# Yet More Optimization

• Nov 15th 2006, 12:46 PM
Jacobpm64
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?
• Nov 15th 2006, 03:41 PM
topsquark
Quote:

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
• Nov 15th 2006, 03:53 PM
Quick
Quote:

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?
• Nov 15th 2006, 04:23 PM
topsquark
Quote:

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