-
1
Introduction
What is AP Calculus BC?
This is a term used by the College Board, which has an
approximate college-level analog; it refers to the material usually
covered towards the end of a college Calculus I course (roughly the
B aspect) and most of the material covered in Calculus II (roughly
the C aspect). A BC course does not encompass a great deal of new
material. It extends the subjects of differentiation and
integration to new functions; such as parametric, polar, and vector
functions; and introduces the concept of infinite series and their
applications to polynomial approximations. Most the material on the
BC exam is actually derived from the AB aspect of calculus. The
format of the AP Calculus BC exam is identical to that of the AP
Calculus AB exam, save for the fact that the former exam tests more
material. There are three parts: a multiple choice section in which
calculator use is not permitted, a multiple choice section in which
the calculator is permitted and a free response section in which
one may use the calculator for the first half but not the
second.
About this book
I have written this book for two reasons. Firstly, while books
on the market dedicated either specifically to AP Calculus AB or
AB/BC are legion, one could undoubtedly count the number of books
concentrating upon AP Calculus BC on one hand. From one point of
view, this is understandable; there are simply more students who
take the AB test than the BC test. However, the dearth of
concentration upon BC material makes it quite difficult, in my
opinion, to obtain information that is equal in profundity to the
AB material that is available. Secondly, while AP Calculus AB
introduces many interesting applications of differentiation and
integration to the physical, life, and social sciences, AP Calculus
BC is almost completely devoid of interesting applications. Even
spinning a curve around an axis and evaluating the volume of the
solid generated is more interesting than blindly approximating
values of functions using Taylor polynomials! While I do include
discussions on the pure-mathematical aspects of the BC material, I
also introduce interesting, perhaps somewhat biased, applications
of the material that is not tested on the AP exam, as the title of
this book shows. While many of the applications that I discuss are
not tested nationally, I believe that they foster a higher degree
of appreciation for the BC material. I apologize to the student
economists and social scientists using this book; those areas are
not my specialty. Nevertheless, I feel that my combination of pure
mathematics and the physical and life sciences in this book will
nurture a greater interest in the material and, perhaps, even a
sense of freedom from the burden of the standardized test.
I would also like to comment on the last chapter of this book,
entitled Vector Calculus and Curves in Space. The AP Calculus BC
exam only tests a very small portion of the material that I
include, mainly differentiating and integrating vector-valued
functions. My discussion goes well beyond this for a good reason;
vector calculus is an excellent bridge into Calculus III, which is
multivariable calculus (i.e. partial derivatives,
-
2
multiple integrals, etc.). I hope that concluding my book in
this manner will encourage the readers research on multivariable
calculus, even if he or she does not plan to take the course in
college.
Before reading the book
I have generally listed the material with which the student
should be familiar
before he or she begins reading the book: I.) Limits and
Continuity: This includes understanding what a limit is and how to
evaluate one algebraically and graphically. One should also
understand how tell whether or not a function is continuous. II.)
Derivatives: One should know the limit definition of the derivative
(difference quotient) and the various rules for finding derivatives
(i.e. the power rule, the product rule, the quotient rule, and the
chain rule). One should also know how to take the derivative of all
trigonometric functions, exponential functions, and logarithmic
functions. One should also understand implicit differentiation.
III.) Applications of Derivatives: It is expected that one knows
how to find the slope of the tangent line and the normal line to
the curve. One must be able to compare the graph of a function to
the graph of its first and second derivatives and vice versa. One
must be able to find local minima and maxima as well as points of
inflection by using derivatives. One should understand the Mean
Value Theorem and Rolles Theorem. One should be able to do
optimization, motion, and related rates problems. IV.) Integrals:
One must know the various rules for antidifferentiation (i.e. the
power rule, u-substitution, the natural logarithm, inverse
trigonometric functions). One should know how to approximate the
area under a curve by using rectangles and the Trapezoidal Rule and
get the exact area under a curve and between curves by using the
2nd Fundamental Theorem of Calculus. One should also be familiar
with the 1st Fundamental Theorem of Calculus and accumulation
functions. VI.) Applications of Integrals: It is expected that one
knows how to apply the integral to problems in motion. One must
know how to find the volume of a solid of revolution by using the
Disc Method, the Washer Method, the Shell Method, and various
cross-sections. One should know how to evaluate the average value
of a function. VII.) Differential Equations: One must be able to
solve differential equations via separation of variables. One
should know how differential equations apply to exponential change
and Newtons Law of Cooling. One should also be able to understand
slope fields and linearization.
-
3
A Note on Technology: Nowadays in advanced mathematics courses,
the graphing calculator is used extensively. While this machine is
incredibly useful, it is important not to let it overcome the power
of the mind. Before using the calculator to solve a problem
graphically or analytically, the student should understand the
theory behind the steps that the calculator took to solve the
problem. It is also important to remember that half of the AP
Calculus BC test prohibits the use of a calculator. Therefore,
wherever possible, the student should do calculus in the
19th-century fashion (even without a slide rule!) by hand. As of
the completion of the manuscript of this book, the TI-83 graphing
calculators have been discontinued. Thus, when discussing solutions
via calculator, I will refer to its contemporary counterpart, the
TI-84. I would also like to mention that throughout the book,
notably in the final chapter (Vector Calculus), I have generated
some of the graphs using the computer program Mathematica. This
program is one of the most powerful computational tool in the
world, and I encourage readers to learn more about its nearly
limitless capabilities.
-
4
Table of Contents Prelude: At the Level of the Infinitesimal.5-7
Chapter 1: LHpitals Rule and Advanced Techniques of
Integration.8-24 Chapter 2: Differential Equations. 25-39 Chapter
3: Infinite Sequences and Series..40-55 Chapter 4: Power Series and
Polynomial Approximations...56-74 Chapter 5: New Coordinate
Systems: Parametric and Polar.75-102 Chapter 6: Vectors and Vector
Calculus......103-131 Practice Test with Answers..132-188 Appendix
A (Essential Pre-Calculus Information)..189-190 Appendix B (Brief
Table of Integrals)..191 References.192 Index193-195
-
5
Prelude: At the Level of the Infinitesimal
In modern vernacular, the word infinity is thoroughly abused,
most likely the result of a misconception of its mathematical
significance. When one wishes to express an extreme, infinitely is
often inserted for dramatic effect: It was infinitely informative,
This performance was infinitely better than the last one, etc. What
is the true meaning of infinity? An attempt to imagine a physical
infinity will yield no greater victory than an attempt to conceive
of more than three dimensions. Mathematics, however, has the power
not only to make sense of these abstractions, but also to put them
to good use. Differential and integral calculus is based upon the
concept of infinity. When one evaluates the derivative of a
function, he or she is essentially finding the slope of a secant
line on a curve, which intersects two points on that curve, when
the difference between those two points is infinitely small. In
effect, by moving from a discrete difference between points to an
infinitely small difference, the secant line has become a tangent
line! The concept of the integral also exploits the power of
infinity. When one uses the integral to find the area under a
curve, for instance, he or she is actually taking the sum of an
infinite number of infinitely small chunks of an area to yield the
entire area underneath the curve.
Why on Earth is any of this useful? An appreciation for the
practicality of
differential and integral calculus requires a little
philosophical experimentation. A story often discussed in the
introduction to calculus courses is that of Zenos paradoxes. Zeno
of Elea was an ancient Greek philosopher who introduced a
philosophical problem that was largely left unsolved before the
advent of calculus. According to Zenos paradoxes, motion is
illusory and impossible. While there are some variations on the
specifics of his argument, it addresses three principle dilemmas.
Firstly, if a particularly fast runner condescends to give a slower
runner a head start, it will always be the case that the slower
runner will win the race. Why? When the fast runner arrives at the
spot at which the slower runner began, the latter has already moved
a certain, albeit short, distance. It will then take the fast
runner a certain period of time to reach that short distance, while
the slower runner has advanced further, ad nauseam. Even more
striking is Zenos proposal that motion is futile. If a body is to
move from point A to point B, which are separated by a length l,
that body must first move a distance of 1/2l. Before that, the body
must move a distance of 1/4l. Before that, 1/16l. Before that
1/32l, ad nauseam again. In fact, not only is it the case that
motion is futile, it is also impossible. According to Zeno, at each
discrete instance of time, a body is momentarily at rest. If this
is the case, then there is no point during an objects journey that
it could be considered in motion. Of course, motion is possible and
a faster runner can overtake a slower runner despite a head
start
-
6
for the latter. Therefore, there must be a solution to Zenos
paradoxes. This is where calculus makes its entrance. While the
mathematical philosophy of moving to an infinitesimal level was
manifest in antiquity, calculus did not emerge as a serious
discipline and practical tool until the early 18th century, when
scientists were rigorously attempting to find solutions to
difficult problems in physics. Two prominent figures during this
period, Isaac Newton and Gottfried Leibniz, are usually credited
with the invention of differential and integral calculus. While
there are some philosophical naysayers who believe otherwise,
calculus very concisely solves Zenos paradoxes. Through calculus,
it can be shown that even an infinite number of distances can have
a finite sum. Furthermore, it is not the case that an infinite span
of time would be required to travel an infinite number of
infinitely small distances. Infinitesimal distance (dx) and
infinitesimal time (dt) have finite significance in differentiation
and integration. For
instance, the expression vdtdx = , which means that the
derivative of the position function
is equal to the velocity function assumes a very helpful form
when integrated:
= vdtdx , which means that an infinite sum of infinitely small
distances yields the infinite sum of products of instantaneous
velocity and infinitely small time spans. This relates infinitely
small changes in position with infinitely small changes in time,
so, despite the notion of the infinitesimal, motion does exist!
