dt - Helfant Professional Development Sessions

Dr. Pais’s Student Research Journal
Volume 1, Spring 2008
Partially Edited Draft 02-05-08
John Pais © 2008, All Rights Reserved
(Do not distribute without permission)
"I hear and I forget.
I see and I remember.
I do and I understand.”
"The value of a problem is
not so much coming up with
the answer as in the ideas
and attempted ideas it forces
on the would be solver."
Chinese Proverb
I.N. Herstein
1
Contents
1. Kepler’s Laws of Planetary Motion, Sarah Van Cleve …………………………... p. 3
2. Stochastic Calculus in the Stock Market, Daniel Cooper and William Mueller .... p. 16
3. Predator vs. Prey: Lotka-Volterra Equations, Michael Ahrens and Megan Gornet p. 26
4. Understanding Renal Clearance through Calculus, Megan Eyunni and
Dena Rothstein …………………………………………………………………… p. 33
5. Lagrangian Mechanics and the Brachistochrone Problem, Cora Bernard ……….. p. 41
6. A Journey Along a Bézier Curve, Michael Sherwood and Frank Trampe ……..... p. 56
7. Newton’s Law of Cooling, Robert Birkenmeier and Brendan Lyss .…………...... p. 67
8. Millennium Force: Calculus and Roller Coasters, Emily German, Rachel Simon,
and Dominique Zotos ……………………………………………………………... p. 73
9. Calculus in Population Growth, Jessica Houghtaling and Carissa LaFebere …..... p. 82
10. Derivatives: Maximizing the Marginal Profit, John Wuestling and
Shannon Shubert…................................................................................................. p. 91
11. The Calculus in Projectile Motion, Christopher Waterbury and Bennett Walsh.. p.100
12. Derivatives in Weather Forecasting, Ashley Ervin and McKenna Morey……... p. 110
13. Calculus in Waves, Emily Clary and Elizabeth Davidson…………………….... p. 119
14. Digital Device Usage Survey in Conjunction with the Flat Classroom Project,
Ariana Mooradian, Michael Rodriquez, and Walker Harbison............................ p. 131
2
Kepler’s Laws of Planetary Motion
Sarah Van Cleve
In the 15th century, when most scientists believed in a geo-centric universe, the
idea that the sun and other planets revolve around the earth, Nicolaus Copernicus shocked
the scientific community with the revelation that, in fact, the earth revolves around the
sun. This part of Copernicus’s theory still holds true. His mistake, however, lies in the
simplicity of the idea. Copernicus proposed a system of planetary motion whereby the
planets revolved in a simple, circular orbit with the sun at the center. Though this idea
was certainly straightforward and easy to comprehend, it was not proven by any of the
data that one of Copernicus’s successors, Tycho Brahe, collected throughout his entire
lifetime. Luckily, upon Brahe’s death, his records were passed on to Johann Kepler, one
of his assistants. It was Kepler who was able to make sense both of Copernicus’s concept
of the helio-centric universe and Brahe’s observations.
Kepler’s explanations of planetary motion are in the form of three laws. The first
is that a planetary orbit sweeps out equal area in equal time. The second law states that a
planetary orbit is an ellipse with the sun at one of the two foci. The third and final law is
that the square of the period of the orbit is directly proportional to the cube of the mean
distance.
As modern mathematicians, we can see that Kepler, who worked in the late 16th
and early 17th centuries, was at a disadvantage. This is because it was not until the late
17th and early 18th centuries that Isaac Newton and Gottfried Wilhelm Leibniz are
credited with inventing calculus. Thus, Kepler was forced to prove his laws solely with
3
the use of extensive observations, simple geometry, and the basic properties of ellipses.
Today, we have the advantage of proving Kepler’s Laws of planetary motion using
differential equations.
One advantage of calculus is that we can describe the position, velocity and
acceleration of a particle in either polar or Cartesian coordinates in terms of local
coordinates. This means that we can designate that a particle in motion, such as a planet,
is always at the origin. Then, by taking the derivative of the position vector, we can find
the velocity vector. In turn, by taking the derivative of the velocity vector, we can find
the acceleration vector. Using the position, velocity and acceleration vectors for a
particular planet, we can prove Kepler’s three laws using calculus.
1. Newton’s Geometric Proof of Kepler’s First Law
In order to provide an idea of the type of argument the 16th century astronomers used
before calculus, we will begin by giving a geometric proof of Kepler’s first law, namely
“equal area in equal time.” It turns out that the proof we give is actually due to Newton,
and in fact, this proof is more general in that it shows that for any particle in motion and
subject only to radial forces (which may be turned on or turned off), the particle must
sweep out equal area in equal time. We will illustrate this proof by examining two cases.
The first case will address the motion of a particle with no forces acting, and the second
case will address the motion of a particle in which initially there are no forces acting, but
after the first time interval an instantaneous radial force is turned on. Note that since the
area of a triangle is one-half the base times the height, in order to show that two triangles
have the same area we need only show that they have the same base and the same height.
4
Case 1. No (radial) forces acting.
A particle located at A and
traveling with constant velocity
(green arrow), moves to B in a
given time t. Similarly, the
particle takes the same time t to
moves from B to C, and so
|AB| = |BC|. Hence, the area of
△ ASB = area of △ BSC, since
these triangles have equal base
lengths |AB| = |BC| and the
Figure 1.
same altitude.
Case 2. Instantaneous radial force acting at B.
As in Figure 1, a particle
located at A and traveling with
constant velocity (green
arrow), moves to B in a given
time t. However, upon reaching
B an instantaneous radial force
is applied to the particle
producing an additional radial
velocity (red arrow). So, the
Figure 2.
resultant velocity of the
5
particle is the vector sum (yellow arrow) of the radial (red arrow) velocity vector and the
horizontal (green arrow) velocity vector.
Next, suppose that the particle moves from B to D in the same time t, that it took to
move from A to B. In order to show that the area of
△ ASB = area of △ BSD (Figure 2),
it is enough to show that the area of △ BSD = area of
already shown in Figure 1 that the area of
△ BSC (Figure 2), since it was
△ ASB = area of △ BSC. Now, by the
parallelogram law for vector addition, the line segment CD through points C and D must
be parallel to BS, as depicted in the diagram (Figure 2). Further, since these line segments
are parallel, the length of an altitude dropped from D to BS must be the same as the
length of an altitude dropped from C to BS. Hence, using BS as base in both triangles, the
area of △ BSD = area of
△ BSC.
Q.E.D.
Similarly, an extension of this argument can be applied to planetary motion around the
sun, which provides a geometrical demonstration of Kepler’s first law that a planet
travels an equal area in two different periods of equal time.
2. Vector Calculus: The Position Vector and Polar Coordinates
Though it is certainly interesting to examine the area a planet has traveled in
terms of geometry, it is also an outdated technique. Today we have the advantage of
using calculus to prove Kepler’s Laws of Planetary Motion. The first step towards doing
so is to define the position, velocity and acceleration vectors.
The position vector, r (t ) , represents a vector emanating from the origin and
extending to (potentially) any other point in the plane. Every vector has both direction
and magnitude, and we will denote the magnitude of the position vector, r (t ) , simply
6
by r (t ) . A unit vector is a vector whose magnitude is one, and, given any nonzero vector,
we can create a unit vector with the same direction as follows:
unit vector =
vector
vector magnitude
r (t )
Hence, we define the unit position vector by ur (t ) =
. In addition, we resolve r (t )
r (t )
into its x- and y-components using the diagram below.
r (t )
y-component
α(t)
x-component
Figure 3.
cosα (t) =
x-component
→ x-component = r(t)cosα (t)
r (t )
sin α (t) =
y-component
→ y -component = r(t)sinα (t)
r (t )
r (t ) = ( x(t ), y (t ) ) = ( r (t ) cos α (t ), r (t ) sin α (t ) )
(2.1)
Our evolving unit vector equation then becomes:
r (t ) ( r (t ) cos α (t ), r (t ) sin α (t ) )
ur (t ) =
=
r (t )
r (t )
Notice that, since the vector magnitude r (t ) is in every term of the position vector, r (t ) can
be canceled out, yielding the following equation:
7
r (t )
ur (t ) =
= ( cos α (t ),sin α (t ) )
r (t )
(2.2)
Next, we must also find a unit vector perpendicular to ur (t ) , or essentially a 90° rotation
of the unit position vector. Using a similar process as in Figure 3 above, we find that
uα (t ) = ( − sin α (t ), cos α (t ) )
(2.3)
Note that the two unit vectors ur (t ) and µα (t ) form a basis for a system of polar
coordinates which will be crucial to our derivation of Kepler’s Laws.
We now use the chain rule to find the derivative of the each of these unit vectors.
dur d
dα
dα  d α

= ( cos α (t ), sin α (t ) ) =  − sin α
, cos α
( − sin α , cos α )
=
dt
dt
dt
dt  dt

duα d
dα
dα 
dα

= ( − sin α (t ), cos α (t ) ) =  − cos α
, − sin α
( cos α ,sin α )
=−
dt
dt
dt
dt 
dt

Hence, it follow from equations 2.2, 2.3, and these last two equations above that
duα
dur dα dα =
uα and
=−
ur
dt
dt
dt
dt
(2.4)
3. Vector Calculus: The Velocity Vector
As we have seen in section 2, a vector is a combination of magnitude and direction.
Below is the equation of the position vector rewritten in terms of both magnitude and
direction, which follows from equations 2.1 and 2.2.
r = (magnitude)(direction)
r = rur
(3.1)
8
According to the rules of differential calculus, the derivative of the position vector is
equal to the velocity vector.
dr
v=
dt
Because the position vector is the product of two terms (magnitude and direction,
equation 3.1), we can use the product rule to find the derivative of the position vector and
thus, we will have found the velocity vector.
dr dur
dr d =
v=
rur = ur + r
dt dt
dt
dt
( )
Now, keeping in mind the equation that we found in the section on position
d µr dα vectors,
=
µα , we can substitute that equation into our developing formula for
dt
dt
velocity.
d µr dr dα dr d dr v=
= r µr = µr + r
= µr + r
µα
dt dt
dt
dt
dt
dt
Thus we have found a definitive equation for the velocity vector.
dα dr v = µr + r
µα
dt
dt
Acceleration Vector
Using the same rules of derivative calculus that dictated that the derivative of the position
vector was equal to the velocity vector, we can also say that the derivative of the velocity
vector (and the double derivative of the position vector) is equal to the acceleration
vector.
dv
a=
dt
9
dv d  dr  d  dα 
a=
r
=
µr +
µα 
dt dt  dt  dt  dt

This is incredibly complex because product rule must be applied to the first term and
triple product rule must be applied to the second term. Let’s first take the derivative of
the first section
d  dr 
µr using product rule.
dt  dt 
[ derivative of the 1st term ][ 2nd term ] + [1st term ][ derivative of the 2nd term ]
 d 2r   dr   dα 
nd
 dt 2  [ µ r ] +  dt   dt µα  Notice that for the derivative of the 2 term, we used the
 



values that we found in the position section.
Now let’s take the derivative of the second section
d  dα 
r
µα  which is more complex
dt  dt

because it has three terms.
[ derivative of the 1st term ][ 2nd term ][3rd term ] + [1st term ][ derivative of the 2nd term ][3rd term ] + [1st term ][ 2nd term ][ derivative of the 3rd term]
 d 2α  
 dr   dα   dα   dα
 dt   dt  [ µα ] + [ r ]  dt 2  [ µα ] + [ r ]  dt  (− dt )(− µr ) 


Now that we have found the derivatives of each section separately, we can combine them
into one long equation for acceleration.
 d 2 r dr dα   dr dα d 2α dα dα

a =  2 µr +
µα  + 
µα + r 2 µα + r
(−
)(− µ r ) 
dt dt
dt
dt
dt
 dt
  dt dt

2
2
d r dr dα dr dα dα dα dα
a = 2 µr +
µα +
µα + r 2 µα + r
(−
)(− µr )
dt
dt dt
dt dt
dt
dt
dt
This can be simplified by separating the terms with µr from the terms with µα
d 2r
d 2α d 2α
dr dα a = ( 2 − r 2 ) µr + (r 2 + 2
) µα
dt
dt
dt
dt dt
10
d 2α 1 d 2 dα d 2r
a = ( 2 − r 2 ) µr + (
r
) µα
dt
dt
r dt
dt
The term on the left is the radial component of acceleration. The term on the right is the
angular or tangential component of acceleration. Newton proved that acceleration in an
ellipse was purely radial so this right component is zero.
Proof of Kepler’s 1st and 2nd Laws
To prove Kepler’s 1st Law, a planet travels equal areas in equal times, we must remember
two basic ideas.
1. Remember from the previous section that if a planet travels in an ellipse, the
acceleration is purely radial so the tangential component of acceleration is equal to zero.
If we know that the tangential component of acceleration is zero, we can set it equal to
zero now.
1 d  2 dα 
=0
r
r dt  dt 
 dα 
Notice that in the above equation, we must take the derivative of the expression  r 2
.
 dt 
In order for the entire expression
1 d  2 dα 
to equal zero, that derivative must be
r
r dt  dt 
 dα 
equal to zero. The derivative of a constant is zero. Therefore, the expression  r 2
is a
 dt 
constant. Because we have just proved that the tangential acceleration is equal to zero,
we have also proved that the planet moves with only radial acceleration. Therefore, the
planet must travel in an ellipse. This is a proof of Kepler’s 2nd Law.
2. The second thing we must remember is also based on the information we found in the
acceleration vector section.
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We know that if A=area then
dA
= the velocity or the rate at which the radial vector
dt
sweeps out a particular area.
dA
= tangential acceleration
dt ∫
dA
d 2α
dα
= ∫ r 2 = 1 r2
2 dt
dt
dt
dA 1 2 dα
=
r
2 dt
dt
∆A = 1 r 2 ∆α
2
We know that r(t) is in between its minimum value r(t1) and its maximum value r(t2) as
defined by the following expression.
r (t1 ) ≤ r ≤ r (t2 )
If the change in area ( ∆A ) is a dependent variable based on the independent variable r(t),
then it follows that ∆A also has a minimum and maximum value based on the minimum
and maximum independent variables r(t1) and r(t2).
r (t1 ) 2
r (t2 ) 2
∆α ≤ ∆A ≤
∆α
2
2
We then divide each formula by ∆t
r (t1 ) 2 ∆α ∆A r (t2 ) 2 ∆α
≤
≤
2 ∆t
∆t
2 ∆t
In the situation that the final time is approaching the initial time, or the limit as final
approaches initial, the two times will eventually each become t instead of t1 and t2.
r (t ) 2 ∆α ∆A r (t ) 2 ∆α
≤
≤
2 ∆t
∆t
2 ∆t
12
The above expression is evidence that we have just proved that planets move equal areas
in equal time!
Proof of Kepler’s 3rd Law
Remember that from the section above, we have the equation
use the information k = 2
dA 1 2 dα
=
r
. We now
2 dt
dt
dA
to make a substitution.
dt
dA 1 2 dα
=
r
2 dt
dt
Multiply both sides by 2.
2
dA
dα
= r2
dt
dt
Substitute k.
k = r2
dα
dt
Now, solve for
dα
dt
dα
= kr −2
dt
Next, we can substitute the above equation into the equation we found for radial
d 2r
d 2α acceleration in the acceleration section; ( 2 − r 2 ) µ r
dt
dt
2  d 2r
a =  2 − r ( kr −2 )  µ r
 dt

