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Math Help - Newton's Method Help...

  1. #1
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    Newton's Method Help...

    I need help with this problem:


    Any ideas?
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  2. #2
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    Hi,
    the first thing to do is to convert the above question to a root finding problem.

    you can do so by first finding distance of an arbituary point on the curve from (-3,0).

    you should get the equation of d = sqroot( (Xo - (-3) )^2 + (f(Xo) - 0 )^2 ) , where (Xo, f(Xo)) is an arbituary point on the curve.

    since distance in this case has an absolute minimum, use Newton's Method to find the root of this derivative, d'(Xo) = 0, to solve for the value of Xo where this absolute minimum occurs, ie, the point closest to (-3, 0).

    Hope this helps!
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  3. #3
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    Quote Originally Posted by BeSweeet View Post
    I need help with this problem:


    Any ideas?
    Also, we can solve for the square of the distance being a minimum.

    (x+3)^2+y^2=minimum

    (x+3)^2+\left[(x+7)^2+3\right]^2=minimum

    Differentiating and equating to zero

    2(x+3)+2\left[(x+7)^2+3\right]2(x+7)=0

    leading to a cubic

    4x^3+84x^2+602x+1462=0

    Using Newton's method

    x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

    f'(x)=12x^2+168x+602

    Finally, choose an initial estimate for x and run through a few iterations
    to find the x co-ordinate of the point of contact of the tangential circle.
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