It is often necessary in the sciences and business to move to
the infinitesimal, or differential, level. Why? Without doing so,
one is restricted to differences between two points and ignores
what is in between! In business, for example, it is necessary to
optimize the volume of production to maximize profit. This is
achieved by finding the number of products at which the difference
between revenue and cost of manufacture (profit) is a maximum. Two
isolated points for revenue and cost is not sufficient; an infinite
number of such points is needed! One must evaluate the derivative
of the profit function and find the point at which it undergoes a
transition from positive to negative values, a method covered in AP
Calculus AB. In the pure and applied sciences, moving to the
differential level is indispensable in modeling phenomena. Similar
to the business example, calculus allows one to consider all of the
points that model an event, not just two isolated points.
Furthermore, analysis of a differential aspect of a discrete body
easily allows for the extension to the whole body; that is, through
the methods of calculus, the mathematical equations that describe
an infinitely small sliver of an object also describe that object
as a whole. This somewhat bottom-up approach is very often more
practical that analyzing the object as a whole. In physics, for
instance, objects have a property known as the center of mass, the
point at which the mass of the object can be considered
concentrated. In effect, one can consider infinitely small chunks
of an object, each having a mass mi. When the average of these
masses is found and weighted (statistically) by each masss position
with respect to a reference point, the center of mass
is found. Mathematically, center of mass = M
rdm
m
rm
i
ni
iii
n
==
=
1lim , where ri is the position
of the ith infinitesimal mass with respect to a reference point
and M is the total mass of the object.
-
7
In this book, the theme of moving to the differential level will
continue to be emphasized as a useful method for understanding both
the pure-mathematical aspect of AP Calculus BC topics and their
applications to the sciences.
-
8
Chapter 1: LHpitals Rule, Advanced Techniques of Integration,
and Improper
Integrals Since AP Calculus AB usually begins with an
introduction to limits and their
algebra, this book for AP Calculus BC will begin with a further
discussion on limits. Oftentimes, one cannot evaluate a limit via
the methods prescribed in AP Calculus AB. For instance, these
methods do not suffice when they result in an indeterminate form, a
mathematical construction that has no immediate arithmetic meaning.
There are seven
instances of these forms: ,,0,1,0,,00 00
and . Each will be discussed separately.
LHpitals Rule
LHpitals Rule was developed by the Frenchman Guillaume de
LHpital
(pronounced loh-pee-tahl) during the 17th century. He published
this rule in his L'Analyse des Infiniment Petits pour
l'Intelligence des Lignes Courbes (1696), which is regarded as the
worlds first textbook on differential calculus! LHpitals Rule can
be expressed as follows:
If Axfax
= )(lim and Bxgax = )(lim , and A and B are either both equal to
infinity or both equal to zero, the following holds:
)()(lim
)('')(''lim
)(')('lim
)()(lim
xgxf
xgxf
xgxf
xgxf
n
n
axaxaxax === The above relation states that if the ratio of two
functions has a limit as x
approaches a of indeterminate form, the derivatives of both the
numerator and the denominator can be found until the limit has a
determinate form. While it will not be presented here, a proof of
LHpitals Rule can be found from Cauchys Mean Value Theorem.
There are certain limitations of LHpitals Rule that one must
take into account.
Firstly, the rule can only be used for limits in the
indeterminate forms of 00 and
; one must not attempt to use LHpitals Rule to evaluate limits
that are not in indeterminate form. Thus, it is imperative that
limits be first evaluated by conventional methods and
then, only if the need arises, be subject to the rule. Note also
that the forms 01 ,
1 , and
0 are not indeterminate; they have meaningful arithmetic
significance. The first fraction yields infinity, the second
fraction yields zero, and the third fraction yields infinity.
-
9
The Indeterminate Form 0/0: If one attempts to take the limit of
a quotient and 00 results,
the derivative of the numerator and denominator may be taken
until a determinate form is achieved.
Ex.) Evaluate x
xx
tanlim0 .
Solution: x
xx
tanlim0 = 0
0)0(
)0tan( = Use LHpitals Rule
x
xx
tanlim0 = 11
)0(seclim1
seclim2
0
2
0== xx
x .
The Indeterminate Form /: If the form
results upon attempting to determine the limit of a quotient, a
similar procedure can be used.
Ex.) Evaluate 3lnlimx
xx .
Solution: 3lnlimx
xx =
=
3)()ln( Use LHpitals Rule.
3lnlimx
xx = 0)(3
131lim
3
1
lim 332 === xxx
xx.
In the introductory discussion of LHpitals Rule, it was stated
that the rule can
only be used to evaluate limits of the forms 00 and
. What of the other five indeterminate forms? In order to apply
LHpitals Rule to these, a little algebraic manipulation is
required. The Indeterminate Form 0: This form can be algebraically
transformed into either of the forms already discussed by dividing
one by the reciprocal of the other:
==
01
0 and .00
100 =
=
Note that evaluating the product of two numbers and dividing one
number by the reciprocal of the other number are different
operations that mean the same thing
arithmetically! For instance, 6
312
21332 === . Thus, while the above technique does
-
10
change the representation of the indeterminate form 0, it does
not change the arithmetic result. Ex.) Evaluate x
xex
2lim .
Solution: xx
ex 2lim = = 0)( )(2 e
Take the reciprocal: xx
ex 2lim =
===
)(222 )(lim
1lim
eex
e
xxx
x
xUse LHpitals
Rule: === )(
2 )(22limlimee
xex
xxxxUse LHpitals Rule again:
0222lim2lim )( ==== eeex
xxxx.
The Indeterminate Forms 1, 00, and 0: Since these indeterminate
forms involve exponents, a logarithm must somehow be applied for a
reversal. If one takes the natural logarithm of these forms, one
can then deal with the problem in the context of the techniques
already discussed. Recall the property of logarithms that abab lnln
= . Once the value of the limit is determined after taking the
natural logarithm and applying the previous techniques, it is
important to remember to exponentiate (take the e of) of the answer
to know the original functions behavior.
Ex.) Evaluate 11
1lim +
xx
x .
(Note that the + subscript denotes a right-hand limit, which,
along with left-hand limits, is discussed in AP Calculus AB. This
is a necessary specification, since one cannot approach a value of
1 from the left because the function does not exist to the
left!)
11
1lim +
xx
x = = 1)1( 1)1(1
Take the natural logarithm of the function:
11
1lim +
xx
x = =
=
+ 0)1ln(1)1(
1ln1
1lim1
xxx
Take the reciprocal:
=
=
=
+++ )1ln(
11)1(
1
lim
ln1
11
limln1
1lim111 xxx
x
xxx
Use LHpitals Rule:
==
=
+ 00
1)1ln()11(
ln)1(
ln1
11
lim1
xx
x
x
xx
Use LHpitals Rule again:
-
11
===
=
=
=
+++ 0
1
101
)1()1ln(
1ln
1limln1
1limln
)1(lim
22
1
2
11
xx
x
xx
xxx
xxxx
. Note
that is not the final answer. Since the problem was manipulated
through the use of the natural logarithm, this must be undone
through exponentiation. Thus, the final answer is .0=e
The Indeterminate Form : Transforming this form into one that is
workable requires an algebraic manipulation of terms to yield two
terms such that either one or both are fractions. Once this is
accomplished, one can multiply each term by a common denominator to
trivialize the subtraction. Since this explanation is undoubtedly
not as clear as the others, it is important to analyze the
following example closely..
Ex.) Evaluate
xxx1
sin1lim
0
In order to make the operation of subtraction obsolete, the
first fraction is multiplied by x and the second fraction is
multiplied by sin x:
xxx1
sin1lim
0= =
)0(1
)0sin(1 Use the technique discussed:
xxx1
sin1lim
0 = == 0
0)0sin()0()0sin()0(
sinsinlim
0 xxxx
xUse LHpitals Rule:
=+=+
= 00
)0sin()0cos()0()0cos(1lim
sincoscos1lim
sinsinlim
000 xxx xxxx
xxxx Use LHpitals
Rule again:
.020
)0cos(2)0sin()0()0sin(
coscossinsinlim
sincoscos1lim
00==+=++=+
xxxx
xxxx
xxx
Advanced Techniques of Integration
AP Calculus AB introduced the concept of the integral and
several general techniques of integration, such as u-substitution.
While these skills are quite valuable, they are often of no use
when confronted with more difficult problems, such as those that
arise in the real world. In fact, the techniques of integration
introduced in this book do not even scratch the surface of the
myriad of techniques that exist. Indeed, integration is much more
difficult, and requires more rigorous mathematics, than
differentiation partly due to the fact that one is working
backwards. In this section two new techniques of integration will
be covered: integration by parts and integration by partial
fractions.