 d 2r k 2  a (t ) =  2 − 3  µr
r 
 dt
13
Leave that information aside for now. We will come back to it after we have explained
how to find the double derivative of r. To do that, we use the fact that the general
equation of an ellipse in polar coordinates is r (1 + ε cos α ) = c where c and ε are
constants. (An equivalent equation, which is, in fact, the more familiar equation, in
Cartesian coordinates, is
x2 y 2
+
= 1 .) Using the first equation, we can find the derivative
a
b
of r using quotient rule.
r=
c
1 + ε cos α
r=
(1 + ε cos α )( 0 ) + ( c )( ε sin α )
2
(1 + ε cos α )
dr −cε (− sin α ) dα
=
dt (1 + ε cos α )2 dt
Substitute the equation
dα
= kr −2
dt
dr cε sin α k
= 2 −2 2
dt
cr
r
Next, we differentiated again in order to find the double derivative of position which we
need to solve the acceleration equation and prove Kepler’s third law.
d 2 r kε
dα k ε c − r k k 2 k 2
α
=
sin
=
=
−
dt 2
c
dt
c rε r 2 r 3 cr 2
Now let’s return to our equation for radial acceleration and substitute in the value for the
double derivative of position that we just found.
 d 2r k 2  a (t ) =  2 − 2  µr
cr 
 dt
14
 k 2 k 2  k 2  a (t ) =  3 − 2  − 3  µ r
cr  r 
 r
−k 2 a (t ) = 2 µ r
cr
So, in an ellipse when the acceleration is only radial, acceleration is inversely
proportional to the square of the distance from the origin. We have just proved Kepler’s
3rd Law!
Combining multiple sources of information—the laws of differential calculus,
polar coordinates, rectangular coordinates, geometry, and finally the rules of physics for
position, velocity and acceleration—we have proved Kepler’s Three Laws of Planetary
Motion; a planetary orbit sweeps out equal area in equal time, the planetary orbit is an
ellipse with the sun at one of two foci, and the square of the period of the orbit is directly
proportional to the cube of the mean distance.
Work Cited
Bressoud, David M. Second Year Calculus. New York: Springer-Verlag, 1991.
Sarah Van Cleve
1st Semester Differential Equations Project
AP BC Calculus
Dr. Pais
December 5, 2007
15
Stochastic Calculus in the Stock Market
Daniel Cooper and William Mueller
The use of calculus is necessary to gain a more in-depth understanding of the
activity of the stock market. A good starting point is with the Markov Property, which
establishes the principle that only present conditions play a role in predicting what the
events of the future may hold: history cannot be considered assuming the Markov
Property holds true. A stock must be initially priced according to its history and current
performance. Stochastic calculus helps to determine future events through the analysis of
probabilities of different random occurrences, and when combined with the Markov
Property is extremely useful in predicting future prices. The Wiener Process, also known
as Brownian Motion, explains the seemingly random way in which stock prices can rise
and fall. To ensure that prices never reach negative values, the Ito Process considers drift
and variance rates as functions affected by time. Through these processes, the graphs of
stock prices begin to make much more sense. Stock prices are essentially random, but
through these processes we have a much better idea as to what their activity will be in the
future and how different events may affect them.
The Stochastic Process is essential for understanding how stock prices move.
Given a starting point, stochastic calculus considers the randomness of the way in which
new points will arise. Continuous time is most helpful in this process because prices do
not change instantly; they change gradually over time. In order to get from one price to
another over a given time period, every price within that interval will be obtained. By
16
beginning with the expected value and adding the standard deviation which takes the
randomness of the process into account, the stochastic variable may be found.
Example courtesy of http://www.puc-rio.br/marco.ind/stoch-a.html
The Markov Property, which basically states that only the present conditions can
be considered in the stochastic movement of a subject, is important to understanding the
movement of a stock. This property explains that the historical records of the
profitability, movement, and rate of return of a particular stock cannot be considered in
the ostensibly random stochastic process the stock undergoes. It assumes that the stock
price is efficiently determined considering all previous stock fluctuations. Obviously, the
Markov property must be true considering that there are so many different investor
watchdogs that continually track the market, looking to make a profit from inefficiently
17
determined stock prices. For example, if a stock price is considered to be too high based
on its history, a broker will sell the stock. If the stock price is too low considering its
history, a broker will buy in bulk, hoping the market will eventually fix this discrepancy
and benefit the broker that invested in the stock. Consequently, the Markov property
must be held true based on the fact that brokers are constantly looking out for their own
self interest, finding discrepancies in stock prices. Similarly, this property holds true for
something as trivial as rolling a set of dice: each role acts independent of the role
preceding it. Rolling a six on the first role will neither increase nor decrease the
probability of rolling a six again on the second roll. The independent nature of the
Markov property as illustrated in both dice and stock prices is essential to understanding
the random nature of the stochastic process.
In addition to the Markov Process, the Weiner Process, also known as Brownian
motion, describes the randomness of stock market prices. One significant property of the
Weiner process is that two prices that occur at two different times are completely
unrelated to each other. Each price (z) is associated to time by the equation ∆z =∈ ∆t .
The increase in the value of z, represented by z(t)-z(0) can be expressed by a summation
N
equation z(t)-z(0) = ∑ ∈i
∆t where N is equivalent to short time intervals along the
i =1
whole time interval ∆t and where i is a representation of the standard deviation. In
considering the generalized wiener process dz =∈ dt , we must also consider the
independent market fluctuations evident in some independent variable a, and the variance
constant rate, b; thus, we formulate the equation dx = adt + bdz , implying that x has an
expected drift rate of a, and without the bdz term the equation is:
18
dx = adt
dx
=a
dt
x = x0 + at
Through the use of the drift rate in addition to the knowledge of the randomness of future
stock prices, market trends may be more successfully predicted.
One-Dimensional Weiner Process
In order to understand the more specific nature of the stochastic process, one must
consider the Ito Process. The Ito process is a generalized Wiener process that considers
the drift rate, a, and the variance rate, b, to be functions of the independent variable, time
19
(t). This means that both a and b are not constant values in the stochastic process hitherto
mentioned. Through this information, we develop the equation
dx = a(x, t)dt + b(x, t)dz
As you can see, the Ito Process is important to determining the most accurate
representation of the stock market as a stochastic process.
The inclusion and acknowledgement of the processes clarified above develop the
common application to derivatives as it relates to predicting stock market trends. As
detailed above in the Ito process explanation, in determining future stock prices, the
constant expected drift rate and constant variance rate are inappropriate considering that
these represent a percentage value of the stock price change rather than a constant value.
Therefore, one must consider these percentage values as variables represented in the
equation: the percentage constant will be represented by the variable “S” as µ represents
the expected rate of return and σ represents the unstable market presence constant. The
common equations represented with these variables are:
∆S
= µ dt + σ dz
S
Or, considering the limiting case of the change in time
∆S
= µ dt + σ ∈ ∆t
S
The equation we will commonly use in our examples is:
St = S 0 e rt
20
Where
R t =(µ -
σ2
2
)t+σ Wt , S0 is equivalent to the initial price, t is equivalent to
time, Wt is equivalent to the Wiener function enumerated above, σ is equivalent to
volatility, and µ is equivalent to the drift rate.
The derivation comes from the common application of the derivative of the
logarithmic value of St . For the stochastic process to hold true Rt = Log
St
S0
follow the derivation below:
dSt 1 (dSt ) 2
−
St 2 St 2
dLogSt =
1
dLogSt = µ dt + σ dWt − σ 2 dt + o(dt )
2
dLogSt = ( µ −
drt = ( µ −
Rt = ( µ −
Or formulated differently St
σ2
2
σ2
= S0e
2
σ2
)dt + σ dWt + o(dt )
2
)dt + σ dWt
)t + σ Wt
(µ−
σ2
2
) ∆t +σ N (0,1) t
Common Examples in the Stock Market:
A representation of the final equation used to find a given stock price, St:
St = S 0 e
(µ −
σ2
2
) ∆t +σ N (0,1) ∆t
S0 represents the initial stock price.
21
. One can
St represents the projected stock price after t years.
µ represents the standard drift rate, .15.
σ represents the volatility of the stock, which in this case we have made .4.
t represents the time between the initial price and the predicted price.
S4 = 50e
(.15−
.42
)×4 +.4(.3413)×2
2
S4 = 50e1.7385
S4 = 86.92
This would indicate that over four years, this stock, with a volatility of .4, would have
increased by $36.92 per share, a 72.58% increase. A larger volatility would have yielded
a smaller increase.
A representation of the stock price Texas Instruments over a four year period:
S 4 = S0 e
S 4 = S0 e
(.15−
(.15−
σ2
2
)4 +σ (.3413)(2)
(.72437)2
)4 + (.72437)(.3413)(2)
2
S 4 = S0 e.6− 2(.72437)
2
+.6826(.72437)
S 4 = 30.18e.6− 2(.72437)
2
+.6826(.72437)
S 4 = 31.57
This illustrates how, after four years, the stock price of Texas Instruments increased by
$1.39 per share, a 4.6% increase. This smaller increase is a result of the larger volatility.
22
The stochastic process plays an important role in both the stock market and its
Brownian motion application in physics. The essentially random properties of the
stochastic process mirror the trends evident in every stock of all worldwide markets.
With the utilization of the stochastic process in the stock market, a balanced, intelligent
approach can be taken to attempting to earn money on an investment. The stochastic
process has been consistently used by investors throughout history and will, in all
likelihood, be used persistently because of accuracy and validity.
Bibliography
Texts
Hull, John. Options, Futures, and Other Derivatives. 3rd. New Jersy: Prentice Hall, 1997.
This book was extremely informative and thorough, offering us several abstract
examples of stochastic processes, as well as detailed explanations of the processes and
several other properties, such as the Markov property.
Baxter, Martin, and Andrew Rennie. Financial Calculus. New York: University of
Cambridge, 1996.
This book gave us an in-depth look into concrete and discrete processes, giving us
a solid foundation for further understanding of stochastic calculus. It also offers insight
into normal distributions and a few different models for evaluating stocks.
Online Sources
23
http://www.puc-rio.br/marco.ind/stochast.html#estimation
This source, Real Options in Petroleum, a website created by Marco Antonio
Guimarães Dias as a component of his dissertation for PUC-Rio, focuses on different
approaches to investments in general and offered us outstanding general information as
well as clear graphs and tables that we were able to incorporate into our presentation.
www.maths.uq.edu.au/~kb/sde_notes/SDEs_introduction(1).ppt
This website, from Kevin and Pamela Burrage of the Math Department of the
University of Queensland Australia, offered a step-by-step process that helped in
understand stochastic calculus. It also included some examples, which aided in
understanding more thoroughly what each variable represented and how they affect the
different equations we use.
http://arxiv.org/PS_cache/cs/pdf/0304/0304009v1.pdf
This file, a thesis paper for the University of Manchester written by Gilles Daniel,
was perhaps the best online resource we utilized. It explained essentially everything our
project entailed and went even more in-depth by explaining more advanced models and
by offering several different examples of typical stock market prediction methods. This
was especially helpful in that it gave us clear-cut models to which we later applied
values.
http://www.quantlet.com/mdstat/scripts/sfe/html/sfenode27.html
24
Though relatively concise, this resource, written by an unknown author for
Methods and Data Quantlets, gave us a brief explanation of the use of stochastic calculus
in predicting stock prices. It also touched on normal distributions, which we later used in
our sample predictions.
http://www.fintools.com/docs/Stochastic_Stock_Price_Modeling.pdf
This particular resource, a paper on modeling stock prices submitted to the firm
Montgomery Investment Technology by Sorin R. Straja, Ph.D., offered a great deal of
information regarding different approaches to models of prediction, showing in detail
many we had not found elsewhere.
http://anson.ucdavis.edu/~rsen/index_files/thesis.pdf
This paper, a dissertation for the University of Chicago on modeling stock prices
by Rituparna Sen, was also one of the best sources we used. Though, at times, it went
into the most advanced techniques of modeling stock prices, there were parts in which
this source was absolutely superb with its detailed explanations of probability and its
simulations of possible values stocks can attain.
http://finance.yahoo.com/q/hp?s=TXN&a=11&b=1&c=2003&d=11&e=1&f=2007&g=d
&z=66&y=0
This site, Yahoo Finance, offered us the necessary information to project prices
for the Texas Instruments stock as well as to determine its value for volatility. This
allowed us to create a more concrete example.
25
Predator vs. Prey: Lotka-Volterra Equations
Michael Ahrens and Megan Gornet
Biological interactions, an essential and fundamental concept in our world, result
from the fact that organisms in an ecosystem interact with each other. In the known
natural world, no organism is an independent body completely isolated from its
environment. It is part of its surroundings, which contains living and non living elements
in abundance, all of which interact with each other in some fashion. An organism’s
interactions with its environment are fundamental to the survival of that organism and the
functioning of the ecosystem as a whole. The Lotka-Volterra equations were created and
utilized to model the “predator vs. prey” relationship, an essential correlation in the
balance of our world and its resources. This association, predation, is mathematically
described as a pair of first order, non-linear, differential equations. More specifically,
they are frequently used to describe the dynamics of biological systems in which two
species interact, one a predator and one its prey. They were proposed independently by
Alfred J. Lotka in 1925 and Vito Volterra in 1926, both highly revered mathematicians
for their contributions to the field of population dynamics and mathematical biology.
As previously defined, biological interactions in ecology are the relationships
between two species in a given ecosystem. These relationships can be categorized into
several different classes of interactions based on either the effects of the interaction or the
means by which the interaction takes place. The associations between two species
contrast greatly in terms of these characteristics, specifically in extent and vigor. Types of
interactions include pollination, when species meet once in a generation, or
26
endosymbiosis, when species “live completely within another;” effects of these
interactions vary from one species eating the other, predation, to mutual benefit of both
parties, mutualism. However, these interactions do not require direct contact.
Ecosystems by nature may affect each other through “intermediaries” (e.g. resources or
familiar enemies). Symbiosis is described as a relationship when one organism lives on
or in another, or when organisms are related by “mutual stereotypic behaviors.” This
relationship is essential because almost every organism has numerous internal parasites
that are necessary for proper function. A particular instance of this is Mycorrhiza fungus
living with the roots of plants. The root tips provide the fungus with a relatively constant
and direct access to carbohydrates, such as glucose and sucrose produced by the plant
through photosynthesis. In return, the plant gains the use of the fungi’s larger surface
area. Mycorrhizae improve the mineral absorption capabilities of the plant roots by
taking up water and mineral from the soil. Another type of interaction, competition, can
be defined as an interaction between species where the “fitness of one is lowered by the
presence of another.” In this specific type, there is a limited resource that both organisms
need to survive. Such physical resources are typically water, food, and territory, but
competition can also occur without contact, in both a direct and indirect fashion.
Competition within a specific species is known as intraspecific competition, and
competition between two different species is interspecific competition. An example of
this interaction is trees in the Bangladesh forest. Because the forest is so dense, trees of
all species compete for light. They adapt through creating wider leaves or growing taller.
In ecology, predation describes a biological interaction, specifically where a
predator organism feeds on another living organism(s); the organism(s) being fed on are
27
known as prey. In these situations, predators may or may not kill their prey prior to
feeding on them. The other main category of consumption is the consumption of dead
organic material, or detrivitory. However, the defining characteristic of predation that
surrounds all of these behaviors is the predator’s direct impact on the prey population. In
other words, the unifying theme in all classifications of predation is the predator lowering
the “fitness” of its prey, or the reduction of the prey’s chances of survival and
reproduction. An example of predation, shown unmistakably below, is the lion and the
cape buffalo.
The derivative equation for the prey population is represented by
the one for predator population is
dx
= x(α − β y ) , and
dt
dy
= − y (γ − δx) . Distributing the x and y-variables
dt
yields the equations:
dx
= α x − β xy
dt
dy
= −γ y + δ xy
dt
Here is an example using rabbits, x, and wolves, y:
28
The “knowns” are:
α = 0.08
β = 0.001
γ = 0.02
δ = 0.00002
x(0) = 1000
y (0) = 40
Substituting these numbers into the given equations yields:
dx
= 0.08 x − 0.001xy
dt
dy
= −0.02 y + 0.00002
dt
These equations are used to graph the fluctuating populations of the rabbits and wolves:
Factoring the x and y-variables back out gives the following equations. Setting these
derivative equations equal to zero gives the equilibrium solutions:
29
dx
= x(0.08 − 0.001 y ) = 0
dt
dy
= y (−0.02 + 0.00002 x) = 0
dt
0.08
= 80
0.001
0.02
x=
= 1000
0.00002
y=
From here the variable t can be eliminated:
dy dy dx
=
dt dx dt
dy
dy dt −0.02 y + 0.00002 xy
=
=
dx dx
0.08 x − 0.001xy
dt
Graphing this function gives the direction field:
The first equation of the Lokta-Volterra equations is the one for prey. This
equation reads
dx
= αx − β xy , where x is the number of prey, and y is the number of
dt
predators. α is a positive growth constant of the prey, and β is a positive constant
determining the rate at which the prey is killed by the predator. The presence of
30
predators and prey represented in the term − β xy decreases the growth of the prey. This
equation is significant because it allows for the incorporation of loss of prey due to
predators in determining the growth of a population.
The second equation is the one for predators:
dy
= −γy + δxy . Once again the
dt
number of prey is represented by x, and predators by y. The positive constants γ and δ
help determine the growth rate of the predators. γ is a natural death constant for the
predators, and δ is a constant determining the rate of growth of predators. The
term + δxy , which incorporates the number of prey and predators present, increases the
growth rate of the predators. The significance of this equation is that it connects
predator growth with the number of predators and prey present in a system.
These equations are a proper fit for this predator-prey relationship because both of
the derivative equations take into account the prey and predator population at a given
time. The negative effect on prey population growth depends both on the predator and
prey population; the same is true for the predators except that the predator and prey
populations positively affect the predator population growth. Also, if there were no
predators, the change in prey population would reflect the derivative equation
dx
= αx ,
dt
thus the prey population would grow exponentially without the presence of predators. On
the other hand, if there was no prey present, the predator population would change by the
derivative equation
dy
= −γy . Therefore, the predator population would exponentially
dt
decline until the population became extinct. These equations go hand in hand: what
31
changes in one, affects the other just as population fluctuations effect other populations in
a system.
Best explained through a mathematical medium, it is obvious that the method of
predation is the most successful biological interaction. Because of the Lotka-Volterra
equations, we are given a reasonable model for this essential relationship, particularly
which demonstrates its efficiency and success. Predation, by definition, is the animal
behavior of pursuit, capture, and killing of animals for food. This method stabilizes
population and makes for a stronger ecosystem, balancing our world and its resources.
Bibliography:
Stewart, James. Calculus Early Transcendentals. 6th ed. Belmont: Thomson, 2003. 608612.
"Predation." Wikipedia. 10 Dec. 2007. 12 Oct. 2007.
<http://en.wikipedia.org/wiki/Predation>.
"Biological Interactions." Wikipedia. 4 Dec. 2007. 12 Oct. 2007
<http://en.wikipedia.org/wiki/Biological_interaction>.
"Trophic Links: Predation and Parasitism." Predator-Prey Relationships. 02 Nov. 2005.
11 Nov. 2007
<http://www.globalchange.umich.edu/globalchange1/current/lectures/predation/pr
edation.html>.
"Lotka-Volterra Equations." Wolfram Mathworld. 29 Apr. 2003. 13 Nov. 2007
<http://mathworld.wolfram.com/Lotka-VolterraEquations.html>.
Bailey, Regina. "Predator Vs. Prey." About.Com. 2007. 2 Dec. 2007
<http://biology.about.com/library/weekly/aa092001a.htm>.
"Predators." Wikipedia. 10 Dec. 2007. 3 Dec. 2007
<http://upload.wikimedia.org/wikipedia/commons/thumb/3/30/Male_Lion_and_C
ub_Chitwa_South_Africa_Luca_Galuzzi_2004.JPG/250pxMale_Lion_and_Cub_Chitwa_South_Africa_Luca_Galuzzi_2004.JPG>.
32
Understanding Renal Clearance through Calculus
Megan Eyunni and Dena Rothstein
Renal Clearance deals with the kidney filtering system. “It measures the volume
of plasma from which a substance is completely removed by the kidney in a given
amount of time (Meyertholen).” The liquid which enters the kidney from the plod is
processed and cleaned. Plasma enters the kidney in order to be filtered, and during this
process the other substance (ex. Inulin) is filtered from the plasma and excreted; therefore
the plasma does not contain any of this substance. Renal clearance is of great importance,
as the screening increases for renal problems associated with cardiovascular disease.
(Traynor). One of the main uses of renal clearance is to measure drug excretion, and by
using these equations the rate can be measured. Accurate measurement is difficult, but
the equations focused on deal with exponential decay, dialysis function, and renal
clearance rate.
An overview of the kidney and its function is necessary to understand the
mathematics of renal clearance in connection to flow rate and glomerular filtration. 625
mL of plasma go to the glomerulus, the capillary tuft that surrounded by the Bowman’s
capsule, per minute. “The kidney has several interlinked functions. These depend on
glomerular filtration rate, the unit measure of kidney function (Traynor).” Of the 625
mL, 125 mL are filtered into the Bowman’s capsule, this forms the filtrate. So the
glomerular filtration rate is 125 mL/min. The other 500 mL of plasma is absorbed in the
blood and goes to the peritubular capillaries, the small blood vessels that wrap around the
tubules of the nephron. The 125mL that entered the Bowman’s capsule and filtered, most
33
of the water is returned to the blood. The only ways a substance can make it into the
urine is if it is filtered by the glomerulus and not reabsorbed or if it the peritubular
capillaries secrete the substance into the tubules.
In order to get the main equation we must take the derivative of the equation
which measures the change of a certain (toxic) substance within the body. To calculate
this change it is necessary to find the sum of the substance intake, release, and generation.
We must keep in mind the release will be negative since there is a decrease in
concentration and we are going to take the derivative.
∆mbody = (− mout + m in + mgen )∆t
We know that the volume times the concentration equals the mass of the substance
present in the body and the clearance rate times the concentration equals the amount of
substance that is being released. Having these two other equations allows us to substitute
into the mass equation in order to take the derivative. The intake and release of the
substance combine into a total “m,” and we end up with the rate of toxin removal. The
volume change that results is irrelevant and therefore we obtain the equation below:
V
dC
= −K • C + m
dt
This particular equation describes the exponential decay that is used for kidney
function as well as the hemodialysis machine function (Clearance).” In order to find the
rate dC/dt, we must know the clearance rate, the mass generation rate of the substance,
the concentration of the substance, and the total volume. It is used to determine how long
and for how often they are to be on dialysis so they can clear all the toxins. Now it is
34
important to understand that the kidney cannot completely filter all of the substance from
the plasma, but this will be a very accurate estimation.
To find this equation we start with the equation to find ∆mbody. To find the mass
of a substance in the body you just multiply concentration of the substance by the volume
of the body so
mbody = C • V
And the amount of toxin removed is the rate of clearance for the substance times
the concentration of the substance so
mout = K • C s
So then in the mass balance equation simply substitute for mbody and mout
∆ (C • V ) = (− K • C + min + m gen )∆t
Then just group min and mgen together as m. Then divide by the change in time.
∆ (C • V )
= −K • C + m
∆t
And when the limit as ∆t approaches zero is applied, it results in a differential equation.
d (C • V )
= −K • C + m
dt
Then using the chain rule it expands to
C
dV
dC
+V
= −K • C + m
dt
dt
And then assume that the change in volume is not significant so it can be set as zero so
the differential equation becomes
35
V
dC
= −K • C + m
dt
The solution to this differential equation is
m
m
C = + (C o − )ℓ
K
K
− K •t
V
The ratio of the dialyzer constant, or the rate at which a hemodialysis filters, (k)
multiplied by the time a person is on the dialysis machine (t) this represents the amount
of fluid that passes through the dialyzer. Then to find V you take 60% of the patient’s
body weight (because only the volume of water in the patient is of interest and a person is
approximately 60% water by weight). The ratio of kt to V is used to compare the amount
of fluid that passes through the dialyzer to the amount of fluid in the patient’s body. This
number is necessary to find out if treatment is effective. The ideal kt/v after hemodialysis
is 1.4 and anywhere between 1.2 and 1.6 is considered acceptable.
For example, if the dialyzer’s clearance is 300 mL/min and the patient weighs 78
kg and is on dialysis for four hours, their kt/v will be 1.5.
To figure that out you simply multiply the renal clearance (k) by the time (min) on
dialysis.
Kt = 300mL / min• 240 min
Kt = 72000mL = 72L
Then find 60% of the patient’s body weight.
78 kg • 0.6 = 46 .8kg
This is the patient’s fluid volume.
Then divide the value of kt by the patient’s fluid volume (v)
36
72 L
= 1.54 L / kg
46.8kg
This fictitious patient’s kt/v is a little high which is not as bad as a below par kt/v
(which indicates dialysis was not entirely successful), but the more fluid lost does not
mean treatment was more successful, the removal of to much fluid has a negative affect
on the heart and circulation. If a patient’s kt/v were too low, the patient would either
have to increase the time he/she dialyze or the dialyzer’s clearance rate. If he or she were
to increase the time spent on dialysis, the increase in time would be found by dividing the
patient’s current kt/v by the target kt/v. So a patient with kt/v of 1 would increase his/her
time by 40%.
1.4
= 1.4
1
So if he/she normally dialyzed for 3 hours he/she would now dialyze for 4.2 hours.
The patient could also increase the dialyzer clearance by increasing the rate of blood flow
through the dialysis machine, to achieve this a good access key (the vascular connection
to the machine) is necessary, and often improving vascular access will increase k.
Beyond that, using a large machine that filters 500 or 600 mL/min as opposed to
300mL/min can increase k, Occasionally a patient is attached to two dialysis machines
simultaneously in order to increase the dialyzer clearance.
The equation for finding the renal clearance is presented above. Cu stands for the
urine concentration, Q stands for the urine flow, and CB stands for the plasma
concentration. It is important to recognize that along with this equation a timed collection
of urine must be recorded, then the values of urine flow and concentration can be
calculated. The normal urine flow for an individual is approximately 1.5 ml/min. f the
37
substance is not at a steady plasma concentration, and we are given the following date,
then we can calculate the renal clearance.
For example, using the sample data below we can calculate a clearance rate for an
individual.
K=
Cu • Q
CB
9.0mg / mL • 2.0mL / min
.15mg / mL
K = 120mL / min
K=
The urinary concentration of inulin must be multiplied by the urine flow rate, and then
divided by the plasma concentration of inulin. The final result for the renal clearance of
this individual is 120mL/min. This will help keep track of the patient’s urinary system
and kidney function when compared to the urine flow rates that are collected over time.
The various calculations above demonstrate the exponential decay used for kidney
function (hemodialysis machine), as well as the renal clearance rate. The renal clearance
rate will help to determine the kidney function depending on the urine flow rate and
concentration. In connection to dialysis, we derived the equation and used the ratio of the
dialyzer constant and multiplied this by the time a person is on a dialysis machine. Using
the equation kt/v we can compare the amount of fluid that passes through the dialyzer to
the amount of fluid in the patient’s body. In medical practice is not the calculus that is
used, it is the ratio kt/v, however the meaning of this ratio, the volume of fluid cleared
versus the patient’s fluid volume, would be unknown because it is the result of the above
derivations.
38
The ratio of the dialyzer constant, or the rate at which a hemodialysis filters, (k)
multiplied by the time a person is on the dialysis machine (t) this represents the amount
of fluid that passes through the dialyzer. Then to find V you take 60% of the patient’s
body weight (because only the volume of water in the patient is of interest and a person is
approximately 60% water by weight). The ratio of kt to V is used to compare the amount
of fluid that passes through the dialyzer to the amount of fluid in the patient’s body. This
number is necessary to find out if treatment is effective. The ideal kt/v after hemodialysis
is 1.4 and anywhere between 1.2 and 1.6 is considered acceptable.
39
Works Cited
"Clearance (Medicine)." Wikipedia. 2007
<http://en.wikipedia.org/wiki/Clearance_(medicine)>.
Hochhaus, Guenther. "Renal Clearance: General Notes." Basic Principles of Dose
Optimization I. 2007
<http://www.cop.ufl.edu/safezone/pat/pha5127/notes/Renal_Clearance_2.htm>.
Meyertholen, Edward. "Renal Clearnace." Austin Community College. Dept. of Bio,
Austin CC. 2007 <http://www2.austin.cc.tx.us/~emeyerth/clearancehtm.htm>.
"Renal Clearance Ratio." Wikipedia. 2007
<http://en.wikipedia.org/wiki/Renal_clearance_ratio>.
Schmedlap, Harley. "Renal Clearance Problems." Mammalian Physiology. 2007
<http://www.life.umd.edu/classroom/bsci440/higgins/renal_problems.html>.
"Standardized Kt/V." Wikipedia. 2007
<http://en.wikipedia.org/wiki/Standardized_Kt/V>.
Traynor, Jamie, Robert Mactier, Colin C. Geddes, and Jonathan G. Fox. "How to
Measure Renal Function in Clinical Practice." BMJ. 2007
<http://www.bmj.com/cgi/content/full/333/7571/733>.
40
Lagrangian Mechanics and the Brachistochrone Problem
Cora Bernard
Purpose
This paper discusses three things in relation to differential equations: the
minimization of a functional, independent of the integrand, and the subsequent
application of this principle to Lagrangian Mechanics and the Brachistochrone Problem.
Before exploring these topics, however, background information and mathematical
explanation is necessary.
Background Information
Derivatives
In calculus, the derivative of y ( x ) is equal to the lim∆x →0
∆y
. As ∆x approaches
∆x
zero, the expression approaches y ' . When ∆x = 0, the derivative is
infinitesimal, and in writing
dy
. “d” is an
dx
dy
, we assume that the limit has already been taken. We
dx
can also assume that ∆y = y ( x + ∆x ) − y ( x ) .
e.g. y = x 2
∆y = ( x + ∆x ) 2 − x 2
∆y = x 2 + 2 x∆x + (∆x ) 2 − x 2
∆y = 2 x∆x + (∆x ) 2
divide both sides by ∆x
∆y
= 2x + ∆x
∆x
∆x =0
dy
= 2x
dx
41
as shown above, only the terms with ∆x to the first power (terms that are linear in regard
to ∆x ) will affect the derivative. We can neglect any term with ∆x to a greater power
than one. After we divide both sides of the equation by ∆x , these terms will still have
∆x as a component, and ∆x goes to when we take the limit and is zero when we write
dy
.
dx
We also know from this that
 dy 
∆y dy
≈
and therefore ∆y ≈  ∆x . Instead of
 dx 
∆x dx
 dy 
writing an approximation, we can write a true statement dy =  dx . dy and dx are
 dx 
infinitesimals, which means, essentially, that a minute change has been made to y and x
respectively, and
dy
is the derivative of y with respect to x . Having now proven the
dx
 dy 
validity of the relationship dy =  dx , we can move onto more complicated calculus.
 dx 
Functionals
In calculus, we are used to working with functions that are essentially machines
into which you plug in a value for an independent variable, x , and get out a value for the
dependent variable, y . In calculus of variation, however, we work with functionals.
Whereas functions give a relationship between two or more variables, functionals give a
relationship between a function and a variable. We plug a function into a functional and
get out a value. In calculus a derivative tells us how a functions changes when we change
the independent variable. A derivative in calculus of variations tells us how our
functional changes when we change the function upon which it depends.
42
A common functional that we use in general calculus is the integral,
I[ f ] =
∫
b
a
f (x)dx . When we integrate, we set up a functional into which we plug a
function, f ( x ) , and when we evaluate between two limits, we get a numerical value for
the functional, I[ f ]. In calculus of variation, we are not limited to integrating just f ( x ) .
We can plug any variation of our function into the functional, just as, in general calculus,
we can derive any form of y ( y ', y '') with respect to x . For instance, we can set up the
functional I[ f ] =
 df

