Rain Unit

UNIT 2—Rain

Teacher Materials


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TEKS Support
Teacher Notes
 Preparation Reading—Rain Falls Mainly on the Plain?
 Activity 1—First Time Passing Through?
 Homework 1—Rain on Me!
 Discussion—"Revised" Model
 Supplemental Activity 1—Continental Divide
 Activity 2—Caution: Under Construction
 Homework 2—In Pursuit of Knowledge
 Supplemental Activity 2—Bird Territories
 Activity 3—Area of Concern
 Homework 3—The Means Justify the End
 Activity 4—Model Gets a Second Chance
 Homework 4—Texas Storms are SO Big…
 Discussion—Modeling Process in this Context (again)
 Activity 5—Get the Point?
 Homework 5—Rain Keeps Falling on My Head
 Activity 6—Power in Those Rules
 Homework 6—Prelude to a Problem Solved
 Activity 7—Third Time’s the Charm
 Unit Project—Rainy Days in Texas
 CENTER TI-83 Program
 PERP TI-83 Program
 POLYAREA TI-83 Program
Annotated Student Materials
 Preparation Reading—Rain Falls Mainly on the Plain?
 Handout—Instructions for Constructing a Perpendicular Bisector
 Activity 1—First Time Passing Through?
 Homework 1—Rain on Me!
 Supplemental Activity 1—Continental Divide
 Activity 2—Caution: Under Construction
 Homework 2—In Pursuit of Knowledge
 Supplemental Activity 2—Bird Territories
 Activity 3—Area of Concern
 Homework 3—The Means Justify the End
 Activity 4—Model Gets A Second Chance
 Homework 4—Texas Storms Are SO Big…
 Activity 5—Get the Point?
 Homework 5—Rain Keeps Falling on My Head
 Activity 6—Power in Those Rules
 Homework 6—Prelude to a Problem Solved
 Activity 7—Third Time’s the Charm
 Dividing the Job
 Unit Project—Rainy Days in Texas
 Unit Summary—Modeling Estimation of Rainfall Totals
 Mathematical Summary
 Key Concepts
Solution to Short Modeling Practice
Solutions to Practice and Review Problems


TEKS Support

This unit contains activities that support the following knowledge and skills elements of the TEKS.

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The mathematical prerequisites for this unit are

The mathematical topics included or taught in this unit are

The equipment list for this unit is



Teacher Notes



Preparation Reading—Rain Falls Mainly on the Plain?

The reading introduces the Texas Natural Resource Conservation Commission (TNRCC) and its task of finding a good estimate for the level of rain that has fallen. Possible follow-up discussion questions based on the reading include What are the local reservoirs in our area? Where is the rain gauge location for our area? How many gauge stations are in the entire state? Make sure that students understand the distinction between average depth and total amount of water before proceeding to the discussion on modeling the situation. (Refer to the mathematical summary.)




Activity 1—First Time Passing Through?
   

As a class, or in groups, students should begin the preliminary aspects of building a model by identifying the problem that is being solved and the key features and assumptions that will be made. Initially, this model should be pretty simple, although possibly not very realistic.

  1. What is the problem that the TNRCC is trying to solve?

    2. This will be a pretty straightforward part of the discussion—they are trying to estimate the amount of rainfall for the entire state. The distinction between average rainfall (which is a depth measurement) and total rainfall (which is a volume measurement) should be part of the statement of the problem. Which interpretation makes more sense for this problem? What time frame are we discussing here? Is specifying a time frame even necessary?

  2. What assumptions should we make about the problem?

    3. This will probably be a lot more difficult to draw out of students in a discussion. Remind them that they want to start out simply—to understand the problem, set up the mathematics, and test whether or not it truly describes the situation.

    As a first attempt to build a model, students could assume that the location of the gauges is either not known or not important, the state is idealized to that of Colorado (approximately 375 mi. by 275 mi.), and the estimate simply boils down to taking the average height of rain in the gauges and multiplying it by the area of the state to get the total rainfall. Try not to lead them too much in the discussion; it’s better to let them struggle with Questions 1 and 2 of the first activity, and bring the class back together for a discussion if necessary. After they have a clear set of assumptions, let them finish working on the rest of the activity. Close the activity with a discussion of two things:




Homework 1—Rain on Me!
   

Question 1 brings students back to the modeling problem as it applies to Texas. The analysis is the same as in Activity 1, so students should be able to answer these questions without trouble. It is a good "gauge" of whether or not they understand the first pass through the modeling process.

Question 2 is a chance for students to visit (or revisit) the idea of a weighted average in this context. The grade example should look familiar to many students. If you use weighted averages in your class, this will be an opportunity to explore some of the nuances of the grading system. For instance, in the example from this homework, it is impossible to get higher than 80% without doing homework. Students may have seen weighted averages before with the weights being probabilities. In this example, the weights reflect the importance of a part of a student’s grade. Point out that the "simple average" used in part a) may be rewritten as 89 (1/4) + 85 (1/4) + 99 (1/4) + 75 (1/4), by "distributing the division." Then it, too, may be viewed as a weighted average, with equal weights of 1/4 for each grade.

Question 3 is designed to make students comfortable with the use of distance to determine which gauge represents a city. In discussion the next day you might ask students how they determined their answers (ruler, compass, Mira, etc.). Wray is equidistant from gauges 3 and 7, while Boulder is equidistant from 2, 5, and 6. These ties are fine. Students may feel uncomfortable about this but should be encouraged to explain the tie. Students may ask about geographical obstacles (such as mountains), but these have been ignored as a simplifying assumption. This assumption may lead to a class discussion about the role of simplifying assumptions in modeling. After ranking the gauges by area, students should discuss the idea that the gauges at the top of the list should count more than the gauges at the bottom of the list.

Question 4 allows students a chance to create a Voronoi diagram using three noncollinear points. Precision is of secondary importance in this exercise; the most important thing is for the students to get a feel for how to create the Voronoi regions. The students’ ability to handle this "simpler-related" problem is a good measure of how well they are ready to tackle the second pass through the modeling process. The discussion outlined in the next set of Teacher Notes explains your role in that determination.




Discussion—"Revised" Model

Modeling Process in This Context

The last several questions of Homework 1 represent a transition in the fundamental way we are examining the problem of estimating rainfall. The modeling process is just that—a process—and students have to develop a habit of mind about problem solving as an ongoing task that ends only when the problem changes. A possible way to discuss the transition between the "simple" and "revised" models could proceed something like this:

Ending that discussion, bring students to understand that, in the revised model, we are changing the assumption that the relative position of the rain gauges doesn’t matter (or isn’t available information). This subtle shift in the fundamental nature of the model forces us to be able to address two additional problems:

  1. resolving a map into its Voronoi diagram, by drawing the perpendicular bisector of the line segment that connects any two rain gauges, and


  2. estimating total rainfall by using a weighted average of (rainfall × area).



Supplemental Activity 1—Continental Divide
   

This is an optional activity that should be done if students are having a difficult time determining the location of the boundary lines or are not really sure how the weighted average calculation fits into the contextual problem.

The first set of problems provides an experience for students of actually drawing the boundaries between regions of influence. They explore methods of dividing a region, and have to estimate the way in which the lines divide the overall area. Students should not spend lots of time working out exact ratios; rough estimates are okay as long as they are plausible. The second set of problems allows students to explore weighted averages with areas of a region, using the areas as weights. Again, students shouldn’t spend too much time estimating the areas of the gauge regions.

Possible followup discussion for the activity:




Activity 2—Caution: Under Construction
   

This activity provides an opportunity for students to experiment with various construction tools, to discover some basic constructions, and to apply what they’ve learned to a simple rain gauge problem. You could proceed through Part 1 of the activity in a couple of ways:


Option 1

Distribute the activity description and one or two of each of the construction tools to each group. Do not give them instruction on how to use the tools. This is an opportunity for them to discover for themselves the properties of each tool and ways of doing some simple constructions. The groups may want to concentrate on one tool at a time or might want to divide the tools among themselves. In the end, every group should have tried the two constructions using compass and straightedge, Mira, and wax paper for each.

If some groups have been successful with the two constructions, have them volunteer to show the entire class how to do those constructions with all three tools. By the end of this part of the activity, each group should be able to show one method of solving one of the problems. If any constructions stumped the entire class, such as using the Mira for the angle bisector, demonstrate the technique for the class yourself.


Option 2

Treat this as a teacher-led part of the learning, and model for the students how to do each of the tasks with all three of the tools: compass/straightedge, Mira, and wax paper. Have students go through the constructions with you, recording the steps as notes in their own words. Let them know ahead of time that you are showing them this one time, and that their notes must make sense to them.

No matter which option you use for Part 1, students must demonstrate their ability to construct perpendicular bisectors in Part 2. In examining the work, look for the lines to be reasonably close to the location, and question how the lines were determined (Miras don’t leave much scratch work, but students should explain how they came up with their lines. Groups might explore the effect of using a particular construction tool, with each member of the group using a different one. But in any case, the group should check each others’ work to make sure that everyone is able to construct the boundary lines correctly.




Homework 2—In Pursuit of Knowledge
   

This set of problems is intended to reinforce the idea of equidistance and to tie in the idea of locating the center of a circle that passes through three given points. Have each student work on this individually; they will need compass and straightedge. Note that the context has changed to helicopter service and helipad location. After exploring how construction of overlapping circles provides the location of the perpendicular bisector, Question 5 asks a key question for future consideration. Then students are led into the construction of the Voronoi vertex from intersecting perpendicular bisectors. To check for understanding after this homework activity, ask students, "How do you know the Voronoi vertex is the same distance from the three centers of influence in its immediate area?

The homework activity finishes with some practice problems, but also answering the question "How does the kind of triangle formed by the location of the centers of influence affect where the Voronoi vertex is determined? You might ask your students that in processing Questions 9–11 the next day. Question 12 is another litmus test for the students, on whether they truly understand the process and are ready to tackle the problem of resolving the map of Colorado into its regions of influence. If so, proceed directly to Activity 3, which tackles the question of what to do with the regions after you’ve made the Voronoi diagram. Otherwise, have students work on Supplemental Activity 2.




Supplemental Activity 2—Bird Territories
   

This activity introduces the method of combining three-point solutions to create a four-point solution, which is crucial to solving any Voronoi problem with many points. You might suggest that pairs of students split up the work on Questions 1–2, and then combine their resources to answer Question 3. If they get really stuck, suggest to them that points A and C from Question 1 and points D and E from Question 2 correspond to points Q and R from Question 3.

Following that, Questions 4–7 should help students think about proximity and the Voronoi diagram. Another very important piece of information is the fact that all Voronoi vertices are centers of circles that go through three centers of influence. This could be another strategy for finding a Voronoi diagram. Notice that the four-point figure in Question 3 is concave, but students must adjust their thinking in Question 8.

As students work through the problems with four points of influence, have them think about the process they use. One strategy they might consider is to break the problems down into smaller, three-point problems. Then they can combine the answers to their three point problems to get the answer to the larger problem. Notice that, in a four-point problem, there are four different ways of choosing three points and finding their Voronoi vertex. One good question to ask is "Which three points should you choose?" If students choose A, B, and C, they will find that the Voronoi vertex is closer to D (the unchosen point) than it is to A, B, or C. Therefore, the vertex for A, B, and C will not be a vertex for A, B, C, and D. Help students to conclude that, if there is an interior point for any three points chosen, the vertex for those three points will not be a vertex for the entire problem. Whether three points have an interior point can be established several ways, including by drawing the triangle that connects the three points and noting whether another center of influence lies inside that triangle.

