Maths Problem

it's Math, not Maths you damn frenchie

Yeah, it's not like mathematics is a plural word, so why would the short from need the 's'?

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Originally Posted by Rich

it's Math, not Maths you damn frenchie

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Ah, why don't you go somewhere and get drunk?

go somewhere _and_ get drunk?

You buncha geeks.

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Originally Posted by kedawa

Yeah, it's not like mathematics is a plural word, so why would the short from need the 's'?

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math is its own fucking word. It is no longer "short" for anything. Get with the times you damn cheese eaters. It is not the 1700s anymore.

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Originally Posted by johnk_

go somewhere _and_ get drunk?

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Fuck. You.

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Originally Posted by Brisco Bold

Ah, why don't you go somewhere and get drunk?

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I would, but I'm not in College Park.

And it's Math.

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Originally posted by Jetman
Ahh shit - product rule - you're right -

let me see:

3x^2 -3y (dy/dx) -3x + (dy/dx) - 3 = 0

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

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

(x^2 -x -1 ) /y = 0

x^2 - x -y^-1 = dy/dx

then dy/dx = 0 since at a stationary point , then solve for x and y?

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I checked with my teacher today and he agreed with my differentiation, so differentiating definately gives:
3x^2 +(3y^2).(dy/dx) -3y -3x.(dy/dx) = 0
--> x^2 + (y^2).dy/dx - y - x.(dy/dx) = 0

Whenever you differentiate f(y), it goes to f'(y)dy/dx, so differentating y^3 gives 3y^2 dy/dx, and when you differentiate -3xy, it goes to (-3x.dy/dx) + (-3y).

After manipulating the differentiated expression, you get:
dy/dx = (x^2 - y)/(x - y^2) = 0 where x can never be equal to y^2
which gives y = x^2 again.
This suggests that every point on this curve is a turning point, which is a load of crap.

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Originally posted by IronPlant
wasn't there some trick where you could keep taking the deravative of x untill you got a value for x, and then plug it back into the orginal eq or first deravative to find y?

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I've never heard of this, but I haven't seen as much maths as you have either. I tried this out, but I might have screwed up.
If you keep differentiating the x term you get 3! = 6. If you put that into the first derivitive equation, you get y = 6^2 = 36, which does not agree with the first equation.
If you put it into the original expression, then y = -6.59... is the only root (using my calculator to solve that)
And I have no idea what that is telling us, but I'm sure I did that thing wrong.

It's the intersections of y=x^2 with the given curve that give the points where dy/dx=0.

Likewise, dx/dy=0 at the intersections of the given curve with x = y^2.