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Orders of Magnitude
Task
It is said that the average person blinks about 1000 times an hour. This is an orderofmagnitude estimate, that is, it is an estimate given as a power of ten. Consider:
 100 blinks per hour, which is about two blinks per minute.
 10,000 blinks per hour, which is about three blinks per second.
Neither of these are reasonable estimates for the number of blinks a person makes in an hour. Make orderofmagnitude estimates for each of the following:
 Your age in hours.
 The number of breaths you take in a year.
 The number of heart beats in a lifetime.
 The number of basketballs that would fill your classroom.
Can you think of others questions like these?
IM Commentary
The purpose of this task is for students to develop a feel for large powers of ten, which is a critical component of working fluently with numbers in scientific notation. This is an extension of the number sense that students work to develop around whole numbers in the early elementary grades, followed by fraction (and decimal) number sense in late elementary and early middle school. Note that this task develops "very large number sense"strategies for helping students understand very small numbers are forthcoming.
The term order of magnitude is used in this task (and elsewhere) to mean the power of 10 closest to a given quantity. The task provides an opportunity for students to get comfortable with the language; often people say, "Your estimate is off by an order of magnitude," which means that the estimate is 10 times bigger or 10 times smaller than is appropriate, or, "There are on the order of a billion people in China" to mean that $10^9$ is within such an error range.
There is a subtle point about what it means to be "closer" to one order of magnitude than another because orders of magnitude increase exponentially. Technically, we should find the middle point between two orders of magnitude on a $\log_{10}$ scale rather than taking the midpoint between the two powers of ten. That would mean, for example, that any number greater than 10 but less than $10\sqrt{10}\approx 31.62$ is closer (in this sense) to 10 than 100, and any number greater than $10\sqrt{10}$ but less than 100 is closer (in this sense) to 100 than 10. Because 8th graders do not have the mathematical tools to deal with this, the numbers in this task are conveniently chosen to fit this criteria without requiring a discussion of this subtle point.
The four parts of this task are ordered in increasing complexity. Finding your age in hours is the most straightforward because students can reasonably be expected to know all the information they need to solve it; the complexity can be increased by asking them to compute how old they are in seconds and minutes as well. For the breath question, there are different ways to present the problem. The students can determine their respiration rate, do research, or the teacher can provide a number. If students do some research, they should find that the respiration rate is dependent on age. This could lead to a great discussion about how to best estimate a rate that changes. Once they solve this problem, students can compare this solution to the solution in the heart beat problem. The students should realize that they need to convert the solutions to the same time frame. Since there are three or four heartbeats per breath, their answers could be the same order of magnitude or differ by an order of magnitude depending on how they made assumptions and rounded as they went. This can lead to a rich discussion of when to express a number as an order of magnitude; students can discuss the usefulness as well as limitations of such an estimate.
The breaths and heartbeats questions are similar in that they both require students to figure out some reasonable respiration and heart rate; the difference is that the breaths are for a given amount of time but students will have to decide what a reasonable length for a “lifetime” would be. The heartbeat problem can be adapted to be more or less openended, depending on time limit and instructional intent. Students could use research to determine the average heart beats per minute and age of a person. Another option is for students to take their own heart rate and determine how long they are expected to live. If time is limited, the teacher can provide the heartbeat rate and the age. The teacher can further the exploration by having students determine how fast a heart would have to beat per minute to change the order of magnitude. From here, the teacher could discuss if it is possible for a human heart to beat that fast. If a human heart cannot do that, is it possible for other animals' hearts to do so? This could allow for connections to science or health classes.
The basketball problem is the most complex because it includes a geometry component in addition to the other aspects that it has in common with the first three parts. It also requires students to make even more decisions, like how to pack the basketballs and whether to account for stuff in the room or assume the classroom would be empty.
