Critical Concepts in AP Calculus and Activities to Promote ... · 8. Participants will understand...

Critical Concepts in AP Calculusand

Activities to Promote Understanding

April 25, 2008

i

AP Calculus BC Institute

The AP Calculus BC Summer Institute is an intensive study of the topics covered in anAP course. There will be special emphasis on the effects of calculus reform to traditionaltopics and new approaches to teaching challenging concepts. The focus of the institute is theCalculus BC syllabus. However, we will also address the Calculus AB topics included in thatsyllabus. The institute will include a review of the 2008 AP Calculus Exam, including thescoring guides and student samples.

Outline of Topics by Day

Day 1

1. Welcome and introductions; discussion of technology issues; participant experience andexpectations; tips for developing a Calculus BC program.

2. Functions and transformations; limits, end behavior and difference quotients.

3. Rates of change: average; approximate; instantaneous; tangent lines; slopes; local linearity;L’Hopital’s Rule.

4. Analysis of the derivative; new rules for sign charts; tabular representation of data.

5. Graphs of functions and their derivatives; extrema.

6. Implicit differentiation.

Day 2

1. Riemann sums and accumulated rates of change.

2. Functions defined by integrals and the fundamental theorem of calculus.

3. Motion, total distance, and displacement.

4. Average value of a function.

5. Applications of integration: volumes of solids.

Day 3

1. Differential equations; separation of variables.

2. Slope fields; Euler’s Method; calculator applications.

3. Logistic Growth and applications.

4. Connecting differential equations to slope fields.

5. Textbooks, resources, technology.

6. Participant ideas and presentations (optional).

7. Everything you always wanted to know about calculus, but were afraid to ask.

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Day 4

1. Series; Taylor polynomials; convergence.

2. A review of the grading process for AP Calculus; preparing for the AP exam.

3. Complete summary of the 2008 AP Calculus examinations.

4. Review of student samples.

5. Summary and assessments.

Objectives

1. Participants will understand the philosophy and aims of the Advanced Placement Calculusprogram.

2. Participants will know how to use technology appropriately as a teaching and learningmedium.

3. Participants will be able to write assessment items that reflect understanding of the APCalculus material.

4. Participants will understand the content of the AP Calculus curriculum.

5. Participants will learn how to use activities to promote understanding of AP Calculustopics.

6. Participants will understand how to start, maintain, and improve Advanced Placementprograms, particularly in calculus.

7. Participants will understand the methodology of assessment in the Advanced PlacementExamination.

8. Participants will understand how to prepare students to be successful on the AP CalculusExamination.

Content

1. Nature of the AP Calculus Program, including recent changes in the syllabus and philos-ophy. Building, maintaining, and expanding an AP Calculus program.

2. Limits, continuity, the derivative, and differentiability.

3. Techniques of differentiation and applications of differentiation.

4. Analyzing functions using derivatives and graphs.

5. Implicit differentiation.

6. Riemann sums and accumulated rates of change.

7. Fundamental theorem of calculus; the integral; functions defined by integrals.

8. Applications of integration: area, volume, motion, quantity from rate, average value.

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9. Differential equations: solution by separation of variables.

10. Slope fields and Euler’s Method.

11. Logistic growth, series, Taylor’s Formula, convergence.

12. The AP Calculus Examination grading process.

Evaluation

Participants enrolled in the AP Calculus Summer Institute program for college credit will berequired to do the following.

1. Actively participate in discussion and group work.

2. Complete all daily assignments.

3. Prepare a scoring rubric for the 2008 AP Examination.

4. Make a brief presentation to the class concerning an activity, idea, or teaching technique.

5. Write assessment questions that reflect the nature of the AP program.

6. Complete a two page paper that details a lesson that could be used to teach a specificconcept in the AP Calculus syllabus, including a sample problem, activity, and assignment.

CHAPTER 1

Limits, Continuity, End Behavior, and

Aspects of the Difference Quotient

1.1 Background

1. In many classes, the topic of limits has been deemphasized.It is no longer a central idea of calculus, but a background issue.

2. However, an understanding of limits remains a critical idea in dealing with the notion ofchange - a critical part of calculus.

3. Calculus reform: we examine limits differently, use technology extensively.

Algebraic computations are deemphasized.Some limit calculation techniques still important.

4. Most important limit concept: If a function f has limit L, then we can get as close to Las we want by manipulating x.

Many investigations with transcendental functions that enable students to explore anddiscover this concept.

5. Need to know limits in order to understand continuity and differentiability.

6. The concept of a limit is hard to teach.

(a) It is a deep idea, but examples tend to be trivial: limx→1

(3x− 5) = −2

(b) Consider examples in which the behavior of a function is not evident from the graph.

7. Limits and continuity: consider graphs of functions without giving explicit rules.

1

2 Chapter 1 Limits, Continuity, End Behavior, and Aspects of the Difference Quotient

1.2 Explorations Involving Limits

Example 1.2.1 Consider limx→2

(3x− 1)

(a) Evaluate this limit.

(b) Is the function f(x) = 3x− 1 defined at x = 2?

(c) Is f continuous at x = 2?


x2 − 1

x− 1

(a) Evaluate this limit.

(b) Is the function f(x) =x2 − 1

x− 1defined at x = 1?

(c) Is f continuous at x = 1?

1.2 Explorations Involving Limits 3


3x − 1

x

(a) Carefully sketch a graph of this function using ZDecimal.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

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x

y

(b) Zoom in on the graph where x = 0. Approximate the limit to three decimal places.

(c) Do you recognize this number? Can you write it analytically? (Hint: Think logarithm.)

Example 1.2.4 Consider limx→∞

3

2 + e−x.

(a) Carefully sketch a graph of this function using ZDecimal. Change the window as neededto approximate this limit.

(b) Can you find the more general limit limx→∞

c

k + e−x



(1 + x)1/x.

(a) Evaluate this limit to three decimal places.

(b) Do you recognize this number? Can you write this number analytically?


5000x2

sin(x) + 5000x2

(a) Carefully sketch a graph of this function using ZDecimal. Find an approximation for thislimit.

(b) Try to narrow the window using ZBox. Find an approximation for this limit using thenew window. Is your answer the same as in part (a)?

(c) Continue to narrow the window. Revise your approximation as necessary.


Example 1.2.7 Consider the expression limx→∞

f(x) = L.

We can make f(x) arbitrary close to l by choosing values for x that are big enough.

If f(x) =3x− 1

4x+ 1and lim

x→∞

f(x) =3

4,

how big must x be so that f(x) is within 0.0001 of 3/4? Use your calculator to find an answer.

Example 1.2.8 Evaluate each limit, if it exists. Use your calculator as necessary.

(a) limx→0

sin(

1

x

)

(b) limx→0

x sin(

1

x

)

(c) limx→0

x2 sin(

1

x

)


Challenge: About 2500 years ago, Archimedes considered the area of a circle by using apolygon with an increasing number of sides. This is one of the first limit problems in history.Consider a regular n-gon, inscribed in a unit circle. It consists of n congruent triangles asshown. Using trigonometry, write an expression, in terms of n, for the total area of the n-gon.Find the limit as n → ∞. You should recognize the answer.


Example 1.2.9 The graphs of the functions f and g are given below. (Calculus Problemsfor a New Centruy, Volume 2, MAA.)

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

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y

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•

•

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��

��

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

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y

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Graph of y = f(x) Graph of y = g(x)

Find each limit, if it exists.

(a) limx→−1

f(x) (b) limx→1

f(x)

(c) limx→−1

g(x) (d) limx→1

g(x)

(e) limx→1

[f(x) + g(x)] (f) limx→0

[2f(x) + 3g(x)]

(g) limx→1

[f(x)g(x)] (h) limx→2

[f(x)g(x)]

(i) limx→0

f(x)

g(x)(j) lim

x→0

g(x)

f(x)

(k) limx→−2

g(f(x)) (l) limx→−1

f(g(x))

(m) limx→−1−

f(x) (n) limx→1+

f(x)

(o) limx→−1−

g(x) (p) limx→−1+

g(x)

(q) limx→0−

f(x+ 2) (r) limx→−1−

f(x2)


Example 1.2.10 Evaluate limx→0

3√1 + cx− 1

xwhere c is a constant.

Example 1.2.11 Evaluate limx→0

|2x− 1| − |2x+ 1|x

CHAPTER 2

Rates of Change: Average, Approximate,

Instantaneous

2.1 Background

1. Average rate of change, a difference quotient:f(b)− f(a)

b− a

Instantaneous rate of change: consider a difference quotient over smaller and smaller in-tervals.

2. Tangent lines and local linearity:

(a) If a function is differentiable on an interval, the graph looks like a line, if we choose asufficiently small interval.

(b) At a point of tangency, f(x) and the tangent line L(x), share two properties:f(c) = L(c) and f ′(c) = L′(c).

3. Estimating rates of change using graphs and tables. For positive h, consider

Right difference quotient Left difference quotient Symmetric difference quotient

f(x+ h)− f(x)

h

f(x)− f(x− h)

h

f(x+ h)− f(x− h)

2h

4. lim∆x→0

∆y

∆x=

dy

dx:

The secant line becomes indistinguishable from the tangent line for small ∆x.

Zoom in on a curve at a point until it appears linear.Zoom in on a curve and its tangent line until they blend together.

9

10 Chapter 2 Rates of Change: Average, Approximate, Instantaneous

Consider the following example.

-2 -1 1 2 3

-2

-1

1

2

3

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x

y

0.5 0.75 1 1.250

0.5

1

1.5

PSfrag replacementsx

y

Note:

1. L’Hopital’s Rule demonstration: a combination of limits and local linearity.

2. Other activities: rate functions graphically, local linearity.

2.2 Constructing and Interpreting the Graph of a Rate Function 11

2.2 Constructing and Interpreting the Graph of a RateFunction

Example 2.2.1 Water is pured into several containers of different shapes at a constant rate(constant volume per unit of time) so that the depth of the water, h, changes with the increasein volume of water. For each of the following containers:

(a) Sketch a graph of the depth of the water h as a function of time.

(b) Sketch a graph of the rate of change in the depth of the water as a function of time.

t

h

t

dh/dt

t

h

t

dh/dt


t

h

t

dh/dt

t

h

t

dh/dt

Example 2.2.2 Suppose a bathtub is full of water, and the drain is opened. As the bathtubstarts to empty, the water exits fastest, and then slows as the tub empties.

(a) Carefully sketch a graph of the volume of water as a function of time.