Question 9 presents a case in which four points lie on a circle. This is not obvious—at first the problem looks like a standard quadrilateral. However, broken down, all of the possible three-point solutions have the same Voronoi vertex, which is the center of the circle containing the four points. The inaccuracy of their drawings may lead students away from the idea that there is only one Voronoi vertex, but looking at symmetries may help them to believe it. Finally, Question 10 challenges students to apply their thinking about a general process by providing a five-point problem.




Activity 3—Area of Concern
   

The other half of the mathematical problem of estimating the total rainfall is addressed in this activity. Students are asked to calculate the areas of the three regions and then estimate the total rainfall from taking a weighted average, using the rain gauge data as the "weights." Again, the modeling process shows up, with students working first with a simplified three-point map and moving on to a four-point map. If students experience difficulties with this activity, expect it to be centered around two issues:




Homework 3—The Means Justify the End
   

Four different ways of calculating the area of these Voronoi regions are introduced in this activity. Questions 1–4 should be self-explanatory, but you may want to discuss how to do the Monte Carlo simulation from Question 4 with the class, if this activity is assigned as homework. Question 5 provides the debriefing discussion with the class; try to get students to identify which method they prefer and why, but make sure they understand the limitations in using their method, and suggest that they master more than one method.




Activity 4—Model Gets a Second Chance
   

Students have the opportunity to revisit the problem of estimating the total rainfall for the state of Colorado. Again, they are asked to apply the steps of the modeling process in answering Questions 1–2, and then use the weighted-average approach to calculate the total rainfall. Finally, students are asked to compare the estimate to that obtained by the simple model (Activity 1), and to test the model for sensitivity.




Homework 4—Texas Storms are SO Big…
   

This is an opportunity for students to begin working on the unit project. It has been broken into three distinct parts, so that the work to be done can be given as homework assignments while finishing the rest of the unit material. As the teacher, you have the choice on whether to give them the project description in stages (Homework 4, Homework 5, and Unit Project) or simply give them the Unit Project at this time and verbally instruct them on which part of the work they should be completing each night




Discussion—Modeling Process in This Context (again)

Use the answers to Activity 4 Question 9 as a springboard for discussing the direction in which the unit will now turn. Hear what the students suggest as the next steps to take in modeling this problem. Remind them that they will be working on a project that actually customizes the problem to their geographic region, and that they may want to incorporate their own suggestions into the project work. No matter what they might suggest, at this point in the unit, students should be comfortable with the idea that the locations have something to do with the precision of the answer. So it won't be difficult to get them to "assume" that having better information about the locations (actual coordinates) would yield even better answers.




Activity 5—Get the Point?
   

The goal of this activity is for students to create formulas for the midpoint of the line segment connecting two Voronoi centers, the slope of the perpendicular bisector, and the equation for the perpendicular bisector, all from just the coordinates of the two Voronoi centers. The flow for this activity is for students to work through a specific example (Questions 1–5) and then follow their own logic in developing the formulas (Questions 6–10). After that, they are asked to practice applying those formulas to a three-point map.

As the teacher, you can choose several options for structuring this activity. You could have students work in groups on the entire activity, and debrief the work at the end of the period. You can have students work through the specific example, and have a full-class discussion of the method of generalization (knowing that they will have the opportunity to develop one for themselves in Activity 6). You can walk through the specific example, making sure students know how to find the slope of the perpendicular bisector from the slope of the segment connecting the two centers and how to go from point-slope form to slope-intercept form, and then let them develop the formula themselves.

At any rate, the students should be able to demonstrate understanding of the algebra involved when they tackle the three-point problem in Part 2. In examining their work on Part 2, either close by reviewing the steps taken in deriving the formulas, or go over how to get one of the perpendicular bisector equations from the coordinates of the centers, and have students work out the remaining two for themselves. Encourage students that the formula is useful, not something to be memorized, and point out that calculator or computer programs could do all our "dirty work," if a formula has been developed. Whether or not the students master the formula development, they all should be able to solve specific numeric examples.




Homework 5—Rain Keeps Falling on My Head
   

This is the second installment on the unit project description. The data provided are annual rainfall totals, which is why the numbers are bigger than the ones students have been using in the unit material. A website is included for students to begin researching the problem for themselves, as is a table for organizing their calculations of the areas of the various regions of influence.




Activity 6—Power in Those Rules
   

This is the second half of the algebraic development for the "refined" model. Students should develop a rule for finding the point of intersection of two lines expressed in slope-intercept form, and then develop another rule for finding the length of a line segment, knowing the coordinates of the endpoints.

The recommended approach for this activity would be to have students work through all the questions in their groups, and have a discussion of their results at the end. This will depend on the level of the students’ ability to apply algebraic thinking, and they may not have had experience at doing this. If possible, avoid developing the rules for the students; at the very worst, let them struggle with the formulas themselves before leading them through the process. And, definitely, when the formulas have been developed, all students need the opportunity to practice applying them independently.




Homework 6—Prelude to a Problem Solved
   

Students use the formulas derived in class to solve a four-point Voronoi diagram problem. Comment on the time saved by using the formulas compared with solving each problem from scratch. Get students to examine their algebraic solutions (from the formulas) with the drawing made on graph paper. Do the answers make sense? Are the intersection points predicted by the formulas consistent with the location determined by construction methods (or eyeball estimation) on the graph paper?




Activity 7—Third Time’s the Charm
   

Here is the final opportunity to solve the problem of estimating the total amount of rainfall in the state of Colorado. An ideal way to proceed on this activity is to have each group of students solve the problem together. One possible approach is to delegate the regions among the group—Student #1 (Regions 1 and 2), Student #2 (Regions 3 and 4), Student #3 (Regions 5 and 6), and Student #4 (Regions 7 and 8). They find all the equations of the perpendicular bisectors for their regions and compare answers to make sure no mistakes have been made. Then they take those equations and find the coordinates of all the intersection points for their regions and compare answers again. Finally, they find the lengths of the segments forming the boundaries of their region, and again compare answers when there is overlap.

At this point, each student finds the area of his/her region, using Heron’s formula and calculates the weighted average for the region. Then the group puts all the work together to come up with its estimate. As a class, the steps can be checked when groups report their findings; since they are all starting with the same rain gauge data and location coordinates, the groups should be fairly close to each other.

This activity is an ideal place to use a graphing calculator technology by allowing students to use one or more of the following three programs:

    PERP: Finds the equation of a perpendicular bisector of a segment from the coordinates of the segment's endpoints. (Also finds the segment's midpoint.)

    CENTER: Finds a Voronoi vertex determined by three centers of influence.

    POLYAREA: Finds the area of a polygon from its vertices.

Here is an example of how a student might use these programs for a part of the Colorado problem. In this example, the student is trying to find the area of Region 1.

The student sketches an approximate Voronoi region about gauge 1, as shown in this figure. The region's five vertices have been labeled for convenience.

The coordinates of vertex A are (0,275) since the dimensions of the rectangle are known.

To find the coordinates of vertices B and E, the student uses the program PERP, followed by substitution. For example, from the coordinates of centers 1 and 4, PERP determines that the equation of the segment connecting B and C is (5/34)x. Since B is on the left boundary, substituting 0 for x gives 11539/68. Thus the coordinates of B are (0, 11539/68).

The student determines the coordinates of E with a similar application of PERP.

To find the coordinates of vertices D and C, the student uses the program CENTER. For example, applying CENTER to centers 1, 2, and 5 gives the coordinates of D.

When the coordinates of all five vertices have been found, the student uses the program POLYAREA to find the area of the region.




Unit Project—Rainy Days in Texas
   

This is the formal written description of the unit project, which is to relate what’s been learned about estimating rainfall totals in Colorado to the students’ local area. The teacher has options on how to proceed with the project. Here are some possibilities—

Discuss with the students your own expectations for the project and how you are going to evaluate their work. Encourage them to go beyond the central problem of estimating rainfall totals; some suggestions for ways to extend the project are included, but the students will have others as well. Require that the report be as formal as you want; feel free to incorporate a public speaking component into it. Students' experience and comfort in mathematical modeling will establish the level of difficulty they want to pursue. Try to "customize" the project work to match the abilities of the individual students. Have fun, and good luck!




CENTER TI-83 Program
   

ClrHome

Input "1ST X ",A

Input "1ST Y ",B

Input "2ND X ",C

Input "2ND Y ",D

Input "3RD X ",E

Input "3RD Y ",F

((B-F)/2-(A--C-)/(2D-2B)+(C--E-)/(2F-2d))/((C-E)/(F-D)-(A-C)/(D-B))_P

((A-C)/(D-B))P+((B+D)/2-(A--C-)/(2D-2B))_Q

Disp "CENTER X",P

Disp "CENTER Y",Q




PERP TI-83 Program
   

ClrHome

Input "1ST X ",A

Input "1ST Y ",B

Input "2ND X ",C

Input "2ND Y ",D

(A+C)/2_E

(B+D)/2_F

(A-C)/(D-B)_M

F-E*M_N

Disp "MID X",EúFrac

Disp "MID Y",EúFrac

Pause

Disp "PRP BIS SLOPE",MúFrac

Disp "PRP BIS Y NTRCP",NúFrac




POLYAREA TI-83 Program
   

ClrHome

Input "NO.SIDES? ",N

Disp "X COORD LIST"

Input L/

Disp "Y COORD LIST"

Input L*

1_K:0_D

While K2N-2

à((L/(1)-L/(1+K))-+(L*K(1)-L*(1+K))-)_A

à((L/(1)-L/(2+K))-+(L*K(1)-L*(2+K))-)_B

à((L/(1+K)-L/(2+K))-+(L*(1+K)-L*(2+K))-)_C

(A+B+C)/2_S

à(S(S-A)(S-B)(S-C))+D_D

1+K_K

End

Disp "AREA IS",D



Annotated Student Materials



Preparation Reading—Rain Falls Mainly on the Plain?

One of the most important duties of the Texas Natural Resources Conservation Commission (TNRCC) is estimating the rainfall for the entire state. This estimate is of vital importance to many people in the state because the state relies on a series of reservoirs for its water needs. A reservoir is a lake that is used to supply water for a region. When there is not enough rainfall, many reservoirs are in danger of falling below safe levels. If this seems likely, Texas may be forced to purchase water from a neighboring state.

Many stations for measuring rainfall have been set up all around the state. Using samples from those stations, the agency has to make a good estimate of the rainfall for the state. The statisticians at the agency need to figure out what to do with these sample data. Maybe they should average the numbers, but what does that answer tell them? They want their estimate to be as accurate as possible, and would put gauges everywhere if they could. Because the agency doesn’t have enough staff or money, that simply isn’t practical.

There are two ways to interpret the phrase "rainfall." First, it might mean: "What is the average depth of the water around the state?" That is, if you pretended that the state is a large wading pool, how deep will the pool be? The other way to interpret the term "rainfall" is to consider it to be the total amount of the water that falls on the state.

As mathematical modelers, you need to ask questions, make assumptions, describe relationships in mathematical terms, come up with answers, and question whether those answers make sense; this is what we call the modeling process. Then you get to test the model to determine which assumptions control the critical behaviors, and you get to change assumptions and go through the entire process over and over again. Ideally, this leads you to "better" answers and more realistic descriptions of the problem in mathematical terms. Good luck!




Handout—Instructions for Constructing a Perpendicular Bisector

Paper

When paper folding is used, it is sometimes easier to work with wax paper or patty paper. The problem and its solution are drawn on the wax paper.

  1. Place the wax paper on top of the diagram for the problem. Trace the boundaries of the domain and the points of influence.
  2. Pick up the wax paper and fold it over in such a way that two of the points of influence (A and B) coincide (are on top of each other). When the points are appropriately lined up, crease the paper along the fold.
  3. Open out the paper and draw a line along the crease. This line is the perpendicular bisector of the segment joining A and B.

