Math Modules (DRAFT)

NSF-ATE: Fuse LAB Mathematics Modules

H¨seyin Y¨ce
u u
Draft date December 10, 2012

Contents

Contents i

Preface 1

1 Mathematical Modeling and Functions Module 3
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Library of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.1 Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.2 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . 6
1.2.3 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . 6
1.2.4 Logistic Functions . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.5 Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . 8
1.2.6 Cubic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.2.7 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 9
1.3 Tangents and Normals . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.4 Regressions and Interpolations . . . . . . . . . . . . . . . . . . . . . . 14
1.5 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.6 Appendix: Review of Solving Triangles . . . . . . . . . . . . . . . . . 19
1.7 Problems, Projects, and Activities . . . . . . . . . . . . . . . . . . . . 22

2 Discrete Mathematics Module 29
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.2 Logic and Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

i

ii CONTENTS

2.2.1 Logic and Truth Tables . . . . . . . . . . . . . . . . . . . . . . 29
2.3 Boolean Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.3.2 Properties of Boolean Algebras . . . . . . . . . . . . . . . . . 35
2.3.3 Boolean Functions . . . . . . . . . . . . . . . . . . . . . . . . 38
2.3.4 Switching Circuits . . . . . . . . . . . . . . . . . . . . . . . . 38
2.3.5 Logic Networks . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2.4 Modular Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2.5 Recurrence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2.6 Problems, Projects, and Activities . . . . . . . . . . . . . . . . . . . . 39

Bibliography 41

List of Figures

1.1 The graph representation of the model for the tunnel . . . . . . . . . 5
1.2 Triangular representation of the tunnel . . . . . . . . . . . . . . . . . 5
1.3 Exponential growth and decay functions . . . . . . . . . . . . . . . . 7
1.4 Logarithmic growth and decay functions . . . . . . . . . . . . . . . . 8
1.5 Logistic growth and decay functions . . . . . . . . . . . . . . . . . . . 8
1.6 Concave up and concave down quadratic functions . . . . . . . . . . . 9
1.7 Cubic functions with a > 0 . . . . . . . . . . . . . . . . . . . . . . . . 10
1.8 Cubic functions with a < 0 . . . . . . . . . . . . . . . . . . . . . . . . 10
1.9 Sine and cosine functions . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.10 Average rate of change vs. instantaneous rate of change [2] . . . . . . 13
1.11 Length of a guy wire . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.12 Length of a guy wire . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.13 Rectangular picnic park . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.14 Fencing function of rectangular picnic park . . . . . . . . . . . . . . . 19
1.15 Angles and sides of a triangle . . . . . . . . . . . . . . . . . . . . . . 20
1.16 Angles and sides of a triangle . . . . . . . . . . . . . . . . . . . . . . 21
1.17 The Gateway Arch in St. Louis . . . . . . . . . . . . . . . . . . . . . 23
1.18 The rectangular box . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.19 Open diagram of cylindrical can . . . . . . . . . . . . . . . . . . . . . 25
1.20 Highway design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.21 Norman window . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
1.22 Cable-stayed bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

iii

iv LIST OF FIGURES

1.23 Animation of the cable-stayed bridge . . . . . . . . . . . . . . . . . . 27

2.1 Primes in 2D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
2.2 Primes in complex plane . . . . . . . . . . . . . . . . . . . . . . . . . 38

Preface

This workbook contains selected mathematical topics that are widely used in ar-
chitecture and civil engineering freshmen courses. The topic selection is based on
the “Fuse LAB” project supported by NSF-ATE grant under the direction of PI
Shelley Smith. Each chapter represents a module to be covered within ARCH 3590
Computation Fabrication course and few civil engineering courses. The module also
make use of computer algebra system Mathematica and provides Mathematica codes
when necessary.

1

Chapter 1

Mathematical Modeling and
Functions Module

1.1 Introduction
The process of translating a real-world problem into a usable mathematical equation
is called mathematical modeling, and the equation is referred to as a model. We
use mathematical models to describe numerical data or verbal information.

Deﬁnition 1.1.1 A function is a rule that takes certain numbers as inputs and
assigns to each a deﬁnite output number. The set of all input numbers is called the
domain of the function and the set of resulting output numbers is called the range
of the function.

Representation of Functions Functions can be represented by tables, graphs,
formulas, and descriptions in words. Example below shows the use of each repre-
sentations.
Example: A civil engineer is planning to dig a tunnel through a mountain. The
tunnel will begin 575 feet above the sea level and will be constructed with a constant
upward slope of 5%; that is, the tunnel will rise vertically 5 feet for every 100 feet of
horizontal distance. Table 1.1 shows the amount of vertical rise of several horizontal
distances [1].

(a) Use the verbal description and the above table to write a model for the elevation
of the tunnel in herms of the horizontal distance from where the tunnel begins
at the base of the mountain.

3

Horizontal increase (feet) Vertical increase (feet)
100 5
200 10
300 15
400 20
500 25

Table 1.1: Vertical rise vs. horizontal distance

(b) Find the elevation of the tunnel at a horizontal distance of 2500 feet from the
starting point.
(c) If the tunnel exits the mountain at a horizontal distance of 7000 feet from
where it began, what is the elevation of the tunnel when it emerges from the
mountain?
(d) If the tunnel will cost $120 per foot to construct, what will be the cost of
building the tunnel?

