# Newton's Method Help...

• Mar 24th 2010, 09:34 AM
BeSweeet
Newton's Method Help...
I need help with this problem:
http://img10.imageshack.us/img10/2761/captureny.png

Any ideas?
• Mar 24th 2010, 09:48 AM
navelwort
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!
• Mar 24th 2010, 12:28 PM
Quote:

Originally Posted by BeSweeet
I need help with this problem:
http://img10.imageshack.us/img10/2761/captureny.png

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

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

$4x^3+84x^2+602x+1462=0$
$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
$f'(x)=12x^2+168x+602$