A tank is being filled with water using a pump that is old and slows down as it runs. The table below gives the rate at which the pump pumps at ten-minute intervals. If the tank initially has 570 gallons of water in it, approximately how much water is in the tank after 90 minutes?
|Elapsed time (minutes)||0||10||20||30||40||50||60||70||80||90|
|Rate (gallons / minute)||42||40||38||35||35||32||28||20||19||10|
And so, integration begins.
Ask your students to do this problem alone. When they are ready (after a few minutes) collect their opinions. They will not all be the same (we hope, because there is more than one reasonable way to approximate the amount). Ask exactly how they got their answers and what assumptions they made. Be sure they always include units (gallons). Here are some points to make in your discussion – points that we hope the kids will make and you can just “underline.”
- Answers between 3140 and 3460 gallons are reasonable. Other answers in that range are acceptable. They will not use terms like “left-sum”, “right sum” and “trapezoidal rule” because they do not know them yet, but their explanations should amount to the same thing. An answer of 3300 gallons may be popular; it is the average of the other two, but students may not have gotten it by averaging 3140 and 3460.
- Ask if they think their estimate is too large or too small and why they think that.
- Ask what they need to know to give a better approximation – more and shorter time intervals.
- Assumptions: If they added 570 + 42(10) + 40(10) + … +19(10) they are assuming that the pump ran at each rate for the full ten minutes and then suddenly dropped to the next. Other will assume the rate dropped immediately and ran at the slower rate for the 10 minutes. Some students will assume the rate dropped evenly over each 10-minute interval and use the average of the rates at the ends of each interval (570 + 41(10) + 39(10) + … 14.5(10) = 3300).
- What is the 570 gallons in the problem for? Well, of course to foreshadow the idea of an initial condition. Hopefully, someone will forget to include it and you can point it out.
- With luck someone will begin by graphing the data. If no one does, you should suggest it; (as always) to help them see what they are doing graphically. They are figuring the “areas” of rectangles whose height is the rate in gallons/minute and whose width is the time in minutes. Thus the “area” is not really an area but a volume ((gal/min)(min) = gallons). In addition to the unit analysis, graphing is important since you will soon be finding the area between the graph of a function and the x-axis in just this same manner.
Follow up: Flying to Integrationland
Be sure to check the “Thoughts on ‘The Old Pump'” in the comments section below.
Revised from a post of November 30, 2012.