I need help with this problem:
Any ideas?
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!
Also, we can solve for the square of the distance being a minimum.
$\displaystyle (x+3)^2+y^2=minimum$
$\displaystyle (x+3)^2+\left[(x+7)^2+3\right]^2=minimum$
Differentiating and equating to zero
$\displaystyle 2(x+3)+2\left[(x+7)^2+3\right]2(x+7)=0$
leading to a cubic
$\displaystyle 4x^3+84x^2+602x+1462=0$
Using Newton's method
$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
$\displaystyle f'(x)=12x^2+168x+602$
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.