∫  dx f ( x ) • f ( x ) • dx , which changes with respect to changes in both
b
a
f ( x ) and its derivative,
df
f ( x ) . To minimize I[ f ], we would find the function whose
dx
integral gives a minimal value between (x a , y a ) and (x b , y b ).
Partial Derivatives
A function can depend on more than one variable. For instance, the function
f (x, y ) is dependent on both an x and y variable. If we look at the general equation of a
circle, a = x 2 + y 2 , we see that here f (x, y ) = x 2 + y 2 . Technically, all polynomials are
functions of multiples variables we just chose to treat many of the variables as constants.
When we take the derivative of the function f ( x, a, b, c ) = ax 2 + bx + c , we take it with
respect to x ; a , b , and c are treated as constants.
In calculus of variation it becomes especially important to take partial derivatives,
meaning that we take the derivative of a function with respect to one variable while
treating all others as constants. For instance, in the case of a circle, we could write
∂f
,
∂x
which means the partial derivative of f (x, y ) with respect to x . It is just like writing
43
df
dx
except in the first we are treating y as a constant. For f (x, y ) = x 2 + y 2 ,
∂f
= 2 x while
∂x
∂f
= 2 y . For a more complicated function such as f (x, y ) = y 2 x 3 + 3y 3 x ,
∂y
∂f
∂f
= 3 y 2 x 2 + 3 y 3 and
= 2 x 3 y + 9 xy 2 .
∂x
∂y
An interesting property of partial derivatives is that
our previous example, we can show this by setting
∂ ∂f ∂ ∂f
=
. If we look at
∂y ∂x ∂x ∂y
∂
(3 y 2 x 2 + 3 y 3 ) equal to
∂y
∂
2 x 3 y + 9 xy 2 ). Sure enough, 6 x 2y + 9 y 2 = 6 x 2y + 9 y 2 . Thus, mixed partial
(
∂x
derivatives are equal as long as the function is continuous. In the rest of the paper,
however, we will only be doing first partial derivatives, so we do not need to further
examine this aspect of them.
The important thing for us to understand about partial derivatives is that, for the
function f (x, y ), df =
∂f
∂f
dx + dy . This means that the derivative of f with respect o
∂x
∂y
x and y is equal to the sum of the partial derivative of f with respect to x and the
partial derivative of f with respect to y . We obtained this formula by looking back at
our general function f ( x ) for which df =
df
dx . When taking the partial derivative, we
dx
follow the same procedure and simply do it with respect to each variable. If we were
making finite changes in the function f (x, y ), then our definition of df would be much
more complicated and involve terms with a derivative of f with respect to both x and y ,
but since we are looking an infinitesimals, we simply take the sum of two single-variable
44
terms. This is because, like we discussed in the derivative section, we make the
assumption that all non-linear ∆x terms are zero when we use the notation d .
Part I: Finding the Extremum of a Functional
If we begin with the function x ( t ) , then we can construct the functional,
I[x ] =
∫ f [x (t), x ' (t)]dt , for which I[x ] depends upon both
t2
t1
x ( t ) and the derivative of
x ( t ) . We set the limits for the functional as t1 and t 2 , where x ( t1 ) = x1 and x ( t 2 ) = x 2 .
We want to minimize I[x ] and find a generic formula for the minimization of any
functional that depends on both x and x' .
To minimize I , we find the extremum and assume that the extremum is a
minimum. We know that in calculus, we find an extremum by taking the derivative of
the function and setting it equal to zero,
Therefore, if
 dx 
dx
= 0 . We also know that dx =  dt .
 dt 
dt
dx
is equal to zero at the extremum, then dx = 0 .
dt
Before we find the extremum mathematically, however, it helps to understand the
following graphical interpretation. In blue, we have graphed x(t) and specified (t1, x1 )
and (t 2, x 2 ) . The graph of x(t) + δx(t) is shown in read, also with the same starting and
ending points, (t1, x1 ) and (t 2, x 2 ) . In calculus of variation we use the lowercase delta, ∂ ,
to signify an infinitesimal change in a function. Thus, while dx would mean a change in
the value of variable x , δx(t) means a change in the function x(t) . The graph illustrates
our initial function, x(t) , and an infinitesimal change to that function, x(t) + δx(t) . We
know that we are at a extremum for the function if an infinitesimal change in x(t) does
not change the value of the functional.
45
δx(t) is an arbitrary infinitesimal function with which we make out new function
x(t) + δx(t) . In calculus, the definition of arbitrary is that an infinitesimal change is
either positive or negative. In calculus of variation, arbitrary means that an infinitesimal
change can be either positive or negative anywhere on the graph. This we illustrated
above, as the curve of x(t) + δx(t) is both greater than and less than x(t) at different
points of t on the graph. The only conditions we set to our arbitrary function are that
δx(t1 ) = 0 and δx ( t 2 ) = 0 . This makes sense because even though we are varying the
function, we do not vary the limits of the functional, or, in other words, the pathway must
still begin and end at the same points.
Now, in order to find the extremum of our functional, we must make an
infinitesimal change in it and set that equal to zero. To begin with, we find an expression
for the change in I [x ]. This we can say is dI [x ] = I f − I o , which is equivalent to
I[x + δx ] − I [x ]. The expanded version is
∫
t2
t1
f (x + δx, x'+δ (x'))dt −
∫
t2
t1
f (x, x')dt .
Before we go on we need to clarify what δ (x') means. In words, it is the change
in the derivative of x(t) . Mathematically, we represent that by writing
δ ( x ') =
d
d
(x ( t ) + δx (t )) − x (t ) . We can then expand the left term and simplify.
dt
dt
δ ( x ') =
d
d
d
x ( t ) + (δx (t )) − x (t ) and thus δ (x') = (δx)'. Q.E.D. The change in the
dt
dt
dt
derivative of x is equal to the derivative of the change in x —a seemingly simple
statement, but one that is important to understand.
Now we can go back to our finding the extremum of our functional. We combine
our two integral terms to get the equation dI [x ] =
46
∫ [ f (x + δx,x'+δ (x'))−
t2
t1
f (x,x')]dt . If
we look at the expression inside the brackets, we notice that this is simply df :
df = f ( x + δx, x '+δ ( x ')) − f ( x, x ') . We also know, from our work with partial
derivatives, that df =
∂f
∂f
∂f
∂f d
δx + δx ' . This can be rewritten as df = δx +
(δx ),
∂x
∂x '
∂x
∂x ' dt
since we showed above that (δx)' is equivalent to
d
(δx ) or δ (x').
dt
The new expression for the derivative of I[x ] when we make our partial
derivatives substitution is dI [x ] =
∫
t2
t1
∂f

∂f d
∂x δx + ∂x ' dt (δx )dt . We can look at the right
hand term as a separate integral which we integrate by parts:
u=
du =
∂f
∂x '
v = δx
d  ∂f 
 
