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Directional Derivatives and Gradient question

  1. Jun 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the surface and point given below:-
    Surface: f(x,y)= 4-x2-2y2
    Point: P(1,1,1)

    a) Find the gradient of f.
    b) Let C' be the path of steepest descent on the surface beginning at P and let C be the projection of C' on the xy-plane. Find an equation of C in the xy-plane.

    2. Relevant equations
    1) ∇f = <fx , fy>
    2) ad92a17b245e925a2bff2e4444a520af.png
    3) 864157ee7a7f4f55c1a7ce58dfb5cbb1.png

    3. The attempt at a solution
    a) ∇f = <fx , fy> = <-2x, -4y>
    b)Descent means -∇f = <2x,4y>
    Subbing in P(1,1,1)
    -∇f = <2,4> ⇔ <1,2>
    Unit vector for descent, u = (1/√5) <1,2>
    ,where <1,2> is the direction vector of the descent gradient.

    I am really stuck here. I am not sure what they mean by the projection on the xy-plane. So are we moving from xyz to xy dimensions? And can I am not sure if I should use the projection formulas in this case or use derivatives to get the projection. Please help!!
     
  2. jcsd
  3. Jun 28, 2012 #2

    LCKurtz

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    I think that sometimes in this type of problem it helps to visualize a physical model. Suppose you have a thin metal plate and your ##f(x,y) =4 - x^2-2y^2## represents the temperature at the point ##(x,y)##, so the temperature at ##(1,1)## is ##1##. You are looking for the path to follow to cool off quickest. You have calculated that the direction to go at each point is ##-\nabla f = \langle 2x,4y \rangle##. From this you can conclude that the slope of the desired curve at ##(x,y)## is ##\frac{dy}{dx} =\frac {4y}{2x}=\frac {2y}{x}##. This is a simple first order differential equation. Do you know how to solve it? If so, the solution through ##(1,1)## is what you are looking for.
     
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