Solution:

(a) The elevation of the tunnel starts at 575 feet and rises 5 feet for every 100
horizontal feet. We can write this as
Elevation of tunnel = 575 + 5 feet for every 100 horizontal feet
Let y be the elevation and x be the horizontal feet in hundreds, then the above
statement can be converted into
y = 575 + 5x feet above sea level
This function is a model for the elevation of the tunnel. Figure 1.1 is the
graphical representation of the model.
(b) Substitute x = 25 into the equation above, y = 575 + 5 ∗ 25 = 700 feet above
sea level.
(c) Using the same substitution method in (b), for x = 70 we have y = 575+5∗70 =
925 feet above sea level.
(d) Using the Pythagorean theorem for the right triangle in Figure 1.2, we obtain
√
d = 70002 + 3502 = 7008.74 ≈ 7009 feet

4

y
1000

800

600

400

200

x
10 20 30 40 50

Figure 1.1: The graph representation of the model for the tunnel

d
350 feet

7000 feet

Figure 1.2: Triangular representation of the tunnel

5

1.2 Library of Functions
In this section, we build a collection of widely used functions and study their prop-
erties.

1.2.1 Linear Functions
Linear functions have a constant rate of increase or decrease. A function is linear
if its slope, or rate of change, is the same everywhere. So, linear functions can be
identiﬁed as functions whose outputs are repeated additions. A linear function has
the form
y = f (x) = b + mx.
Its graph is a straight line such that m is the slope (rate of change of y with respect
to x), b is the vertical intercept.
Note that if the slope is zero, m = 0, we have y = b, a horizontal line. The
equation of a line of slope m through a point (x0 , y0 ) is
y − y0 = m(x − x0 ).
The tunnel problem in section 1.1 is a linear model.

1.2.2 Exponential Functions
Exponential function has a constant percentage change, that is to say it is a func-
tion whose output is the result of repeated multiplication by a constant at regular
intervals. An exponential function has the form
y = f (x) = abx
where a = 0 and b > 0. If b > 1, f (x) is exponential growth and If 0 < b < 1, f (x)
is exponential decay as shown in Figure 1.3, which is plotted by computer algebra
system Mathematica
Plot[Exp[x], {x, -3, 3}, PlotRange -> {-1, 8}, AxesLabel -> {"x", "y"},
PlotStyle -> Thick]

1.2.3 Logarithmic Functions
A logarithmic (log) function has a vertical asymptote (the line x = 0) and continues
to grow or decline as x becomes large. A log function has the form
y = f (x) = a + b ln x

6

y y
8 8

x x
y y
6 6

4 4

2 2

x x
3 2 1 1 2 3 3 2 1 1 2 3

Figure 1.3: Exponential growth and decay functions

where b = 0. If b > 0, f (x) is a growth and If b < 0, f (x) is decay function as shown
in Figure 1.4.
A note on inverse relationship:
If we have data whose input/output relationship can be modeled by an exponential
function, then the inverse (output/input) relationship can be modeled by a log
function, and vice versa. This relationship in its simplest form can be stated:
If f (x) = ln x and g(x) = ex , then f (g(x)) = ln (ex ) = x and g(f (x)) = eln x =
x as long as x is positive.
The inverse relationship for the model f (x) = a + b ln x becomes f −1 (x) = AB x
where A = e−a/b and B = e1/b .

1.2.4 Logistic Functions
Sometimes it is unrealistic to believe that exponential growth can continue forever.
The constrains that slows down the growth results S-shaped behavior. A mathe-
matical function with such an S-shaped curve is called logistic function and it has
the form
L
y = f (x) =
1 + Ae−Bx
where L is the limiting value of the function. If B > 0, f (x) is a growth and If
B < 0, f (x) is decay function as shown in Figure 1.5. Logistic functions have two
types of curvatures, concave up and concave down. The point at which the graph
of f (x) changes its concavity is called an inﬂection point.

7

y
y
5

2 4
y=ln(x)
3
y=-ln(x)
x
1 1 2 3 2

1
2
x
1 1 2 3

1
4

2

Figure 1.4: Logarithmic growth and decay functions

y y
5 5

4 4
5 5
y x
y 1 x
1 3 3

2 2

1 1

x x
5 5 5 5

1 1

Figure 1.5: Logistic growth and decay functions

1.2.5 Quadratic Functions

When the first differences in data are constant, the model is linear. When the second
differences are constant, the data can be modeled by the quadratic function and it
has the form
y = f (x) = ax2 + bx + c

as long as a = 0. The graph of a quadratic function is a parabola. If a > 0, f (x) is
concave up at all times and If a < 0, f (x) is concave down at all times as shown in
Figure refquad.

8

y y
5 1
2
y x
4 x
3 2 1 1 2 3

3 1

2 2

1 3

x 4
y x2
3 2 1 1 2 3

1 5

Figure 1.6: Concave up and concave down quadratic functions

1.2.6 Cubic Functions
When the third differences are constant, the data can be modeled by the cubic
function and it has the form

y = f (x) = ax3 + bx2 + cx + d

as long as a = 0. If the scatter plot of a set of data fails to exhibit an inflection
point, then it is not appropriate to fit a cubic equation to the data. Figure 1.7
and Figure 1.8 show the cases when the leading coefficient a positive and negative
respectively.

1.2.7 Trigonometric Functions
Definition 1.2.1 Functions that repeat their values at regular intervals are called
periodic.

Periodic functions repeat exactly the same cycle forever.

