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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 ....

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)

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...

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.

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..

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.

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

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.

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).

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)

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?