# Maths Problem That I Really Should Know!

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#1
Any help much apprieciated with this differential equation problem:

Find the stationary points of the following functions determining in each case whether they are maxima, minima or saddle points:
iii) f = x^3 + xy + y^2

i said the derivitives with respect to...
fx = 3x^2 +y
fy = x + 2y

they must both equal zero for a stationary point. therefore are equal. Therefore should be able to work it out from there but my brain's gone dead!

I just need help to work out the co-ordinates. I can work out if it is max etc.
0
16 years ago
#2
3x^2 = -y = x/2
=> x(6x - 1) = 0
=> x = 0 or x = 1/6

Case 1: x = 0.
y = 0.

Case 2: x = 1/6.
y = - x/2 = -1/12.
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#3
(Original post by Jonny W)
3x^2 = -y = x/2
=> x(6x - 1) = 0
=> x = 0 or x = 1/6

Case 1: x = 0.
y = 0.

Case 2: x = 1/6.
y = - x/2 = -1/12.
ok, thanks.

so how do i find the co-ordinates?
0
16 years ago
#4
(Original post by Ollie)
ok, thanks.

so how do i find the co-ordinates?
Surely those are the co-ords?
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#5
(Original post by Nylex)
Surely those are the co-ords?
ummm... shouldn't there be more?
0
16 years ago
#6
There are stationary points at (0, 0) and (1/6, -1/12).
0
16 years ago
#7
(Original post by Ollie)
ummm... shouldn't there be more?
Eh? How come?
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#8
(Original post by Jonny W)
There are stationary points at (0, 0) and (1/6, -1/12).
ok, thanks, i thought there might be 4.
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#9
(Original post by Ollie)
ok, thanks, i thought there might be 4.
ok, so why does this one have 4:

f = x^3 - 3x + xy^2
fx = 3x^2 - 3 + y^2 = 0 (1)
fy = 2xy (2)

2 is satisfied by x=0, y=0
x=0 into (1) give y^2 = 3 therefore y = +- root3.
therefore (0, root3) and (0, -root3) are stationery

y=0 into (1) gives x^2 =1, x = +-1
therefore (1,0) and (-1,0) are stationery

Thats what he wrote up in the example.

I'm not entirely sure whats going on!
0
16 years ago
#10
(Original post by Ollie)
Any help much apprieciated with this differential equation problem:

Find the stationary points of the following functions determining in each case whether they are maxima, minima or saddle points:
iii) f = x^3 + xy + y^2

i said the derivitives with respect to...
fx = 3x^2 +y
fy = x + 2y

they must both equal zero for a stationary point. therefore are equal. Therefore should be able to work it out from there but my brain's gone dead!

I just need help to work out the co-ordinates. I can work out if it is max etc.
Set the partial derivatives equal to zero and solve them simultaneously:

fx = 3x^2 +y = 0
fy = x + 2y = 0 => x = -2y (*)

Substitute -2y for x in fx to obtain:

fx = 3(-2y)^2 + y = 0
fx = 12(y)^2 + y = 0 => y(12y + 1) = 0 => y=0 or y=-1/12

Substitute y values in (*) to obtain corresponding x values: x = 0 or x = 1/6

Thus, the stationary points are: (0,0),(1/6,-1/12).

As for your example, not all functions have 4 stationary points!! Some have 1, some have none, some have 4, some have 3 etc...

~Darkness~
0
16 years ago
#11
now we know why you're not a moderator here
0
#12
(Original post by elpaw)
now we know why you're not a moderator here
lol, easter's made my mind go so blank!

Yeah i know not all have 4 stationary points, but i couldn't see why the other one did.
0
16 years ago
#13
(Original post by Ollie)
Yeah i know not all have 4 stationary points, but i couldn't see why the other one did.
Okay. Let's set the partial derivatives to zero and find the corresponding x and y values:

fx = 3x^2 - 3 + y^2 = 0 (1)
fy = 2xy = 0 (2)

As before, we want to solve these equations simultaneously. Now (2) is true (i.e. LHS = RHS) if x = 0 OR y = 0. In the case where x = 0, we can find the corresponding y value by substituting x into (1):

fx = -3 + y^2 = 0 => y^2 = 3 => y = +- sqrt3. In other words we have TWO values of y associated with this value of x, and thus two stationary points when x = 0: (0,sqrt3) AND (0, -sqrt3).