Integration by Parts: Integration by parts was developed by the
18th-century English mathematician Brook Taylor, for whom Taylor
Series are named, which are studied in Chapter Four of this book.
The technique can be described as the integral analog of the
product rule used in differentiation. In fact, the formula for
integration by parts is derived from the product rule for
differentiation:
-
12
+=+=+=+= vduudvuvvduudvuvdvduudvuvddxduvdxdvuuvdxd )()(
The last equation can be rearranged into the form: = vduuvudv .
This equation allows one to decompose an integral into two
manageable parts. The most renown example of the necessity for
integration by parts lies in the integration of the natural
logarithm.
Ex.) Evaluate xdxln . The form of this integral is deceptively
simple. While it may seem that the integration of this function is
relatively straightforward, the techniques of AP Calculus AB cannot
be used here. Instead one must use the technique of integration by
parts. One
chooses u to represent ln x and dv to represent dx. In this way,
x
dxdu = (because dxdu , in
this case, is the derivative of ln x) and v = x. Now the formula
for integration by parts
can be applied: +=== Cxxdxxxxdxxxxxdx 1ln))((ln))((lnln . In
this last problem, it was relatively simple to determine what to
choose as u and
what to choose as v. This determination, however, is not always
so clear. One should not blindly choose these values, for making an
incorrect choice could cost a great deal of time and energy. While
it may be intuitive to the reader which functions to choose, it is
easier to follow a heuristic (rule of thumb), conveyed by the
acronym LIPET (Logarithmic functions, Inverse trigonometric
functions, Polynomials, Exponential functions, and Trigonometric
functions). Basically, if one encounters an integral that must be
solved via integration by parts, he or she should first look for a
logarithmic function to represent u. If a logarithmic function is
not present, one should choose an inverse trigonometric function to
represent u, and so on, in accordance with the LIPET algorithm.
Also note that integration by parts may need to be performed
several times
before a manageable integral is achieved. That is, the vdu term
might require the technique to be performed on it. When integration
by parts must be performed on multiple occasions, it is imperative
not to switch back and forth between choices of functions for u and
dv. For instance, if 3x is chosen as u in the first round of
integration by parts, 23x should be chosen as u in the second
round. Otherwise, one would wind up undoing the first step!
Ex.) Evaluate dxex x222 . The integrand contains a polynomial
and an exponential function. In accordance
with LIPET, 22x is chosen as u and dxe x2 is chosen as dv.
Therefore, xdxdu 4= and .
21 2xev = While it should be obvious in the context of the
material of AP Calculus
AB, v was obtained through integration by u-substitution
(Remember that
-
13
+= .Cedue uu In this case, xu 2= and .2dxdu = Therefore, while
the integrand must be multiplied by 2 to obtain the form dueu , the
integral must be multiplied by -1/2 in order to compensate for this
algebraic manipulation. Again, this is AP Calculus AB
material).
dxex x222 +== .421)21)(2(421)21)(2( 222222 xdxeexxdxeex xxxx The
integral is still not in a manageable form. Integration by parts
must be performed upon this last integral. In this case, xu 4= ,
dxedv x2= , dxdu 4= , and
.21 2xev =
= xdxe x 42 = dxeexdxeex xxxx 2)21)(4(421)21)(4( 2222 = Ceex xx
+ 22 )21)(4( .
The last phrase is not the final answer. Recall that this is the
solution to the
integral in the first round of integration by parts. That is, it
is the solution to xdxe x 42 . The solution to this integral must
be multiplied by 1/2 and added to )
21)(2( 22 xex . The
final answer is determined in the following way:
dxex x222 = =
+
+=+= Ceexexxdxeex xxxxx 2222222 21)4(21)21)(2(421)21)(2(
Ceeex xxx21
21 2222 + = Ceex xx + 222
21 . This is the final answer.
Note that, since C is merely an unknown constant, 1/2C is merely
another C. Some integration by parts problems require addition of
integral expressions, as the following example shows:
Ex.) Evaluate .cos xdxe x Using LIPET, xe is chosen as u and
xdxcos is chosen as dv. Thus, dxedu x= and .sin xv = xdxe x cos =
dxexxe xx ))((sin))(sin( Another round of integration by parts must
be performed on the new integral, where xeu = , xdxdv sin= , dxedu
x= , and xv cos= = dxex x ))((sin dxexxe xx ))(cos()cos)(( . Note
that this last mathematical phrase is the solution to the integral
yielded by the first round of integration
-
14
by parts. That is, it is the solution to .))((sin dxex x Thus,
this solution appears in the solution to the overall (original)
integral:
= dxexxexexdxe xxxx ))(cos()cos)(())(sin(cos += xdxexexexdxe
xxxx cos))(cos())(sin(cos . Notice that the integral on the far
right is the same as the integral that one wishes to solve. This is
where addition comes in:
+= xdxexexexdxe xxxx cos))(cos())(sin(cos ++ xdxexdxe xx coscos
Cxxexdxexexexdxe
xxxxx ++=+= 2 )cos(sincos))(cos())(sin(cos2 .
Integration by Parts Using the Tabular Method: While the
previous problems were undoubtedly mathematically tedious, imagine
having to perform integration by parts three, four, or more times
for a single problem! There is, in fact, a method for integrating
by parts that is far more elegant (and easy!). This is known as the
tabular method. The following algorithm explains how to employ this
useful technique: 1.) Make a table with three columns, labeled u,
dv, and 1. 2.) Choose a value for u such that it will eventually
yield zero when differentiated enough (e.g. polynomials), put it in
the first row of the u column and differentiate down the column
until zero is reached. 3.) Choose a value for dv and put it in the
first row of the dv column and integrate down the column. Constants
of integration are not necessary here. 4.) Put a +1 in the first
row of the 1 column and alternate between positive and negative
signs down the length of the column. 5.) Draw a diagonal line from
each row in the u column such that it passes through the
appropriate row in the dv and 1 columns. Once the zero in the u
column is reached, cease drawing the diagonal. 6.) Multiply all
three terms connected by a given diagonal and add all of these
multiplied terms to yield the final answer. The algorithm sounds
confusing, but it usually takes less than a minute to compute the
integral. The tabular method is best conveyed through an
example:
Ex.) Evaluate .cos5 xdxx . The polynomial ( 5x ) will be chosen
as u and xdxcos will be chosen as dv. With these values, one may
generate the table to solve the integral:
-
15
u dv 1 5x xcos +1 45x xsin -1 320x xcos +1 260x xsin -1 x120 cos
x +1
120 xsin -1 0 xcos +1 -1
Notice that the first diagonal connects the terms 5x , xsin ,
and -1; the second diagonal connects the terms ,cos,5 4 xx and 1 ;
the third diagonal connects the terms
,sin,20 3 xx and +1; and so on. The terms of each diagonals are
multiplied and these multiplied terms are then added: .cos5 xdxx
=
Cxxxxxxxxxxx ++++ cos120sin120cos60sin20cos5sin 2345 . The
length of this answer indicates that solving the problem without
the use of the tabular method is quite rigorous. Indeed, the
calculation takes up a couple of pages if one uses the formula for
integration by parts alone! Note, however, that the tabular method
can only be used when the u term is not infinitely differentiable,
i.e., when it is a polynomial. Integration by Partial Fractions:
Upon adding fractions with dissimilar denominators, one looks for
common denominators to yield one equivalent fraction. For
instance,)24)(13(
))(13()24)(2(2413
2+++++=+++ xx
xxxxxx
xx . This is elementary algebra.
Working backwards is slightly more difficult, and can often be
much more difficult. How can one decompose a fraction into a sum of
fractions? To do this, one would use the method of partial fraction
decomposition. In this method, a fraction is decomposed into a sum
of fractions with denominators that are the factors of the
denominator of the original fraction and with numerators that
contain unknown constants. For instance, the
fraction, xxx
xx++++
23
2
223 , has a denominator with factors of x and 122 ++ xx
(i.e.
)12(2 23 ++=++ xxxxxx ). Thus the decomposition is as
follows:
xxxxx++++
23
2
223
122 ++++=xxCBx
xA , where A, B, and C are unknown constants.
Notice how the degree of the numerator is one less than the
degree of the denominator. In the denominator of the first
fraction, x is linear, so the numerator must be of zero order, with
a constant of A. In the denominator of the second fraction, 122 ++
xx is quadratic, so the numerator must be linear, with Bx+C. Note
that it is possible for factors
of the denominator to be repeated, as in the fraction
5)1()2(
++
xx , which is decomposed as,
-
16
.)1()1()1()1()1()1(
)2(54325 +++++++++=+
+x
Ex
Dx
Cx
Bx
Axx Here, x+1 is known as the
repeated factor, and repeats five times in the decomposition
process, with the value of the exponent increasing from 5 to 1.
Notice that the denominators are actually linear; this is why all
of the numerators are constants. The denominators are still linear
despite their being raised to a power. This is simply a repetition
of linear factors. This discussion of partial fractions could be
proven by a corollary to the Fundamental Theorem of Algebra.
