## anonymous 5 years ago Find the minimum distance from the surface x^2+y^2-z^2=49 to the origin.

1. nowhereman

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

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

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

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

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

6. anonymous

*So (the minimum distance)...