Finding the curve on a surface with a specific curvature I wish to find a curve $\langle x(t), y(t), z(t) \rangle$ on the surface $x + y + z = (x-y)^2 + (y-z)^2$ whose curvature is 1/2 for all $t$.
I am struggling with how to proceed. I initially tried expanding the right side and attempted to parametrize but because of the $yz$ and $xy$ terms, I wasn't able to do so. I would greatly appreciate any approaches to this problem. Thank you!
 A: By inspection, if the given surface is cut by a plane where $y$ is constant, the equation of the resulting curve is a circle (possibly degenerate) parallel to the $xz$-plane.

Leaving $y$ as an unknown constant and completing the square in the usual way, the equation
$$
x+y+z = (x-y)^2+(y-z)^2
$$
can be rewritten as
$$
(x-h)^2+(z-k)^2=r^2
$$
where
$$
\left\lbrace
\begin{align*}
h&=y+\frac{1}{2}\\[4pt]
k&=y+\frac{1}{2}\\[4pt]
r^2&=3y+\frac{1}{2}
\end{align*}
\right.
$$
For the circle to have curvature ${\large{\frac{1}{2}}}$, we want $r=2$.

Solving the equation $r^2 = 4$ for $y$ yields $y={\large{\frac{7}{6}}}$, hence $h=k={\large{\frac{5}{3}}}$.

Thus the curve satisfying the system
$$
\left\lbrace
\begin{align*}
&y=\frac{7}{6}\\[4pt]
&\Bigl(x-\frac{5}{3}\Bigr)^2+\Bigl(z-\frac{5}{3}\Bigr)^2=4\\[4pt]
\end{align*}
\right.
$$
is a circle of radius $2$, hence curvature ${\large{\frac{1}{2}}}$, lying on the given surface.

The circle can be expressed parametrically as
$$
\left\lbrace
\begin{align*}
x&=\frac{5}{3}+2\cos(t)
\qquad\qquad\;\,
\\[4pt]
y&=\frac{7}{6}\\[4pt]
z&=\frac{5}{3}+2\sin(t)\\[4pt]
\end{align*}
\right.
$$
where $0\le t < 2\pi$.
A: Hint:
Let $u:=\dfrac{x-y}{\sqrt2}$ and $v:=\dfrac{y-z}{\sqrt2}$, which defines a rotation. The surface is
$$\frac{u-v}{\sqrt2}+\frac32y=u^2+v^2,$$ a family of circles of radius
$$\frac32y+\frac14$$ and center $$\left(\frac1{2\sqrt2},-\frac1{2\sqrt2}\right)$$ in the $(u,y,v)$ coordinates. (The surface is a cone of revolution.)