However, since this is not an advanced algebra textbook, it will
not be presented here. Where does integration play a role in all of
this? Similar to the philosophy of integration by parts, certain
expressions that are not integrable by conventional methods can be
broken down into more accessible components. In the method of
integration by partial fractions, a fraction that is not integrable
through the use of the methods already learned is decomposed into
smaller fractions that can be integrated through the use of
conventional methods. Remember that these fractions contain unknown
constants that must be determined. This determination is first
carried out by multiplying both sides of the equation (i.e. the
whole fraction side and the partial fractions side) by the
denominator of the whole fraction. This operation cancels the
denominators on both sides of the equation. When this is completed,
one can solve for a system of equations.
Ex.) Evaluate dxxxx
xx++++ 232 2 23 .
xxx
xx++++
23
2
223 =
122 ++++xxCBx
xA
Multiply both sides of the equation by xxx ++ 23 2 :
++
++++=++++++
12)2(
223)2( 2
2323
223
xxCBx
xAxxx
xxxxxxxx Remember that
)12(2 223 ++=++ xxxxxx . =++ 232 xx xCBxxxA )()12( 2 ++++
CxBxAAxAxxx ++++=++ 222 223 AxCAxBAxx ++++=++ )2()(23 22
This last equation is a standpoint from which one can solve a
system of equations. Since the coefficients on both sides of the
equation must be equal,
1)( =+ BA 3)2( =+CA
A = 2 This simple system of equations is rather easy to solve.
Since the constant A is now known, this value can be substituted
into the other equations that contain A to find values for B and C.
When this is done, it is determined that B = -1 and C = -1. With
the decomposition completed, the integral can be easily solved:
dxxxx
xx++++ 232 2 23 = ++ + dxxx xdxx 12 )1(2 2 = Cxxx +++ 12ln2ln2 2
.
-
17
The system of equations in the previous problem was not at all
daunting. Oftentimes, however, the system of equations is so
horrendous that one must use matrix algebra to determine the
unknown constants, which is not part of the AP Calculus BC
curriculum. In the field of mathematics known as linear algebra,
matrices, arrays of numbers that have a myriad of uses, can be used
to solve a complex system of equations. While a thorough discussion
of matrix algebra will not be presented here, this point in the
book allows an excellent opportunity to use a graphing calculator
to solve difficult problems. In the case of a complex system of
equations, the TI-84 graphing calculator can represent a system of
equations as a matrix and can then transform this matrix into one
of reduced row echelon form (RREF). A matrix in RREF has the
following properties: each non-zero row has a greater number of
leading zeros than the previous row, the first non-zero number in a
row is 1, and the initial 1 of row is the only non-zero element in
the column in which the 1 appears. Observe the matrix below to
clarify these requirements:
9342
1000010000100001
Notice how each row has one more has one more leading zero than
the previous one, though it is not a requirement that each row
differ by just one zero. Note also that whenever a value of 1
appears for the first time in a row, there are only zeros in the
column in which the 1 appears. Through the methods of linear
algebra, namely a method known as Gaussian elimination, a matrix
that represents a system of equations can be reduced to RREF so
that the elements in the final column represent the values of the
unknowns. To understand how this is done, see the following
example. Again, this technique is not tested on the AP Calculus BC
exam, but is very useful nonetheless.
Ex.) Represent ++ ++++ 22 234 )3)(2( 9201643 xx xxxx as the sum
of three integrals. The first step is to decompose the fraction
into a sum of partial fractions. Note that the term 32 +x is a
repeated factor. 22
234
)3)(2(9201643
++++++
xxxxxx
222 )3()3()2( ++++
+++= xEDx
xCBx
xA
[ ] 22 2342 )3)(2( 9201643)3)(2( ++ ++++++ xx xxxxxx = [ ]22222
)3)(2()3()3()2( +++++++++= xxx EDxx CBxx A
)2)(()3)(2)(()3(9201643 222234 ++++++++=++++ xEDxxxCBxxAxxxx
CCxCxCxBxBxBxBxAAxAxxxxx 623623969201643 2332424234
++++++++++=++++
EExDxDx 222 ++++
-
18
)236()236()2()(9201643 234234 EDCBxDCBAxCBxBAxxxxx
+++++++++++=++++ )269( ECA +++
92692023616236
423
=++ =+++=+++ =+
=+
ECAEDCBDCBA
CBBA
This is the system of equations that must be solved through
matrix algebra. The matrix will have as many rows as there are
powers of x, including the power of zero for the constant, and as
many columns as there are unknowns plus one for the equality. Since
the system of equations in this example is associated with a
fourth-order polynomial, there will be five rows, and since there
are five unknown constants, there will be six columns. After the
matrix is set up, each row is treated like one of the equations in
the system. For example, 3=+ BA , in matrix language, appears as (
)300011 , because the coefficient of A is one, the coefficient of B
is one, there are no Cs, Ds, or Es, and the whole equation is equal
to 3. The final matrix is:
92060920123601601236400120300011
This matrix can be set up in the TI-84 calculator as follows: Go
to 2ND + x-1, which will open up the MATRIX menu. Go to EDIT and
choose a matrix to edit (e.g. [A], [B], etc.). The main screen of
the calculator will ask for the number of rows and columns. For
this example, in put 56. Fill in the matrix. This matrix must now
be reduced to RREF. To do this via calculator, go to the MATRIX
menu again and go to the MATH option. Scroll down to B: rref( and
select this option. On the main screen of the calculator will
appear the operation rref(. Input the matrix (either [A], or [B],
etc.) into this operator by choosing it from the MATRIX menu. Press
the ENTER key to calculate the RREF matrix. For this example, it
should
be:
010000401000000100200010100001
Therefore,
04021
=====
EDCBA
and the partial fractions are 222 )3(4
)3(2
)2(1
+++++ xx
xx
x.
-
19
The problem merely asked for the setting up of the integral as a
sum of integrals. Thus, the final answer is:
=++++++ 22 234 )3)(2( 9201643 xx xxxx
.)3(07317.353659.5
)3(60976.119512.1
)2(60976.0
222 ++++++= x xxxx
Improper Integrals
Recall the Second Fundamental Theorem of Calculus: =ba bFaFdxxf
)()()( . This theorem states that a definite integral is evaluated
by finding the antiderivative of a function, substituting the lower
and upper limits of integration into this expression, and
subtracting the substitution of the upper limit from the
substitution of the lower limit. In the cases in which this theorem
is employed, f(x) is bounded between a and b and the function is
continuous between and at these points (i.e. on a closed interval).
A definite integral with these properties is known as a proper
integral. Whenever one or both of these properties is not
fulfilled, the expression is an improper integral. For instance,
the
integral +0 2dxx is an improper integral because the upper limit
of integration is infinity. Therefore, there are no restrictions on
the interval. In another case, both limits may be finite, but the
function could exhibit a discontinuity within the interval in
which
integration is to take place. For instance, the integral + 11 1
dxx is an improper integral because there exists an infinite
discontinuity at x = 0. Interestingly, it may very well be the case
that an improper integral represents a finite area despite an
infinite interval or an infinite discontinuity. Such improper
integrals are said to be convergent. Improper integrals that do not
represent a finite area are said to be divergent. In order to apply
the Second Fundamental Theorem of Calculus to improper integrals,
it is necessary to define the limit(s) of integration or the value
of discontinuity that makes for impropriety as a limit. This is the
major theme of calculus yet again thinking on an infinitesimal
level. Indeed, it is not possible to think of an infinite geometric
representation as having a finite area. Nevertheless, by applying
the definition of the limit to the problem, it is very much
possible!
Ex.) Is the area under the curve x
y31= from
-
20
+= aa dxxA 1 31lim , where a is some value that approaches
infinity. [ ] ===== +++ 0)1ln()ln(lim31lnlim3131lim 11 axdxxA xaaaa
Therefore, no finite area exists under this curve in the
interval
-
21
into play. While our intuition maintains that an infinite
geometric figure cannot possibly have a finite area or volume
(after allit is infinite!), the methods of calculus beg to differ.
This is philosophically quite interesting. If one considers the
non-Platonistic premise that mathematics is a tool created by human
beings, it is remarkable to note that the human mind has birthed
something that can explain something that it itself cannot explain!
Consider the case of Gabriels Horn, for instance. This geometric
figure, conceived by the Italian mathematician Evangelista
Torricelli, has a finite volume, but an infinite surface area. It
can be modeled as the solid of revolution generated when a
hyperbola (e.g. x
y 1= ) is revolved about the x-axis:
Generated on Mathematica
To calculate the volume of this figure, on can use the methods
of AP Calculus AB. In this case, the disc method is the most
appropriate:
=+=
+=
==
=
)10()1(1)(1lim1limlim1 11 21 2 axxdxdxxV aaa aa .