(b) Carefully sketch the rate of change in the volume of water as a function of time.

t

Vt

dV/dt

2.2 Constructing and Interpreting the Graph of a Rate Function 13

Example 2.2.3 When a model rocket is launched, the booster burns for a short time tosend the rocket upward. It continues upward for a period of time and, when it approachesthe ground on its return, a parachute opens to soften the landing. A graph of the rocket’svelocity is given below. (Source: Demana, Waits, Finney, Thomas)

1 2 3 4 5 6 7 8 9 10 11 12

-50-40-30-20-10

0102030405060

PSfrag replacements

t

v

(a) How fast was the rocket climbing when the engine stopped?

(b) For how many seconds did the engine burn?

(c) At what time did the rocket hit its highest point?

(d) When did the parachute open? How fast was the rocket falling?

(e) How long did the rocket fall before the chute opened?

(f) When was the acceleration greatest? When was it constant?


2.3 Tangent Line Approximations and Local Linearity

Example 2.3.1 Sketch the following functions on your graphing calculator in the window[−1, 1] × [−1, 1]: Y1(x) = sin(x), Y2(x) = x. The graphs appear very similar near x = 0.Consider the same two graphs in the window [−0.1, 0.1]× [−0.1, 0.1].

[−1, 1]× [−1, 1] [−0.1, 0.1]× [−0.1, 0.1]

For x sufficiently small, the line y = x approximates y = sin x, or x ≈ sin x.

In general, we can define the tangent line approximation, L(x), to f at x = a:

L(x) = f(a) + f ′(a)(x− a) or f(x) ≈ f(a) + f ′(a)(x− a).

The Error Function

1. The tangent line approximation is only useful if the error in the approximation is small.

Except at the point of tangency, there will be some error in the approximation.

2. Error function: E(x) = f(x)− L(x) = f(x)− f(a)− f ′(a)(x− a)

3. E(a) = 0 E(a+ h) = f(a+ h)− f(a)− f ′(a)h

4. Note:

(a) As h → 0, E → 0.

(b) The error depends on f ′(a) and h.

A graph with a steep slope: larger errors near a, even when h is small.

2.3 Tangent Line Approximations and Local Linearity 15

Error function illustration:PSfrag replacements

x

y

a a+ h

f(a)

f(a+ h)

E(x)

}

f ′(a)h

Measuring and Controlling Error in a Linear Approximation

Use y = x as a linear approximation to y = sin x. Suppose we want the error to be no morethan 0.01.

Consider the following graphs:

-0.75 -0.5 -0.25 0.25 0.5 0.75

-0.15

-0.1

-0.05

0.05

0.1

0.15

PSfrag replacements

x

y

-0.75 -0.5 -0.25 0.25 0.5 0.75

0.005

0.01

0.015

PSfrag replacementsx

y

y = E(x) = x− sin x y =| E(x) |, y = 0.01

Find the bounds for x such that |E(x)| ≤ 0.01.


Example 2.3.2 Let f(x) = x3 − x.

(a) Find a linear approximation to f(x) near x = 1.

(b) For what interval about x = 1 is the linear approximation within 0.1 units?


Limits and the Error Function

limx→a

E(x) = 0 and

limx→a

E(x)

x− a= lim

x→a

f(x)− f(a)− f ′(a)(x− a)

x− a= lim

x→a

[

f(x)− f(a)

x− a− f ′(a)

]

= f ′(a)− f ′(a) = 0

Interpretation

If E(x) is a polynomial, then E(x) always has at least a double root at x = a.

E(x) is divisible by x− a (E(a) = 0). Let E(x)/(x− a) = Q(x). Q(a) = 0.

Therefore, Q(x) is divisible by x− a.

Example 2.3.3 For each function and the value of a:

(a) Find the linear approximation to f at x = a.Carefully sketch a graph of y = f(x) and y = L(x) in a symmetric window about x = a.

(b) Find the error function, E(x), and an interval about a such that |E(x)| < 0.1.

(c) Verify that limx→a

E(x)

x− a= 0

(a) f(x) =√x; a = 4.


(b) f(x) = x + sin x; a = 0.

(c) f(x) = 2x2 + 1; a = 1


Example 2.3.4 This activity involves your knowledge of the Fibonacci sequence. Recall, aFibonacci sequence has the following properties: given a1 and a2, ai = ai−1 + ai−2.

When the signal is given, write down the Fibonacci sequence beginning with a1 = 1 anda2 = 1. At each 20-second interval, mark your point in the calculations. Certainly, as thenumbers get larger, it will take more time, and you may need to use some scratch paper. Thisactivity will continue for 3 minutes.

When time has expired, you will be asked to summarize your data in the table below, showingthe cumulative number of Fibonacci numbers listed after each 20-second interval. We will thenplot the points, consider a smooth curve through these points, and analyze these results.

i ai Time i ai Time

1 1 16

2 1 17

3 18

4 19

5 20

6 21

7 22

8 23

9 24

10 25

11 26

12 27

13 28

14 29

15 30


Data Summary:

Interval (sec.) (x 20 40 60 80 100 120 140 160 180

Number of Fibs. (y)

Construct a scatter plot of your data on the axes below. Note that the point (0, 0) is thestarting point on your graph.

Draw a smooth curve through the data points.

0 20 40 60 80 100 120 140 160 1800

10

20

30

40

50

60

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x

y

(a) Find the average rate at which you computed Fibonacci numbers over the entire intervalof 180 seconds. Give your answer in Fibonaccis per second.

(b) Find an estimate of the instantaneous rate at which you wrote Fibonacci numbers at the40 second mark in three ways.

(i) Find the average rate on the interval [20, 40].

(ii) Find the average rate on the interval [40, 60].

(iii) Find the average rate on the interval [20, 60].


(c) Of the methods you used in part (b), which do you believe is the most accurate? Why?

(d) Assume that your data fits a curve of the form f(x) = k√x, where f is the number of

Fibonacci numbers and x is the number of seconds. Use any observation in your data setto obtain a value for k.

(e) Enter your data set in your calculator using the lists L1 and L2. Enter the function fas Y1. Graph y = f(x) and a scatter plot in the same viewing window. Use ZoomStat.Comment on your results.

(f) Find an equation of the tangent line to your function f at the point x = 140 seconds.Use the tangent line to estimate how many Fibonacci numbers you wrote at x = 160seconds. Find the error in your estimate.

(g) A curve of different form may provide a better fit to your data. Go to STAT; CALC andexperiment with other regression forms until you find one that fits well. Consider aquadratic, cubic, quartic, linear, logarithmic, and exponential regression.


L’Hopital’s Rule - A Visual Investigation

L’Hopital’s Rule provides a method for evaluating limits that are indeterminate, even whenthe expression cannot be simplified by algebraic means.

For example: limx→1

sin(πx)

x2 − 1

Direct substitution yields 0/0, an indeterminate form.

There is no algebraic technique to simplify and evaluate this limit.

L’Hopital’s Rule

Suppose f and g are differentiable and g′(x) 6= 0 on an open interval I that contains a(except possibly at a). Suppose that

limx→a

f(x) = 0 and limx→a

g(x) = 0

or that limx→a

f(x) = ±∞ and limx→a

g(x) = ±∞

Then

limx→a

f(x)

g(x)= lim

x→a

f ′(x)

g′(x)

if the limit on the right side exists (or is ∞ or −∞).

Note:

1. The limit of a quotient of functions is equal to the limit of the quotient of their derivatives.The conditions must be satisfied.

2. Valid for one-sided limits and for limits at infinity.

3. Proof: try the special case where f(a) = g(a) = 0.

4. May be used repeatedly, provided the conditions are met at each step.

5. The conclusion may be that the limit does not exist.


Geometric Interpretation of L’Hopital’s Rule

For values of x close to a, f and g are approximated by their tangent lines (f(a) = g(a) = 00:

f(x): y = f ′(a)(x− a) g(x): y = g′(a)(x− a)

Replace f and g by their tangent lines:

limx→a

f(x)

g(x)= lim

x→a

f ′(a)(x− a)

g′(a)(x− a)=

f ′(a)

g′(a)

Illustration:

0.5 1. 1.5

-0.5

0.

0.5

PSfrag replacements

x

y

0.5 1. 1.5

-0.5

0.

0.5

PSfrag replacements

x

y

Example 2.3.5 Use L’Hopital’s Rule to find the limit, if it exists.

(a) limx→1

2x− 2

1− x2


(b) limx→1

5x2 − 1

3x2 + 2x+ 2

(c) limx→0

sin(x2)

x

(d) limx→0

2x − 1

x


Example 2.3.6 Suppose f ′′ is continuous. Find limh→0

f(x+ h)− 2f(x) + f(x− h)

h2

Note: L’Hopital’s Rule allows us to establish which kinds of functions dominate, or growfaster, in a quotient as x → ∞. Consider the following table:

Slowest Fastest

sin x, cos x ln x xn, n > 0 ax, a > 1

Example 2.3.7 Use the principle of dominance to find each limit, if it exists. Use L’Hopital’sRule to verify your conclusion.

(a) limx→∞

x5

2x

(b) limx→0

ln x√x


(c) limx→∞

ln x

sin x

(d) limx→∞

x10

ex

(e) limx→∞

(0.9)x

x

CHAPTER 3

Analysis of the Derivative and

Applications to Local Extrema and

Points of Inflection

3.1 Background

1. One of the most important concepts in differential calculus: interpretation of the graph ofa function’s derivative (predates reform).

2. Lots of these problems on the AP Calculus exam.

3. Consider the graph of a derivative and a sign chart, simultaneously.

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

PSfrag replacements

xy

x

y

0 1−2

− 0 + 0 − 0 + f ′(x)

f(x)

These two graphs convey the same information about existence and type of extrema.

The two-dimensional graph gives us more information about the slope of the tangent lineat various points as well as concavity and points of inflection.

4. Student work on analysis of extrema and points of inflection often relies on two false

propositions:

(a) All critical points indicate the existence of extrema.

(b) Only zeros of f(x) result in critical points.

27

28 Chapter 3 Analysis of the Derivative and Applications to Local Extrema and Points of Inflection

Example 3.1.1 For each graph of y = f ′(x), sketch a possible graph of y = f(x) andy = f ′′(x).

y = f(x) y = f ′(x) y = f ′′(x)

1.

PSfrag replacementsxy



2.




3.




4.




5.




3.1 Background 29

y = f(x) y = f ′(x) y = f ′′(x)

6.




7.




8.




9.




10.





