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,

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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.

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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.

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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

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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.

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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

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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:

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+=+=+=+= 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

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+= .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