1. Optimizing Distance

Hi,
Here is my question:
Find the point P on the parabola y=(x^2) closest to the point (3,0).
I solved it... but I am not sure what my domain restrictions should be...
Thanks,

D = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

Let (x1, y1) = (3, 0) and (x2, y2) = (x, y). Then

D = sqrt( (x - 3)^2 + (y - 0)^2 )
D = sqrt( (x - 3)^2 + y^2 )

*y = x^2

D = sqrt( (x - 3)^2 + (x^2)^2 )
D = sqrt( (x - 3)^2 + x^4 )

We want to minimize this by making dD/dx = 0.
We *could* differentiate this directly, but when we have to deal with derivatives with roots followed by the chain rule. I will instead square both sides, and then differentiate implicitly.

D^2 = (x - 3)^2 + x^4

Differentiate implicitly with respect to x,

2D (dD/dx) = 2(x - 3) + 4x^3

Make dD/dx = 0, to get

0 = 2(x - 3) + 4x^3

Solve for x.

0 = 2x - 6 + 4x^3
0 = 4x^3 + 2x - 6
0 = 2x^3 + x - 3

x=1... sub into original -> y=1. Therefore (1,1)

2. $R = \sqrt{(x - 3)^2 + y^2} = \sqrt{(x - 3)^2 + (x^2)^2} = \sqrt{x^4 + x^2 - 6x + 9}$

$R' = \frac{4x^3 + 2x - 6}{2\sqrt{x^4 + x^2 - 6x + 9}}$

$4x^3 + 2x - 6 = 2x^3 + x - 3 = 0$

whoop, thunderstorm, gotta check out. =p I know I didn't help, but hey, at least you know it wasn't any more effort not to do it implicitly.

Edit: Well, that brushed off quickly. Anyway, all your asking is what your "domain" restrictions should be? You already established a relationship between x and y when you substituted for x^2 and unlike a lot of optimization problems, this one doesn't require to be answers along a certain interval.

3. Alright, so no specific domain... that makes it easier
Thanks,