3.2 Explorations Involving Extrema of Functions

Example 3.2.1 The values of a differentiable function f for some specific values of x aregiven in the following table.

x -3 -2 -1 0 1 2 3 4 5

f(x) -4 3 5 2 -3 0 6 3 0

(a) Where would you expect f(x) to have a local extremum? Indicate whether each is a localmaximum or a local minimum.

(b) One way to estimate the value of f ′(a) is to evaluate the difference quotient:

f(a+ 1)− f(a− 1)

(a+ 1)− (a− 1)=

f(a+ 1)− f(a− 1)

2

Use the formula to complete the following table.

x -2 -1 0 1 2 3 4

f(x) 4.5

(c) Carefully sketch a graph of y = f(x) and y = f ′(x) on the Axes below. What conclusionscan you make about f ′(x) where f has a local extreme value?

PSfrag replacements

xy

3.2 Explorations Involving Extrema of Functions 31

Example 3.2.2 A graph of y = f ′(x) is given below. Use this graph to answer the followingquestions.

-4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12

-8

-6

-4

-2

2

4

6

8

PSfrag replacements

xy

x

y

(a) At which value(s) of x does f have a relative maximum?

(b) At which value(s) of x does f have a relative minimum?

(c) On which interval(s) is f increasing?

(d) On which interval(s) is f decreasing?

(e) Suppose f(x) represents distance traveled in x hours. Interpret the points where f ′(x)changes direction.

(f) Draw a possible graph of y = f(x) on the axes above.


Example 3.2.3 The figure below is a graph of y = f ′(x), the derivative of f . The functionf has domain [−4, 4], f ′(−x) = f(x), and f ′(−3) = f(3) = 0.

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

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xy

x

y

(a) For what value(s) of x does f have a relative minimum? Justify your answer.

(b) For what value(s) of x does f have a relative maximum? Justify your answer.

(c) For what value(s) of x is f concave upward? Justify your answer.

(d) For what value(s) of x does f have a point of inflection? Justify your answer.

(e) Carefully sketch a possible graph of y = f(x), given f(0) = 0.


Example 3.2.4 Consider the function v(t) = t2 − e5t, where v is the velocity of an objectmoving along a horizontal line, measured in meters per second.

(a) Carefully sketch v(t) in the window [−2, 10]× [−5, 20].

-1 1 2 3 4 5 6 7 8 9

-4-2

2468

1012141618

PSfrag replacements

xyt

v

(b) On which interval(s) is the object moving to the right? To the left?

(c) At which time in the interval [−2, 10] has the object moved furthest to the right? Justifyyour answer.


Example 3.2.5 Consider the function f(x) = x + sin(kx), where k is a positive number.

(a) Carefully sketch a graph of y = f9x) for k = 0.5, k = 1, and k = 2 in a standard viewingwindow (ZStandard).

k = 0.5 k = 1 k = 2

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8


-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8


-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8


(b) How many critical points do there appear to be for k = 0.5? k = 1? k = 2?

(c) Use nDeriv to sketch a graph of y = f ′(x) for each value of k. How many local extremaare there for k = 0.5? k = 1? k = 2?

(d) Find f ′ analytically. What must be true of k if f(x) is to have any local extrema?


Example 3.2.6 This problem involves curve fitting or the method of least squares. Supposewe want to form an open box from a rectangular sheet of cardboard that measures 8 inchesby 12 inches, by cutting a square from each corner and folding up the sides. For example, ifwe cut a 1 inch square from each corner, the dimensions of the box are 6× 10× 1, and thevolume is 60 in3.

(a) Let x be the length of the square cut from each corner, and let v be the volume of theresulting box. Complete the following table.

x 0.5 1 1.5 2 2.5

v 60

(b) Enter this data in two lists, L1 and L2.

(c) The volume of the box requires three dimensions: length, width, and height. What is thedegree of the polynomial that represents the volume of this box?

(d) Use the calculator functions under STAT; CALC to find the best fit for your data.

(e) Construct a graph of your best fit function and a scatter plot on the same coordinateaxes.

(f) Use your best fit function to estimate the maximum volume of the box.


(g) Write an analytical formula for the volume of the box in terms of the variable x. Findv′(x) and find the maximum volume of the box.

CHAPTER 4

Implicit Differentiation and Its Meaning

4.1 Background

1. A paradox: students do this well (it is purely analytical), but understand little.

2. Consider an example to understand why implicit differentiation is practical and feasible.

x2 + y2 = 4 a circle.

Can be written using two functions: f1(x) =√4− x2, f2(x) = −

√4− x2

Let y = f(x). We can rewrite the original equation: x2 + [f(x)]2 = 4

Consider three derivatives:

x2 + [f(x)]2 = 4 f1(x) =√4− x2 f2(x) = −

√4− x2

2x+ 2[f(x)]f ′(x) = 0 f1(x) = (4− x2)1/2 f2(x) = −(4− x2)1/2

f ′(x) =−x

f(x)=

−x

yf ′

1(x) =1

2(4− x2)−1/2(2x) f ′

2(x) = −1

2(4− x2)−1/2(−2x)

f ′

1(x) =−x

(4− x2)1/2f ′

2(x) +x

(4− x2)1/2

Since f1(x) =√4− x2 and f2(x) = −

√4− x2,

f ′

1(x) =−x

yand f ′

2(x) =−x

y

Implicit differentiation: compactly states the derivative of all functions embedded in theequation.

37

38 Chapter 4 Implicit Differentiation and Its Meaning

Illustration: the circle with two tangent lines.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

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y

m > 0 m < 0

3. Analyzing implicit derivatives:

Consider finding a slope for a given point (x, y), or finding a point on the curve where theslope is zero or undefined.

The function and its derivative are inter-substitutible.

Considerdy

dx=

x− 2y

y + 2x

x = 2y can be used in the original equation to find points where the slope is zero.

Implicit differentiation: differentiate both sides of the relation with respect to x and thensolve the resulting equation for y′.

Challenge: Why can you take the derivative of both sides and maintain the equality?

4.1 Background 39

Example 4.1.1 Consider the function y =x

x2 + 1

(a) Find the derivative, y′, by using the quotient rule.

(b) Rewrite the original function by multiplying both sides of the equation by x2 + 1.

(c) Differentiate implicitly.

(d) Replace y withx

x2 + 1and simplify.

How does this compare with your answer in part (a)?


(e) Using your answer from part (a) (the explicit derivative), find the points on the curvewhere the slope is 0.

(f) Using the answer obtained via implicit differentiation, find the points on the curve wherethe slope is 0.

(g) Carefully sketch a graph of y = f(x), and indicate the points on the curve where theslope is 0.

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-0.5

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4.1 Background 41

Example 4.1.2 Consider the equation x3 + y3 = 3xy.

Note: Also given parametrically: x(t) =3t

1 + t3y(t) =

3t2

1 + t3

-2 -1 1 2

-2

-1

1

2

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x

y

(a) Find an equation of the tangent line to the point when t = p.

(b) Find the equation of the asymptote.

(c) Try plotting this one on your calculator.


Example 4.1.3 Consider the curve defined by the equation√x+

√y =

√c

Find the sum of the x- and y-intercepts of any tangent line to this curve.

CHAPTER 5

Riemann Sums and Accumulated Rates

of Change

5.1 Background

1. Calculus reform has profoundly affected the way in which we teach integration.

Techniques of integration =⇒ conceptual understanding of integration.

2. Free response questions: students need to understand Riemann sums, what they mean,how to write them.

A regular partition:n∑

i=1

f(xi)∆x

The definite integral of f from a to b:∫ b

af(x) dx = lim

∆x→0

n∑

i=1

f(xi)∆x

3. Introduce the connection between Riemann sums and area: the idea of motion.

4. Using technology, we can explore multiple techniques for approximating a definite integral.

AB and BC Exam: how to apply the Trapezoidal Rule.

Simpson’s Rule: powerful, let technology do the work.

5. Applications of definite integrals: entirely new source of problems involving accumulatedrates and geometric products.

43

44 Chapter 5 Riemann Sums and Accumulated Rates of Change

5.2 Accumulated Rates of Change

Example 5.2.1 In a bicycle race, one of the contestants travels with increasing speed overan interval of 12 seconds. The speed of the racer is recorded at 2 second intervals and issummarized in the following table.

Time (sec) 0 2 4 6 8 10 12

Speed (ft/sec) 10 14 20 27 32 34 35

(a) Plot the racer’s speed against time.

0 2 4 6 8 10 120

5

10

15

20

25

30

35

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s

(b) Consider the time interval [0, 2].

(i) What is the least distance (lower bound) the racer could have traveled?

(ii) What is the greatest distance (upper bound) the racer could have traveled?

(c) For each two second interval, compute the lower and upper bounds on distance traveled.

Interval [0, 2] [2, 4] [4, 6] [6, 8] [8, 10] [10, 12] Total

Lower

Upper

(d) Based on this table, what can you conclude about the distance traveled over the entire12 second interval?

5.2 Accumulated Rates of Change 45

(e) Using the data, construct rectangles that represent the upper and lower estimates of thedistance traveled.

0 2 4 6 8 10 120

5

10

15

20

25

30

35

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s

(f) What would you need to know in order to improve your estimate of the distance traveledover the entire time interval?

(g) Draw a smooth curve through the data points. How is the total distance related to thiscurve?

(h) In general, we can find the average speed: total distance / total time.Use your results to obtain upper and lower estimates for the average speed.

(i) Let v(t) be a function that describes the speed of the racer. Write an expression for thefollowing that involves a Riemann sum.

(i) Total distance

(ii) Average spped

46 Chapter 5 Riemann Sums and Accumulated Rates of Change

(j) The table below summarizes the speed of a second racer on the same 12 second interval.

Time (sec) 0 2 4 6 8 10 12

Speed (ft/sec) 12 15 22 28 31 33 37

(i) Plot this data on the graph in part (a). Draw a smooth curve through these datapoints.

(ii) Can you determine from the graph which of the racers was ahead after the 12 secondinterval? Why or why not?

(iii) Compute a lower and upper bound for the distance traveled by the second racer. Dothese results confirm your answer to (ii)? Why or why not?

(iv) Suppose a third racer traveled at a constant speed for the entire 12 second interval.What speed would the racer have to maintain to ensure being ahead after the 12 seconds?

5.2 Accumulated Rates of Change 47

Example 5.2.2 Evaluate∫ 2

1x3 dx by using the partition

x0 = 1, x1 = 21/n, x2 = 22/n, . . ., xi = 2i/n, . . ., xn = 2n/n = 2. Let x∗

i = xi.

