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Find Steepest Climb of this Hyperbolic Paraboloid

  1. Oct 11, 2009 #1
    Delete please.
     
    Last edited: Oct 12, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    lanedance

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    if you define F(x,y,z) = x^2-y^2+z

    then the paraboloid is defined by the level surface
    F(x,y,z) = 0

    the gradient direction represents the direction of maximum change, and will by defintion be perpindicular to any level surface.

    how about this.... consider the function
    f(x,y) = z = y^2-x^2

    the calculate the 2D gradient
    [tex]\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x})[/tex]

    this will represenet the direction of greatest change of z = f(x,y), with x & y and i think it should be easy to relate the slope to the magintude of the gradient
     
  4. Oct 11, 2009 #3
    Thank you for the help, but one more question...
    Do I need to multiply the gradient of F(x,y) by the unit vector to find the direction of greatest ascent?
     
  5. Oct 11, 2009 #4

    lanedance

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    no worries, but the question you asked doesn't really make sense, as the gradient is a vector, so what do you mean multiplying a vector by a vector?

    the steps i outlined, you will give you the x,y direction, which direction of greatest ascent & the value of that slope in that direction

    if you want to find the unit vector representing the direction on surface, use the x,y direction with the slope to find a vector in that direction (in 3D), then normalise by divding by the vector magnitude

    note as a check, the vector you find will be perpindicular to the gradient you gave in you first post
     
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