Critical Concepts in AP Calculus
and
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 an
AP course. There will be special emphasis on the effects of calculus reform to traditional
topics and new approaches to teaching challenging concepts. The focus of the institute is the
Calculus BC syllabus. However, we will also address the Calculus AB topics included in that
syllabus. The institute will include a review of the 2008 AP Calculus Exam, including the
scoring guides and student samples.
Outline of Topics by Day
Day 1
1. Welcome and introductions; discussion of technology issues; participant experience and
expectations; 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 Calculus
program.
2. Participants will know how to use technology appropriately as a teaching and learning
medium.
3. Participants will be able to write assessment items that reflect understanding of the AP
Calculus 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 Calculus
topics.
6. Participants will understand how to start, maintain, and improve Advanced Placement
programs, particularly in calculus.
7. Participants will understand the methodology of assessment in the Advanced Placement
Examination.
8. Participants will understand how to prepare students to be successful on the AP Calculus
Examination.
Content
1. Nature of the AP Calculus Program, including recent changes in the syllabus and philosophy. 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 be
required 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 specific
concept 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 of
change - 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 L
as we want by manipulating x.
Many investigations with transcendental functions that enable students to explore and
discover 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: lim (3x − 5) = −2
x→1
(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
1.2
Chapter 1 Limits, Continuity, End Behavior, and Aspects of the Difference Quotient
Explorations Involving Limits
Example 1.2.1 Consider lim (3x − 1)
x→2
(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
Example 1.2.2 Consider lim
x→1 x − 1
(a) Evaluate this limit.
(b) Is the function f (x) =
x2 − 1
defined at x = 1?
x−1
(c) Is f continuous at x = 1?
1.2
Explorations Involving Limits
3
3x − 1
x→0
x
Example 1.2.3 Consider lim
(a) Carefully sketch a graph of this function using ZDecimal.
y
2
1
-4
-3
-2
1
-1
2
3
4
x
-1
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-2
-3
(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 lim
x→∞
3
.
2 + e−x
(a) Carefully sketch a graph of this function using ZDecimal. Change the window as needed
to approximate this limit.
(b) Can you find the more general limit lim
x→∞
c
k + e−x
4
Chapter 1 Limits, Continuity, End Behavior, and Aspects of the Difference Quotient
Example 1.2.5 Consider lim (1 + x)1/x .
x→0
(a) Evaluate this limit to three decimal places.
(b) Do you recognize this number? Can you write this number analytically?
5000x2
x→0 sin(x) + 5000x2
Example 1.2.6 Consider lim
(a) Carefully sketch a graph of this function using ZDecimal. Find an approximation for this
limit.
(b) Try to narrow the window using ZBox. Find an approximation for this limit using the
new window. Is your answer the same as in part (a)?
(c) Continue to narrow the window. Revise your approximation as necessary.
1.2
Explorations Involving Limits
5
Example 1.2.7 Consider the expression lim f (x) = L.
x→∞
We can make f (x) arbitrary close to l by choosing values for x that are big enough.
If f (x) =
3x − 1
3
and lim f (x) = ,
x→∞
4x + 1
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) lim sin
x→0
(b) lim x sin
x→0
2
1
x
(c) lim x sin
x→0
1
x
1
x
6
Chapter 1 Limits, Continuity, End Behavior, and Aspects of the Difference Quotient
Challenge: About 2500 years ago, Archimedes considered the area of a circle by using a
polygon 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 as
shown. 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.
1.2
7
Explorations Involving Limits
Example 1.2.9 The graphs of the functions f and g are given below. (Calculus Problems
for a New Centruy, Volume 2, MAA.)
y
y
3
•
-4
-3
-2
2
•
•
1
-1
•
1
-1
•
3
2
3
•
•
4
x
-4
-3
-2
-3
1
-1
-2
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-3
Graph of y = g(x)
Find each limit, if it exists.
(a) lim f (x)
(b) lim f (x)
(c) lim g(x)
(d) lim g(x)
(e) lim [f (x) + g(x)]
(f) lim [2f (x) + 3g(x)]
(g) lim [f (x)g(x)]
(h) lim [f (x)g(x)]
x→−1
x→1
x→1
f (x)
x→0 g(x)
x→1
x→1
x→0
x→2
g(x)
x→0 f (x)
(i) lim
(j) lim
(k) lim g(f (x))
(l) lim f (g(x))
x→−2
(m)
(o)
x→−1
lim f (x)
(n) lim+ f (x)
lim g(x)
(p)
x→−1−
x→−1−
(q) lim− f (x + 2)
x→0
2
-1
Graph of y = f (x)
x→−1
•
1
-2
g replacements
2
x→1
(r)
lim g(x)
x→−1+
lim f (x2 )
x→−1−
3
4
x
8
Chapter 1 Limits, Continuity, End Behavior, and Aspects of the Difference Quotient
√
3
1 + cx − 1
Example 1.2.10 Evaluate lim
where c is a constant.
x→0
x
|2x − 1| − |2x + 1|
x→0
x
Example 1.2.11 Evaluate lim
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 intervals.
2. Tangent lines and local linearity:
(a) If a function is differentiable on an interval, the graph looks like a line, if we choose a
sufficiently 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 0 (c) = L0 (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
dy
∆y
=
:
∆x→0 ∆x
dx
4. lim
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.
3
y
y
1.5
2
1
-2
1
1
-1
2
3
x
0.5
-1
g replacements
-2
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0
0.5
0.75
1
1.25
Note:
1. L’Hopital’s Rule demonstration: a combination of limits and local linearity.
2. Other activities: rate functions graphically, local linearity.
x
2.2 Constructing and Interpreting the Graph of a Rate Function
2.2
11
Constructing and Interpreting the Graph of a Rate
Function
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 increase
in 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.
h
dh/dt
t
h
t
dh/dt
t
t
12
Chapter 2 Rates of Change: Average, Approximate, Instantaneous
h
dh/dt
t
t
h
dh/dt
t
t
Example 2.2.2 Suppose a bathtub is full of water, and the drain is opened. As the bathtub
starts 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
V
dV /dt
t
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 to
send the rocket upward. It continues upward for a period of time and, when it approaches
the ground on its return, a parachute opens to soften the landing. A graph of the rocket’s
velocity is given below. (Source: Demana, Waits, Finney, Thomas)
60
v
50
40
30
20
10
0
-10
1
2
3
4
5
6
7
8
9
10
11
12
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-30
-40
-50
(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?
t
14
Chapter 2 Rates of Change: Average, Approximate, Instantaneous
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 0 (a)(x − a)
or
f (x) ≈ f (a) + f 0 (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 0 (a)(x − a)
3. E(a) = 0
E(a + h) = f (a + h) − f (a) − f 0 (a)h
4. Note:
(a) As h → 0, E → 0.
(b) The error depends on f 0 (a) and h.
A graph with a steep slope: larger errors near a, even when h is small.
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15
2.3 Tangent Line Approximations and Local Linearity
Error function
illustration:
PSfrag
replacements
y
f (a + h)




E(x)
f 0 (a)h



f (a)
a
x
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 more
than 0.01.
Consider the following graphs:
y
y
0.15
0.015
0.1
0.05
-0.75
-0.5
0.25
-0.25
0.5
0.75
0.01
x
-0.05
0.005
-0.1
-0.15
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-0.75
y = E(x) = x − sin x
Find the bounds for x such that |E(x)| ≤ 0.01.