dt  ∂x ' 
dv =
d
(δx )dt
dt
We write the derivative of our u term in a way helpful for our later calculations but
uncommon in calculus. If we say that u = x 2 , then we would say that du = 2 xdx , but we
d 
could also write this as  x 2 dx : the derivative of the function being integrated with
 dx 
respect to the variable of integration, a technique that we have chosen to use above.
Therefore, using integrations by parts, we see that
∫
t2
t1
∂f d
∂f
δx +
(δx )dt =
∂x' dt
t ∂x'
t2
1
∫
t2
t1
d  ∂f 
  • δx • dt . We can substitute this back into our
dt  ∂x' 
t
 t2  ∂f
d  ∂f    2 ∂f
initial expression for dI[x ] and we get dI[x ] =  ∫  δx −  δx dt+
δx . We
dt  ∂x'    t1 ∂x'
 t1  ∂x
t2
know that
t1
∂f
δx = 0 because, as we said before, δx(t1 ) = 0 and δx ( t 2 ) = 0 .
∂x'
47
Our expression for dI[x ] is even further simplified by factoring out a δx :
dI[x ] =
0=
∫
t2
t1
∫
t2
t1
 ∂f d  ∂f 
 −   •δx • dt . Now, to find the extremum, we say that dI = 0 and
 ∂x dt  ∂x' 
 ∂f d  ∂f 
 −   • δx • dt . This means that for any function of δx , the integral is
 ∂f dt  ∂x' 
equal to zero. In order for this to be true, it must always be true that
∂f d  ∂f 
−   = 0 for
∂f dt  ∂x ' 
any value of t . It helps to look at this mathematical relationship in simpler terms.
Suppose we say that the integral of the product of two functions, g(y) and h( y ) , is equal
to zero, 0 =
∫
y2
y1
g(y) • h(y) • dy . If we say that with any variation of h( y ) , the integral
still equals zero, in other words, h is arbitrary, then we know that g(y) must always, in
every case, be zero.
∂f d  ∂f 
−   = 0 is an extremely important equation, one that will help us
∂f dt  ∂x ' 
minimize complicated functionals in Parts II and III. In Physics, we call it the EulerLagrange equation.
Part II: Applying Part I to Lagrangian Mechanics
Background
Joseph Louis Lagrange is credited with creating Lagrangian mechanics, which is
essentially a modification to classical mechanics, in 1788. Lagrangian mechanics uses
both the principles of the conservation of mechanical energy as well as principles from
the conservation of momentum. Lagrange posited that objects move in paths that
minimize their action, action here meaning the difference between total kinetic energy
( KE ) and total potential energy ( PE ). Lagrangian mechanics is an alternative to
48
Newtonian mechanics, a method to solve problems that can be exceptionally and
unnecessarily complex in certain problems. For instance, electro-magnetic fields do not
follow Newton’s Laws and thus Lagrangian mechanics can be used instead. Or consider
a scenario in which a ball rolls down a slope as the slope slides along a frictionless track.
This is a non-inertial frame of reference and once again Newton’s Laws do not apply.
In the following mathematical explanation, we will validate Lagrangian
mechanics by showing that Lagrange’s principle that particles move to minimize their
action is equivalent to Newton’s Second Law, F = ma. Thus, we will show that using
either Lagrangian or Newtonian mechanics will give us the same final result.
Mathematical Explanation
Let’s begin by looking at the path of a particle from point A, (x a , y a ,z a ), to point
B, (x b , y b ,z b ). We want to know how this particle will move and what function will
describe its motion. We will choose time, t , as our independent variable. The path is
thus x(t) or x, y,z(t). To simplify, we will chose x(t) . The functional of this path
through time is the action, and particles move to minimize the action.
We also know another thing about the action, A[x ]: it is equal to the integral of
the Lagrangian, L , between time t1 and t 2 . We also know that the Lagrangian is equal
difference of the kinetic energy and the potential energy, L = KE − PE. Thus,
A[x ] =
∫ [(KE ) − (PE )]dt .
t2
t1
Now, if we examine KE and PE closer, we see that they are both functions of x ,
which of course makes sense since we know that the action is the functional of x or some
version of x . We know that KE = 12 mv 2 and the v =
49
dx
, or, as we write it in physics,
dt
v = xÝ, the dot above the x here marking it as derivative with respect to time. Thus,
KE = 12 mxÝ2 . We also know that PE = 12 kx 2 for a mass on a spring or = mgx if x is a
vertical height, and, in general, it can be written as U(x). And since L = KE − PE, we
see that L is a function of both x and xÝ. Therefore, A [x ] =
∫
t2
t1
L(x,xÝ)dt . This of course
looks very familiar to our original functional from Part I except before the function
dependent on x and xÝwas annotated by f and is now L , and the functional, previously
I for integral is now A for action.
Now we can use the Euler-Lagrange Equation that derived in Part I:
∂L d ∂L 
−   = 0 where L = 12 mxÝ2 − U ( x ) . The derivative of L with respect to x is
∂x dt ∂xÝ
simply −U(x) , and the derivative of L with respect to xÝis mÝ
x or momentum. Now we
have −
dU d
− [mxÝ] = 0 . The slope of the graph of a potential energy curve is always
dx dt
d2x
Ý= 2 . We are thus left with
−F and the derivative of velocity is acceleration, a = xÝ
dt
F − ma = 0 or F = ma, Newton’s Second Law. F = ma is a second order differential
equation. Like all second order differential equations it requires two additional pieces of
information to specify the problem completely. Usually, these are v o and x o , but in this
case we are more concerned with x o and x f .
50
Part III: Applying Part I to the Brachistochrone Problem
Background
The Brachistochrone problem asks a relatively simple intuitive question, but one
that can only be proven using calculus of variations. Suppose I roll a ball along a 2D
track so that the ball will only move horizontally and vertically. Assuming that the initial
velocity is zero, gravity is the only force that acts on the ball, and that the track is
frictionless, what track will allow the ball to get from the starting point to ending point in
the least amount of time. In order to
solve this, we will have to set up a
functional since we are trying to find
the function (not the variable) that
will minimize time.
First, we can look at the problem graphically. To start, we mark two points,
A (x a , y a ) and B (x b , y b ), in Quadrant I of the Cartesian Plane. The track we draw
between them here is the curve of some
9
function, f ( x ) , upon which the time, in
8
7
this case our functional, depends.
6
Path 1
5
Path 2
4
Intuitively, we know that if we draw a
Path 3
straight line, the ball will eventually get
3
2
from A to B, but, using principles of the
1
Conservation of Potential Energy, we
0
0
2
4
6
8
know that a steeper slope at the
51
beginning will give the ball a greater KE and thus greater velocity: the greater the
velocity, the less the time. However, if we make the curve too steep, then we will end up
increasing the distance since the ball must not only travel down but back up as well: the
greater the distance, the greater the time. Thus, we can assume that the graph we want is
somewhere between a line with constant slope and a curve with incredibly steep slope.
This is illustrated in the example to the right in which our curve of choice is drawn in
yellow.
The optimal curve is officially named the Brachistochrone curve, also known as
the curve of fastest descent. We can solve the problem now with calculus of variations,
but it caused much confusion in the middle ages and even up until the eighteenth century.
In 1638, Galileo asserted in his Two New Sciences, that the optimal curve was the arc of a
circle. This is incorrect. The actual curve is the path of cycloid, a curve generated by
the tracing of the radius of circle as it rolls horizontally.
In 1696, Johan Bernoulli solved the problem after closely analyzing the solution
to the tautochrone curve, a
problem somewhat similar to
the brachistochrone one for
which the solution is also a
cycloid. After solving it, Bernoulli posed it as a question in the journal Acta Eruditorum.
Four prominent mathematicians of Bernoulli’s time submitted solutions: Isaac Newton,
Bernoulli’s brother, Jakob, Gottfried Leibniz, and Guillaume de l’Hopital. l’Hopital’s
excepted, the solutions were published in 1697, also in Acta Eruditorum. Newton and
52
Leibniz, famous for their rivalry, laid similar claims to having solved the problem first
and bickered over the calculus used to find the solution as well.
Jakob Bernoulli hoped to outdo his brother and thus created a new, more difficult
version of the Brachistochrone problem. In order to solve it, Jakob Bernoulli was forced
to develop a new branch of calculus, one that was later enhanced by Leonhard Euler into
calculus of variations in 1766. Joseph-Louis Lagrange, mentioned in Part II, further
developed Bernoulli’s and Euler’s work into the infinitesimal calculus we use today.
Mathematical Explanation
While the actual path of the ball in the Brachistochrone problem is located in
Quadrant I, it is mathematically
0
1
0
easier to place it in Quadrant IV
1
2
3
4
5
-0.002
while using calculus of variation
-0.004
to find the solution. We assume
-0.006
that x increases as we move
right and that y increases as we
+y
1.3
y(x)
-0.008
4.5
-0.01
2
2.5
-0.012
3
4
3.5
move down. The graph to the
-0.014
x
right shows a possible pathway,
y(x).
Using conservation of energy, we can find the velocity at a specified point, (x, y),
as a function of the height, y . If we know the function, y(x), then we can also find
velocity as a function of x . We know from simple kinematics equations that
∆x
= t,
v
distance over velocity equals time. Using what we know, we can use this formula to write
53
ds
= dt where ds is the infinitesimal distance along the pathway, v is the ball’s velocity,
v
and dt is an infinitesimal amount of time in which some function of x . We know that
2
ds = 1+ ( dy
dx ) dx and that v is some function of the derivative of x . If we take the
integral of both side,
∫
ds
=
v
∫ dt , we obtain the relationship T [y ] = ∫
2
1+ ( dy
dx )
, where
v
T [y ] is total time, a functional of the function y(x).
Once again we want to minimize a functional because we want to the find what
function will give us the minimum amount of time to get from ( x1, y1) to ( x 2 , y 2 ) . Since
dy
= y' , we can rewrite the infinitesimal distance traveled as ds = 1+ ( y ') 2 dx . We can
dx
also fin an expression for the velocity. We know, according to the law of conservation of
energy that at the point (x, y) between ( x1, y1) and ( x 2 , y 2 ) , ∆E = 0 , and therefore
∆(PE) + ∆(KE) = 0. Thus, −gy = − 12 v 2 and v = 2gy .
Our functional is now T [y ] =
∫
x2
x1
1+ (y') 2
dx . We write the limits as x1 and x 2
2gy
instead of coordinate points since we are integrating with dx in regard to x . Once again
we have the integral of a function that is dependent on both y and y' so we can use the
Euler-Lagrange equation. We set it up as
factor out
1
from both terms:
2gy
∂
1+(y' )2
2gy
∂y
 1+(y')2
d  ∂ 2gy
−
dx  ∂y'

 1+(y')2
 1+(y')2
1 ∂
d
y
y
∂
−


∂y
dx
∂y'
2g


through both sides of the equation to get rid of
54

 = 0 and can then



 = 0. We can divide


1
and we are left with
2gy
∂
1+(y' )2
y
∂y
 1+(y')2
d ∂
y
−

dx
∂y'