Mira

A Mira is a plastic device that is a helpful tool when working with the reflection of an image.

  1. Hold the Mira upright so that the long, beveled edge is on the diagram. In the face of the Mira you should see a reflection of the image on the paper.
  2. Place the Mira between the two points for which you want a perpendicular bisector. Look through the red plastic until you see one of the points of influence (point A). Slide the Mira about the diagram until the point visible through the Mira is lined up with the reflection of a point (point B) in front of the Mira.
  3. When the two points appear to lie on top of each other (coincide), draw a line along the beveled edge of the Mira. This line segment is on the perpendicular bisector of the segment joining A and B.

Compass and Straightedge

  1. Draw the segment connecting two points of influence (A and B).
  2. Set the pointed end of the compass on point A. Open the compass a little more than half the distance between A and B.
  3. With the pointed end on A, swing an arc in the region between A and B.
  4. Without changing the opening in the compass, put the pointed end on point B and swing another arc in the region between A and B until the two arcs intersect in two places (points X and Y).
  5. Use a straightedge to draw a line containing X and Y. This line is the perpendicular bisector of segment AB.



Activity 1—First Time Passing Through?
   

 = 10, 11, 12, 15

The table below shows the rainfall depths (in inches) recorded for eight gauges located in Colorado.

Table 1

X1

X2

X3

X4

X5

X6

X7

X8

2.11

2.15

1.21

4.45

2.67

2.51

3.11

2.43


  1. What is the problem that you want to investigate?

    2. We are trying to estimate the total rainfall for the state of Colorado.

  2. What assumptions are you going to build into your model? Do you need any additional information? (Remember that it might be easiest to keep things simple for now!)

    3. These are the only rain gauges for the state, and they are representative of how rain falls over the entire region.

    Either the locations are not known or it isn’t important to know where they are placed relative to each other.

    The state is shaped like a rectangle, with length 375 mi. and width 275 mi.

  3. What is your plan? Describe what are you going to do to solve the problem.

    4. I’m going to average the eight rain gauges together to represent the rainfall behavior for the entire state. Then, I’ll multiply the area of the state by this average depth to estimate the total volume of water that fell on the state.

  4. Carry out your plan and solve the problem. What are the units for your answer? Convert your answer to a standard unit of ft3.

    5. Average depth: (x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 ) / 8 = 2.58 in.

    Total volume: (Area of Colorado) * (Average depth of the rain gauges) = 266,000 in.-mi2 = 22,200 ft.-mi.2 = 14,300,000 ft-acres = 6.18 × 1011 ft3

  5. Take a few minutes to test your model.


    1. If your average rain gauge calculation was off by 0.01 in. (either high or low), how would that affect your answer?
    2. If the distance measurements for the length and width of Colorado were off by 1 mile, how would that affect your answer?

      3. A change in rain gauge depth of 0.01 in. would produce a change in the total volume estimate of roughly 2.4 billion ft3. A change in the distance measurement for the length and width of Colorado would produce a change in the total volume estimate of roughly 3.9 billion ft3.

  6. Look back over your work from this activity. Is this a good answer? Does the way in which the answer was found make sense in the real-world problem? Why? Is the limitation due to the process you used, or based upon your assumptions?

    7. The answer is a beginning estimate for the total rainfall. The process implies that rainfall falls equally as much on every point in the state; even if we used a thousand times as many gauges, a lot of places would still be unrepresented. The limitation of our answer is based on the assumption that the gauge locations weren’t important. In Colorado, as in Texas, geography plays a big role in determining where the rainfall will land.

  7. Think about where you might want to go next in modeling this situation. What do you think is the key feature that should change so that the answer better describes the real-world situation? How would you modify the activity to improve the model?

    8. Since it is impossible to represent every point in the state with its own unique rain gauge, I would want to begin to look at the relative position of the gauges, and see if I could use them to represent their regions somehow. Evidence to support that direction comes from the fact that there is variability in the data to begin with, plus the commonsense understanding of weather being a localized phenomenon.




Homework 1—Rain on Me!
   
  1. Seventeen "first-order" stations are spread across the state of Texas. In addition, one station in Shreveport, Louisiana, is used quite often. Here is the total rainfall data for the 1997 calendar year:

    2. Table 2

    X1

    X2

    X3

    X4

    X5

    X6

    X7

    X8

    X9

    27.08

    43.27

    24.95

    46.79

    36.21

    44.10

    36.16

    23.01

    45.00

    X10

    X11

    X12

    X13

    X14

    X15

    X16

    X17

    X18

    9.63

    76.79

    22.67

    17.10

    23.38

    33.92

    65.05

    23.78

    69.20


    1. What is the average depth of the rain in the gauges used in Texas?

      2. (27.08 + 43.27 + 24.95 + 46.79 + 36.21 + 44.10 + 36.16 + 23.01 + 45.00 + 9.63 + 76.79 + 22.67 + 17.10 + 23.38 + 33.92 + 65.05 + 23.78 + 69.20) / 18 = 37.12 in.

    2. Do you think the measurements used in Activity 1 were yearly totals? If so, argue why that would be the case. If not, estimate the time interval in which the data were collected. Is time an important factor in the modeling of this situation?

      3. Probably not, because the numbers are so low. Colorado doesn’t get gulf storms, but it gets storms from the Arctic, has lots of snow during the winter, and isn’t considered a desert region. A more reasonable estimate for the time interval would be one month.

    3. Using the same method as in Activity 1, estimate the 1997 total rainfall for Texas. You may need to look in an atlas or encyclopedia to get some information needed to answer this question.

      4. The area of Texas is listed in the New American Desk Encyclopedia as being 267,229 mi2. Therefore, the yearly total rainfall for Texas would be around 9,920,000 in.-mi2, 827,000 ft.-mi2, 5.29 × 108 acre-ft or 2.31 × 1013 ft3.

  2. Gerry is trying to figure out her average for math. Her grades were 89% for tests, 85% for quizzes, 99% for homework, and 75% for class participation.


    1. Compute a simple average for these grades.

      2. (89 + 85 + 99 + 75) / 4 = 87%

    2. Gerry remembers that the teacher said that tests counted 50%, quizzes counted 20%, homework counted 20%, and class participation counted 10%. Now compute the weighted average (using the percentages above as weights).

      3. (89)(0.50) + (85)(0.20) + (99)(0.20) + (75)(0.10) = 88.8%

    3. Which particular grade has the most impact on the weighted average? Explain your answer. If you’re not sure, try changing each grade (one at a time) by ten points to find out which affects the average the most.

      4. The test grade. Changing the test grade by 10 points affects the grade by 5%, while similar changes to the others affect the grade by 2%, 2%, and 1%.

    4. Which grade had the least impact on the weighted average?

      5. Class participation

    5. How does this problem relate to the problem in Activity 1 of estimating the rainfall for Colorado? In what way might weighted averages help make a better estimate?

      6. Using the area of each region to figure out how much its gauge should be counted toward the whole is a problem of calculating a weighted average.

  3. The following is a map of Colorado, with the eight rain gauges that were used in Activity 1 shown, as well as some cities.

    4. Figure 1


    1. Identify the gauge or gauges that best measure each of the cities listed in the table. Explain the method that you’re using, and why it’s harder to find a gauge for some cities than for others.

      2. Table 3

      City

      Gauge #

      Denver

      6

      Trinidad

      8

      Sterling

      3

      Wray

      3, 7

      Boulder

      2, 5, 6


      The gauge that is closest to each city is the one that represents it. Sometimes a city is the same distance to more than one gauge. Note: geographical obstacles (like the Rocky Mountains) have been ignored.

    2. For each gauge, list the cities for which the gauge is the best measure of rainfall. Be sure to place all fourteen cities.

      3. Table 4

      Gauge

      Cities

      Gauge

      Cities

      1

      Craig, Meeker

      5

      Boulder

      2

      Fort Collins, Boulder

      6

      Denver, Boulder, Colo. Springs

      3

      Sterling, Wray

      7

      Burlington, Lamar, Wray

      4

      Grand Junction, Durango

      8

      Monte Vista, Trinidad


    3. Using your answers and the map:

      4. Rank the gauges by the amount of areas for which they are the best representative. Put the gauge that corresponds to the largest area at the top of the list, and the smallest area at the bottom.


      Table 5

      Rank

      Gauge

      1

      7

      2

      8

      3

      4

      4

      1

      5

      3

      6

      6

      7

      2

      8

      5


      These answers are one possiblity. Student ranking may differ somewhat since area estimates are imprecise. Be sure answers are based on apparent sizes.

      As you start to look at these areas around gauges, some terminology will help clarify what is meant. Centers of influence (gauges in our rainfall example) are the points around which we are building boundaries. The regions around the centers are called regions of influence.

    4. When you take the map into consideration, how would you revise your method of determining the rainfall totals from Activity 1?

      5. I would consider dividing the map into regions based on the locations of the rain gauges. The dividing lines would be determined in such a way that a location that is closest to one gauge would be in the region represented by that same gauge.

  4. Try your revised method on the following simplified map of Colorado, in which only three rain gauge locations are indicated. Show how you would divide the state so that each part of the state is represented by only one gauge. Explain your method.

    5. Answer: See figure below.

    Figure 2

    I drew a line that went directly between two gauges and extended it to the edge of the rectangle (state border). I did the same thing with a different pair of gauges, and again with the third set of gauges. Then I extended the three lines until they came to a single point somewhere in the middle of the three gauges.




Supplemental Activity 1—Continental Divide
   

Figures 3-8 are diagrams describing the positions of two or three rain gauges. Each gauge is responsible for the area of the state that is closest to it. The domain for any problem is the area that is being divided; for these problems, the domain is the total area of the state. For each of the diagrams:

  1. Draw the divider between the regions for which the rain gauges are responsible.


  2. Estimate the percentage (or fraction) of the area of the state for which each gauge is responsible.


    3. Figure 3

     

    % of total area for A:

    % of total area for B:

    Figure 3 answer

     

    50%

    50%


    Figure 4

     

    % of total area for A:

    % of total area for B:

    Figure 4 answer

     

    25%

    75%


    Figure 5

     

    % of total area for A:

    % of total area for B:

    Figure 5 answer

     

    50%

    50%


    Figure 6

     

    % of total area for A:

    % of total area for B:

    % of total area for C:

    Figure 6 answer

     

    20%

    20%

    60%


    Figure 7

     

    % of total area for A:

    % of total area for B:

    % of total area for C:

    Figure 7 answer

     

    15%

    25%

    60%


    Figure 8

    % of total area for A:

    % of total area for B:

    % of total area for C:

    Figure 8 answer

    50%

    40%

    10%

    Returning to the original Activity 1 problem with the eight given gauges, how would you divide the state? If you could divide it into regions around centers of influence (the gauges in the original problem) so that everything in the region is closer to its center of influence than to any other center of influence, you would have a Voronoi diagram. A Voronoi diagram is also called a Voronoi tiling. Voronoi diagrams are named for the Russian mathematician who invented them in 1908.

    To simplify the rain problem, you could split the state into small regions. Consider the regions below. For each region, create the Voronoi diagram that will divide the region so that every point in each region is closest to the gauge that is at the center of influence of the region. When you have the Voronoi diagram, use a weighted average to find an estimate for the rainfall in the region. Compare the weighted average to the simple average of the height readings for each region.

    For the following regions, use the following rain gauge heights:

    Table 6

    X1

    X2

    X3

    X4

    X5

    X6

    X7

    X8

    2.11

    2.15

    1.21

    4.45

    2.67

    2.51

    3.11

    2.43


    Northwest region: Gauges 1 and 2

    Figure 9

    % of area for gauge #1: 70%

    % of area for gauge #2: 30%

    Figure 9 answer

    Simple average height: 2.13 in.