The question of whether to allow calculators depends on the instructional goals of the teacher. If it would benefit students to practice hand computations, then a calculator isn’t needed. If the teacher wants to focus on the more complex modeling aspects of this task (like how students make and justify assumptions) then a calculator would be helpful.
This task was designed for an NSF supported summer program for teachers and undergraduate students held at the University of New Mexico from July 29 through August 2, 2013 (http://www.math.unm.edu/mctp/). MCTP participants helped write the solutions and edit the task.
Solution
For this problem, the students must convert an age in years to an age in hours. For example, a 13yearold student would calculate their age in hours in the following way: $$ 13 \text{ years} \times 365 \frac{\text{days}}{\text{year}} \times 24 \frac{\text{hours}}{\text{day}} = 113,880 \text{ hours}. $$ Now we note that 113,880 hours is between 100,000, and 1,000,000 and is quite close to 100,000. To verify this, let's determine how many years 100,000 hours is: $$100,000\div24\div365\approx 11$$ How many years is 1,000,000 hours? $$1,000,000\div24\div365\approx 114$$ Which reaffirms that $100,000=10^5$ is the best orderofmagnitude estimate for the age in hours of a 13yearold student.

The solution is presented in steps.
 Step 1: Approximate the number of breaths per minute for a person. Teachers can suggest students measure their own breathing rate. Teachers may also provide data. For an example, we use 12 breaths per minute.
 Step 2: Multiply the number of breaths per minute by 60 to find the number of breaths per hour. So: $$ 12 \frac{\text{breaths}}{\text{minute}} \times 60 \frac{\text{minutes}}{\text{hour}} = 720 \frac{\text{breaths}}{\text{hour}} $$
 Step 3: Multiply by 24 to convert breaths per hour to breaths per day: $$ 720 \frac{\text{breaths}}{\text{hour}} \times 24 \text{ hours} = 17,280 \frac{\text{breaths}}{\text{day}} $$
 Step 4: Multiply by 365 to convert breaths per day to breaths per year. So: $$ 17,280 \frac{\text{breaths}}{\text{day}} \times 365 \text{ days} = 6,307,200 \frac{\text{breaths}}{\text{year}} $$
 Step 5: Determine the order of magnitude. According to our example, the order of magnitude is $10^7$. Again, students should explore why 10$^6$ or 10$^{8}$ are not reasonable estimates. Working backwards, 10$^6$ corresponds to about 2 breaths per minute, which is well below a healthy range, whereas 10$^{8}$ yields a breathing rate of about 190 breaths per minute, which is well above a healthy range.
 Students can use the fact that there is a direct correlation between breaths and heart rate. Once this relationship is determined, they can use the calculations from part b to determine the solution. For instance, a healthy human heart beats approximately 6 times per breath. Students can multiply the number of breaths per year and the number of heartbeats per breath to find the number of heartbeats per year. For our example: $$ 6,307,200 \frac{\text{breaths}}{\text{year}} \times 6 \frac{\text{heartbeats}}{\text{breath}} = 37,843,200 \frac{\text{heartbeats}}{\text{year}} $$ Finally, to determine the number of heart beats per lifetime, they can multiply by the number of years in a lifetime. For our example, we place the lifespan at 80 years: $$ 37,843,200 \frac{\text{heartbeats}}{\text{year}} \times 80 \text{ years} = 3,027,456,000 \frac{\text{heartbeats}}{\text{lifetime}} $$ This is between $10^9$ and $10^{10}$, but is closer to $10^9$. Again, students should explore why 10$^8$ or 10$^{10}$ are not reasonable estimates. Working backwards, 10$^8$ corresponds to about 2.4 heartbeats per minute (well below a healthy range), whereas 10$^{10}$ corresponds to about 240 heartbeats per minute (well above a healthy range).