-0.5
-0.25
0.25
0.5
y =| E(x) |, y = 0.01
0.75
x
16
Chapter 2 Rates of Change: Average, Approximate, Instantaneous
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?
2.3 Tangent Line Approximations and Local Linearity
17
Limits and the Error Function
lim E(x) = 0
and
x→a
"
#
f (x) − f (a)
f (x) − f (a) − f 0 (a)(x − a)
E(x)
=
lim
=
lim
− f 0 (a) = f 0 (a) − f 0 (a) = 0
lim
x→a
x→a
x→a x − a
x−a
x−a
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.
E(x)
=0
x→a x − a
(c) Verify that lim
(a) f (x) =
√
x;
a = 4.
18
Chapter 2 Rates of Change: Average, Approximate, Instantaneous
(b) f (x) = x + sin x;
(c) f (x) = 2x2 + 1;
a = 0.
a=1
2.3 Tangent Line Approximations and Local Linearity
19
Example 2.3.4 This activity involves your knowledge of the Fibonacci sequence. Recall, a
Fibonacci 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 and
a2 = 1. At each 20-second interval, mark your point in the calculations. Certainly, as the
numbers get larger, it will take more time, and you may need to use some scratch paper. This
activity will continue for 3 minutes.
When time has expired, you will be asked to summarize your data in the table below, showing
the cumulative number of Fibonacci numbers listed after each 20-second interval. We will then
plot the points, consider a smooth curve through these points, and analyze these results.
i
ai
Time
i
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
ai
Time
20
Chapter 2 Rates of Change: Average, Approximate, Instantaneous
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 the
starting point on your graph.
Draw a smooth curve through the data points.
y
60
50
40
30
20
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0
0
20
40
60
80
100
120
140
160
180
x
(a) Find the average rate at which you computed Fibonacci numbers over the entire interval
of 180 seconds. Give your answer in Fibonaccis per second.
(b) Find an estimate of the instantaneous rate at which you wrote Fibonacci numbers at the
40 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].
2.3 Tangent Line Approximations and Local Linearity
21
(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 set
to obtain a value for k.
(e) Enter your data set in your calculator using the lists L1 and L2. Enter the function f
as 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 = 160
seconds. Find the error in your estimate.
(g) A curve of different form may provide a better fit to your data. Go to STAT; CALC and
experiment with other regression forms until you find one that fits well. Consider a
quadratic, cubic, quartic, linear, logarithmic, and exponential regression.
22
Chapter 2 Rates of Change: Average, Approximate, Instantaneous
L’Hopital’s Rule - A Visual Investigation
L’Hopital’s Rule provides a method for evaluating limits that are indeterminate, even when
the expression cannot be simplified by algebraic means.
sin(πx)
x→1 x2 − 1
For example: lim
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 0 (x) 6= 0 on an open interval I that contains a
(except possibly at a). Suppose that
lim f (x) = 0
and
lim f (x) = ±∞
and
x→a
or that
x→a
lim g(x) = 0
x→a
lim g(x) = ±∞
x→a
Then
f (x)
f 0 (x)
lim
= lim 0
x→a g(x)
x→a 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.
2.3 Tangent Line Approximations and Local Linearity
23
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 0 (a)(x − a)
g(x): y = g 0 (a)(x − a)
Replace f and g by their tangent lines:
f 0 (a)
f 0 (a)(x − a)
f (x)
= lim 0
= 0
x→a g (a)(x − a)
x→a g(x)
g (a)
lim
Illustration:
y
y
0.5
0.5
0.
0.5
g replacements -0.5
1.
1.5
x
0.
0.5
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Example 2.3.5 Use L’Hopital’s Rule to find the limit, if it exists.
(a) lim
x→1
2x − 2
1 − x2
1.
1.5
x
24
Chapter 2 Rates of Change: Average, Approximate, Instantaneous
5x2 − 1
x→1 3x2 + 2x + 2
(b) lim
sin(x2 )
(c) lim
x→0
x
2x − 1
x→0
x
(d) lim
2.3 Tangent Line Approximations and Local Linearity
25
f (x + h) − 2f (x) + f (x − h)
h→0
h2
Example 2.3.6 Suppose f 00 is continuous. Find lim
Note: L’Hopital’s Rule allows us to establish which kinds of functions dominate, or grow
faster, in a quotient as x → ∞. Consider the following table:
Slowest
sin x, cos x
Fastest
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’s
Rule to verify your conclusion.
(a) x→∞
lim
x5
2x
ln x
(b) lim √
x→0
x
26
Chapter 2 Rates of Change: Average, Approximate, Instantaneous
ln x
x→∞ sin x
(c) lim
x10
x→∞ ex
(d) lim
(0.9)x
x→∞
x
(e) lim
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 of
a 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.
y
3
2
1
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-3
-2
1
-1
2
3
x
-1
x
y
-2
-3
−
0
−2
+
0
0
−
0
+
1
f 0 (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 line
at 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 0 (x), sketch a possible graph of y = f (x) and
y = f 00 (x).
y = f (x)
y = f 0 (x)
y = f 00 (x)
1.
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x
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y
x
y
x
2.
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y
x
y
x
y
x
3.
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y
x
y
x
y
x
4.
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y
x
y
x
y
x
5.
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y
x
y
x
y
x
3.1 Background
y = f (x)
y = f 0 (x)
y = f 00 (x)
6.
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x
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y
x
7.
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x
y
x
y
x
8.
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x
y
x
y
x
9.
Sfrag replacements
y
x
y
x
y
x
10.
Sfrag replacements
y
x
y
x
y
x
29
30
Chapter 3 Analysis of the Derivative and Applications to Local Extrema and Points of Inflection
3.2
Explorations Involving Extrema of Functions
Example 3.2.1 The values of a differentiable function f for some specific values of x are
given 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 local
maximum or a local minimum.
(b) One way to estimate the value of f 0 (a) is to evaluate the difference quotient:
f (a + 1) − f (a − 1)
f (a + 1) − f (a − 1)
=
(a + 1) − (a − 1)
2
Use the formula to complete the following table.
x
-2
f (x)
4.5
-1
0
1
2
3
4
(c) Carefully sketch a graph of y = f (x) and y = f 0 (x) on the Axes below. What conclusions
can you make about f 0 (x) where f has a local extreme value?
Sfrag replacements
x
y
3.2 Explorations Involving Extrema of Functions
31
Example 3.2.2 A graph of y = f 0 (x) is given below. Use this graph to answer the following
questions.
y
8
6
4
2
PSfrag replacements
-4 -3 -2 -1
-2
1
2
3
4
5
6
7
8
9
10 11 12
x
-4
x
y
-6
-8
(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 0 (x)
changes direction.
(f) Draw a possible graph of y = f (x) on the axes above.
32
Chapter 3 Analysis of the Derivative and Applications to Local Extrema and Points of Inflection
Example 3.2.3 The figure below is a graph of y = f 0 (x), the derivative of f . The function
f has domain [−4, 4], f 0 (−x) = f (x), and f 0 (−3) = f (3) = 0.
y
3
2
1
PSfrag replacements
x
y
-3
-2
1
-1
2
3
x
-1
-2
-3
(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.