    Weighted average height: (0.70)(2.11) + (0.30)(2.15) = 2.12 in.


    Northeast region: Gauges 3 and 7

    Figure 10

    % of area for gauge #3: 25%

    % of area for gauge #7: 75%

    Figure 10 answer

    Simple average height: 2.16 in.

    Weighted average height: (0.25)(1.21) + (0.75)(3.11) = 2.6 in.


    Central region: Gauges 5 and 6

    Figure 11

    % of area for gauge #5: 60%

    % of area for gauge #6: 40%

    Figure 11 answer

    Simple average height: 2.59 in.

    Weighted average height: (0.6)(2.67) + (0.4)(2.51) = 2.61 in.



Activity 2—Caution: Under Construction
   

 = 1, 2

You may recall that a point equidistant from two given points is on the perpendicular bisector of the segment joining the two points. To be perpendicular, the line must form right angles with the segment. To be a bisector, the line must pass through the midpoint of the segment. The midpoint is halfway between the endpoints of the segment. Creating accurate diagrams of perpendicular bisectors can be quite a challenge.

People often try to draw geometric figures. When they do, they usually try to approximate the figure by looking at it carefully and perhaps using a straightedge. Geometers, people who study and use geometry, use construction to create precise sketches of geometric figures. The circles of a compass, the reflections of a Mira, or the folds of a piece of wax paper can help to create accurate figures. Try discovering some methods, using these tools, for completing the following tasks:

  1. Construct an angle bisector for an arbitrary angle.
  2. Construct a perpendicular bisector for a segment.

When you have become comfortable with one or more of the construction tools and have completed the two tasks, you are ready to move on.



Part 2

The figure below shows a region that contains 3 rain gauges. It is 15 miles wide and 20 miles long. Determine the region that each rain gauge represents, by resolving the figure into a Voronoi diagram. Use a straightedge to indicate boundaries clearly. Use one or more of the tools you explored in Part 1 of this activity to complete the task. Detailed instructions are included in case you get stuck.

Answer: Figure below

Figure 12


Homework 2—In Pursuit of Knowledge
   
  1. The drawings below represent the locations of three competing helicopter services and their helipads. In Figure 13, draw a boundary for all points that are within two miles of helipad A and a boundary for all points that are within two miles of helipad B. (Assume the grid marks are 1 mile in distance.)


    		2.

    Figure 13

    Figure 13 answer

    Figure 14

    Figure 14 answer


  2. On Figure 14, draw boundaries for all points that are within three miles of helipad A and those that are three miles of helipad B. Is it clear which helipad would service people living within three miles of helipad A? Explain.

    3. Some of the people within three miles of Helipad A are clearly closer to A’s helicopters. Likewise, some of the people within three miles of Helipad B are clearly closer to B’s helicopters. However, those people on the segment that connects the two intersection points of the circles are the same distance from A and B. Therefore, some people within three miles of A are closer to B, and vice versa.

  3. In Figure 15, circles of radii of four miles and five miles are drawn around points A and B. Mark and label the points of intersection. How far is each of these points from the helicopters at A and B?


    		4.

    Points I1 and I4 are both 5 miles from points A and B, while Points I2 and I3 are both 4 miles from points A and B.

    Figure 15 answer

    		  


  4. Connect the points of intersection in Figure 15. What kind of pattern do you notice? How would you describe the complete set of points as it relates to where people live and to helicopters located at A and B?

    5. The points all lie on a line, which is the perpendicular bisector of AB. People living at these intersection points are equidistant from both A and B, and could be serviced by either helicopter company.

  5. If you wanted equal-sized circles to touch, rather than cross over, each other, what would their radii have to be? Where is the point at which they would intersect exactly once?

    6. About 2.2 miles. At the midpoint of the segment joining A and B.

  6. In Figure 16, draw a pair of circles to find the perpendicular bisector of A and B. Then do the same for A and C. Repeat the process for points B and C.

    7. Figure 16

    Figure 16 answer

  7. The three perpendicular bisectors that you have drawn intersect at a point. Mark that point V and describe the location of that point.

    8. If this were a first-quadrant graph, V would be located near (11,9). Other descriptions may also be used, e.g., V is about five miles east of A.

  8. In Figure 17, plot the point V and draw the circle centered at V and going through one of the centers of influence (helipad at A, B, or C). What do you notice about the position of V relative to the other centers of influence? Draw the segments AB, AC, and BC. A circle is said to circumscribe a polygon if the circle contains the vertices of the polygon. What do you notice about the circle you’ve just drawn?

    9. Figure 17 answer

    The point V is a center of the circle that circumscribes the triangle ABC. Thus, V is equidistant from A, B, and C.

    In each of the following problems, construct perpendicular bisectors (by the most convenient means) to determine the center of the circle that contains the points A, B, and C. Draw the triangle ABC.


  9. Figure 18


    Figure 18 answer



  10. Figure 19


    Figure 19 answer


  11. How does the type of triangle affect the location of the center of the circumscribed circle for that triangle?

    12. Question 9 shows an obtuse triangle; the center of the circle is outside the triangle. Question 10 is a right triangle; the center of the circle is on the triangle.

  12. Are you ready to tackle the rain gauge locations in Colorado (or Texas)? Let’s see! The figure below should be broken into a Voronoi diagram, with the labeled points the centers of influence for each of the regions.

    13. Figure 20


    Figure 20 answer



Supplemental Activity 2—Bird Territories
   

Voronoi diagrams help determine what regions are best associated with specific rain gauges. Voronoi tilings also help define territories of specific birds that are guarding areas around their nests. In each domain indicated below, birds’ nests are located at points A, B, C, D, E, and F. Your task is to find the territories of domination for the birds in each of the following problems.


  1. Figure 21


    Figure 21 answer



  2. Figure 22


    Figure 22 answer


  3. Use the results from Questions 1 and 2 to help you determine the boundaries for the territories of the birds that have nests located at points P, Q, R, and S in the figure below.

    4. Figure 23


    Figure 23 answer


  4. A Voronoi vertex is the point at which Voronoi boundaries meet. One of the interior vertices of a Voronoi region in the solution to the above problem is equidistant from points P, Q, and R. Identify that vertex.

    5. The vertex is at point X.

  5. Why wasn’t the perpendicular bisector of PS used?

    6. Because all the points on that line are closer to either point Q or point R.

  6. To which three centers of influence is the other interior vertex equidistant?

    7. Q, R, and S.

  7. Every interior vertex of a Voronoi diagram is at the center of a circle on which at least three of the points of influence must lie. Why must this be true?

    8. Because the vertex is at the intersection of two perpendicular bisectors; it is equidistant from the three points that define the two perpendicular bisecting segments.

Determine the Voronoi tiling for each of the following diagrams. For each interior vertex of a Voronoi region, identify which centers of influence are equidistant from that vertex.


  1. Figure 24


    Figure 24 answer

    P is equidistant from A, B, and D. R is equidistant from B, C, and D.
    Q is equidistant from A, C, and D.


  2. Figure 25


    Figure 25 answer

    P is equidistant from A, B, C, and D.



  3. Figure 26


    Figure 26 answer

    Point P is equidistant from points A, C, and E. Point Q is equidistant from points A, B, and E. Point R is equidistant from points B, D, and E.




Activity 3—Area of Concern
   

 = 8, 9, 14

One of the two hurdles to cross in the development of our second model is now out of the way. Given a domain with the position of some rain gauges known, it is possible to subdivide the areas into regions of proximity. These are regions in which the closest gauge represents the entire region. From this, we still have to be able to estimate total rainfall, however. Let’s look at a simplified problem; the map below represents a 15-miles by 10-miles county with three gauges located in it. The Voronoi regions for the map are shown.

Figure 27
  1. What is the area of region 1, the shape defined by the points A, B, D, M, and F? Explain the method you used to come up with your answer.

    2. I broke the area into two rectangles and two triangles. The areas of the rectangles are (6.6)(3) and (2.8)(7), while the areas of the triangles are 0.5(.5)(3) and 0.5(4.2)(7). The total area is 54.85 mi2.

  2. If you compare your answer with other students in your class, do you expect that there will be one "right" answer? Explain.

    3. Probably not; answers will vary with the method used to determine them.

  3. Do you think you have found the correct area for region 1? If not, explain how you could do better.

    4. Probably not. Very good estimates are possible, but someone could always get a more precise answer than mine by using a better way to determine area.

  4. Using your method from Question 1 again, find the areas of regions 2 and 3.

    5. Region 2: 0.5(7)(4.2) + 0.5(6)(8) + (1)(8) = 46.7 mi2

    Region 3: 0.5(0.5)(3) + 0.5(8)(6) + (8)(3) = 48.75 mi2

  5. The depth readings, in inches, for the gauges are given in the table at right. Estimate the total rainfall for the county. Explain your method.

    6. Table 7

    1

    2

    3

    2.11 in.

    2.15 in.

    1.21 in.


    I used the method of weighted averages, in which the amount of area became the "weight." The calculations look like this:

    (54.85 mi2)(2.11 in.) + (46.7 mi2)(2.15 in.) + (48.75 mi2)(1.21 in.) = 275 in.-mi2 = 22.9 ft.-mi2 = 14,700 acre-ft. = 638,000,000 ft3

  6. Here is another problem, featuring 4 gauge locations. What is the estimate for the total rainfall for this county?

    7. Gauge

    Gauge

    Gauge

    Gauge

    1

    2

    3

    4

    2.11 in.

    2.15 in.

    1.21 in.

    4.45 in.



    Figure 29

    Figure 28


    Areas

    Region 1: (7)(7.8) – 0.5(5.7)(5.5) – .5(1.3)(7.8) – 0.5(7)(2.3) = 25.8 mi2

    Region 2: (2.5)(5.5) + 0.5(5.7)(5.5) + 0.5(0.9)(2.5) = 30.6 mi2

    Region 3: (5.5)(7.8) + 0.5(1.3)(7.8) + (2.2)(4.2) + 0.5(2.2)(1.3) = 58.6 mi2

    Region 4: (2.5)(3.7) + 0.5(2.5)(0.9) + (7)(2.2) + 0.5(2.2)(1.3) + 0.5(7)(2.3) = 35.3 mi2

     

    Rainfall Estimate:

    (25.8)(2.11) + (30.6)(2.15) + (58.6)(1.21) + (35.3)(4.45) = 348 in.-mi2 = 29.0 ft-mi2 = 18,600 acre-ft = 809,000,000 ft3




Homework 3—The Means Justify the End
   

Let’s examine how a method for obtaining an area can affect the answer. The Voronoi region in Figure 30 below is going to be used for each of the methods explored.

Figure 30

  1. Work with Region 3; divide it into a rectangle and two triangles by drawing a line straight down from point P and another line going straight across from point E to your first line.


    1. Find the area of each part of Region 3 then find the total area.

      2. Rectangle: 16 mi2 Triangles: 2 mi2 and 12.5 mi2

      Total Area: 30.5 mi2

    2. Using this method, does the way in which you divide the original region affect your answer?

      3. No, but the work may be easier or more difficult depending on the way in which you break up the region.

  2. Work with Region 4 now, and apply a rule called Pick’s formula. Pick’s formula says that, if the vertices of a polygon are points of a lattice (grid), the area of the polygon is Apoly = 0.5b + c – 1, where b is the number of lattice points on the boundary of the polygon and c is the number of lattice points in the polygon’s interior. (Note: The polygon can be either concave or convex, but no side can cross another.) What does Pick’s formula predict for the area of Region 4?