Thus, the solid has a finite volume. But what of the surface
area? While a topic of neither the AP Calculus AB curriculum nor
the AP Calculus BC curriculum, the surface area of revolution (SA)
for a curve yxf =)( about the x-axis bounded on the closed interval
bxa is described as: += ba dxdxdyySA 212 . For Gabriels Horn:
+
=
+= a
adx
xx
dxxx
SA1
4
1
2
2
11lim2112
=
=
+= 2211 23 )1(2 11ln)(2 1lnlim221lnlim211lim2 aaxxdxxx aaa
aa
==
= )(2210)0(2 . Thus, the surface area is infinite. This is
an
interesting scenario. If Gabriels horn were to hold paint, it
would hold cubic units of
-
22
it. Obviously, this is not enough paint to cover its own
surface, which is infinite! While this rather philosophical problem
has little practical significance (actually none), the theory
behind this problem plays an integral role (no pun intended) in the
sciences in the form of probability distributions. Note that the
following section is not part of the AP Calculus BC curriculum, but
should be read to gain an appreciation of the material. Probability
Distributions in the Sciences: Very often in the physical and life
sciences, one must apply the laws of probability and statistics to
solve problems. A probability density function (often denoted as
p(x)) is a mathematical representation that yields the probability
of the occurrence of a random variable x. For a probability
distribution on the interval +
-
23
into play? Within the container described are many gas
particles, perhaps on the order of 1023. The kinetic-molecular
theory maintains that these particles move randomly in straight
lines and transfer energy and momentum during collisions but do not
lose it. All of the particles are moving at different speeds; one
may be moving extremely quickly, while another extremely slowly and
these speeds change quite often. However, at a given temperature
there will be a most probable speed for the collection of gas
molecules. This is shown graphically as a Maxwell-Boltzmann
distribution:
Note that the probability never decreases to zero as one moves
farther along the x-axis. Indeed, it is possible for a gas particle
at 273 K (0C) to have a speed of, say, 50101 m/s, but it is not
very probable! Notice the characteristics of the Maxwell-Boltzmann
distribution when the temperature increases:
The curve is flattening out such that the most probable speed is
greater but is less probable than the most probable speed at a
lower temperature (i.e. the peak is lower). Why is this case?
Recall that for a probability density function the improper
integral,
+ = 1)( dxxp , holds true. Thus, in order to keep the area equal
to unity when the most probable speed is greater in magnitude, the
probability of the most probable speed must be lowered. What would
the Maxwell-Boltzmann distribution look like at absolute zero (0
K)? What about at a very, very high temperature? Note that the
Maxwell-Boltzmann distribution does not solely apply to gases, but
to all matter. This distribution is also important in understanding
such processes as diffusion, which is quite important to life on
Earth. The last application of the probability density function to
be considered in this chapter concerns the field of quantum
mechanics. While this is, indeed, a broad field, it isgenerally
based upon the assumption that matter at the submicroscopic level,
especially electrons, exhibit the characteristics of waves more so
than particles. As a result of these wave characteristics, it is
impossible to know with certainty both the position and momentum of
a particle simultaneously, a concept known as Heisenbergs
uncertainty
-
24
principle. An electron, for instance, cannot be modeled as a
particle, but as a wave function, denoted as ),,,( tzyx , which is
a function of three-dimensional space and time. Squaring this
function results in the probability density function. As the
distance from the nucleus increases, the probability of locating an
electron greatly decreases, but never reaches zero. For instance,
for the hydrogen atom, its one electron can be considered a cloud
of probable positions in a spherical space:
While it is very unlikely that an electron will be located far
away from the nucleus, it is possible nonetheless. An electron
associated with an atom in this sheet of paper could be somewhere
on the moon, but it is not very probable!
Concluding Remarks This first chapter introduced several new
techniques in differential and integral calculus. LHpitals Rule,
integration by parts, integration by partial fractions, and
improper integrals are all very useful tools when faced with more
difficult problems that cannot be solved through the use of the
techniques of AP Calculus AB. While these techniques are powerful,
it is important to realize that they are not omnipotent. There are
many other techniques that can be used to evaluate troublesome
limits and integrals, and most are far beyond the scope of AP
Calculus BC. One very important application of improper integrals,
the probability density function, was also discussed in the context
of its significance in modeling systems in physical chemistry and
quantum mechanics. The next chapter will apply many of the
techniques discussed in this chapter to approach various scientific
problems. Key Terms: indeterminate form divergent LHpitals Rule
Gabriels Horn integration by parts probability density function
LIPET random variable tabular method normalized partial fraction
decomposition Gaussian function repeated factor normal distribution
integration by partial fractions kinetic-molecular theory reduced
row echelon form (RREF) Maxwell-Boltzmann distribution improper
integral Heisenberg uncertainty principle convergent wave
function
-
25
Chapter 2: Differential Equations Differential equations are
introduced in AP Calculus AB. These equations
contain the derivative of a function as a variable. For
instance, the equation yxdxdy 2= is
differential equation that relates the derivative of y with
respect to x to those variables themselves. The differential
equations studied in AP Calculus AB and BC are known as first-order
differential equations, because they involve the first derivative
of a function. Differential equations of higher order require more
sophisticated techniques to solve and will not be discussed here,
bute will be briefly discussed in the context of chapter 4.
Differential equations are one of the most important (arguably, the
most important) expressions used in mathematical modeling, mostly
due to the fact that, by their very nature, they express a change.
This chapter will discuss two ways in which one may solve
differential equations: the method of separation of variables (an
analytical method) and Eulers method (a numerical method). While
the AP Calculus AB exam only tests the first method, the AP
Calculus BC exam tests both. In addition to the presentation of
these methods, various applications of differential equations to
the physical and life sciences will be discussed. As a final note,
it is important to appreciate the shear profundity of the study of
differential equations. This chapter only covers two techniques for
solving these equations, which is fine for the AP exam. However,
new theories on differential equations are constantly being
developed, and even a course that is specifically devoted to
differential equations will probably not do them justice.
The Method of Separation of Variables
This section should be a review of material from AP Calculus AB.
The method of separation of variables basically separates a
differential equation into two sides of an
equation that have the same variables. For instance, xy
dxdy 2= is a differential equation
that can be easily solved through algebraically rearranging the
equation so that the y-variables appear on one side and the
x-variables appear on the other side:
.2
2x
dxy
dyxy
dxdy == Once this is completed, one may integrate both sides to
determine
y as a function of x: +==== .lnln212 2ln2ln Cxyeexyxdxydy xy In
order to determine the value of C, it would be necessary to be
given an initial condition, such as .3)0()3(3)0( 2 =+== CCy
Eulers Method
Eulers Method, pronounced oi-ler, is a numerical method used to
approximate solutions to differential equations. The method was
developed by the Swiss mathematician Leonhard Euler in 1768.
Numerical methods for solving differential
-
26
equations are necessary when solutions cannot be found
analytically, such as when one does not explicitly know the
algebraic structure of the differential equation, but knows certain
values for variables and slopes. Given the initial values of the
variables and the slope, a discretized (i.e. non-continuous) form
of the limit definition of the derivative can be used and
rearranged to yield the formula for using Eulers method. Recall
from AP
Calculus AB the limit definition of the derivative:h
xyhxyxyh
)()(lim)('0
+= . If h is not
infinitely small (i.e. it has a finite value), the equation
becomes: .)()()('h
xyhxyxy + For the purposes of Eulers Method, let h be called x ,
let )( xxy + be denoted as 1+ny and let )(xy be denoted as ny . The
discretized form of the limit definition for the
derivative now becomes: .)(' 1x
yyxy nn
+ This is rearranged to yield the formula for Eulers Method:
.)('1 nnn xyxyy +=+ What is the significance of this formula? If
one knows initial values for a function and its derivative, while
not necessarily knowing what those functions are, and chooses a
certain increment x , known as the step size, for the numerical
analysis, one can approximate the value of y(x) for a certain x.
See the example below. Ex.) Approximate the value of y(1.5) by
using increments of 0.1 if y(1) = 4 and y(x) = yx 2 .
Notice that in this problem, the actual algebraic structure for
the differential
equation is known. This sort of problem often appears on the AP
exam, primarily as a free-response problem. The rationale for a
problem of this type will become clear at the end of this
example.
Since the starting x-value is 1 and the step size is 0.1, there
must be six steps involved to approximate a value for the function
at x = 1.5. The first conditions given are
10 =x and 40 =y . Thus, the initial slope is 4)4()1( 2 = . The
formula for Eulers Method may now be used: .4.4)4)(1.0()4()(' 11
=+=+=+ yxyxyy nnn Five more steps to go! The next x-value is found
by adding the step size to the previous x-value: xxx += 01
1.1)1.0()1( =+= . Now 1y can be found:
.324.5)4.4()1.1( 212
11 === yxy The remainder of the problem proceeds as follows:
Find y2:
9324.4)324.5)(1.0()4.4(112 =+=+= yxyy Find y3:
2.1)1.0()1.1(12 =+=+= xxx 102656.7)9324.4()2.1( 22
222 === yxy
2035056.5)102656.7)(1.0()49324.4(223 =+=+= yxyy Find y4:
3.1)1.0()2.1(23 =+=+= xxx 793924464.8)2035056.5()3.1( 23
233 === yxy
-
27
082898046.6)793924464.8)(1.0()2035056.5(334 =+=+= yxyy Find
y5:
4.1)1.0()3.1(34 =+=+= xxx 92248017.11)082898046.6()4.1( 24
244 === yxy
275146063.7)92248017.11)(1.0()082898046.6(445 =+=+= yxyy Find
y6:
5.1)1.0()4.1(45 =+=+= xxx 36907864.16)275146063.7()5.1( 25
255 === yxy
912053927.8)36907864.16)(1.0()275146063.7(556 =+=+= yxyy Since
this last number is the value of y when x = 1.5, this is the
approximation for which the problem asked. The reason that this
sort of problem actually supplies the algebraic structure of the
differential equation is so that one can compare this approximate
value with the actual value yielded from an analytical solution.
