Emptying a container
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 m3/s. If V(x) m3 is the volume lost, then dV/dt = dV/dx dx/dt = A(x) dx/dt where A(x) m2 is the cross-sectional area. >Huh? Take my word for it, okay? Anyway, since dV/dt = C we have: A(x) = C/(dx/dt). 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 vo, attains a speed v m/s after falling a distance x m, where::
v2 = vo2 + 2gx     g, measured in m/s2, 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 / {vo2 + 2gx}1/2
• However, for our container,
A(x) = π y2
• Hence we get:
y = {C / π }1/2 / { vo2 + 2gx }1/4

>Yeah, so what does such a container look like?
Did you notice Figure 1?