Example 5.2.3 Find∫ 2

1

1

x2dx

Use a regular partition and let x∗

i =√xi−1xi (the geometric mean).

CHAPTER 6

Functions Defined by Integrals

6.1 Background

1. Many problems on the AP Calculus exam involve functions define by integrals.

These problems test many concepts: the integral as an accumulator, the Fundamental The-orem of calculus, properties of definite integrals, analysis of a function from its derivative,function domains, the chain rule for differentiation.

These types of problems rarely appear in applied settings.

2. 1998 Exam: AB and BC students were required to work with functions defined by integralsin which the upper limit was (possibly) a function of x.

Complications: Must consider the chain rule when finding the derivative of an integral.May have to deal with domain issues (worksheets to follow).

3. Two basic problems involving functions defined by integrals.

A function definition involves an integral in which the integrand is given graphically.Questions involve both the analytical properties of the curve and the geometric propertiesof the curve related to area and to the integral as an accumulator.

A function definition involves an integral over a variable interval in which the integrandis a known function. Usually, the given function has no closed form antiderivative. Thestudent must use the Fundamental Theorem of calculus to answer questions about ex-trema, intervals on which the function is increasing or decreasing, concavity, and pointsof inflection.

4. A difficulty part: students must write verbal arguments that are logical and support as-sertions.

Common errors:

Using the phrase the function when there are 2 or 3 functions in the problem.

Use of the generic f to denote any function, even when it is named g in the problem.

The difference between a function and its integral.

The use of geometry, instead of an analytical approach.

49

50 Chapter 6 Functions Defined by Integrals

6.2 Problems

Each of the following problems involves functions that are defined by integrals.

Use the Fundamental Theorem of calculus and the concept of the integral as an area accu-mulator to answer each questions.

Express justifications in your own words or by using appropriate notation.

Example 6.2.1 Let F (x) =∫ x

0f(t) dt for 0 ≤ x ≤ 3, where the graph of y = f(t) is shown

below.

1 2 3

1

2

3

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x

y

(a) Find F (0).

(b) At what value of x does F have an absolute maximum? Explain.

(c) At which value of x does F have an absolute minimum? Explain.

6.2 Problems 51

Example 6.2.2 Let F (x) =∫ x

0sin(t2) dt for 0 ≤ x ≤ 3.

(a) On what interval(s) is F increasing? Justify your answer.

(b) Find the x-coordinate of all points of inflection of F in the interval (0, 3).

(c) Carefully sketch a graph of y = F (x).

1 2 3-0.25

0.25

0.5

0.75

1.

1.25

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52 Chapter 6 Functions Defined by Integrals

Example 6.2.3 Let h(x) =∫ 2x

1f(t) dt where the graph of y = f(t) is piecewise linear and

shown below.

1 2 3 4 5 6

-2

-1

0

1

2

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x

y

(a) Find the domain of the function h.

(b) Find the value(s) of x for which h(x) = 0. Justify your answer.

(c) Find the value of h(2).

(d) At which value(s) of x does h have a local minimum? Justify your answer.

(e) At which value(s) of x does h have a local maximum? Justify your answer.

(f) At which value(s) of x does h have a point of inflection? Justify your answer.

CHAPTER 7

The Average Value of a Function

7.1 Background

1. A minor topic perhaps, but pervasive on the exam.

Trouble for students: so many contexts in which to use the word average in calculus andother areas of mathematics.

2. Average value is often confused with:

(a) The average rate of change of a function.

(b) The numerical average (mean), obtained by taking two endpoints of an interval andaveraging the function values.

3. Here are some ways to explain and clarify this concept.

(a) Average always implies accumulation. When we average a set of test scores, we firstaggregate, and then divide by the total number of scores. Implicit in average, therefore,is the concept of antidifferentiation.

(b) Rate of change always implies a difference quotient, that is, a slope. Implicit in rateof change, therefore, is the concept of differentiation.

(c) When we say average rate of change, we are referring to the aggregate of the slopes,or the integral of the derivative, two inverse operations. Therefore, the operation isapplied to the function itself.

(d) When we say average value, we are referring to integration, with the given function asthe integrand.

4. Suggestion: introduce average value when first discussing Riemann sums.

53

54 Chapter 7 The Average Value of a Function

5. There are many applications in which the meaning of average value is made clear.

(a) To compute average velocity, we use the distance function. We divide the total distancetraveled by the elapsed time. We know distance is the integral of velocity, and elapsedtime is the length of the interval [a, b], or b− a.

(b) Estimate the average temperature over a 24 hour period. Take a number of temperaturereadings, multiply by the length of each interval, sum, and divide by the length of theinterval. (Activity to follow.)

7.2 Average Value Experiment

1. When a function grows at a constant rate, as a linear function does, computing an averagevalue for a given interval is straightforward.

2. The average value is at the midpoint of the interval and is midway between the left andright endpoint function values as well.

3. For example: if f(x) = 2x + 3 on [0, 10].

The average value is f(5) =f(0) + f(10)

2= 13

Consider the figure below, and finding the average value via the calculus rule.

0 2 4 6 8 100

5

10

15

20

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4. When a function grows at a variable rate, computing the average value means, theoretically,averaging an infinite number of values.

7.2 Average Value Experiment 55

Example 7.2.1 Suppose $1000 is invested at 5% annual interest compounded continuously.The value of the investment after x years is given by f(x) = 1000e0.05x. This activity involvesfinding the average value of the investment over the first 50 years.

(a) Carefully sketch a graph of y = f(x) on the axes below.

0 5 10 15 20 25 30 35 40 450

2000

4000

6000

8000

10 000

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x

y

(b) To estimate the average, compute the value of the investment at the beginning of eachten year period, and then average the results.

Complete the following table, and compute this rough estimate of the average.

x 0 10 20 30 40 50

f(x)

(c) To improve this estimate, average 50 years, starting with x = 0 and ending with x = 49(or beginning with x = 1 and ending with x = 50). Use your calculator for this one!


(d) Someone might suggest the average value is [f(0) + f(50)]/2. Explain why this methodwould be incorrect.

(e) Using the axes below, carefully sketch the rectangular regions used to obtain the estimatein part (b).

0 5 10 15 20 25 30 35 40 450

2000

4000

6000

8000

10 000

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(f) This suggests a method to obtain the true average value of the function using the limitof a Riemann sum.

Compute the average analytically and confirm your answer using the numerical integra-tion feature on your calculator.

7.3 Riemann Sums, Average Values, and Regression Curves 57

7.3 Riemann Sums, Average Values, and RegressionCurves

Example 7.3.1 Consider the following temperature data (in ◦ F) for a typical day inPhoenix, Arizona.

Time 12 AM 4 AM 8 AM 12 PM 4 PM 8 PM 12 AM

Temperature 62 68 77 85 81 72 62

Let t = 0 represent the first 12 AM and let t = 24 represent 12 AM the following day.

(a) Write a Riemann sum to represent the average temperature over the 24 hour period usingleft endpoint values. Compute this average.

(b) Enter the time and temperature data in you calculator using the lists L1 and L2. itemFind a sine regression for this data and save the function in Y1.

(c) Find the average value of the temperature using the numerical integration feature onyour calculator. How does this answer compare with part (a)?


(d) Construct a histogram with Xlist: L1 and Freq: L2.Use the following window parameters:

Xmin = 0 Xmax=24 Xscl=4 Ymin=60 Ymax=87

Graph the histogram and the regression equation.

0 4 8 12 16 2060

65

70

75

80

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(e) Use your calculator to display the right endpoint Riemann sum.

0 4 8 12 16 2060

65

70

75

80

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CHAPTER 8

Differential Equations

8.1 Background

1. Many students learn about growth and decay in pre-calculus, in advance of studying dif-ferential equations.

Sometimes hard to stop these students from going directly to a (memorized) solution,without all the intermediate steps.

2. Recent AP problems: focus on solving separable differential equations

Consider a typical growth and decay equation:dy

dt= ky

Solution steps:

Separate the variables.

1

ydy = k · dt Caution 1: Students who see k as joined to y often divide

both sides of the equation by ky. This results in a complicatedintegration on the left side and further trouble the student solvesfor y.

Integrate both sides.

ln |y| = kt+ C Caution 2: It is common for students to omit the absolute valuesymbol, since growth and decay problems rarely require thisnotation. See 1997, AB/BC 6 for an example of how dangerousthis omission can be.

Exponentiate both sides.

|y| = ekt+C = eC · ekt Caution 3: The right side of this equation is positive. Therefore,y must be within an absolute value symbol. Often overlooked.This omission could cost students a point.

Simplify.

y = Aekt where A = ±eC There is a subtle transition here. y = ±ekt+C .

59

60 Chapter 8 Differential Equations

Note:

1. Students in AP Calculus should be able to solve separation of variable differential equationsin a variety of settings.

Students must know the technique separation of variables.

2. There are no other required solution methods on the exam for differential equations.

3. BC students must also be able to solve logistic growth problems (partial fractions).

WHY DOES SEPARATION OF VARIABLES WORK?

We take the method, separation of variables, for granted. We are really solving in one differ-ential on one side and a different one on the other side. Does this really work?

Consider the differential equation g(y) dy = f(t) dt where y = h(t)

dy

dt= h′(t) and dy = h′(t) dt

Therefore: g(h(t)) · h′(t) dt = f(t) dt

Integrate both sides with respect to t:∫

g(h(t)) · h′(t) dt =∫

f(t) dt

And, G(h(t)) = F (t) + C or G(y) = F (t) + C

TO USE ABSOLUTE VALUE OR NOT?

Students struggle with absolute value. It rarely seems necessary, and there are two differentscenarios in which natural logs may arise in a differential equation solution.

Scenario 1

1

ydy = k · dt

There is no restriction on y in the original equation. Therefore, the solution should not restricty either. Use absolute value to keep the range of y unrestricted.

Scenario 2

ey = 2t+ 1

Suppose this equation arises in the context of solving a differential equation, and we need tosolve for y explicitly.

Take the natural log of both sides: y = ln(2t+ 1)

If we place absolute values symbols around (2t+1), the we will artificially extend the domain.(2t+ 1) must be positive. To preserve this, we do not use absolute value.

8.2 Solving Differential Equations by Separation of Variables 61

8.2 Solving Differential Equations by Separation ofVariables

Solve the following differential equations. In each case, evaluate the limit of the solutionfunction as t → ±∞, if it exists. Carefully sketch the solution curve in an appropriatewindow. Check you solution using a calculator program for slope fields.

