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Math Help - shortest distance question

  1. #1
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    shortest distance question

    Find the point on the surface y^2=9+xz that are closest to the origin.

    so far I have

     d= \sqrt{x^2+y^2+z^2} = \sqrt{x^2+9+xz+z^2}

     d^2=x^2+9+xz+z^2

    F_x = 2x+z \ \mbox{and} \ F_z = 2z+x

    F_x = 2x+z = 0\ \mbox{and} \ F_z = 2z+x = 0

    x = 0\ \mbox{and} \ z = 0

    F_{xx} = 2, \ F_{zz} = 2, \ F_{xz} = F_{zx} = 1

     D=(2)(2) -1^1 = 3 > 0 \ mbox{and} \ F_{xx} = 2 > 0 \therefore \mbox{minimum}

    and the shortest distance would be \sqrt{0^2+9+(0)(0)+0^2} = 3

    is this correct?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Hmm

    Quote Originally Posted by lllll View Post
    Find the point on the surface y^2=9+xz that are closest to the origin.

    so far I have

     d= \sqrt{x^2+y^2+z^2} = \sqrt{x^2+9+xz+z^2}

     d^2=x^2+9+xz+z^2

    F_x = 2x+z \ \mbox{and} \ F_z = 2z+x

    F_x = 2x+z = 0\ \mbox{and} \ F_z = 2z+x = 0

    x = 0\ \mbox{and} \ z = 0

    F_{xx} = 2, \ F_{zz} = 2, \ F_{xz} = F_{zx} = 1

     D=(2)(2) -1^1 = 3 > 0 \ mbox{and} \ F_{xx} = 2 > 0 \therefore \mbox{minimum}

    and the shortest distance would be \sqrt{0^2+9+(0)(0)+0^2} = 3

    is this correct?
    At a precursory glance...it looks great!
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