3.2 Explorations Involving Extrema of Functions
33
Example 3.2.4 Consider the function v(t) = t2 − e5t , where v is the velocity of an object
moving along a horizontal line, measured in meters per second.
(a) Carefully sketch v(t) in the window [−2, 10] × [−5, 20].
18
v
16
14
12
10
8
6
PSfrag replacements
x
y
4
2
-1
-2
1
2
3
4
5
6
7
8
9
t
-4
(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? Justify
your answer.
34
Chapter 3 Analysis of the Derivative and Applications to Local Extrema and Points of Inflection
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 = f 9x) for k = 0.5, k = 1, and k = 2 in a standard viewing
window (ZStandard).
k = 0.5
-8
-6
-4
k=1
8
8
8
6
6
6
4
4
4
2
2
2
-2
4
6
8
-8
-2
-4
g replacements
y
x
k=2
-6
PSfrag replacements
-8
y
x
-6
-4
2
2
-2
4
6
8
-8
-6
-4
2
-2
-2
-2
-4
-4
-6
-6
PSfrag replacements
-8
y
x
4
6
8
-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 0 (x) for each value of k. How many local extrema
are there for k = 0.5? k = 1? k = 2?
(d) Find f 0 analytically. What must be true of k if f (x) is to have any local extrema?
3.2 Explorations Involving Extrema of Functions
35
Example 3.2.6 This problem involves curve fitting or the method of least squares. Suppose
we want to form an open box from a rectangular sheet of cardboard that measures 8 inches
by 12 inches, by cutting a square from each corner and folding up the sides. For example, if
we cut a 1 inch square from each corner, the dimensions of the box are 6 × 10 × 1, and the
volume is 60 in3 .
(a) Let x be the length of the square cut from each corner, and let v be the volume of the
resulting box. Complete the following table.
x
v
0.5
1
1.5
2
2.5
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 the
degree 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 coordinate
axes.
(f) Use your best fit function to estimate the maximum volume of the box.
36
Chapter 3 Analysis of the Derivative and Applications to Local Extrema and Points of Inflection
(g) Write an analytical formula for the volume of the box in terms of the variable x. Find
v 0 (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 + y 2 = 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:
√
√
f2 (x) = − 4 − x2
x2 + [f (x)]2 = 4
f1 (x) =
2x + 2[f (x)]f 0 (x) = 0
f1 (x) = (4 − x2 )1/2
f2 (x) = −(4 − x2 )1/2
1
f10 (x) = (4 − x2 )−1/2 (2x)
2
1
f20 (x) = − (4 − x2 )−1/2 (−2x)
2
f 0 (x) =
−x
−x
=
f (x)
y
4 − x2
−x
(4 − x2 )1/2
√
√
Since f1 (x) = 4 − x2 and f2 (x) = − 4 − x2 ,
f10 (x) =
f10 (x) =
−x
y
and
f20 (x) =
f20 (x) +
x
(4 − x2 )1/2
−x
y
Implicit differentiation: compactly states the derivative of all functions embedded in the
equation.
37
38
Chapter 4 Implicit Differentiation and Its Meaning
Illustration: the circle with two tangent lines.
y
3
2
1
m>0
PSfrag replacements
-4
-3
x
y
-2
m<0
1
-1
2
3
4
x
-1
-2
-3
3. Analyzing implicit derivatives:
Consider finding a slope for a given point (x, y), or finding a point on the curve where the
slope is zero or undefined.
The function and its derivative are inter-substitutible.
Consider
dy
x − 2y
=
dx
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 then
solve the resulting equation for y 0 .
Challenge: Why can you take the derivative of both sides and maintain the equality?
4.1 Background
Example 4.1.1 Consider the function y =
x2
x
+1
(a) Find the derivative, y 0 , by using the quotient rule.
(b) Rewrite the original function by multiplying both sides of the equation by x2 + 1.
(c) Differentiate implicitly.
x
and simplify.
+1
How does this compare with your answer in part (a)?
(d) Replace y with
x2
39
40
Chapter 4 Implicit Differentiation and Its Meaning
(e) Using your answer from part (a) (the explicit derivative), find the points on the curve
where the slope is 0.
(f) Using the answer obtained via implicit differentiation, find the points on the curve where
the slope is 0.
(g) Carefully sketch a graph of y = f (x), and indicate the points on the curve where the
slope is 0.
y
0.75
0.5
0.25
PSfrag replacements
x
y
-4
-3
-2
1
-1
-0.25
-0.5
-0.75
2
3
4
x
4.1 Background
Example 4.1.2 Consider the equation x3 + y 3 = 3xy.
Note: Also given parametrically: x(t) =
3t
1 + t3
y(t) =
3t2
1 + t3
y
2
1
-2
-1
1
PSfrag replacements
-1
x
y
-2
2
x
(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.
41
42
Chapter 4 Implicit Differentiation and Its Meaning
Example 4.1.3 Consider the curve defined by the equation
√
x+
√
y=
Find the sum of the x- and y-intercepts of any tangent line to this curve.
√
c
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
X
f (xi ) ∆x
i=1
The definite integral of f from a to b:
Z
b
a
f (x) dx = lim
∆x→0
n
X
f (xi ) ∆x
i=1
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 accumulated
rates 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 over
an interval of 12 seconds. The speed of the racer is recorded at 2 second intervals and is
summarized in the following table.
Time (sec)
Speed (ft/sec)
0
2
4
6
8
10
12
10
14
20
27
32
34
35
(a) Plot the racer’s speed against time.
s
35
30
25
20
PSfrag replacements 15
10
x
y
5
0
0
2
4
6
8
10
12
t
(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 entire
12 second interval?
5.2 Accumulated Rates of Change
45
(e) Using the data, construct rectangles that represent the upper and lower estimates of the
distance traveled.
s
35
30
25
20
PSfrag replacements 15
10
x
y
5
0
0
2
4
6
8
10
12
t
(f) What would you need to know in order to improve your estimate of the distance traveled
over the entire time interval?
(g) Draw a smooth curve through the data points. How is the total distance related to this
curve?
(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 the
following 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)
Speed (ft/sec)
0
2
4
6
8
10
12
12
15
22
28
31
33
37
(i) Plot this data on the graph in part (a). Draw a smooth curve through these data
points.
(ii) Can you determine from the graph which of the racers was ahead after the 12 second
interval? Why or why not?
(iii) Compute a lower and upper bound for the distance traveled by the second racer. Do
these 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
Example 5.2.2 Evaluate
Z
2
1
x3 dx by using the partition
x0 = 1, x1 = 21/n , x2 = 22/n , . . ., xi = 2i/n , . . ., xn = 2n/n = 2.
2
Let x∗i = xi .
1
dx
1 x2
√
Use a regular partition and let x∗i = xi−1 xi (the geometric mean).
Example 5.2.3 Find
Z
47
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 Theorem 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 integrals
in 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 properties
of 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 integrand
is a known function. Usually, the given function has no closed form antiderivative. The
student must use the Fundamental Theorem of calculus to answer questions about extrema, intervals on which the function is increasing or decreasing, concavity, and points
of inflection.
4. A difficulty part: students must write verbal arguments that are logical and support assertions.
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
6.2
Chapter 6 Functions Defined by Integrals
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 accumulator to answer each questions.