 = 0 . Knowing that d y = y' and d y'= y'' , we can simplify the

dx
dx

left hand side of the equation to 2y • y''+(y') 2 + 1 = 0. This is an equation for a cycloid
that is equivalent to the more traditional cycloid function, (y') 2 =
C
−1. This tells us that
y
when y = 0, y'= ∞. This of course fits with the curve of cycloid, which has an infinite
slope where the radius of the circle touches the x -axis.
Works Consulted
Boas, Mary L. Mathematical Methods in the Physical Sciences. 3rd. Hoboken, NJ: John
Wiley & Sons, 2006, as explained by Claude Bernard.
Wikipedia. 2007. 1 Jan 2008 <http://www.wikipedia.org/>.
Cora Bernard
Dr. Pais
AP BC Calculus
January 7, 2008
55
A Journey Along a Bézier Curve
Michael Sherwood and Frank Trampe
A Bézier curve is a parametric curve that is very important in computer
graphics. The curve itself is a parametric curve defined using a certain
amount of given points. For example, imagine three points P1, P2, P3 with a
straight line drawn between P1 and P2 and a straight line connecting P2 and
P3. A line would be constructed with points Q1 and Q2 beginning on points
P1 and P2 respectively. Point Q1 would then move along the line P1P2 and
point Q2 would move along the line P2P3 respective to time, hence
parametric. An imaginary straight line between Q1 and Q2 would be drawn
with a time-dependent slope; it is from this slope that the curve is drawn. If
more initial points were added, then another set of points would be added to
the first line derived, the same method would be applied to find the timedependent slope of this new line, and the curve would be drawn from the
newly calculated slope. In short, 3 points yields one slope line and is called
a quadratic, 4 points yield two slope lines (the first being the slope of the
initial 4 points and the second being the slope of the first slope line) and is
called a cubic, and so on and so forth. Looking at the graphical depiction of
a Bezier curve, we determined that a function produced from integrating the
slope of the final line would produce a function identical to that produced
from scaling a point along that line. Thus the derivative of the curve should
be equal to the slope of the final line; and thus the final line is always
tangent to the curve.
Born in Paris, Pierre Bézier received degrees in both electrical and
mathematical engineering. Following this, he obtained a PhD in
56
mathematics from the University of Paris. Leading a relatively uninteresting
life, Bézier began working as an engineer for Renault. It was at this point
that Pierre Bézier designed a system for modeling smooth curves
mathematically. Bézier curves (and Bézier surfaces) were very helpful in
designing automobiles. Bézier and some friends designed an automated
design system called UNISURF CAD CAM which employed Bézier
surfaces and curves for facilitating the creation of complex designs.
Following a 42-year career at Renault, Bézier was recognized by ACM
SIGGRAPH with a “Steven A. Coons” award for his outstanding lifetime
contribution to computer graphics. Pierre Bézier died 11 years after he had
reached the peak of his fame. The application of this is very simple: Bézier
curves are generally used when smooth curves must be drawn. The very
nature of the Bézier curve allows the creator to manipulate the curve
infinitely in an intuitively simply way, resulting in a curve that is perfect for
each independent application.
The following function scales a value
between two specified values for a scaling value
between 0 and 1.
This function scales between two points on a Cartesian plane.
First we are building a scaling-based model of the curve.
These are the 5 base points of a quartic Bezier curve:
57
The scale function is applied to the five points to get a scaled set of
four points; these four points form the second tier of points which will be
scaled to form the final equation for a Bezier curve.
The previous four points are then scaled to yeild three points which
form the third tier of points, brining us ever closer to the final function.
The following is the fourth tier of points.
58
The fifth and final tier gives the function for the Bezier curve.
59
The previous funtion was then condensed to form a parametric
representation of the Bezier curve.
The x and y-coordinates of the parametric Bezier curve were
differentiated in order to find the slope at any time, t. Dividing the x and y
parts of the differentials yield the final differential equation for the Bezier
curve.
60
We hypothesized that a curve produced from the slope of a line from
the fourth tier points would produce a curve a curve equivalent to scaling a
point between the fourth tier points, i.e. the fifth tier.
61
Close examination of the two equations above show that they are
indeed equal, proving our hypothesis. Below is a graph of a Bezier curve
produced by five random points (0,8) (4,8) (0,0) (8,4) (8,0). The Bezier
curve is shown as a bolded line, and the 9 other lines represent the other four
tiers in the Bezier curve. The points in each tier are scaled according to the
points in the previous tier.
62
63
In order to check our proof, the same process was used to compare the
slope of the line between the fourth tier points and the equation of the fifth
tier, produced by scaling the fourth tier points.
To find the slope, we divide the differential of the y component by the
differential of the x component.
64
This is the differential equation for the slope of the final line for a
given set of points and a given t-value:
This is the differential equation for the Bezier curve:
65
The two expressions are equal. If based upon the same points, and
both have the same starting point, the curves will be identical.
Works Consulted
Lancaster, Peter. Curve and Surface Fitting: An Introduction. London:
Academic Press, 1986.
Global community. Bezier Curve. 14 December 2007. 1 December 2007
<http://www.en.wikipedia.org/wiki/B%C3%A9ZA_curve>.
66
Newton’s Law of Cooling
Robert Birkenmeier and Brendan Lyss
Sir Isaac Newton was a man of science and math that always had a curious mind
and will power to figure world physics with little resources. His ability to construct
equations and laws during his time that still are applicable to today’s society is simply,
remarkable. His famous three laws and less known law of heating and cooling are
scientific and mathematic wonders that help students in their classrooms today. How he
developed his law of heating and cooling is not apparent and scrutinized. Despite the
complex equation that we associate with Newton’s law is in fact derived from his original
theory not developed by Newton himself.
He was the first to believe the relationship for the
energy flow rate due to convective heat transfer at the
interface between a solid object and a fluid environment.
This means that he studied the loss of an objects heat is
proportional to the temperature of the objects
surroundings, further meaning that the temperature will
most likely rise and fall in a proportional rate until it is at
or in the proximity of the environment. The experiment that lead to this finding is still
investigated, however is assumptions were correct and poring to make Newton a pioneer
of math and science in the past and today.
67
The rate of change in temperature (T) in an object is proportional to the difference in
temperature between the object and the surrounding temperature (S):
∆ T is proportional to the difference (T – S); as (T – S) decreases the rate of change
lessens
Also the limit of T as time ∞ is S.
Graphically this appears as an exponential decaying line:
Temp
(T)
∆T
would give the slope of the tangent line at any point, describing the rate
∆t
of temperature change at a given time. This would also be defined as the first
derivative of the equation for T(t), T'(t). We know that T'(t) is proportional to (T –
S).
Given some constant k, such that k>0 and since we the object cools with time:
Thus
T'(t) = -k (T(t) – S)
Or
T'
= -k
(T(t) - S)
68
so long as T(t) – S ≠ 0
Again
T'
= -k
(T(t) - S)
so long as T(t) – S ≠ 0
This equation can be solved to
ln T(t) - S = -kt + C
Which equals
At To T(t)
T(t) - S = (e-kt) (ec)
– S = ec therefore:
T – S = (To – S) e-kt or T = S + (To – S) e-kt
We have arrived at Newton's Law of Cooling:
T = S + (To – S) e-kt
Another way to derive this is equation is using the exponential decay equation:
dT
= -ky which solves to y(t) = y(0) e-kt
dt
if y = T(t) – S then by substitution
T(t) – S = (To – S) e-kt or
T = S + (To – S) e-kt
Still another way to look at the equation is:
dT
= -k (T - S) since y = (T – S) can be rewritten as T = (y + S) we can
dt
substitute (y + S) for T yielding:
dT d(y + S)
dy dS
=
=
+
but S is a constant so dS = O therefore
dt
dt
dt
dt
dT dy
=
which is the exponential decay equation so that we get:
dt
dt
69
dy
1
= -ky or ( ) dy = -k (dt) this solves to :
dt
y
ln T(t) - S = -kt + C
as above ln T(t) - S = -kt + C the same equation shown in the first example
which solves to:
T = S + (To – S) e-kt
Newton’s law of heating and cooling has been applied to many aspects of modern
day life. We learn about it in classrooms, apply it to every day jobs, even forensic
science. Without Newton’s law being able to determine how long a dead body has been
dead for would be a significant problem leaving a gaping hole in an investigation.
Forensic scientists and investigators can find out when a person was killed by, taking the
body’s internal temperature, than comparing it to the outside external temperature while
simultaneously keeping the time at which such tests are completed. Then using what we
know about normal human temperatures (approximately 98.6 degrees Fahrenheit) and
plugging the finding into Newton’s equation T(t)= Te + (To- Te) e-kt. Where T(t) is the
temperature of the body at time t and Te is the constant temperature of the environment,
To is the initial temperature and k is a constant that depends on the material properties of
the body. Without Newton’s law of Heating and cooling forensic scientists would have a
much harder time at evaluating the time of death for any one person.
70
There are many more examples of Newton’s law of cooling. For example how
long it will take for ice on one’s driveway or roadways to melt based on the temperature
outside. To solve this problem it is the same technique for the dead body and a sheet of
ice.
An example of the use of Newton's Law to determine the time of death.
There is a crime where a man is found murdered in his apartment. The coroner
measures the temperature of the body at 7 a.m. and records 72.5o, 1 hour later the body is
72.0o. The thermostat in the house is set to keep the air temperature at 70.0o
We have then: To = 98.6o; T(t) = 72.5o, T(t+1) = 72.0o, S = 70.0o
Plugging these values into T = S + (To – S) e-kt we get:
72.5 = 70 + (28.6) e-kt and
72.0 = 70 + (28.6) e-k(t +1) simplify and divide these equations yields:
2 .5
thus k = - 0.233 plug this into 72.5 = 70 + (28.6) e-kt and solve
2 .0
we find that approximately 10.92 hrs have elapsed since the time of death, therefore the
e-k =
man died around 8:00 pm the previous night.
71
Works Consulted
http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html
UBC Calculus Online- This source gives a description of the definition of Newton’s law
along with some applications of the law. It gives problems for the reader in order to
explain in full derivations of Newton’s law of heating and cooling.
http://www.ncsec.org/team3/history.htm
National Computational Science Leadership Program- This gives a short explanation on
what we know about Sir Isaac Newton and how he discovered his Law of heating and
cooling.
http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/MethodesMaths/white/deqn/a1/deadtime/deadtime.html
Dr. J R White UMass-Lowell- Explanation of the application of forensic science and how
Newton’s law can help determine how long a person has been dead for.
Themes for Advanced Calculus- Text book explaining Newton’s law of cooling but also
allowing learning of how the explanation of the law is derived into an equation.
http://www.rose-hulman.edu/fc/lc/functions/exp/newton.html
Rose Hulman Institute of Technology- Explanation of functions derived from Newton’s
law of heating and cooling.
72
Millennium Force: Calculus and Roller Coasters
Emily German, Rachel Simon, and Dominique Zotos
For our senior calculus final project, we decided to find a real roller coaster,
discover its equation, and create a graph in order to utilize the calculus skills we obtained
this semester. We chose to analyze a famous Cedar Point roller coaster: Millennium
Force. We knew from the beginning that we were not going to be able to recreate the
entire coaster since there was simply not enough information available about all of its
peaks and valleys; the only information that we could obtain through the amusement
park’s website were the heights of the first three hills and the length of the entire roller
coaster, all measured in feet. After analyzing a miniature picture of the coaster’s layout
(see image below), we decided to graph and analyze just the first three hills of
Millennium Force.
We began the process by sketching a qualitative graph of this picture, first
“flattening” the coaster to make it twodimensional; we then entered in the yvalues of the three peaks and two valleys.
Then we were confronted with a serious
problem: we could not find an equation
for this graph until we obtained the xvalues of its peaks and valleys. In order to
find the x-values, we would have to find
the length of the graphed portion of the coaster in comparison to the length of the entire
73
coaster. In order to do this, we took the miniature picture of Millennium Force, broke it
down into small pieces, and measured its length in centimeters. We then measured the
portion of the graph we wanted to analyze and found its length in relation to the entire
track, and then converted this value into feet. Next we broke down the portion of the
graph we were going to analyze into each of the three hills, then found the length of each
hill as a percentage of the entire graphed portion. Finally, we converted all of these
values into feet. Once we had the length of each bump of the track, we divided each
bump in half and used the Pythagorean Theorem to find the approximate horizontal x
value for each half bump.
Finally, we found the
lengths of x and matched
each with its corresponding
y-value. We now had all
the coordinates of the peaks
and valleys needed to graph
the equation.
The graph that we
created from the three
peaks and two valleys of
Millennium Force,
however, did not translate into one complete function, but an amalgam of seven different
functions. Dr. Pais, therefore, took the coordinates we obtained and, through a program
called Maple Graphing Calculator, divided the graph into seven individual piecewise
74
functions. He then carefully connected the functions so that they created one continuous
graph with all the correct peaks and valleys of our initial drawing of Millennium Force.
Despite its similarities to the real Millennium Force, our graph and equations are
obviously only a two-dimensional approximation of the roller coaster, and thus lack the
coaster’s physical characteristics. As a warning, therefore, nobody should try to ride this
version of Millennium Force!
Once we had obtained the seven equations, we were then free to use what we had
learned in calculus and apply it to our graph and seven functions (see attached chart). We
first found the derivative of each equation by using the sum of functions rule. We then
found the second derivative of each equation
by again using the sum of functions rule. At
first we encountered problems while graphing
the derivatives: when we graphed the original
equations and the derivatives on a graphing
calculator, only f(x) would appear on the
graph. After discussing the problem with Dr.
Pais and using Maple, however, we discovered
that the derivative did not show up on the calculator because of a huge discrepancy in
size: while the highest point of f(x) was over 200 feet, the highest point of f’(x) was
under 5 feet! With a more sophisticated program, however, the graph of the derivative
became clearly apparent, as is shown in the picture above.
We then found the bump points of f(x) by setting f’(x) to zero and finding its xintercepts, for the x-intercepts of f’(x) are equal to the x-values of the bump points of
75
f(x). We then found the inflection points of f(x) by setting f”(x) to zero and finding its xintercepts, for the x-intercepts of f”(x) are equal to the x-values of the inflection points of
f(x). What we encountered, however, was that a number of the x-intercept values of f’(x)
and f”(x) were not part of the roller coaster graph, for the inflection points and bump
points were not necessarily within the graphed portion of each function; therefore, we
noted in the data chart which x-values appeared on the graph and which were outside of
the restricted domain of each function.
Lastly, we found a tangent line for each equation, or a linear y=mx+b equation
that touches f(x) at one point; we did this by substituting an x-value into f(x) to find the
slope, then substituting the slope, x-value, and y-value into y=mx+b to find b, or the yintercept.
Rachel, Dominique, and I chose to study roller coasters because of our interest in
coasters and their strong association with the mathematics of calculus. Learning about the
structure of a real-world roller coaster was an eye-opening experience because it showed
us that calculus is not just a theoretical subject, but an important tool that is frequently
used in the real world.
Millennium Force Roller Coaster Data
Function 1:
F1 f(x): -.0000005891304713x^3 – .003189401306x^2 + 2.042977339x
1) Finding Derivative and Second Derivative:
f´(x): 3(-.0000005891304713)x^2 – 2(.003189401306x) + 2.042977339
f"(x): 6(-.0000005891304713)x – 2(.003189401306)
2) Finding Bump Points of f(x):
x = -3905.162386 and 296 296 is a bump point of the function!
3) Finding Tangent Line at Point: (400, 269.18)
76
-.0000005891304713(400)^3 – .003189401306(400)^2 + 2.042977339(400)
m = 269.1823765
269.18 = (400) (269) + b
b = -107403.7706
y= 269x - 107404
4) Finding Inflection Points of f(x):
f"(x) = -.00000353478283x -.0063788026
.0063788026 = -.00000353478283x
x = -1804.581193 Not in graphed portion of function!
Function 2:
F2 f(x): -.00002312230474x^3 + .05902030372x^2 – 45.93884257x + 11320.52204
1) Finding Derivative and Second Derivative:
f´(x): 3(-.00002312230474)x^2 + 2(.05902030372)x - 45.93884257
f"(x): 6(-.00002312230474)x +2(.05902030372)
2) Finding Bump Points of f(x):
323.5239497 and -2028.208422
3) Finding Tangent Line at x=550
(550, 60.826989)
y=-2.000000026x+1160.827003
4) Finding Inflection Points of f(x): -916.9353634
Function 3:
f(x): -.000006807053813x^3 + .01509277567x^2 – 10.48992579x + 2338.776589
1) Finding Derivative and Second Derivative:
f´(x): 3(-.000006807053813)x^2 + 2(.01509277567x) – 10.48992579
f"(x): 6(-.000006807053813)x + 2(.01509277567)
2) Finding Bump Points of f(x):
x = -.290.4449983 and 1768.593706 not in graphed portion of function!
3) Finding Tangent Line at Point x =750
F’’(x)=3(-.00002312230474)7502+2(.05902030372)
-38.90084859=3(-.00002312230474)7502+2(.05902030372)
Y=m(x)+b
Y= -38.90084859 (x)+b
F(750)=-.00002312230474(750)3+.05902030372(750)2-45.93884257(750)+11320.52204
77
310.5886428=-.00002312230474(750)3 + .05902030372(750)2-45.93884257(750) +
11320.52204
Y=m(x)+b
310.5886428=-38.90084859 (750)+b
29486.22509=b
y=-38.90084859x+29486.22509
4) Finding Inflection Points of f(x):
x = 739.075283 This value is in graphed portion of function!
Function 4:
F4 f(x): .00004224244702x^3 - .1338617050x^2 + 140.2850129x – 48531.14164
1) Finding Derivative and Second Derivative:
f´(x): 3(.00004224244702)x^2 – 2(.1338617050)x + 140.2850129
f"(x): 6(.00004224244702)x – 2(.1338617050)
2) Finding Bump Points of f(x):
2439739355 and .2914728845 Not in graphed portion of function!
3) Finding Tangent Line at Point ( ):
4) Finding Inflection Points of f(x): 1056.29692
Function 5:
F5 f(x): -.000000000091567626x^3 + .00000038048674x^2 - .00051202493x +
10.22475788
1) Finding Derivative and Second Derivative:
f´(x): 3(-.000000000091567626)x^2 + 2(.00000038048674)x - .00051202493
f"(x): 6(-.000000000091567626)x +2(.00000038048674)
2) Finding Bump Points of f(x): 1151.553536 and 1618.615748
3) Finding Tangent Line at Point ( ):
4) Finding Inflection Points of f(x): 1928.000004
Function 6:
F6: f(x): -.00008721448078x^3 + .5044485581x^2 – 970.9439060x + 621990.4426
78
1) Finding Derivative and Second Derivative:
f´(x): 3(-.00008721448078)x^2 + 2(.5044485581)x - 970.9439060
f"(x): 6(-.00008721448078)x + 2(.5044485581)
2) Finding Tangent Line at Point: (1939, 113.85)
-.00008721448078(1939)^3 + .5044485581(1939)^2 – 970.9439060(1939) +
621990.4426
m = 4.193324991 10^9
113.85 = (1939) (4.19E9) + b
b = -8.13085716E12
y = 4.19x +.000000000008
3) Finding Bump Points of f(x):
4) Finding Inflection Points of f(x):
Function 7:
F7: -.000004780806792x^3 + .03390016224x^2 – 80.07293542x + 63013.97310
1) Finding Derivative and Second Derivative:
f´(x): 3(-.000004780806792)x^2 + 2(.03390016224)x - 80.07293542
f"(x): 6(-.000004780806792)x + 2(.03390016224)
2) Finding Tangent Line at Point: (2300, 7.0036)
-.000004780806792(2300)^3 + .03390016224(2300)^2 – 80.07293542(2300) +
63013.97310
m = 10.004
7.0036 = (10.004)(2300) + b
b = -23001.38
y= 10x -23001
3) Finding Bump Points of f(x): 5705.737255 and -11881.36532
4) Finding Inflection Points of f(x):
79
80
Works Cited
Hammer, Patricia W., Jessica A. King, Steve Hammer. "B. Design and Thrill of
One Coaster Drop Using a Trig Function." MathDL. 2004. The Mathematical
Association of placecountry-regionAmerica. 5 Dec 2007
<http://mathdl.maa.org/images/upload_library/4/vol5/coaster/exportfiles/trigcoaster1hill.
html>.
Hiob, Eric. "The Roller Coaster or Brachistochrone Problem." addressStreetBCIT
Math Square :: Entertainment : Roller Coaster. 22 Dec 1996. British Columbia Institute
of Technology, Math Department. 4 Dec 2007
<http://commons.bcit.ca/math/entertainment/coaster/index.html>.
McVeen, Kieth. "Millennium Force Overview." Coasterbuzz. 2001. 5 Dec
2007<http://badnitrus.coasterbuzz.com/Coasters/MillenniumForce/MillenniumForce_Ov
erviewLo.jpg>
"Millennium Force, Roller Coasters." Cedar Point. 2007. Cedar Fair
Entertainment Company. 5 Dec 2007
<http://www.cedarpoint.com/public/park/rides/coasters/millennium_force/index.cfm>.
Weisstein, Eric W.. "Cycloid." Mathworld. 1999. Wolfram Research, Inc.. 5
Dec2007 <http://mathworld.wolfram.com/Cycloid.html>.
Weisstein, Eric W.. "Brachistochrone Problem." Mathworld. 1999. Wolfram
Research, Inc. 5 Dec 2007
<http://mathworld.wolfram.com/BrachistochroneProblem.html>.
81
Calculus in Population Growth
Jessica Houghtaling and Carissa LaFebere
We chose to do our project on the exponential growth of population in various
countries. Population growth is a relevant topic to today’s world with the population
expanding at an exponential
rate. The only way to find the
growth rate of a population at
one given point is to calculate
the tangent to the curve at that
point. Through this calculus we
can determine how fast the
population is growing at a specific point of time, and how much it has changed over the
years. Population changes follow a simple J shaped curve; the reasoning for this is that
people tend to multiply exponentially. The equation for this is Pt+1 = r × Pt , where t
represents the time period, Pt represents the population at one given period (Pt+1 being the
next period), and r, referred to as the Malthusian factor, is the multiple that determines
growth rate. This is the most simple equation for estimating the population for the next
period; however it relies entirely on the population of the current period. The Malthus
factor or Malthusian ideas consist of the belief that overpopulation is possible based on
82
the capacity of the environment. Thomas Malthus believed that poverty was directly
connected to the population problems and he wrote a series of essays on this topic. What
this Malthusian equation neglects to acknowledge are the factors of birth rate, death rate,
environmental impact, and the carrying capacity. Carrying capacity is the idea that any
given area can only hold so many people before all its resources are depleted. This
changes a population rate from the unlimited J-shaped curves, to a more logistic curve (Sshaped) which shows the
steady and rapid increase,
and then the eventual
plateau. Once it stables out
the population would
ideally stay at this level.
The only problem with this
idea is the suggestion of Age structure dependencies. This is basically the idea that not
everyone in the population will reproduce, and that the reproduction rate is likely to
change depending on age. For example, there are some countries where the majority of
the population is under 15. This would have a dramatic influence on the birth rate simply
because most people would not be getting pregnant at that age. The simple equation for
this is Pt+1 = f ( Pt ) , meaning that the population of the next period is a function of the
population now, but this function is related to the population from before which gives
you Pt+1 = f ( Pt , Pt −1 , Pt − 2 ...Pt − n ) . Each population is based on a function from the year
before; therefore this data is fairly inconclusive.
83
In all these equations, the Malthusian factor (value r) plays a big role. When the value of
r is between one and three it will increase in a sigmoid (s-shaped) curve. This graph
shows three different cases, and how they can increase at different speeds towards a
stable value. The stable values can also be considerably different. The yellow line (value
of r being larger than 2.4) the line will oscillate before it hits a steady value. This sort of
oscillation is also very normal in population growth. In fact it is possible for a stable
population to reach a point in which it oscillates between several stable amounts
repeatedly. It seems as though the values jump up and down randomly, however this
behavior is called “chaos”. It still repeats a pattern and jumps between alternate values.
84
Another one of the stranger-looking population graphs is known as the “Boomand-Bust” curve. It is possible for
a population to go through a time
period in which their numbers
will go dramatically up and then
dramatically back down. The low
point of this graph is the stable
population. This type of dramatic population change is most common in insects but it
wouldn’t be unheard of in other species. The different models of population growth need
to be able to incorporate the various factors that would give irregularities in the
population growth or decline. These factors can vary from anything to natural diseases or
natural disasters that kill off large numbers, to a drop in fertility of the species, or even to
a change in climate or geography. This is what makes population growth so difficult to
determine. In order to calculate approximately a probable number for any given time,
you would have to be able to take everything into consideration, which is nearly
impossible, so any population estimates have to be seen as such; just estimates.
The formula used by the US census bureau to find the population predictions
 P( t+1) 
in this table was r(t) =ln 
 , t being the year, r(t) being the growth rate from year to
 P(t ) 
mid-year, P(t) being population at midyear t. The data was put into maple to find a
function of the population in order to find the inflection point of the growth rate. By
finding this through the population growth rate now, we can predict the year when the
population will peak. While an attempt to find the function was made, the Graphical
analysis software could not format the function to a polynomial equation that it
85
should have been. Because we cannot find a graph that correlates to the predicted rate
of change of population, we cannot use
the second derivative to find the point in
time when the growth of the population
is predicted to decline.
When the graph and function
derived from the graph is used to
determine the year in which the
population will reach its carrying
capacity on Earth, the year is predicted through a smooth curve, which cannot fully
represent any form of instantaneous population growth as it is only represented in years.
The population growth is currently following the idea of the “S” curve as it should be to
reach population equilibrium. Population is predicted to reach equilibrium in
approximately 2048 at around 11 billion people which can be found by setting the
derivative of the function for population equal to zero, thus finding the bump points for
the population function, predicting when the populations, globally, or in certain countries
will reach equilibrium or have a peak. For example, Russia, when the population growth
and data is applied to a polynomial fit, peaks around 1996 at 148,695,000 whereas the
actual value for 1996 is 148,311,699 showing that the smooth curve is fairly accurate, but
can be inaccurate by thousands. The peak is found through solving for the zeros of the
derivative of the function. The function for the population of Russia is
f (t ) = −0.0004420678314t 5 − 16.80126182t 4 + 1.500541913 • 105 t 3
−4.618074015 • 108 t 2 + 6.195846046 • 1011 t − 3.094066261 • 1014
86
When the derivative of this function is taken, the derivative is found to be
f '(t ) = −0.002210339157t 4 − 67.20504728t 3
+4.501625739 • 105 t 2 − 9.236148030 • 108 t + 6.195846046 • 1011
When f '(t ) is set equal to zero, and the equation is solved for t, the peak of population
can be determined. Determining the peak of the population is key because in population
prediction, it allows scientists to determine when the largest population will be achieved,
and at what point the population will begin to decrease. By finding inflection points using
the second derivative of the function, and focusing in on a smaller section of the data, you
can monitor changes in the population growth rate. This model of data, a set taken from
the population graph of Russia in the years 1960-1985, allows us to further see what
changes can occur.
Through using the second derivative of the
function for this applied curve, we can
determine when the slope changes by
setting f ''( x) equal to zero.
f ''( x) = 7452.744874t − 1.470440741 • 107
is the expression for the function of this
curve. When set equal to zero, the
inflection point for the curve is at 1973.016111, meaning that there was a change in the
growth rate of the population in 1973.
Also through using the data taken from censuses and finding a function for
population, the population growth rate in any given year can be found. Through using
87
the derivative of the original function, we can determine the near-instantaneous
growth rate of the population. For example, if we use the original function for the
population of Russia, in the year 1960, we can solve the derivative, using 1960 as t in
the equation
f '(t ) = −0.002210339157t 4 − 67.20504728t 3
+4.501625739 • 105 t 2 − 9.236148030 • 108 t + 6.195846046 • 1011
Substituting 1960 in for t, we solve this equation for f '(t )
f '(t ) = −0.002210339157(1960) 4 − 67.20504728(1960)3
+4.501625739 • 105 (1960) 2 − 9.236148030 • 108 (1960) + 6.195846046 • 1011
From this, we learn that the growth rate is 1368358.3 people per year in 1960, compared
to the growth rate of only 522006.6, when the population had already begun to decline.
Using the population of the United States, we can find the same data in the same
ways as was done for Russia. Like the graph for Russia, a fifth degree polynomial curve
can be used to find a good fit, resulting in the equation of
f (t ) = 0.006338014053t 5 − 74.29339209t 4 + 3.403892806 •105 t 3
−7.663213875 •108 t 2 + 8.509646831•1011 t − 3.738799536 •1014
While this is, graphically, a more linear
curve, when extended out further, the
graph becomes more parabolic, and using
a fifth degree polynomial curve gives the
best fit to this data, demonstrating a
change in slope and a moderate s-curve
through the t values of 1970-1995.
88
While mathematically, it is possible to predict the trends in population through
calculus, it is not entirely possible to completely find the trend of population growth
and expansion without factoring in certain non-mathematical, unpredictable factors
such as famine, disease, and any other natural deaths, but through calculus, the
government can more easily determine population, and how to deal with its
repercussions especially when using a smooth curve fitted to scattered data points.
89
Bibliography
1. Human Population and its Limits <http://wwwformal.stanford.edu/jmc/progress/population.html>
2. United States Census Bureau <http://www.census.gov/main/www/cen2000.html>
3. US Census bureau Population Projections
<http://www.census.gov/population/www/projections/popproj.html>
4. Wolfram Mathworld—Population growth
<http://mathworld.wolfram.com/PopulationGrowth.html>
5. Mathematical Models Population growth
<http://www.arcytech.org/java/population/facts_math.html>
6. Population, Sustainability, and Earth's Carrying Capacity: A framework for
estimating population sizes <http://dieoff.org/page112.htm>
7. Birth and Death Rates <http://www.os-connect.com/pop/p3.htm>
8. Total Midyear Population for the World
<www.census.gov/ipc/www/idb/worldpop.html>
90
Derivatives: Maximizing the Marginal Profit
John Wuestling and Shannon Shubert
Derivatives play a major role in many aspects of economics. Specifically,
intertwining derivatives with economics, an individual or company can find the most
efficient way for maximizing his or her marginal profit or predict future values of
securities that don’t pay dividends. Maximizing the marginal profit is the primary goal of
companies and can be accomplished through minimizing total cost and maximizing total
revenue. The maximum marginal profit is often depicted in the form of a vertex of a
downward opening parabola. This parabola is formed by using an equation that includes
current profit per unit multiplied by the change in the original price: This quantity is
multiplied by the amount sold of the product by the original price which was added to the
potential profit from the decrease of the first price. With calculus applications, like
derivatives, companies and individuals can more easily determine the most effective
prices to sell products with the intent of making the highest profit possible. These
applications mesh studies of economics and math and help develop a more advanced
conceptual and analytical understanding of both subjects.
To understand how changes in price affect consumption it is important to use real
life models. For example, the Beasley third grade class is trying to raise money to buy a
new gerbil, thus they decide to have a lemonade stand. For their supplies, they purchase
an industrial size of Country Time Lemonade mix which costs $4.50 as well as several
large packs of Dixie cups; these together should yield approximately 150 cups. This puts
their marginal cost at $.0717 or, for our purposes, 7 cents. They initially agree to charge
91
$0.50 per cup. On the first day, they are disappointed because they only sell 80 cups;
however, they hypothesize that if they reduce the price to $0.45, they could sell 20 more
cups. In order to test the class’s hypothesis, it will be necessary to create a parabolic
equation. Please see equation and verbal steps below.
f ( x) = ((0.50 − 0.07) − 0.05 x))(80 + 20 x)
f ( x) = 34.4 − 4 x + 8.6 x − x 2
f ( x) = − x 2 + 4.6 x + 34.4
f '( x) = −2 x + 4.6
f '(0) = −2 x + 4.6
−4.6 = −2 x
2.3 = x
2.3* 0.05 = 0.115
.45 − .115 = .385
The derivative in this graph only reveals to us the amount times the current price
reduction must be multiplied.
92
1. The first step is to form a quantity with the original price, subtracting the marginal
cost, followed by the subtraction of the proposed decrease of the original price.
The variable x represents the amount of times it is necessary to reach the desired
maximum marginal profit. This quantity is then multiplied by a quantity that
contains the original amount sold (80), followed by the desired increase in sales
(20). Again, the x represents the amount of times necessary to reach the desired
maximum marginal profit.
2. The second step is to foil the equation.
3. The third step is to gather like terms
4. The fourth step is to take the derivative of the equation from step 3, which will
later provide the slope of the tangent line.
5. The fifth step and remaining ones is to set the f’ equation equal to zero, which
will allow us to solve for the location of the peak. The answer is 2.3, which
reveals that when multiplied by the 0. 05 decrease, the maximum marginal profit
will be calculated (0.115). Then it is necessary to subtract this from the old price
(0.50), which equals 0. 385. Finally, this represents the maximum marginal profit.
This equation contains a very important point, the maximum or peak. This simple
