2010 Excellence in Mathematics Contest
Team Project Level II
(Below Precalculus)
School Name:
Group Members:
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Reference Sheet
Formulas and Facts
You may need to use some of the following formulas and facts in working through this project. You may not need
to use every formula or each fact.
A bh
Area of a rectangle
C 2l 2w
Perimeter of a rectangle
1
bh
2
Area of a triangle
C 2 r
A
Circumference of a circle
A
r2
Area of a circle
m
y2 y1
x2 x1
Slope
a 2 b2 c 2
Pythagorean Theorem
5280 feet = 1 mile
3 feet = 1 yard
16 ounces = 1 pound
2.54 centimeters = 1 inch
100¢ = $1
1 kilogram = 2.2 pounds
1 ton = 2000 pounds
1 gigabyte = 1000 megabytes
1 mile = 1609 meters
1 gallon = 3.8 liters
1 square mile = 640 acres
1 sq. yd. = 9 sq. ft
1 cu. ft. of water = 7.48 gallons
1 ml = 1 cu. cm.
V
r 2h
V
Area of Base height
Volume of cylinder
Lateral SA = 2
Volume
r h
Lateral surface area of cylinder
b 2 4ac
2a
Quadratic Formula
x
b
4 3
r
3
Volume of a sphere
V
tan
sin
cos
References: www.usbr.gov
www.enchantedlearning.com
www.associatedcontent.com
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TEAM PROJECT Level II
2010 Excellence in Mathematics Contest
____________________________________________________________________________________________
The Team Project is a group activity in which the students are presented an open ended, problem situation
relating to a specific theme. The team members are to solve the problems and write a narrative about the theme
which answers all the mathematical questions posed. Teams are graded on accuracy of mathematical content,
clarity of explanations, and creativity in their narrative. We encourage the use of a graphing calculator.
Flying Under the Gateway Arch in St. Louis, Missouri – A Hypothetical Situation
The St. Louis Visitors and Convention Bureau has contacted you about a new idea that is being pitched to the
Bureau next month. A local airplane pilot, Charles, has requested permission to conduct tourist flights through the
Gateway Arch. According to www.enchantedlearning.com, “the St. Louis Gateway Arch is an elegant monument
to westward expansion in the USA. Located on the banks of the Mississippi River in St. Louis, Missouri, the 630foot tall stainless steel arch rises above the city skyline.” Charles would like to take tourists on a thrilling flight
passing through the Gateway Arch. The Bureau, considering the safety of such a stunt, wants to be sure that the
plane Charles will be using will be able to safely pass through the arch at the altitude proposed by Charles. Charles
has indicated that he will fly through the arch at an altitude of 600 feet, safely passing through the arch. Can this
be done? To answer this question, complete the following activities.
Part I – Graphical Investigation
1. Using the image of the Gateway Arch below to develop a function that models the shape of the arch. The input
variable, h, is the horizontal distance from the left edge of the arch and the output variable, v, is the vertical
height of the arch. For this part of the project, assume that the arch follows a parabolic path and express your
function in standard form ( v f (h) a h2 b h c ).
v
We anticipate that students will use quadratic
regression using the 3 points (0, 0), (315, 630), and
(630, 0). If students do not use calculators, they can
solve a system of 3 equations and 3 unknowns.
v
f (h)
0.006349h2
4h
h
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2. Sketch a graph of your function. Be as accurate and neat as you can be. Include appropriate numeric scales and
label the axes.
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3. Using the graph you just created and the fact that the airplane that Charles will fly through the arch has a
wingspan of 36 feet, determine if the flight can be performed safely. Remember that Charles would like to fly
through the arch at an altitude of 600 feet. Write several sentences describing your ideas. Show your ideas
visually using the graph you just created.
Since the plane will be flying at a height of 600 feet
above the ground, we can see that the horizontal
distance between the legs of the arch at that height
is more than enough to accommodate the 36 foot
wingspan of the plane. In fact, we can estimate the
horizontal distance between the legs of the arch to
be about 150 feet.
Roughly, 150 feet
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Part II – Algebraic Investigation
1. Since pilot Charles has indicated that he plans to fly the plane through the arch at an altitude of 600 feet, write
an equation that would determine the values of h on the parabola that would produce a v-value of 600.
600
0.006349h2 4h
2. Solve the equation found you created above. Describe what your result means in the context of this problem
situation. Express your final result as a decimal number rounded to the hundredths place.
0.006349h 2
4h 600
h
h
0
16 4(0.006349)(600)
4
2(0.006349)
2(0.006349)
246.26,386.74
When the horizontal distance from the left leg of the arch is 246.26 feet and again when the horizontal distance
is 386.74 feet, the height of the arch above the ground is 600 feet.
3. Use the results from number 2 above to determine the horizontal distance across the Gateway Arch at the 600
foot height. Express your final result as a decimal number rounded to the hundredths place.
383.74 246.26
137.48 feet
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4. Charles’s plane, a Cessna 180, has a wingspan of 36 feet. Do you think that if he is to fly his plane through the
arch at an altitude of 600 feet that there will be enough clearance for the plane to make it through? Explain.
