AP Calc

AP Calculus AB

S. BuddLamar High School

pdf

September 29, 2010

ii

Mr. Budd, compiled September 29, 2010

Contents

1 Area and Slope 11.1 Using Graphs to Multiply: Definite Integrals . . . . . . . . . . . 3

1.1.1 What is Calculus? . . . . . . . . . . . . . . . . . . . . . . 31.1.2 What is a Definite Integral? . . . . . . . . . . . . . . . . . 41.1.3 Approximating Area: Counting Squares . . . . . . . . . . 51.1.4 What is “Signed” Area . . . . . . . . . . . . . . . . . . . 61.1.5 Using Known Shapes to Evaluate Definite Integrals . . . . 61.1.6 Using Symmetry . . . . . . . . . . . . . . . . . . . . . . . 8

1.2 Approximating Definite Integrals from Tables . . . . . . . . . . . 131.2.1 Using Tables of Data . . . . . . . . . . . . . . . . . . . . . 131.2.2 Rectangular Approximation Method (RAM) . . . . . . . 171.2.3 Trapezoidal Approximation: Quasi-RAM . . . . . . . . . 181.2.4 Streamlining Calculations for Equal Widths . . . . . . . . 191.2.5 Finding a Range of Values . . . . . . . . . . . . . . . . . . 201.2.6 Midpoint RAM . . . . . . . . . . . . . . . . . . . . . . . . 211.2.7 Unequal Subdivisions . . . . . . . . . . . . . . . . . . . . 23

1.3 Approximating Definite Integral from Formulas . . . . . . . . . . 311.3.1 Definite Integrals from Known Shapes . . . . . . . . . . . 311.3.2 Approximating Definite Integrals . . . . . . . . . . . . . . 321.3.3 Using Symmetry . . . . . . . . . . . . . . . . . . . . . . . 34

1.4 Slope and Rate of Change . . . . . . . . . . . . . . . . . . . . . . 371.4.1 Instantaneous Rate of Change . . . . . . . . . . . . . . . . 371.4.2 Definition and Notation . . . . . . . . . . . . . . . . . . . 381.4.3 Approximating Derivatives from Tabular Data . . . . . . 39

1.5 IROC as a limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471.5.1 Approximating Rate of Change from a Formula . . . . . . 471.5.2 Kinematics: Displacement, Velocity, Acceleration . . . . . 48

1.6 Slope and Area: Pulling It Together . . . . . . . . . . . . . . . . 53

2 Limits 552.1 Introduction to Limits . . . . . . . . . . . . . . . . . . . . . . . . 57

2.1.1 Graphic Introduction to Limits . . . . . . . . . . . . . . . 572.1.2 Step Discontinuities & One-Sided Limits . . . . . . . . . . 602.1.3 Limits from a Table . . . . . . . . . . . . . . . . . . . . . 61

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iv CONTENTS

2.1.4 Limits from an Expression . . . . . . . . . . . . . . . . . . 622.1.5 Substitution and Properties of Limits . . . . . . . . . . . 63

2.2 Limits at Cancelable Discontinuities . . . . . . . . . . . . . . . . 672.2.1 Limits at Cancelable Discontinuities . . . . . . . . . . . . 672.2.2 De-rationalizing with Conjugates . . . . . . . . . . . . . . 682.2.3 Derivative at a Point . . . . . . . . . . . . . . . . . . . . . 692.2.4 De-denominatorizing with LCDs . . . . . . . . . . . . . . 71

2.3 Limit Definition of Derivative as a Function . . . . . . . . . . . . 752.3.1 Derivative as a Function . . . . . . . . . . . . . . . . . . . 752.3.2 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . 77

2.4 Basic Calculus of Polynomials . . . . . . . . . . . . . . . . . . . . 812.4.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812.4.2 Basic Properties of Derivatives . . . . . . . . . . . . . . . 822.4.3 Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 832.4.4 Higher Order Derivatives . . . . . . . . . . . . . . . . . . 842.4.5 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . 84

3 Basic Differentiation 893.1 Antidifferentiation of Polynomials . . . . . . . . . . . . . . . . . . 91

3.1.1 Notation of Antiderivatives . . . . . . . . . . . . . . . . . 913.1.2 Anti-Power Rule . . . . . . . . . . . . . . . . . . . . . . . 913.1.3 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . 933.1.4 General vs. Particular Solutions . . . . . . . . . . . . . . 94

3.2 Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . . 993.2.1 Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . 993.2.2 Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . 102

3.3 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1073.3.1 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 107

3.4 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1133.4.1 Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . 1133.4.2 Horizontal Tangents . . . . . . . . . . . . . . . . . . . . . 1143.4.3 Vertical Tangents . . . . . . . . . . . . . . . . . . . . . . . 1153.4.4 Normal Lines . . . . . . . . . . . . . . . . . . . . . . . . . 1163.4.5 Tangent Line Approximations . . . . . . . . . . . . . . . . 1163.4.6 Introduction to Slope Fields . . . . . . . . . . . . . . . . . 118

4 Curve Sketching 1234.1 Relating Graphs of f and f ′ . . . . . . . . . . . . . . . . . . . . . 125

4.1.1 Relative Extrema . . . . . . . . . . . . . . . . . . . . . . . 1254.1.2 First Derivative Test . . . . . . . . . . . . . . . . . . . . . 126

4.2 Second Derivative Sketching . . . . . . . . . . . . . . . . . . . . . 1354.2.1 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . 1354.2.2 Points of Inflection . . . . . . . . . . . . . . . . . . . . . . 137


CONTENTS v

5 Trigonometrics 1415.1 Differentiation of Trigonometric Functions . . . . . . . . . . . . . 143

5.1.1 Special Limits . . . . . . . . . . . . . . . . . . . . . . . . . 1435.1.2 Trigonometric Derivatives . . . . . . . . . . . . . . . . . . 144

5.2 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 1495.2.1 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . 149

5.3 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1535.3.1 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . 1535.3.2 Differentiating Inverse Functions . . . . . . . . . . . . . . 155

5.4 Related Rates (Triangles) . . . . . . . . . . . . . . . . . . . . . . 1575.4.1 Introduction to Related Rates . . . . . . . . . . . . . . . . 1575.4.2 Related Rates w/ Triangles . . . . . . . . . . . . . . . . . 158

5.5 Antidifferentiating Trig . . . . . . . . . . . . . . . . . . . . . . . 1615.5.1 Antidifferentiating to Inverse Functions . . . . . . . . . . 1615.5.2 Antidifferentiation of Trigonometric Functions . . . . . . 162

6 Exponentials 1656.1 Antidifferentiation by Simplification . . . . . . . . . . . . . . . . 167

6.1.1 u-Simplification . . . . . . . . . . . . . . . . . . . . . . . . 1676.1.2 Simplification with Trigonometrics Inside . . . . . . . . . 1686.1.3 Simplification with Trigonometrics Outside . . . . . . . . 169

6.2 The Happy Function . . . . . . . . . . . . . . . . . . . . . . . . . 1736.2.1 Differentiating the Exponential Function . . . . . . . . . . 1736.2.2 Antidifferentiating the Exponential Function . . . . . . . 1756.2.3 Skippable u-Simplification . . . . . . . . . . . . . . . . . . 176

6.3 Inverse of the Happy Function . . . . . . . . . . . . . . . . . . . . 1796.3.1 Inverse of the Exponential Function . . . . . . . . . . . . 1796.3.2 Implicit Differentiation with ln . . . . . . . . . . . . . . . 1816.3.3 Antidifferentiating Reciprocals . . . . . . . . . . . . . . . 1826.3.4 Antidifferentiating Fractions . . . . . . . . . . . . . . . . . 184

6.4 Separable Differential Equations . . . . . . . . . . . . . . . . . . 1876.4.1 Separable Differential Equations . . . . . . . . . . . . . . 1876.4.2 Separable Differential Equations with Logs . . . . . . . . 188

6.5 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . 1916.5.1 Proportional Growth . . . . . . . . . . . . . . . . . . . . . 1916.5.2 Other Applications of Differential Equations . . . . . . . . 192

7 Existence Theorems 1957.1 Quasi-Limits: One-Sided and Infinite . . . . . . . . . . . . . . . . 197

7.1.1 Step Discontinuities & One-Sided Limits . . . . . . . . . . 1977.1.2 One-Sided Derivatives . . . . . . . . . . . . . . . . . . . . 1987.1.3 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 200

7.2 Limits at Infinity and Horizontal Asymptotes . . . . . . . . . . . 2057.2.1 Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . 2057.2.2 Horizontal Asymptotes . . . . . . . . . . . . . . . . . . . . 207

7.3 Continuity and Differentiability . . . . . . . . . . . . . . . . . . . 209


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7.4 Some Basic Calculus Theorems . . . . . . . . . . . . . . . . . . . 2137.4.1 Intermediate Value Theorem . . . . . . . . . . . . . . . . 2137.4.2 Extreme Value Theorem . . . . . . . . . . . . . . . . . . . 2137.4.3 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 214

7.5 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 2197.5.1 Average Rate of Change . . . . . . . . . . . . . . . . . . . 2197.5.2 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . 219

7.6 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2277.6.1 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . 2277.6.2 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . 2277.6.3 Evaluating Definite Integrals Exactly . . . . . . . . . . . . 2287.6.4 Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2297.6.5 Evaluating Definite Integrals . . . . . . . . . . . . . . . . 231

