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?

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