Definition 1.2.2 For any periodic function of time:

• The amplitude is the half the difference between its maximum and minimum
values.

• The period is the time for the function to execute one complete cycle.

9

y y
3 10

y x3 y x3 8x
2
5
1

x x
2 1 1 2 4 2 2 4

1
5
2

3 10

Figure 1.7: Cubic functions with a > 0

y y
3 20
y x3 y x3 4 x2
2
15

1
10

x
2 1 1 2
5
1

x
2 2 2 4 6

3 5

Figure 1.8: Cubic functions with a < 0

Many periodic functions are represented using the function called sine and
cosine shown in Figure 1.9. In general, the functions y = A sin (Bx) + C and
y = A cos (Bx) + C are periodic with amplitude |A|, period 2π/|B|, and vertical
shift C.

1.3 Tangents and Normals
Primary goal of this section is to describe change, average rate of change, and
instantaneous rate of change.

10

y y
2 2
y=sin(x) y=cos(x)

1 1

x x
5 5 5 5

1 1

2 2

Figure 1.9: Sine and cosine functions

Definition 1.3.1 Average rate of change of a function f (x) between x = a and
x = b is given by
∆y f (b) − f (a)
= .
∆x b−a

Definition 1.3.2 The instantaneous rate of change of a function f (x) at point
x = a is defined to be the instantaneous rates of change of f (x) at x = a and denoted
by f (a).

Visualizing average rate of change and instantaneous rate of change: Fig-
ure 1.10 shows the average rate of change of a function represented by the slope of
the secant line joining points A and B. Instantaneous rate of change is found by
taking the average rate of change over smaller and smaller intervals.

Definition 1.3.3 The derivative of a function f (x) at point x = a is defined to
be the limit of the average rates of change of f (x) over shorter and shorter intervals
around a. Then the instantaneous rate of change of f is the slope of the tangent
line at x = a.

Some useful derivative rules are given in Table 1.2.
Example: One tower is 50 ft high and another tower is 30 ft high. The towers are
150 ft apart. A guy wire is to run from Point A to the top of each tower. Locate
Point A so that the total length of guy wire is minimal [2].

11

Function Derivative

dy
y=b =0
dx
dy
y = mx + b =m
dx
dy
y = xn = nxn−1
dx
dy
y = ex = ex
dx
dy
y = bx = (ln b)bx
dx
dy 1
y = ln x =
dx x
dy
y = sin x = cos x
dx
dy
y = cos x = − sin x
dx
dy
y = kf (x) = kf (x)
dx
dy
y = f (x) ± g(x) = f (x) ± g (x)
dx
dy
y = f (g(x)) = f (g(x))g (x)
dx
dy
y = f (x)g(x) = f (x)g(x) + f (x)g (x)
dx
dy f (x)g(x) − f (x)g (x)
y = f (x)/g(x) =
dx g(x)2

Table 1.2: Derivative rules

12

Figure 1.10: Average rate of change vs. instantaneous rate of change [2]

Figure 1.11: Length of a guy wire

Solution: Let x be the distance between point A and 30 ft high tower. Then the
length of guy wire is the sum of the hypothenuses of two triangles in Figure 1.11,
√
L(x) = 900 + x2 + 2500 + (150 − x)2 for 0 ≤ x ≤ 150. The graph of the the
length function L(x) is given in Figure 1.12.
We set L (x) = 0 and solve for x:

x (150 − x)
L (x) = √ − =0
900 + x2 2500 + (150 − x)2

. Let θA be the angle at point A facing to the 30ft high tower and θB be the angle at
point A facing to the 50ft high tower. Since √900+x2 = cos θA and √ (150−x) 2 =
x
2500+(150−x)
cos θB , then L (x) = 0 when cos θA = cos θB , or θA = θB . Since two triangles are

13

y
210

200

190

180

170

x
0 20 40 60 80 100 120 140

Figure 1.12: Length of a guy wire

similar,
x 150 − x 225
= =⇒ x = = 56.25 feet.
30 50 4

1.4 Regressions and Interpolations
In section 1.2, we discussed several types of models/functions by building a library
of functions. It is important that we choose the right model/function for our data
set. Here are some guidelines for determining which model to use:
Examine the scatter plot of the data

1. If the scatter points appear to lie in a straight line, use linear function.

2. If the scatter points curved with no inﬂection point, try a quadratic, an expo-
nential, or a logarithmic functions.

3. If the scatter plot appears to have an inﬂection point, try a cubic and/or a
logistic function. Check the end behavior of the plot to distinguish cubic from
logistic.

4. If the scatter plot appears to be periodic, try trigonometric functions.

5. If above trials fail, try combining two or more functions.

14

If the input values are equally spaced,

1. data is linear if the first differences are equal.
2. data is exponential if the consecutive ratios are equal.
3. data is quadratic if the second differences are equal.
4. data is cubic if the third differences are equal.

