Question

Find the distance between the point (2,-2,-9) and the paraboloid \[z = -\frac{1}{4}(x^2+y^2). \]

Answer 1

There are a number of ways to do this, might I ask what course you are taking?

Answer 2

\[distance^2= \left( x-2 \right)^2+(y+2)^2+(z+9)^2\]
Substitute z and find minimum

Answer 3

teaching AIME prep, so anything is fair game

Answer 4

Well, you can either substitute for z in the distance formula, take partial derivatives and set them to zero, find the critical point, and substitute it into the distance formula, or you could do a nice application of Lagrange Multipliers if you're familiar with that. I would personally prefer the latter just because I like the technique, but it's up to you. We can help with whichever you'd prefer.

Answer 5

Lagrange Multipliers looks easier. Proofs that avoid calculus entirely are also welcomed.

Answer 6

Fair enough. Lagrange multipliers:

The function to minimize is
\[D^2 = (x-2)^2 + (y+2)^2 + (z+9)^2 \]
subject to the constraint
\[F(x,y,z) = \frac{1}{4}(x^2+y^2) + z = 0 \]
We'll say that
\[ \nabla D^2 = \lambda \nabla F \]
which yields the following equations:
\[2x-4 = \lambda \frac{x}{2} \]
\[2y+4 = \lambda \frac{y}{2} \]
\[2z + 18 = \lambda \]

Answer 7

We can add the first equation to the second and multiply by two to give
\[4(x+y) = \lambda(x+y) \]
so either x= -y or lambda = 4. If we try lambda = 4 in either of the first two equations, though, we find immediately that it doesn't work. Oh well, such is life.

Answer 8

So apparently x = -y. If we plug that into our constraint equation, we find that
\[z = \frac{-1}{2} x^2 \] But if we plug this into our third equation, we get
\[-x^2 + 18 = \lambda \]
Almost done. Let's substitute that into equation one to yield
\[2x - 4 = \frac{-x^3}{2} + 9x \]
rearranging and multplying by 2 yields
\[x^3 - 14x - 8 = 0 \] which yields three solutions:
\[x = \{4, \sqrt{2}-2, -\sqrt{2} - 2\} \]
With dreams of simplicity, lets try x = 4...
Then from equation 1, lambda = 2. Wonderfully, this also works in the second equation. Thus x = 4, y = -4, and z = 1/4(16+16) = 32/4 = 8.

Answer 9

Confirmation: http://www.wolframalpha.com/input/?i=minimize+%28x-2%29^2+%2B+%28y%2B2%29^2+%2B+%28z%2B9%29^2+subject+to+1%2F4%28x^2%2By^2%29%2Bz+%3D+0

Answer 10

Oh, oops. And therefore our answer is
\[D = \sqrt{(4-2)^2 + (-4+2)^2 + (-8+9)^2} = \sqrt{4 + 4 +1} = 3 \]
Forgive the typo, that should be z = -8 above...

Answer 11

^= 3....