Since the wingspan of the plane is only 36 feet, there is plenty of space for Charles to fly the plane through
the arch at a height of 600 feet above the ground.
5. Carefully analyze the relationship between the solutions to your equation from number 2 above and the vertex
of the parabola (i.e. the arch). What relationship to you notice? That is, explain how the vertex can be used to
establish the symmetry found in the arch.
The vertex, (315, 630), is exactly in the middle of the horizontal line segment connecting the two points on the
arch where the height above the ground is 600 feet. There is symmetry about this vertex.
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Part III – Analyzing Rates of Change
1. Complete the following chart using your quadratic function from Part I. The rate of change column is to be
created by finding the rate of change from the values in the row above to the values in the row containing the
rate of change you are focusing on finding.
Horizontal distance from
left edge h (in feet)
Vertical height ground to arch
V (in feet)
Rate of Change
0
0
XXXXXXXXXXXX
100
336.51
336.51
3.3651 vertical feet for each horizontal foot
100
200
546.03
209.52
100
2.0952 vertical feet for each horizontal foot
300
628.57
82.54
100
0.8254 vertical feet for each horizontal foot
400
584.13
44.44
100
0.4444 vertical feet for each horizontal foot
500
412.70
171.43
100
1.7143 vertical feet for each horizontal foot
600
114.29
298.41
100
2.9841 vertical feet for each horizontal foot
2. Choose one of the rates of change computed in the table and write a description of what this rate of change
means in the context of this problem situation. That is, if you were going to help Charles understand the
significance of this particular rate of change, what would you tell him? Be sure to include appropriate units in
your description.
For example, the rate of change of 2.0952 vertical feet for each horizontal foot means that as you move closer to
the center of the arch horizontally, the height above the ground increases by 2.0952 feet.
Charles might be more interested in understanding the reciprocal relationship. As he increases his altitude by
say 2.0952 feet, the horizontal distance between the legs of the arch is changing by 2 feet (1 foot on each side of
the vertex).
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3. Work to create a graph of the rate of change of v as a function of the horizontal distance, h. Write several
sentences explaining what information this graph displays in the context of the situation. Like in number 2
where you described one particular rate of change computation, you will now explain what the rate of change
function means in the context of this situation.
This is a bit open ended and students might respond differently and each way might be appropriate. They may
decide to plot the rate of change values as a function of the horizontal distance at the right end of the subinterval
given in the table.
Or, students might assume an average rate of change over each 100 foot change in horizontal distance and
represent the rate of change graph as a step function.
It is possible that students would create a linear relationship using the left end of the subinterval or even using
something in the middle. But, the table pushes students to do one of the two graphs above.
Students might tell Charles that the rate of change function is showing that as the horizontal distance from the
right leg of the arch increases, the change in the vertical distance, while positive initially, gets less and less
positive until the rate of change is zero (at the vertex). Then, as the horizontal distance continues to increase, the
rate of change in the vertical distance becomes negative and gets more and more negative as the horizontal
distance increases.
Charles might again be more interested in the reciprocal relationship. It takes smaller and smaller changes in the
vertical distance to create a 2 foot change in the distance between the legs of the arch (again, we have to
consider a 1 foot change on either side of the vertex). For example, when h = 300, the change in the vertical
distances is less than 1 foot (0.8254 vertical feet for each increase in 1 horizontal foot). This means that if
Charles increases his altitude by only 0.8254 feet, his “cushion” on either side of the wings of his plane decrease
by 2 feet! The closer he is to the vertex, the smaller the change in altitude needed to shrink this “cushion” by 2
feet.
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4. What is the maximum altitude that Charles could fly and still make it safely through the arch? Show all the
work that supports your answer. It is advisable to include graphs to help support your reasoning and your final
result. Also discuss any real world considerations that might be made as you think about the mathematics
involved in this problem situation.
Students should take advantage of the symmetry established earlier. Since the wingspan is 36 feet and assuming
that Charles will fly the plane through the center of the arch, he will have to be 315 – 18 = 297 horizontal feet
from the center (vertex).
v
f (297)
627.94
0.006349(297) 2
4(297)
If Charles flies at a height of 627 .94 feet, the horizontal distance between the legs of the arch will be 36 feet. Of
course, this will be a tight squeeze since the wingspan of the plane is 36 feet! We might suggest that Charles fly
at a height of 627 feet.
Furthermore, we have not considered the height of the plane. At a height of 627 feet above the ground, there
will be less than 3 feet to the bottom of the arch! We want to be careful not to scrape the top of the plane across
the bottom of the arch as we fly through!
Consider students thinking on such issues when judging their response.
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Part IV – Perspective Regarding the Size of the Arch
In this part, you will approximate the volume of the arch. That is, imagine that you could fill the arch with water.
How much water would it take to fill the arch? Below are several facts that may prove to be useful for this part of
the project.
Arch consists of 142 equilateral triangular sections.
The equilateral triangles at the base have sides of length 54 feet.
The equilateral triangles at the top of the arch have sides of length 17 feet.