coordinate point reveals the optimal price for selling our lemonade. Without
incorporating mathematics with economics, one could spend hours tweaking their prices
and still not be close, so instead of all the guess work, there is a simple and easy way to
determine the peak. We know that from calculus we can use the derivative. The
derivative will help us determine the slope of the tangent line to the original curve. And
93
since we know that peaks have a slope of zero, we can set the F’ equation equal to zero,
and in literally no time we can derive that optimal price for selling the lemonade.
Recently, this notion of using derivatives to maximize the marginal profit can be
seen with the iPhone. When the iPhone was released in late June by Apple and AT&T, it
was priced at approximately $600.00, selling an average of 9,000 phones per day. The
objective of the two companies manufacturing the iPhone, however, was to sell one
million phones by September 9th. In order to accomplish this goal, surveys and studies
had to be conducted using derivates to find the most appropriate price for selling the
phone and achieving the maximum marginal profit.
Since the companies were selling around 9,000 phones a day, in order to reach the
one millionth sale by September 9th, about 27,000 phones would have to be sold per day
instead. This meant that there would have to be a 200% increase in the phone sales! The
companies lowered the price of the phone to about $400.00, close to a 33.33 % decrease
or 1/3 of the original price of the phone. Through decreasing the price, many more
phones were sold and Apple and AT&T reached their aim by September 9th. While the
companies were able to reach their goal in selling one million iPhones by this date, the
drastic cut in price left many consumers of the original price angry for being what some
think exploited by the manufactures. Even though, there is a tendency for prices of
electronic products to decrease overtime after their release, the iPhone had a rather
significant decrease in price: 1/3.
In order to demonstrate, finding the maximum marginal profit and the exponential
growth of the sales for the iPhone, an equation must be formed similar to the one used for
the lemonade sales. Again,
94
f ( x) = ((600 − 250) − 200 x)(9,000 + 18,000 x)
f ( x) = (350 − 200 x)(9,000 + 18,000)
f ( x) = 3,150,000 − 1,800,000 x + 6,300,000 x − 3,600,000 x 2
f ( x) = −3,600,000 x 2 + 4,500,000 x + 3,150,000
f ( x) simplified = −3.6 x 2 + 4.5 x + 3.15
f ' ( x ) = −7.2 x + 4.5
f ' (0) = −7.2 x + 4.5
x = .625
.625 * 200 = 125
600 − 125 = 425
The derivative graph represented in red is not entirely representative of the actual
marginal price. Instead it tells us the amount of times the current reduction in price has to
be multiplied. Therefore it is just one piece of the puzzle mathematically and not the final
answer.
95
1. The first step is to form a quantity with the original price, subtracting the marginal
cost, followed by the subtraction of the proposed decrease of the original price. The
variable x represents the amount of times it is necessary to reach the desired
maximum marginal profit. This quantity is then multiplied by a quantity that contains
the original amount sold (9,000), followed by the desired increase in sales (18,000).
Again, the x represents the amount of times necessary to reach the desired maximum
marginal profit.
2. The second step is to foil the equation.
3. The third step is to gather like terms
96
4. The fourth step is to take the derivative of the equation from step 3, which will
later provide the slope of the tangent line.
5. The fifth step and remaining ones is to set the f’ equation equal to zero, which
will allow us to solve for the location of the peak. The answer is .625, which
reveals that when multiplied by the 200 decrease, the maximum marginal profit
will be calculated (125). Then it is necessary to subtract this from the old price
(600). Finally, this represents the maximum marginal profit.
Lastly, we want to discuss another mathematical application using the function e
that is connected with derivatives and economics. The equation involves predicting the
future values of non-dividend paying common stocks or bonds. John C Hull demonstrates
this notion in Options, Futures, and Other Derivatives with his general
formula F = Ser (T −t ) . “S” stands for the amount of dollars borrowed by the investor, r
represents the interest rate, T stands for the final time when the stock was stock was sold
and the t stands for the time when the stock was bought. The overall profit at the selling
time T is defined by F- Ser (T −t ) and Ser (T −t ) is the amount of money required to pay off the
loan.
Similar to the formula above, suppose you recently bought a three-year bond,
currently valued at $800.00. Then the bond makes a coupon payment of $50.00 in six
months and an additional $50.00 after twelve months. Assume that the interest is
compounded at 10%.
I = 50e −0.1/ 2 + 50e −0.1
I = 47.56 + 45.24
I = 92.80
97
The variables are as followed: I represents the present value, while the interest rate is
represented by the numerator of the exponent, and the denominator was the amount of
coupon payments in a given time. Finally, the next term represents the original price of
the bond.
Between the Beasley’s lemonade stand and the sales of the iPhone, it is easy to
see the advantages with using derivates in order to find the maximum marginal profit.
Along with this, using the function e in an exponential equation, an individual can find
the future values of their stocks. Individuals and companies can form equations
incorporating the original price designated to the product, the marginal cost, the
production of the cost, the proposed decrease in price of the product, and finally, the
desired increase in the amount of sales that will maximize the total revenue. With all
these components considered in a derivative equation, the maximum marginal price can
be solved. Without using this equation, individuals and companies would have to do a lot
of guess and check work in order to find the maximum marginal profit. The Beasley
third class used derivates and discovered they could increase their sales by twenty cups of
lemonade if they lowered the price their price from $0.50 to $0.45. Their increase in sales
would be 25%. In a recent example, the AT&T and Apple companies hired someone to
conduct surveys and use mathematics, particularly derivatives, to find the appropriate
price to sell the iPhone in order to reach the one millionth sale by September 9th. The
companies lowered the phone from $600.00 to $400.00, a 33.33% decrease; however,
their sales rose dramatically, and they were able to sell one million phones by their
desired date. Through using derivates and intertwining mathematics with economics, both
98
the Beasley third grade class and the AT&T and Apple companies were able to maximize
their marginal profit.
Bibliography
Elmer-Dewitt, Philip. Apple 2.0. Business 2.0 Beta. 2007. 03 Dec. 2007
<http://blogs.business2.com/apple/2007/09/report-price-cu.html>.
IPhone At $599. 2007. Ian L McQueen. 03 Dec. 2007 <http://www.techcrunch.com/wpcontent/iphonemoney.jpg>.
McConnel, Campbell R., and Stanley L. Brue. Economics: Principles, Problems, and
Policies. 7th ed. New York: McGraw-Hill /Irwin, 2008.
2007. Technabob. 03 Dec. 2007 <http://technabob.com/blog/wp
content/uploads/2007/09/399_iphone.jpg>.
"Futures and Forward Contracts." 13 Dec. 2007
<www.bus.ucf.edu/cschnitzlein/Course%20Pages/Graduate%20Notes/Futures%2
0and%20Forward%20Con
"Futures and Forward Contracts." 13 Dec. 2007
<www.bus.ucf.edu/cschnitzlein/Course%20Pages/Graduate%20Notes/Futures%2
0and%20Forward%20Contracts-MBA.ppt>.
99
The Calculus in Projectile Motion
Christopher Waterbury and Bennett Walsh
Calculus can be used to examine the path of a projectile, how far it will travel,
where it will land, and how long it has been in the air. For purposes of simplicity, air
resistance will be insignificant in the equations and calculations that will be used. The
general equations for the path of projectiles will be represented by a parabola. Once the
projectile has been launched, the only force that is acting upon it is that of gravity, -9.8
meters per second squared. Thus its acceleration in the y direction is always -9.8 m/s2,
while the motion in the x direction is positive. [2]
With and object moving in the y direction or vertically, gravity is the only acting
force. For the following two equations, θ will be equal to the initial angle and the initial
time will be 0 with an initial velocity of vi. These can be used to determine the initial
velocities in the x and y direction.
cos θ i =
vxi
vi
sin θ i =
v yi
vi
[2]
These equations can be used to find both the x any y elements of the velocity.
For the initial x velocity: v xi = vi cosθi
For the initial y velocity: v yi = vi sin θ i
To find the change in distance in the x direction this equation can be used:
x = vcosθ t
100
To find the change in the vertical motion, a series of equations can be used. Once
the projectile is launched, gravity takes over and acts upon the object with a constant
negative acceleration of 9.8m/s2. Acceleration is an instantaneous rate of change of the
velocity of the object and the force of gravity is found by using this equation:
dv
=a=g
dt
When the variables of velocity and time are known, these can be used to find the
acceleration.
∫ dv = ∫ gdt
Velocity is equal to the rate of change of distance over time so this equation is
used.
dd
=v
dt
The given information can be used to find a more concrete equation for a change
in distance in the x direction using a constant acceleration of gravity and time:
1 2
gt
2
∫ dx = ∫ gtdt
d=
1 2
gt
2
Note: this equation is assuming that the initial velocity of the projectile is 0.
d=
The general equation of a projectile in motion is that of a quadratic. Assuming
there is no air resistance, the general path of the object is that of a parabola. This can be
seen in the graph below: [3]
101
Example problem:
A ball is fired from the ground with an initial speed of 1.02 x 103 m/s at an initial angle of
470 to the horizontal. (no air resistance) Find the amount of time the ball is in the air.
To find the time, use the equation: △h = v yi (t ) +
1
g (t 2 ) . For this, the change in height is
2
zero because the ball is landing at the same elevation that it was fired. Gravity is -9.8 and
the initial y velocity can be found using the equation: v yi = vi sin(θ ) .
v yi = vi sin(θ )
v yi = 1.02 x103 sin(47)
v yi = 746 m/s Using this number,
there is now one equation with one unknown.
△h = v yi (t ) +
1
g (t 2 )
2
1
0 = 746t + (−9.8)t 2
2
t = 152.2 seconds.
Bibliography for above
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Vectors/ProjectilesMotion.html
102
http://en.wikibooks.org/wiki/Physics_with_Calculus:_Part_I:_Projectile_Motion_&_Rela
tive_Velocity
When a gun is fired, a ball is thrown, or a projectile is launched into the air, the
firing/throwing/launching mechanism is irrelevant. This means that different types of
guns or different people throwing an object have no effect on the trajectory of the
projectile. The only things that really matter, concerning physics, are the (multiple)
magnitudes and the vector of the projectile. This is assuming, of course, that all other
environmental factors (such as changing winds, wind speeds and directions, and
humidity) are naught: this examination deals with a hypothetical “ideal projectile” that
has no other outside factors acting on it. [1]
It should also be noted that while gravity can be assumed to be a constant factor in
determining trajectory (at 9.8m/s2), latitude does have a minimal effect on gravity that
can change it. For example, while in northern South America and in Africa at the
equator, gravity is 9.78 m/s2, whereas in Antarctica at the South pole, as well as at the
North pole, gravity is 9.83 m/s2. This effect, however, is negligible considering
trajectory, so it really does not have to be considered: 9.8m/s2 can be used.
Another question to consider, dealing with projectile motion and BB guns
specifically, is whether or not the mass of the BB has any effect on the speed of the bullet
or the overall energy of the bullet itself. The answer is that for large distances (around
125 feet and more), heavier bullets will reach their targets a little faster than lighter
bullets, but this difference is almost so small as to be considered negligible. As for the
question of different energies, lighter bullets will start off faster than heavier bullets when
fired from the muzzle of the gun, but because of this they will lose their energy (have
103
their energy be dissipated) much faster than heavier bullets. This means that over long
distances, lighter bullets will lose energy faster and therefore have less energy than
heavier bullets, assuming that both kinds of bullets have the same shape and diameter,
and assuming that all other environmental factors are naught. Heavier bullets will not
only retain their energy longer, but because of this they will experience less drift when
fired. Acceleration is inversely proportional to mass, so the heavier the bullet gets the
less lateral acceleration (drift) the bullet will experience.
When it comes to diameter as opposed to mass in BB bullets, however, bigger is
not always better. For example, bullets with larger diameters (8mm) dissipate energy
faster than bullets with smaller diameters (6mm), and therefore the smaller bullets have a
greater firing range. And considering wind resistance as well, the smaller diameter
bullets have the advantage: the larger 8mm bullets are less dense and occupy a greater
amount of three-dimensional space than the 6mm bullets, so it is easier for the wind to
blow it and accelerate it laterally (drag), and there is more space for cross winds to hit
and blow it in an errant direction, respectively. This difference can be somewhat
staggering considering the percentage of difference, as an 8mm bullet can have more than
200% times the amount of drag or lateral deflection as a 6mm bullet. The heavier the
bullet, the less the lateral deflection, but even an 8mm bullet with more than twice the
mass of a 6mm bullet can have more drag and sway off course more than its smaller
(volume-wise) counterpart.
It is a common misconception that wind resistance is greatly affected by the
velocity of the projectile, but the truth— concerning BB bullets at least— is that highvelocity bullets do not dissipate or fight off wind resistance any better than low-velocity
104
bullets. The key is that because they are moving faster, they reach their targets sooner, so
problematic cross winds have less time to unleash their harmful (direction-wise) effects
on bullets with higher velocities; bullets with lower velocities, on the other hand, are
airborne for a longer period of time, and assuming the same cross winds exist, there is
more time for the cross winds to blow these bullets off course. For example, if two
bullets of equal masses and diameters were fired from guns, one traveling at 300 feet per
second and the other traveling at 175 feet per second, both will be affected by winds
equally, but they will have different amounts of total time in the air (before they hit a
target, presumably) for the wind to have this effect on them. After a half of a second
after both guns have been fired, the faster bullet will have traveled 150 feet and the
slower bullet will have traveled 87.5 feet. The wind will affect them the same, so if the
cross winds blow the faster bullet 4 inches off course, the slower bullet will also be
blown 4 inches off course, but after covering less distance. If the faster bullet hits the
target (a haystack 300 feet away, for example) after the next half of a second, and gets
blown off course another 4 inches, the slower bullet will travel another 87.5 feet and get
blown off course another 4 inches as well, but it will not have reached the target yet and
will have more chances to get blown off course even more (this is assuming that the
bullets do not lose velocity as they shoot, which they will). Therefore, it is not that
lighter bullets get blown off course more by cross winds, but that they have more chances
to be blown off course because they are in the air for a greater amount of time.
All of the above facts that have to do with BB bullets also apply to airsoft bullets,
paintballs, and any other projectiles that are fired from a source, such as a cannon ball
fired off a cliff. There are numerous factors that affect trajectory, however, such as drag
105
force, velocity, distance, magnus force, terminal velocity, spin decay, drag coefficient, lift
coefficient, gravity. These all have equations that can be applied to them: drag force, for
example, is equal to one half times the drag coefficient times the air density, times the
cross-sectional area of the projectile, times the instantaneous velocity squared. The most
mathematically interesting equation of all of these, however, was for the terminal
velocity, which is the maximum velocity possible that a projectile can reach while in free
fall. This equation is as follows: the terminal velocity is equal to the force of gravity
divided by the quantity of (one half times the drag coefficient times the air density times
the cross sectional area of the projectile) to the one half power.
For a specific example, this is the equation for the projectile of a BB gun bullet
with a 20 mph headwind:
y = (−.0066481994) x 2
The y variable represents the height of the bullet in inches above the ground, and
the x variable represents the distance in feet that the bullet goes with the headwind
factored in.
This is how we got the equation:
I started by taking two points from the graph on the “cybersloth” website, which
were (0,0) and (95,-60). I then plugged the first set of points into the equation
y=ax2+bx+c:
0=a(0)2+b(0)+c
c=0
If I plugged in the other point, I would hit a roadblock:
(-60)=a(95)2+b(95)+0
106
I had still had two variables, so I found the derivative of the original quadratic
equation:
y=ax2+bx+c
D(y)=2ax+b
And then I plugged in the first point (0,0):
0=2a(0)+b
0=0+b
b=0
So after this, I knew that both “c” and “b” were equal to zero, so I could go back
to my equation with two variables and solve for “a”:
(-60)=a(95)2+b(95)+0
(-60)=a(95)2+(0)(95)+0
(-60)=a(95)2
a=(-60)/(95)2
a=(-0.0066481994)
With this knowledge, I could write out the final equation:
y=ax2
y=(-0.0066481994)x2
If we wanted to do a sample problem with this equation, we could figure out the
height that the fired bullet has reached based on how far it has traveled, all with the same
20 mph headwind:
If we say that the bullet has traveled 50 feet through the air so far, we would plug
in 50 for x and solve for y:
107
y=(-0.0066481994)(50)2
y=(-16.62049861)
The answer is negative because the bullet has dropped in the air from the position
it was fired from. This represents the y coordinate of the point (50,-16.62049861)
between the original points we used to get the equation: (0,0) and (95,-60). This is how
you would find where the bullet is with regards to height and distance. It should also be
noted that when the bullet had a 10 mph headwind pushing it back, it landed about 2.5
feet farther than the bullet with a 20 mph headwind. Similarly, the bullets with no
headwind or tailwind, a 10 mph tailwind pushing it along, and a 20 mph headwind
pushing it along, landed about 5, 7.5, and 10 feet farther than the original bullet. This
seems to indicated that for this example at least, adding 10 mph in wind to help/hinder
the speed of the bullet will affect the length of its trajectory and final landing spot by
about 2.5 feet.
Bibliography for above
Allen, Nathan. “ATP: The Airsoft Trajectory Project.” Cybersloth.org. 2007.
http://cybersloth.org/airsoft/trajectory/index.htm
Graph: Allen, Nathan. “ATP: The Airsoft Trajectory Project.” Cybersloth.org. 2007.
http://cybersloth.org/airsoft/trajectory/05-A-01.htm
References
[1] http://cybersloth.org/airsoft/trajectory/index.htm
[2] http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Vectors/ProjectilesMotion.html
108
[3]http://en.wikibooks.org/wiki/Physics_with_Calculus:_Part_I:_Projectile_Motion_&_R
elative_Velocity
Works Consulted (and Fun) but not extensively used
http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/ProjectileMotion/jarapp
let.html Application: A projectile motion.
http://jersey.uoregon.edu/newCannon/nc5.htmlGame: Shoot the monkey.
http://www.jstor.org/view/0025570x/di021201/02p00063/4?frame=noframe&userID=4b
[email protected]/01c0a80a6900501d1164c&dpi=3&config=jstor Graphs: JSTOR book
with graphs and explanations and equations.
109
Derivatives in Weather Forecasting
Ashley Ervin and McKenna Morey
Meteorology is the study of the atmosphere and atmospheric phenomena; it is the focus
of the weather and prediction analysis. In meteorology, instruments are used to measure
temperature, wind speed, pressure and several other elements of the environment. Scientific
theory is used to understand how these aspects of the earth all relate together. Mathematics is
included for graphs, charts, and numerically modeling. Meteorology goes beyond just simple
mathematics, but in addition focuses on equations. These equations enable forecasters to pull
together, average, and overall determine information. Specifically, meteorologist use partial
derivatives to determine the change in speed when moving in one direction. A partial derivative
is the function of many variables, and all but the variable at hand remain constant during the
differentiation. This differs from the normal derivative where all variables are able to change. A
specific example of mathematics used is the lapsed rate. This is the negative rate of change in a
variable such as speed or temperature. The varying in an environments lapse rate is significant to
the study of the atmosphere. Calculus is also highly used in the study of cyclones within
meteorology. The study of this aspect of the weather is very significant not only because it
forecast severe storms, but also because it is the source of the transportation of energy,
momentum, and water vapor. Therefore it plays a vital role in mean-wave flow interactions.
Thus, finding the derivative is also very important in this element. By using calculus it allows for
a more accurate ways of tracking cyclones. Understanding weather forecasting can also help to
price weather derivatives. Weather derivatives are a kind of weather insurance for companies
that insure for low risk high probability weather, instead of high risk low probability weather.
110
By using their understanding of forecasting along with probability and derivatives, companies
can price weather derivatives accurately. Overall mathematics has proven to be imperative for
forecasting and making predictions in weather more precise.
Cyclones are significant synoptic systems that have important effect on the weather and
society as a whole. It is due to cyclones and anticyclones that account for majority of the
atmospheric variability in mid-latitudes. There are several aspects of cyclones that are
misunderstood; but one specific element that is increasingly difficult is the differentiation
between tropical and mid-latitude systems. In tropics, “convective instabilities are more
important whereas in mid-latitudes the cyclones are driven rather by baroclinic instabilities.”
However in both latent heat conversion is a necessity because is it a primary energy source.
Cyclones in mid-latitude were manually identified and investigated using synoptic
weather maps during the 1850’s. Studies on cyclones began years ago; however, it has taken
years to figure out a better systematic analysis. This took place in the mid-twentieth century. In
order to track the cyclones, they used the advancement and general application of the digital sealevel pressure (SLP) data, therefore regular tracking algorithms became needed. The algorithms
were created to facilitate the descriptions of cyclone climatology. Recent studies show that
cyclone elements vary on different timescales. Researchers have found it complicated to
determine trends in cyclone statistics because of the variance due to inter-annual and decadal
variations. Meanwhile some studies indicate there are no considerable trends in mid-latitude;
others have found long term trends in some regions. For clarity, there is a difference between
storm tracks diagnostics and identifying the storm centres or feature point. A highly sophisticated
feature tracking device can be utilized for analyzing storm tracks; however, the detection of the
111
definite storm centres is considered more archaic and less high tech. In general, there is a variety
of ways to track cyclones and as time progresses more and more techniques are created.
One of the more complicated methods that researchers have discovered is calculus-based
cyclone identification (CCI). It is more advanced than some previous methods, because it is
more computationally challenging. The CCI approach is similar to the B-splines and the
Laplacian of pressure surface; however, the CCI technique “offers an advantage of providing
explicit analytical solutions for the gradients estimated through a regression analysis.” One
significant distinction between methods is the CCI recognizes the local minima as the points that
cross zero of the derivatives. The CCI method is more mathematical than many other methods
and it seeks answers steady with the form eikx, which is universal in several differential
equations. This process of identifying cyclones “involves multiple least-squares regression to a
truncated series of sinusoids in order to estimate the values of the coefficients,” instead of
evaluating the grid-point value with those points around it. The local minima are discovered
along the north, south, east and west profiles of the SLP. This is done by evaluating for the first
and second order derivatives, and looking for points for prior and positive values for the latter
cross zero. The method used to calculate the derivative is a slight variation of the method used by
Benestad and Hanssen-Bauer.
This is the original equation used to predict the cyclone:
Nθ
P(ө) = Po +
∑ [a (i) cos(ω (i)θ ) + b (i) sin(ω (i)θ )]
Θ
θ
θ
i =1
Nφ
P( φ ) = Po +
θ
∑ [aφ (i) cos(ωφ (i)φ ) + bφ (i) sin(ωφ (i)φ )]
i =1
112
These two equations show how to find derivatives of the original equation:
•
P0 is a constant; therefore, when taking the derivative its value equates to zero.
•
aθ (i ) and bθ (i ) both represent coefficients, so when taking the derivative, their values
are unaffected.
•
The Cosine function when taking the derivative changes to negative Sine, so ultimately
the negative sign is placed in front of the coefficient.
•
The Sine function when taking the derivative changes to Cosine.
•
Finally the chain rule in finding the derivative is applied because there is a function
inside of the parenthesis of both the sine and cosine functions.
dp (θ ) Nθ
= ∑ ωθ (i )[ − aθ (i ) sin(ωθ (i )θ ) + bθ (i ) cos(ωθ (i )θ )]
dθ
i =1
dp (φ ) Nθ
= ∑ ωφ (i )[− aφ (i ) sin(ωφ (i )φ ) + bφ (i ) cos(ωφ (i )φ )]
dφ
i =1
In a more simplified version this is how the derivative is derived using θ , and the same
process it used when finding the derivative of φ . Allow: aθ (i ) = a, ωθ (i ) = w, bθ (i ) = b
D[ acos(w) + bsin(w)] = a D[cos (w)] + b D[sin(w)]
= a[-sin(w) • (w) + b(cos (w) • (w)]
= w[ -asin (w) + bcos (w)]
113
In this graph is the test of the
calculus-based cyclone identification
process for the local minima. On the graph
the thick grey area represents the data and
the lines indicate the Fourier
approximation. “Fourier studied the
mathematical theory of heat conduction.
He established the partial differential
equation governing heat diffusion and
solved it by using infinite series of
trigonometric functions.” With periodic
functions found in the atmosphere, it can
be articulated “as the sum of a number of
sine and cosine terms.” This is known as
the Fourier series. This approximation is
applied to the north and south, east and
west lines, which then enables the
cyclones to be identified. They are discovered where the local minima in the north and south
lines meet the local minima of the east and west lines. For both, these are on the first derivative
at the zero crossing point. The smaller the θ and φ the more precise the identification of the
cyclone would be. Specifically on this graph, the dotted lines represent the points of inflexion,
and the grey area represents the range of the points of inflexion that are the closest on both ends
114
of the storm centre. Then the radius of the cyclone is discovered by taking the minimum distance
of the storm centre location and the inflexion point.
In essence, the Fourier series is the combination of cosine and sine functions. Here is a
simplistic example using the basics of the original equation given, designating the value for the
variable “a” as the number 5 and “b” as the number 6.
The first graph: (5cos( θ ) + 6sin( θ ))
The next graph: (5cos( θ ) + 6sin( θ )) + (10cos(2 θ ) + 12sin(2 θ ))
The last graph: (5cos( θ ) + 6sin( θ )) + (10cos(2 θ ) + 12sin(2 θ )) + (15cos(3 θ ) + 18sin(3 θ ))
115
Another way derivatives are used in meteorology is weather derivatives, which are
relatively a new idea. It is the idea that companies should be paid money if certain weather
anomalies occur resulting in a change in sales. For example, an ice cream company might buy a
weather derivative for if there is a cool summer. Cool summers usually result in a decline of ice
cream sales, and purchasing a weather derivative is a way to insure that they will be paid even if
there is a weather anomaly such as a cool summer. Unlike insurance, weather derivatives do not
require you to file a claim, and weather derivatives take into account comparatively good
weather in other areas. Any company whose sales could possibly be influenced by weather is a
candidate for weather derivatives. Farmers, energy companies, insurance companies, ski resorts,
and cruise lines are some of the types of companies that often purchase weather derivatives.
The reason for purchasing a weather derivative is its ability to prevent weather surprises
and anomalies from hurting the income of a business. The main problems in weather derivatives
lie in its pricing. To price a weather derivative, one must first determine what the risk is of a
weather anomaly, also known as weather noise. Without weather models and a deep
understanding of weather forecasting, this would be an impossible question to answer. Most
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forecasting is short term or “short horizon”, but weather derivatives rely on long term or “long
horizon” weather. The main variables that long term forecasting focuses on predicting is daily
temperature, heating degree days, and cooling degree days. Only ten cities in the countries data
is used to price derivatives including Philadelphia, Atlanta, Las Vegas, and Chicago. The
graphs below show the total daily average weather, one of the components in understanding
weather derivatives.
Using these models, one can find a formula to predict weather that proves the basic
knowledge that temperature change between seasons is gradual not sudden.
p
seasonalt = ∑ (σ c, p cos(2π p
p =1
d (t )
d (t )
) + σ s, p sin(2π p
)),
365
365
In this equation, the function “d(t)” is a step function for days 1 through 365 of the year.
Another equation can be written to account for the daily trends and not just seasonal
trends of the cities,
Q
σ t2 = ∑ ( yc, q cos(2π q
P =1
d (t )
d (t )
+ Ys, q sin(2π q
))
365
365
R
s
r =1
s =1
+ ∑ α r (σ t − r ε t − r )2 + ∑ β sσ t2− s
ε t ∼ iid (0,1),
Once these two equations are combined, the resulting equation is as follows:
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L
Tt = Trend t + Seasonalt + ∑ ρ t − l + σ t ε r ,
l =1
Where
in this equation, L=25, M=1, P= 3, q=3, and r=1
M
Trend t = ∑ β mt ,
m
m=0
Both seasonal and daily trends need to be accounted for in a weather derivative to be
accurate and priced correctly. Another thing to account into predicting long term weather, are
anomalies, also known as residuals. These residuals have been found to be less changing than
the daily weather, making it somewhat easier to predict anomalies and factor this into the
weather derivative price.
Overall it has become apparent the need for mathematical equations and specifically
calculus in weather forecasting. It facilitates the analysis of different types of weather and is vital
in insuring companies against unexpected changes in the weather. Using techniques such
calculus-based cyclone identification and weather derivatives make predications and pricing
more accurate. This accuracy and proficiency in weather forecasting makes the study of the
atmosphere more of a security for society. Focusing on the phenomenon within weather eases the
unexpected and alleviates unanticipated.
Sources:
•
“The use of a calculus-based cyclone identification method for generating storm
statistics” by R. E. Benestad and D. Chen
•
“Weather Forecasting for Weather Derivatives” by Sean D. Campbell and Francis X.
Dielbold
•
http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Fourier.html
•
http://www.britannica.com/eb/topic-215122/Fourier-series
118
Calculus in Waves
Emily Clary and Elizabeth Davidson
Waves are a mathematical phenomenon, which can be unraveled through the use
of Calculus. They can be found in many different forms: water waves, seismic waves,
sound waves, and light waves. Waves travel through a variety mediums, depending on
the type of wave, and the form patterns of waves within that substance. Matter waves
must travel through a medium and are produced by electrons and particles. Airwaves and
water waves are examples of mechanical waves that also require a medium. Certain
waves don’t require a medium to travel through. For example electromagnetic waves
don’t require a medium: they include light waves and radio waves. Once traveling, a
wave will go on forever if it is not obstructed. The study of waves falls into the category
of both physics and calculus. Waves can be studied in two different ways. It can be
studied in the one-dimensional form of the wave and the three dimensional form of a
wave. Calculus helps describe how a wave travels through its specific medium. Without
calculus mathematicians and physicists would be lost because this is the basis for the
study of waves; they utilize these mathematics in order to discover and to study the
patterns, which waves create in their travel. Mapping out these patterns is crucial in the
study of waves and understanding the math behind the phenomenon.
Waves are generally recognized as sine and cosine functions. The sine function
is expressed with the equation f ( x) = sin x . This graph is a continuous function, which
means that there are no holes or breaks in the function; there are also no vertical or
horizontal asymptotes, and the domain is all real numbers. Because of this, the graph of
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the function extends to infinity in both the positive and negative directions on the x-axis.
The graph is symmetrical in relationship to the origin and the function crosses the origin
at an inflection point as the function travels concave to convex. This function repeats in
periodic waves in alternating peaks and valleys. It is a bounded function because it has
an absolute minimum of -1 and an absolute maximum of 1. Unlike the cosine graph, this
one crosses the origin at (0,0). Below is a sine graph.
The cosine function is similar to the sine function in that they are both continuous
and have no asymptotes. The cosine function also is a bounded function because it has a
constricting range of [-1, 1]: it also has a domain of all real numbers. This graph
continually oscillates between the range points in a periodic pattern of valleys and peaks.
This function differs from the sine function because it crosses the origin at the local max
point of [0,1] and is symmetrical to the y-axis. Below is a picture of a cosine graph with
the equation f ( x) = cos x .
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Both the cosine and sine function are examples of sinusoids. Sinusoids are
written as f ( x) = a sin(bx + c) + d or f ( x) = a cos(bx + c) + d . The constant “a”
represents amplitude. On a graph the amplitude is half the height of the wave. The
period of the function is the length that the wave takes to complete one full cycle. The
frequency of a sinusoid is the number of complete cycles the wave creates in a set
interval. The constant “d” in the equation above represents the vertical translation and the
constant “c” represents the phase shift. Understanding the math used with sinusoids is
crucial in the measurement of waves.
To demonstrate the application of a sine function I decided to graph the function
π

