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Thread: Shortest distance Point - surface

  1. #1
    Klo
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    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, ($\displaystyle D^2 = (x - 3)^2 + y^2 + (x^2 + y^2)^2$)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!
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    Quote Originally Posted by Klo View Post
    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, ($\displaystyle D^2 = (x - 3)^2 + y^2 + (x^2 + y^2)^2$)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
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  3. #3
    Klo
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    Perfect, thanks!
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