Example 8.2.1dA

dt= 0.2A A(0) = 10

A(t) =

limt→∞

A(t) =

limt→−∞

A(t) =-4 -2 2 4 6 8

5

10

15

20

25

30

35

40

45

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A


Example 8.2.2dy

dt+ y = 0 y(0) = 1

y(t) =

limt→∞

y(t) =

limt→−∞

y(t) =-4 -3 -2 -1 1 2 3 4

5

10

15

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t

y


Example 8.2.3dP

dt= 100− P P (2) = 50

P (t) =

limt→∞

P (t) =

limt→−∞

P (t) =

1 2 3 4

-80

-60

-40

-20

20

40

60

80

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P


Example 8.2.4dy

dt= −2y − 6 y(1) = −10

y(t) =

limt→∞

y(t) =

limt→−∞

y(t) =

-1.5 -1. -0.5 0.5 1. 1.5

-4

-3

-2

-1

1

2

3

4

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y


Example 8.2.5dy

dt= t(4− 2y) y(0) = −2

y(t) =

limt→∞

y(t) =

limt→−∞

y(t) =

-4 -3 -2 -1 1 2 3 4

-2

-1

1

2

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y

y


Example 8.2.6dM

dt=

M

tM(1) = −3 and M(−t) = −M(t)

M(t) =

limt→∞

M(t) =

limt→−∞

M(t) =

-4 -3 -2 -1 1 2 3 4

-10

-5

5

10

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M


Example 8.2.7dy

dt=

ty + 3t

t2 + 1y(2) = 2

y(t) =

limt→∞

y(t) =

limt→−∞

y(t) =

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

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CHAPTER 9

Sample Review Problems

Example 9.0.8 Functions and Transformations The graph of a function y = f(x)is shown below. On the accompanying set of axes, carefully sketch a graph of each of thefollowing:

(a) y = |f(x)| (b) y = f(|x|) (c) y = 2f(x) (d) y = f(2x) (e) y = f(2|x|)

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

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(a) y = |f(x)| (b) y = f(|x|)

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4


x

y

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4


x

y

69

70 Chapter 9 Sample Review Problems

(c) y = 2f(x) (d) y = f(2x)

-4 -3 -2 -1 1 2 3 4

-8

-6

-4

-2

2


x

y

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4


x

y

(d) y = f(2|x|)

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4


x

y

Chapter 9 Sample Review Problems 71

Example 9.0.9 Rates of Change (Source: Ostebee and Zorn, page 107, No. 21) Supposef is a continuous function such that f(1) = −2 and f ′(x) ≤ 3 ∀x ∈ [−10, 10].

(a) Show that f(2) ≤ 1.

(b) Is f(−5) > −23? Justify your answer.

(c) Suppose g(x) = 4x− 6? Explain why f and g do not intersect for 1 < x ≤ 10.


Example 9.0.10 Rates of Change from Tabular Data (Source: Harvard Calculus, page107, Ex. 3)The table below gives the values of C(t), the concentration of a drug in mg/cm3 after tminutes in the bloodstream.

t 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

C(t) 0.84 0.89 0.94 0.98 1.00 1.00 0.97 0.90 0.79 0.63 0.41

(a) Find an estimate for C ′(0.2), the rate of change in concentration at time t = 0.2 minutes,using a left, right, and midpoint difference quotient.

(b) Construct a table of estimates for C ′(t) for 0 ≤ t ≤ 0.9 using a right-hand differencequotient with ∆t = 0.1.

t 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

C ′(t)

(c) Given that C(t) is a differentiable function, is there a time t when C ′(t) = 0? Justifyyour answer.


Example 9.0.11 The Racetrack Principle (Source: Ostebee and Zorn, page 107, No. 29)The graph of y = g′(x), the derivative of g(x), is shown below.

-1 1 2 3 4 5 6 7 8 9

-2

-1

1

2

3

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x

y

(a) Is g(2) < g(5)? Explain your answer.

(b) Rank the numbers the following numbers in increasing order. Explain your reasoning.

0, 1, g(2)− g(1), g(8)− g(7), g(9)− g(8).


Example 9.0.12 Extreme Values of a Function (Source: David Loman, University ofArizona)A graph of y = f(x) is shown below.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

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x

y

Consider a differentiable function g(x) = [f(x)]2 − 1 on the interval −3 ≤ x ≤ 4.

(a) Find all critical numbers of g.

(b) Classify each critical point in part (a) as a maximum, minimum, or neither.

(c) Find the absolute extrema of g(x) on −3 ≤ x ≤ 4.


(d) Answer questions (a)-(c) for a function h(x) if:

(i) h(x) = f(x2 − 1)

(ii) h(x) = cos(f(x))

(iii) h(x) = f(cos x)


Example 9.0.13 Implicit DifferentiationGiven the equation x2 − xy + y2 = 3.

(a) Finddy

dx

(b) Find the x- and y-coordinates of all points on the graph where the slope of the tangentline is 0.


(c) Suppose L is a line tangent to the curve with slope 1. Find the x- and y-coordinates ofall points on the curve that have tangent lines perpendicular to L.

(d) Is there a line tangent to the curve at a point where y = 3? If so, find it. If not, explainwhy not.


Example 9.0.14 Riemann Sums and AreaA region in the first quadrant is bounded by the x and y axes and a curve y = f9x). Thefollowing values of x and f(x) are known:

x 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

f(x) 6.0 5.979 5.916 5.810 5.657 5.454 5.196 4.873 4.472 3.969 3.317 2.398 0.0

(a) Use a left Riemann sum with ∆x = 0.5 to estimate the area under the curve.

(b) Use a midpoint Riemann sum with 6 subintervals to estimate the area under the curve.


(c) Construct a scatter plot of the points. Can you suggest an analytical expression forf(x)? Use your guess to compare your answers in parts (a) and (b) to the analytical, orgeometric, solution.

1 2 3 4 5 6

1

2

3

4

5

6

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x

y


Example 9.0.15 Functions Defined by IntegralsA graph of y = f(x) on the interval [−5, 5] is given below. The graph consists of line segmentsand a semi-circle.

-5 -4 -3 -2 -1 1 2 3 4 5

-4

-3

-2

-1

1

2

3

4

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x

y

Let F (x) =∫ x

0f(t) dt

(a) Find the following: F (0), F (1), F (−1), F (−3), F (3), F (5)

(b) Find the absolute minimum and absolute maximum value of F (x). Justify your answer.


(c) At which value(s) of x does F (x) have a point of inflection? Justify your answer.

(d) Using the information from parts (a)-(c), carefully sketch a graph of y = F (x) on theaxes below.

-5 -4 -3 -2 -1 1 2 3 4 5

-4

-3

-2

-1

1

2

3

4

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x

y


Example 9.0.16 Differential EquationsThe function f(x) is twice differentiable, and has the following properties:

f ′(x) = 1 +f(x)

x, f(1) = 0, f ′(1) = 1

(a) Write an equation of the tangent line to the graph of y = f(x) at x = 1.

(b) Find f ′′(x) in terms of x.

(c) Find f(x) in terms of x.


Example 9.0.17 Series and Differential EquationsSuppose an infinite geometric series, {sn}, has s1 = 1 and common ratio x, where |x| < 1.

(a) Write an expression for s, the sum of the infinite series, in terms of x.

(b) Using your answer from part (a), write the first four nonzero terms of the infinite series

for f(x) =1

1 + x2

(c) Write the first four nonzero terms of the series for F (x) =∫ x

0f(t) dt


(d) Write a closed form expression for F (x). Use this expression to evaluate limx→∞

F (x).

(e) Show that y = f(x) is a solution to the differential equationdy

dx= −2xy2 by

(1) Differentiating.

(2) Solving the differential equation.

CHAPTER 10

Slope Fields and Euler’s Method

10.1 Background

1. In 2004, the AP Calculus AB Exam tested the topic of slop fields for the first time.

Most of the content can be taught and mastered very quickly.

Flexibility: the BC Exam now has a substantial AB subcomponent (AB subscore).

2. Slope fields: allow us to visualize the family of solutions to a differential equation withoutactually solving the equation.

Samples of slopes at various points computed from the derivative formula.

The slopes are visualized as small line segments, when viewed together, often reveal apattern.

Can be done by hand, but tedious and long.

Many graphing calculator programs. Some have a program built-in.

Window selection is important.

3. Slope fields lead to Euler’s Method.

Numerical approximations to the solutions of a differential equation.

Use an initial point and the slope at that point to walk to the next point.

This repetitive process produces an estimate of the solution curve.

Calculator programs for this method also.

4. Both of these topics provide an intuitive and visual basis for the study of differentialequations.

85

86 Chapter 10 Slope Fields and Euler’s Method

10.2 Slope Fields

1. A slope filed is a graphical representation that shows the flow of tangent lines to thefamily of solutions to a differential equation.

2. The slope field often reveals information about the nature of the solution curve: polynomial,exponential, logarithmic, trigonometric, etc.

3. Constructed in the plane by substituting points, (x, y), into the differential equation tocompute the slope of the tangent line (represent as segments).

CONSTRUCTING A SLOPE FIELD

It is often helpful to create a table of values for a specific grid of values (x and y).

Example 10.2.1 Supposedy

dx=

xy

4for x ∈ [−2, 2] and y ∈ [−2, 2].

(a) Complete the grid below. The body of the table contains the computed values of dy/dx.

y value

dy/dx −2 −1 0 1 2

−2

−1

xvalue

0

1

2

10.2 Slope Fields 87

(b) Draw the slope field for the values in the grid above.

-2 -1 1 2

-2

-1

1

2

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y

(c) What kind of function do you think the solution family represents?


READING A SLOPE FIELD

1. Reading a slope field: look for clues about the behavior of the differential equation andthe family of solutions.

2. Difficult to read a slope field one segment at a time.

3. Look for trends in the slope field that indicate the relationship between x and y.

Here are some techniques.

Examine slope field segments along vertical lines. If the segments along each vertical linehave the same slope, then the differential equation does not depend on y, because, as yvaries, the slope does not change.

Examine slope field segments along horizontal lines. If the segments along each horizontalline have the same slope, then the differential equation does not depend on x.

Examine slope field segments in the first quadrant. If the segments have positive slope,then there are likely no negatives in the expression of the differential equation. If theslopes become larger as x increases, then dy/dx varies directly with x. Similarly for y.Otherwise, we can determine that the slop is inversely related to one, or both, variables.

If the slope field suggests a curve that looks familiar, check by differentiating the equationassociated with the curve to see if its slope fits the graphical data.

