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Contour map of the function showing several level curves

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  1. Feb 15, 2016 #1
    1. The problem statement, all variables and given/known data
    ƒ(x,y) = ln(x2+4y2)
    2. Relevant equations
    I'm not really sure but I solved for y

    3. The attempt at a solution

    New_Note.png
     
    Last edited: Feb 15, 2016
  2. jcsd
  3. Feb 15, 2016 #2

    haruspex

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    What do you get for y at x=0 from x2+4y2=1? Where is/are that/those on your diagram?

    I'm not sure whether you are supposed to plot curves for equal increments in f, as one would on a cartographic map.
     
  4. Feb 17, 2016 #3

    HallsofIvy

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    "Level curves" are curves on which the expression has a specific value. I would not solve for y. Instead, I would look at the equation [itex]ln(x^2+ 4y^2)= C[/itex] and recognize that, taking the exponential of both sides, [itex]x^2+ 4y^2= e^C[/itex] which is equivalent to [itex]\frac{x^2}{e^C}+ \frac{y^2}{e^C/4}= 1[/itex]. You should immediately see that, for every C, this is an ellipse, centered at the origin, with x intercepts at [itex]\left(-e^{C/2}, 0\right)[/itex] and [itex]\left(e^{C/2}, 0\right)[/itex], y intercepts at [itex]\left(0, -\frac{e^{C/2}}{2}\right)[/itex] and [itex]\left(0, \frac{e^{C/2}}{2}\right)[/itex].
     
  5. Feb 17, 2016 #4

    haruspex

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    Yes, that was a bit confusing in the OP, but judging from the drawing I would say cosmos42 understood about different values for C.
    I was trying to guide cosmos into realising that the curves were centred wrongly.
     
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