8 Integral Theorems 2358.1 MVT for Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 237

8.1.1 Substitution of Variables . . . . . . . . . . . . . . . . . . . 2378.1.2 Properties of Definite Integrals . . . . . . . . . . . . . . . 2388.1.3 Average Value . . . . . . . . . . . . . . . . . . . . . . . . 2388.1.4 Mean Value Theorem for Integrals . . . . . . . . . . . . . 241

8.2 Accumulation Functions . . . . . . . . . . . . . . . . . . . . . . . 2478.2.1 Accumulation Functions . . . . . . . . . . . . . . . . . . . 2478.2.2 Fundamental Theorem of Calculus, part II . . . . . . . . . 2478.2.3 Curve Sketching with Accumulation Functions . . . . . . 250

8.3 Quick, Cheap Antiderivatives . . . . . . . . . . . . . . . . . . . . 2578.3.1 Creating Quick, Cheap Antiderivatives . . . . . . . . . . . 257

9 Area and Volume 2619.1 More Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . 263

9.1.1 Definite Integral . . . . . . . . . . . . . . . . . . . . . . . 2639.1.2 Area: Slicing dx . . . . . . . . . . . . . . . . . . . . . . . 2639.1.3 Total Distance . . . . . . . . . . . . . . . . . . . . . . . . 2649.1.4 Other Applications . . . . . . . . . . . . . . . . . . . . . . 266

9.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2719.2.1 High and Low y Switch . . . . . . . . . . . . . . . . . . . 2719.2.2 Area: Slicing dy . . . . . . . . . . . . . . . . . . . . . . . 2729.2.3 Total Distance . . . . . . . . . . . . . . . . . . . . . . . . 272

9.3 Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2779.3.1 Volumes of Rotation . . . . . . . . . . . . . . . . . . . . . 277

9.4 Volume: Slicing with Washers . . . . . . . . . . . . . . . . . . . . 2819.4.1 Slicing with Washers . . . . . . . . . . . . . . . . . . . . . 281

9.5 Non-Circular Slicing . . . . . . . . . . . . . . . . . . . . . . . . . 2859.6 Area and Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . 2899.7 Related Rates with Volume . . . . . . . . . . . . . . . . . . . . . 295

9.7.1 Volume problems . . . . . . . . . . . . . . . . . . . . . . . 295


CONTENTS vii

10 Extrema and Optimization 29910.1 Absolute Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . 301

10.1.1 Absolute Extrema . . . . . . . . . . . . . . . . . . . . . . 30110.1.2 Absolute Extrema from the Derivative . . . . . . . . . . . 30210.1.3 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . 304

10.2 First Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30910.2.1 First Derivative Test . . . . . . . . . . . . . . . . . . . . . 309

10.3 Second Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . 31110.3.1 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . 31110.3.2 Second Derivative Test . . . . . . . . . . . . . . . . . . . . 313

11 Review 31911.1 Separable Differential Equations . . . . . . . . . . . . . . . . . . 321

11.1.1 Separable Differential Equations . . . . . . . . . . . . . . 32111.1.2 Separable Differential Equations with Logs . . . . . . . . 323

11.2 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . 32711.2.1 Proportional Growth . . . . . . . . . . . . . . . . . . . . . 32711.2.2 Other Applications of Differential Equations . . . . . . . . 328

11.3 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33111.4 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33711.5 Integral as Accumulator . . . . . . . . . . . . . . . . . . . . . . . 34511.6 Particle Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35111.7 Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35711.8 Extrema and Optimization . . . . . . . . . . . . . . . . . . . . . 36311.9 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 36711.10Differential Equations Again . . . . . . . . . . . . . . . . . . . . . 37111.11Related Rates Again . . . . . . . . . . . . . . . . . . . . . . . . . 37711.12Area and Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . 38111.13Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38711.14Miscellany . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39311.15More Miscellany . . . . . . . . . . . . . . . . . . . . . . . . . . . 405

12 Makeup 40712.1 MU: Differential Equations . . . . . . . . . . . . . . . . . . . . . 40812.2 MU: Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . 40912.3 MU: Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41012.4 MU: Integral as Accumulator . . . . . . . . . . . . . . . . . . . . 41712.5 MU: Linear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 41912.6 MU: Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42312.7 MU: Extrema and Optimization . . . . . . . . . . . . . . . . . . 42412.8 MU: Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . 42812.9 MU: Area and Volume . . . . . . . . . . . . . . . . . . . . . . . . 42912.10MU: Tangent Lines . . . . . . . . . . . . . . . . . . . . . . . . . . 435


viii CONTENTS


Unit 1

Introduction to Calculus:Area and Slope

1. The Definite Integral as Area

2. Approximating Definite Integral by Riemann Slicing

3. Rate of Change

4. Approximating Rate of Change from Graph, Table, or Equation

5. Slope and Area: Pulling It Together

Advanced Placement

Concept of the derivative.

• Derivative presented geometrically, numerically, and analytically.

• Derivative interpreted as an instantaneous rate of change.

Interpretations and properties of definite integrals.

• Computation of Riemann sums using left, right, and midpoint evaluationpoints.

• Basic properties of definite integrals (Examples include additivity and linear-ity.)

1

2 AP Unit 1 (Area and Slope)

Numerical approximations to definite integrals. Use of Riemann and trapezoidalsums to approximate definite integrals by functions represented algebraically, geo-metrically, and by tables of values.

International Baccalaureate

(MM 8.6) The estimation of the numerical value of a definite integral using thetrapezium rule. Included: an appreciation of the effect of doubling the number ofsub-intervals.


AP Unit 1, Day 1: Using Graphs to Multiply: Definite Integrals 3

1.1 Using Graphs to Multiply: Definite Inte-grals

Advanced Placement


• Basic properties of definite integrals (Examples include additivity and linear-ity.)

Numerical approximations to definite integrals.

Textbook We won’t be following the book too closely the first couple weeks,so that there is limited correspondence to a section in the book. The closestsection in content would be §4.3 Area or §4.4 The Definite Integral. [16]

Resources §5.1 Areas and Integrals in Ostebee and Zorn [17]. §1-3 One Typeof Integral of a Function in Foerster [10]. Explorations 1-3a:“Introduction toDefinite Integrals” and 1-4a:“Definite Integrals by Trapezoidal Rule” in [9].

1.1.1 What is Calculus?

Ostebee and Zorn describe the focus of calculus as follows: “The tangent-lineproblem and the area problem are the two main geometric problems of calculus.”[17] The tangent-line problem is an issue of slope, so that our main concernsin calculus are slope and area. Our interest in slope and area is not purelygeometric, however. Slope is our codeword for rate of change, which can be rateof people entering AstroWorld, or the rate at which oil leaves a gash in an oiltanker. Likewise, area can be a whole range of things from people who haveentered Super Happy Fun Land in a six-hour time period to the amount of oilthat has bled out of a shipwrecked tanker. Area, as we shall see, can even beused to represent the volume of an object.

Foerster, whose materials we will see much of this year, describes calculus asconsisting of four things: limits, derivatives, integrals, and integrals [10]. Thesethings probably have no meaning for you, and you are probably wondering whyintegrals is listed twice. Understanding these things is what we will seek todo over the next nine months. As a brief introduction, I will tell you thatderivatives are related to slope, and one of the integrals (definite integrals) isrelated to area. Both derivatives and definite integrals are limits, and the othertype of integral (indefinite) is also related to both derivatives and indefiniteintegrals.



1.1.2 What is a Definite Integral?

Example 1.1.1 Begin with Exploration 1-3: Introduction to Defi-nite Integrals.

Figure 1.1: Velocity, v(t), as a function of the number of seconds, t, since youstarted slowing. [9]

Distance is Velocity times Time. Looking at the graph in Figure 1.1, we can seethat after 30 seconds, the velocity is basically constant at 60 feet per second.If we were to find the distance traveled between 30 and 50 seconds, we wouldmultiply 20 seconds by 60 feet per second, yielding 1200 feet.

Notice that the same product (20 seconds × 60 feet per second) is representedby the area under the curve of v(t) from 30 seconds to 50 seconds. The shapeof this region is a rectangle, and the formula for finding the area of a rectangleis A = l × w. If you look at the region under the curve of v(t) from 30 secondsto 50 seconds, you should notice that it is a rectangle with width of 20 secondsand height of 60 feet per second. The area of this rectangle is 1200 feet, or 20seconds × 60 feet per second.

The distance traveled from a starting time to a stopping time is the area underthe velocity curve between the two times. This area is called the definite inte-gral. This is true for the simple case between 30 and 50 seconds, but it is alsotrue for the less simple case between 0 and 20 seconds. To find the distance trav-eled between 0 and 20 seconds, we would need to multiply the time difference(20 seconds) by the velocity, except this is not so straightforward because thevelocity is changing with time. So instead of finding the product by multiplyingtwo numbers, I need a different approach.



When we multiplied two constants, we were doing the same arithmetic as findingthe area of a simple rectangle of constant height. In the less straightforwardcase of multiplying a varying v(t) by t, the distance is still the area under thecurve of v(t) from the starting time to the ending time. But since we don’t havea simple formula (because we don’t have a simple shape) we need a differentapproach to finding the area.

