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Math Help - scalar field question

  1. #1
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    scalar field question

    hey people, i have a question asking me to sketch the level curves of this scalar field:

    T(x,y) = (2x-y)/(x^2+y^2), for T = -1,-.5,0,.5 and 1.

    The theory we were taught was to let T equal a constant and then rearrange the equation to find y in terms of x, but you cannot simply rearrange this equation like that. I know i must have done something like this before, it seems pretty familiar, but it's bugging me :P can someone lend a hand?
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  2. #2
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    Quote Originally Posted by ben08 View Post
    hey people, i have a question asking me to sketch the level curves of this scalar field:

    T(x,y) = (2x-y)/(x^2+y^2), for T = -1,-.5,0,.5 and 1.

    The theory we were taught was to let T equal a constant and then rearrange the equation to find y in terms of x, but you cannot simply rearrange this equation like that. I know i must have done something like this before, it seems pretty familiar, but it's bugging me :P can someone lend a hand?

    If T=1

    we get...

    1=\frac{2x-y}{x^2+y^2} \iff x^2+y^2=2x-y \iff x^2-2x+y^2+y=0
    if you complete the square we will get an equation of a circle.

    x^2-2x+1+y^2+y+\frac{1}{4}=1+\frac{1}{4} \iff (x-1)^2+(y+\frac{1}{2})^2=\frac{5}{4}

    I hope this helps.

    Good luck
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  3. #3
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    Quote Originally Posted by ben08 View Post
    hey people, i have a question asking me to sketch the level curves of this scalar field:

    T(x,y) = (2x-y)/(x^2+y^2), for T = -1,-.5,0,.5 and 1.

    The theory we were taught was to let T equal a constant and then rearrange the equation to find y in terms of x, but you cannot simply rearrange this equation like that. I know i must have done something like this before, it seems pretty familiar, but it's bugging me :P can someone lend a hand?
    I'll do one of them - it should be very obvious how to do the others.

    1 = \frac{2x-y}{x^2 + y^2} \Rightarrow x^2 + y^2 = 2x - y \Rightarrow x^2 - 2x + y^2 + y = 0.

    Complete the square in x-terms and y-terms and re-arrange to get the standard form of a circle.
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  4. #4
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    thanks

    thanks for the help, i knew it was something simple that i hadn't used in a while
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