This problem stems from something I have seen a lot of on my cross-country motorcycle trip: irrigation systems. Basically they use a long arm that rotates around a center pole, with sprinklers spaced at even intervals. I studied them as I rode along, trying to figure out how much water should be dumped from each sprinkler, since the sprinklers on the far end of the arm cover a lot more area than the ones near the middle.If a sprinkler arm is 450 feet long, and the sprinklers are 10 yards apart, for each gallon of water dumped by the sprinkler 10 yards from the center (which would be the second sprinkler, since the first is attached to the center pole), how much water needs to be dumped by the last sprinkler?
- Annie Fetter
Annie says: Solutions
There are a number of ways to solve this. You can use circumference, or area, or a couple of other things. Greg Pack did it the "thorough" way, also known as "the way Annie solved it" :-), as did Leif Linden. Brian Gordan was also thorough, and pointed out very clearly why the "proportion" method works. Justin Lam pointed out a different interpretation along with the "conventional" answer. I hadn't thought of it, and at first thought Justin was wrong, but indeed if you look at it his way, there is a different answer. He provided it along with the "right" answer, and I would have given him credit in either case, since his explanations were so complete.
Below are some of the more thorough solutions received. A list of all the people who got this problem right and most of the solutions are also available.
Leif Linden Grade: 9 School: Germantown Academy, Fort Washington, Pennsylvania The first weekly problem is quite complicated. I listed the following steps as best I could: Since the irrigation arm is 450 feet long, it is 150 yds long. The sprinklers are 10 yards apart (thus there are 16 sprinklers total). Each sprinkler needs to water halfway between itself and the next sprinkler. Thus, each has a reach of five yards as the arm traverses. The area covered by the first sprinkler is pi*5yds squared (78.5 square yds). The area covered by the second sprinkler as the arm rolls is pi*15yds squared minus the area of the first sprinkler (628 sq yds). The answer: 628 sq. yards defines the amount of area covered by the second sprinkler. This will be your standard measurement for the amount covered by another sprinkler. Now find the area covered by the last sprinkler. Since the arm is 150 yds long (i.e. where the last sprinler sits) the distance from the center reached by water from the last sprinkler is between 145 and 155 yds. In order to find the area that the last sprinkler covers, find the area of the inner circle (the one 145yds from the center) and subtract it from the outer circle: pi*155yds squared minus pi*145yds squared = 3.14(24025-21025). 9420 sq. yds is the area covered by the outermost sprinkler. For every gallon used by the second sprinkler, how many gallons are needed by the last sprinkler? (we are assuming that each sq. yd. covered by each sprinkler will receive the same amount of water). Set up the ratio: 1 gal / 628 sq yds = x gal / 9420 sq yds. Then x = 15 gallons.
Justin Lam Grade: 8 School: Sequoia Middle School, Oakland, California It depends on which direction the sprinklers spray. (Solution 1) Suppose each sprinkler sprays straight upward covering the ground 5 yards outside and 5 yards inside from the sprinkler, that is, each sprinkler (except the 1st one) covers the ground in the shape of a doughnut ring with an outside radius, from the center, (location of that sprinkler)+5 and an inside radius of (location of that sprinkler)-5. Sprinkler | Area Covered |Gallons Needed| ----------|-------------------------------|--------------| 2 | PI(15^2-5^2) = 200PI | 1 | ----------|-------------------------------|--------------| 3 | PI(25^2-15^2) = 400PI | 2 | ----------|-------------------------------|--------------| 4 | PI(35^2-25^2) = 600PI | 3 | ----------|-------------------------------|--------------| n | PI((5*(2n-1))^2-(5*(2n-3))^2) | n-1 | ----------|-------------------------------|--------------| 16 | PI(155^2-145^2) = 3000PI | 15 | ----------|-------------------------------|--------------| (Solution 2) Suppose each sprinkler sprays outward covering the ground in the shape of a doughnut ring between the next sprinkler and this sprinkler. Then the outer radius is (location + 10) and the inner radius is (location). Sprinkler| Area Covered | Gallons Needed| ---------|----------------------------------|---------------| 2 | PI(20^2-10^2) = 300PI | 1 | ---------|----------------------------------|---------------| 3 | PI(30^2-20^2) = 500PI | 500/300 = 5/3 | ---------|----------------------------------|---------------| 4 | PI(40^2-30^2) = 700PI | 700/300 = 7/3 | ---------|----------------------------------|---------------| n |PI((10n)^2-(10n-10)^2)=100PI(2n-1)| (2n-1)/3 | ---------|----------------------------------|---------------| 16 | PI(160^2-150^2) = 3100PI |3100/300 = 31/3| ---------|----------------------------------|---------------|
Gregory Pack Teacher, Pensacola Catholic HS, Pensacola, Florida Assuming exactly full coverage, each sprinkler head covers a diameter of ten yards. Hence, the first sprinkler, located at 10 yards, waters the difference between two circles: the inner circle is 5 yards radius and the outer is 15 yards. Using A = pi times radius squared, this area is 225 - 25 = 200 square yards. Assuming the last sprinkler covers the section between circles of radii 145 and 155 yards, this area is 24025 - 21025 = 3000. Therefore, assuming water pumped is proportional to area covered, the last sprinkler must pump 3000/200 or 15 times as much water as the first sprinkler. In other words, for each gallon dropped by the first sprinkler, 15 gallons needs to be dropped by the last sprinkler.
Brian Gordon Dartmouth '92 School: Wethersfield, Connecticut This problem presented me with two ideas, based on how we interpret the sprinkling action. My first idea is this: Each sprinkler on the arm is responsible for a ten yard wide ring. The second sprinkler, ten yards from the center, is responsible for a ring with outer radius 15 yards and inner radius 5 yards. That's an area of 225pi - 25pi = 200pi square yards. The last sprinkler, 150 yards from center, is responsible for a ring with radii 155 and 145 yards, so that area is 24025pi - 21025pi = 3000pi square yards. Since this is 15 times the area, you'd need 15 gallons for every gallon on the second sprinkler. Now, if the sprinkler only traces out a linear path, it doesn't sway from side to side, covering a two-dimensional ring, but only irrigates below itself like a leaking hose. Then the inner sprinkler traverses a path of 2 * pi * radius = 20pi yards, while the outer path is 2 * pi * 150 yards = 300pi yards. Again, this is 15 times as long, so we still need 15 gallons for every gallon inside. The reason the answer is the same is simple. The area of the ring, although a two-dimensional measurement, is proportional to the radius, not its square: let R be the radius of the center of the ring, and let r be the displacement from that line for the width of the ring (in other words, the inner radius is R - r, while the outer radius is R + r). Then the area of the ring is pi*( (R+r)^2 - (R-r)^2) = pi*(R^2 + 2Rr +r^2 - R^2 +2Rr - r^2) = 4Rr*pi. That's proportional to R linearly, and R is the radius you'd use to compute the corresponding circumference as discussed in the second approach. --bri
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