Question

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

Answer 1

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.

Answer 2

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\[\nabla D^2 = 2D \nabla D = 2D(2x,2y,2z)\]

Answer 3

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).

Answer 4

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.

Answer 5

Note, I assumed this was the minimum. You can check this by plugging in values either side.

Answer 6

*So (the minimum distance)...