Definition 1.1 (Definite Integral). The definite integral of the function f fromx = a to x = b [written

∫ baf(x) dx] gives a way to find the product of (b− a)

and f(x), even if f(x) is not a constant. [10]

This definition of the definite integral tells us why we need the definite integral.It doesn’t tell us how to get it. Here’s a geometric explanation from a differentauthor:

Definition 1.2 (The Integral as Signed Area). Let f be a function defined fora ≤ x ≤ b. ∫ b

a

f(x) dx

denotes the signed area bounded by x = a, x = b, y = f(x), and the x-axis. [17]

1.1.3 Approximating Area: Counting Squares

If f(x) is always positive, and a is less than b, then∫ baf(x) dx is the area under

f between a and b. We will talk more about the qualifier “signed” in a bit.

One way to approximate an area is the “Counting Squares” approach.[10] Thisapproach is fairly basic; you superimpose a grid on your graph and lightlyshade the area you are approximating (or imagine the shading in your head).You count the number of whole squares that are shaded. Sometimes I like tocount all the whole squares in vertical strips, and this approach will be helpfulto us in the future. Then you count the partial squares, rounding the shadedportion of each square to the nearest tenth, 0.1. After you have added the totalnumber of whole and partial squares, you multiply the number of squares bythe area of each square, being conscious of your units. The number of squaresmultiplied by the area of each square is the area of the shape.

Looking at the example Figure 1.1, we can see that the area under the curvefrom 30 to 50 seconds gives 24 squares. At 50 feet per square (10 feet per secondtimes 5 seconds), that gives a displacement of 1200 feet. How many squares arebetween 0 and 20 seconds? Therefore what is the change in your position in thefirst twenty seconds?



1.1.4 What is “Signed” Area

To find the definite integral of f(x) from x = a to x = b, you are basicallylooking for the signed area under the curve of f between a and b. What do wemean by signed area? If the area is above the x-axis, it is counted as positive;if the area is below the x-axis it is counted as negative (i.e., a negative amountabove the axis). Think about why this is important. What’s happening whenthe velocity is negative? If the velocity were negative, how should you countthe area/ distance?

[Another way to make the signed area negative is if b is less than a, so thattaking you from a to b means that you go right to left on the graph, i.e., thechange in x, ∆x ≈ dx, is negative.]

Figure 1.2: [10]

Example 1.1.2 The graph in Figure 1.2 shows v(t) centimeters persecond as a function of t seconds after an object starts moving. Atwhat time does the object change direction? How far is the objectfrom its starting point when t = 9 sec? What is the total distancetraveled by the object? [adapted from [10]]

[Ans: 5 sec, 7.1 cm]

Note the distinction between displacement and total distance.

1.1.5 Using Known Shapes to Evaluate Definite Integrals

Example 1.1.3 Several areas are shown in Figure 1.3, labeled asintegrals. Use familiar area formulas to evaluate each integral. [17]

[Ans: 6; k (b− a); 9

4π + 32

]Mr. Budd, compiled September 29, 2010


Figure 1.3: [17]

Example 1.1.4 (adapted from AB ’03) Let f be a function defined

Figure 1.4: From 2003 AP Calculus AB exam

on the closed interval −3 ≤ x ≤ 4. The graph of f ′, a function thatis related to f , but different from f , known as the derivative of f ,consists of one line segment and a semicircle, as shown in Figure 1.4Find

(a)∫ 0

−3f ′(x) dx

(b)∫ 4

0f ′(x) dx

[Ans: - 3

2 ; −8 + 2π]



1.1.6 Using Symmetry

Example 1.1.5 (adapted from AB ’01) A car is traveling on astraight road with velocity 40 ft/sec at time t = 0. For 0 ≤ t ≤ 18seconds, the car’s acceleration a(t), in ft/sec2, is the piecewise linearfunction defined by the graph in Figure 1.5.

Figure 1.5: A car’s acceleration

(a) How fast is the car going at time t = 0?

(b) How much does the car’s velocity increase during the first sec-ond? During the first two seconds?

(c) What is the car’s velocity at time t = 2? At time t = 6?

(d) What change takes place to the car’s velocity at time t = 6?

(e) At what time does the velocity of the car return to 40 ft/sec?

[Ans: 40 (ft/sec); 15, 30 (ft/sec); 70, 100 (ft/sec); v decreases; 12 s]

Problems

big giant blue-green Calculus book p. 380: Writing Exercises # 1,2; # 41-44

1.A-1 The online supplement for AP Calculus AB during the academic year hasbeen migrated to Lamar’s new Moodle site, so take the following steps toenroll. Go to http://moodle.houstonisd.org/lamarhs/ and follow theinstructions for creating a new account. You will then need to search forAP Calculus AB to enroll. When asked for it, the enrollment key for thisclass will be area for Mr. Budd’s class, or thompson# for Mr. Thompson’sclass, where # represents the period. This is a vital online supplement towhat happens in class.


http://moodle.houstonisd.org/lamarhs/login/index.php


Figure 1.6: [10]

1.A-2 In Figure 1.6, a car is slowing down from a speed of v = 60 ft/sec. Estimatethe distance it goes from time t = 5 sec to t = 25 sec by finding the definiteintegral. [10] [Ans: about 680 feet]

1.A-3 In Figure 1.7, a car speeds up slowly from v = 55 mi/hr during a long trip.

Figure 1.7: [10]

Estimate the distance it goes from time t = 0 hr to t = 4 hr by findingthe definite integral. [10] [Ans: about 266 miles]

1.A-4 In the previous two problems, you found a distance using a definite inte-gral. Suppose you use the formula d = vt, rearranged to v = d

t . Whatvelocities do you get in the previous two problems when you divide thedistance by the change in time? What do you think this represents?

1.A-5 The rate at which people enter an amusement park on a given day ismodeled by the function E of time t. E(t) is measured in people per hourand time t is measured in hours after midnight. When the park opens at 9a.m., there are no people in the park. Explain the meaning of

∫ 17

9E(t) dt.

Is this equal to the number of people in the park? Why or why not?[Ans: The number of people who entered the park by 5 p.m.; no]

1.A-6 A blood vessel is 360 millimeters (mm) long with circular cross sections ofvarying diameter. If x represents the distance from one end of the blood



vessel, and B(x) is a function that represents the diameter at that point,

then using correct units explain the meaning of∫ 360

0π

(B(x)

2

)2

dx, and

∫ 275

125π

(B(x)

2

)2

dx.

1.A-7 Let g be the function shown graphically in Figure 1.8. When asked to

Figure 1.8: Graph of g [17]

estimate∫ 2

1g(x) dx, a group of calculus students submitted the following

answers: −4, 4, 45, and 450. Only one of these responses is reasonable;the others are “obviously” incorrect. Which is the reasonable one? [17][Ans: 45]

1.A-8 The graph of a function f is shown in Figure 1.9. [Adapted from [17]]

Figure 1.9: Graph of f [17]

(a) Which of the following is the best estimate of∫ 6

1f(x) dx: −24, 9,

20, 38? Justify your answer.

(b)∫ 8

6f(x) dx ≈ 4. Does this approximation overestimate or underesti-

mate the exact value of the integral? Justify your answer.

(c) Explain a quick way to tell that 12 ≤∫ 7

3f(x) dx.

[Ans: 20; underestimate]




1.A-9 The graph of a function f (shown in Figure 1.10) consists of two straightlines and two one-quarter circles. Evaluate each of the following integrals.

(a)∫ 2

0f(x)dx

(b)∫ 5

2f(x)dx

(c)∫ 5

0f(x)dx

(d)∫ 9

5f(x)dx

(e)∫ 4

4f(x)dx

(f)∫ 15

0f(x)dx

(g)∫ 15

0|f(x)| dx [

Ans: 4, 9π4 , 4 + 9π

4 , −4π, 0, −8− 7π4 , 16 + 25π

4

]1.A-10 Suppose Mr. Budd is driving to the Utah Shakespearean Festival in Cedar

City, UT. Once he gets on the road, he sets his cruise control for 55 mph.Let t be the number of hours since he started driving on cruise control.

(a) How far has he gone during the first half hour on cruise control? thefirst hour? the first two hours?

(b) Write an equation for the velocity, i.e., v(t) =(something).

(c) Graph the velocity versus time.

(d) Find the area under the curve of v(t) from t = 0 to t = 0.5. Also,find the area from t = 0 to t = 1 and also to t = 2.

(e) What shape are these areas in? If I look at the area from t = 0 tot = tstop, what is the width of the figure (as an expression with tstop

in it)? the height? the area (as an expression of tstop)? Call yourexpression for area A(tstop).

(f) Plot a graph of distance traveled versus time. Use the points (0.5,distance for 0.5),(1,distance for 1), and (2,distance for 2). Look for a pattern, anduse your result for A(tstop) to connect the dots.



(g) On your graph of distance versus time, what is the slope at t = 0.5?at t = 1? at t = 2? Indicate units.

1.A-11 (from Explorations 1-3a [9]) As you drive on the highway you accelerateto 100 feet per second to pass a truck. After you have passed, you slowdown to a more moderate speed. Table 1.1 shows your velocity, v(t), as afunction of the number of seconds, t, since you started slowing.

Table 1.1: Your velocity after passing a truck

t v(t)(s) ft/s0 1005 77.375510 67.5477

(a) How fast are you going at t = 0? How fast are you going at t = 5?Why is it not so straightforward to ask how fast you were going for0 ≤ t ≤ 5?

