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Grad f and grad g

  1. Mar 27, 2008 #1
    Hi,

    I have some problems to solve this:

    a=8 , b=5

    Let f (x, y)=xye^(− ((x−a)^2)− ((y−b)^ 2)). In which direction, one will go from the point (a, b) in the definition amount if one wants to that the function values will increase so quickly as possible? Decide an equation for tangent plane to the surface z=f (x, y) in the point (a, b, ab). Use differential to f in order to calculate a close value to f (9a/10, 6b/5). In which points, key planet is to the surface z=f (x, y) horizontal?

    I derivated and got:

    df/dx = ye^(− ((x−a)^2)− ((y−b)^ 2)) * (-2x+2a) + ye^(− ((x−a)^2)− ((y−b)^ 2))

    df/dy = xe^(− ((x−a)^2)− ((y−b)^ 2)) * (-2y+2a) + ye^(− ((x−a)^2)− ((y−b)^ 2))

    and grad f(a,b) = (y,x) = (5,8) , Is this correct?
    ______________________________________________________________

    And for the tangent plane:

    g(x,y,z)=f(x,y) - z

    grad g = (5,8,-1)

    And the tangent plane: 5(x-8)+8(y-5)+(1)(z-40)=0 => 5x+8y-z=40 , Is this correct?

    _____________________________________________________________

    And for the close value:

    z=5x+8y-40 close to (8,5). In (7.2,6) is z=5*7,2+8*6-40=44 , Is this correct?

    _________________________________________________________

    The are horizontal when df/dx = df/dy = 0 , Is this correct? I dont know if i can get a value to this?
     
  2. jcsd
  3. Mar 30, 2008 #2

    Shooting Star

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    (I presume you mean [tex]\frac{\partial}{\partial x}[/tex].)

    There should be an 'x' in the first term.

    [tex]\frac{\partial f}{\partial x} = [/tex] xye^(− ((x−a)^2)− ((y−b)^ 2)) * (-2x+2a) + ye^(− ((x−a)^2)− ((y−b)^ 2)).

    The same missing 'y' in [tex]\frac{\partial f}{\partial y}[/tex].

    Correct these first, in order to get the correct answers for the rest of the problems.
     
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