Thread: 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

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,

Thus,




Call this a function,

Thus, how do you minimize a differenciable function?
You make its derivatives zero,

Thus,

Thus,

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

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.

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:



EB

oh! thanks =D