# Thread: Shortest distance Point - surface

1. ## Shortest distance Point - surface

Im supposed to find the point B(x,y,z) which is a point on the surface z = x^2 + y^2 and is closest to the point A(3,0,0,)?

HOW?

I was said to differentiate the distance D between B and A, ( $D^2 = (x - 3)^2 + y^2 + (x^2 + y^2)^2$)then the shortest distance will be for dD/dx = 0 & dD/dy = 0. Can anyone tell me why it is so?

Also i believe that posing y=0 is a good thought... then the question would be to find the distance from A to the curve z=x^2

Thanks!

2. Originally Posted by Klo
Im supposed to find the point B(x,y,z) which is a point on the surface z = x^2 + y^2 and is closest to the point A(3,0,0,)?

HOW?

I was said to differentiate the distance D between B and A, ( $D^2 = (x - 3)^2 + y^2 + (x^2 + y^2)^2$)then the shortest distance will be for dD/dx = 0 & dD/dy = 0. Can anyone tell me why it is so?

Also i believe that posing y=0 is a good thought... then the question would be to find the distance from A to the curve z=x^2

Thanks!
A generic point on that surface has the form $(x,y,x^2+y^2)$, so the square of the distance from such a point to $(3,0,0)$ is the expression you wrote, and now you've a minimum problem of a two-variable function.
The reason you can work with $D^2$ instead of $D$ is that both these functions have their minimal-maximal points exactly at the same points (proof? It's easy...), so:

$f(x,y):= (x-3)^2+y^2+(x^2+y^2)^2\Longrightarrow f_x=2(x-3)+4x(x^2+y^2)=0$

$f_y=2y+4y(x^2+y^2)=0$ , and we get the system of non-linear eq's.:

$4x^3+(4y^2+2)x=6$

$4y^3+(4x^2+2)y=0\iff y=0\,\,\,or\,\,\,4y^2+4x^2+2=0$ . Since this last eq. has

no real solutions (sum of positive terms on the left), it must be $y=0$, and then from the 1st

eq. we get

$0=2x^3+x-3=(x-1)(2x^2+2x+3)$ , whose only real solution is $x=1$.

I hope now you know how to check whether the critical point $(1,0)$ we get for our function is a maximal, minimal, saddle point or none of these (it, is a minimum, of course), and thus you get the point $(1,0,1)$ on your surface.

Tonio

3. Perfect, thanks!