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#1 Re: Help Me ! » Finding and classifying stationary points » 2008-08-24 09:04:42

ok i had a go at dz/dy:

dz/dy=8xy² + 4y = 0

=2x² + 1 = 0

y= √2x²  + 1

Is the above correct?

#2 Re: Help Me ! » Finding and classifying stationary points » 2008-08-24 08:40:24

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 ??

#3 Re: Help Me ! » Partial Derivative help... » 2008-08-24 02:43:15

no one can answer these ?

#4 Help Me ! » Partial Derivative help... » 2008-08-24 01:33:20

boombastictiger
Replies: 2

I have a basic question I am stuck on...

Given the function Z= ln(y /x) Find the partial derivatives:

(A) dz/dx

(B) dz/dy

(C) Given the function, z=x³ + xy + y², show that (0,0) and (1/6, - 1/12) the only stationary points and classify them (that is, decide whether these points are maximum or minimum or saddle points).

Edited i added a comma after 1/6 it makes a big difference sorry...

#5 Re: Help Me ! » Finding and classifying stationary points » 2008-08-24 00:56:46

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

#6 Re: Help Me ! » Finding and classifying stationary points » 2008-08-22 09:54:52

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..

#7 Re: Help Me ! » Finding and classifying stationary points » 2008-08-22 09:16:42

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

#8 Help Me ! » Finding and classifying stationary points » 2008-08-22 08:26:38

boombastictiger
Replies: 8

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

Man I knew these questions were to hard...

Hi again...Ok I have some more questions for you!!!!!! These are mixed diffrentiation and intergration and are pretty hard........ i will state whther to diffrentiate or intergrate..

Diffrentiate the following with respect to x:

(a)y=5x+1

(b)f(x)=6x² -ln2x

(c)y=2(x+cos2PIEx)³

(d)f(x)=x^4e^-³x

(e)y=sin²x/2x+3

Evaluate the following integrals:

(a) ∫(2x+1/x+2)dx

(b)^4∫^°(1+2/ √t)dt
{the small zero is at the bottom of the curvy line}

(c) ^PIE/4∫^°4sin2θdθ
{the small zero is at the bottom of the curvy line}

(d) ∫x(x-3)^4dx       Use the substitution u=x-3

HAHAH lets see how quick and accurate you can respond to these questions!!!!!!

boombastictiger
Replies: 12

hi guys its me again, i wrote a post previously reguarding diffrentiation, but i believe it was deleted in the hack attack....sad sad people.... I now have some more questions for the clever people here(yes im picking ur brian again) lol..

4. A curve has the following equation: y=x³ -12x+7

(a)
(i) Find the gradient function ,dy?dx, of the curve.   (2 marks)

(ii) Find the coordinates of the points on the curve at which the gradient is zero.  (6 marks)

(iii) Describe the nature of the curves turning points.  (4 marks)

(b)
A box, with a lid, has a square base.  The length of each side of the square is x cm and the height of the box is h cm.

(i) Sketch a diagram of this box  (1 mark)  <<<<<<<<VERY BASIC I KNOW HOW TO DO LOL

The total surface area of the box is 2040cm²

(ii)Show that its volume, V cm³, is given by:

V=510x- x³/2               (6 marks)

(iii) Find the value of x that gives the maximum volume of the box.  Give your answer correct to 3      significant figures..     (6 mark)

These are rock solid questions, can anyone help me answer these using common sense AND using the neccesary equations such as the chain rule, product rule and quotient rule???

Challenging....