- #1
number0
- 104
- 0
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?