Sector of a circle transformed into a cone Referencing this and this question, when considering the transformation from a sector to a cone, is it fair to ask how $\frac{d\alpha}{dt}$ and $\frac{dh}{dt}$ are related, where $\alpha$ is the central angle of the missing sector and $h$ is the height of the resulting cone, while the sector is "rising" to create the cone?  
Let:
$R$ be the radius of the sector and slant height of the cone
$r$, $h$ be the radius and height of the cone
$\theta$ be the base angle formed by $r$ and $R$
$\alpha$ be the central angle of the missing sector of the circle
$r'$, $h'$ and $\theta'$ denote the "interim" radius, height and base angle as the sector is "lifted" from the center
$\theta' \to \theta$, $h' \to h$ and $r' \to r$ and the relationship $R^2 = (r')^2 + (h')^2$ always holds as the sector is lifted into the cone.  But, I am not sure if $\alpha$ has any meaning outside the flat sector.  
If not, is there a way to relate the angle between the two radii of the sector as the cone forms to the height of the cone, perhaps spherical coordinates?
 A: Let's consider a sector lying flat, with a missing central angle of $\alpha_0$. If we now linearly (with respect to angle) draw this missing sector closed, then $\alpha(t) = \alpha_0 - \frac{d\alpha}{dt} t$ is the missing sector angle when viewing the rising cone from above, on-axis. Let $\alpha(1) = 0$ (when $t=1$, we make the final cone), so $\alpha(t) = \alpha_0(1-t)$.
At any instant, we need to relate $\alpha(t)$ and the height $h(t)$ of the partly formed cone.
$$ [h(t)]^2 = R^2-[r(t)]^2 $$
So now, we need to know how the base radius changes. The radius is linearly proportional to circumference, which is linearly proportional to the central angle of the non-missing part. Therefore, $r(t) = R - (R - \frac{2\pi-\alpha}{2\pi}R)t$. This is enough now to relate the quantities of interest. Taking derivatives on both sides of the above equation,
$$ 2h\frac{dh}{dt} = -2r\frac{dr}{dt} = 2r (R - \frac{2\pi-\alpha}{2\pi}R)\frac{dr}{dt} = 2rR\frac{\alpha}{2\pi}\frac{dr}{dt}$$
$$ \frac{dh}{dt} = R\frac{\alpha}{2\pi}\frac{r}{h}\frac{d\alpha}{dt}= R\frac{\alpha}{2\pi}\frac{R - (R - \frac{2\pi-\alpha}{2\pi}R)}{\sqrt{R^2-[R - (R - \frac{2\pi-\alpha}{2\pi}R)]^2}}\frac{d\alpha}{dt} $$
$$ \frac{dh}{dt} = R\frac{\alpha}{2\pi}\frac{1-\frac{\alpha}{2\pi}}{\sqrt{1-(1 - \frac{\alpha}{2\pi})^2}}\frac{d\alpha}{dt} $$
To get an explicit time dependence, substitute the expression for $\alpha(t)$ everywhere you see $\alpha$ in the last equation.
