Vector Calculus - Closest point to the origin

[number0]

Homework Statement

A curve C in space is defined implicitly on the cylinder x^2+y^2=1 by the additional equation: x^2-xy+y^2-z^2=1. Find the point or points on C closest to the origin.

Homework Equations

d = ((x-x0)+(y-y0)+(z-z0))^(1/2) - This is the distance formula.
Please note that I did NOT learn Lagrange Multipliers, yet - it is the next section in my math book.

The Attempt at a Solution

First, I used the distance formula: ((x-x0)^2+(y-y0)^2+(z-z0)^2)^(1/2).
where (x0,y0,z0) = (0,0,0) - the origin.

I solved for z from the additional equation: x^2-xy+y^2-z^2=1.

So the equation now looks like this: ((x^2+y^2+(x^2-xy+y^2-1))^(1/2)

I then square the entire equation (so that I can derive easier) because the extrema points remained the same if the equation were not squared.

I solved the partials for x and y. And for some reason, I got fx=4x-y and fy=4y-x. Thus the point of intersection, which I believe to be incorrect, is (0,0).

I plugged (0,0) back into the additional equation and I have square root by a negative number!

And now... I am stuck.

Can anyone please show me how to approach this problem using the derivation of the distance formula and NOT using Lagrange Multipliers?

 

[tiny-tim]

Hi number0! Welcome to PF!

(have a square-root: √ and try using the X² and X₂ tags just above the Reply box )

Hint: you know x² + y² = 1, so you really only need to minimise z², and you can use x² + y² = 1 in the definition of C.

 

[number0]

Thanks tiny-tim!