I mentioned to my (high school) students today that the intersection of a plane and a cone gives a conic section.
One asked whether if you 'unroll the cone' the conic section becomes a straight line on the resulting circular sector.
I can find examples that show this is false in general, but are there instances where it is true?
More specifically, given the cone defined by $\alpha\rho=z$ in cylindrical polars and the points $A=(\rho_0,0,\alpha\rho_0), B=(\rho_0,\phi_0,\alpha\rho_0)$, is the geodesic from $A$ to $B$ ever a planar curve?
Let line $PH$ be a geodesic line on the lateral surface of a cone (I'll suppose $PH$ does not pass through the vertex $O$ of the cone), and $H$ its point nearest to $O$. Let $OH=d$ and $t=\angle POH$. Then: $OP=d/\cos t$.
If $a=OM=ON$ is the slant height of the cone, then (see figures): $$ \text{arc}\ MN = ta=\theta a\sin\alpha, \quad\text{and:}\quad \theta={t\over\sin\alpha}, $$ where $\alpha$ is the semi-aperture of the cone. It follows that we can write the coordinates of point $P$ as: $$ P(t)=\left({d\over\cos t}\sin\alpha\cos\left({t\over\sin\alpha}\right), {d\over\cos t}\sin\alpha\sin\left({t\over\sin\alpha}\right), {d\over\cos t}\cos\alpha\right). $$ The above equation defines then the curve in space corresponding to geodesic $PH$; parameter $t$ can take any value in $(-\pi/2, \pi/2)$.
This is a plane curve only if its torsion $\tau(t)$ vanishes for all $t$. But a straightforward calculation gives: $$ \tau(t)={1\over d}{\cot\alpha \sin t \cos^2 t}, $$ which vanishes only for $t=0$. Hence it cannot be a plane curve.