Indeed, the differential equation given can be solved
analytically:
( ) 333322 333
3ln
xC
xCx
CeeeeyCxydxxy
dyyxdxdy =
==+===
+ .
Recall that C is merely an unknown constant, so Ce is just
another C. This constant can be determined based upon the initial
values given:
3/13)1(
3 4)4(33
eCCeCey
x
===
828287262.84 3)5.1(
3/1
3
=
= ee
y
This analytical solution differs from the numerical solution by
0.0837666646. There are several key points to take away from this
Eulers Method problem. Firstly, the numerical method never yields
exact solutions to differential equations. Secondly, as the step
size becomes infinitely small, the solution becomes exact. Thus,
the smaller the step size that one uses, the less error is involved
in the calculation. However, notice that even this problem with the
relatively large step size of 0.1 was quite tedious to solve. One
must weigh the precision of the solution needed against the cost of
actually solving the equation numerically. Thirdly, Eulers Method
is only one of many numerical methods of solving first-order
differential equations, and is generally the least powerful. In
fact, Eulers Method really only has historical significance, since
no one uses this method any more.
The Law of Exponential Change
Very often in the physical and life sciences, one encounters
natural logarithms in mathematical models of phenomena. This is the
case because many aspects of nature seem to change in direct
proportion to an amount of something present. Whether this amount
refers to number of organisms in a population or the number of
molecules in a
-
28
mixture of chemicals in a beaker, the change in these entities
often takes on the form of
the following differential equation: kNdtdN = . This
mathematical statement means that
the change in some amount N over time is proportional to that
amount. That is, if there is more of something present, its rate of
change is greater. To understand why so many natural processes are
modeled mathematically with natural logarithms (in fact, its
natural prevalence is one of the reasons for its name!), it is
necessary to solve the differential equation:
( )( ) ===+=== + .ln )( ktCktCkt CeeeeNCktNkdtNdNkNdtdN Let the
initial value of N be N0: CCeN k == )0(0 . The solution to this
differential equation, ,0
kteNN = is known as the law of exponential change. The law of
exponential change, its derivation, and its applications are
covered in both the AP Calculus AB and BC curricula.
Logistic Growth While many processes seem to exhibit exponential
growth, given enough time, most of these processes would occur less
rapidly once some limiting value is reached. How can the
differential equation for exponential growth be modified to account
for this limiting value? In order to answer this question, it is
necessary to translate the meaning of this logistic growth into a
mathematical statement. In the logistic growth model, the rate of
change of a quantity is proportional to both the amount present and
the amount relative to the limiting value, or carrying capacity
(K). This is represented by the
following differential equation: ( )NKkNdtdN = . This means that
the rate of change of
a quantity N is jointly proportional to the amount present and
to the amount relative to the carrying capacity that is still
available for growth. While this differential equation can be
solved by the method of separation of variables, it requires a
technique discussed in the previous chapter integration by partial
fractions:
( ) ==== ktkdtNKN dNkdtNKN dNNKkNdtdN )()( +C Partial fraction
decomposition:
NKB
NA
NKN += )(1
AKABNBNANAKBNNKA +=+=+= )()(1 Since there are no Ns on the left
side of the equation, there cannot be any on the right side either.
Therefore, .0= AB Also, to make the constants on both sides of the
equation equal to 1, A must equal 1/K, which means that B must
equal 1/K as well. Thus:
.)(
11)(
1NKKKNNK
BNA
NKN +=+= Integration may now be
-
29
carried out:
( ) == +==+ kCKktNK NNKNKdNNKNKNKN dNCkt lnlnln1111)((As always,
kC will just be considered another C)
.1
1 KktKktKkt
kC
Kkt
CeKNCe
KNCe
ee
NKN
+===
= This is logistic growth
equation. Note some important characteristics of its algebraic
structure. The carrying capacity is located in the numerator, the
exponential is multiplied by a constant C, and e is raised to the
negative power of the product of the carrying capacity, another
constant k, and time. This structure puts a limit on the growth
that the equation models. Compare the exponential growth model to
the logistic growth model:
Exponential Growth Model Logistic Growth Model
Note that for the logistic curve g(t), Ktgt
= )(lim . The practical applications of these curves will be
discussed shortly. Note that logistic growth is frequently tested
on the AP exam either as a multiple-choice or free-response
question.
The Learning Curve In certain cases, the rate of change of a
process is directly proportional to the difference between the
endpoint of a process and the degree to which the process has
already ensued. Mathematically, this is expressed as the
differential equation:
),( NAkdtdN = where A is the endpoint of the process. Suppose
that the process in question is the typing of written statement on
a word processor. Let the total number of letters in the statement
be 1,000 and let the efficiency be determined by the number of
correct letters typed relative to the 1,000 letters. If the process
follows a learning curve, or bounded growth, the efficiency will
increase with time. What is the nature of this increase? To answer
this question, the differential equation must be solved:
=== kdtNAdNkdtNAdNNAkdtdN )()()( ( )( ) ktCktCkt CeeeeNACktNA
===+= )(ln ktCeAN = . In the typewriting example, A would be 1,000
and C and k could be determined from a set of initial conditions.
Notice the graphical nature of the learning curve:
-
30
The rate of change is most rapid near the beginning and
decreases to zero at N=A. What does this mean? Basically, the
greatest degree of learning takes place near the beginning and
decreases as the endpoint is reached. Thus, the more one learns (or
the more of a process that is carried out), the lesser the rate of
increase of learning. Note also that the efficiency, measured in
this case as the difference between A and N, increases with the
time invested in the process. The learning curve has many
implications in economic and educational strategy. If it is the
case that as the amount of time invested in a project increases the
efficiency increases, then corporations and classrooms should
concentrate on those strategies that increase the experience of
those involved. Problems concerning the learning curve appear on
both the AP Calculus AB and BC exams. Ex.) An elementary school
student must memorize the capitals of all fifty United States in
thirty minutes. The rate of memorization is directly proportional
to the difference between the number of capitals that must be
memorized and the number of capitals that have been memorized so
far. Assuming that the child can flawlessly memorize two state
capitals during the first three minutes, and assuming that the
child knew none of the state capitals at the beginning of the study
session, will he be able to memorize all fifty state capitals in
thirty minutes?
The problem did not require solving of the differential
equation, so it would be helpful in a problem like this to commit
the resulting equation to memory. The same applies to the
exponential and logistic growth models as well, unless, of course,
the problem does ask for the derivation.
ktkt CeCeAN == 50 It is assumed that the child knew none of the
state capitals before beginning memorization, so 5050)0( )0( == CCe
. It is given that at t = 3, N = 2: .0136073.0
3)50/48ln(5050)2( )3( ==
ke k
Now, the value of N when t = 30 may be found: .1775834.165050
)30)(0136073.0( == eN Unfortunately, thirty minutes is not
enough for the youngster to memorize all of the state
capitals.
Mathematical Models of Population Ecology
-
31
Ecology is a rather broad field of biology that studies the
relationships between organisms and their environment. One branch
of this field, population ecology, focuses on these relationships
at the level of a group of the same species inhabiting the same
area, a group referred to as a population. Population ecology is
perhaps the most quantitative of the subfields of ecology, since it
often studies the changes in the number of individuals in a
population over time, the strategies that different species use to
proliferate, and how biotic (living) and abiotic (non-living)
factors affect a certain population. Populations are generally
modeled mathematically by the exponential growth model or the
logistic growth model. The Exponential Growth Model in Population
Ecology: The exponential growth model for populations, specifically
human populations, was devised by English economist Thomas Malthus
in his An Essay on the Principle of Population (1798). This
treatise exclaimed certain danger for the human race in that while
the global food supply grows linearly, the human population grows
exponentially. Population ecologists model this
exponential growth as NrdtdN
max= , which is essentially the same equation as the one
discussed earlier, but instead of using the constant k, rmax is
used. This constant is known as the intrinsic rate of increase, a
measure of the capacity of a population to grow at its maximum
potential.
Ex.) A population of fruit flies (genus Drosophila) exhibits
exponential growth. There are initially 10 flies in the population.
After three days, the population has grown to 67 individuals. What
is the intrinsic rate of increase of this population? How many
individuals will be present after one week has elapsed? The problem
did not require a derivation of the formula for the law of
exponential change, so it suffices just to commit the formula to
memory. For an example concerning population ecology, the formula
is .max0
treNN = It is given that 100 =N and that at .67,3 == Nt From
this information, the intrinsic rate of increase can be
determined:
634036.03
)10/67ln()10()67( max)3(max == rer .
Now the value of N when t = 7 can be found: 846268.84610
)7)(634036.0( == eN flies!
