A cylindrical container starts empty and has a base area of \(200\,\text{cm}^2\). It is filled at a constant rate of \(500\,\text{mL}\) per minute.
a) How many centimeters per minute does the water height increase?
b) Write a function for the water height \(h\), in centimeters, after \(t\) minutes.
c) How would the graph change for a container with a base area of \(100\,\text{cm}^2\) filled at the same rate? Compare the slopes.
d) Suppose a container consists of two stacked cylinders. The lower cylinder has base area \(200\,\text{cm}^2\) and height \(10\,\text{cm}\). The upper cylinder has base area \(100\,\text{cm}^2\) and height \(10\,\text{cm}\). Describe the water-height graph and what happens after \(4\) minutes.
Hints
- Use the cylinder volume relationship \(V=Bh\).
- A constant inflow gives a constant increase in volume.
- Compare how the same volume changes height when the base area is smaller.
Solution
1. Since \(500\,\text{mL}=500\,\text{cm}^3\), use \(V=Bh\): \(500=200h\). Thus, the height increases by \(2.5\,\text{cm}\) per minute.
2. Starting empty, the height function is \(h(t)=2.5t\).
3. For base area \(100\,\text{cm}^2\), \(500=100h\), so the height increases by \(5\,\text{cm}\) per minute. The slope doubles.
4. In the lower section, the graph has slope \(2.5\). The water reaches \(10\,\text{cm}\) after \(10\div2.5=4\) minutes. In the upper section, the slope is \(5\). Therefore, the graph has a corner at \((4, 10)\) and becomes twice as steep after \(4\) minutes.
Answer
a) \(2.5\,\text{cm}\) per minute
b) \(h(t)=2.5t\)
c) The slope doubles to \(5\,\text{cm}\) per minute.
d) The graph has slope \(2.5\) until \((4, 10)\), then slope \(5\); the water rises twice as fast in the narrower upper section.