Express justifications in your own words or by using appropriate notation.
Example 6.2.1 Let F (x) =
below.
Z
x
0
f (t) dt for 0 ≤ x ≤ 3, where the graph of y = f (t) is shown
y
3
2
PSfrag replacements
1
x
y
1
2
3
x
(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
Example 6.2.2 Let F (x) =
Z
x
0
sin(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).
y
1.25
1.
0.75
PSfrag replacements
0.5
0.25
x
y
1
-0.25
2
3
x
51
52
Chapter 6 Functions Defined by Integrals
Example 6.2.3 Let h(x) =
shown below.
Z
2x
1
f (t) dt where the graph of y = f (t) is piecewise linear and
y
2
1
PSfrag replacements
x
y
0
1
2
3
4
5
6
x
-1
-2
(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 and
other 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 and
averaging 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 first
aggregate, 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 rate
of 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 is
applied to the function itself.
(d) When we say average value, we are referring to integration, with the given function as
the 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 distance
traveled by the elapsed time. We know distance is the integral of velocity, and elapsed
time 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 temperature
readings, multiply by the length of each interval, sum, and divide by the length of the
interval. (Activity to follow.)
7.2
Average Value Experiment
1. When a function grows at a constant rate, as a linear function does, computing an average
value for a given interval is straightforward.
2. The average value is at the midpoint of the interval and is midway between the left and
right 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)
= 13
2
Consider the figure below, and finding the average value via the calculus rule.
y
20
15
PSfrag replacements 10
x
y
5
0
0
2
4
6
8
10
x
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 involves
finding the average value of the investment over the first 50 years.
(a) Carefully sketch a graph of y = f (x) on the axes below.
y
10 000
8000
6000
PSfrag replacements
x
y
4000
2000
0
0
5
10
15
20
25
30
35
40
45
x
(b) To estimate the average, compute the value of the investment at the beginning of each
ten 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!
56
Chapter 7 The Average Value of a Function
(d) Someone might suggest the average value is [f (0) + f (50)]/2. Explain why this method
would be incorrect.
(e) Using the axes below, carefully sketch the rectangular regions used to obtain the estimate
in part (b).
y
10 000
8000
6000
PSfrag replacements
x
y
4000
2000
0
0
5
10
15
20
25
30
35
40
45
x
(f) This suggests a method to obtain the true average value of the function using the limit
of a Riemann sum.
Compute the average analytically and confirm your answer using the numerical integration feature on your calculator.
7.3 Riemann Sums, Average Values, and Regression Curves
7.3
57
Riemann Sums, Average Values, and Regression
Curves
Example 7.3.1 Consider the following temperature data (in
Phoenix, Arizona.
Time
Temperature
◦
F) for a typical day in
12 AM
4 AM
8 AM
12 PM
4 PM
8 PM
12 AM
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 using
left endpoint values. Compute this average.
(b) Enter the time and temperature data in you calculator using the lists L1 and L2. item
Find 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 on
your calculator. How does this answer compare with part (a)?
58
Chapter 7 The Average Value of a Function
(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.
y
80
75
PSfrag replacements 70
x
y 65
60
0
4
8
12
16
20
x
(e) Use your calculator to display the right endpoint Riemann sum.
y
80
75
PSfrag replacements 70
x
y 65
60
0
4
8
12
16
20
x
CHAPTER
8
Differential Equations
8.1
Background
1. Many students learn about growth and decay in pre-calculus, in advance of studying differential 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
= ky
dt
Solution steps:
Separate the variables.
1
dy = k · dt
y
Caution 1: Students who see k as joined to y often divide
both sides of the equation by ky. This results in a complicated
integration on the left side and further trouble the student solves
for y.
Integrate both sides.
ln |y| = kt + C
Caution 2: It is common for students to omit the absolute value
symbol, since growth and decay problems rarely require this
notation. See 1997, AB/BC 6 for an example of how dangerous
this 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 equations
in 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 differential 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
= h0 (t) and dy = h0 (t) dt
dt
Therefore: g(h(t)) · h0 (t) dt = f (t) dt
Integrate both sides with respect to t:
And, G(h(t)) = F (t) + C
Z
0
g(h(t)) · h (t) dt =
Z
f (t) dt
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 different
scenarios in which natural logs may arise in a differential equation solution.
Scenario 1
1
dy = k · dt
y
There is no restriction on y in the original equation. Therefore, the solution should not restrict
y 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 to
solve 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
8.2
61
Solving Differential Equations by Separation of
Variables
Solve the following differential equations. In each case, evaluate the limit of the solution
function as t → ±∞, if it exists. Carefully sketch the solution curve in an appropriate
window. Check you solution using a calculator program for slope fields.
Example 8.2.1
dA
= 0.2A A(0) = 10
dt
A
45
A(t) =
40
35
30
25
lim A(t) =
t→∞
20
PSfrag replacements
15
10
x
y
5
lim A(t) =
t→−∞
-4
-2
2
4
6
8
t
62
Chapter 8 Differential Equations
Example 8.2.2
dy
+ y = 0 y(0) = 1
dt
y
y(t) =
15
10
lim y(t) =
t→∞
PSfrag replacements
5
x
y
lim y(t) =
t→−∞
-4
-3
-2
-1
1
2
3
4
t
8.2 Solving Differential Equations by Separation of Variables
Example 8.2.3
dP
= 100 − P
dt
63
P (2) = 50
P
P (t) =
80
60
40
20
lim P (t) =
t→∞
PSfrag replacements
1
-20
-40
x
y
lim P (t) =
t→−∞
-60
-80
2
3
4
t
64
Chapter 8 Differential Equations
Example 8.2.4
dy
= −2y − 6 y(1) = −10
dt
y
y(t) =
4
3
2
1
lim y(t) =
t→∞
PSfrag replacements
-1.5
-1.
0.5
-0.5
-1
-2
x
y
lim y(t) =
t→−∞
-3
-4
1.
1.5
t
65
8.2 Solving Differential Equations by Separation of Variables
Example 8.2.5
dy
= t(4 − 2y) y(0) = −2
dt
y
y(t) =
2
1
lim y(t) =
t→∞
PSfrag replacements
-4
-3
-2
1
-1
-1
x
y
lim y(t) =
t→−∞
-2
2
3
4
y
66
Chapter 8 Differential Equations
Example 8.2.6
M
dM
=
dt
t
M (1) = −3 and M (−t) = −M (t)
M
M (t) =
10
5
lim M (t) =
t→∞
PSfrag replacements
-4
-3
-2
1
-1
-5
x
y
lim M (t) =
t→−∞
-10
2
3
4
t
67
8.2 Solving Differential Equations by Separation of Variables
Example 8.2.7
ty + 3t
dy
= 2
dt
t +1
y(2) = 2
y
y(t) =
4
3
2
1
lim y(t) =
t→∞
PSfrag replacements
-4
-3
-2
1
-1
-1
-2
x
y
lim y(t) =
t→−∞
-3
-4
2
3
4
t
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 the
following:
(a) y = |f (x)|
(b) y = f (|x|)
(c) y = 2f (x)
(d) y = f (2x)
(e) y = f (2|x|)
y
4
3
2
1
-4
PSfrag replacements
-3
-2
1
-1
2
3
4
x
-1
-2
x
y
-3
-4
(a) y = |f (x)|
(b) y = f (|x|)
y
-4
-3
-2
y
4
4
3
3
2
2
1
1
1
-1
-1
2
3
4
x
-4
g replacements
y
x
-4
-2
1
-1
-1
-2
-3
-3
2
3
4
x
-2
PSfrag replacements
y
x
-3
-4
69
70
Chapter 9 Sample Review Problems
(c) y = 2f (x)
(d) y = f (2x)
y
y
4
2
3
2
-4
-3
-2
1
-1
2
3
4
x
1
-2
-4
-4
PSfrag replacements
y
x
(d) y = f (2|x|)
y
4
3
2
1
-4
-3
-2
1
-1
-1
-2
g replacements
y
x
-3
-4
1
-1
-2
-8
y
x
-2
-1
-6
g replacements
-3
2
3
4
x
-3
-4
2
3
4
x
Chapter 9 Sample Review Problems
71
Example 9.0.9 Rates of Change (Source: Ostebee and Zorn, page 107, No. 21) Suppose
f is a continuous function such that f (1) = −2 and f 0 (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.