In the case where y=0, we again substitute into (1) to get:

fx = 3x^2 - 3 = 0 => x^2 - 1 = 0 => x^2 = 1 => x = +-sqrt1 = +-1. Therefore this value of y is associated with TWO x values, so we have a further TWO stationary points: (1,0) AND (-1,0).

HTH

~Darkness~
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#14
(Original post by Darkness)
Okay. Let's set the partial derivatives to zero and find the corresponding x and y values:

fx = 3x^2 - 3 + y^2 = 0 (1)
fy = 2xy = 0 (2)

As before, we want to solve these equations simultaneously. Now (2) is true (i.e. LHS = RHS) if x = 0 OR y = 0. In the case where x = 0, we can find the corresponding y value by substituting x into (1):

fx = -3 + y^2 = 0 => y^2 = 3 => y = +- sqrt3. In other words we have TWO values of y associated with this value of x, and thus two stationary points when x = 0: (0,sqrt3) AND (0, -sqrt3).

In the case where y=0, we again substitute into (1) to get:

fx = 3x^2 - 3 = 0 => x^2 - 1 = 0 => x^2 = 1 => x = +-sqrt1 = +-1. Therefore this value of y is associated with TWO x values, so we have a further TWO stationary points: (1,0) AND (-1,0).

HTH

~Darkness~
Thanks alot for this.
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#15
hey guys, come back round to this topic, finding it still hard to get the coordinates. I've got an answer but don't think it's right.

for example:
f(x,y) = x^2y - y + 3x^2 + 4y^2

fx = 2xy + 6x = 2x(y+3) = 0
fy = x^2 -1 + 8y = 0

now, I reckon there are 3 stationary points, where do you make them to be?
0
16 years ago
#16
(Original post by Ollie)
f(x,y) = x^2y - y + 3x^2 + 4y^2

fx = 2xy + 6x = 2x(y+3) = 0 (1)
fy = x^2 -1 + 8y = 0 (2)
Cheat! Bouncing your thread back to the top by deleting and reposting the same thing!

I've never done partial differentiation, but:
From (2): 2y = (1 - x^2)/4
Sub into (1): x(1 - x^2)/4 + 6x = 0
x(1 - x^2) + 24x = 0
x^3 - 25x = 0
x(x+5)(x-5) = 0 so stationary points occur at x = -5,0,5
0
16 years ago
#17
(Original post by Ollie)
hey guys, come back round to this topic, finding it still hard to get the coordinates. I've got an answer but don't think it's right.

for example:
f(x,y) = x^2y - y + 3x^2 + 4y^2

fx = 2xy + 6x = 2x(y+3) = 0
fy = x^2 -1 + 8y = 0

now, I reckon there are 3 stationary points, where do you make them to be?
f=x²y - y + 3x² + 4y²

f'x = 2xy + 6x
f'y = x² - 1 + 8y

f'x = 2xy + 6x = 0
2x(y + 3) = 0
x=0, y= -3
=======

x=0
==
f'y = x² - 1 + 8y = 0
0 - 1 + 8y = 0
y = 1/8
====

y= - 3
====
f'y = x² - 1 + 8y = 0
x² - 1 - 24 = 0
x² = 25
x = ± 5
=====

Stationary points are
(0, 1/8), (5, -3), (-5, -3)
================
0
#18
(Original post by Bezza)
Cheat! Bouncing your thread back to the top by deleting and reposting the same thing!

I've never done partial differentiation, but:
From (2): 2y = (1 - x^2)/4
Sub into (1): x(1 - x^2)/4 + 6x = 0
x(1 - x^2) + 24x = 0
x^3 - 25x = 0
x(x+5)(x-5) = 0 so stationary points occur at x = -5,0,5
thanks, (and fermat), this agrees with my answer so have some rep for being 1st.
0
16 years ago
#19
(Original post by Ollie)
thanks, (and fermat), this agrees with my answer so have some rep for being 1st.
Always the bridesmaid
0
#20
(Original post by Fermat)
Always the bridesmaid
maybe 2mrw
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