Note: There are certainly anomalies in the appearance of slope fields due to the way in whichthey are generated on the calculator. Small discrepancies, or inconsistencies, in the appearanceof segments can usually be discounted.


Example 10.2.2 Consider the slope field below, in the window [−2.5, 2.5]× [−2.5, 2.5]

-2 -1 1 2

-2

-1

1

2

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x

y

(a) What can you deduce from reading / interpreting the slope segments?

(b) Which of the following is most likely the differential equation?

(A)dy

dx= 0.5xy (B)

dy

dx=

x2

y(C)

dy

dx= 0.5x2


Example 10.2.3 Which of the following differential equations has the solution slope fieldpicture to the right?

(A)dy

dx= 0.5y

(B)dy

dx=

0.2x

y

(C)dy

dx= xy

(D)dy

dx= x + y

(E)dy

dx=

1

x

-2 -1 1 2

-2

-1

1

2

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x

y


(A)dy

dx= x2

(B)dy

dx=

y

x

(C)dy

dx= −y

(D)dy

dx= −x

y

(E)dy

dx= x2 + y2

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

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x

y



(A)dy

dx= x + y

(B)dy

dx= x− y

(C)dy

dx= x2

(D)dy

dx= 2y

(E)dy

dx= y/x

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

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x

y


10.3 Euler’s Method

1. The basic idea behind slope fields can be used to find numerical approximations to solutionsof differential equations.

2. Letdy

dx= f(x, y) be a differential equation in x and/or y.

Let (x0, y0) be an initial condition.

Suppose the differential equation cannot be solved by our (only) known method: separationof variables. It could even be unsolvable by any method.

Goal: estimate points on the solution curve using the given information.

Example 10.3.1 Consider the initial value problem :dy

dx= x + y y(0) = 1

dy

dx(0, 1) = 0 + 1 = 1 The solution curve has slope 1 at (0, 1).

As a first approximation to the solution, use the linear approximation L(x) = x + 1.This is the tangent line at (0, 1) as a rough approximation to the solution curve.

0.25 0.5 0.75 1.

1

2

3

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x

y

Solution curve

y = L(x)

Proceed a short distance along this tangent line, and then make a midcourse correction.

Change direction as indicated by the slope field.

10.3 Euler’s Method 93

Suppose we start out along the tangent line, and stop at x = 0.5.The horizontal distance traveled is the step size.

L(0.5) = 1.5 =⇒ y(0.5) ≈ 1.5

Take (0.5, 1.5) as the starting point for a new line segment.

dy

dx(0.5, 1.5) = 0.5 + 1.5 = 2 =⇒ y = (2(x− 0.5) + 1.5 = 2x+ 0.5 Approx for x > 0.5.

Note: A smaller step size leads to a better Euler approximation.

0.25 0.5 0.75 1.

1

2

3

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y

0.25 0.5 0.75 1.

1

2

3

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y

General Method

In words:

1. Start at the point given by the initial value.

2. Proceed in the direction indicated by the slope field.

3. Stop after a short time, consider the slope at the new location, and proceed in that direc-tion.

4. Continue to stop, and change direction accordingly.

Note: Euler’s Method does not produce the exact solution to an initial value problem, onlyan approximation. Decreasing the step size (increasing the number of midcourse corrections),leads to a better approximation.


In symbols:

As before:dy

dx= f(x, y) y(x0) = y0

1. Find approximate values for the solution at equally spaced numbers:

x0, x1 = x0 +∆x, x2 = x1 +∆x, . . . where ∆x is the step size.

2. The slope at (x0, y0) is dy/dx = f(x0, y0).

The approximate value of the solution when x = x1 is y1 = y0 +∆xf(x0, y0)

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x

y

∆x

∆x f(x0, y0)

(x1, y1)

x0 x1

Therefore:

y1 = y0 +∆x f(x0, y0)

y2 = y1 +∆x f(x1, y1)

In general: yn = yn−1 +∆x f(xn−1, yn−1) = yn−1 +∆y


Example 10.3.2 Supposedy

dx= x + y (x0, y0) = (1, 1).

Use Euler’s Method with ∆x = 0.1 to complete the following table. Give your estimatesaccurate to three decimal places.

n xn yn dy/dx ∆y n xn yn dy/dx ∆y

1 1.1 1.2 2.3 0.23 6 1.6

2 1.2 1.43 2.63 0.263 7 1.7

3 1.3 8 1.8

4 1.4 9 1.9

5 1.5 10 2.0

Example 10.3.3 Givendy

dx= 2y with initial condition (x0, y0) = (0, 1). Use Euler’s Method

to estimate the solution using ∆x = 0.2 with five steps.

n xn yn dy/dx ∆y

0 0.0

1 0.2

2 0.4

3 0.6

4 0.8

5 1.0


Example 10.3.4 Consider the differential equation in the previous example:dy

dx= 2y with

(x0, y0) = (0, 1).

(a) Solve the differential equation analytically. Use the initial condition to resolve the con-stant.

(b) Complete the following table in order to compare values of y obtained using Euler’sMethod with values of y obtained using the analytical solution.

x 0 0.2 0.4 0.6 0.8 1.0

Euler y

Actual y

Difference

(c) Plot the actual solution curve and the points your found in part (b).

0. 0.2 0.4 0.6 0.80

1

2

3

4

5

6

7

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y

0. 0.2 0.4 0.6 0.80

1

2

3

4

5

6

7

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x

y


Example 10.3.5 For each of the following situations, indicate whether the Euler’s Methodapproximation is less than, greater than, or equal to the actual solution of the differentialequation for positive values of ∆x.

(a) The solution curve is concave up.

(b) The solution curve is concave down.

(c)dy

dx= k, where k is a constant.

Example 10.3.6 In general, what can you do to improve the accuracy of an Euler’s Methodapproximation to the solution of a differential equation?


Example 10.3.7 Consider the initial-value problemdy

dx= 6x2 − 3x2y y(0) = 3

(a) Use Euler’s method to compute y(1) with step size (i) ∆x = 1 (ii) ∆x = 0.1(iii) ∆x = 0.01 (iv) ∆x = 0.001

(b) Verify that y = 2 + e−x3 is a solution to the differential equation.

(c) Find the errors in using Euler’s Method to compute y(1) with each step size. Whathappens to the error when the step size is divided by 10?

10.4 Euler’s Method Postscript 99

10.4 Euler’s Method Postscript

The Fundamental Theorem with Euler’s Method

The Problem:

Given a function f ′, find an approximation to f with the initial condition (x), y0) and evaluateit on the interval [x0, xn].

The Solution:

For small ∆x, ∆y ≈ f ′(x0)∆x

f(x1) ≈ f(x0) + f ′(x0)∆x

f(x2) ≈ f(x1) + f ′(x1)∆x = f(x0) + f ′(x0)∆x+ f ′(x1)∆x

Continue in this manner, to obtain:

f(xn) ≈ f(x0) + f ′(x0)∆x+ f ′(x1)∆x + · · ·+ f ′(xn−1)∆x

Using the left endpoint approximation (Riemann sum), the area bounded by f ′ from x0 toxn can be approximated by:

A ≈ f ′(x0)∆x+ f ′(x1)∆x+ · · ·+ f ′(xn−1)∆x

But, f(xn) ≈ f(x0) + f ′(x0)∆x + f ′(x1)∆x+ · · ·+ f ′(xn−1)∆x

Or, f(xn)− f(x0) ≈ f ′(x0)∆x + f ′(x1)∆x+ · · ·+ f ′(xn−1)∆x

By substitution, we obtain:

A ≈ f(xn)− f(x0) and lim∆x→0

A ≈ f(xn)− f(x0)

This is the Fundamental Theorem of Calculus!

CHAPTER 11

Logistic Growth

11.1 Background

1. The logistic growth model is more sophisticated and realistic than exponential growth.Can use lots of tools to explore and solve these problems: slope fields, Euler’s Method, andseparation of variables.

2. Differential equations with a logistic equation solution have been on the BC syllabus for along time.

Consider BC6, 1991: a differential equation that represented the spread of a rumor.

3. Logistic growth and the new syllabus: reflects the power of technology. We can modellogistic growth using statistical plots, and we can use logistic regression to obtain anestimate of the curve.

4. Logistic growth describes many real-world phenomena: the growth of an animal populationsubject to finite resources, learning curves, the spread of a disease, the popularity of a newproduct, popcorn popping in a microwave oven, and even approximating π from a polygon.

5. Graphically: logistic curves have an inflection point at the location of the greatest rate ofgrowth: good for slope field investigations.

6. The differential equation: quadratic in y.

101

102 Chapter 11 Logistic Growth

11.2 The Model

1. Suppose a population increases exponentially in its early stages, but levels off eventually,and approaches a carrying capacity because of limited resources.

2. Let P (t) be the size of the population at time t, and N is the carrying capacity.

The differential equation:dP

dt= kP

(

1− P

N

)

3. Some logistic growth curve facts:

Suppose an initial condition: P (0) = P0.

The solution: P (t) =N

1 + Ae−ktwhere A =

N − P0

P0

The point of inflection occurs when dP/dt achieves its greatest value, when P = N/2.

N

2=

N

1 + Ae−kt=⇒ t =

lnA

k

If P is small compared to N :dP

dt≈ kP

As P → N , P/N → 1 and dP/dt → 0.

If 0 < P < N then dP/dt > 0. If P > N then dP/dt < 0

limt→∞

P (t) = N

Solve the differential equation.

11.3 Logistic Growth Activity 103

11.3 Logistic Growth Activity

One morning, a student in your Calculus BC class hears a rumor before school. Starting withthis one person, the rumor spread hour by hour to others in the class, until all, or nearly all,of the class had heard the rumor.

Our task is to track how the rumor spread by modeling it in a simulation and then analyzingthe results. Each time the rumor is passed to a new person, the number of rumor spreaders

increases by one. At the same time, the number of new rumor hearers decreases by one.Assuming that the rumor spreads randomly, that is, each person in the class has an equalchance of hearing the rumor, we can model the process using randomly generated numberson a graphing calculator.

Some specifics:

N = the number of students in the class.

t = the number of hours since the first class member heard the rumor.

P = the number of students who have heard the rumor at time t.

Directions

1. Assign each member of the class a number from 1 to N .

2. Use int(N*rand)+1 or randInt(1,N) to generate the first rumor hearer.

3. Repeat this command. If the output is a different number, there are now two rumor hearers

that have become rumor spreaders.

4. For each additional time period, repeat the command a number of times equal to thenumber of rumor spreaders from the previous step.