72
Chapter 9 Sample Review Problems
Example 9.0.10 Rates of Change from Tabular Data (Source: Harvard Calculus, page
107, Ex. 3)
The table below gives the values of C(t), the concentration of a drug in mg/cm3 after t
minutes in the bloodstream.
t
C(t)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
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 (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 0 (t) for 0 ≤ t ≤ 0.9 using a right-hand difference
quotient 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 0 (t)
(c) Given that C(t) is a differentiable function, is there a time t when C 0 (t) = 0? Justify
your answer.
Chapter 9 Sample Review Problems
73
Example 9.0.11 The Racetrack Principle (Source: Ostebee and Zorn, page 107, No. 29)
The graph of y = g 0 (x), the derivative of g(x), is shown below.
y
3
2
1
PSfrag replacements
x
y
1
-1
2
3
4
5
6
7
8
9
x
-1
-2
(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).
74
Chapter 9 Sample Review Problems
Example 9.0.12 Extreme Values of a Function (Source: David Loman, University of
Arizona)
A graph of y = f (x) is shown below.
y
3
2
1
PSfrag replacements
-4
-3
x
y
-2
1
-1
2
3
4
x
-1
-2
-3
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.
Chapter 9 Sample Review Problems
(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)
75
76
Chapter 9 Sample Review Problems
Example 9.0.13 Implicit Differentiation
Given the equation x2 − xy + y 2 = 3.
(a) Find
dy
dx
(b) Find the x- and y-coordinates of all points on the graph where the slope of the tangent
line is 0.
Chapter 9 Sample Review Problems
77
(c) Suppose L is a line tangent to the curve with slope 1. Find the x- and y-coordinates of
all 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, explain
why not.
78
Chapter 9 Sample Review Problems
Example 9.0.14 Riemann Sums and Area
A region in the first quadrant is bounded by the x and y axes and a curve y = f 9x). The
following values of x and f (x) are known:
x
0.0
f (x) 6.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
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.
Chapter 9 Sample Review Problems
79
(c) Construct a scatter plot of the points. Can you suggest an analytical expression for
f (x)? Use your guess to compare your answers in parts (a) and (b) to the analytical, or
geometric, solution.
y
6
5
4
3
PSfrag replacements
x
y
2
1
1
2
3
4
5
6
x
80
Chapter 9 Sample Review Problems
Example 9.0.15 Functions Defined by Integrals
A graph of y = f (x) on the interval [−5, 5] is given below. The graph consists of line segments
and a semi-circle.
y
4
3
2
1
-5
-4
-3
-2
1
-1
2
3
4
5
x
-1
PSfrag replacements
x
y
-2
-3
-4
Let F (x) =
Z
x
0
f (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.
Chapter 9 Sample Review Problems
81
(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 the
axes below.
y
4
3
2
1
-5
-4
-3
-2
1
-1
-1
PSfrag replacements
x
y
-2
-3
-4
2
3
4
5
x
82
Chapter 9 Sample Review Problems
Example 9.0.16 Differential Equations
The function f (x) is twice differentiable, and has the following properties:
f 0 (x) = 1 +
f (x)
,
x
f (1) = 0,
f 0 (1) = 1
(a) Write an equation of the tangent line to the graph of y = f (x) at x = 1.
(b) Find f 00(x) in terms of x.
(c) Find f (x) in terms of x.
Chapter 9 Sample Review Problems
83
Example 9.0.17 Series and Differential Equations
Suppose 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
1
for f (x) =
1 + x2
(c) Write the first four nonzero terms of the series for F (x) =
Z
x
0
f (t) dt
84
Chapter 9 Sample Review Problems
(d) Write a closed form expression for F (x). Use this expression to evaluate lim F (x).
x→∞
(e) Show that y = f (x) is a solution to the differential equation
(1) Differentiating.
(2) Solving the differential equation.
dy
= −2xy 2 by
dx
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 without
actually 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 a
pattern.
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 differential
equations.
85
86
10.2
Chapter 10 Slope Fields and Euler’s Method
Slope Fields
1. A slope filed is a graphical representation that shows the flow of tangent lines to the
family 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 to
compute 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 Suppose
dy
xy
=
for x ∈ [−2, 2] and y ∈ [−2, 2].
dx
4
(a) Complete the grid below. The body of the table contains the computed values of dy/dx.
y value
dy/dx
x value
−2
−1
0
1
2
−2
−1
0
1
2
10.2 Slope Fields
(b) Draw the slope field for the values in the grid above.
y
2
1
-2
1
-1
2
x
-1
PSfrag replacements
x
y
-2
(c) What kind of function do you think the solution family represents?
87
88
Chapter 10 Slope Fields and Euler’s Method
READING A SLOPE FIELD
1. Reading a slope field: look for clues about the behavior of the differential equation and
the 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 line
have the same slope, then the differential equation does not depend on y, because, as y
varies, the slope does not change.
Examine slope field segments along horizontal lines. If the segments along each horizontal
line 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 the
slopes 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 equation
associated 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 which
they are generated on the calculator. Small discrepancies, or inconsistencies, in the appearance
of segments can usually be discounted.
10.2 Slope Fields
Example 10.2.2 Consider the slope field below, in the window [−2.5, 2.5] × [−2.5, 2.5]
y
2
1
-2
1
-1
2
x
-1
PSfrag replacements
x
y
-2
(a) What can you deduce from reading / interpreting the slope segments?
(b) Which of the following is most likely the differential equation?
(A)
dy
= 0.5xy
dx
(B)
dy
x2
=
dx
y
(C)
dy
= 0.5x2
dx
89
90
Chapter 10 Slope Fields and Euler’s Method
Example 10.2.3 Which of the following differential equations has the solution slope field
picture to the right?
(A)
dy
= 0.5y
dx
(B)
dy
0.2x
=
dx
y
(C)
dy
= xy
dx
(D)
dy
=x+y
dx
(E)
dy
1
=
dx
x
y
2
1
-2
1
-1
2
x
-1
PSfrag replacements
x
y
-2
Example 10.2.4 Which of the following differential equations has the solution slope field
picture to the right?
