Question: A hemispherical bowl is being filled with water at a constant volumetric rate. The level of water in the bowl increases
- in direct proportion to time,
- in inverse proportion to time,
- faster than direct proportion to time,
- 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.
The red circled triangle has sides $R$ (the radius of the sphere), $r$ (the radius of the water surface) and $R-h$ (where $h$ is the water height). Pythagoras tells you that $R^2 = r^2 + (R-h)^2.$
Best Answer
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.