(b) If your speed is constantly decreasing, give an upper estimate of howfar you traveled in the first 5 seconds. Give a lower estimate of yourdisplacement in the first 5 seconds.

(c) If you had to give one number for your distance in feet for the first 5seconds, what might it be? Give a reason for how you obtained youranswer.

(d) How far did you travel for 5 ≤ t ≤ 10? For 0 ≤ t ≤ 10?

[Ans: 100, 77.3755, ; 500, 386.878 (ft); 443.439 ft; 362.308, 805.747 ft]

1.A-12 Read the handout “How to Succeed in Calculus.”

(a) Give examples of three things on the list that you already do.

(b) Name one thing on the list that you will try to improve this year.Describe specifically what actions you will take this week.

(c) Submit your answer in the appropriate place on the moodle site.

1.A-13 Start Exploration 1-4a: “Definite Integrals by Trapezoidal Rule”; do prob-lems 1 through 3.


AP Unit 1, Day 2: Approximating Definite Integrals from Tables 13

1.2 Approximating Definite Integrals from Ta-bles

Advanced Placement




Textbook §4.7 Numerical Integration [16]

Resources §5.1 Areas and Integrals in Ostebee and Zorn [17]. §1-4 DefiniteIntegrals by Trapezoids, from Equations and Data in Foerster [10]. Exploration1-4a:“Definite Integrals by Trapezoidal Rule” in [9].

1.2.1 Using Tables of Data

Example 1.2.1 (from Explorations 1-3 [9]) As you drive on thehighway you accelerate to 100 feet per second to pass a truck. Afteryou have passed, you slow down to a more moderate speed. Table1.2 shows your velocity, v(t), as a function of the number of seconds,t, since you started slowing.

Table 1.2: Your velocity after passing a truck

t v(t)(s) (ft/s)0 1005 77.375510 67.547715 63.278620 61.424225 60.618730 60.2687



(a) What’s the fastest you went in the first five seconds? Theslowest? Give an upper and lower range for the displacementin the first 5 seconds.

(b) How might one obtain a single best estimate for the change inposition for the first 5 seconds?

(c) On Figure 1.11, show that each estimate, upper and lower, isrepresented graphically by a rectangle with a width of 5 sec-onds. Graphically visualize why the upper-estimate rectangleincludes too much area, and the lower-estimate rectangle doesnot include enough area.

(d) If, instead of rectangles that are either too big or too small,suppose we represent the area with one trapezoid, with a widthof 5 seconds. The formula for the area of a trapezoid is

AT = b

(h1 + h2

2

)i.e., the base times the average of the two heights. For a trape-zoid that best represents the area of the graph between t = 0and t = 5, what are the two heights, and what is the area?

(e) Estimate your change in position for each of the subintervals[5, 10], [10, 15], [15, 20], and for the overall interval [0, 20].

Figure 1.11: Velocity, v(t), as a function of the number of seconds, t, since youstarted slowing. [9]

Example 1.2.2 (adapted from Finney, et al. [8]) Try this in yourmighty, mighty groups of four. A power plant generates electricity by



burning oil. Pollutants produced by the burning process are removedby scrubbers in the smokestacks. Over time the scrubbers becomeless efficient and eventually must be replaced when the amount ofpollutants released exceeds government standards. Measurementstaken at the end of each month determine the rate at which pollu-tants are released into the atmosphere as recorded in the Table 1.3.

Table 1.3: [8]

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov DecPollutantRelease Rate 0.20 0.25 0.27 0.34 0.45 0.52 0.63 0.70 0.81 0.85 0.89 0.95(tons/day)

(a) What is an upper estimate for the total tonnage of pollutantsreleased in the month of January? February? June?

(b) Suppose you plotted the data on a graph of Pollution Rate(tons/day) vs. Time (day). Describe how the total tonnagereleased for each of those months represents a rectangle, onefor each month.

(c) What are the lower estimates for these months?

(d) Give an upper estimate of the total tonnage of pollutants re-leased from the beginning of January to the end of June. As-suming that new scrubbers allow only 0.05 ton/day released,what is a lower estimate? Why would this problem be easier ifthe scrubbers didn’t decline, and the pollution rate stayed at0.05 tons/day? [Ans: 61.32, 47.04]

(e) In the best case, approximately when will a total of 125 tons ofpollutants have been released into the atmosphere? [Ans: Oct 27]

(f) The upper and lower approximations give a range of reason-able values for the definite integral, but neither one of them isnecessarily very reliable. Graphically, instead of having a rect-angle at the highest possible y-value for each subinterval, orthe lowest possible y-value, what might be a more reasonableapproach. Numerically, rather than using the upper or lowerapproximations, what might be a more reasonable approach?

For this problem, the Upper Rectangular Approximation also happens to be aRight-Endpoint Rectangular Approximation. An Upper RAM will be the RightRAM so long as the function is always increasing. For this problem, the Lower



Rectangular Approximation also happens to be the Left-Endpoint RectangularApproximation. The Left RAM will happen to be the Lower RAM wheneverthe function is increasing. Were the function decreasing, the left RAM wouldbe the upper approximation.

Terms

In using a rectangular approximation to estimate∫ baf(x) dx:

• The interval starts at a and ends at b. For this problem, it is the beginningof January to the end of June. The interval width is b−a, e.g., 181 or 182days depending on leap-hood.

• The interval is divided into subintervals. For the pollution problem, thesubintervals are the months. The subinterval widths would be 31 days, 28(or 29) days, etc.

• We will pretend that Riemann sum is German for RAM.

With a rectangular approximation method, the actual area for each subintervalis replaced with the area of a rectangle. The rectangle will have the samewidth as the subinterval, and the height is determined by whichever RectangularApproximation Method is chosen. There are countless types of RectangularApproximation Methods, but five which you need to know:

• Left endpoint Rectangular Approximation Method (RAM)[8] - the heightof each approximating rectangle is the height of the left side of the corre-sponding subinterval (e.g., beginning of the month).

• Right endpoint RAM - the height of each approximating rectangle is theheight of the right side of the corresponding subinterval (e.g., end of themonth).

• Midpoint RAM - the height of each approximating rectangle is the heightin the middle of the subinterval (e.g., the sixteenth of the month).

• Upper RAM - the height of each approximating rectangle is the maximumheight in the corresponding subinterval.

• Lower RAM - the height of each approximating rectangle is the minimumheight in the corresponding subinterval.

In addition to the Rectangular Approximation Methods, there is also:



• Trapezoidal Approximation Method, in which the actual area of each subin-terval is replaced by the area of a trapezoid. The base of the trapezoid isthe same as the subinterval width, just as for rectangles. For each trape-zoid, the two heights used are the two heights on the left and right of eachsubinterval.

Example 1.2.3 Why did I not ask for the Midpoint Approximationfor the pollution problem? In what cases could I ask for a MidpointRAM?

1.2.2 Rectangular Approximation Method (RAM)

There are many physical situations where we must multiply two quantities, oneof which is not a set constant, but a continuously changing variable. In orderto multiply two things, one of which is changing, we utilize the definite integral,which is nothing more than the signed area under a curve. Previously, we haveestimated the area under a curve via the “counting squares” approach. Anotherway to estimate the area of a funky shape is to approximate the shape with aseries of vertical rectangles, which, together, form a blocky or pixelized versionof the original shape.

Our approach here is to divide the shape or region into a number of funky stripsthat have three straight sides and a curved top (or bottom if below the x-axis)that follows the function whose definite integral we are finding, i.e., that we areintegrating.

Once you have divided the shape into a number of strips, the next thing to dois replace the strip with a rectangle of approximately the same size. The ideais that we are approximating the area of the strip, which we don’t know (sincewe don’t have a formula to find the area of a funky strip) with the area of arectangle, for which we do have a formula. The area of a rectangle is base timesheight.

• The width of each rectangle is the subinterval width, which is usuallydetermined to some extent either by the way the problem is asked, or bythe data itself.

• The constant height of each approximating rectangle is based on the vary-ing height of the funkily-shaped strip.

There are several different rules for assigning a height to each rectangle. Hereare two, but we will discuss others a little later:



• Left endpoint Rectangular Approximation Method (RAM)[8] - the heightof each approximating rectangle is the height of the left side of the corre-sponding strip.

• Right endpoint RAM - the height of each approximating rectangle is theheight of the right side of the corresponding strip.

Figure 1.12: Left endpoint and right endpoint rules for the Rectangular Ap-proximation Method [8]

The left endpoint rectangular approximation is designated by Ln, where n isthe number of rectangles (or strips or slices or subintervals). The right endpointrectangular approximation is designated by Rn.

1.2.3 Trapezoidal Approximation: Quasi-RAM

What is usually better than taking a constant height at the right endpoint, or aconstant height at the left endpoint is joining the left and right endpoints of eachfunky strip with a line segment, creating a trapezoid instead of a rectangle. TheTrapezoidal Approximation Method finds the funky area by adding up the areasof multiple replacement trapezoids, just like the Rectangular ApproximationMethod added the areas of multiple rectangles.

Example 1.2.4 (adapted from AB ’98) A table of values for thevelocity v(t), in ft/sec, of a car traveling on a straight road, at 5second intervals of time t, for 0 ≤ t ≤ 50, is shown in Table 1.4.