Example: Consider following three sets of data representing distributions of loads
on a cantilever beam in a certain structure.
x y x y x y
-2 9.3 0 12 1 2
0 9.1 2 48 2 3
2 8.9 4 192 3 6
4 8.7 6 768 4 11
6 8.5 8 3072 5 18

Table 1.3: Linear Table 1.4: Exponential Table 1.5: Quadratic

Linear: The first differences are 9.1−9.3 = −0.2, 8.9−9.1 = −0.2, 8.7−8.9 =
−0.2, 8.5−8.7 = −0.2 so this is a linear model. It is of the form y = mx+b where m is
the slope and b is the y-intercept (when x = 0). The slope is m = −0.2/2 = −0.1 and
the y-intercept is 9.1 (from the table). Thus the model y = −0.1x + 9.1 represents
the linear distribution of the loads on the cantilever beam. Use of computer algebra
system is recommended for the data does not exactly fit the data. Below is the
Mathematica code for the linear case:
lindata = {{-2, 9.3}, {0, 9.1}, {2, 8.9}, {4, 8.7}};
line = Fit[lindata, {1, x}, x]
Exponential: The first ratios are 48/12 = 4, 192/48 = 4, 768/192 = 4,
3072/768 = 4 so this is an exponential model. It is of the form y = abx where a is
the initial value (when x = 0) and b is the rate. The initial value is a = 12 (from
the table). To find the rate b, we substitute one of the points in the table, choose
(2, 48). Then

y = 12bx
48 = 12b2

Dividing both side by 12, b2 = 48/12 = 4. Taking square root of both sides b = ±2.
Since base (rate) can not be negative, b = 2. Thus the model y = 12 · 2x represents

15

the exponential distribution of the loads on the cantilever beam. Below is the
Mathematica code for the exponential case:
expdata = {{0, 12}, {2, 48}, {4, 192}, {6, 768}};
exp = FindFit[expdata, a*b^ x, {a, b}, x]
Quadratic: The first differences and the second differences are given in Table
1.6. The constant second differences imply a quadratic model. It is of the form
y = ax2 + bx + c.
x y First differences Second differences
1 2 − −
2 3 3−2=1 −
3 6 6−3=3 3−1=2
4 11 11 − 6 = 5 5−3=2
5 18 18 − 11 = 7 7−5=2

Table 1.6: First and second differences

We substitute the first three points (1, 2), (2, 3), (3, 6) into the equation y =
2
ax + bx + c. We obtain the 3 × 3 linear system.
2 = a+b+c
3 = 4z + 2b + c
6 = 9a + 3b + c
From the first two, we eliminate c and obtain 3a + b = 1. From the last two, we
eliminate c and obtain 5a + b = 3. Now we have a 2 × 2 system to solve:
3a + b = 1
5a + b = 3
Solution of the 2 × 2 system is a = 1 and b = −2. Substituting a and b into the first
equation, we receive c = 3. Thus the quadratic model is y = x2 − 2x + 3. Even in
tho simple example we had to solve a 3 × 3 system which may be tedious. Also use
of computer algebra system is recommended for the data does not exactly fit the
data. Below is the Mathematica code for the quadratic case:
quaddata = {{1, 2}, {2, 3}, {3, 6}, {4, 11}};
parabola = Fit[quaddata, {1, x, x^ 2}, x]
Now, one can interpolate, by asking ”what is the linear load on the beam at
location x = 1?” which is not listed in the tabular data. Or we can extrapolate by
asking ”what is the quadratic load at the location x = 0?” which is out side the
range of the data set. These questions can be answered by simply substituting the
x values into the corresponding models.

16

1.5 Optimization
Most of the real-world problems are given as data sets as it is presented in the pre-
vious section. Then after finding appropriate model for the data (using regression),
we use the methods of calculus to determine the largest (maximum) or smallest
(minimum) value of the model/function.

Definition 1.5.1 Let f be a function defined on an interval I that contains the
number c. Then

• f (c) is the absolute maximum of f on I if f (c) ≥ f (x) for all x in I

• f (c) is the absolute minimum of f on I if f (c) ≤ f (x) for all x in I

Both absolute minima and absolute maxima are called absolute extrema .

How to find absolute extrema:
In order to be able to find the absolute extrema of a continuous functions f on a
closed interval a ≤ x ≤ b:

• Find all critical numbers, c, of f (f (c) = 0 or undefined) in the open interval
(a, b).

• Compute f (x) at the critical numbers found in the previous step.

• Compute f (x) at the end points of the interval a and b.

• The largest and smallest values found in the previous two steps are the absolute
maximum and absolute minimum values of f on the closed interval a ≤ x ≤ b
respectively.

The second derivative test for absolute extrema:
Suppose that f (x) is continuous on an interval I where x = c is the only critical
number and that f (c) = 0. Then,

• if f (c) > 0, the absolute minimum of f (x) on I is f (c),

• if f (c) < 0, the absolute maximum of f (x) on I is f (c).

17

x

y Picnic park y

Highway

Figure 1.13: Rectangular picnic park

Example: The highway department is planning to build a picnic park for motorist
along a major highway. The park is to be rectangular with an area of 5,000 square
yards and is to be fenced oﬀ on the three sides not adjacent to the highway shown
in Figure 1.13. What is the least amount of fencing required for this job? How long
and wide should the park be for the fencing to be minimized [3]?
Solution: Since the park is to have area 5,000 square yards, we have xy = 5, 000.
The length of the fencing is L = x + 2y, where x > 0 and y > 0 (otherwise we can
5, 000
not have a picnic park area). Since xy = 5, 000 or y = , we can eliminate y
x
from the formula for L. Then L will be a function of x:

5, 000 10, 000
L(x) = x + 2y = x + 2 =x+ for x > 0
x x

The derivative of L(x) is

10, 000
L (x) = 1 − .
x2
18

We ﬁnd the critical numbers by solving L (x) = 0 for x:
10, 000
L (x) = 1 − = 0
x2
2
x − 10, 000
= 0
x2
x2 = 10, 000
x = 100, reject x = −100, since x > 0.