y = −2sin  x −  + 1 and analyze it using calculus. First I found the x-intercepts of the
4

original function. I learned that the one x-intercept 1.3 and the period is 2 π because this
is the original period of the sine function. To find the next x-intercept you must add 2.1
to that intercept. Then you add 4.2 to the second x-intercept to find the third x-intercept.
This makes sense because 1.3 plus 4.2 is equal to 6.3 which is the approximate numerical
value of 2π (the period of the function). This pattern of a 2.1 space between the
intercepts and then a 4.2 space between the intercepts is a repeating pattern for the entire
π

function. After I found the derivative of y = −2sin  x −  + 1 and found the derivative
4

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π

to be y = −2 cos  x −  and then set it to zero to find its x-intercepts. I got the intercept
4

to be 2.356. Because the domain is all real numbers and the function oscillates there is
more than one x-intercept. The other x-intercepts are found by adding π to the
preceding x-intercept. These intercepts are the peak and valley points of the original sine
function. I then found the double derivative of the original sine function to be
π
π

y = 2sin  x −  and its x-intercept to be . Like the first derivative the other x4
4

intercepts can be found by adding π to the preceding term because the period is 2 π and
the function crosses the x-axis twice in one period, once when on the way up to a peak
and once on the way down to a valley. These intercept points are the possible inflection
points of the original sine function.
The cosine equation is comprised of amplitude, vertical shift, and period. Cosine
equations are useful in calculating numerous wave situations, especially the ebb and flow
of tides. For example in an equation, in which high tide occurs at 9:00 am causing the
water to be 3 meters deep, and low tide occurs at 4:00 pm resulting in 2 meter water
depth, the depth of the water is a sinusoidal function of time with a period of half a lunar
day, or 12 hours and 24 minutes. Constructing the model with a depth, D as a sinusoidal
function of time, t, the depth of the water varies from 3 meters to 2 meters, so the
amplitude is A =
3.0 − 2.0
3.0 + 2.0
= .5 . The vertical shift would be C =
= 2.5 . The
2
2
period is 12 hours and 24 minutes, which is converted into 12.4 hours, so b in the
sinusoidal function is B =
2π
π
=
. Because the maximum height is reached at 9:00
12.4 6.2
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am, 9:00 is converted to 9 hours after midnight, 0. The resulting cosine model from this
 π

information is D(t) = 0.5cos
(t − 9.0) + 2.5 .
 6.2

To discover further information about the graph, calculus rules can be applied
such as finding the derivative of the equation and examining the bump points or local
maximum and minimums. The derivative of the ebb and flow equation above is
 π

D(t )′ = −sin
(t − 9.0)(0.5) . By taking the derivative of the original equation, the
 6.2

cosine function is transformed into a sine equation that proves the perpetual relationship
between the sine and cosine sinusoidal functions. The derivative of the cosine function
reveals the x-coordinates of the inflections point, or where the graph changes from
concave to convex, on the original equation. In conjunction with the inflection points of
the graph, the derivative also reveals the peaks, valleys, and plateaus, or the local
maximum and minimum points of the graph. In application to the wave equation, the
cosine function depicts at what point in time the water waves change from concave up to
concave down and visa versa as the tide changes periodically.
Graphing of a sinusoid is useful in the measurement of sound waves because it
gives us a visual representation of what we hear. Sound waves differ from other waves
because it is a mechanical wave and cannot travel through a vacuum. Other waves
classified as electromagnetic waves can travel trough a vacuum, common example of this
type of wave is the light wave. Most people think that medium sound travels through is
air; however, the medium can be any type of substance, “the medium is simply a series of
interconnected and interacting particles” (b1). When an object such as a tuning fork
vibrates the air particles are disturbed, and so they collide. Because the collision is
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mechanical and forces the particles out of place the wave is labeled as a mechanical one.
As the sound wave travels one particle is moved from its equilibrium position and then
collides to displace another particle in a domino effect. The equilibrium position is the
original place of the particle in the medium. After the particle collides, the particle
returns to its equilibrium position. In this matter the wave travels through the medium.
One phenomenon that can be observed in relation to sound waves is the Doppler
Effect. The Doppler Effect is used to describe the different frequencies of the sound of an
object in motion. The Physics Classroom Tutorial (b2) gives an example of a police car
and its siren. As the car moves closer and closer to a person, the sound emitted from the
siren is a high pitch, but as the car moves away, the siren sound begins to get lower. Here
is a diagram to help explain this effect.
This relationship between distance and projection of sound happens because
longer sound waves have lower frequencies and smaller sound waves have higher
frequencies. No matter where a person stands is in relation to the siren the siren still
omits the same number of sound waves. So when the siren is moving away from a
person, the sound waves travel a longer distance and therefore make a lower pitched
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sound, and when the car is moving toward the person the sirens travel a shorter distance
and make a high pitched sound. Conceptually for sound waves, the Doppler Effect
describes the pitch of the sound in relationship to the distance it travels and the length of
the wave.
Mathematically there is an equation that describes the relationship between the
frequency, which is emitted and the frequency, which is observed. The following
 v
equation demonstrates this effect: f ' = 
 v ± vs

