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Gradient operator of a function

  1. Dec 11, 2011 #1
    (1) Let [itex]f(x)=x^3+y^3+z^3-3xyz[/itex], Find grad(f).

    [itex]grad(f)=(3x^2-3yz, 3y^2-3xz, 3z^2-3xy)[/itex].

    (2) Identify the points at which grad(f) is
    a) orthogonal to the z-axis
    b) parallel to the x-axis
    c) zero.


    I have managed to solve for (1), but don't have a clue how to solve for the second part. I have not come across about the theory of "orthogonal to z-axis" and "parallel to x-axis", tried to look up on the internet but still quite confused.

    However, for (c) I have come up with something like [itex]grad(f)=(3x^2-3yz, 3y^2-3xz, 3z^2-3xy)=(0,0,0)[/itex], so the points at which grad(f)=0 are (0,0,0). Is that right?
     
  2. jcsd
  3. Dec 11, 2011 #2

    LCKurtz

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    Do you know how to test whether two vectors are parallel or perpendicular?

    Certainly that point works. But without seeing what you did, I don't know if you know whether or not that is the only point that works.
     
  4. Dec 11, 2011 #3
    Yes, I know that if
    1) a and b are parallel, axb=0
    2) a and b are perpendicular, axb=|a||b|n

    But in the past I have been given specific vectors like a=(1,2,3) b=(5,3,2). This time, I've only been given statements such as "orthogonal to z-axix", and I really don't have a clue what that actually means.

    Let's say, for example, if grad(f) is vector a, so what will vector b be? This is where I'm stuck.


    I actually didn't quite understand the question, but guessed that if grad(f)=0, then it's pretty obvious that
    [itex]3x^2−3yz=0[/itex]
    [itex]3y^2−3xz=0[/itex]
    [itex]3z^2−3xy=0[/itex]

    simplify them and you get
    [itex]x^2−yz=0[/itex]
    [itex]y^2−xz=0[/itex]
    [itex]z^2−xy=0[/itex]

    therefore x=0, y=0, z=0.

    This is what I thought, but I might be wrong though.
     
  5. Dec 12, 2011 #4

    LCKurtz

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    (2) isn't a test for perpendicularity. What is the dot product test?

    Can you find a vector in the direction of the z axis? Then use the dot product test and see what works.

    You can't say "therefore x=0, y=0, z=0" unless you know (0,0,0) is the only solution. Just because it obviously works doesn't mean it is the only solution. You have to work with those equations to either show that is the only solution or find any others that there might be. In this case, you will find others.
     
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