Since x = 100 is the only critical number in the interval x > 0, we can apply
the second derivative test. The second derivative of L(x) is
20, 000
L (x) = .
x3
So, L (100) > 0 and an absolute minimum of L(x) occurs where x = 100 as shown
in Figure 1.14.
y
600

500

400

300

200

100

x
100 200 300 400

Figure 1.14: Fencing function of rectangular picnic park

The minimal amount of fencing is L(100) = 100 + 10, 000/100 = 200 yards
which is achieved when the park is x = 100 yards long and y = 5, 000/100 = 50
yards wide.

1.6 Appendix: Review of Solving Triangles
Solving a triangle means to ﬁnd all missing sides and angles. a, b, and c are sides.
A, B, and C are angles as shown in Figure ??. Side a faces angle A, side b faces

19

angle B and side c faces angle C. First, we review few laws/rules.

A

b
c

B C
a

Figure 1.15: Angles and sides of a triangle

• The angles always add to 180◦ . When you know two angles you can find the
third.
• Law of sines:
a b c
= =
sin A sin B sin C
• Low of cosines
a2 = b2 + c2 − 2bc cos (A)
b2 = a2 + c2 − 2ac cos (B)
c2 = a2 + b2 − 2ab cos (C)

We have the following possible cases:

1. AAS (Angle, Angle, Side): This means that two angles and one side are known.
Note that the known side is not between the angles. We use law of sine to find
the other two sides.
2. ASA (Angle, Side, Angle): This means that two angles and one side are known.
Note that the known side is between the angles. We use law of sine to find
the other two sides.

20

3. SAS (Side, Angle, Side): This means that two sides and one angle are known.
Note that the known angle is between the sides. We use law of cosine to find
the third side, and law of sine for the other angles.

4. SSA (Side, Side, Angle): This means that two sides and one one are known.
Note that the known angle is not between the sides. We use law of sine to find
one of the other two angles (therefore the third angle is also known). Then
use law of sine again to find the third side.

5. SSS (Side, Side, Side): This means that all three sides are known. We use
law of cosine twice to find two of the angles (therefore the third angle is also
known).

Example: This is an example of SAS case. In this triangle, we know angle A = 49◦
and sides b = 5, c = 7 as shown in Figure 1.16. Solve the triangle in Figure 1.16.

C

5 a

49 B
A
7

Figure 1.16: Angles and sides of a triangle

Solution: We use law of cosines first to find side a:

a2 = 52 + 72 − 2 · 5 · 7 · cos (49◦ )
a2 = 25 + 49 − 70 cos (49◦ )
a2 = 74 − 45.924 = 28.075
√
a = 28.075
a = 5.298

21

Now we use the the law of sines to find one of the other two angles:
sin B sin A
=
b a
sin B sin (49◦ )
=
5 5.298
which gives sin B = 0.7122 then B = sin−1 (0.7122) = 45.4◦ . Thus the the third
angle is C = 180◦ − 49◦ − 45.4◦ = 85.6◦ .

1.7 Problems, Projects, and Activities
1. A rectangular-shaped garden has one side along the side of a house. The other
three sides are to be enclosed with 60 feet of fencing. What is the largest
possible area of such garden [2]?

2. A florist uses wire frames to support flower arrangements displayed at wed-
dings. Each frame is constructed from a wire of length 9 feet that is cut into
6 pieces. Vertical edges of the frame consists of four of the pieces of wire that
are each 12 inches long. One of the remaining pieces is bent into a square to
form the base of the frame; the final piece is bent into a circle to form the top
of the frame [2].

(a) How should the florist cut the wire of length 9 feet in order to minimize
the combined area of the circular top and the square base of the frame?
(b) Verify that the answer to part (a) minimizes the combined area.

3. You need to design a display booth for a company. Because the company
generally must pay for the amount of square footage your booth requires, you
want to limit the floor size to 300 square feet. The booth is to be 6 feet tall
and three-sided, with the back of the booth a display board and the two sides
of the booth made of gathered fabric. The display board for the back of the
booth costs $30 per square foot. The fabric costs $2 per square foot and needs
to be twice the length of the side to allow for gathering. Find the minimum
cost of constructing a booth according to these specifications. What should
be the dimensions of the booth?

4. (Project) A popular historical site in Missouri is the Gateway Arch. Designed
by Eero Saarinen, it is located on the original riverfront town site of St. Louis
and symbolizes the city’s role as gateway to the West. The stainless steel
Gateway Arch (also called the St. Louis Arch) is 630 feet (192 meters) high
and has an equal span.

22

In honor of the 200th anniversary of the Louisiana Purchase, which made St.
Louis a part of the United States, the city has commissioned an artist to design
a work of art at the Jefferson National Expansion Memorial National Historic
Site. The artist plans to construct a hill beneath the Gateway Arch, located
at the Historic Site, and hang strips of mylar from the arch to the hill so as
to completely fill the space. The artist has asked for your help in determining
the amount of mylar needed [1].

Figure 1.17: The Gateway Arch in St. Louis

(a) If the hill is to be 30 feet tall at its highest point, find an equation for the
height of the cross-section of the hill at its peak. Refer to Figure 1.17.