    3. b = 11 c = 33 Area = 37.5 mi2

  3. Work with Region 1 now; we will apply Heron’s formula for finding the area of a triangle from the length of its sides. The formula is

    4. Atriangle =

    where a, b, and c are the lengths of the sides of the triangle and s = a + b + c / 2. Here’s a step-by-step breakdown on how to use it the formula.

    1. Divide Region 1 into two triangles. It doesn’t matter which way, but, for this activity, let’s agree that the two triangles will be DEBP and DBPF.


    2. Find the length of side EP. If you measure, don’t forget to scale the measurement into distances of miles (each grid represents one mile). How could you find the length of side EP without measuring?

      3. EP » 4.1 mi. Use Pythagoras’ theorem if you don’t want to measure.

    3. Find the lengths of the other sides that form DEBP.

      4. EB = 6 mi. BP » 6.4

    4. Add all the lengths, then divide by two. This is the value of the triangle’s semiperimeter, s. In this example, what do you get as the value for s?

      5. s » 8.25

    5. For each length, subtract it from the value of s. You are finding the values of the formula terms (s – a), (s – b), and (s – c). Multiply all those answers together, then multiply that by s. Finally, take the square root of that answer. What does Heron’s formula predict for the area of that answer. What does Heron’s formula predict for the area of the first triangle?

    6. Repeat the steps for the other triangle. Then add the two areas together. What is the total area of Region 1?

      7. Area of DBPF = 20 mi2 Total area of Region 1 » 31.94 mi2


  4. Finally, work with Region 2. Try your luck at a Monte Carlo simulation by generating a random number [0..15) on your calculator, then a random number [0..10). The pair of numbers forms an (x, y) coordinate on your map. Keep track of the number of times you run the simulation and the number of times the point lies in Region 2. The ratio (Number of random points in Region 2) : (Number of trials) gives you an estimate of the percent of the entire area that lies within Region 2. The more trials you do, the better the estimate becomes.


    5. Number of trials

    Number of Random Points in Region 2

    50

    14


    Since the number of times the dot "landed" in Region 2 was 28% of the time, an estimate for the area of Region 2 would be 28% of the total area of the county. Since the county is 15 miles by 10 miles, 28% of 150 is about 42 mi2.

  5. Of the various methods used, which one might be:


    1. the most accurate?

      2. Heron’s formula, since a polygonal region can always be broken down into triangular pieces, and Pythagoras’ Theorem gives good distance calculations if the coordinates are fairly accurate.

    2. the most limited in use?

      3. Pick’s formula can be used only when the vertices are on the grid dots.

    3. the most time-consuming?

      4. The answer depends on the availability or absence of technology. The Monte Carlo method could be quite tedious if one had to do each trial singly and determine if the dot is in the correct region. The first approach could also be fairly tedious, since it didn’t use an algorithm for breaking the region into rectangles and triangles, and, if the region was more irregularly shaped, it would require a lot more work.

    4. the one that lends itself best to the use of technology?

      5. Either the Monte Carlo method (how many times must we run the simulation to get reasonably accurate results?) or Heron’s formula.

    5. the one to use when the number of regions gets as large as Texas?

      6. Since Monte Carlo isn’t a suburb of Dallas-Ft. Worth, we’ll probably want to use repeated applications of Heron’s formula.



Activity 4—Model Gets a Second Chance
   

We’ve been hip-deep in mathematics, but progress has definitely been made on the improved version of the model. It’s time to apply what you’ve learned so far to the problem of estimating the rainfall for the state of Colorado.

  1. In considering this newer version of the mathematical model, what key assumption changed?

    2. The assumption that the gauge locations didn’t matter. Now we’re assuming that the gauge locations determine which gauge represents a particular part of the state.

  2. The plan described in Activity 1 on how to proceed in solving the problem also had to be changed.


    1. What condition determines the region covered by each gauge?

      2. The gauge that is closest to a particular part of the state represents it.

    2. What mathematical process did we apply to determine which regions were best represented by each of the gauges?

      3. Voronoi diagrams, based upon perpendicular bisectors of the segments connecting two gauges.

    3. What other major area of consideration did we have to resolve before we could put our model back together?

      4. How to find the area of each of the Voronoi regions.

  3. Use the map of Colorado Figure 1b, that goes with this activity. In your group, determine which parts of the state should be represented by each of the rain gauges. The next steps are affected by the work done here, so be careful!

    4. See handout on next page.

  4. Now, in your group, figure out how much of the state is represented by each of the rain gauges. Remember that Colorado is 375 mile × 275 mile. Each person in the group should do at least two regions (you might consider doing one or two of your partners’ regions, to double check the work). Record your final answers Table 8.

    5. Table 8

    Gauge No.

    Area of Region

    Percent of Entire State

    1

    11,600 sq mi

    11.2

    2

    5920 sq mi

     5.7

    3

    10,400 sq mi

    10.1

    4

    17,400 sq mi

    16.9

    5

    5730 sq mi

     5.6

    6

    7470 sq mi

     7.2

    7

    24,700 sq mi

    24.0

    8

    20,100 sq mi

    19.5


  5. Here are the rainfall depths (in inches) recorded for the eight gauges located in Colorado again.

    6. Table 9

    X1

    X2

    X3

    X4

    X5

    X6

    X7

    X8

    2.11

    2.15

    1.21

    4.45

    2.67

    2.51

    3.11

    2.43


    Using your work from Question 4, determine the total amount of rainfall that fell on Colorado according to your new model. Be sure to show all your work.

    (2.11)(11600) + (2.15)(5920) + (1.21)(10400) + (4.45)(17400) + (2.67)(5730) + (2.51)(7470) + (3.11)(24700) + (2.43)(20100) = 287,000 in.-mi2 = 23,900 ft-mi2 = 15,300,000 ft-acres

  6. Convert your answer to a standard unit of ft3, then compare your answer to the one found in Activity 1, when you were using your simplified model. Does this new answer make sense? Explain.

    7. 666,000,000,000 ft3. The previous estimate was 618,000,000,000 ft3. The higher estimate can be attributed to the fact that the largest and third-largest regions had the highest rain gauge values.

  7. Test the model again.


    1. If the gauge depths were off by 0.01 in, how would that affect your answer?

      2. A change in the depths of 0.01 in. would produce a change in the volume estimate of roughly 2.4 × 109 ft3.

    2. If each of the Voronoi regions had areas that were off by 1 square mile, how would that affect your estimate of the total rainfall?

      3. A change in each of the areas of 1 square mile would produce a change in the volume estimate of roughly 5.2 × 107 ft3.

  8. Look back over what you’ve done in this second pass through the modeling process. Describe the steps you took and the mathematical basis for your answer.

    9. In changing the assumption that the relative locations of the gauges made a difference, I treated those gauges as centers of influence for the regions they represent. I broke the state into a Voronoi diagram, found the area of each part of the state, and took a weighted average of the rainfall estimates that fell on each part.

    Activity 4 Map (solution)
    Figure 31




Homework 4—Texas Storms Are SO Big…
   

The final project for this unit is going to be to estimate the amount of rainfall in Texas during the 1997 calendar year from actual data. Here are the locations of the 17 "first-order" stations across the state. An 18th gauge, located in Shreveport, Louisiana, is added to the list; scientists often include it since no first-order station is located in East Texas. As you are working through the project stages, just remember: There are actually over 600 weather observing stations across the state, so this project could have been a whole lot worse!

Abilene

Houston

Amarillo

Lubbock

Austin

Midland/Odessa

Brownsville

San Angelo

Beaumont/Port Arthur

San Antonio

Corpus Christi

Victoria

Del Rio

Waco

Dallas/Ft. Worth

Wichita Falls

El Paso

Shreveport, Louisiana

You need to locate a map of Texas with which to work. It can be an actual road map, like the type that AAA provides, or it can be a photocopy of an atlas map of the state, but it should be fairly big so you can actually work with it. Mark the places where the rainfall gauges are located. Then divide the state into regions of proximity by finding all the Voronoi boundaries. Depending on the size of the map, you may want to know just where the gauge is located in that region. The data-collection stations are located at the airports in their vicinities.

Yes, you will have to find the areas of those regions, so you can start doing that as soon as you’ve divided the state into its various regions of influence. A table in which to record the area calculations, and the data on the rain gauge depths, will be provided in a few days. The map will be part of the work you turn in, so be neat and accurate! Good luck!




Activity 5—Get the Point?
   

In our latest attempt to refine the model, we are going to work with exact locations for the rain gauges in the hope that this will determine the boundary lines exactly, and will allow us to calculate the lengths of line segments and the area of regions just as precisely.

As before, it makes sense to look at a simpler problem before tackling the Colorado map.

Part 1

 = 3, 4, 19, 23

The figure at right has two rain gauges shown, with the given coordinates. We want to find the equation of the line that forms the perpendicular bisector of line segment AB. If we knew a point on that line and its slope, we’d be able to write the equation for that line. Since we know only points A and B, the slope will have to be found from the coordinates; we’ll also use a special point on the line.

Figure 32

  1. Imagine a circle centered at point A and another one the same size centered at point B, with the circles just touching each other in one point. What do we know about that point’s position with respect to the perpendicular bisector?

    2. It will be on the perpendicular bisector, because any point the same distance from both gauges (circles being the same size guarantees this) has to be on the perpendicular bisector.

  2. What are the coordinates of that point? How did you determine that?

    3. M(5, 5.5). It’s halfway between A and B, so the x- and y-coordinates are also halfway between.

  3. What is the slope of line segment AB? What is the slope of the perpendicular bisector to segment AB?

    4. Slope of line segment m1 = –3/4
    Slope of ^ bisector m2 = 4/3

  4. What is the equation for the boundary line of the Voronoi region (perpendicular bisector to segment AB)?

    5. y – 5.5 = 4/3(x – 5)

  5. Rewrite your answer in slope-intercept form.

    6. y = 4/3x – 7/6

Now, let’s go through the same steps, only this time we will generalize the process to find the perpendicular bisector from any two sets of coordinates P(x1, y1) and Q(x2, y2).

  1. What would be the coordinates of the midpoint of the line segment PQ? Generalize the method used to answer Question 2.

    2. The midpoint M would have coordinates (x1 + x2)/2, (y1 + y2)/2.

  2. What’s the relationship between the line segment connecting points A and B and the perpendicular bisector of that line segment? What’s the relationship between the slopes of those lines?

    3. The lines are perpendicular. The product of their slopes must equal –1; another way to express this is to say m2 = –1/m1.

  3. What would be the slope of the boundary line (perpendicular bisector of line segment PQ)? Generalize the method used to answer Question 3.

    4. Slope of PQ = (y2 – y1) / (x2 – x1), so the slope of ^ bisector = (x1 – x2) / (y2 – y1)

  4. What would be the equation (in point-slope form) for the boundary line? Generalize the method used to answer Question 4.

    5. Y – ((y1 + y2) / 2) = ((x1 – x2) / (y2 – y1))(X – (x1 + x2) / 2)

  5. Now, simplify your answer to Question 9, so that the equation for the boundary line is expressed in slope-intercept form.

    6. y = ((x1 – x2) / (y2 – y1)) x + ((y1 + y2)(y2 – y1) – (x1 – x2)(x1 + x2)) / (2(y2 – y1)) = ((x1 – x2) / (y2 – y1)) x + (y22 – y12 – x12 + x22) / (2(y2 – y1))



Part 2

Find the equations of the three perpendicular bisectors defined by these three centers of influence.

Figure 33

^ bisector to AC: y = –7x + 53
^ bisector to AB: y = 0.75x + 1.625
^ bisector to BC: y = – 0.8x + 11.9





Homework 5—Rain Keeps Falling on My Head
   

Here is the other information you need to complete the unit project—the rainfall data from the 18 primary measuring stations for the 1997 calendar year.

