = 3

6x((y^2)+1)^-1 - (6yx^2((y^2)+1)^-2 + 1)dy/dx = 3

6x((y^2)+1)^-1 - 3 = (6yx^2((y^2)+1)^-2 + 1)dy/dx

(6x / ((y^2) + 1) - 3) / ((6yx^2/((y^2)+1)^2) + 1)  =  dy/dx

it's been re-corrected.  dy/dx didn't have the '3' factor to factor out.

Title: Re: is anyone on this board competent in calculus ?
Post by: nibs on April 28, 2006, 11:22:33 AM
!
Title: Re: is anyone on this board competent in calculus ?
Post by: K A I N on April 28, 2006, 11:23:23 AM
Calculus II will kill you if you having problems with this... ;D

It's scary how I forgot all this shit  :-\

http://www.mathematicshelpcentral.com/lecture_notes/calculus_1.htm


btw fuck geometry
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 11:40:20 AM
yeah nibs, i was wondering wtf you changed the 3 to a 1 before... ima go eat some soup then put down on paper your workings... the solution is different in the text  :-\
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 11:41:54 AM
Calculus II will kill you if you having problems with this... ;D

It's scary how I forgot all this shit  :-\

http://www.mathematicshelpcentral.com/lecture_notes/calculus_1.htm


btw fuck geometry

yeah this is only my first year... its basic stuff and i'm struggling  >:( might need a tutor next year.. hot blonde will do  ;D

props for the web-site !
Title: Re: is anyone on this board competent in calculus ?
Post by: coola on April 28, 2006, 12:08:49 PM
heres the solution... i'm hittin the sack, too much for one night..

[ 3((y^2) + 1)((y^2) - 2x + 1) ] / [ ((y^2) + 1)^2 - 6(x^2)y ]

fuck knows hot to get that  :'(
Title: Re: is anyone on this board competent in calculus ?
Post by: nibs on April 28, 2006, 12:50:07 PM
i factored wrong:

3x^2((y^2)+1)^-1 + y = 3x + 1

6x((y^2)+1)^-1 - 3x^2((y^2)+1)^-2(2ydy/dx) + dy/dx = 3

6x((y^2)+1)^-1 - 6x^2(ydy/dx)((y^2)+1)^-2 + dy/dx = 3

6x((y^2)+1)^-1 + (1 - 6yx^2((y^2)+1)^-2)dy/dx = 3

6x((y^2)+1)^-1 - 3 = (6yx^2((y^2)+1)^-2 - 1)dy/dx

(6x / ((y^2) + 1) - 3) / (6yx^2((y^2)+1)^-2 - 1)  =  dy/dx

then they simplified this further:

(6x / ((y^2) + 1) - 3) / (6yx^2((y^2)+1)^-2 - 1)  =  dy/dx

(6x((y^2)+1)^2 / ((y^2) + 1) - 3((y^2)+1)^2) / ( 6yx^2 - ((y^2)+1)^2)  =  dy/dx

(6x((y^2)+1) - 3((y^2)+1)^2) / ( 6yx^2 - ((y^2)+1)^2)  =  dy/dx

(6x((y^2)+1) - 3((y^2)+1)((y^2)+1)) / ( 6yx^2 - ((y^2)+1)^2)  =  dy/dx

[ 3((y^2)+1)(2x - (y^2) + 1) ] / [  6yx^2 - ((y^2)+1)^2 ]  =  dy/dx

factoring out the minus sign

[ 3((y^2)+1)((y^2) - 2x + 1) ] / [ ((y^2)+1)^2 - 6y(x^2) ]  =  dy/dx


Title: Re: is anyone on this board competent in calculus ?
Post by: Doggystylin on April 28, 2006, 01:33:47 PM
thanks for that nibs and scarab..

Ask my Muslim brothers nibs and scarab (Kassem Fawzy), they'll take care of you.  The world has the Muslims to thank for Algebra (invented by Al-Jabeer), Arabic numerals, and the number 0.

well fuck him then cause i hate algebra
Title: Re: is anyone on this board competent in calculus ?
Post by: Shallow on April 28, 2006, 08:33:54 PM
thanks for that nibs and scarab..

Ask my Muslim brothers nibs and scarab (Kassem Fawzy), they'll take care of you.  The world has the Muslims to thank for Algebra (invented by Al-Jabeer), Arabic numerals, and the number 0.

well fuck him then cause i hate algebra

He was just joking around. "Al-Jabeer", "Al-ge-bra". While the name algebra is devired from the arabic word "al-ğabr" or "al jebr", which I'm told stands for "reunion of broken parts", the idea of agebra as we know it goes as far back as the Babylonian times at least 4000 years ago. 2400 years before anyone was calling themselves a Muslim. Now Calculus as we know it today came about from Newton and his colleagues in Europe about 300 years ago.
Title: Re: is anyone on this board competent in calculus ?
Post by: Sikotic™ on April 29, 2006, 01:10:00 AM
I just failed my calculus test today.  ;D
Title: Re: is anyone on this board competent in calculus ?
Post by: herpes on April 29, 2006, 06:47:43 AM
I can count to 12