# Math Help - Yet More Optimization

1. ## Yet More Optimization

Find the coordinates of the point on the parabola $y = x^2$ which is the closest to the point (3,0).

I'm just wondering if I did this correctly:
a point on the curve would be $(x,x^2)$

So, i found the distance between $(3,0)$ and $(x,x^2)$.
$d = \sqrt{(0-x^2)^2+(3-x)^2}$
$d = \sqrt{x^4 + 9 - 6x + x^2}$

Then, I took the derivative to minimize the distance.
$d' = \frac{1}{2} (x^4 + 9 - 6x + x^2)^{-\frac{1}{2}} * (4x^3 - 6 + 2x)$

Set it equal to zero to find where the minimum occurs.
$\frac{4x^3 - 6 + 2x}{2 \sqrt{x^4 + 9 - 6x + x^2}} = 0$
$4x^3 - 6 + 2x = 0$
$x = 1$

Plug it back in to get the y-value
$y(1) = 1^2 = 1$

The point on $y = x^2$ which is closest to the point $(3,0)$ is $(1,1)$.

Is this correct?

2. Originally Posted by Jacobpm64
Find the coordinates of the point on the parabola $y = x^2$ which is the closest to the point (3,0).

I'm just wondering if I did this correctly:
a point on the curve would be $(x,x^2)$

So, i found the distance between $(3,0)$ and $(x,x^2)$.
$d = \sqrt{(0-x^2)^2+(3-x)^2}$
$d = \sqrt{x^4 + 9 - 6x + x^2}$

Then, I took the derivative to minimize the distance.
$d' = \frac{1}{2} (x^4 + 9 - 6x + x^2)^{-\frac{1}{2}} * (4x^3 - 6 + 2x)$

Set it equal to zero to find where the minimum occurs.
$\frac{4x^3 - 6 + 2x}{2 \sqrt{x^4 + 9 - 6x + x^2}} = 0$
$4x^3 - 6 + 2x = 0$
$x = 1$

Plug it back in to get the y-value
$y(1) = 1^2 = 1$

The point on $y = x^2$ which is closest to the point $(3,0)$ is $(1,1)$.

Is this correct?
I didn't check the numbers, but the format is correct.

-Dan

3. Originally Posted by topsquark
I didn't check the numbers, but the format is correct.

-Dan
You do realize that you can insert points into the graph through the graph program?

4. Originally Posted by Quick
You do realize that you can insert points into the graph through the graph program?
Thank you. I figured there had to be a way, but I didn't want to bother looking it up while answering a question.

-Dan