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Reply 1
joe_the_smoe
Hi,

Can someone help me on this question I really can't do it. Thanks.

+ 3XY² - = 3

Find the co-ordinates of the turning points.

I tried differentiating it and I got

3Y²(dy/dx) + 6XY(dy/dx) + 3Y² -3X² = 0

Then factorised out dy/dx so it came to

dy/dx(3Y² + 6XY) = 3X² - 3y²

Then made dy/dx the subject:

dy/dx = [3(X²-Y²)]/[3(y² + 2XY) =

(X² - Y²)/(y² + 2XY) = 0

Which just leaves you with = :eek:

Someone please help me!!!

You have x=+-y.
You can substitute these, in turn, into the original equation to find the values of x and y.
Reply 2
Gaz031
You have x=+-y.
You can substitute these, in turn, into the original equation to find the values of x and y.


Ah me should have thought of that. Thanks I should be able to do it now.
Is this off an AEA paper? I remember doing something like this and it caught me out, too.
Reply 4
ljfrugn
Is this off an AEA paper? I remember doing something like this and it caught me out, too.

I'm not sure whether this is from an AEA paper, but if you're doing AEA, have you looked at the book of STEP questions by Dr Silkos?
He goes through questions in a quite useful way, sharing his thoughts. One question teaches you how to do things like ππsinx+cosx\int_{-\pi}^{\pi} \left| \sin x + \cos x \right|
Gaz031
I'm not sure whether this is from an AEA paper, but if you're doing AEA, have you looked at the book of STEP questions by Dr Silkos?
He goes through questions in a quite useful way, sharing his thoughts. One question teaches you how to do things like ππsinx+cosx\int_{-\pi}^{\pi} \left| \sin x + \cos x \right|

Yeah, I've had a glance at the book, but it seems a bit too advanced for me. I don't really intend to start having a look at some STEP stuff until the exams are over and I have a bit of spare time. However, I do like the way Dr Siklos goes through the way he would approach the question - it gives me an idea of the way I should think about problems.

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