(A)
dy
= x2
dx
(B)
dy
y
=
dx
x
(C)
dy
= −y
dx
(D)
dy
x
=−
dx
y
(E)
dy
= x2 + y 2
dx
y
4
3
2
1
-4
-3
-2
1
-1
-1
PSfrag replacements
x
y
-2
-3
-4
2
3
4
x
91
10.2 Slope Fields
Example 10.2.5 Which of the following differential equations has the solution slope field
picture to the right?
(A)
dy
=x+y
dx
(B)
dy
=x−y
dx
(C)
dy
= x2
dx
(D)
dy
= 2y
dx
(E)
dy
= y/x
dx
y
4
3
2
1
-4
-3
-2
1
-1
-1
PSfrag replacements
x
y
-2
-3
-4
2
3
4
x
92
Chapter 10 Slope Fields and Euler’s Method
10.3
Euler’s Method
1. The basic idea behind slope fields can be used to find numerical approximations to solutions
of differential equations.
2. Let
dy
= f (x, y) be a differential equation in x and/or y.
dx
Let (x0 , y0 ) be an initial condition.
Suppose the differential equation cannot be solved by our (only) known method: separation
of 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
(0, 1) = 0 + 1 = 1
dx
dy
=x+y
dx
y(0) = 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.
y
3
PSfrag replacements
Solution curve
2
x
y
y = L(x)
1
0.25
0.5
0.75
1.
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
(0.5, 1.5) = 0.5 + 1.5 = 2 =⇒ y = (2(x − 0.5) + 1.5 = 2x + 0.5
dx
Approx for x > 0.5.
Note: A smaller step size leads to a better Euler approximation.
y
y
3
replacements
x
y
3
2
PSfrag replacements
1
x
y
0.25
0.5
0.75
1.
2
1
x
0.25
0.5
0.75
1.
x
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 direction.
4. Continue to stop, and change direction accordingly.
Note: Euler’s Method does not produce the exact solution to an initial value problem, only
an approximation. Decreasing the step size (increasing the number of midcourse corrections),
leads to a better approximation.
94
Chapter 10 Slope Fields and Euler’s Method
In symbols:
As before:
dy
= f (x, y) y(x0 ) = y0
dx
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 )
y
PSfrag replacements
x
y
(x1 , y1 )
∆x f (x0 , y0 )
∆x
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
x
10.3 Euler’s Method
95
dy
= x + y (x0 , y0 ) = (1, 1).
dx
Use Euler’s Method with ∆x = 0.1 to complete the following table. Give your estimates
accurate to three decimal places.
Example 10.3.2 Suppose
n
xn
yn
dy/dx
∆y
n
xn
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
yn
dy/dx
∆y
dy
Example 10.3.3 Given
= 2y with initial condition (x0 , y0 ) = (0, 1). Use Euler’s Method
dx
to estimate the solution using ∆x = 0.2 with five steps.
n
xn
0
0.0
1
0.2
2
0.4
3
0.6
4
0.8
5
1.0
yn
dy/dx
∆y
96
Chapter 10 Slope Fields and Euler’s Method
Example 10.3.4 Consider the differential equation in the previous example:
(x0 , y0 ) = (0, 1).
dy
= 2y with
dx
(a) Solve the differential equation analytically. Use the initial condition to resolve the constant.
(b) Complete the following table in order to compare values of y obtained using Euler’s
Method 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).
y
y
7
7
6
6
5
5
4
4
PSfrag replacements3
PSfrag replacements 3
x
y
2
x
y
1
0
0.
0.2
0.4
0.6
0.8
x
2
1
0
0.
0.2
0.4
0.6
0.8
x
10.3 Euler’s Method
97
Example 10.3.5 For each of the following situations, indicate whether the Euler’s Method
approximation is less than, greater than, or equal to the actual solution of the differential
equation for positive values of ∆x.
(a) The solution curve is concave up.
(b) The solution curve is concave down.
(c)
dy
= k, where k is a constant.
dx
Example 10.3.6 In general, what can you do to improve the accuracy of an Euler’s Method
approximation to the solution of a differential equation?
98
Chapter 10 Slope Fields and Euler’s Method
Example 10.3.7 Consider the initial-value problem
dy
= 6x2 − 3x2 y
dx
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
3
(b) Verify that y = 2 + e−x is a solution to the differential equation.
(c) Find the errors in using Euler’s Method to compute y(1) with each step size. What
happens to the error when the step size is divided by 10?
10.4 Euler’s Method Postscript
10.4
99
Euler’s Method Postscript
The Fundamental Theorem with Euler’s Method
The Problem:
Given a function f 0 , find an approximation to f with the initial condition (x) , y0 ) and evaluate
it on the interval [x0 , xn ].
The Solution:
For small ∆x, ∆y ≈ f 0 (x0 ) ∆x
f (x1 ) ≈ f (x0 ) + f 0 (x0 ) ∆x
f (x2 ) ≈ f (x1 ) + f 0 (x1 ) ∆x = f (x0 ) + f 0 (x0 ) ∆x + f 0 (x1 ) ∆x
Continue in this manner, to obtain:
f (xn ) ≈ f (x0 ) + f 0 (x0 ) ∆x + f 0 (x1 ) ∆x + · · · + f 0 (xn−1 ) ∆x
Using the left endpoint approximation (Riemann sum), the area bounded by f 0 from x0 to
xn can be approximated by:
A ≈ f 0 (x0 ) ∆x + f 0 (x1 ) ∆x + · · · + f 0 (xn−1 ) ∆x
But,
Or,
f (xn ) ≈ f (x0 ) + f 0 (x0 ) ∆x + f 0 (x1 ) ∆x + · · · + f 0 (xn−1 ) ∆x
f (xn ) − f (x0 ) ≈ f 0 (x0 ) ∆x + f 0 (x1 ) ∆x + · · · + f 0 (xn−1 ) ∆x
By substitution, we obtain:
A ≈ f (xn ) − f (x0 ) and
lim A ≈ f (xn ) − f (x0 )
∆x→0
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, and
separation of variables.
2. Differential equations with a logistic equation solution have been on the BC syllabus for a
long 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 model
logistic growth using statistical plots, and we can use logistic regression to obtain an
estimate of the curve.
4. Logistic growth describes many real-world phenomena: the growth of an animal population
subject to finite resources, learning curves, the spread of a disease, the popularity of a new
product, 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 of
growth: 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.
dP
P
The differential equation:
= kP 1 −
dt
N
3. Some logistic growth curve facts:
Suppose an initial condition: P (0) = P0 .
The solution: P (t) =
N
1 + Ae−kt
where A =
N − P0
P0
The point of inflection occurs when dP/dt achieves its greatest value, when P = N/2.
N
N
ln A
=
=⇒
t
=
2
1 + Ae−kt
k
If P is small compared to N :
dP
≈ kP
dt
As P → N , P/N → 1 and dP/dt → 0.
If 0 < P < N then dP/dt > 0.
lim P (t) = N
t→∞
Solve the differential equation.
If P > N then dP/dt < 0
11.3 Logistic Growth Activity
11.3
103
Logistic Growth Activity
One morning, a student in your Calculus BC class hears a rumor before school. Starting with
this 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 analyzing
the 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 equal
chance of hearing the rumor, we can model the process using randomly generated numbers
on 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 the
number 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
104
Chapter 11 Logistic Growth
The Questions
P
dP
dP
= kP 1 −
where
dt
N
dt
is the rate at which student hear the rumor, P is the number who have heard the rumor
at time t, and N is the number of students in the class.