Table 1.4: Velocity of a car traveling on a straight roadt (seconds) 0 5 10 15 20 25 30 35 40 45 50v(t) (ft/sec) 0 12 20 30 55 70 78 81 75 60 72

(a) Approximate∫ 50

0v(t) dt with a left and right rectangular and

trapezoidal approximations, each with five subintervals.



(b) Draw rectangles or trapezoids on the graphs in Figure 1.13 todemonstrate each of the three methods of estimation. From thegraphs, which seems to be the most accurate?

(c) Find L10, R10 and T10.

(d) Using correct units, explain the meaning of∫ 50

0v(t) dt.

(e) How could you estimate the average velocity of the car?

Figure 1.13: Draw the appropriate rectangles or trapezoids for L5, R5, and T5

1.2.4 Streamlining Calculations for Equal Widths

If all of the subinterval widths are the same, the calculations for rectangularand trapezoidal approximations can be simplified.

Left RAM

∫ b

a

f(x) dx ≈ Ln = ∆x (y0 + y1 + y2 + · · ·+ yn−1)

where ∆x = b−an .

Right RAM

∫ b

a

f(x) dx ≈ Rn = ∆x (y1 + y2 + · · ·+ yn−1 + yn)



Trapezoidal Approximation

∫ b

a

f(x) dx ≈ Tn = ∆x(y0

2+ y1 + y2 + · · ·+ yn−1 +

yn2

)=

∆x2

(y0 + 2y1 + 2y2 + · · ·+ 2yn−1 + yn)

Example 1.2.5 Refer again to Table 1.2.

(a) What is the width of each subinterval?

(b) Use the above formulas to find T4, R4, and L4 to approximate∫ 20

0v(t) dt on your calculator in one input step, using values

from the table.

Example 1.2.6 Refer once again to the data in Table 1.4 on page18. Estimate

∫ 40

5v(t) dt. What does this represent?

1.2.5 Finding a Range of Values

There may be a situation in which you want to find a range of values for thedefinite integral, i.e., what is the best case scenario, and what is the worst casescenario.

Here are two more rules for approximating the height of each funky shape.

• Upper RAM - the height of each approximating rectangle is the maximumheight in the corresponding strip.

• Lower RAM - the height of each approximating rectangle is the minimumheight in the corresponding strip.

The upper Riemann sum is designated Un, where n is the number of subintervals.The lower sum is designated Ln. That’s right, the lower and left sums have thesame designation. You will have to tell which is which from context. If you’reasked to find L8 and U8, you should find a lower and upper Riemann sum, nota left and upper.

The importance of these rules are in giving a range of values. The upper RAMis always an overestimate, whereas the lower RAM always underestimates theactual integral. If I is the actual value of the definite integral, then

Ln ≤ I ≤ Un



Example 1.2.7 Refer once again to the data in Table 1.4 on page18. Find lower and upper rectangular approximations, using fivesubintervals. Then find U10 and L10. What’s happening to therange as you increase the number of subintervals? Draw rectangleson Figure 1.14 to demonstrate these approximation methods.

Figure 1.14: Draw the appropriate rectangles or trapezoids for U5 and L5.

Key Questions

1. When is a lower sum always the same as a left sum?

2. What happens to the range between the upper and lower rectangularapproximations as the number of subintervals increases?

1.2.6 Midpoint RAM

Recall that we have discussed the left and right rectangular approximation meth-ods. A third, similar, method is the midpoint approximation, which typicallygets confused with the trapezoidal approximation.

• Midpoint RAM - the height of each approximating rectangle is the heightof the strip in the middle. If you were to take the funky strip and fold itso that the right side and the left side touch, the creased side would bethe midline, and the length of that folded side would be used as the heightof the approximating rectangle. This would be done for each funky strip.

Example 1.2.8 Refer yet again to the data in Table 1.4 on page18. Estimate the area under the graph by using the Midpoint Rect-angular Approximation with n = 5. Draw rectangles on the graphsin Figure 1.16 to show that you understand the midpoint approxi-mation.



Figure 1.15: Left endpoint, right endpoint, and midpoint rules for the Rectan-gular Approximation Method[8]

Figure 1.16: Draw the appropriate rectangles for M5

Key Questions

1. What is the difference between Mn and Tn?

Accuracy

Upper and Lower are obviously the worst methods in terms of accuracy, as theygive us extreme values.

Right and Left cannot be considered much more reliable than upper and lower,and frequently give upper and lower. (Why?)

Trapezoidal approximations can be considered more accurate than right, left,upper, or lower approximations. But what about the midpoint rule?

If you consider a trapezoidal and a midpoint approximation with the samenumber of subintervals, then generally the midpoint approximation is abouttwice as accurate. However, think about using a table of data. If I have nineequally spaced data points, i.e., eight subintervals, then I can use all eight datapoints to calculate T8. I would not, however, be able to find M8. (Why not?)The best I could do would be M4. For midpoint and trapezoidal approximations,



doubling the number of subintervals generally quadruples your accuracy. Sothat:

• Tn is accurate

• Mn is roughly twice as accurate as Tn

• T2n is roughly four times as accurate as Tn and therefore roughly twice asaccurate as Mn

It is worth noting that if Mn overestimates the actual integral, then Tn under-estimates, and vice versa. Since the midpoint approximation is about twice asaccurate as the trapezoidal approximation, we can make a super approximationthat is a weighted average of the midpoint and trapezoid, with the midpointbeing weighted twice as much as the trapezoid:

S2n =2Mn + Tn

2 + 1

Why does the number of subintervals double?

1.2.7 Unequal Subdivisions

Recall that the data you have may be such that it is not evenly spaced. In thesecases, you must calculate each rectangle or trapezoid separately, and add themat the end.

Example 1.2.9 (adapted from AB ’03) The rate of fuel consump-tion, in gallons per minute, recorded during an airplane flight isgiven by function R of time t. A table of selected values of R(t), forthe time interval 0 ≤ t ≤ 90 minutes is shown.

(a) Approximate the amount of fuel consumed in the first 30 min-utes using left and right endpoint and trapezoidal methods.Indicate units.

(b) Approximate the amount of fuel consumed in the time inter-val 30 ≤ t ≤ 40 minutes, using left and right endpoint andtrapezoidal methods. Indicate units.

(c) If we know that R(t) is what we call strictly increasing, i.e.,R always increases and never decreases, then what would bea range for the amount of fuel consumed by the plane in thefirst 90 minutes? Why is it important that we know that R isstrictly increasing?



Table 1.5: Rate of fuel consumption of a plane

t R(t)(minutes) (gallons per minute)

0 2030 3040 4050 5570 6590 70

(d) Draw rectangles on Figure 1.17 to demonstrate that you under-stand the upper and lower approximation methods.

(e) Approximate the value of∫ 90

0R(t) dt using the five subintervals

indicated by the data in the table. What is the most appro-priate method: left, right, trap, upper, lower, or mid? Draw ap-propriate polygons on Figure 1.17 to demonstrate which methodyou used.

(f) For 0 < b ≤ 90 minutes, explain the meaning of∫ b

0R(t) dt in

terms of fuel consumption for the plane.

(g) What do you think is the physical meaning of 190

∫ 90

0R(t) dt?

Of 1b

∫ b0R(t) dt

Figure 1.17: from 2003 AP Calculus AB exam

Problems

1.B-1 Refer to the graph which is repeated in Figure 1.18.

(a) Find a range of values for the area under the graph using upper andlower rectangular approximation methods, with n = 4.



Figure 1.18: [20] Draw the appropriate rectangles or trapezoids for M4, T4 andR4

(b) Approximate the area under the graph using the Left, Right, andMidpoint Rectangular Approximation Methods and the TrapezoidalApproximation Method, with n = 4. Are these values within yourupper/lower range?

(c) Draw appropriate rectangles or trapezoids to demonstrate your un-derstanding of M4, T4 and R4.

(d) Approximate the area under the graph using the Left, Right, andMidpoint Rectangular Approximation Methods and the TrapezoidalApproximation Method, with n = 8.

(e) What happens to the discrepancies between the various methods asyou increased the number of subintervals?

(f) How might you make all four methods of approximation get closerand closer to the same number?

1.B-2 Refer to Figure 1.19. By counting squares, find an approximation forthe definite integral of f(x) from x = 2 to x = 14. Find an estimateof the definite integral of f(x) from x = 2 to x = 14, using rectangularapproximation method with:

(a) 3 subintervals and a midpoint method for finding the height of therectangle.

(b) 3 subintervals and a left-point method for finding the height of therectangle.

(c) 3 subintervals and a right-point method for finding the height of therectangle.

(d) 6 subintervals and a left-point method for finding the height of therectangle.

(e) 6 subintervals and a right-point method for finding the height of therectangle.

(f) 6 subintervals and an upper-point method for finding the height ofthe rectangle.



Figure 1.19: [10]

(g) 6 subintervals and a lower-point method for finding the height of therectangle.

[Ans: 308, 252, 356, 280, 332, 346, 266]

1.B-3 (adapted from AB ’04) A test plane flies in a straight line with positivevelocity v(t), in miles per minute at time t minutes, where v is a functionof t. Selected values of v(t) are shown in Table 1.6.

Table 1.6: Test plane velocitiest (minutes) 0 5 10 15 20 25 30 35 40

v(t) (miles per minute) 7.0 9.2 9.5 7.0 4.5 2.4 2.4 4.3 7.3

(a) Use a midpoint Riemann sum and values from the table, what isthe most number of subintervals that could be used to approximate∫ 40

0v(t) dt?

