# Shortest distance Point - surface

• Jun 25th 2010, 04:08 AM
Klo
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? (Thinking)

I was said to differentiate the distance D between B and A, (\$\displaystyle 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!
• Jun 25th 2010, 04:33 AM
tonio
Quote:

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? (Thinking)

I was said to differentiate the distance D between B and A, (\$\displaystyle 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 \$\displaystyle (x,y,x^2+y^2)\$, so the square of the distance from such a point to \$\displaystyle (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 \$\displaystyle D^2\$ instead of \$\displaystyle D\$ is that both these functions have their minimal-maximal points exactly at the same points (proof? It's easy...), so:

\$\displaystyle 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\$

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

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

\$\displaystyle 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 \$\displaystyle y=0\$, and then from the 1st

eq. we get

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

I hope now you know how to check whether the critical point \$\displaystyle (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 \$\displaystyle (1,0,1)\$ on your surface.

Tonio
• Jun 25th 2010, 05:12 AM
Klo
Perfect, thanks!