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is anyone on this board competent in calculus ?

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nibs:

--- Quote from: Chief 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


--- End quote ---

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

MidoriHaze:
Damn, questions like these were the reason why i stuck to business maths at school


What field is this applicable to, overall ?

nibs:
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.

coola:
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:
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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