5. Continue in this manner until all N or at least n− 1 have heard the rumor.

6. Use your calculator to construct a scatter plot.

t 0

P 1


The Questions

1. Assume the spread of the rumor follows a logistic model:dP

dt= kP

(

1− P

N

)

wheredP

dtis the rate at which student hear the rumor, P is the number who have heard the rumorat time t, and N is the number of students in the class.

Solve the differential equation for P in terms of t. Do not resolve the constants, yet.

2. Using the data points we generated at t = 0 and t = 3, find the value of each constant and

write the solution in the form P (t) =N

1 + Ae−kt


3. Use your solution found in part (2) to compute the value of P at t = 5. How well does thisequation value approximate the experimental (table) value at t = 5? Comment.

4. Explain the role (affect) of the constants N and A in the solution. How is each of theserelated to the original conditions of the problem?

5. Suppose at t = 0 there were two people who heard the rumor, rather than one. Explainhow this might affect the logistic curve.


6. Use your calculator to find the point of inflection of P (t).

7. Using the function P (t), at what time t did one-half of the class hear the rumor? Showthe equations used to find this value.

8. Explain any connections between the answers to (6) and (7).


9. How does the value of k affect the logistic curve? Experiment with different values of kand report the results.

10. Describe several phenomenon (other than those discussed already) that might be describedby a logistic curve.


11.4 Extensions

Example 11.4.1 (Stewart) There is evidence to suggest that for some species there is aminimum population m such that the species will become extinct if the size of the populationfalls below m. This condition can be incorporated into the logistic equation by introducingthe factor (1−m/P ), The new model becomes

dP

dt= kP

(

1− P

N

)(

1− m

P

)

(a) Using this differential equation, describe the behavior of the solution if m < P < N . Andfor ) < P < m.

Suppose k = 0.08, N = 1000, and m = 200. Carefully sketch a direction field and use itto sketch several solution curves. Describe what happens to the population for variousinitial populations. What are the equilibrium solutions?

11.4 Extensions 109

(b) Solve the differential equation analytically.

(c) Use your solution in part (c) to show that if P0 < m, then the species will become extinct.(Hint: Show that the numerator in P (t) is 0 for some value of t.)


Example 11.4.2 (Stewart) In a seasonal-growth model, a periodic function of time is usedto account for seasonal variations in the rate of growth. Such variations could be caused byseasonal changes in the variations in the availability of food.

(a) Find the solution to the seasonal-growth modeldP

dt= kP cos(rt− φ), P (0) = P0, where

k, r, and φ are constants.

(b) Graph the solution for several values of k, r, and φ. Explain how these values affect thesolution. What can you say about lim

t→∞

P (t)?

11.5 π From a Polygon - An Exploration 111

11.5 π From a Polygon - An Exploration

Archimedes (287-212 B.C) was a prolific mathematician. He used an algorithm (Method of

Exhaustion, Eudoxus, 408-355 B.C.) to demonstrate properties of circles. The most funda-mental property, that the ratio of the circumference of a circle to its radius is 2π, can beshown using a recursion technique based on the perimeter of regular n-gons inscribed in aunit circle.

The Method

Begin with an inscribed triangle.It is fairly easy to establish that the lengthof each side is

√3. The perimeter is 3

√3.

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Consider a formula for the length of a side of a 2N -gon, S2N , in terms of the length of a sideof an N -gon, SN .

Using this formula, we will be able to relate the perimeters, PN and P2N .

SN = length of a side of an N -gon.

S2N = length of a side of a 2N -gon.

Use the figure to the right to obtain thefollowing two equations.

(1) 12 = (1− x)2 +(

Sn

2

)2

(2) (S2N)2 = x2 +

(

Sn

2

)2

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11

Sn/2

S2Nx

Solve for x in Equation (1): Solve for x in Equation (2):

x = 1−√

1−(

Sn

2

)2

x =

√

S22N −

(

Sn

2

)2


Set these two expressions for x equal, and solve for S2N : S2N =

√

2−√

4− S2N

Since Pn = N · SN P2N = 2N

√

2−√

4− S2N for N ≥ 3.

We know S3 =√3, therefore, S6 =

√

2−√

4− (√3)2 = 1 and P6 = 6.

Similarly, S12 =√

2−√4− 12 =

√

2−√3 and P12 = 12

√

2−√3

Using Technology

Using a TI-83/84, consider the followinginitial values and calculation.

Successive estimates are obtained bypressing ENTER. Each time, the value of Pis updated.

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For a real technological thrill ride, consider the following.

Initial values, initial calculations. The new values are stored in a list. LetL2 = 2π−L1. Using STAT, EDIT, EDIT (theStat List Editor), we can see the error ofeach successive estimate.

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11.5 π From a Polygon - An Exploration 113

The Growth of P

Construct a scatter plot of the estimates of 2π versus N . That is, plot the points (N,PN).

Use your calculator to find a logistic regression equation.The initial condition is (3, 3

√3).

Caution: Do not use too many data points. The logistic regression feature on the TI-83 mayreturn a memory error.

Here are some results:

The points with the logistic regressioncurve.

The logistic regression equation.

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CHAPTER 12

Series and Taylor Polynomials

12.1 Background

1. Series: includes series of constants, power series and Taylor polynomials.

Longest unit in Calculus BC, 6-8 weeks.

Sequences are assumed known from pre-calculus.

2. Technology has changed the way in which we teach this topic, and the questions we askabout series.

Convergence of Taylor Polynomials: demonstrate visually.

3. Textbooks, authors, teachers, differ in how they classify convergence tests.

Seven tests: (1) Geometric with |r| < 1, (2) p-series, (3) Alternating series, (4) Integral,(5) Comparison, (6) Limit comparison, (7) Ratio.

nth term test: implicit in series of constants.

nth root test: uncommon, probably not necessary.

4. Using technology we can visualize series graphically, in function mode and sequence mode.

Numerically: using sum(seq(. Issues: harmonic series.

Consider a proof of the divergence of the harmonic series:

115

116 Chapter 12 Series and Taylor Polynomials

1 +1

2+

1

3+ · · · >

∫

∞

1

1

xdx

= limb→∞

(ln b) = ∞

0 1 2 3 4 5 60.

0.5

1.

1.5

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y

5. AP Exam questions: not necessarily complex calculations of derivatives.

Reason: calculators can produce Taylor polynomials.

Focus: constructing polynomials given the essential parts, knowing when it is appropriateto use an approximation to a function f(x).

6. Important facts about power series:

The tangent line to a curve at a point: the first degree Taylor polynomial.

f(x) ≈ f(a) + f ′(a)(x− a)

Many power series representations can be produced using simple manipulations of exist-ing series. This includes multiplication, division, differentiation, integration. Note whichmanipulations affect the domain.

12.2 Power Series Explorations 117

12.2 Power Series Explorations

For |x| < 1, the expression1

1− xis equal to a polynomial with infinitely many terms. Consider

showing this in two ways.

1. 1− xn = (1− x)(1 + x+ x2 + x3 + · · ·+ xn−1)

When x < 1, xn becomes close to zero as n becomes large.

Therefore: 1− xn is close to 1.

Divide both sides by (1− x), and let n become arbitrarily large.

1

1− x= 1 + x+ x2 + x3 + · · ·+ xn−1 + · · ·

2. We can write the fraction1

1− xin a different from, repeatedly.

1

1− x=

1− x + x

1− x= 1 +

x

1− x= 1 +

x− x2 + x2

1− x= 1 + x +

x2

1− x= . . .

Each step generates an additional term in the polynomial.

The remainder after each step isxn

1− x

As n becomes large, xn tends to zero, and hence, so does the remainder.

1

1− x= 1 + x + x2 + x3 + · · ·+ xn + · · · =

∞∑

n=0

xn for |x| < 1

This is called a power series.

This is an infinite geometric series with ratio x.


There are many infinite series that can be derived from this initial power series. Consider thefollowing operations.

1. Replace x by a different term or expression, for example −x or x2.

2. Differentiate both sides of the power series.

3. Integrate both sides of the power series on an interval.

4. Multiply or divide both sides by a quantity.

5. Add or subtract on both sides of the power series.

Example 12.2.1 To obtain an infinite series for1

1 + x, replace x with −x:

1

1− (−x)=

1

1 + x= 1 + (−x) + (−x)2 + (−x)3 + · · ·+ (−x)n + · · ·

= 1− x+ x2 − x3 + · · · =∞∑

n=0

(−x)n

Note: This is a geometric series with ratio −x and initial term 1.

Example 12.2.2 Use the power series above to demonstrate how the approximation processworks.

Let x = 1/2.1

1 + (1/2)=

2

3

x s1 s2 s3 s4 s5 s6 s7 s8

0.5 1 0.5 0.75 0.625 0.6875 0.65625 0.671875 0.6640625

Calculator command: sum(seq((-X)^N,N,0,K,1)) for X=0.5 and various K.

12.3 Exercises 119

12.3 Exercises

Example 12.3.1 Find a power series representation (first five terms) for each and write theseries using summation notation.

(a)1

1 + x2

(b) ln(1 + x)

(c) tan−1 x


(d)1

1− x2

(e)1

(1 + x)2

(f) ln(1− x)

12.3 Exercises 121

(g)x

1− x

(h)x2

1 + x2

(i)x

1 + x2


(j) ln(1 + x2)

Example 12.3.2 For each power series in the previous example, compare the value of theexpression at x = 0.5 with the value of s4, the sum of the first four terms of the series.Calculate each difference.

Power series (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

Expression value 0.800

s4 0.797

Difference 0.003

12.3 Exercises 123

Example 12.3.3 Assume there is a power series representation for ex in the form

ex = a0 + a1x+ a2x2 + a3x

3 + · · ·+ anxn + · · · =

∞∑

n=1

anxn

(a) Use the fact that e0 = 1 to find a0.

(b) Use the fact that the derivative of ex is ex to find a1. (Differentiate both sides of theexpression and use the same technique as in part (a)).

(c) Repeat the process in part (b) to find a2, a3, a4, and a5.

(d) Find the general expression for ex. Give the first 5 terms and summation form.


Example 12.3.4 Enter the following functions on your calculator:

Y1(x) =1

1 + xY2(x) = P11 = the series up to the x11 term

(a) Carefully sketch a graph of Y1 and Y2 in a ZDecimal window.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

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x

y

(b) Use the TRACE or TABLE feature to find the value of each function and the difference foreach value of x.

x 0 0.1 0.3 0.5 0.7 0.9 -0.1 -0.3 -0.5 -0.7 -0.9

Y1

Y2

Diff

(c) Describe the behavior of the differences as x varies.