Completed in October 1965, the cost to build the Arch was $13 million.
The usual sway of the Arch is 1/2" and the maximum sway in a 150 mph wind is 9" each way.
Nine hundred tons of stainless steel was used to build the Arch and it weighs 17,246 tons.
1. Estimate the total length of the arch (as if we could straighten it out and measure the total length of the arc
formed by the arch) by using straight line segments from the bottom to the top (see figure).
v
d
2 652 507 2
1580 feet
2 2502 1232
(65, 507)
h
2. At the base of the arch (at the origin), what is the area of the triangular cross section?
A
1
54 542 27 2
2
1262.67 square feet
3. If the triangular cross sections don’t decrease from bottom to top, determine the volume of the arch using the
result from #2 above.
V
1262.67 1580 1,995,018.6 cubic feet
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4. At the top of the arch (the final section being placed in the photo), what is the area of the triangular cross
section?
A
1
17 17 2 8.52
2
125.14 square feet
5. If the triangular cross sections don’t decrease from bottom to top, determine the volume of the arch using the
result from #4 above.
V
125.14 1580 197,721.2 cubic feet
6. According to www.usbr.gov, the Hoover Dam was constructed using 3.25
million cubic yards of concrete. How many times could you fill the St. Louis
Arch using the amount of concrete used to build the Hoover Dam? You may
take an average of the larger volume and lesser volume found previously.
We will take the volume of the arch to be 1,096,369.9 cubic feet. This is
121,818.9 cubic yards.
How many copies of the volume of the arch are contained in the volume of the
Hoover Dam?
3, 250, 000 cubic yards
26.7
121,818.9 cubic yards
You could fill the arch almost 27 times with the concrete used to make the
Hoover Dam!
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Part V – Actually a Catenary
Because of the physics involved in the St. Louis arch, we can model its shape more accurately as a catenary rather
than as a parabola. The simplest way to think about a catenary is to imagine holding the ends of a string and
allowing the string to hang.
This catenary can be modeled using exponential growth (the right side of the catenary) and exponential decay (the
left side of the catenary).
This type of function is used often enough that it is given its own name…it is a hyperbolic cosine (cosh) function
and is defined as follows:
cosh(ax)
eax
e
2
ax
The St. Louis Arch is actually in inverted catenary and is best modeled by the function
v
f (h)
68.8cosh(0.01h 3.15) 700 .
Use this hyperbolic cosine function to confirm that Charles can safely fly the airplane through the arch at an
altitude of 600 feet. Remember, the wingspan of the airplane is 36 feet. How do your results here compare to what
you found when you modeled the arch using a parabola? You may use technology to help you.
Using the hyperbolic cosine function as a model, it appears as
though the horizontal distance from leg to leg at a height of 600 feet
above the ground is about 180 feet! This is about 40 feet greater than
when we examined the situation using the quadratic model.
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Part V Continued…
The horizontal distance is 406.961 – 223.039 = 183.922 feet.
Using technology to solve, we do indeed find that the horizontal distance between the legs of the arch is greater
than when using the quadratic model.
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Technology Tip
The hyperbolic trigonometric functions are defined as follows:
e ax e ax
2
ax
e
e ax
sinh(ax)
2
These functions may be graphed using the TI-84 graphing calculator as follows:
cosh(ax)
Technology Tip: Hyperbolic Trigonometric Functions
1. Bring up the graphing list by pressing the Y= button.
2. Type in the function(s) using the X,T, ,n button for the variable
and the ^ button to place an expression in an exponent. Make sure
you use parentheses as needed.
3. The cos( and sinh( functions can be accessed by first pressing
CATALOG (2nd 0). You can then press the letter C (PRGM) to
quickly move to functions beginning with C or S (LN) to quickly
move to function beginning with S. Note that the calculator is
already in ALPHA mode so there is no need to first press the
ALPHA button. However, you can access the alphabet anytime
you need by pressing ALPHA followed by the key corresponding
to the letter you need.
4. Enter in the desire arguments for the sinh( function (for example).
5. Draw the graph by pressing the GRAPH button.
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GRADING RUBRIC FOR St. Louis Arch Project
High School Mathematics Contest - 2010
WORK/RESULTS
Correct computations throughout
o Part I
 Quadratic Function
 Accurate Graph
o Part II
 Set up equation
 Solve equation
 Determine horizontal distance
 Correct vertex
o Part III
 Correct entries in ROC table
 Correct interpretation of ROC
 Accurate ROC function graph
 Correct maximum height computation
o Part IV
 Correct total length of arch
 Correct area of larger triangle
 Correct larger volume
 Correct area of smaller triangle
 Correct smaller volume
 Correct comparison to Hoover Dam
o Part V
 Correct work with the hyperbolic cosine function
 Appropriate comparison to the parabolic model
Accurately written interpretations of mathematical ideas
o Part I – explain how graph is used to address problem situation
o Part II – explain relationship between vertex and vertical intercepts
o Part III – accurate description of rate of change value and function/graph
TOTAL……….
____/5
____/10
____/10
____/15
____/5
____/15
____/60
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