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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.

This is an optimization problem. I tried constraining the distance $d = (x^2 + y^2 + z^2)^{1/2}$ to $x^2 - xy + y^2 - z^2 = 1$ by substituting $z^2$ into the distance equation and then finding the partial derivatives but I get $x=0, y=0$ which seems incorrect. Alternatively, I tried plugging $x^2 + y^2 = 1$ into the additional equation and then tried the same approach. I still got $x=0$ and $y=0$.

Zev Chonoles
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iuppiter
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1 Answers1

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The square of the distance from the origin is $x^2+y^2+z^2=1+z^2$ for any point on the cylinder. Any point on the curve satisfies $xy+z^2=0$, or $z^2=-xy$, so the square of the distance to the origin is $1-xy$, where $x^2+y^2=1$.

Brian M. Scott
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