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Question about graphing L(x,y) against f(x,y)

  1. Apr 15, 2010 #1
    The question I'm stuck on is this:
    Let [tex]f(x,y) = (x+1)^2 + y^2[/tex]
    I'm asked to find the partial derivatives and then evaluate them at (1,2). From there, find L(x,y), the linear approximation to f(x,y) at (1,2).
    That part I'm ok with. I got the following:
    [tex]f_x(x,y) = 2(x+1)[/tex]
    [tex]f_y(x,y) = 2y[/tex]
    [tex]f_x(1,2) = 4[/tex]
    [tex]f_y(1,2) = 4[/tex]
    [tex]f(1,2) = 8[/tex]

    [tex]L(x,y) = 8 + 4(x-1) + 4(y-2)[/tex]

    Next part of the quesion was just filling in a table that shows that as f(x,y) moves away from (1,2), L(x,y) becomes less accurate approximation of f(x,y).

    So where am I stuck?
    On the last question, which is:
    "On a set of xy-axes, mark the point (1,2) and draw the level curves f(x,y) = k and L(x,y) = k for k = 5, 6, 7, 8, 9, 10.
    Draw the curves into the square [0,3]x[0,3] and label each curve with the value of k. How do the level curves reflect L' approximation of f near (1,2)?"

    This is likely a very easy question to do, but for the life of me I can't see how! Any (and all) help much appreciated.
     
  2. jcsd
  3. Apr 16, 2010 #2
    Level curves f(x,y) = k are the sets { (x,y) | f(x,y) = k}. Now think geometrically: for a fixed k, what is the graph of k = (x + 1)^2 + y^2 ? (hint: rewrite k = (sqrt(k))^2 and this should start to look like a very familiar object). You could always try plotting a few points to get the gist of it.

    Level curves L(x,y) = k are the same thing, except with less complicated formulas. Do a little algebra and you will find another very familiar object.

    Once you put all of these lines on the xy plane, a pattern will emerge and I think you will have no trouble with the last part.
     
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