Turn on thread page Beta

Maths Problem That I Really Should Know! watch

    • Thread Starter
    Offline

    1
    ReputationRep:
    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.
    Offline

    2
    ReputationRep:
    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.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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?
    Offline

    10
    ReputationRep:
    (Original post by Ollie)
    ok, thanks.

    so how do i find the co-ordinates?
    Surely those are the co-ords?
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Nylex)
    Surely those are the co-ords?
    ummm... shouldn't there be more?
    Offline

    2
    ReputationRep:
    There are stationary points at (0, 0) and (1/6, -1/12).
    Offline

    10
    ReputationRep:
    (Original post by Ollie)
    ummm... shouldn't there be more?
    Eh? How come?
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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!
    Offline

    0
    ReputationRep:
    (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~
    Offline

    15
    ReputationRep:
    now we know why you're not a moderator here
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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.
    Offline

    0
    ReputationRep:
    (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~
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    1
    ReputationRep:
    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?
    Offline

    2
    ReputationRep:
    (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
    Offline

    8
    ReputationRep:
    (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)
    ================
    • Thread Starter
    Offline

    1
    ReputationRep:
    (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.
    Offline

    8
    ReputationRep:
    (Original post by Ollie)
    thanks, (and fermat), this agrees with my answer so have some rep for being 1st.
    Always the bridesmaid
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Fermat)
    Always the bridesmaid
    maybe 2mrw
 
 
 
Poll
Which accompaniment is best?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.