calculus forum
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?
- 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*, is the acceleration of gravity^{2} - 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?
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