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Find the formula of the level curve (scalar) for the graph T(x,y) = (2x+y)/(x^2 -y^2)

  1. Mar 30, 2009 #1
    Hi this is the question that I am unsure about....

    Find and then sketch the level curves of the scalar field T(x,y) = (2x+y)/(x^2 -y^2) for;
    T = -1
    T = -0.5
    T = 0
    T = 0.5
    T = 1

    I am unsure about these answers which I got by subbing each of the values into T(x,y)

    1) T = -1

    After gathering x and y values on one side of the equation and completeing the square I got;

    (x + 1)^2 -(y - 0.5)^2 = 3/4 ....

    I am unsure about what kind of graph this is because I know that a hyperbola should be equal to 1 and this isn't.

    T = -0.5

    For this one I got (x + 2)^2 -(y - (0.5))^2 = (15/4)

    T = 0
    Finally one I could do :)

    y = 2x

    T=0.5
    (x - 2) ^2 - (y +1)^2 = 3

    T = 1
    (x - 1)^2 - (y - 1/2)^2 = 3/4

    I am pretty sure that I got the equations correct but I would appreciate if someone could help me with;

    a) Explaining how I draw this since it is not in the form of a hyperbola ie not = 1.

    b) What these mean?
     
  2. jcsd
  3. Mar 31, 2009 #2

    CompuChip

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    Re: Find the formula of the level curve (scalar) for the graph T(x,y) = (2x+y)/(x^2 -

    What is the equation for a circle of radius r and center point (a, b)?
     
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