1. Assume the spread of the rumor follows a logistic model:
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
N
write the solution in the form P (t) =
1 + Ae−kt
11.3 Logistic Growth Activity
105
3. Use your solution found in part (2) to compute the value of P at t = 5. How well does this
equation 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 these
related to the original conditions of the problem?
5. Suppose at t = 0 there were two people who heard the rumor, rather than one. Explain
how this might affect the logistic curve.
106
Chapter 11 Logistic Growth
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? Show
the equations used to find this value.
8. Explain any connections between the answers to (6) and (7).
11.3 Logistic Growth Activity
107
9. How does the value of k affect the logistic curve? Experiment with different values of k
and report the results.
10. Describe several phenomenon (other than those discussed already) that might be described
by a logistic curve.
108
Chapter 11 Logistic Growth
11.4
Extensions
Example 11.4.1 (Stewart) There is evidence to suggest that for some species there is a
minimum population m such that the species will become extinct if the size of the population
falls below m. This condition can be incorporated into the logistic equation by introducing
the factor (1 − m/P ), The new model becomes
dP
P
= kP 1 −
dt
N
1−
m
P
(a) Using this differential equation, describe the behavior of the solution if m < P < N . And
for ) < P < m.
Suppose k = 0.08, N = 1000, and m = 200. Carefully sketch a direction field and use it
to sketch several solution curves. Describe what happens to the population for various
initial 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.)
110
Chapter 11 Logistic Growth
Example 11.4.2 (Stewart) In a seasonal-growth model, a periodic function of time is used
to account for seasonal variations in the rate of growth. Such variations could be caused by
seasonal changes in the variations in the availability of food.
dP
= kP cos(rt − φ), P (0) = P0 , where
(a) Find the solution to the seasonal-growth model
dt
k, r, and φ are constants.
(b) Graph the solution for several values of k, r, and φ. Explain how these values affect the
solution. What can you say about lim P (t)?
t→∞
11.5 π From a Polygon - An Exploration
11.5
111
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 fundamental property, that the ratio of the circumference of a circle to its radius is 2π, can be
shown using a recursion technique based on the perimeter of regular n-gons inscribed in a
unit circle.
The Method
Begin with an inscribed triangle.
It is fairly easy√
to establish that the length
√
of each side is 3. The perimeter is 3 3.
PSfrag replacements
x
y
Consider a formula for the length of a side of a 2N -gon, S2N , in terms of the length of a side
of 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.
1
Use the figure to the right to obtain the
following two equations.
PSfrag replacements
2
Sn
(1) 12 = (1 − x)2 +
x
y
2
(2) (S2N )2 = x2 +
Sn
2
x=1−
s
Sn
1−
2
2
Sn /2
2
Solve for x in Equation (1):
1
x
Solve for x in Equation (2):
x=
s
2
S2N
Sn
−
2
2
S2N
112
Chapter 11 Logistic Growth
Set these two expressions for x equal, and solve for S2N : S2N =
r
Since Pn = N · SN
We know S3 =
√
Similarly, S12 =
P2N = 2N 2 −
3, therefore, S6 =
q
q
4 − SN2
r
2−
r
2−
q
4 − SN2
for N ≥ 3.
√
4 − ( 3)2 = 1 and P6 = 6.
q
q
q
√
√
√
2
2 − 4 − 1 = 2 − 3 and P12 = 12 2 − 3
Using Technology
Using a TI-83/84, consider the following
initial values and calculation.
Successive estimates are obtained by
pressing ENTER. Each time, the value of P
is updated.
g replacements
PSfrag replacements
x
y
x
y
For a real technological thrill ride, consider the following.
Initial values, initial calculations.
The new values are stored in a list. Let
L2 = 2π − L1. Using STAT, EDIT, EDIT (the
Stat List Editor), we can see the error of
each successive estimate.
g replacements
PSfrag replacements
x
y
x
y
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 may
return a memory error.
Here are some results:
The points with the logistic regression
curve.
The logistic regression equation.
g replacements
PSfrag replacements
x
y
x
y
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 ask
about 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
y
Z ∞
1
1 1
dx
1+ + +··· >
2 3
x
1
1.5
= lim (ln b) = ∞
b→∞
PSfrag replacements
x
y
1.
0.5
0.
0
1
2
3
4
5
6
x
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 appropriate
to 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 0 (a)(x − a)
Many power series representations can be produced using simple manipulations of existing series. This includes multiplication, division, differentiation, integration. Note which
manipulations affect the domain.
12.2 Power Series Explorations
12.2
117
Power Series Explorations
For |x| < 1, the expression
showing this in two ways.
1
is equal to a polynomial with infinitely many terms. Consider
1−x
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 + x2 + x3 + · · · + xn−1 + · · ·
1−x
2. We can write the fraction
1
in a different from, repeatedly.
1−x
1
1−x+x
x
x − x2 + x2
x2
=
=1+
= 1+
=1+x+
= ...
1−x
1−x
1−x
1−x
1−x
Each step generates an additional term in the polynomial.
The remainder after each step is
xn
1−x
As n becomes large, xn tends to zero, and hence, so does the remainder.
∞
X
1
= 1 + x + x 2 + x3 + · · · + x n + · · · =
xn
1−x
n=0
This is called a power series.
This is an infinite geometric series with ratio x.
for
|x| < 1
118
Chapter 12 Series and Taylor Polynomials
There are many infinite series that can be derived from this initial power series. Consider the
following 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 for
1
, replace x with −x:
1+x
1
1
=
= 1 + (−x) + (−x)2 + (−x)3 + · · · + (−x)n + · · ·
1 − (−x)
1+x
= 1 − x + x 2 − x3 + · · · =
∞
X
(−x)n
n=0
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 process
works.
Let x = 1/2.
1
2
=
1 + (1/2)
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
12.3
119
Exercises
Example 12.3.1 Find a power series representation (first five terms) for each and write the
series using summation notation.
(a)
1
1 + x2
(b) ln(1 + x)
(c) tan−1 x
120
Chapter 12 Series and Taylor Polynomials
(d)
1
1 − x2
(e)
1
(1 + x)2
(f) ln(1 − x)
12.3 Exercises
(g)
x
1−x
(h)
x2
1 + x2
(i)
x
1 + x2
121
122
Chapter 12 Series and Taylor Polynomials
(j) ln(1 + x2 )
Example 12.3.2 For each power series in the previous example, compare the value of the
expression 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)
Expression value
0.800
s4
0.797
Difference
0.003
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
12.3 Exercises
123
Example 12.3.3 Assume there is a power series representation for ex in the form
x
2
3
n
e = a0 + a1 x + a2 x + a3 x + · · · + a n x + · · · =
∞
X
a n xn
n=1
(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 the
expression 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.
124
Chapter 12 Series and Taylor Polynomials
Example 12.3.4 Enter the following functions on your calculator:
Y1 (x) =
1
1+x
Y2 (x) = P11 = the series up to the x11 term
(a) Carefully sketch a graph of Y1 and Y2 in a ZDecimal window.
y
8
6
4
2
PSfrag replacements
-8
-6
-4
2
-2
4
6
8
-2
x
-4
x
y
-6
-8
(b) Use the TRACE or TABLE feature to find the value of each function and the difference for
each 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 approximate
1
well?