We need to use a measurement system that allows us to easily compare the size of a basketball to the size of a room. In this case, we will measure both the room and the basketball in terms of feet. First let us consider the size of a basketball. We can determine the circumference by simply using a tape measure. We can then use the circumference formula to find the diameter. We should find that the diameter of a standard basketball is approximately 9 in, or 0.75 feet. We can estimate that a basketball will fit into a cube that measures 0.75 feet $\times$ 0.75 feet $\times$ 0.75 feet. First, assume that the basketball almost fills the cube, as if we were stacking the balls directly on top of each other in a rectangular array. The basketball fits into a cube with volume of about $0.75^3\text{ ft}^3$, or $0.421875 \text{ ft}^3$.
Now, let’s consider the size of a classroom. Students can use whatever is at their disposal to measure their own classroom, such as yardsticks or measuring tape. As an example, we will use a room that measures 20 feet by 20 feet by 10 feet, since this is a reasonable estimate of an average classroom. This example gives us a classroom with a volume of $4000\text{ ft}^3$ (using the volume formula mentioned above). To compute about how many basketballs fit in the room, we will divide the volume of the room by the volume of the cube that contains the basketballs: $$ 4000 \text{ ft}^3 \div {0.75}^3\text{ ft}^3 \approx 9481 $$ so about 9500 basketballs will fit in the room.
So as an order of magnitude estimate, we find that we can fit about 10,000 basketballs in the room. We might ask ourselves what would happen if we packed the basketballs more snugly together. We could imagine that there is no gap between the balls and divide by the actual volume of a basketball with a diameter of $\frac34$ feet. The radius is half of that, or $\frac38$, and the volume is $$\frac43 \pi \frac38^3\approx 0.22$$ Again, the volume of the room is 4,000 cubic feet, so we can divide to find the number of basketballs: $$ 4000 \text{ ft}^3 \div 0.22\text{ ft}^3 \approx 18,108 $$ So with this estimate, about 18,000 basketballs will fit in the room. This still seems closer to 10,000 than 100,000.
Let's check: How much space would $10^3$ basketballs take up? Imagine they are loosely packed (and so taking up a lot of space); each one takes up the space of a containing cube which has a volume of about $0.42$ cubic feet. The total volume would be about 420 cubic feet. Even if the ceiling in the classroom is only 8 feet tall, that would leave only 52.5 square feet for the area, which would be a very tiny classrooma square with 7 foot sides, for example.
How about $10^5$ basketballs? Imagine $10^5$ tightly packed without any gaps between them, so each one takes up the space of a containing cube which has a volume of about $0.22$ cubic feet. The total volume would be about 22,000 cubic feet. If the ceiling in the classroom is only 10 feet tall, the area of the classroom would be 2200 square feet, which is more like the size of a gym. It is possible to have a classroom that large, but not likely, and any real packing of the basketballs would take up more space anyway.
Examples of others questions like these:
The number of hours they spend in school each year
The number of hours they spend doing homework per day, per week, per month, per year, per elementary school, and/or per lifetime
The number of hours they spend sleeping in their lifetime as compared to the number of hours they spend awake/studying/eating/exercising in their lifetime
Orders of Magnitude
It is said that the average person blinks about 1000 times an hour. This is an orderofmagnitude estimate, that is, it is an estimate given as a power of ten. Consider:
 100 blinks per hour, which is about two blinks per minute.
 10,000 blinks per hour, which is about three blinks per second.
Neither of these are reasonable estimates for the number of blinks a person makes in an hour. Make orderofmagnitude estimates for each of the following:
 Your age in hours.
 The number of breaths you take in a year.
 The number of heart beats in a lifetime.
 The number of basketballs that would fill your classroom.
Can you think of others questions like these?
Comments
Log in to commentCam says:
about 3 yearsVery neat  especially in conjunction with a discussion on "order of magnitude" estimates, and the surrounding issues with rounding and approximation. Thanks for sharing!
Seth says:
about 3 yearsWhen completing this task, we discussed adding in the concept of a leap year. Some students used 365.25 instead of 365.
yuridia says:
almost 5 yearsremoved