anonymous
  • anonymous
Find the minimum distance from the surface x^2+y^2-z^2=49 to the origin.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
nowhereman
  • 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.
anonymous
  • 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\[\nabla D^2 = 2D \nabla D = 2D(2x,2y,2z)\]
anonymous
  • 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).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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.
anonymous
  • anonymous
Note, I assumed this was the minimum. You can check this by plugging in values either side.
anonymous
  • anonymous
*So (the minimum distance)...

Looking for something else?

Not the answer you are looking for? Search for more explanations.