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Math Help - Multivariate Functions

  1. #1
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    Multivariate Functions

    I am having trouble with this problem, directions indicate to sketch the indicated level curve f(x,y) = C for each choice of constant C.

    f(x,y)= x+2y ; C=1, C=2, C=-3

    I am not too concerned with the actual sketching, but I am having trouble finding the points to graph. The answer in the book shows 3 separate lines with line C=1 having points of (0,1/2), (1,0) line C=2 having points of (0,1), (2,0) and line C=-3 having points of (-3,0), (0,-3/2). My professor showed us an example where we set the C number equal to the function. So I did for C=1 1=x+2y. I then isolated Y by itself as he did in the example and divided by 2, and got Y= -x+1 Is this right? How do I find the other coordinates? Thanks...
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  2. #2
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    Quote Originally Posted by kdogg121 View Post
    I am having trouble with this problem, directions indicate to sketch the indicated level curve f(x,y) = C for each choice of constant C.

    f(x,y)= x+2y ; C=1, C=2, C=-3

    I am not too concerned with the actual sketching, but I am having trouble finding the points to graph. The answer in the book shows 3 separate lines with line C=1 having points of (0,1/2), (1,0) line C=2 having points of (0,1), (2,0) and line C=-3 having points of (-3,0), (0,-3/2). My professor showed us an example where we set the C number equal to the function. So I did for C=1 1=x+2y. I then isolated Y by itself as he did in the example and divided by 2, and got Y= -x+1 Is this right? How do I find the other coordinates? Thanks...
    Setting f(x,y)=C gives

    x+2y=C \iff y=-\frac{1}{2}x+\frac{C}{2}

    This gives us a family of lines with slope -1/2 and y intercepts C/2

    so if we plug in different values of C we get all of the level curves of the function.

    I hope this helps.
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  3. #3
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    Thank You for your help. So I would apply the same technique to this problem: f(x,y) = x^2-4x-y ; C= -4 , C=5

    Y= x^2-4x-C ?? How do I get points out of this?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by kdogg121 View Post
    Thank You for your help. So I would apply the same technique to this problem: f(x,y) = x^2-4x-y ; C= -4 , C=5

    Y= x^2-4x-C ?? How do I get points out of this?
    Complete the square and you get

    y=x^2-4x+4 -4 -C \iff y= (x-2)^2-(C+4)


    So our level curves are Parabola's with the vertex at (2,-(C+4))
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