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Grad F question verification

  1. May 24, 2007 #1
    1. The problem statement, all variables and given/known data

    For the function f(x, y, z) = z^4(x^2 − y^2), find the coordinates of
    those points in the plane z = 1 at which the direction of the greatest rate of
    change of f is parallel to the vector i − 2j + 3k.

    2. Relevant equations

    3. The attempt at a solution

    i just want to verify that i am approaching this question correctly.

    1.First, i substitute, z =1 into the equation.
    2.so i get f(x,y,z) = x^2- y^2.
    3. Then i find grad f(x,y,z). cos i need the direction of greatest rate of change. (i mean i find grad f of the equation x^2-y^2.)
    4. To find the coodinates parallel to the vector i-2j+3k, i substitute
    these values into my grad f equation.


    As z= 1
    f(x,y,z) = x^2 - y^2

    grad f = 2xi - 2yj

    Subsitute, x =1,y=-2 ->2xi-2yj

    Therefore coordinates are: (2,4,0)

    Could someone please tell me if this is correct.

    Last edited: May 24, 2007
  2. jcsd
  3. May 24, 2007 #2


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    Take the gradient BEFORE you substitute for z! The gradient of x^2-y^2 is NEVER parallel to the given vector.
  4. May 24, 2007 #3
    So i'll get grad f = (2xz^4)i-(2yz^4)j - (4z^3 x^2 - 4z^3 y ^2)k

    And i substitute, x = 1, y=-2 and z=3 into grad f:

    and i get, the coordinates, (162, 324,972). <----Is this right????

    does that fact that it is on the plane z= 1 hav anything to do with the solution?

  5. May 24, 2007 #4


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    Take the gradient, then substitute z=1. 2xi-2yj+4(x^2-y^2)k. Now try to find values of x and y that make this parallel to your given vector. Do you know how to determine if two vectors are parallel. BTW - I can't find any such x and y. There may be no solution.
    Last edited: May 24, 2007
  6. May 24, 2007 #5

    D H

    Staff: Mentor

    Dick, check the sign on your gradient. I get [itex]2x\hat i - 2y\hat j + 4(x^2-y^2)\hat k[/itex]. There are still no solutions.
  7. May 24, 2007 #6


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    Check. Thanks.
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