(b) Find M4.

(c) Find T4. Does T4 use the same data points as M4?

(d) Find the super-approximation S8 = 2M4+T43 .

1.B-4 Oil is leaking out of a tanker damaged at sea. The damage to the tankeris worsening as evidenced by the increased leakage each hour, recorded inTable 1.7. [8]

(a) Give an lower and upper estimate of the total quantity of oil that hasescaped after 5 hours.



Table 1.7: [8]Time (h) 0 1 2 3 4 5 6 7 8Leakage (gal/h) 50 70 97 136 190 265 369 516 720

(b) Repeat (a) for the quantity of oil that has escaped after 8 hours.

(c) The tanker continues to leak 720 gal/h after the first 8 hours. Ifthe tanker originally contained 25,000 gal of oil, approximately howmany more hours will elapse in the worst case before all the oil hasleaked? in the best case?

[Ans: 543–758 gal, 543 gal; 1693–2363 gal; 31.4 more hours, 32.4 hours]

1.B-5 (adapted from Acorn book) Table 1.8 gives the values for the rate (ingal/sec) at which water flowed into Lake Lamar, with readings taken atspecific times.

Table 1.8: Water Flow into Lake LamarTime (sec) 0 10 25 37 46 60Rate (gal/sec) 500 400 350 280 200 180

(a) Give a range of values for the total amount of water that flowed intothe lake during the time period 0 ≤ t ≤ 60.

(b) Find a trapezoidal approximation to the amount of water that flowedinto the lake during that time period.

(c) Does your trapezoidal approximation fall within the range you gave?

[Ans: 16930–20520 gal, 18725 gal, yes]

1.B-6 An object is dropped straight down from a helicopter. The object fallsfaster and faster but its acceleration (rate of change of its velocity) de-creases over time because of air resistance. The acceleration is measuredin ft/sec2 and recorded every second after the drop for 5 sec, as shown inTable 1.9.

Table 1.9: Acceleration of a falling object [8]t 0 1 2 3 4 5

a(t) 32.00 19.41 11.77 7.14 4.33 2.63

(a) Use L5 to find an upper estimate for the speed when t = 5.



Figure 1.20: [20]

(b) Use R5 to find a lower estimate for the speed when t = 5.(c) Use upper estimates for the speed during the first second, second

second, and third second to find an upper estimate for the distancefallen when t = 3.

[Ans: 74.65 ft/sec; 45.28 ft/sec; 146.59 ft]

1.B-7 Let I =∫ 4

0f(x) dx, where f is the function whose graph is shown in

Figure 1.20. [20]

(a) Use the graph to find L2, R2, and M2.(b) Are these underestimates or overestimates of I?(c) Use the graph to find T2. How does it compare with I?(d) For any value of n, list the numbers Ln, Rn, Mn, Tn, and I in

increasing order.

[Ans: 6, 12, 9.6; L2: u, R2: o, M2: o; 9 < I; Ln < Tn < I < Mn < R]

1.B-8 As the fish and game warden of your Buddville, you are responsible forstocking the town pond with fish before the fishing season. The averagedepth of the pond is 20 feet. Using a scaled map, you measure the distancesacross the pond at 200-foot intervals, as shown in the diagram in Figure1.21. [8]

(a) Use the Trapezoidal Rule to estimate the volume of the pond.(b) You plan to start the season with one fish per 1000 cubic feet. You

intend to have at least 25% of the opening day’s fish population leftat the end of the season. What is the maximum number of licensesthe town can sell if the average seasonal catch is 20 fish per license?

[Ans: 26.36 million cubic feet; 988]



Figure 1.21: [8]




AP Unit 1, Day 3: Approximating Definite Integral from Formulas 31

1.3 Approximating Definite Integral from For-mulas

Advanced Placement




Textbook §4.3 Area, §4.4 The Definite Integral, and §4.7 Numerical Integration[16]

Resources §5.9 Approximate Integration in Stewart [20]. §5.1 Estimating withFinite Sums and §5.5 Trapezoidal Rule in Finney, et al. [8]. §1-4 DefiniteIntegrals by Trapezoids, from Equations and Data by Foerster [10]. Exploration1-4: “Definite Integrals by Trapezoidal Rule” in [9].

1.3.1 Definite Integrals from Known Shapes

We have looked at calculating definite integrals using graphs and using tables.Many times, however, instead of having a graph or a table, we have a formulaor expression which we are integrating.

Sometimes, we can find these definite integrals by looking at a graph of theexpression.

Example 1.3.1 Find:

(a)∫ 50

30

60 dt

(b)∫ 0

−3

(−x− 2) dx

(c)∫ 4

0

(√4− (x− 2)2 − 2

)dx

[Ans: 1200, - 3

2 , −8 + 2π]



Example 1.3.2 Find∫ 1

−3

√16− (x+ 3)2

dx.

1.3.2 Approximating Definite Integrals

Unfortunately, very often, when we graph the integrated expression, we are notgiven a graph with nice shapes, for which we have area formulas. Although wemay not be able to calculate the exact area under these curves (yet), we stillcan use the other techniques which we’ve already learned.

Graphing

Example 1.3.3 Let f(x) = 1−x2. Estimate a value for the integralI1 =

∫ 2

0f(x) dx. To graph use an xstep of 1 and a ystep of 1. [17]

Example 1.3.4 Let g(x) = x3. Estimate∫ 1

0g(x) dx. To graph use

an xstep of 0.5 and a ystep of 0.5. [17]

Riemann slicing

Remember that when we had data points (or graphs), we approximated definiteintegrals by dividing the overall interval into subintervals. A definite integralwas approximated for each subinterval, using area formulas for rectangles ortrapezoids, and then the individual areas were added together. For tabulardata, the subintervals were usually predetermined by what data was available.

Graphically, this meant dividing the funky shape into several funky strips, eachof which was replaced with a rectangle or trapezoid of similar area. The area ofall the funky strips were added together, to get the area of the funky shape. Forgraphs, the number of subintervals was limited by our resolution to distinguishsmall changes in height or width.

If we are given an expression to integrate, our approach will be similar. Wedivide the overall shape into smaller strips, and then replace the smaller stripswith rectangles or trapezoids, whose areas we then add together. The advantageof using formulas is that we don’t have restrictions on which or how manysubintervals to use. The number of strips can be a few, if we are going tocalculate he areas by hand, or infinitely many, in a theoretically ideal case. Ifyou are looking at an interval from t = 0 min to t = 8 min, you might naturallypick 8 strips, each of width 1 min. You might also pick a factor of 8, such as 2



or 4, to give subinterval widths of 4 and 2 minutes, respectively. Alternatively,you might choose a multiple of 8, such as 16 (subinterval widths of 30 seconds).It is usually best to make all the strips of equal width.

When we divide the shape into several strips we determined the width or baseof each rectangle.

wslice =total interval width

number of subintervals

or

∆x =b− an

When the areas of all the rectangles are added together, the sum of the areas iscalled a Riemann sum. The Riemann sum is an approximation to the definiteintegral. It is named after a German guy who apparently invented rectangles.

Definition 1.3 (Riemann sum). A sum of the form∑f(x)∆x where each term

of the sum represents the area of a rectangle of altitude f(x) and base ∆x. ARiemann sum gives an approximate value for a definite integral.[10]

Example 1.3.5 Return to Exploration 1-4. (You should have al-ready completed problems 1 through 3 for homework.)

(a) Use your graphing calculator to make a table of values for theequation v(t) = t3 − 21t2 + 100t + 110 for the even values of tfrom t = 0 to t = 8.

(b) Do problem 4 on Exploration 1-4

(c) Find a way to determine the answer for problem 4 on yourgraphing calculator in one line of input that doesn’t requireyou to copy data from a table.

Example 1.3.6 Using a program, calculate rectangular and/ortrapezoidal approximations to the integral

∫ 20

0

(60 + 40 (0.92)2x

)dx

using 4, 8, and 16 subintervals.

Example 1.3.7 Without a program, find M3 for∫ 2

0

(1− x2

)dx.

Check with a program, and then use more and more subintervals tofind the actual value.

[Ans: − 16

27 ; − 23



Example 1.3.8 Without a program, find T4 for∫ 1

0x3 dx. Check

with a program, and then use more and more subintervals to findthe actual value.

[Ans: 17

64 ; 14

]Example 1.3.9 In your mighty, mighty groups of four: Approxi-mate

∫ π0

2 sin2 x dx using:

(a) M2

(b) T2

(c) M3

(d) T3

(e) R4

1.3.3 Using Symmetry

Example 1.3.10 Let f(x) = 1− x2. Find (or estimate) values forthe integrals I1 =

∫ 2

0f(x) dx and I2 =

∫ 2

−2f(x) dx. [17]

[Ans: − 2

3 , − 43

]Note that f(x) is an even function.

Example 1.3.11 Let g(x) = x3. Find or estimate∫ 1

0g(x) dx and∫ 1

−1g(x) dx. [17]

[Ans: 0.25, 0]

Note that g(x) is an odd function. (Why?)

