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#1 2008-08-22 08:26:38

boombastictiger
Member
Registered: 2005-08-28
Posts: 10

Finding and classifying stationary points

Hello

I am trying to work through an example i have and am having difficulty!!

Here is the question

Given the function z= x^4 + 4x²y² - 2x² + 2y² - 1 , find and classify the stationary points.

(please note the ^4 is to the power of 4, there was no symbol to copy from above)


I didnt get far in trying to solve:

z = 4x³ + 8xy² - 4x + 2y²

Update 21:41 - Ok i tried some more - i attemted to simplify...

z=x²+8xy²+2y²

This is where im stuck, i dont know what the next step is or what to do, do i diff it again, and then what sad....

Would be great if someone could solve it and show thier workings out.

Help!!

Thanks Sabir

Last edited by boombastictiger (2008-08-22 08:43:02)

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#2 2008-08-22 09:16:42

boombastictiger
Member
Registered: 2005-08-28
Posts: 10

Re: Finding and classifying stationary points

If it helps the questions is from a past exam paper from Technology Mathematics that i am revising from...

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#3 2008-08-22 09:28:06

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Finding and classifying stationary points

You need to partially differentiate with respect to x and y, each time treating any appearances of the other variable as a constant.

You had the right idea with finding dx/dx, but should have dropped the 2y² because it's entirely a constant and so differentiates to nothing.

dz/dx = 4x³ + 8xy² - 4x.
Equate this to 0 and solve:

4x³ + 8xy² - 4x = 0
So x = 0 is one solution, and the rest are found by solving 4x² + (8y² - 4) = 0.

x² + (2y² - 1) = 0
x = ±√(1 - 2y²), by the quadratic equation.

Note that this will only be real when y is small enough to make the root non-negative.


Now you do a similar kind of thing by finding dz/dy (treating all x's as constants while differentiating), and then setting that to 0. That part should be a bit easier than the bit I did, since y only goes up to quadratic order in the expression for z and so dz/dy should be linear.

Once you have the solution set for that, find all (x,y) pairs that make dz/dx = dz/dy = 0.
Those are your stationary points.


Why did the vector cross the road?
It wanted to be normal.

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#4 2008-08-22 09:54:52

boombastictiger
Member
Registered: 2005-08-28
Posts: 10

Re: Finding and classifying stationary points

So x = 0 is one solution, and the rest are found by solving 4x² + (8y² - 4) = 0.

x² + (2y² - 1) = 0
x = ±√(1 - 2y²), by the quadratic equation.

Thank you very much for the reply.  I am a bit confused at the above, firstly i thought it would have been 4x³+ (8y² - 4) = 0 but you have 4x². 

Also im guessing you solved 4x³+ (8y² - 4) = 0 by dividing everything by 4? ( which i have to substitute back in later to find the nature of the stationary points?) 

Sorry if i seem a but dumb but maths is not my strongest subject heh..

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#5 2008-08-22 10:33:17

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Finding and classifying stationary points

After differentiating, it's 4x³ + (8y² - 4)x.
The first step is to recognise that x=0 is a solution and then look at when x≠ 0.
With that constraint, dividing by x is now allowed and so you can turn it into 4x² + (8y² - 4).

After that I divided by 4, yes.
But you find the nature of the stationary points by differentiating dz/dx and dz/dy again and checking whether they're positive or not, so nothing you do to dz/dx in order to find its roots will affect that.


Why did the vector cross the road?
It wanted to be normal.

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#6 2008-08-24 00:56:46

boombastictiger
Member
Registered: 2005-08-28
Posts: 10

Re: Finding and classifying stationary points

After differentiating, it's 4x³ + (8y² - 4)x.
The first step is to recognise that x=0 is a solution and then look at when x≠ 0.
With that constraint, dividing by x is now allowed and so you can turn it into 4x² + (8y² - 4).

I didnt understand that part.  How did you come to 4x²? I know you said you divided but why did you divide and is it ths standard/only procedure? 

Thanks

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#7 2008-08-24 02:38:14

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Finding and classifying stationary points

Maybe it'd help if you thought of it as factorising instead of dividing.

You start with 4x³ + 8xy² - 4x.
All terms here have a common factor of 4x, which means it can be factorised to give
4x(x² + 2y⊃ - 1).

Since this is equated to 0, then either 4x = 0 (meaning x=0), or x² + (2y²-1) = 0, which is where we got to before.

That's technically not the only way to solve cubic equations (there's a horrendously complicated formula that lets you solve them directly, like with quadratics), but it's certainly the easiest and most common method.


Why did the vector cross the road?
It wanted to be normal.

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#8 2008-08-24 08:40:24

boombastictiger
Member
Registered: 2005-08-28
Posts: 10

Re: Finding and classifying stationary points

mathsyperson wrote:

After differentiating, it's 4x³ + (8y² - 4)x.
The first step is to recognise that x=0 is a solution and then look at when x≠ 0.
With that constraint, dividing by x is now allowed and so you can turn it into 4x² + (8y² - 4).

After that I divided by 4, yes.
But you find the nature of the stationary points by differentiating dz/dx and dz/dy again and checking whether they're positive or not, so nothing you do to dz/dx in order to find its roots will affect that.

I still dont understand what you did or divided to get 4x².  If x = 0 then 4x³ / 0 cannot be done? this is the stage at where im getting confused.  you said that with the constraint dividing by x is allowed, but what value did you use x as to divide by?

I think i get it now did you divide everything by 4x?
as in (4x³/4x) + (8xy²/4x) - (4x/4x) = x² + 2y² - 1 ??

Last edited by boombastictiger (2008-08-24 08:50:09)

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#9 2008-08-24 09:04:42

boombastictiger
Member
Registered: 2005-08-28
Posts: 10

Re: Finding and classifying stationary points

ok i had a go at dz/dy:

dz/dy=8xy² + 4y = 0

=2x² + 1 = 0

y= √2x²  + 1

Is the above correct?

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