'
 , the f is the observed frequency and the

f is the emitted frequency. The v stands for the speed that the waves travel through the
air (or any other medium) and the vs is the velocity of the source generating the sound
waves. As the object emitting the sound moves closer to the observer the vs is subtracted
because this distance between the source and observer is getting smaller. When the
object emitting the sound is traveling away from the observer the vs is added because the
distance between the source and observer is getting larger. This equation helps
mathematicians and physicists to understand different relationships with waves.
In the same category as sound waves, water waves are mechanical waves that are
created by the effect wind and global patterns, outside sources, have on water. Wind has
three effects on waves: first, wind speed; second, the distance or amount of water that
wind has effected; and third, the duration of time the wind has blown. All of these
factors determine the sizes and shapes of waves. Following the model of the sinusoidal
functions, water waves are measured in terms of height, wavelength, and period. Water
waves proliferate between the surface of the water and the air, in which the restoring
force is gravity; thus, water waves are often referred to as surface gravity waves. As
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wind blows across the water’s surface, pressure and friction disturb the water’s
equilibrium culminating in waves. The effect of wind on water to create waves is
represented by this pictorial diagram “Waves and Wind”:
Waves in deep water create a circular path making water waves a balance
between the longitudinal, or the back and forth happening of the water, and the
transverse, or vertical motion of the waves. However, as the water becomes shallower,
the circular motion of the waves is compressed and the motion becomes more elliptical.
An object floating on the water’s surface in deep water only moves transversely as the
waves moves in a circular direction, but as the depth of the water becomes shallower, and
the waves become more elliptical, the particle on the wave’s surface is slowly pushed
forward longitudinally. This wave motion is known as Stokes Drift. This motion is
depicted in the following diagram “Wave Motion & Depth”:
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Waves are measured in harmonic or periodic motion. Utilizing amplitude, or the
maximum disturbance the wave causes to the medium; wavelength, or the distance
between two crests; period, or the time a wave takes to perform a complete oscillation;
frequency, how many units per time; and velocity, the rate of the wave’s motion.
The wave equation is a differential equation that relates the progession of the
harmonic motion over time. The water wave specific formula depends on the depth of
the water and the speed at which the wave progress; this equation is represented by
C=
gλ
2π d
tanh(
) , in which “c” is phase speed, “λ” is wavelength, “d” is water depth,
2π
λ
and “g” is acceleration due to the effect of gravity on the Earth. Water waves obey the
basic wave relationship C = f λ , where “c” represents celerity or wave speed. In deeper
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water, d ≥
1
λ ; therefore, this expression proves that waves at different wavelengths
2
travel at different speeds because depth is greater than one half wavelength.
There is a general wave equation for all types of wave; however, the wave
equation differs slightly depending on how the wave is transmitted and which medium it
is travelling through. The general one-dimensional and three-dimensional models can be
derived from the variables of wavelength, period, amplitude, and frequency. The onedimensional model is represented by the equation:
this equation becomes:
1 ∂ 2u ∂ 2 u
=
. In three-dimensions,
v 2 ∂t 2 ∂x 2
1 ∂ 2u
= ∇ 2u .
v 2 ∂t 2
Even though waves are found in numerous forms, they share the general form of
the sinusoidal graphs. Water and sound waves while differing in tangibility and
formation, are both mechanical waves that share the property of having to travel through
a medium by means of similar equations. In society, wave equations help predict patterns
in tidal waves and solve tsunami disasters. The Doppler Effect helps us better understand
how we detect different sound frequencies and improves sound technology. Without
calculus to aid physicists and mathematicians the wave phenomenon would go
unexplained.
128
Bibliography
Information Websites
http://www.thephysicsfront.org/static/unit.cfm?sb=8&pc_coursetype=5
http://www.rodenburg.org/theory/y800.html
http://www3.open.ac.uk/courses/bin/p12.dll?C01MS324
http://en.wikibooks.org/wiki/Physics_with_Calculus/Part_I/Harmonic_Motion,_Waves,_and_Soun
ds
http://en.wikipedia.org/wiki/Ocean_surface_wave
http://en.wikipedia.org/wiki/Waves
http://www.scienceaid.co.uk/physics/waves/properties.html
http://en.wikipedia.org/wiki/Water_waves
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/watwav2.html
http://www.seafriends.org.nz/oceano/waves.htm
http://www.mos.org/oceans/motion/wind.html
http://www.glenbrook.k12.il.us/gbssci/phys/Class/waves/wavestoc.html (b2)
http://www.physicsclassroom.com/Class/sound/soundtoc.html (b1)
http://en.wikipedia.org/wiki/Doppler_effect
Demana, Waits, Foley, and Kennedy. Precalculus: Graphical, Numerical, Algebraic. 7th
ed. Boston: Pearson, 2006. 384-387.
Picture Credits
http://www.nipissingu.ca/calculus/tutorials/trigonometrygifs/cosine_graph.gif
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2002.Fall/Dufera/Sine.Expl/image3.gif
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2002.Fall/Dufera/Sine.Expl/image3.gif
http://www.nipissingu.ca/calculus/tutorials/trigonometrygifs/cosine_graph.gif
http://www.glenbrook.k12.il.us/gbssci/phys/Class/sound/u11l3b.html
129
http://www.seafriends.org.nz/oceano/waves.htm.
http://www.seafriends.org.nz/oceano/waves.htm.
http://www.allaboutspace.com/subjects/tsunami/
130
Digital Device Usage Survey
In Conjunction with the Flat Classroom Project
Ariana Mooradian, Michael Rodriquez, and Walker Harbison
Introduction
Although it doesn’t seem likely, most people nowadays would be completely lost
without technology. Whether it be cell phones, computers, or video games, people all use
or have access to all types of technology. The purpose of this survey was to find out how
much people actually use their various electronic devices. We specifically asked about
certain habits that people have formed and the amount of time that an individual has
spent using a device. The devices we asked about were cell phones, video games,
computers, television, and MP3 players. The subjects, 66 in total (n), were polled as part
of the Flat Classroom Project. The project was made in order to connect students across
the world through the internet and other applications. Thus the survey will be able to give
a rather accurate representation of technological habits of the teenagers across the world.
Cell Phone Use
Questions Eight through Thirteen
As expected, the results of the questions relating to cell phone use indicated most people
regardless of gender own and operate cell phones. The only aspect we found that gender
affected was the camera function. More males use their cell phone camera as their
primary picture-taking utility, while an overwhelming majority of females do not.
Question eight asked the total number of calls made or received on a daily basis.
The data was then split into four ranges: the first between zero and five, the second
between six and ten, the third between eleven and twenty, the fourth more than twenty. A
preliminary evaluation of the graphs indicates that the majority of those polled engage in
ten or less phone calls (29 for G, 27 for B). When observing the data on a gender-specific
basis, responses from females seem to be more than that for males in the second range;
therefore, overall females use their cell phones more. However, the responses from males
show that their cell phone use is concentrated in the first range, which constitutes 61% of
their total responses, whereas the females’ first range is only 37%. In both cases it is clear
that very few people use their cell phone more than twenty times a day (5% of all
responses). The p value in this set is 0.15, which does not meet the 0.05 test, so we
cannot reject the null hypothesis, meaning there is a chance of a demonstrated correlation
131
between gender and cell phone use. The critical value calls for at least a 7 statistic, but
this data only has a 5.
Question nine asked for the total number of test messages sent or received on a
daily basis. It uses the same four ranges as listed in question eight. Across the both
genders, most people use five or less text messages (48.5%). Surprisingly, the same
number of people fall in the second and third ranges, using anywhere from 6 to twenty
text messages (39% range). The data for male responses in completely skewed towards
the first range, which is 61% of the responses. The next three ranges are each around 4%
of the total. The females’ data is much more even. While the first range takes a lead with
37%, the subsequent columns are closer than with the males. These columns each are
around 15%. The p value in this set is 0.26 which does not meet the 0.05 test, so we
cannot reject the null hypothesis, meaning there is not a definite demonstrated correlation
between gender and text message use. There is, however, more of a relationship between
gender and text messaging than between gender and phone calls.
Question ten asked about the total number of cell phone photos that were either
sent or received in a day. There were four data sets from which the group could choose
from. The first between 0-5, second between 6-10, third between 11-20, and fourth was
all above 20. Looking at the general population, 95% said they sent 0-5 picture messages
a day. There was also only one person in each data sets 2,3, and 4. In the gender specific
graphs only one male was out of the first data set and he was placed in the 4th data set. In
the female graph there were two outside data set one. Both were either in data set 2 or 3.
The p value is .405 therefore this does not meet the .05 test, so we cannot reject the null
hypothesis meaning that there is a chance that there will be no difference between the
expected results.
Question eleven asked if a cell phone camera was the camera that was most often
used to take photos. It uses different ranges then the previous questions. If the person
answered yes then they would be in column one. If they answered no then they would
fall under column 2. Across both genders, over twice as many people answered no to the
question (about 69%). 48% of males said that their cell phone was their primary camera
where the other 52% said it was not: pretty evenly split. 86% of females, however, said
that they did not use a cell phone for their primary camera. The p value for this set of
data was .0026. This means that we can reject the null hypothesis at the .05 level. This
shows that there is a distinct difference between males and females with respect to a
primary camera.
132
Question 12 asked whether or not at least one person in their family talked and
drove on the phone. When looking at the general population, it was noted that most know
at least one person who talks on the phone. Also, the gender specific graphs yielded the
same results. The p value is .25 and this does not meet the .05 test. So we cannot reject
the null hypothesis.
Question 13 asked if one person in his or her family sometimes drive while
sending or receiving text messages. The ranges for this question are as follows: a one on
the survey means that they do receive or send text messages while driving and a two
means that no one in his or her family does that. Across both genders, more people said
that they do not send or receive text messages while driving (59%). There were similar
results when the data was broken down between males and females. All three graphs
were nearly identical. Percentage wise, almost the same number of males participate in
this activity (42%) as females (40%). Because the p value for this data was .873, we fail
to reject the null hypothesis at the .05 level. This means that there is no difference
between males and females for this study.
Video Game Play
Questions Fourteen through Eighteen
With little surprise, many more mores use video games than females. If given a choice,
most people would choose not to play any video game, though we can assume this
statistic is skewed from the females’ data. We confirmed our guess that males would play
more games online, though were proven wrong in the assumption that video games make
you violent from the inconclusive data in question eighteen.
133
Question fourteen marked the beginning of the set of questions pertaining to video
game play. It asked the subjects whether or not they had access to a video game console
at home. The graph of the general population was generally at a ratio of 2:1 in favor of
owning a console. When looking at the female graphs there was a few more people that
did not have a console while the males tended to have more access to a consol. The p
value was .48 and therefore it rejects the null hypothesis at the .05 level.
Question fifteen asked what genre of video game the subjects preferred to play.
The data sets that they could choose from were violent fighting or conflict, sports, racing,
adventure, role playing, and none. The genre which was chosen the most was actually to
play no video games at all, with violent fighting or conflict as the next highest and the
rest being relatively equal. However, when looking at the gender-specific graphs there is
a major difference. The males’ graph has the violent fighting and conflict choice as the
largest value and no video games as the second while the females mostly chose no video
games and all other columns were relatively equal. The p value for this set is .08 which
fails to reject the null hypothesis at the .05 level, although there seems to be a trend
leaning toward more use by males.
The next question asked about the amount of time the subjects played during an
average week. The data sets they were able to choose from were 0-1 hours, 2-4 hours, 510 hours, 11-20 hours, and over 20 hours. The non-gender specific graphs indicate that
most of them played 0-1 hours. But looking at the gender-specific graphs, the males
tended to be more spread out in a linear fashion but the females mostly said 0-1 hours and
then tapered off greatly. The p value for this question is.04 and therefore it did in fact
reject the null hypothesis and there is a correlation between gender and time spent
playing games.
Question seventeen asked the subjects what percent of their games they play
against online opponents. The data sets offered were 0% to 10%, 11% to 30%, 31%,
50%, and over 50%. The general population said they played anywhere between 0 to 10
percent against online opponents. The males had a higher percentage of online play than
the female subjects. The p value is .01 and therefore it rejects the null hypothesis.
134
This next question asked the subjects whether or not they know an individual who
has become more violent as a result of their video game play. This is an important aspect
of the survey because this claim is often used to criticize video gaming. The majority of
the subjects said no but about 15% percent said that they did know at least one person
who became more violent because of video games. The females had a lower percentage
of people who said that they knew someone who became more violent because of video
games. The p value is .37 and therefore rejects the null hypothesis.
Internet Use
Questions Nineteen through Twenty-One
As expected, this set of data confirmed that an overwhelming majority of the sample set
has access to the internet regularly. This makes sense, because so often we are expected
to receive or complete assignments and notifications online, with e-mail dominating the
lines of communication. The females’ data indicated that they prefer to use social
networks to communicate online, while the males responded with a majority for instant
messaging, which could potentially be due to the majority of online gamers. We had
suspected that females would spend more time communicating online, though a
breakdown of data indicated otherwise.
Question nineteen started the collection of questions dealing with internet use. As
expected, the majority of all responders use the internet regularly. This particular
question asked the survey pool if they had access to the internet at home. This was a yes
or no question, so only two bars appear on the graph. Overall, the Yes replies dominated
the statistic, constituting 98.5% of the data. All females that responded said they had
internet access, so the discrepancy came from the male data, which says approximately
3.23% do not. The p value for this experiment is 0.284, which does not meet the 0.05
level, so we fail to reject the null hypothesis. This means that statistically there may or
may not be a relationship between gender and internet access at home. The statistic of
1.146 also fails to meet the 3.84 critical value. Therefore we cannot assume a
relationship. This makes sense because theoretically we wouldn’t relate someone’s
gender to their access to the internet.
Question twenty asked the preferred method of communication when using a
computer. The responses were split into four utilities: instant messaging, social networks,
e-mail, and other. The data for all responses shows that 98% of internet uses chose the
first three, which indicates very little use of alternative modes of communication. Instant
messaging and e-mail each are about 30% while social networks are 40%, which shows a
135
fair amount of distribution. There are notable differences between the genders. The male
group had a majority of use with instant messaging (45%), while the female group had a
majority using social networks, about 55%, i.e. Facebook. The p value is 0.0224, which
meets the 0.05 level. At this point we can reject the null hypothesis and assume a
correlation between gender and preferred method of online communication. The statistic
also exceeds the critical value (9.58 to 7.81), which also shows a relationship.
Question twenty one asked how many hours were spent communicating online.
The responses were split into 5 ranges, between 0 and 1, between 1.5 and 2, between 2.5
and 3, between 3.5 and 4, and more than 4. Overall, the highest concentration was in the
first range (less than or equal to 1 hour) with 37%. The male only response graph showed
an ever higher percent with 45%. Very few responses in either were in the fourth range,
between 3.5 and 4, but surely this is coincidence and due to people rounding. In the
female group the data is a little bit more evenly distributed. The p value is 0.2759, not
meeting the 0.05 level, so we fail to reject the null hypothesis. The statistic is also less
than the critical value, so we cannot draw any definite conclusions as to a relationship
between gender and amount of time spent communication online.
TV Use
Questions Twenty-Two through Twenty-Four
The data regarding TV use had similar trends between the genders as the online
communications. Overwhelmingly, most responders have access to TV at home. When
observing a break down of specific TV genres, no female responders chose sports as their
favorite, though it constituted a significant portion of the male data. Fortunately, people
do not seem to watch as much TV as many would suspect of teenagers, with a majority in
the 0-1 hour range. Though our numbers do not indicate strict correlations between
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gender, logically we can assume a distinction in type of show, even if there is none in
time spent watching it.
Question twenty two began the series of questions asking about TV use. The first
question asked survey takers if they had a TV with cable or satellite channels available in
their home. The ranges for this question are as follows: a one on the survey means that
they do have cable or satellite TV at home and a two means that he or she does not have
access to cable or satellite at home. Across both genders, 86% of people said that they
did have access to one of these services. 90% of males and 83% of females said that they
had either satellite or cable channels at home. Because the p value for this data was .377,
we fail to reject the null hypothesis at the .05 level. This means that there is no indicated
correlation between males and females for this study.
Question twenty three asked what was the favorite type of TV show was. The
ranges for this data is as follows: 1: Movie 2: Sports 3: Soap Opera 4: Drama/Comedy
Series 5: Game Show 6: Reality Show 7: Other. Across both genders, the most popular
type of TV show was Drama/Comedy Series (44%). The two least favorite TV shows
were Soap Operas and Game Shows, both showing just over 3%. There were no males
that chose Soap Operas as there favorite program type but they were the third favorite
program among females. While 16% of males chose sports channels to be their favorite
program type, there were zero females that fell into this category. The p value for this
data was .105 which means that we would fail to reject the null hypothesis at the .05
level. This shows that there is little correlation between the shows that males and females
watch; however, note the discrepancy in the sport genre responses below (bar 2).
Question twenty four asked how many hours per day the subjects spent watching
TV. The ranges for this data was as follows: 1: Between 0 and 1 hours 2: Between 1.5
and 2 hours 3: Between 2.5 and 3 hours 4: Between 3.5 and 4 hours 5: More than 4
hours. Across both genders, the vast majority (54.54%) watch between 0 and 1 hour of
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TV a day. The graph indicates decreasing values. There were no males that watched
more than 4 hours of TV and there were 5.7% of females that watched between 3.5 and 4
hours. The p value for this data .487 which means that we would fail to reject the null
hypothesis at the .05 level. This shows that there is little correlation between males and
females for this data.
Mp3 Use
Questions Twenty Five and Twenty Six
Questions twenty five and twenty six relate to Mp3 use. No surprises here: the vast
majority own one, and gender has no effect on time spent. Like TV use, most people
spend less than an hour using their Mp3 device daily.
The next question began the last individual category: music players. Question
twenty five asked whether or not the subjects used an Mp3 player to listen to music (Yes
or No). On all three graphs, there was a dominant response of yes. With a p value of
0.3477, we cannot reject the null hypothesis, so cannot assume a correlation between
gender and owning an Mp3.
Question twenty six asked how many hours per day were spent using an Mp3.
The ranges were the same for amount of time spent communication online: 0 to 1, 1.5 to
2, 2.5 to 3, 3.5 to 4, and over 4. Again, there was a dominant amount of responses falling
in the first category across the board. The p value was 0.56, so there is no relationship
between gender and amount of time spent listening to music on an Mp3. (As graph
shows, both genders responded in a vast majority for the first range).
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Multi-Tasking
Questions Twenty Seven through Twenty Nine
These questions were unique in that they related to multi-tasking and therefore required
double coding. Though in all cases we fail to reject the null hypothesis, there seemed
somewhat of a relationship between genders and devices used.
Questions twenty seven through twenty nine asked about multi tasking. They used
double coding to split responses into multiple categories. The first question asked about
multi-tasking: how many of these devices are used simultaneously for five minutes. The
major bar in all data responses tell us that most use a combination of cell phone,
computer, and TV, which is also the highest bar for the females Most guys combine cell
phone and computer. The p value of this data is 0.17, so we fail to reject the null
hypothesis, meaning we cannot relate gender to devices with which responders multitask. (Graph below shows distribution of different combinations of devices used to multitask. Data was split into sixteen categories seen below the graphs of male and female
data).
{1 = 305, 2 = 340, 3 = 345, 4 = 2305, 5 = 2340, 6 = 10000, 7 = 10005, 8 = 10040, 9 =
10300, 10 = 10305, 11 = 10340, 12 = 10345, 13 = 12000, 14 = 12305, 15 = 12340, 16 =
12345}
Question twenty eight asked the same general question as twenty seven, except
only relating to multi-tasking during homework. The biggest responses were cell phone
and computer, and computer and TV, each about 7% of responses. For males, the
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majority was clearly the latter combination, with, again, 7%. The females’ data was tied
for the most in three categories, each with 5%. The first was Mp3 and computer, the
second computer, TV, and Mp3, and the third cell phone and computer. While this might
suggest a connection between gender and multi-tasking during homework, the p value is
0.41, which does not meet the 0.05 level, so we fail to reject the null hypothesis. The
statistic of 13.44 also fails to meet the critical value of 22.36, so we cannot decide a
definite relationship in this case.
Question twenty nine asked the subjects which three personal entertainment or
communication devices they would choose. The options where cell phones, video game
consoles, computers (non-video games), TV (non-video game), MP3 player. The most
chosen option was the combination of cell phone, computer games, and TV. The Males
second choice was the combination of cell phone, video game consoles and a computer.
The females tended to choose the non video game options. The p value is .10 which fails
to reject the null hypothesis.
Conclusion
In a world of constant technological development, it seems that only teenagers are
able to keep up with all the new uses for new devices. The flat classroom project was a
way of connecting teens across the world to discuss and research the emerging trends in
technology. For this project we completed a survey that looked at different electronic
media and how often they were used, dividing the data further by the gender of those
polled. The survey inquired about cell phones, videogames, computers, and television
use. While our data did not produce sufficient values to draw general conclusions about
males and technology or females and technology, it is possible to draw conclusions in a
topic based manner. For example, females tend to send and receive more calls per day
then males, but males tend to play more videogames than females. Technology growth is
an ever evolving matter; it is necessary to be aware of and utilize its benefits for the
advancement of society.
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