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Calculus problem with directional vectors

  1. Apr 5, 2010 #1
    At the point (2,-1,2) on the surface z = x(y^2), find the direction vector for the direction of greatest decrease of z.

    A) i - 2j
    B) i - 4j
    C) (i - 4j)/sqrt(17)
    D) -i + 4j
    E) i + j

    Do I need a function f(x,y,z)? If so then f(x,y,z) = z = xy^2.

    Then the gradient vector would be <Fx, Fy, Fz>.

    This would be give me. < -y^2, -2xy, 1>

    At the point (2,-1,2) we get < -1, 4, 1>

    But how do i get the direction of the vector of greatest decrease?

    From the choices though it seems they only find the gradient vector of z = F(x,y) = xy^2

    In which case the gradient vector is < fx, fy>

    < y^2 , 2xy> .
    At the point (2, -1 ,2) we get < 1, -4>
    Again i am not sure how to find the vector of greatest decrease.

    It says in my book the direction of minimum increase is - norm( grad f(x,y,z)). But i am not getting any of these answers can some one help?
  2. jcsd
  3. Apr 5, 2010 #2


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    No, it's z=f(x,y). You're given a function from R2 to R, not R3 to R.
    I doubt your book says that. The norm of the gradient is a number, not a vector; it doesn't have a direction.

    How are the gradient and the rate and direction of maximum increase related?
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