23

(b) Estimate the height of the arch in at least ten different places. Use the
estimated heights to construct a model for the height of the arch.
(c) Estimate the area between the arch and the hill.
(d) The artist plans to use strips of mylar 60 inches wide. What is the
minimum number of yards of mylar that the artist will need to purchase?
(e) Repeat Task 4 for strips 30 inches wide.
(f) If the 30-inch strips cost half as much as the 60-inch strips, is there any
cost benefit to using one width instead of the other? If so, which width?
Explain.

5. A square sheet of cardboard 18 inch on a side is made into an open box (i.e.,
theres no top), by cutting squares of equal size out of each corner (see Figure
1.18) and folding up the sides along the dotted lines. Find the dimensions of
the box with the maximum volume [2].

Figure 1.18: The rectangular box

6. A cylindrical can is to hold 12 fluid liters (see Figure 1.19). Find the dimensions
that will minimize the amount of material used in its construction, assuming
that the thickness of the material is uniform [2].

7. (Project) The state wants to build a new stretch of highway to link an existing
bridge with a turnpike interchange, located 8 miles to the east and 8 miles to
the south of the bridge. There is a 5-mile-wide stretch of marsh land adjacent
to the bridge that must be crossed (see Figure 1.20). Given that the highway

24

r

h

Figure 1.19: Open diagram of cylindrical can

costs $10 million per mile to build over the marsh and only $7 million to build
over dry land, how far to the east of the bridge should the highway be when
it crosses out of the marsh [2]?

Figure 1.20: Highway design

25

8. A showroom for a department store is to be rectangular with walls on three
sides, 6-ft door openings on the two facing sides and a 10-ft door opening on
the remaining wall. The showroom is to have 800 square feet of ﬂoor space.
What dimensions will minimize the length of wall used [2]?

9. (Project) A Norman window has the outline of a semicircle on top of a rect-
angle, as shown below. Suppose there is 8 + π feet of wood trim available.
Discuss why a window designer might want to maximize the area of the win-
dow. Find the dimensions of the rectangle (and, hence, the semicircle) that
will maximize the area of the window (see Figure 1.21) [2].

Figure 1.21: Norman window

10. (Project) You are designing a cable-stayed bridge, illustrated in Figure 1.22.
The objective of this problem is to estimate the optimal height of the tow-
ers, using the simple idealization shown in Figure 1.23. Both the cable and
column will be idealized as cylinders with uniform cross section. Here is the
information you need to make the decision.
(http://www.engin.brown.edu/courses/en3/notesframe.htm)

• The force in the cable is Pcable = W/(2 sin θ), where W is the weight of
the roadbed.

26

Figure 1.22: Cable-stayed bridge

• The force in the column is Pcolumn = −W .
• The cable will break if the force per unit area in the cable exceeds σ0cable .
• The column will collapse if the magnitude of the force per unit area in
the column exceeds σ0column .

Figure 1.23: Animation of the cable-stayed bridge

(a) Write down an expression for the height of the column in terms of distance
d and the angle θ.
(b) Write down an expression for the length of the cable in terms of d and
the θ.
(c) Find the minimum allowable cross sectional area and volume of the cable
and column.

27

Chapter 2

Discrete Mathematics Module

2.1 Introduction
Discrete mathematics is dealing with objects that can assume only distinct, sepa-
rated values. The term “discrete mathematics” is used in contrast with “continuous
mathematics”. The objects that are considered are integers, propositions, sets, and
relations, which are all discrete. The concepts associated with these objects, their
properties, and relationships among them are the content of this module. Since
Discrete mathematics encompasses a very wide range of mathematical topics, this
module presents some selected choice of materials based on the FUSE-Lab Project.

2.2 Logic and Sets

2.2.1 Logic and Truth Tables
Logic is used to establish the validity of arguments. We are not concerned about
what the argument is but interested providing rules so that the general form of the
argument can be judged as sound or unsound.
A proposition is a declarative statement which is either true or false, but not
both. Here are few examples of propositions:

1. This rose is red.

2. 5 < 21.

3. Pentagons have four vertices.

29

4. 3 + 1 = 5.

Exclamations, questions and demands are not propositions since they cannot
be declared true or false. For example “don’t be late”, “how are you?” etc. The
truth (T) or falsity (F) of a proposition is called truth value.
Let p be a proposition then p (or∼ p or −p or ¬p) symbolizes the negation of
¯
p. Table 2.1 shows the relationship between the truth values of p and the negation
of p.
¯

p p
¯
T F
F T

Table 2.1: The truth table for the negation of a proposition

A table which summarizes truth values of propositions is called a truth table.
There are several alternative ways of stating the negation of a proposition. If
we consider the proposition “All roses are red”, some examples of its negation are:
“It is not the case that all roses are red”, “Not all roses are red”, “Some roses are
not red”. Note that the proposition “No roses are red” is not the negation of “All
roses are red”. Remember that to be the negation, the second statement must be
false in all circumstances that the ﬁrst is true and vice versa.
Let p and q be two propositions. The conjunction of p and q, denoted by
p ∧ q, is the proposition “p and q”. We say that the sentence p ∧ q is true if the two
sentences p, q are both true, otherwise conjunction is false. Table 2.2 is the truth
table of conjunction.
The sentence “2 + 2 = 4 and 2 + 3 = 5” is true.
The sentence “2 + 2 = 4 and π is rational” is false.

