Maths Problem That I Really Should Know!

Watch
This discussion is closed.
Ollie
Badges: 2
Rep:
?
#1
Report Thread starter 16 years ago
#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
Jonny W
Badges: 8
Rep:
?
#2
Report 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.
0
Ollie
Badges: 2
Rep:
?
#3
Report Thread starter 16 years ago
#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
Nylex
Badges: 10
Rep:
?
#4
Report 16 years ago
#4
(Original post by Ollie)
ok, thanks.

so how do i find the co-ordinates?
Surely those are the co-ords?
0
Ollie
Badges: 2
Rep:
?
#5
Report Thread starter 16 years ago
#5
(Original post by Nylex)
Surely those are the co-ords?
ummm... shouldn't there be more?
0
Jonny W
Badges: 8
Rep:
?
#6
Report 16 years ago
#6
There are stationary points at (0, 0) and (1/6, -1/12).
0
Nylex
Badges: 10
Rep:
?
#7
Report 16 years ago
#7
(Original post by Ollie)
ummm... shouldn't there be more?
Eh? How come?
0
Ollie
Badges: 2
Rep:
?
#8
Report Thread starter 16 years ago
#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.
0
Ollie
Badges: 2
Rep:
?
#9
Report Thread starter 16 years ago
#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
Darkness
Badges: 0
Rep:
?
#10
Report 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
elpaw
Badges: 15
Rep:
?
#11
Report 16 years ago
#11
now we know why you're not a moderator here
0
Ollie
Badges: 2
Rep:
?
#12
Report Thread starter 16 years ago
#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
Darkness
Badges: 0
Rep:
?
#13
Report 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~
0
Ollie
Badges: 2
Rep:
?
#14
Report Thread starter 16 years ago
#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.
0
Ollie
Badges: 2
Rep:
?
#15
Report Thread starter 16 years ago
#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
Bezza
Badges: 2
Rep:
?
#16
Report 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
Fermat
Badges: 8
Rep:
?
#17
Report 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
Ollie
Badges: 2
Rep:
?
#18
Report Thread starter 16 years ago
#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
Fermat
Badges: 8
Rep:
?
#19
Report 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
Ollie
Badges: 2
Rep:
?
#20
Report Thread starter 16 years ago
#20
(Original post by Fermat)
Always the bridesmaid
maybe 2mrw
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

If you do not get the A-level grades you want this summer, what is your likely next step?

Take autumn exams (240)
47.06%
Take exams next summer (69)
13.53%
Change uni choice through clearing (111)
21.76%
Apply to uni next year instead (53)
10.39%
I'm not applying to university (37)
7.25%

Watched Threads

View All