Problems

big giant blue-green Calculus book p. 368: #33; p. 380 #41-44; p. 413, #5

1.C-1 Quickly draw a graph of the appropriate functions, then calculate eachdefinite integral. [17]



(a)∫ 3

−3(x+ 2) dx

(b)∫ 3

−3|x+ 2| dx

(c)∫ 3

−3(|x|+ 2) dx

[Ans: 12, 13, 21]

1.C-2 Evaluate∫ 1

0

√1− (x− 1)2

dx. [Hint: Sketch a graph of the integrand,

i.e.,√

1− (x− 1)2.] [17][Ans: π

4

]1.C-3 Evaluate

∫ 3

1

(6−

√4− (x− 3)2

)dx exactly. [17] [Ans: 12− π]

1.C-4 Evaluate∫ 3

0

√4− (x− 1)2

dx exactly.[Ans:

√3

2 + 4π3

]1.C-5 Go to

http://math.furman.edu/~dcs/java/NumericalIntegration.html

and estimate∫ 2

−1

21 + 4t2

dt.

(a) Start with a left-hand rule, using four subintervals. While doublingthe number of subdivisions, watch what happens to the error (i.e.,the difference between the estimate and the actual value). The “suc-cessive error ratio” that is reported if you use the “Double” buttongives the ratio of the new error to the error from before, i.e., the onewith half as many subintervals. What value does the successive errorratio approach as you continue to double the number of subintervals,while using the left-hand rule? [Ans: 0.5]

(b) Now using the trapezoidal rule, what value does the successive errorratio approach as you continue to double the number of subintervals,i.e., as n→∞? [Ans: 0.25]

(c) Now using the midpoint rule, what value does the successive errorratio approach as you continue to double the number of subintervals?[Ans: 0.25]

(d) For midpoint or trapezoidal methods, if you tripled the numberof subintervals, what would you expect to happen to the error?[Ans: one-ninth of what it was before]

(e) For a large number of subintervals, compare the absolute value ofthe error for the midpoint approximation and for the trapezoidal ap-proximation. Which one is bigger? By roughly what percentage?(Make sure your approximations use the same number of subinter-vals) [Ans: error for trapezoidal is roughly double that of midpoint]


http://math.furman.edu/~dcs/java/NumericalIntegration.html


1.C-6 Go to

http://math.hws.edu/javamath/config_applets/RiemannSums.html

and set f(x) = ex+1. Let xmin be 0 and xmax be 1. Set ymin and ymax sothat you can see the graph. As you increase the number of subintervals,what does the sum appear to be approaching? Do you recognize thisnumber? [Ans: 2.718 = e]

1.C-7 Play around with

http://www.plu.edu/~heathdj/java/calc2/Riemann.html

1.C-8 Estimate∫ π

0sinx dx using trapezoids with 2 subintervals, 3 subintervals,

and 4 subintervals. Give exact and decimal answers to three places afterthe decimal. Do your trapezoidal approximations over- or underestimatethis definite integral? What’s happening to the values as you increase thenumber of subintervals? Make a conjecture as to what the exact answermight be.

[Ans: π

2 = 1.571, π√

33 = 1.814, π

4

(1 +√

2)

= 1.896; under;]


http://math.hws.edu/javamath/config_applets/RiemannSums.html

http://www.plu.edu/~heathdj/java/calc2/Riemann.html

AP Unit 1, Day 4: Slope and Rate of Change 37

1.4 Slope and Rate of Change

Advanced Placement


• Derivative presented geometrically and numerically.


Textbook §1.1 A Brief Preview of Calculus and §2.1 Tangent Lines and Ve-locity [16]

Resources §1-1 The Concept of Instantaneous Rate in Foerster [10]. Explo-ration 1-1: “Instantaneous Rate of Change of a Function” in [9].

1.4.1 Instantaneous Rate of Change

Recall: average velocity and slope.

Example 1.4.1

(a) A car driving due east away from Houston is 20 miles from thecity limits at 1 p.m. and 130 miles from the city limits at 3p.m. What is the car’s average velocity between 1 p.m. and 3p.m.

(b) A ball thrown up into the air has an average velocity, between 3seconds and 5 seconds after it was thrown, of −14 feet per sec-ond. If, 5 seconds after it was thrown, the ball was 20 feet abovethe ground, how high was it 3 seconds after it was thrown?

(c) A ball thrown straight up into the air has a height above theground of s(t) = −16t2 +96t feet, t seconds after it was thrown.Find the average velocity of the ball during the time periodbetween 1 and 3 seconds after it was thrown.

Slope = Rate of Change

1. Average Rate of Change. This is the Algebra I version of slope. The slopeof a secant line between two points. It is rise over run; change in y overchange in x.



2. Instantaneous Rate of Change. This is the Calculus version of slope. It isthe slope of the tangent line at one point. In a sense, it is an oxymoron,because there is no change in an instant.

1.4.2 Definition and Notation

Definitions and Notation

The derivative is another way of saying instantaneous rate of change. It isdenoted by a ‘prime’ after the function, i.e., the derivative of f(x) is writtenf ′(x).

Definition 1.4 (Derivative). The derivative of a function at a particular valueof the independent variable is the instantaneous rate of change of the dependentvariable with respect to the independent variable.[10]

We’ve already noted that the rate of change is essentially the slope, so that theinstantaneous rate of change is the slope at a point.

Notation

The derivative of f is denoted by f ′. f ′(3) is the slope of the curve of f at thepoint where x = 3.

The second derivative is the derivative of the derivative, and is denoted by f ′′.

Approximating the derivative given a graph

Example 1.4.2 Graph s(t) = −16t2 + 96t.

(a) Draw the line tangent to the graph of s(t) at t = 1.

(b) Estimate s′(1) by finding the slope of your tangent line.

(c) What is the physical meaning of s′(1), the rate of change ofheight, with respect to time, at t = 1?

(d) Write an equation of the line tangent to the graph of s(t) ats = 1, and use it to approximate s(1.1). Compare this approx-imation to the actual value of s(1.1).

(e) Estimate s′(3). Why is s′ easy to find at t = 3?



[Ans: ; 128; v(3); ; 0]

Example 1.4.3 Graph f(x) = x2 + x.

(a) Draw the line tangent to the graph of f(x) = x2 + x at x = 2.

(b) Estimate the slope of this line, i.e., f ′(2)

(c) Write the equation of this line. Plug 2.5 into the formula foryour tangent line, and compare it to the actual value of 2.5.When do you suppose the tangent line approximation is a goodapproximation?

[Ans: ; 5; y = 5 (x− 2) + 6, 8.5, 8.75]

1.4.3 Approximating Derivatives from Tabular Data

Rate of Change = Difference Quotient

• Rate from ratio is a quotient.

• Change is the difference.

Difference Quotients

• Forward: an interval to the right of the point of interest. This intervalstarts at the point at which you are estimating the derivative (i.e., instan-taneous rate of change), and ends at some very slightly higher x-value.

• Backward: an interval to the left of the point of interest. This inter-val starts at some point very slightly lower x-value than where you areestimating the derivative, and ends at the point of interest.

• Symmetric: an interval to the left and right of the point of interest. Thisinterval starts with a very slightly lower x-value than where you are ap-proximating the derivative, and ends at a very slightly higher x-value.

Note that any difference quotient can be a forward, backward, or symmetricdifference quotient. The point of interest helps decide which it is.

Example 1.4.4 Use the table of values to answer the followingquestions [19].



x −1.5 −1 −0.5 0 0.5 1y 2.027 0.632 −0.357 −1 −1.399 −1.718

(a) What is the best approximation of f ′(1), the derivative at 1?Is this difference quotient forward, backwards, or symmetric?

(b) What is the best estimate of f ′(−1.5)? Name the type of in-terval.

(c) Find the best estimate of f ′(−1.25). Name the type of interval.

(d) What is the best estimate of f ′(−1)? Name the type of interval.

(e) Extension Find the best estimate of f ′′(−1), i.e., the secondderivative at −1, i.e., the rate of change of the rate of change.

[Ans: −0.638, b; −2.79, f ; −2.79, s; −2.384, s; 1.624]

Key Questions

1. Before doing any calculations, how can you determine whether the deriva-tive should be positive or negative?

Example 1.4.5 Suppose that f is a function for which f ′(2) exists.Use the values of f given in the table to estimate f ′(1.9), f ′(2), andf ′(2.02). Name the type of difference quotient used. [17]

x 1.9 1.97 2.0 2.02 2.2f(x) 6.6 6.905 7 7.059 7.5

[Ans: 4.357 f, 2.95 f, 2.95 b]

Note that difference quotients might be used for graphs and expressions as well.

Problems

big giant blue-green Calculus book p. 155, WE #3; #1-7 odd, 37

1.D-1 The position, s(t) (measured in inches), at any time, t (measured in sec-onds), of an object is described in Figure 1.22. Use the graph to determine:

(a) s(0)

(b) s(1)



Figure 1.22: Displacement, s(t) [15]

(c) v(2)

(d) Is v(3) > 0?

(e) Is v(1) > 0?

[Ans: 1, 0, 0, Y, N]

1.D-2 The graph of a position function in Figure 1.23 represents the distance inmiles that a person drives during a twelve minute drive to school. Make

Figure 1.23: Displacement, s(t) [15]

a sketch of the corresponding velocity function. [15]



Ans:

1.D-3 The graph of a function f is shown in Figure 1.24. Rank the values of

f ′(−3), f ′(−2), f(0), and f ′(4) in increasing order. [17] [Ans: f ′(4), f ′(−3), f ′(−2), f ′(0)]


1.D-4 Suppose that f(x) = x3 − 5x2 + x− 1 and that g(x) = x3 − 5x2 + x+ 4.Explain why f ′(x) = g′(x) for every x. [Hint: How are the graphs of fand g related?] [17]

1.D-5 The graph of the derivative of a function f appears in Figure 1.25. [17]

(a) Suppose that f(1) = 5. Find an equation of the line tangent to thegraph of f at (1, 5).

