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Math Help - Using Calculus to find a Point on a circle

  1. #1
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    Using Calculus to find a Point on a circle

    Hello, I am trying to find the closest point on the circle x^2+y^2=1 to (1,2) using calculus, not geometry. However, when I try, I get:

    x^2+y^2=1
    y=sqrt(1-x^2)
    trying to find the distance between (x,sqrt(1-x^2)) and (1,2) = (1-x, 2-sqrt(1-x^2))
    d=sqrt((1-x)^2+(2-sqrt(1-x^2))^2)
    d=sqrt(1-x^2+4-1+x^2)
    d=sqrt(5)

    Simple geometry tells me this is the distance between the two points, so how do I solve for x and y from here?
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  2. #2
    MHF Contributor chisigma's Avatar
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    You have to minimize the function f(x,y)= (x-1)^{2} + (y-2)^{2} with the constraight x^{2} + y^{2} = 1... a classical problem involving Lagrange's multipliers...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by Dudealadude View Post
    Hello, I am trying to find the closest point on the circle x^2+y^2=1 to (1,2) using calculus, not geometry. However, when I try, I get:

    x^2+y^2=1
    y=sqrt(1-x^2)
    trying to find the distance between (x,sqrt(1-x^2)) and (1,2) = (1-x, 2-sqrt(1-x^2))
    d=sqrt((1-x)^2+(2-sqrt(1-x^2))^2)
    d=sqrt(1-x^2+4-1+x^2)
    d=sqrt(5)

    Simple geometry tells me this is the distance between the two points, so how do I solve for x and y from here?

    Don't write y as a function of x in the beginning. you want the minimum of

    (x-1)^2+(y-2)^2=-2x-4y+6 , after we substitute x^2+y^2=1. Now you write y=\sqrt{1-x^2} ,


    observing that the wanted point must be in the first quadrant and thus both x,y>0:

    -2x-4\sqrt{1-x^2}+6 . This is a func. of x, derivate and equal to zero and get \displaystyle{x=\frac{1}{\sqrt{5}}\,,\,y=\frac{2}{  \sqrt{5}}

    Tonio
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