Author Topic: is anyone on this board competent in calculus ?  (Read 678 times)

coola

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is anyone on this board competent in calculus ?
« 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
 

Kassem

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Re: is anyone on this board competent in calculus ?
« Reply #1 on: April 28, 2006, 07:15:57 AM »
wat does the whole q say
« Last Edit: April 28, 2006, 07:18:59 AM by Za ScArab »
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Re: is anyone on this board competent in calculus ?
« Reply #2 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.
 

coola

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Re: is anyone on this board competent in calculus ?
« Reply #3 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..  >:(
 

Kassem

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Re: is anyone on this board competent in calculus ?
« Reply #4 on: April 28, 2006, 08:25:24 AM »
1st one   

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

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Re: is anyone on this board competent in calculus ?
« Reply #5 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
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MidoriHaze

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Re: is anyone on this board competent in calculus ?
« Reply #6 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 ?
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nibs

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Re: is anyone on this board competent in calculus ?
« Reply #7 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.
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coola

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Re: is anyone on this board competent in calculus ?
« Reply #8 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
 

coola

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Re: is anyone on this board competent in calculus ?
« Reply #9 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...
 

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Re: is anyone on this board competent in calculus ?
« Reply #10 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
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Re: is anyone on this board competent in calculus ?
« Reply #11 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

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Kassem

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Re: is anyone on this board competent in calculus ?
« Reply #12 on: April 28, 2006, 10:11:13 AM »
cool i didn't check it
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Re: is anyone on this board competent in calculus ?
« Reply #13 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.
« Last Edit: April 28, 2006, 10:31:17 AM by Hajj Abdul Infinite Ibrahim Islam Abu Muhammad Uthman »
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coola

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Re: is anyone on this board competent in calculus ?
« Reply #14 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 ?
 

coola

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Re: is anyone on this board competent in calculus ?
« Reply #15 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...
 

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Re: is anyone on this board competent in calculus ?
« Reply #16 on: April 28, 2006, 10:38:59 AM »
 ???
 

coola

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Re: is anyone on this board competent in calculus ?
« Reply #17 on: April 28, 2006, 10:40:32 AM »
try this one !!

(3x^2) / ((y^2)+1) + y = 3x + 1
 

nibs

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Re: is anyone on this board competent in calculus ?
« Reply #18 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.

« Last Edit: April 28, 2006, 11:22:19 AM by nibs »
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nibs

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Re: is anyone on this board competent in calculus ?
« Reply #19 on: April 28, 2006, 11:22:33 AM »
!
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K A I N

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Re: is anyone on this board competent in calculus ?
« Reply #20 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
« Last Edit: April 28, 2006, 11:25:17 AM by Doggie KAIN »
 

coola

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Re: is anyone on this board competent in calculus ?
« Reply #21 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  :-\
 

coola

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Re: is anyone on this board competent in calculus ?
« Reply #22 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 !
 

coola

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Re: is anyone on this board competent in calculus ?
« Reply #23 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  :'(
 

nibs

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Re: is anyone on this board competent in calculus ?
« Reply #24 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


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