(b) Suppose that f(−3) = −6. Find an equation of the line tangent tothe graph of f at (−3,−6).

[Ans: y − 5 = 2 (x− 1); y = −6]

1.D-6 (adapted from AB ’06) The rate, in calories per minute, at which a personusing an exercise machine burns calories is modeled by the function f ,shown in Figure 1.26.

(a) Find f ′(22). Indicate units of measure.



Figure 1.25: Graph of f ′ [17]

Figure 1.26: Graph of f , Burning Calories

(b) For the time interval 0 ≤ t ≤ 24, at approximately what time t doesf appear to be increasing at its greatest rate? Why?

(c) Find the total number of calories burned over the time interval 6 ≤t ≤ 18 minutes.

(d) What do you suppose is the meaning of1

(18− 6) min

∫ 18

6

f(t) dt

[Ans: −3 cal/min/min; t = 2; 132; ]

1.D-7 The graph shows how the price of a certain stock varied over a recenttrading day. [17]



(a) For each time interval below, find the total change in price and theaverage rate of change of price. (Be sure to indicate units used tomeasure these quantities.)

i. 8:00 to 11:00ii. 9:00 to 1:00

iii. 9:30 to 2:00iv. 11:00 to 1:00 [

Ans: 1, 0, 53 , −1 $/hr

](b) Estimate the instantaneous rate of change of the stock’s price at each

of the following times. (Be sure to indicate units with your answers.)

i. 9:15 a.m.ii. 10:30 a.m.

iii. 12:15 p.m.iv. 1:45 p.m.

[Ans: −4, 4, −2, 8 $/hr]

1.D-8 Go to

http://math.hws.edu/javamath/basic_applets/SecantTangentApplet.html

(a) For f(x), put in the function x2 + x, and hit ‘New Function’.

(b) Put Tangent at x=2

(c) Change the window so that you can see the parabola, along with thered dot, the red tangent line, and the green secant line.

(d) The green secant line is anchored at the point x = 2. You controlthe placement of the other point. Drag the green circle along theparabola, and notice how the slope of the secant line changes.

(e) Pay careful attention to what happens to the green secant line as youdrag the green dot closer and closer to the red dot. Make a table ofthe ‘Secant at x=’ values with the ‘Secant Slope =’ values. What doyou notice? If the ‘Secant at x=’ value could be 2, what would the‘Secant Slope =’ value be?





1.D-9 (adapted from AB ’06) Rocket A has positive velocity v(t) after beinglaunched upward from an initial height of 0 feet at time t = 0 seconds.The velocity of the rocket is recorded for selected values of t over theinterval 0 ≤ t ≤ 80 seconds, as shown.

t (seconds) 0 10 20 30 40 50 60 70 80v(t) (feet per second) 5 14 22 29 35 40 44 47 49

(a) Find the average rate of change of the velocity of Rocket A over thetime interval 0 ≤ t ≤ 80 seconds. Indicate units of measure.

(b) Approximate the instantaneous rate of change of the velocity ofRocket A at t = 0. Repeat for t = 80 seconds, t = 20, and t = 55seconds. Indicate units of measure.

(c) Using correct units, explain the meaning of∫ 70

10v(t) dt in terms of the

rocket’s flight. Use a midpoint Riemann sum with 3 subintervals ofequal length to approximate

∫ 70

10v(t) dt.

(d) What do you suppose is the meaning of1

(70− 10) sec

∫ 70

10

v(t) dt?

[Ans: 11

20 ft/s2; 0.2, 0.75, 0.4 ft/s2; , 2020 ft;]

1.D-10 A differentiable function f has values shown. Estimate f ′(1.5). [14]

x 1.0 1.2 1.4 1.6f(x) 8 10 14 22

[Ans: 40]

1.D-11 Suppose that f is a function for which f ′(2) exists. Use the value of fgiven below to estimate f ′(1.99), f ′(2), f ′(2.01), and f ′(2.1). Explain howyou obtained your estimates. [17]

x 1.9 1.99 1.999 2.0 2.001 2.01 2.1f(x) 25.34 33.97 34.896 35 35.104 36.05 46.18

[Ans: 102.889 fdq, 104 (sdq), 105.111 (bdq), 112.556 (bdq)]




AP Unit 1, Day 5: IROC as a limit 47

1.5 Instantaneous Rate of Change from a Lim-iting Process

Advanced Placement


• Derivative presented geometrically, numerically, and analytically.


Textbook §1.1 A Brief Preview of Calculus and §2.1 Tangent Lines and Ve-locity [16]

Resources §1-2 Rate of Change by Equation, Graph, or Table in Foerster [10].§2.5 Average and Instantaneous Rates: Defining the Derivative in Ostebee andZorn [17].

1.5.1 Approximating Rate of Change from a Formula

Graphs

Example 1.5.1 In your mighty, mighty groups of four: Do Explo-ration 1.5: “Instantaneous Rate of Change”

Tables

Example 1.5.2 Refer to f(x) = x2 + x.

(a) Write a formula for the forward difference quotient used toestimate f ′(2) using an interval of [2, 2 + h].

(b) Put your formula into the calculator, using a variable step-size.Evaluate the forward difference quotient for h = 0.1, h = 0.01,and h = 0.001. Continue using smaller step sizes (h) until thereis no change in the thousandths place of your slope.

(c) What would be the appropriate interval for a backward differ-ence quotient of width h to approximate f ′(2)?

(d) As you did with the forward difference quotient, use smaller andsmaller h’s until you see no change in the thousandths place ofyour slope.



Preview of a formal definition of derivative using formulas

Example 1.5.3 Refer to f(x) = x2 + x.

(a) Write an expression for the average rate of change of f(x) from2 to 2.1.

(b) Write an expression for the average rate of change of f(x) from2 to 2.01.

(c) Write an formula for the average rate of change of f(x) from 2to x. If we want an instantaneous rate of change, we want x tobe as close as possible to what?

(d) Make a table of values for the AROC from 2 to x for differentvalues of x that get closer and closer to 2 from both sides.

(e) Rewrite your formula, using the expression for f(x) and thevalue for f(2).

(f) Factor the numerator in your above expression. Cancel. Whenwould canceling not be allowed? What happens to the remain-ing expression as x gets closer and closer to 2?

(g) Write an expression for the difference quotient for f(x) from 2to 2 + h. If we want this difference quotient to represent thef ′(2), what do we want h to approach?

(h) Multiply the numerator out and combine like terms. Cancel.When would canceling not be allowed? What happens to theremaining expression as h get closer and closer to 0?

Look at the work for simplifying f(2+h)−f(2)h in the above example. Try to see

that you can replace the 2 with x and you could still do the problem.

1.5.2 Kinematics: Displacement, Velocity, Acceleration

Velocity is the instantaneous rate of change of displacement, i.e., velocity is thederivative of displacement. v(t) = d′(t) or v(t) = x′(t).

Speed is the magnitude of velocity. In one dimension, speed is the absolutevalue of velocity. speed = |v(t)|

Acceleration is the instantaneous rate of change of velocity, i.e., acceleration isthe derivative of velocity. a(t) = v′(t)

Acceleration is the derivative of the derivative of displacement, i.e., accelerationis the second derivative of displacement. a(t) = d′′(t)



Problems

big giant blue-green Calculus book p. 156: # 11, (31 or 33), 37, 39

1.E-1 Visit

http://www.slu.edu/classes/maymk/SecantTangent/SecantTangent.html

and use the following settings:

• Let f(x) be x2 − x.

• Let X0 be 3. This is the reference point at which we will be estimatingthe derivative.

• Let dX be 1 for now.

• Don’t worry about guessing f ′(x) for now.

• Make sure the first scroll-down menu is on ‘Right Secant’.

• Change the second scroll-down menu to ‘X1 click’.

• Make sure the two boxes are unchecked.

• Change your window as appropriate.

(a) Try to make some sense out of all the information that you’re beinggiven, e.g., what are the meanings of X0, dX, and X1? Can youidentify the backward, symmetric, and forward difference quotients?What do you suppose a negative dX means?

(b) Use your mouse pointer or the ‘dX In’ and ‘dX Out’ buttons to movethe second point on the secant line closer to and farther away fromthe reference point at x = 3. What’s happening to the line? What’shappening to the difference between the three slopes (left, balanced,and right)?

(c) Write down the left, balanced, and right slope for X0= 3.0, anddX= 0.1, 0.01, and 0.001.

(d) Play around: use some different functions, different points, etc. Gonuts!

(e) Can you find some other, similar, websites? Anything better that weshould know about?

1.E-2 Refer to g(x) = x2 − x.

(a) Write an expression for the difference quotient for g(x) from 3 to 3.1.Find the value.

(b) Write an expression for the difference quotient for g(x) from 3 to3.01. Find the value.





(c) Write a formula for the difference quotient for g(x) from 3 to x, usingthe expression for g(x) and the value for g(3).

(d) