Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - partial derivative please help

  1. #1
    tak
    tak is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    29

    partial derivative please help

    find the stationary point for  f(x,y) =x^4 +64y^4-2(x+8y)^2
    i got my fx =3x^3 -4x-32y
    and fy= 256y^3-32x-256y

    I don't know how to get the three points....any one can highlight to me??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hello

    Hint : \frac{32}{4}=\frac{256}{32} It should help you getting rid off x and y and you'll be able to solve for x^3 and y^3
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    fx would rather be 4x^3-\dots

    Stationary points are points (x,y) where \frac{\partial f}{\partial x}=0 and \frac{\partial f}{\partial y}=0

    So solve for x and y in

    \bigg\{ \begin{array}{ccc} 4x^3-4x-32y & = & 0 \\ 256y^3-32x-256y & = & 0 \end{array}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    tak
    tak is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    29

    stuck

    y^3-32x/256-y=0
    x^3-x-32y/4=0 how to proceed?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    I think that the better way is, as Flyingsquirrel mentioned, to see that 256/32=32/4=8

    So if you multiply the first equation by 8, you will be able to get rid of x and y, and keep y^3, while substracting term by term the two equations
    Follow Math Help Forum on Facebook and Google+

  6. #6
    tak
    tak is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    29

    stuck again

    2y=x i got it after some manipulation........what next????
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    What about substituting x=2y in one of the equation and solving for y ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    tak
    tak is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    29

    stuck again again

    sub x=2y into the f(x,y)
    i got 80y^4 - 200y^2
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    "Substituting in one of the equations" meant substituting either in f_x=0 or in f_y=0 but not in f(x,y), which is not an equation.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    tak
    tak is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    29
    i got y=+-(5/4)^(1/2)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    No, these are not the solutions for y.

    4x^3-4x-32y=0 \Rightarrow 4\times8y^3-8y-32y=0 \Rightarrow 16y(2y^2-3)=0 \Rightarrow  \left\{ \begin{array}{l}y=\ldots\\ \mbox{or} \\y=\ldots\\ \mbox{or} \\y=\ldots\end{array}\right.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    tak
    tak is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    29

    solve

    ehhh....really??
    i better look at it....
    Follow Math Help Forum on Facebook and Google+

  13. #13
    tak
    tak is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    29

    equation correct?

     8y^3 - x- 8y

    x^3-x-8y

    two equation correct?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Yes
    Follow Math Help Forum on Facebook and Google+

  15. #15
    tak
    tak is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    29

    solve

    solve....=)
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Derivative of arctan in a partial derivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 12th 2010, 01:52 PM
  2. Partial Derivative?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2010, 05:56 PM
  3. how do you take this partial derivative
    Posted in the Calculus Forum
    Replies: 10
    Last Post: July 26th 2009, 07:09 PM
  4. Partial derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 16th 2009, 11:21 AM
  5. partial derivative...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 9th 2009, 05:06 PM

Search Tags


/mathhelpforum @mathhelpforum