Rainfall Measuring Station

Measurement (in.)

  

Rainfall Measuring Station

Measurement (in.)

Abilene

27.08

  

Houston

76.79

Amarillo

24.95

  

Lubbock

22.67

Austin

46.79

  

Midland/Odessa

17.10

Brownsville

36.21

  

San Angelo

23.38

Beaumont/Port Arthur

44.10

  

San Antonio

33.92

Corpus Christi

36.16

  

Victoria

65.05

Del Rio

23.01

  

Waco

43.27

Dallas/Ft. Worth

45.00

  

Wichita Falls

23.78

El Paso

9.63

  

Shreveport

69.20


In the project, you may want to incorporate more gauges, or compare total rainfall from one or two different years. The following web site will provide you with as much information as you can swallow:

http://www.ncdc.noaa.gov/ol/climate/climatedata.htm

The other thing you need to be working on is determining the areas of the various regions of proximity surrounding these measuring stations. When you’ve calculated those values, record them in the table below, but be sure to hold on to the work; it must be turned in as part of the project!

Rainfall Measuring Station

Proximity Area (mi2)

  

Rainfall Measuring Station

Proximity Area (mi2)

Abilene

    

Houston

  

Amarillo

    

Lubbock

  

Austin

    

Midland/Odessa

  

Brownsville

    

San Angelo

  

Beaumont/Port Arthur

    

San Antonio

  

Corpus Christi

    

Victoria

  

Del Rio

    

Waco

  

Dallas/Ft. Worth

    

Wichita Falls

  

El Paso

    

Shreveport

  





Activity 6—Power in Those Rules
   

A general rule for finding the equation of the perpendicular bisector seemed hard to develop at first. However, solving a problem in this way makes future problems of the same kind as easy as plugging in numbers. This suggests that we might want to look for ways to develop general rules for the other parts of our rain gauge problem. If we know the equations for two boundary lines, can we determine where the intersection point is? If we know the locations of the endpoints of our boundary lines, can we determine their lengths?

Part 1

 = 5, 6, 7, 21

  1. Use the substitution method to determine the coordinates of point P.

    2. Figure 34

    0.4X + 3.5 = –2X + 22.5
    2.4X = 19
    X = 19/2.4 » 7.92

    4(7.92) + 3.5 » 6.67 or
    –2(7.92) + 22.5 » 6.66

  2. Now generalize what you’ve done in solving that system of equations. Let the equations be: y1 = m1x + b1 and y2 = m2x + b2.


    1. Set the two expressions for y equal to themselves.

      2. m1x + b1 = m2x + b2

    2. Subtract m2x from both sides of the equation and simplify.

      3. m1x + b1 – m2x = m2x + b2 – m2x
      (m1 – m2) x – b1 = b2

      Subtract the constant term, b1, from both sides of the equation. Then simplify by combining b2 and b1.

      (m1 – m2) x + b1 – b1 = b2 – b1
      (m1 – m2) x = b2 – b1

    3. Divide by the coefficient of x to solve for x.

      4. x = (b2 – b1) / (m1 – m2)

    4. Substitute the result of step d) for x in either of the original equations. Solve for y.

      5. y = m1(b2 – b1) / (m1 – m2) + b1 or y = m2 (b2 – b1) / (m1 – m2) + b2.

  3. Practice: Find the point of intersection of the lines y = x + 2 and y = –3x – 1.

    4. x = (–1 – 2) / (1 – –3) = –3/4

    y = (–3/4) + 2 = 5/4

    Solution: (–3/4, 5/4)


Part 2

 = 5, 6

  1. Use Pythagoras’ theorem to find the length of the line segment AB shown in Figure 35.

    2. Figure 35

    The height of the triangle is 2 and the base is 8. Pythagoras’ theorem says to square those numbers, add them together, and then take the square root of the answer. So the length of AB is around 8.25.


 = 7, 8

  1. How could you have done those same calculations, using just the coordinates (instead of knowing that you were working with the sides of a triangle)?

    2. Take the x-coordinates and subtract them; the base was 8 because the x-coordinates were 3 and 11. Take the y-coordinates and subtract them. Those are the numbers that you then square and add together. Take the square root of that answer.

  2. Now, generalize what you’ve done in finding the length of a line segment. Let the endpoints be A (x1, y1) and B (x2, y2).


    1. Write an expression for the base and height of the triangle in terms of the coordinates of points A and B.

      2. base: x2 – x1 height: y2 – y1

    2. Now, square those numbers and add them together.

      3. (x2 – x1)2 + (y2 – y1)2

    3. Finally, to represent the distance between points A and B (the length of AB), you need to square root.

      4.
  3. Now, put it all together. Points A (14, 30), B (20, 20), C (30, 30) and D (36,22) are locations for four rain gauges. What is the distance between V1 (Voronoi vertex central to gauges A, B and C), and V2 (Voronoi vertex central to gauges B, C, and D)?

    4. ^ Bisector of AB: y = 0.6x + 14.8    ^ Bisector of BD: y = –8x + 245

    ^ Bisector of BC: y = –1x + 50        ^ Bisector of CD: y = 3/4x + 5/4


    Coordinates of V1: x = (50 – 14.8) / (0.6 – –1) » 22
                                  y = ((0.6)(50) – (–1)(14.8))/(0.6 – –1) » 28 (22, 28)

    Coordinates of V2: x = (1.25 – 245) / (–8 – 0.75) » 28
                                  y = ((–8)(1.25) – (0.75)(245))/(–8 – 0.75) » 22 (28, 22)


    Distance of V1V2:





Homework 6—Prelude to a Problem Solved
   

On a piece of graph paper, create a rectangular domain with (0, 0) and (10, 10) as corner points, then plot the four points below. Find the Voronoi diagram for the four points. Be sure to find the equations of the line segments and the coordinates of the vertices that make up the Voronoi diagram.

A: (1, 3)    B: (7, 4)    C: (2, 6)    D: (9, 5)

Figure 36




Activity 7—Third Time’s the Charm
   

 = 22

We’re going to revisit the problem of estimating the rainfall in Colorado one last time. Activity 1 was where we first set up our model for solving the problem, and Activity 4 dealt with the "new, improved" model based on more realistic assumptions. Take a few minutes to look over that work, especially the assumptions that were made each time and the mathematical approaches used in getting the answers.

Figure 37



GaugeNumber

Location (miles)

1

  (83, 234)

2

  (155, 249)

3

  (265, 235)

4

  (98, 132)

5

  (150, 158)

6

  (197, 173)

7

  (280, 173)

8

  (159, 36)


This time, we want to calculate the amount of rainfall from knowing the precise location of each of the gauges. Assume that Colorado is a 375-mi by 275-mi rectangle, with the origin (0, 0) located in the lower left corner. The coordinates given in the table above were determined using that same reference system. Your task is to find the coordinates of the vertices of the Voronoi diagram, then calculate the area of each region. Finally, use the rainfall data below to estimate the total amount of rainfall received by the state. Good luck!


Table 10

X1

X2

X3

X4

X5

X6

X7

X8

2.11

2.15

1.21

4.45

2.67

2.51

3.11

2.43


Answers are based on the labels provided in the picture bwlow.



Equations

FP: y = 1.5x + 169.1
PQ: y = 8.8x + 93.5
GQ: y = –48x + 813.7
QR: y = –0.6x + 213
RS: y = 5.5x + 113.4
HS: y = 83.3x – 1508.5
ST: y = –11x + 458.2
IT: y = 2.4x + 138
TU: x = 238.7
UJ: y = –8.8x + 298.2
VU: y = –2.8x + 153.6
RV: y = –31.4x + 709.8
WV: y = 0.8x + 85
PW: y = –19.4x + 385.1
EW: y = 6.3x + 2.5


Segment Lengths (mi)

  

Intersection Coordinates

  

Area of Proximity Regions

FP

104.97

  

E

(0.0, 2.5)

  

1 (FBGQP)

11,560 mi2

PQ

31.25

  

F

(0.0, 169.1)

  

2 (GHSRQ)

5,910 mi2

GQ

71.31

  

G

(112.2, 275.0)

  

3 (HCITS)

10,340 mi2

QR

34.46

  

H

(214.2, 275.0)

  

4 (EFPW)

17,400 mi2

RS

53.69

  

I

(375.0, 229.2)

  

5 (PQRVW)

5,710 mi2

HS

46.47

  

J

(338.3, 0.0)

  

6 (RSTUV)

7,450 mi2

ST

44.87

  

P

(103.6, 184.6)

  

7 (TIDJU)

24,650 mi2

IT

140.22

  

Q

(126.8, 205.0)

  

8 (AEWVUJ)

20,420 mi2

TU

108.17

  

R

(161.5, 202.8)

    

UJ

133.01

  

S

(208.7, 229.0)

    

VU

46.47

  

T

(238.7, 196.1)

  

Total Area of Colorado:

RV

108.17

  

U

(238.7, 87.8)

  

103,125 mi2

WV

45.67

  

V

(194.2, 100.1)

    

PW

99.36

  

W

(149.1, 96.6)

    

EW

176.28

          

Rainfall Estimate Calculation:

(2.11)(11560) + (2.15)(5910) + (1.21)(10340) + (4.45)(17400) + (2.67)(5710) + (2.51)(7450) + (3.11)(24650) + (2.43)(20420) = 287,267



Dividing the Job
   

Peter is a forest ranger responsible for a rectangular region of forest, as shown in Figure 38. His lookout station is situated on top of a hill, marked P in the figure. Three other hilltops in the area are marked A, B, and C.

Figure 38

The grid shows the exact locations of the four hilltops: P (3, 6), A (11, 8), B (9, 3), and C (15, 2). The origin of this grid, O (0, 0), is in the lower left corner.

The forest is too big for just one ranger. A second lookout will be stationed on one of the three other hilltops (A, B, or C) in this region. Peter and Amy (the second ranger) will be responsible for the part of the forest that is nearest to his or her lookout stations. The hilltop that divides the forest into two parts of about the same area will be chosen for the second lookout station.

Figures 39-41 are copies of Figure 38. Use them to work on Items 1-4.


    1. On Figure 39, draw the regions into which the forest is divided when hilltop A is chosen for the second lookout station.


    2. Estimate the percentage of the forest for which Peter is responsible.


  2. On Figures 40 and 41, mark the division of the forest when hilltop B and hilltop C are chosen for the second lookout station.


  3. Which one of the three choices is the best? Explain.

You chose the location of the second lookout based on which location divides the forest into two approximately equal regions. Another fair way to divide the forest is to choose the hilltop location that minimizes the sight distance from a lookout to the most distant region boundary.


    1. On Figures 39-41, draw the longest sight distance for any of the three possible choices.

      2. See the dotted lines on Figures 39-41.

    2. Using this additional criterion, what is the best choice for the position of the second lookout station? Explain your answer.

      3. Hill A has coordinates (11, 8). The most distant point in the region is (17, 0). Then

        Sa =
        Sa = 10

      Hill C has coordinates (15, 2) and the most distant point in the region is (10.9, 10)

        Se =
        Se = 9.0

      Based on the additional criteria, C is chosen over A. The final choice is hill C.

      Figure 39

        Ap = 10((6.5 + 8.9)/2) = 77
        Ap = 10((10.5 + 8.1)/2) = 93
        Atotal = 170
        %p = 77/170 × 100 = 45%

      Figure 40

      Using B

        Ap = 10((3.9 + 8.5)/2) = 62
        %p = 62/170 × 100 = 36%

      Figure 41

      Using C

        Ap = 10((7.7 + 10.9)/2) = 93
        %p = 93/170 × 100 = 55%



Unit Project—Rainy Days in Texas
   

The central problem in this unit is to estimate the total amount of rainfall received in a region from the amount of rain collected at certain points within that region. Each rain gauge becomes the center of influence for the region around it, and representative of the entire region’s behavior. You are going to estimate the total rainfall in Texas for the 1997 calendar year, with data compiled from 18 primary measuring stations scattered across the state (one of them even in Louisiana).

