motivated by a discussion on a calculus forum
Here's an interesting problem:
 A container (such as is illustrated in Figure 1) is filled with water.
 The water is removed from the bottom at a constant rate of C m^{3}/s.
 If V(x) m^{3} is the volume lost, then dV/dt = dV/dx dx/dt = A(x) dx/dt where A(x) m^{2} is the crosssectional area.
>Huh?
Take my word for it, okay?
Anyway, since dV/dt = C we have:
 Figure 1 
Here's the problem:
Find a container shape such that the speed of the water level (that's dx/dt) is the same as the speed of a body falling freely under the influence of gravity.
>Huh?
Sorta like this. Now, pay attention.
 A freely falling body, starting with an initial speed of v_{o}, attains a speed v m/s after falling a distance x m, where::
v^{2} = v_{o}^{2} + 2gx
g, measured in m/s^{2}, is the acceleration of gravity
 Setting dx/dt = v (so the speed of the water level equals the speed of the falling body) gives:
A(x) = C/(dx/dt) = C / {v_{o}^{2} + 2gx}^{1/2}
 However, for our container,
A(x) = π y^{2}
 Hence we get:
y = {C / π }^{1/2} / { v_{o}^{2} + 2gx }^{1/4}
>Yeah, so what does such a container look like?
Did you notice Figure 1?