(d) For what values of x does P11 approximate1

1 + xwell?

(e) Enter the following function on you calculator: Y2=sum(seq((-X)^N,N,),98,1)). GraphY1 and Y2 in a ZDecimal window. Wait patiently. Does this graph support your answerin part (d)?

12.3 Exercises 125

Example 12.3.5 Previously, we developed a power series expression for tan−1 x. We knowthat tan−1 1 = π/4.

(a) Write an infinite series expression for π/4.

(b) Use the result in part (a) to write an infinite series expression for π.


Example 12.3.6 Construct a fourth degree polynomial P (x) = a0+a1x+a2x2+a3x

3+a4x4

with the following properties:

P (0) = 1; P ′(0) = 2; P ′′(0) = 3; P ′′′(0) = 4; P (4)(0) = 5

That is, find the constants a0, a1, a2, a3, a4, a5

12.3 Exercises 127

Example 12.3.7 Construct a fifth degree polynomial with the following properties:

P (0) = sin(0); P ′(0) = sin′(0); P ′′(0) = sin′′(0); P ′′′(0) = sin′′′(0); P (4)(0) = sin(4)(0);

P (5)(0) = sin(5)(0)


Example 12.3.8 Enter the following functions on your calculator:

Y1(x) = sin x Y2(x) = P5 (from the previous example)

(a) Carefully sketch a graph of Y1 and Y2 in a ZDecimal window.

-8 -6 -4 -2 2 4 6 8

-8

-6

-4

-2

2

4

6

8

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x

y

(b) What do you observe about the graphs?

(c) What characteristics do you observe about the terms in P (x)? Add two more terms toP (x), and enter this as Y3. Consider a all three graphs on the same coordinate axes.What, if any, differences do you observe?

12.3 Exercises 129

Example 12.3.9 In the previous example, we derived a polynomial that can be used toapproximate sin x. Use a typical calculus operation to produce a polynomial to approximatecos x.

Example 12.3.10 Recall: i =√−1, i2 = −1, i3 = −i, i4 = 1. Use this fact and the power

series for ex to write a power series for a complex expression, written as eix.


Example 12.3.11 Use the previous examples to write an equation that links eix with sin xand cos x.

The resulting expression is called Euler’s Formula, in recognition of the great Swiss mathe-matician who developed it. Given this expression, we can evaluate eiπ by substitution, andreveal one of the most unbelievable identities in all of mathematics. This links e, i, π, 1, and0. Find eiπ

12.3 Exercises 131

Example 12.3.12 Challenge: Use the power series for tan−1 x to prove the following ex-pression for π as the sum of an infinite series:

π = 2√3

∞∑

n=0

(−1)n

(2n+ 1)3n


12.4 Measuring Error in a Taylor Polynomial

The nth degree Taylor polynomial for f(x) = ex about x = 0 is

Pn(x) = 1 +x

1!+

x2

2!+

x3

3!+ · · ·+ xn

n!

As n → ∞, Pn(x) converges to f(x), for all real numbers x.

Consider the behavior of the error for finite values of n. Is it possible to measure and/orpredict the error? Consider two approaches.

1. Enter the following functions on your calculator:

Y1 = ex; Y2 = 1 +x

1!+

x2

2!+

x3

3!; Y3 = Y1 − Y2

Turn off Y1 and Y2, and graph only Y3 in a ZDecimal window. This produces a visualizationof the error for P3(x). We can also consider this error analytically.

2. Let f(x) = Pn(x) +Rn(x), where Rn(x) is the error, or remainder, term. Then

Rn(x) = f(x)− Pn(x)

=

(

1 +x

1!+

x2

2!+ · · ·+ xn

n!+

xn+1

(n+ 1)!+ · · ·

)

−(

1 +x

1!+

x2

2!+ · · ·+ xn

n!

)

=xn+1

(n+ 1)!+

xn+2

(n+ 2)!+ · · ·

The error is an (n + 1)th degree polynomial. For P3(x), the error is a quartic.

Consider some properties of R.

1. How large will this error be on a given interval?

Let x ∈ [0, 1]. As x increases, the error increases. Therefore, a bound on the error (for thisinterval) is at x = 1.

R3(x) ≤1

4!+

1

5!+

1

6!+ · · ·

Estimate this bound on your calculator: sum(seq(1/N!,N,4,20,1))

Compare this with Y3 (above). Error bound ≈ 0.0516

12.4 Measuring Error in a Taylor Polynomial 133

2. How large an interval can we use for a given error bound?

Suppose we would like the bound on R3(x) to be 0.1.

Consider a graph of Y3 and Y4 = 0.1, and the intersection points: [−1.323, 1.168]

Illustrations:

A graph of Y3.

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

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xy

x

y

A graph of Y3 and Y4.

-1.5 -1. -0.5 0.5 1. 1.5

0.05

0.1

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xy

xy

x

y


12.5 Connecting Error to the Lagrange Form byIntegration

The Lagrange form of the error is an analytical tool for bounding the error in a Taylorpolynomial.

Consider a third degree Taylor polynomial for a function about a = 0. The fourth degreeterm of the polynomial has the form:

f (4)(0)x4

4!

Consider f (4)(x) on an arbitrary interval [0, b].Suppose this function has an upper bound on this interval, M . Then:

f (4)(x) ≤ M =⇒∫ x

0f (4)(t) dt ≤

∫ x

0M dt

Integrate and simplify:

f ′′′(t)]x0 ≤ M(x − 0) =⇒ f ′′′(x)− f ′′′(0) ≤ Mx =⇒ f ′′′(x) ≤ f ′′′(0) +Mx

Integrate both sides of this expression, from 0 to x:

f ′′(t)]x0 ≤ f ′′′(0)(x− 0) +

M

2!x2

Repeat this procedure, to eventually arrive at:

f(x)− f(0) ≤ f ′(0)x +f ′′(0)

2!x+

f ′′′(0)

3!x3 +

M

4!x4

Rearrange the terms and substitute:

f(x)− P3(x) ≤M

4!x4 The Lagrange error.

The original example: f(x) = ex.Since ex is increasing on [0, 1], the largest value, M , is e.

R4(x) ≤e

4!= 0.113

Note: This is larger than the bound we obtained empirically.

12.5 Connecting Error to the Lagrange Form by Integration 135

Example 12.5.1 Find an error bound for each Taylor polynomial. Use both a numericalapproach and the Lagrange form.

(a) f(x) = e−x on [0, 1]; 4th degree polynomial.

(b) f(x) =1

1− xon [0, 0.5]; 3rd degree polynomial.


12.6 Exploration of Series

Example 12.6.1 Consider the infinite series∞∑

n=1

2n2

2n

(a) Intuitively, how does this infinite sum behave?

(b) How does the presence of the coefficient 2 (in the numerator) affect the series sum? Doesit influence convergence? What would happen if this coefficient were larger or smaller?

(c) Use your calculator (sum(seq() to estimate the sum of the following infinite series (99terms):

(i)∞∑

n=1

2n3

2n(ii)

∞∑

n=1

2n4

2n(iii)

∞∑

n=1

2n5

2n

Explain the pattern.

12.6 Exploration of Series 137

Example 12.6.2 Consider the infinite series∞∑

n=1

(n+ 1)(n+ 2)

n!

(a) Estimate the sum by using sum(seq( for n from 1 to 50.

(b) Approximate each of the following sums using the same calculator technique. In eachcase, relate your answer to a well know constant.

Series Approximation Sum

∞∑

n=1

n

n!

∞∑

n=1

n2

n!

∞∑

n=1

1

n!

(c) Using the results from part (b), find a value for the sum in part (a) in terms of a wellknow constant.

(d) Find a value for the infinite series∞∑

n=1

2n2 − 3n+ 4

n!


Example 12.6.3 Summarize, in your own words, your understanding of the hierarchy offunctions when finding the value of an infinite sum. Explain your strategy for determiningwhether a series converges or diverges.

CHAPTER 13

Parametric, Polar, and Vector-Valued

Functions

13.1 Background

1. Very little change in the syllabus regarding this topic.

Problems: analogous to rectangular coordinates.

2. Finding the first and second derivatives of a function defined parametrically.

3. Use the chain rule to establish:dy

dx=

dy

dt

dx

dt

See 1998 BC6 and 2001 BC1.

4. Vector-valued functions:

speed = |v(t)| =√

v21 + v22

Vector in component form: v = 〈v1, v2〉 or v = (v1, v2)

5. Use parametric equations to simulate motion. This allows students to see functions rep-resented in a variety of ways. Also confirms that rectilinear motion really is motion in astraight line.

Example: s = 10t− 4.9t2 versus x(t) = 10t− 4.9t2, y(t) = 2

6. Integration and parametric and polar form:

Always possible, consider arc length (parametric), area (polar).

Surface area problems less likely, but should still be covered.

139

140 Chapter 13 Parametric, Polar, and Vector-Valued Functions

13.2 Selected Problems

Example 13.2.1 Both x and y are functions of t, t ≥ 0, and x(0) = y(0) = 0.dy

dx= 4t2 and

dx

dt= 2t.

(a) Finddy

dtin terms of t.

(b) Find y in terms of x.

(c) Set up and evaluate an integral expression in t for the length of the curve from t = 0 tot = 1.

13.2 Selected Problems 141

Example 13.2.2 Given the polar curve defined by r = 2 cos(2θ) on [0, 2π]:

(a) Write an equation for the curve in rectangular coordinates.

(b) Finddy

dxin terms of θ and evaluate

dy

dxat θ = π/3.

(c) Find the area of one leaf of the curve.

142 Chapter 13 Parametric, Polar, and Vector-Valued Functions

Example 13.2.3 A particle in the xy plane moves so that, at any time t ≥ 0, x = 2t2 andy = 2t− 1.

(a) Find the velocity vector in terms of t.

(b) Find the magnitude of the speed when t = 1.

(c) Find the unit vector tangent to the curve when t = 1.

13.2 Selected Problems 143

Example 13.2.4 Consider a curve defined by x(t) = t4 − t2 and y(t) = t+ ln t.

(a) Carefully sketch this curve.

-1. -0.5 0.5 1. 1.5 2. 2.5 3. 3.5 4. 4.5

-1.5

-1.

-0.5

0.5

1.

1.5

2.

2.5

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x

y

(b) Estimate the coordinates of the leftmost point on the curve.

(c) Find the coordinates (analytically) of the leftmost point on the curve.

Critical Concepts in AP Calculus and Activities to Promote ... · 8. Participants will understand...

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