In Homework 4, you were introduced to the various locations in which the rainfall is measured. Your task there was to locate an appropriately sized map and begin the process of resolving the state of Texas into Voronoi regions. In Homework 5, you were provided the actual rain gauge data and asked to begin calculating the area of each of those regions. You may have run into some difficulty, since Texas isn’t shaped like a rectangle.

In any case, in this project you will provide an estimate of the total rainfall for Texas during the 1997 calendar year. Materials to turn in for completion of the project include the map you used, the Homework 4 and Homework 5 assignment pages, and all the work you did in calculating your estimate. You should write a complete description of the steps you took in doing the project, identify problems you had along the way, and describe how you overcame them. Also, if possible, suggest ways to improve the process, or get a better answer.

Here are some other ideas that you might want to include in your project work.

  1. Locate more rain gauge data on the Internet and include it in your calculations.


  2. Contact someone who works with data collection to find out how the instruments work. Report your findings.


  3. Contact someone who works with the analysis of data to see if he or she uses a different method of calculation.


  4. Research other fields/careers that might need information on total rainfall, besides the Texas Natural Resources Conservation Commission.


  5. Expand the model to include the water-storage capacity of the lakes/ground and the changes in volume from stream flow, consumption, and evaporation.


Have fun, and good luck!




Unit Summary—Modeling Estimation of Rainfall Totals

In this unit, the problem of estimating the total amount of rainfall that falls on a region was examined. The first, and simplest, model for estimating something like rainfall for an entire region is found by taking an average. This process is based on the unrealistic assumption that everywhere behaves the same—an assumption that is contradicted by the data. If the data points are far away from each other, or if they represent different amounts of the region, it makes more sense to "weight" the different data points. In the rainfall example, we changed the assumption to include the relative position that each gauge has with each other. This meant that each gauge should have its reading weighted by the percent of the state that is closer to it than to any other gauge, or by the total amount of area assigned to the gauge through a Voronoi diagram.

Voronoi boundaries are the lines that divide regions associated with centers of influence. The boundary between two regions of influence is the perpendicular bisector of the segment joining the centers of the two regions. In studying how to determine the location for these lines, construction tools were used, including the Mira, wax-paper folding, and compass/straightedge. It was found that a Voronoi vertex, the place where three boundary lines come together, was also the center for a circle that circumscribes the triangle formed by the three gauges surrounding that vertex. One strategy for determining Voronoi diagrams for complicated regions was to break them into smaller, three-point problems.

When you have defined Voronoi regions, the areas of those regions are used to determine a weighted average of rainfall. Several methods were explored for finding the area of each of the regions of influence. The estimation for the total volume of rain is found by summing the products of the area of the region times the amount of rainfall in the gauge that represents that region.

A final approach through the modeling process was to further refine the work done by incorporating the additional information of the exact locations for the rain gauges. This gain in precision allowed us to move away from construction tools, hand drawings, and estimations of area. By developing equations for the boundary lines, we could determine the exact locations for intersection points, distances between those points, and areas of proximity regions, thereby improving dramatically the estimation of total rainfall. We derived formulas that assisted us in the computations, and we completed the model. After this, to get a "better" answer, you’d need to process a lot more rain gauge data, but the way in which you calculate the answer wouldn’t change, and a decent computer program could handle the 600+ weather observing stations in Texas with relative ease. Who knows, maybe you’ll be writing that program!




Mathematical Summary

The modeling problem from which the mathematics in this unit arises is that of estimating the rainfall for the entire state of Colorado from readings taken at eight rain gauges scattered around the state.

The solution is geometric: Divide the state into eight regions so that the points in a region are closer to its gauge than to any other gauge. Weight the rainfall measured at each region’s gauge according to the portion of the state’s area in that region. The weighted average of the rainfall at the eight gauges estimates the rainfall for the state.

Proximity problems like the Colorado rain gauge problem involve Voronoi diagrams, which are named after the mathematician Georgii Voronoi. To use Voronoi diagrams, you must determine the boundaries of regions from their centers of influence.

The boundaries can be drawn roughly by hand, but answers obtained from rough drawings lack precision. Therefore, the boundaries should be constructed. There are several means of constructing the boundaries. Every Voronoi boundary is the perpendicular bisector of a segment joining two centers. Perpendicular bisectors can be constructed by several methods:

  1. by folding a piece of paper and creasing it so that two centers coincide;


  2. by placing a Mira so that the reflection of one center coincides with the other center;


  3. by striking intersecting compass arcs from the centers and joining the two points of intersection;


  4. or by using the segment, midpoint, and perpendicular construction features of a drawing utility.

Perpendicular bisectors are lines, but Voronoi boundaries are either rays or line segments. Moreover, the perpendicular bisector for some pairs of centers is not a boundary in the Voronoi diagram. Therefore, when perpendicular bisectors are constructed, they must be analyzed carefully to determine which portions to keep. Algorithms for establishing Voronoi boundaries often divide the problem into several smaller problems of, say, three or four vertices, then combine the diagrams that result.

Voronoi regions are usually polygons. (An exception occurs when the boundary of the domain is curved.) Many modeling problems, including the Colorado rain gauge problem, require determination of each region’s area. One way to find a region’s area is to divide it into triangles and apply Heron’s formula, which finds the area of a triangle from the lengths of its sides.

where s = (a + b + c)/2

Another method is to apply Pick’s formula, which finds the area of any polygon whose vertices are points of a grid from the number of grid points that are on the polygon’s border or inside the polygon. If vertices fall on a grid

Ab = 0.5b + c – 1

where b = ± grid points on boundary, c = ± grid points interior to boundary.

Areas can be measured by using a drawing utility’s measuring features or estimated by designing and running a simulation.

Another approach to the problem of finding Voronoi boundaries and the areas of Voronoi regions involves coordinate geometry. If coordinates of the centers are known, the equations of boundary lines can be found.

  1. Find slope of connecting segment.
  2. Find midpoint.
  3. Write the slope of a perpendicular and the mid point in the point-slope form of a linear equation given slope = mid point (x0, y0).

y – y0 = m(x – x0)

After equations of two intersecting perpendicular bisectors are found, the coordinates of the Voronoi vertex at which they intersect are found by solving the system of their two equations. The coordinates of the vertices can be used to determine the area of each region. One way to do this is to divide each region into triangles, use the coordinate geometry distance formula to find the lengths of any unknown sides, then use Heron’s formula to find each triangle’s area.

Since coordinate geometry develops algebraic formulas for geometric objects from known properties of those objects, the results can be used to develop computer or calculator programs that find equations of boundaries and areas of regions. Since the formulas of coordinate geometry give exact results, the precision of answers obtained from coordinate methods is limited by only the precision of the measurements used in the formulas.




Key Concepts

Acute triangle: A triangle in which all of the angles measure less than 90°, but more than 0°.

Center of influence: A point used to establish boundaries of regions of influence. All points in a region are close to that region's center than to any other region's center.

Concave polygon: A polygon in which some of its sides, when extended, intersect other sides.

Convex polygon: A polygon in which none of its sides, when extended, intersect other sides. For every pair of points in the interior of a convex polygon, the segment connecting the points is completely in the interior.

Domain: A region in which centers of influence are located. The domain is the area that is being divided into regions of influence.

Heron's formula: The area of a triangle is
where a, b, and c are the lengths of the triangle's sides and s = 1/2(a + b + c).

Iteration (Iterative procedure): A procedure that repeats the same sequence of steps over and over. Each cycle is considered one iteration.

Midpoint: A point that is halfway along a segment (equidistant from the segment's two endpoints). In coordinate geometry, the coordinates of a midpoint are found by averaging the coordinates of the two endpoints.

Obtuse triangle: A triangle with one angle that measures more than 90 degrees, but less than 180 degrees.

Perpendicular bisector: A line that passes through the midpoint of a given line segment and forms right angles with it.

Pick's formula: If the vertices of a polygon are points of a grid, then the area of the polygon is A = 0.5b + i – 1 where b is the number of grid points on the polygon's border, and I is the number of points in its interior.

Region of influence: A region in which each point is closer to the region's center of influence than to any other center of influence.

Voronoi boundary: A boundary between two centers of influence.

Voronoi center: A center of influence.

Voronoi diagram: A diagram composed of several centers of influence and their regions of influence.

Voronoi region: A region of influence.

Voronoi vertex: A point at which Voronoi boundaries intersect.

Weighted average: The average found by multiplying each category by the decimal weight attached to that category and find a total.




Solution to Short Modeling Practice

Solution for Modeling Center of Gravity

The center of gravity problem is not usually presented as a weighted average in physics or engineering texts. The usual defining equation for weights at discrete locations is

The justification for describing this as a weighted average is seen if the total weight is distributed across the summation.

Each distance in the summation is weighted by the ratio of the weight at that location to the total weight.

This is an exact analogy to a rain gauge reading being weighted by the ratio of its service area to the total area of a state.

Try to lead the students to the conclusion that the math in the rain problem and the calculation of the location of the CG are the same.

The location of the camping gear that would balance the canoe is found as follows:

W = 120 + 60 + 200 + 300 = 680

5 = (120/680)(2) + (60/680)(5) + (200/680)(8) + (300/680)xgear

xgear = 4.21 ft.


Solutions to Practice and Review Problems

Exercise 1

  1. The students’ constructions should be similar to the following figure.


  2. 3 cm


Exercise 2

12/5


Exercise 3

4/50


Exercise 4

c2 = a2 + b2

602 + 912 = 11,881


Exercise 5

c2 = a2 + b2

162 = 42 + b2

256 = 16 + b2

240 = a2


Exercise 6

Use the distance formula


Exercise 7

Use the distance formula


Exercise 8

  1. The two triangles are FAB and FBC; the trapezoid is FCDE.


  2. To calculate the area of triangle FAB, you need the base, which is 245.5', and the height 479.6'. For triangle FBC, you need the base 538.8' and the height 424.9'. For the trapezoid FCDE, you need the height 134', base1 which is 231', and base2 which is 932.5'.


  3. Area of a triangle = 1/2 × Base × Height
    Area of triangle FAB = 1/2 × 245.5' × 479.6'
    Area of triangle FAB = 58,863.9 ft2
    Area of triangle FBC = 1/2 × 538.8' × 424.9'
    Area of triangle FBC = 114,468.06 ft2
    Area of a trapezoid = 1/2 × (Base1 + Base2) × Height
    Area of trapezoid FCDE = 1/2 × (231' + 932.5') × 134'
    Area of trapezoid FCDE = 77,954.50 ft2
    Total area of field = 58,863.9 ft2 + 114,468.06 ft2 + 77,954.50 ft2
    Total area of field = 251,286.46 ft2

  4. Acres = Square feet ÷ 43,560 square feet per acre
    Acres = 251,286.46 ft2 ÷ 43,560 ft2 per acre
    Acres = 5.77 acres (rounded)

Exercise 9

The surface area of a lake can be approximated by summing the area of the individual trapezoids. For ease in working the problem, number the trapezoids from left to right.