At this point, it is probably a good idea to discuss the units
in these problems. The AP Calculus tests are usually not very
strict in regards to including units in calculations, as long as
they are provided along with the final answer. Nevertheless,
especially when dealing with functions such as logarithms and
exponents, it is important to discuss the units of each element
used in these mathematical models. In the formula,
,max0treNN = N and N0 have units of number of individuals, in
the case of the previous
example, flies. The element t has units of time (seconds,
minutes, hours, etc.), days in the previous example. What about
rmax? This element is not unitless. In the exponential growth
equation, it has units of reciprocal time, 1/days or days-1 in the
previous example.
-
32
This is because exponentials; along with logarithms,
trigonometric functions, and inverse trigonometric functions; are
known as transcendental functions, which are functions that do not
satisfy a polynomial expression. One of the consequences of the
failure to satisfy this expression is that the arguments of
transcendental functions must be unitless. Thus, in the example of
the exponential growth model, the product of t and rmax must be
unitless. Since t has units of time, rmax must have units of
reciprocal time. The Logistic Growth Model in Population Ecology:
While certain populations exhibit exponential growth under certain
conditions, no population can ever truly attain its intrinsic rate
of increase. A population could only truly grow in an exponential
fashion if it had unlimited access to resources such as food,
water, and space. Obviously, these resources are limited, as
Malthus noted, meaning that there is a certain limit to the amount
of individuals in a population that an area can maintain. Indeed,
if every population underwent exponential growth, the total mass of
existing organisms would far exceed the mass of the Earth! Thus,
the logistic growth model, discussed earlier in this chapter, is a
far more realistic way to gauge the trends in population growth.
Population
ecologists often model logistic growth as:K
NKNrdtdN )(
max= , which is essentially the
exponential growth model with another term added to tame the
equation. This term,
KNK )( , represents the fraction of the population relative to
the carrying capacity that is
still available for growth. Note that this is different from the
logistic model already discussed, in which the term was merely (K
N), or the number of individuals that can still undergo growth.
While both differential equations yield a logistic function, the
one discussed first should be used for the purposes of AP Calculus
BC. Ex.) A population of grizzly bears in a preserve exhibits
logistic growth defined
by the following differential equation:
=14
16
NNdtdN . What is the carrying capacity of
this population? How many grizzly bears are there when the rate
of increase in the population begin to decrease? If when
grizzliesNyearst 2,0 == and when
,10,5 grizzliesNyearst == determine the time at which this
decrease occurs. For the first part of the problem, recall that the
carrying capacity is reached when the rate of change of the
population (the derivative) is equal to zero. Thus, the carrying
capacity can be found directly by setting the differential equation
equal to zero are
solving for N: 14014
16
==
NNN . This is the carrying capacity for the population. Upon
reading the second part of the problem, the reader may realize that
this question is essentially testing the ability to analyze the
relationships between functions and their derivatives, an AP
Calculus AB skill. When the rate of increase (the derivative)
begins to decrease, the derivative will have a relative maximum. To
determine the value of N at which this occurs one could either
graph the derivative on the TI-84 and determine where its maximum
lies, or use the second derivative test. To save time (assuming
that this is a question in which one is permitted to use a
calculator!), one should use the former method. Graphing the
derivative, it is found that the derivative has
-
33
a relative maximum at N = 7. Thus, the rate of change of the
population begins to decrease when there are 7 grizzly bears. While
it may seem that this last problem requires solving the
differential equation
(notice that it has a slightly different algebraic structure
from ))( NKkNdtdN = , the
problem stated that the population follows the logistic growth
model. So, again, if this formula is committed to memory, and if
the problem does not ask for the derivation, one can simply use the
formula: Two initial conditions are given. These will be used
to
determine the constants k and C in the equation for logistic
growth .1 KktCe
KN +=
61
141
14)2( )0(14 =+=+= CCCe k
( ) 038686.070
60/4ln6010146114)10( 70)5(14 =+=+=
kee
kk
30825.3541604.0
)42/7ln(4271461
14)7( 541604.0)038686.0)(14( =+=+=
teet
t .
Thus, the rate of change in the population will begin to
decrease after a little over three years.
Mathematical Models of Reaction Kinetics While the application
of exponential and logistic growth will appear on the AP exam, this
following material is not a part of the AP Calculus BC curriculum.
However, it still gives the reader a chance to appreciate a very
common application of simple differential equations. Reaction
kinetics is a branch of physical chemistry that studies the rates
and mechanisms of chemical reactions. This field is of the utmost
importance to practical chemistry because it allows one to
determine how quickly a desired reaction will take place, how the
reaction takes place, and how one might speed up (or slow down) the
reaction. Consider the following generic chemical equation:
dDcCbBaA ++
where the capital letters refer to particular chemical species
and the lower-case letters refer to their relative numbers in the
reaction (called a stoichiometric coefficient). The left side of
the equation is referred to as the reactants and the right side of
the equation is referred to as the products. As the reaction takes
place, the amount of reactants will decrease and the amount of
products will increase. How can one express the rate at which this
reaction takes place? In essence, one can take the derivative of
the equation. There is a problem, however; not all species involved
may change at the same rate. For instance, the coefficient b may be
twice as great as coefficient a, meaning B will decrease twice as
fast as A. One can compensate for this by multiplying the rate of
change of the concentration of each species (concentrations are
denoted with brackets [ ]) by the inverse of its stoichiometric
coefficient:
-
34
.][1][1][1][1dtDd
ddtCd
cdtBd
bdtAd
a+=
The negative signs on the left side of the equation signify a
decrease in reactants while the positive sign on the right
signifies an increase in products. For instance, consider the
combustion of glucose ( )6126 OHC :
.666 2226126 OHCOOOHC ++ The expression that conveys the
reaction rate of this reaction would be:
.][61][
61][
61][ 2226126
dtOHd
dtCOd
dtOd
dtOHCd +=
The actual nature of reaction rates can be explained by simple
differential equations that can be solved through separation of
variables. Reaction rates are usually gauged by the disappearance
of reactants. Depending upon the chemical reaction in question,
these rates are dependant upon reactant concentrations in certain
ways. For instance, in a certain chemical reaction (AB), in which
the species A is monitored, the
reaction rate might be expressed by the differential equation:
],[][ AkdtAd = in which
the rate of decrease of A is proportional to the amount of A
present. This equation should look familiar; its solution is the
law of exponential change, but rather than expressing growth, it
expresses decay: .][][ 0
kteAA = A reaction that proceeds in this manner is said to be of
first order because the rate of the reaction is directly
proportional to the concentration of reactants to the first power.
In general, a reaction order refers to the term n in the
expression: krate = [reactants] n , which is known as the rate law
of the reaction. If the rate of decrease of A were proportional to
the square of the reactants, the
reaction would follow second-order kinetics (i.e. 2][][ AkdtAd =
). Reaction rates of
higher order are certainly possible (some biochemical reaction
orders are greater than 10), as are fractional orders, zero orders,
and negative orders. All of these aspects of the reaction are
determined from laboratory experimentation.
Radiometric Dating
The atom is modeled as a collection of three subatomic
particles: protons,
neutrons, and electrons. The electrons are distributed
probabilistically (see chapter 1) around an incredibly dense
nucleus, which is composed of protons and neutrons. While neutral
(uncharged) atoms of the same element have the same number of
protons and electrons, they may very well differ in their number of
neutrons. Atoms of the same element that differ in their number of
neutrons are referred to as isotopes. The stability of an isotope
is related to the ratio of its number of neutrons to its number of
protons. An unstable isotope is said to be radioactive, and will
undergo a mode of radioactive decay to achieve a more stable
neutron-to-proton ratio. Scientists have learned to take advantage
of this submicroscopic phenomenon in the process of radiometric
dating, the determination of the approximate age of an object based
upon the amount of a radioisotope (i.e. radioactive isotope) that
it contains, the known decay rate of the radioisotope, and some
reference object that also contains the radioisotope. One of
the
-
35
most revolutionary forms of radiometric dating employs the
isotope carbon-14. In this technique, invented by Nobel-Prize
winning chemist Willard Libby in 1949, one compares the amount of
carbon-14 in a sample (measured by the amount of radioactivity) to
the amount in living systems. This is done based upon the premise
that the amount of carbon-14 in carbon dioxide has been relatively
constant for many thousands of years and that the ratio of carbon
dioxide with carbon-14 in its molecular structure to carbon dioxide
that does not contain radioactive carbon is the same in the
atmosphere as it is in living things. When a living thing dies, it
ceases to assimilate carbon-14 into its structure, and radioactive
decay ensues. Thus, if one measures the amount of carbon-14 in an
organically derived sample, such as something made from plant
material, and compares it to the amount of carbon-14 in a living
thing, which represents how much carbon-14 was originally present
in the sample, he or she may determine the approximate amount of
time elapsed since the organism whose matter was used to make the
object died. The rate of decay of carbon-14 is not constant, but
follows the law of exponential change. This was determined by the
observation that a radioisotopes half-life, the amount of time
elapsed before half of the material has decayed, is independent of
any external factors such as temperature or concentration. The
formula for the half-life of an object can be derived from the law
of exponential change:
= kteNN 0 When half of the material has decayed, half of the
material still remains, so .
