# closest point on graph

• Dec 17th 2006, 04:54 PM
jeph
closest point on graph
Find the Point(s), P(x,y) on the graph of y=4-x^2 so that the distance between two points P and Q(0,2) is a minimum.

i think what you do is

d=squareroot[(x-0)+(y-2)]

d^2=[(x-0)+(y-2)]

then you plug in either x or y and differentiate?

the answer is supposed to be sqareroot(3/2), -squareroot(3/2), and 5/2
• Dec 17th 2006, 07:18 PM
ThePerfectHacker
Quote:

Originally Posted by jeph
Find the Point(s), P(x,y) on the graph of y=4-x^2 so that the distance between two points P and Q(0,2) is a minimum.

i think what you do is

d=squareroot[(x-0)+(y-2)]

d^2=[(x-0)+(y-2)]

then you plug in either x or y and differentiate?

the answer is supposed to be sqareroot(3/2), -squareroot(3/2), and 5/2

Correct, you need to minimize,
$\displaystyle x^2+(y-2)^2$
Thus,
$\displaystyle x^2+(4-x^2-2)^2$
$\displaystyle x^2+(2-x^2)^2$
$\displaystyle x^2+4-4x^2+x^4$
$\displaystyle x^4-3x^2+4$
Call this a function,
$\displaystyle f(x)=x^4-3x^2+4$
Thus, how do you minimize a differenciable function?
You make its derivatives zero,
$\displaystyle f'(x)=4x^3-6x$
Thus,
$\displaystyle 2x(2x^2-3)=0$
Thus,
$\displaystyle x=0$
$\displaystyle 2x^2-3\to x=\pm \sqrt{3/2}$
• Dec 17th 2006, 07:42 PM
jeph
ok I understand that now, but how did he get the 5/2?
• Dec 17th 2006, 07:48 PM
ThePerfectHacker
Quote:

Originally Posted by jeph
ok I understand that now, but how did he get the 5/2?

I do not know.
I did not get it when I did the problem.
• Dec 17th 2006, 09:31 PM
earboth
Quote:

Originally Posted by jeph
ok I understand that now, but how did he get the 5/2?

Hello, Jeph,

you were looking for a point. Therefore you need two numbers: the x-value, which TPH has already calculated for you and the y-value.
Plug in the x-value into the equation of your function:

$\displaystyle y=4- \left( \sqrt{\frac{3}{2}} \right)^2 = \frac{5}{2}$

EB
• Dec 18th 2006, 11:29 AM
jeph
oh! thanks =D