Maths Problem

I have a problem with this question here:

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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.

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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?

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.

Ok, so substituting in gives (x^3+6)(x^3-8) = 0
Giving x = 2, or x = -6^(1/3)
Thank you very much, that's great.

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Originally Posted by Burky

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.


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.

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I'm probably thinking of something else then. It all kind of smooshes togehter by the 4th semester.