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Question: A hemispherical bowl is being filled with water at a constant volumetric rate. The level of water in the bowl increases

  1. in direct proportion to time,
  2. in inverse proportion to time,
  3. faster than direct proportion to time,
  4. slower than direct proportion to time

To solve this problem, let us consider a hemisphere with radius $r$ and volume $V$. Then $V=\frac{2}{3}\pi r^3$. This implies $\frac{dV}{dt}=2\pi r^2 \frac{dr}{dt}$. Is it possible to solve the problem in this way? I don't know how to find the condition. Please suggest.

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  • $\begingroup$ The radius of the hemisphere does not change or does it? $\endgroup$ – zoli Dec 16 '15 at 18:04
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What is important is not so much the radius of the hemisphere as the radius of the top surface of the water. Let's call that $R(t)$ If you add a volume $v$ of water, it raises the depth by about $\frac v{\pi R(t)^2}$ (ignoring the small increase in the radius). As $R(t)$ is increasing, the rate the depth increases is decreasing. This will be true in any bowl that has the sides going outward.

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The volumetric rate ( say in $ft^3/s$ ) is at every moment $\frac{d V}{dt} = \frac{d h}{dt} \cdot \text{Area of cross section}$.

If the volumetric rate is constant then the rate of change of the height at a certain moment is inverse proportional to the cross section at that moment.

Since the cross section is increasing as $h$ is increasing ( hemisphere), the rate of change of $h$ is decreasing ( while still being positive), so $h$ is growing slower than direct proportional to time.

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