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anonymous

  • 5 years ago

Find the minimum distance from the surface x^2+y^2-z^2=49 to the origin.

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  1. nowhereman
    • 5 years ago
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    The distance to the origin is \[\sqrt{x^2+y^2+z^2}\] so you can minimize this or the square of it. But \[x^2 + y^2 + z^2 = x^2 + y^2 - z^2 + 2z^2 = 49 + 2 z^2\] so all you have to do is find the smallest possible z for which a point lies in the surface.

  2. anonymous
    • 5 years ago
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    You can use Lagrange multipliers to solve this. You need to find the minimum of the distance function from (x,y,z) on the surface to the origin; that is, minimise \[D^2=x^2+y^2+z^2\]subject to the constraint,\[g(x,y,z)=x^2+y^2-z^2-49=0\]by finding\[\nabla D^2=\lambda \nabla g\]The left-hand side is\[\nabla D^2 = 2D \nabla D = 2D(2x,2y,2z)\]

  3. anonymous
    • 5 years ago
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    The right-hand side is\[\lambda \nabla g = \lambda (2x,2y,-2z) \]Equating the two we find the following:\[x=\frac{\lambda }{2D}x\]\[y=\frac{\lambda}{2D}y\]\[z=-\frac{\lambda}{2D}z\]Looking at the last expression, either z=0 or lambda = 0 (D not zero). If lambda = 0, then all of x and y will be zero also. This cannot happen, since x=0, y=0, z=0 does not lie on the surface. The other alternative is for \[\frac{\lambda }{2D}=1\]for then\[x=x,y=y, z=0\] (lambda = -1 will yield a similar conclusion that will lead to the same answer in the end).

  4. anonymous
    • 5 years ago
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    Do the minimum distance is \[D^2=x^2+y^2+0^2=x^2+y^2\]But from the constraint, we know that\[g(x,y,0)=x^2+y^2-49=0 \rightarrow x^2+y^2=49\]So the minimum distance is given by the set of all (x,y) lying on the circle of radius 7. That is, the distance is 7.

  5. anonymous
    • 5 years ago
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    Note, I assumed this was the minimum. You can check this by plugging in values either side.

  6. anonymous
    • 5 years ago
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    *So (the minimum distance)...

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