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Math Help - Find the critical points of a multivariable equation

  1. #1
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    Find the critical points of a multivariable equation

    Find the critical points of a multivariable equation-picture.png
    Picture is attached, may be small
    I can't get past x^2 + y^2 = 5
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  2. #2
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    I don't even get to x^2+ y^2= 5!

    The critical points of a multivariable function are where it is not differentiable or where \nabla f= 0.

    Here, the function is f(x,y)= x^3- 3x- y^3+ 12y so the function is always differentiable and \nabla f= (3x^2- 3)\vec{i}- (3y^2- 12)\vec{j}. That will be 0 when both components are 0: 3x^2- 3= 0 and 3y^2- 12= 0.

    Those should be easy to solve. Where in the world did you get " x^2+ y^2= 5"?
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  3. #3
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    Ok, sorry for the delayed response, I moved towards C++ for a bit.

    I was just wondering. When you get to:
    3x^2 - 3
    and
    -3y^2 + 12

    Do you then solve for x as 1,-1 and then y as 2,-2
    so the critical points are:
    1,2
    1,-2
    -1,2
    -1,-2?

    I'm sorry I was following a youtube video, didn't know what the heck I was doing. Thanks.
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  4. #4
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    Notice that I did NOT just get "3x^2- 3" and "3y^2- 12"! I said that, in order to have a critical point, we must have \nabla f(x,y)= 0 and from that I got the equations "3x^2- 3= 0" and "3y^2- 12= 0". Yes, x= \pm 1 and y= \pm 2 satisfy those equations. Since any pairing makes the gradient 0, the critical points are (1, 2), (1, -2), (-1, 2), and (-1, -2).
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