Maths Problem

I have a problem with this question here:

Quote:

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Find the stationary points of the curve with equation: x^3 + y^3 -3xy = 48

And determine their nature.

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[^ means to the power of, . means multiplied]
This is only high-school, so it shouldn't be too hard for most of you guys but I can't seem to do it.

Where I've gone so far:

If I differentiate the equation, I get: 3x^2 +(3y^2).(dy/dx) -3y -3x.(dy/dx) = 0
Correct me if I'm wrong on that one.
Now since at stationary points, dy/dx = 0 we can substitute this in giving:
x^2 = y
And that seems to be no help at all.

I'd appreciate any help. I might post a few more questions now and again, as I'm going through a lot of past exam papers right now.

Thanks a lot.

remind me what stationary points are

The turning points on a curve when the gradient = 0.

E.g, y = x^2 +1, stationary point (minimum) is at x=-1

that doesn't really refresh my mind enough. Link me a page with a full worked out problem.

I got a B in calc 1 so I should be able to figure this out for you if you can remind me what you're doing.

Here

I think your differentation is wrong. Shouldnt the derivative of -3xy be :

-3 dy/dx ? or negative 3 times the derivative of y?

I'm fairly sure that's right.
You use explicit differentiation as that is a product, so you differentiate that bit using the product rule:
d(uv)/dx = vu' + uv'
d(-3xy)/dx = -3y - 3(dy/dx)

If you want to test that idea further, look at rectangular parabolas (or maybe hyperbolas, I can't remember which), where xy = c^2.
Differentiating like that works perfectly.

Quote:

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Originally Posted by Burky

I'm fairly sure that's right.
You use explicit differentiation as that is a product, so you differentiate that bit using the product rule:
d(uv)/dx = vu' + uv'
d(-3xy)/dx = -3y - 3(dy/dx)

If you want to test that idea further, look at rectangular parabolas (or maybe hyperbolas, I can't remember which), where xy = c^2.
Differentiating like that works perfectly.

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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?

Where in the hell is Diffx when you need him?

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?

Are we looking for inflection points, or just any extrema?