(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant 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.

3. 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Vector Calculus - Closest point to the origin

**Physics Forums | Science Articles, Homework Help, Discussion**