p q p∧q
T T T
T F F
F T F
F F F

Table 2.2: The truth table for the conjunction of two propositions

Let p and q be two propositions. The disjunction of p and q, denoted by p ∨ q,
is the proposition “p or q”. We say that the sentence p ∨ q is true when either or

30

both of its components are true and is false otherwise. Table 2.3 is the truth table
of disjunction.
The sentence “2 + 2 = 2 or 1 + 3 = 5” is false.
The sentence “2 + 2 = 4 or π is rational” is true.

p q p∨q
T T T
T F T
F T T
F F F

Table 2.3: The truth table for the disjunction of two propositions

The exclusive disjunction of p and q is denoted by p⊕q. This compound propo-
sition is true when exactly one (i.e. one or other, but not both) of its components
is true. The truth table for p ⊕ q is given by Table 2.4

p q p⊕q
T T F
T F T
F T T
F F F

Table 2.4: The truth table for the “exclusive or” of two propositions

Let p and q be two propositions. The conditional statement p → q is the
proposition “if p, then q”. The sentence p → q is true if the sentence p is false or if
the sentence q is true or both, and is false otherwise.
It is convenient to realize that the sentence p → q is false precisely when the
sentence p is true and the sentence q is false. To understand this, note that if we
draw a false conclusion from a true assumption, then our argument must be faulty.
On the other hand, if our assumption is false or if our conclusion is true, then our
argument may still be acceptable. The truth table for p → q is given by Table 2.5
The sentence “if 2 + 2 = 4, then π is rational” is false.
The sentence “if π is rational, then 2 + 2 = 4” is true.
Let p and q be two propositions. The biconditional statement p ↔ q is the
proposition “p if and only if q”. The sentence p ↔ q is true if the two sentences p,
q are both true or both false, and is false otherwise. The truth table for p ↔ q is
given by Table 2.6

31

p q p→q
T T T
T F F
F T T
F F T

Table 2.5: The truth table for the conditional statement of two propositions

The sentence “2 + 2 = 4 if and only if π is irrational is true.
The sentence “2 + 2 = 4 if and only if π is rational is also true.

p q p↔q
T T T
T F F
F T F
F F T

Table 2.6: The truth table for the biconditional statement of two propositions

Example: Construct the truth table for the sentence (p ∨ q) ∧ (p ∧ q).
Solution:

p q p∧q p∨q p∧q (p ∨ q) ∧ (p ∧ q)
T T T T F F
T F F T T T
F T F T T T
F F F F T F

Example: Construct truth tables for the following compound propositions [?].

1. p ∨ q

2. p ∧ q

3. q → p

4. p ↔ q

32

Solution:

1. Note that the truth table below is built up in stages. The ﬁrst two columns
give the usual combinations of possible truth values of p and q. The third
column gives, for each truth value of p, the truth value of p. When p is true, p
is false and vice versa. The last column combines the truth values in columns
3 and 2 using the inclusive disjunction connective. The compound proposition
p ∨ q is true when at least one of its two components is true. This is the case
in row 1 (where q is true), row 3 ( p and q are both true) and row 4 ( p is
true). In the second row, p and q are both false and hence p ∨ qis false.

p q p p∨q
T T F T
T F F F
F T T T
F F T T

2. Here we ﬁrst obtain truth values for p and q by reversing the corresponding
truth values of p and q respectively. Now p ∧ q is only true when both p and
q are true, i.e. in row 4. In all other cases p ∧ q is false.

p q p q p∧q
T T F F F
T F F T F
F T T F F
F F T T T

3. The truth table of q → p:

p q q q→p
T T F T
T F T T
F T F T
F F T F

4. The truth table of p ↔ q:

33

p q p q p↔q
T T F F T
T F F T F
F T T F F
F F T T T

2.3 Boolean Algebra

2.3.1 Introduction
A Boolean algebra consists of a set B together with following three operations deﬁned
on that set:

1. a binary operation denoted by ⊕ referred to as the sum (or join);
2. a binary operation denoted by ∗ referred to as the product (or meet);
3. for any element b ∈ B, the element ¯ ∈ B where the operation ¯ is called the
b b
complement of b.

The following axioms apply to the set B together with the operations ⊕, ∗, and¯
.

A1. Distinct identity elements belonging to B exist for each of the binary opera-
tions ⊕ and ∗ and we denote these by 0 and 1 respectively. Thus we have
b⊕0 = 0⊕b=b
b∗1 = 1∗b=b
for all b ∈ B.
A2. The operations ⊕ and ∗ are associative, that is
(a ⊕ b) ⊕ c = a ⊕ (b ⊕ c)
(a ∗ b) ∗ c = a ∗ (b ∗ c)
for all a, b, c ∈ B.
A3. The operations ⊕ and ∗ are commutative, that is
a⊕b = b⊕a
a∗b = b∗a
for all a, b ∈ B.

34

A4. The operation ⊕ is distributive over ∗ and the operation ∗ is distributive over
⊕, that is

a ⊕ (b ∗ c) = (a ⊕ b) ∗ (a ⊕ c)
a ∗ (b ⊕ c) = (a ∗ b) ⊕ (a ∗ c)

for all a, b, c ∈ B.

