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Directional derivative Q, stuck.

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data


    In what directions at the point (2, 0) does the function f(x, y) = xy have rate of change -1?


    [itex]D_{u}(f)(a,b) = \bigtriangledown f(a,b)\cdot (u_{1}, u_{2}) [/itex]

    [itex]f(x,y) = xy[/itex]

    [itex](a,b) = (2,0).[/itex]




    3. The attempt at a solution
    [itex]\frac{\partial f}{\partial x} = y[/itex]

    [itex]\frac{\partial f}{\partial y} = x[/itex]

    [itex]\bigtriangledown f(2,0) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})= (y, x) = (0, 2)[/itex]

    plugging in:

    [itex]D_{u}(xy)(2,0) = \bigtriangledown f(0,2)\cdot (u_{1},u_{2}) = -1[/itex]

    [itex](0,2)\cdot (u_{1},u_{2}) = -1[/itex]

    [itex]u_{2} = -0.5[/itex]

    [itex]u_{1}[/itex] has infinitely many values.

    the last line above is the part im confused about. Are there infinitely many values for u1? thanks.
     
    Last edited: Feb 22, 2015
  2. jcsd
  3. Feb 22, 2015 #2

    Ray Vickson

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    Homework Helper

    Do you want/need ##\vec{u} = (u_1,u_2)## to be a unit vector? If so, there are only two possible values of ##u_1##; if not, there will be infinitely many values of ##u_1##. As before, you need to use additional information if you have it.
     
  4. Feb 22, 2015 #3
    in the equation:

    [itex]D_{u}(f)(a,b) = \bigtriangledown f(a,b)\cdot (u_{1}, u_{2}) [/itex]

    the (u1, u2) is a unit vector (according to my lecture notes). So you suggest that u1 has two possible values. How so? and I posted all the information there is in the Q in the thread, so no missing links. im quite confused,

    edit: I think I know now, from definition of a unit vector: (u1)^2 + (u2)^2 = 1^2

    plug in and you'll get u1 = +/- (root3)/2
     
    Last edited: Feb 22, 2015
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