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Extrema in 3D problem

  1. Dec 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the surface defined by


    (a) Evaluate grad(F)

    My Solution:
    grad(F)=(3x+y+2z, x+3y+2z, 2x+2y+4z)

    (b) Find the highest and lowest points on the surface (i.e. the points where z attains a maximum or minimum).

    So I can solve for z=g(x,y) using the quadratic formula and get a nasty expression (see attached) and
    then go about finding the extrema, but this is ugly.

    If I set each component of the gradient to zero, then the only solution is (x,y,z)=(0,0,0) which is NOT an
    extrema (see attached image), rather there are two extrema.

    How do I do this the "easy" way?

    (c) The surface is illuminated from far above by light rays that are directed parallel to the z-axis. Find the
    shape of its shadow in the plane below the surface parallel to the (x,y) coordinate plane.


    So now this is projecting the surface onto the (x,y)-plane. I have no idea how to do this.

    Please help.

    Attached Files:

  2. jcsd
  3. Dec 25, 2011 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Use the equation F(x,y,z) = 0 to evaluate [itex] z_x = \partial z / \partial x[/itex] and [itex] z_y = \partial z / \partial y[/itex] at a given point [itex] (x_0,y_0,z_0)[/itex] on the surface; that is, you need to figure out [itex] \Delta z[/itex] such that
    [tex] F(x_0 + \Delta x, y_0 + \Delta y, z_0 + \Delta z) = 0. [/tex] You don't need the exact[itex] \Delta z[/itex]; you just need the ``first-order'' expression that is linear in [itex] \Delta x \mbox{ and } \Delta y.[/itex] Then you need [itex] z_z = 0[/itex] and [itex] z_y = 0. [/itex]

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