1+x
(e) Enter the following function on you calculator: Y2=sum(seq((-X)^N,N,),98,1)). Graph
Y1 and Y2 in a ZDecimal window. Wait patiently. Does this graph support your answer
in part (d)?
12.3 Exercises
125
Example 12.3.5 Previously, we developed a power series expression for tan−1 x. We know
that 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.
126
Chapter 12 Series and Taylor Polynomials
Example 12.3.6 Construct a fourth degree polynomial P (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4
with the following properties:
P (0) = 1;
P 0 (0) = 2;
P 00 (0) = 3;
P 000(0) = 4;
That is, find the constants a0 , a1 , a2 , a3 , a4 , a5
P (4) (0) = 5
12.3 Exercises
127
Example 12.3.7 Construct a fifth degree polynomial with the following properties:
P (0) = sin(0);
P 0 (0) = sin0 (0);
P (5) (0) = sin(5) (0)
P 00 (0) = sin00 (0);
P 000(0) = sin000(0);
P (4) (0) = sin(4) (0);
128
Chapter 12 Series and Taylor Polynomials
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.
y
8
6
4
2
PSfrag replacements
-8
-6
-4
2
-2
-2
4
6
8
x
-4
x
y
-6
-8
(b) What do you observe about the graphs?
(c) What characteristics do you observe about the terms in P (x)? Add two more terms to
P (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 to
approximate sin x. Use a typical calculus operation to produce a polynomial to approximate
cos 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 .
130
Chapter 12 Series and Taylor Polynomials
Example 12.3.11 Use the previous examples to write an equation that links eix with sin x
and cos x.
The resulting expression is called Euler’s Formula, in recognition of the great Swiss mathematician who developed it. Given this expression, we can evaluate eiπ by substitution, and
reveal one of the most unbelievable identities in all of mathematics. This links e, i, π, 1, and
0. Find eiπ
12.3 Exercises
131
Example 12.3.12 Challenge: Use the power series for tan−1 x to prove the following expression for π as the sum of an infinite series:
∞
√ X
π=2 3
(−1)n
n
n=0 (2n + 1)3
132
12.4
Chapter 12 Series and Taylor Polynomials
Measuring Error in a Taylor Polynomial
The nth degree Taylor polynomial for f (x) = ex about x = 0 is
Pn (x) = 1 +
x2 x3
xn
x
+
+
+···+
1! 2!
3!
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/or
predict the error? Consider two approaches.
1. Enter the following functions on your calculator:
Y1 = e x ;
Y2 = 1 +
x2 x3
x
+
+ ;
1! 2!
3!
Y3 = Y 1 − Y 2
Turn off Y1 and Y2 , and graph only Y3 in a ZDecimal window. This produces a visualization
of 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)
=
=
x
x2
xn
xn+1
x
x2
xn
1+ +
+···+
+
+··· − 1+ +
+···+
1! 2!
n! (n + 1)!
1! 2!
n!
!
!
xn+2
xn+1
+
+···
(n + 1)! (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 this
interval) is at x = 1.
R3 (x) ≤
1
1
1
+ + +···
4! 5! 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
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 .
y
2
1
g replacements
-4
-3
-2
1
-1
2
3
4
x
-1
x
y
-2
-3
A graph of Y3 and Y4 .
y
0.1
g replacements
x
y
0.05
x
y
-1.5
-1.
-0.5
0.5
1.
1.5
x
133
134
Chapter 12 Series and Taylor Polynomials
12.5
Connecting Error to the Lagrange Form by
Integration
The Lagrange form of the error is an analytical tool for bounding the error in a Taylor
polynomial.
Consider a third degree Taylor polynomial for a function about a = 0. The fourth degree
term 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
=⇒
Z
x
0
f (4) (t) dt ≤
Z
x
0
M dt
Integrate and simplify:
x
f 000(t)]0 ≤ M (x − 0)
=⇒
f 000 (x) − f 000(0) ≤ M x
=⇒
f 000 (x) ≤ f 000 (0) + M x
Integrate both sides of this expression, from 0 to x:
x
f 00 (t)]0 ≤ f 000(0)(x − 0) +
M 2
x
2!
Repeat this procedure, to eventually arrive at:
f (x) − f (0) ≤ f 0 (0)x +
f 00 (0)
f 000 (0) 3 M 4
x+
x +
x
2!
3!
4!
Rearrange the terms and substitute:
f (x) − P3 (x) ≤
M 4
x
4!
The Lagrange error.
The original example: f (x) = ex .
Since ex is increasing on [0, 1], the largest value, M , is e.
e
= 0.113
R4 (x) ≤
4!
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 numerical
approach and the Lagrange form.
(a) f (x) = e−x on [0, 1]; 4th degree polynomial.
(b) f (x) =
1
on [0, 0.5]; 3rd degree polynomial.
1−x
136
Chapter 12 Series and Taylor Polynomials
12.6
Exploration of Series
Example 12.6.1 Consider the infinite series
2n2
n
n=1 2
∞
X
(a) Intuitively, how does this infinite sum behave?
(b) How does the presence of the coefficient 2 (in the numerator) affect the series sum? Does
it 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 (99
terms):
(i)
2n3
n
n=1 2
∞
X
(ii)
2n4
n
n=1 2
∞
X
Explain the pattern.
(iii)
2n5
n
n=1 2
∞
X
12.6 Exploration of Series
Example 12.6.2 Consider the infinite series
137
∞
X
(n + 1)(n + 2)
n!
n=1
(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 each
case, relate your answer to a well know constant.
Series
Approximation
Sum
∞
X
n
n=1 n!
n2
n=1 n!
∞
X
∞
X
1
n=1 n!
(c) Using the results from part (b), find a value for the sum in part (a) in terms of a well
know constant.
(d) Find a value for the infinite series
2n2 − 3n + 4
n!
n=1
∞
X
138
Chapter 12 Series and Taylor Polynomials
Example 12.6.3 Summarize, in your own words, your understanding of the hierarchy of
functions when finding the value of an infinite sum. Explain your strategy for determining
whether 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
dy dx
=
dx
dt dt
See 1998 BC6 and 2001 BC1.
4. Vector-valued functions:
speed = |v(t)| =
q
v12 + v22
Vector in component form: v = hv1 , v2 i or v = (v1 , v2 )
5. Use parametric equations to simulate motion. This allows students to see functions represented in a variety of ways. Also confirms that rectilinear motion really is motion in a
straight 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
= 2t.
dx
dt
(a) Find
dy
in terms of t.
dt
(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 to
t = 1.
13.2 Selected Problems
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) Find
dy
dy
in terms of θ and evaluate
at θ = π/3.
dx
dx
(c) Find the area of one leaf of the curve.
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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 and
y = 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
Example 13.2.4 Consider a curve defined by x(t) = t4 − t2 and y(t) = t + ln t.
(a) Carefully sketch this curve.
y
2.5
2.
1.5
PSfrag replacements
1.
0.5
x
y
-1.
x
y
x
y
-0.5
0.5
1.
1.5
2.
2.5
3.
3.5
4.
4.5
x
-0.5
-1.
-1.5
(b) Estimate the coordinates of the leftmost point on the curve.
(c) Find the coordinates (analytically) of the leftmost point on the curve.
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