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Homework Help: Partial differention hard questions

  1. Oct 15, 2005 #1
    HI GUYS
    i just have some hard problems on partial differentiation... hope you guys can help out

    1. find "day z / day y" its the funny backwards 6 symbols instead of the normal differntiation symbole

    if z=4y^3 * sqrt(x) - ln(4xy^3) + 2x

    i did this:
    i diffenriated 4y^3 * sqrt(x) with respect to y, and since sqrt(x) stays constant, it remains hence i get
    12y^2*sqrt(x) . i then diffed ln(4xy^3) + 2x
    i did

    ln(u)
    u=4xy^3 u`= 12xy^2
    =1/u * u`
    =(1/4xy^3) * 12xy^2
    = 3/y

    and the 2x disappears

    hence my ultimate answer is
    "day z/ day y"=12y^2*sqrt(x) - 3/y


    2. FInd the stationary point of
    z= 5x^2 + 23x + y^2 + 8y + 3xy + 32 amd determine the nature of this point

    day z / day x = 10x + 23 + 3y
    day z / day y = 2y + 8 + 3x
    use elimination hence
    10x + 23 + 3y *2
    2y + 8 + 3x * 3
    -----------------
    20x + 46 + 6y 3)
    6y + 24 + 9x 4)
    ---------------
    11x -22 = 0
    x = 2 sub this into eqn 3)
    y= -43/3
    now sub x=2, y=-43/3 into z=
    hence z=5(2)^2 + 23(2) + (-43/3)^2 + 8(-43/3) + 3*(2)(-43/3) + 32
    z=925/9


    hence stationary points i think are (2, -43/3 , 925/9)

    ok assuming thats right i have to determine the nature of the point, to my knowledge i would have to


    use the formula g= (day^2 z/day*x^2)*(day^2 z/ day*y^2) - ( day^2*z / day*x*day*y )
    day^2 z / day^2 x^2 = 10
    day^2 z / day^2 y^2 = 2
    day^2*z / day*x*day*y = 3

    hence g = (10)*(2) - 3
    G= 17

    HENCE 17 > 0 , AS (day^2 z / day^2 x^2) IS > 0, AND G> 0, THEN THE STATIONARY POINT IS A MINUMUM POINT

    I THINK I DID IT WRONG HAVE NO IDEA
     
  2. jcsd
  3. Oct 15, 2005 #2
    isnt der of z wrt y of sqrt(x) = 0
     
  4. Oct 15, 2005 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Yes, it is that's why the derivative of [tex]4y^3\sqrt{x}[/tex] is
    [tex]12y^2\sqrt{x}[/tex] just as dagg3r says- even if you use the "product rule". What did you think the derivative should be?
    The partial derivative of [tex]z= 4y^3\sqrt{x} - ln(4xy^3) + 2x[/tex]
    with respect to y is:
    [tex]\frac{\partial z}{\partial y}= 12y^2\sqrt{x}- \frac{1}{4xy^3}\left(12xy^2)\right)[/tex]
    = [tex]12y^2\sqrt{x}- 3y[/tex] exactly what dagg3r had!
    (Click on the TEX above to see the code I used- your "day" is driving me mad! I would have preferred you just use "d" with the notation that it is partial differentiation.)
    Okay, you got
    [tex]\frac{\partial z}{\partial x}= 10x+23+ 3y[/tex]
    and
    [tex]\frac{\partial z}{\partial y}= 2y+ 8+ 3x[/tex]
    Of course, at a stationary point, the partial derivatives are 0:
    10x+ 3y+23= 0 and
    3x+ 2y+ 8= 0 (You did not write "= 0". While I understood what you were doing, the "= 0" makes it clearer. Also I have written the two equations with x first, then y. That also makes it a little simpler to see what to do.
    Yes, multiply the first equation by 2 and the second by 3, to get "6y" in both and then subtract:
    20x+ 6y+ 46= 0
    9x+ 6y+ 24= 0
    which gives
    11x+ 22= 0.
    You subtracted wrong! (That might be because of the way you had written the equations.)
    x= -2, not 2, and then, from the second initial equation, -6+ 2y+ 8= 0 so
    2y= -8+ 6= -2, y= -1. The stationary point is at (2, -1).
    z(2, -1)= 5(4) + 23(2) + (-1)^2 + 8(-1) + 3(2)(-1) + 32
    = 20+ 46+ 1- 8- 6+ 32= 85.
    NOW determine the nature of the stationary point.
     
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