Function to describe teardrop shape If I fill a plastic ziploc-shaped bag with water, the cross section profile should be sort of teardrop shaped (assuming we ignore the edge effects of the bag being sealed on the sides as well as the top and bottom). The bag should "sag"/get wider until to get the center of gravity as low as possible. Initially, getting wider will let more water towards the bottom but eventually this is offset by the bottom of the bag moving up (because the sides are fixed length).
Is there a common function that describes the shape the cross-section of the bag makes?
I would guess the bottom is a parabola, since gravity likes to make parabolas. Then I would guess the top is linear because its under tension. But I have no idea what the transition region might look like and whether you could put those two together into a nice function.
 A: This involves physics of equilibrium to form an ODE which decides its shape.
Assumptions are:  Bag is full, does not stretch, too long compared to height, force per unit length $N$ a.k.a. surface tension is proportional to height of water (of  density $\gamma$) gravity load column above it, the hydraulic pressure is proportional to vertical depth (y) only, $\kappa$ membrane curvature.. and the like. Equilibrium requires
$$  N \kappa = p = \gamma \, y, \; \; \; \frac {y''}{(1+y'^2)^{3/2}} =  k y $$
Integration leads to  $Elastica$ shape. At the support point there is no depth of water weighing down on it so it is straight as you expected. Deeper down shape is described by elliptic integrals. At the deepest symmetry point $y^{'} $ is zero, so is a parabola locally.
The ODE gives insight as to its shape. Where the water level stops there and above that the bag is straight.
In Mechanics of materials text-book  Den Hartog mentions shapes of  heavy mercury drops once adopted for construction of large water tanks.
EDIT1:
To an extent, the static equilibrium balance between a helium filled lighter than air balloons floating in air is quite similar. They accordingly have an upside down shape of a teardrop.
EDIT 2:
The balloon gets  meridian shape  below when curvature changes along axis of symmetry per $ \kappa = 6 ( z -0.7): $

A: If you want something that looks like a drop but is not physically based, try the implicit equation $ax^2-(1-y)^3(1+y)=0$. The shape varies a little with $a$. Here it is for $a=4$, courtesy of WA.