A5. For all b ∈ B, b ⊕ ¯ = 1 and b ∗ ¯ = 0.
b b

Note that A5 may lead you to conclude that ¯ is the inverse of b. This is not
b
true. Remember that, if b?1 is the inverse of b, then b ∗ b?1 gives the identity with
respect to the operation ∗. However, b ⊕ ¯ gives the identity with respect to ∗ and
b
b ∗ ¯ gives the identity with respect to ⊕, so that ¯ is not the inverse of b with respect
b b
to either operation.
Note also that note that 0 and 1 are used here as symbols for the two identity
elements and not for the numbers which they conventionally symbolize. We must
therefore be careful not to make assumptions which are true for the integers 0 and
1 but not necessarily so for identities in general.
Example: The simplest Boolean algebra (and also the one of most interest to com-
puter scientists) consists of the set B = {0, 1} together with the binary operations
⊕ and ∗ and complement operation¯deﬁned by the following tables.

⊕ 0 1 ∗ 0 1 b ¯
b
0 0 1 0 0 0 0 1
1 1 1 1 0 1 1 0
(a) Join. (b) Meet. (c) Com-
plemet.

2.3.2 Properties of Boolean Algebras
Duality: Given any proposition about a Boolean algebra, we deﬁne its dual to be
the proposition obtained by substituting ⊕ for ∗, ∗ for ⊕, 0 for 1, and 1 for 0.
The principle of duality is that for any theorem about a Boolean algebra, the
dual is also a theorem. Properties of Boolean algebras are listed as theorems below:

Theorem 2.3.1 The identity elements 0 and 1 are unique.

35

Theorem 2.3.2 Given an element b ∈ B, there is only one element ¯ ∈ B such
b
that b ⊕ ¯ = 1 and b ∗ ¯ = 0.
b b

Theorem 2.3.3 Idempotent laws: For all b ∈ B, b ⊕ b = b and b ∗ b = b.

Theorem 2.3.4 Identity laws: For all b ∈ B, 1 ⊕ b = b ⊕ 1 = 1 and 0 ∗ b = b ∗ 0 = 0.

Theorem 2.3.5 Absorption laws: For all b1 , b2 ∈ B, b1 ⊕ (b1 ∗ b2 ) = b1 and b1 ∗
(b1 ⊕ b2 ) = b1 .

Theorem 2.3.6 Involution laws: For all b ∈ B, ¯ = b.
b

Theorem 2.3.7 De Morgan’s laws: For all b1 , b2 ∈ B, (b1 ⊕ b2 ) = b1 ∗ b2 and
(b1 ∗ b2 ) = b1 ⊕ b2 .

Theorem 2.3.8 ¯ = 1 and ¯ = 0.
0 1

Example: Evaluate the following for the Boolean algebra (0, 1, ⊕, ∗,¯ 0, 1)
,

1. (0 ⊕ 1) ∗ 0

2. 0 ∗ ¯
1

3. (1 ∗ 1) ⊕ (0 ∗ ¯
0)

4. ¯ ⊕ [(0 ∗ 1) ∗ 1]
1

5. [(0 ∗ 1) ∗ 1) ∗ (¯ ⊕ 1)] ⊕ 1
1

6. [1 ⊕ (¯ ∗ 1)] ∗ (¯ ⊕ 0)
1 0

7. [(1 ∗ 1) ⊕ ¯ ∗ [(1 ⊕ 0) ∗ 1]
0]

Solution: Text here Text here Text here Text here Text here Text here Text here
Text here Text here Text here
Example: Following is the basic Mathematica commands for Boolean algebras:
{Boole[False], Boole[True]} produces {0, 1}.
{Boole[{True, False, True, True, False}] produces {1, 0, 1, 1, 0}.
Example: Following is the Mathematica code for a truth table for Boolean expres-
sion x ∗ (y ⊕ z):

36

f1 = x && (y || z);
t1 = BooleanTable[x, y, z -> f1, x, y, z]
that will produce the following result:
{{True, True, True} -> True, {True, True, False} -> True, {True, False,
True} -> True, {True, False, False} -> False, {False, True, True} -> False,
{False, True, False} -> False, {False, False, True} -> False, {False, False,
False} -> False}
Example: Following code produces a scatter plot of primes. PrimeQ yields True if
the expression is a prime number, and yields False otherwise.
test[x ] := Boole[PrimeQ[x]];
data = ParallelTable[test[x + y], {x, -50, 50}, {y, -50, 50}];
ArrayPlot[data]

Figure 2.1: Primes in 2D

Following produces primes in complex plane:
test[x ] := Boole[PrimeQ[x]];

37

data = ParallelTable[test[x + y I], {x, -50, 50}, {y, -50, 50}];
ArrayPlot[data]

Figure 2.2: Primes in complex plane

2.3.3 Boolean Functions

Boolean variable is a variable whose range of possible “values” is the underlying set
B of a Boolean algebra (B, ⊕, ∗,¯ 0, 1).
,
Do we need it?

2.3.4 Switching Circuits

Do we need it?

38

2.3.5 Logic Networks
Do we need it?

2.4 Modular Arithmetic
Text here Text here Text here Text here Text here Text here Text here Text here
Text here Text here

2.5 Recurrence Relations
Text here Text here

2.6 Problems, Projects, and Activities
Text here Text here

39

Bibliography

[1] D. LaTorre, et al. Calculus Concepts, Houghton Miﬄin, 2005.

[2] G. Thomas, et al. Calculus, Addison Wesley, 2009.

[3] L. D. Hoﬀmann, G. L. Breadley Applied Calculus, McGraw Hill, 2007.

41

Math Modules (DRAFT)

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