21
0NN = The time at which this occurs is the half-life,
denoted
as .2ln)2ln()1ln(21ln
21
2/1002/1 ktktkteNNt kt ====
Notice that the half-life only depends upon the constant k,
which is a characteristic only of the specific isotope in question;
other factors do not affect half-life. This holds for other natural
systems that follow the law of exponential change as well. The
half-life of carbon-14 is about 5700 years. Note that carbon dating
can only be applied to relatively young samples that are organic in
origin. After about 36,000 years, so much of the original
radioactive sample has decayed that it becomes quite difficult to
detect, and the accuracy of the experiment is substantially
reduced. Problems concerning radiometric dating are common on both
the AP Calculus AB and BC exams.
Ex.) A medieval art collector plans to buy a flag that is
claimed to have been used
in the Battle of Hastings in 1066. Before paying a hefty price
of $50,000 for this supposed relic, the collector demands that the
seller have the flag carbon-dated. Analysis shows that the
radioactivity of carbon-14 detected in living plants from which
such a cloth may be made is 20.9 disintegrations per minute per
gram and that the radioactivity of the material in the flag is 18.6
disintegrations per minute per gram. Note that the half-life of
carbon-14 is 5700 years. Is the flag authentic?
Let the variable A represent radioactivity. The radioactive
decay of carbon-14
follows the law of exponential change: kteAA = 0 . While the
initial radioactivity of the sample (taken to be equivalent to the
radioactivity of the living plant) and the final
-
36
radioactivity of the sample are known, the constant k is not.
This constant may be found
from the formula for half-life: 42/1 10216048.157002ln2ln ===
k
kt .
The approximate time elapsed may now be determined: yearsteeAA
tkt 7416.958)9.20()6.18( )10216048.1(0
4 === . Since the Battle of Hastings occurred 940 years ago,
within a certain margin of error, the flag is probably
authentic.
Newtons Law of Cooling
Unlike the complex computational physics of today, the physics
of Sir Isaac
Newton was known for its brevity and elegance. One example of
this conciseness is manifest in a simple law of heat transfer known
as Newtons Law of Cooling, though the model applies to heating as
well. This model states that the rate of change of an objects
temperature is directly proportional to the difference between the
objects temperature and the temperature of the immediate
environment, assuming that this environment is large enough that it
does not experience a substantial change in temperature.
Mathematically, ),( envTTkdtdT = where T is the temperature of
the object at a certain
time and Tenv is the temperature of the environment, which is
assumed to be constant. If an object is heating up, the value of k
will be positive. If an object is cooling down, the value of k will
be negative. Note the important difference between this equation
and the equation for bounded growth; in Newtons Law of Cooling, the
endpoint (the temperature of the environment) is subtracted from
the temperature of the object, while in bounded growth, the
variable in question is subtracted from the endpoint. This
different order makes a difference in terms of the final equation,
as will be evident in the following solution to the differential
equation for Newtons Law of Cooling:
=== kdtTT dTkdtTT dTTTkdtdT envenvenv )()()( .))(()(ln
ktCktCktenvenv CeeeeTTCktTT ===+= +
Note that this is a variation on the law of exponential growth;
the only difference here is that a constant (Tenv) is subtracted
from the variable (T). What is the significance of the constant C?
This can be determined through an analysis of initial conditions.
At an initial time t = 0, .envTTC = This T is the initial
temperature, or T0. Thus, the equation for Newtons Law of Cooling
becomes: .)()( 0
ktenvenv eTTTT = Again, if the
object is heating up, k is found to be positive, and if it is
cooling down, it is found to be negative. Problems concerning
Newtons Law of Cooling appear on both the AP Calculus AB and BC
exams. Ex.) A blacksmith removes a piece of iron from a furnace
with a temperature of 266C. The temperature of the blacksmiths
workshop is 24C. When the iron has been allowed to cool for one
minute, its temperature is 235C. The blacksmith may work with the
iron when it cools to a temperature of 75C. Assuming that the rate
at which the iron cools is proportional to the difference between
the temperature of the iron and the
-
37
temperature of the environment, how many minutes must the
blacksmith wait until he can handle the iron? It is given that
24=envT and that .2660 =T To determine the value of k, one may use
the condition that when 235,1 == Tt :
1370796.0242211ln)24266()24235()()( )1(0
=== keeTTTT kktenvenv
It can now be determined at what time the iron will cool to the
desired temperature: ( ) .35918.11
1370796.0242/51ln)24266()2475()()( )1370796.0(0 ==
tktenvenv eeTTTT
Therefore, the blacksmith must wait approximately 11 minutes
before handling the iron.
Motion with Air Resistance
According to legend, the famous astronomer Galileo Galilei once
conducted an experiment in which he compared the rate at which two
balls fell from the Leaning Tower of Pisa. One was made of iron and
the other was made of wood. Which one would hit the ground first?
If both started at the same height and were released with the same
initial velocity, they should both hit the ground at the same time
because they both experience the same acceleration due to gravity,
which is approximately 9.8 m/s2. Galileo discovered, however, that
the iron ball hit the ground before the wooden ball did. This is
because the force of gravity is not the only force acting on these
objects; air resistance also plays a role. However, unlike gravity,
air resistance is not a constant force, but one that is
proportional to the velocity of the object. For an object moving at
a relatively slow velocity, such as a feather through the air, the
force of air resistance, called drag (denoted here as )dF , is
directly proportional to the velocity of the object:
cvFd = , where c is a constant that depends upon the properties
of the air (or another fluid). An object moving through the air at
a relatively fast velocity, such as a speeding bullet, experiences
a drag that is directly proportional to the square of its velocity.
While a thorough discussion of vectors must wait until chapter 6,
it is important to introduce them in the context of this section.
Force is a vector quantity, meaning it has both magnitude and
direction. While the choice of a frame of reference is arbitrary,
it will be defined here in an intuitive sense: a downward direction
corresponds to a negative vector and an upward direction
corresponds to a positive vector. With this frame of reference, an
object falling through the air is acted upon by gravity (a negative
vector) and air resistance (a positive vector), and is accelerating
downward. The following free-body diagram, which is used in
classical physics to depict the forces acting upon an object
(represented here by a dot), should make this clear:
dF
gF
-
38
From this free-body diagram, one can apply Newtons Second Law of
Motion, which states that the net vector sum of the forces acting
on an object is equal to that objects mass times its velocity: maF
= . In the scenario of an object falling through the air with a
relatively slow velocity: == mamgcvF , where m is the mass of the
object and g is the acceleration due to gravity, which is
approximately 9.8 m/s2 . Recall from AP Calculus AB that
acceleration is the derivative of velocity. Thus, Newtons
Second Law of Motion is essentially a differential equation:
mgcvdtdvmma == .
While this differential equation is not tested on the AP exam,
it still makes for a good exercise in solving differential
equations by the method of separation of variables while also
providing a meaningful application. This differential equation may
be solved as follows:
=== .dtcvmgmdvdtcvmgmdvcvmgdtdvm These indefinite integrals can
be made into definite integrals to simplify the calculations.
Assume that the initial velocity is zero (at time t = 0) and that
any velocity at time t is v. Then:
===v tt ttdtcvmgmdv0 00 . The integral on the left can be solved
through u-substitution. Let u = mg cv.
Then, du = -cdv, and the integrand must be multiplied by mc in
order to yield the form
,ln/ Cuduu += meaning the outside of the integral must be
multiplied by .cm Now: [ ])ln()ln(ln
0mgcvmg
cmcvmg
cmt v ==
).1(1)(
)(ln // mctmct ec
mgvmgcv
mgcvmge
mgcvmg
mct ====
This is a function of velocity with respect to time for the
object as it falls through the air. Notice the characteristics of
this function when it is graphed:
-
39
The rate of change of velocity (the acceleration) decreases
until a constant velocity is reached. This constant velocity is
known as the terminal velocity. Note that drag may also be
proportional to the square of the objects velocity if this velocity
is of relatively large magnitude. The differential equation that
results from this scenario requires integration that yields a
certain kind of function (an inverse hyperbolic tangent, or )tanh 1
. While it is not terribly difficult to solve, it is somewhat
beyond the scope of this book.
Concluding Remarks This chapter can be considered a brief
introduction to differential equations and their power as
mathematical models of natural phenomena. Note that while this
chapter only discussed first-order differential equations and two
methods for solving them (the method of separation of variables and
Eulers Method), there are many more analytical and numerical
approaches. Hopefully this chapter also instilled in the reader a
greater appreciation for the applications of differential equations
to the physical and life sciences. Key Terms: first-order
differential equations isotopes method of separation of variables
radioactive decay Eulers Method radiometric dating step size
half-life Law of Exponential Change Newtons Law of Cooling logistic
growth drag carrying capacity free-body diagram learning curve
Newtons Second Law of Motion bounded growth terminal velocity
population ecology intrinsic rate of increase transcendental
functions reaction kinetics reaction order rate law
-
40
Chapter 3: Infinite Sequences and Series Thus far, this book has
been dedicated to derivatives and integrals, whether it be using
LHpitals Rule, integrating using new techniques, or solving
differential equations. For many students, the next two chapters
contain material that is quite puzzling and quite boring puzzling
because it seems to relate very little to other topics in AP
Calculus AB and BC, and boring because, most of the time, it
introduc