Area of a trapezoid = 1/2 × (Base1 + Base2) × Height
Area1 = 1/2 × (1 mi + 4 mi) × 2 mi
Area1 = 5 mi2


Area2 = 1/2 × (4 mi + 8 mi) × 2 mi
Area2 = 12 mi2


Area3 = 1/2 × (8 mi + 7.5 mi) × 2 mi
Area3 = 15.5 mi2


Area4 = 1/2 × (7.5 mi + 9 mi) × 2 mi
Area4 = 16.5 mi2


Area5 = 1/2 × (9 mi + 10 mi) × 2 mi
Area5 = 19 mi2


Area6 = 1/2 × (10 mi + 7 mi) × 2 mi
Area6 = 17 mi2


Area7 = 1/2 × (7 mi + 1 mi) × 2 mi
Area7 = 8 mi2


Total area = Area1 + Area2 +...Area7

Total area = 5 mi2 + 12 mi2 + 15.5 mi2 + 16.5 mi2 + 19 mi2 + 17 mi2 + 8 mi2

Total area = 93 mi2


Exercise 10

  1. Volume = Length × Width × Height
    Volume = 48' × 40' × (18" × 1'/12")
    Volume = 2880 ft3


  2. Gallons = Volume in cubic feet × 7.48 gallons per cubic foot
    Gallons = 2880 ft3 × 7.48 gal per ft3
    Gallons = 21,542.4 gal


  3. Time to empty = Volume ÷ Rate
    Time to empty = 21,542.4 gal ÷ [2 pumps × (30 gal per min per pump)]
    Time to empty = 359 min (rounded)

    4. or,

    Time to empty = 359 min ÷ 60 min per hr
    Time to empty = 6.0 hr (rounded)


Exercise 11

  1. Volume of tub = Length × Width × Height
    Volume of tub = 54" × 22" × 11"
    Volume of tub = 13,068 in3


  2. Gallons in tub = Volume of tub ÷ 231 in3 per gal
    Gallons in tub = 13,068 in3 ÷ 231 in3 per gal
    Gallons in tub = 56.57 gal (rounded)


  3. Day’s reserve = Gallons in tub ÷ Gallons used per day
    Day’s reserve = 56.57 gal ÷ 3 gal per day
    Day’s reserve = 18.86 days (rounded)

Exercise 12

A hemisphere and a cone are two geometric figures that could be used to approximate the shape of the hot-air balloon shown. The volume would then be the sum of the volumes of these two shapes.

From the illustration, the hemisphere has a diameter of 50 feet (or a radius of 25 feet). The cone’s base has the same radius of 25 feet, and a height of 35 feet.

  1. Volume of hemisphere = 1/2 volume of sphere, or 2/3 p r3, r = d/2
    Volume of hemisphere = 2/3 × 3.14 × (50' ÷ 2)3
    Volume of hemisphere = 32,708 ft3 (rounded)

    2. Volume of cone = 1/3 pr2h, r = d/2
    Volume of cone = 1/3 × 3.14 × (50' ÷ 2)2 × 35'
    Volume of cone = 22,896 ft3 (rounded)

    Volume of balloon = Volume of hemisphere + Volume of cone
    Volume of balloon = 32,708 ft3 + 22,896 ft3
    Volume of balloon = 55,604 ft3

  2. Amount of lift = Volume of hot air × Lift per ft3 of hot air
    Amount of lift = 55,604 ft3 × 0.020 lb of lift per ft3
    Amount of lift = 1112 lb


Exercise 13

  1. Convert 15" to feet d = 15"(1 ft/12") = 1.25 ft

    2. Volume of rain = Area × Depth
    Volume of rain = 43,560 ft2 × 1.25 ft
    Volume of rain = 54,450 ft3

  2. The unit conversion factor is (7.48 gal/1 ft3)

    3. Gallons of rain = Cubic feet × Gallons per cubic foot
    Gallons of rain = 54,450 ft3 (7.48 gal/1 ft3)
    Gallons of rain = 407,286 gal (rounded)


Exercise 14

  1. A pipeline is a long cylinder.

    2. To obtain cubic feet, convert measurements to feet:

    Diameter = 8" × 1'/12" = 0.667'
    Length = 10 mi × 5280'/1 mi = 52,800'

    Volume of pipeline = pr2h, r = d/2
    Volume of pipeline = 3.14 × (0.667' ÷ 2)2 × 52,800'
    Volume of pipeline = 18,400 ft3 (rounded)

  2. At reduced pressure, you have seven times the volume.

    3. Volume at customer pressure = Volume at transmission pressure × 7
    Volume at customer pressure = 18,400 ft3 × 7
    Volume at customer pressure = 129,000 ft3 (rounded)


Exercise 15

The strategy for this exercise is to find the area to be covered with topsoil, and then multiply by the uniform depth of soil to find the volume of soil.

  1. The total area of the lot is the width times the length.

    2. Alot = W × L
    Alot = 100 feet × 125 feet
    Alot = 12,500 ft2

  2. To find the area of the house requires some deduction of dimensions not given. The L-shaped house with the remaining dimensions is shown here. The area covered can be determined many ways.

    3. One way is to identify two rectangular pieces that make up the house. One is 60' × 28'. The other 22' × 22'.

    Ahouse = (60 ft × 28 ft) + (22 ft × 22 ft)
    Ahouse = 2164 ft2

  3. The area of the driveway is probably the trickiest part of this exercise. The students should determine the circular area of the outer circle (60' diameter). Then subtract the area of the smaller circular area (40' diameter). Then, since the driveway is semicircular, we want only half the area.

    4. Adrive = 1/2[p(60'/2)2 – p (40'/2)2
    Adrive = 785 ft2 (rounded)

    The 15' × 10' walkway can be considered rectangular, though there will be a tiny error in this assumption since the meeting of the walkway and the driveway is curved.

    Awalk = 15 ft × 10 ft
    Awalk = 150 ft2

  4. The topsoil area is then the total lot area less all the areas that won’t be covered.

    5. Asoil = Alot – Ahouse – Adrive – Awalk
    Asoil = 12,500 ft2 – 2164 ft2 – 785 ft2 – 150 ft2
    Asoil = 9401 ft2

  5. The depth of soil must be converted to the same basic unit used for the area: feet. In other words, 4 inches = 1/3 foot. (Or the students can convert the ft2 value for Asoil to in2.) Then multiply the depth by Asoil to find the volume of soil needed.

    6. Vsoil = Depth × Asoil
    Vsoil = 1/3 foot × 9401 ft2
    Vsoil = 3134 ft3 (rounded)

  6. Use the fact that 27 ft3 = 1 yd3.

    7. Vsoil = 3134 ft3 × (1 yd3/27 ft3)
    Vsoil = 116.1 yd3 (rounded)

  7. So, the number of trucks of soil, at 5 yd3 per truck is...

    8. Ntruck = Vsoil ÷ 5 yd3/truck
    Ntruck = 116.1 yd3 ÷ 5 yd3/truck
    Ntruck = 23.2 trucks

    Or, if you want every bit of dirt, you’ll round up to 24 trucks.


Exercise 16

  1. Since 1 acre = 43,560 ft2 and 9 ft2 = 1 yd2...

    2. Asoil = 1.62 acres × (43,560 ft2/1 acre) × (1 yd2/9ft2)
    Asoil = 7840.8 yd2

  2. Since a uniform depth of 18 inches will be removed, the volume will be the product of the area and the depth. The depth must be converted to yards, using 36 inches = 1 yard.

    3. d = (18 in.)(1 yd/36 in.) = 0.5 yd
    Vsoil = Asoil × Depth
    Vsoil = 7840.8 yd2 (0.5 yd)
    Vsoil = 3920.4 yd3

  3. Each truck can haul 10 yd3 of soil, so the number of trucks needed is...

    4. Ntrucks = Vsoil ÷ 10 yd3 per truck
    Ntrucks = (3920.4 yd3/(10 yd3/truck))
    Ntrucks = 392 trucks (rounded)


Exercise 17

  1. Solving this exercise is mainly a matter of dividing the tank into manageable parts. The illustrations in the text pretty much lead the students through this process.

    2. The first part is the rectangular "box" that is given as 40' × 28' × 8'.

    Vbox = L × W × H
    Vbox = (40 ft)(28 ft)(8 ft)
    Vbox = 8960 ft3

  2. The diameter of the cylindrical sides is the same as the height of the tank, 8 feet. Thus, the radius of each cylinder is 4 feet. The length of the cylinders is the same as the length (or width) of the rectangular portion. Thus, there are two 40-ft half-cylinders and two 28-ft half-cylinders, or a total of one each.

    3. Vcyl = V40 + V28
    Vcyl = pr2h40 + pr2h28
    Vcyl = p (4 ft)2 (40 ft) + p (4 ft)2 (28 ft)
    Vcyl = 3418.1 ft3 (rounded)

  3. Each spherical segment is 1/4 of a sphere and there are four identical segments, so the total volume of these segments is equal to the volume of one sphere of 8-foot diameter, or 4-foot radius.

    4. Vsph = 4/3 pr3
    Vsph = 4/3 p (4 ft)3
    Vsph = 268.1 ft3 (rounded)

  4. The total volume is the sum of the parts.

    5. Vtot = Vbox + Vcyl + Vsph
    Vtot = 8960 ft3 + 3418.1 ft3 + 268.1 ft3
    Vtot = 12,646.2 ft3

    Since 1 ft3 = 7.48 gal...

    Vtot = 12,646.2 ft3 × (7.48 gal/1 ft3)
    Vtot = 94,594 gal (rounded)


Exercise 18

s = (a + b + c)/2

s = (6 + 9 + 9)/2 = 24/2 = 12


Exercise 19

  1. (x – 2)2 + (y – 2)2 = 25

  2. The engineer will input the coordinates (2,2) and a range of 75 miles. Cities A, B, H, and I will be added to the viewing area with a total viewer increase of 77,000.

  3. (x – 2)2 + (y – 2)2 = 7.52




Exercise 20

A reduced version of the traced map is shown here, where "W" represents Waco, "H" is Hillsboro, "G" is Gatesville, "T" is Temple, and "M" is Marlin.

A portion of the perpendicular bisectors of the line's joining each office are shown forming the Voronoi region.

The Voronoi region can be broken into 4 triangles, as shown below, having a total area of 1038 mi2.

The area of the 40-mile radius circle = p (40 mile)2 = 5027 mi2. Thus the Waco-office area is 1038 mi2/5027 mi2 = 21% of the entire advertised region.

It would appear that a branch office in the northeast section would significantly reduce the burden of the Hillsboro and Marlin offices. This could be defended by calculating the percent coverage of these two offices and showing how they would be favorably reduced by locating another branch office, perhaps in Dawson or Hubbard.

Sides Miles Sq Miles
AW 19  
WD 16  
DA 16 126
 
WB 27  
BA 27
AW 19 242
 
WB 27  
BC53 
CW35423
 
CW35 
WD16 
CD46247
 
TOTAL Area1,038


Exercise 21

  1. tw + te = 8


  2. Dw = 235tw


  3. De = 205te


  4. 235tw = 205te


  5. 440tw = 1640; tw ~ 3.7; te ~ 4.3; 3.7 + 4.3 = 8


  6. Susan can fly about 875.8 miles in a westerly direction.


Exercise 22

  1. x + y = 3 or y = 3 – x


  2. 0.05x + 0.09y = 0.24


  3. 0.05x + 0.09(3 – x) = 0.24
       0.04x = 0.03
       x = 0.75 and y = 2.25

Shon should mix 0.75 liter of 5% solution and 2.25 liters of 9% solution to obtain 3 liters of 8% solution.


Exercise 23

  1. 32 = a(0) + b or b = 32


  2. 212 = a(100) + b or 100a = 212 – b


  3. 100a = 212 – 32
       100a = 180
       100 = 1.8


  4. F = 1.8(0) + 32    F = 1.8(100) + 32
    F = 32    F = 212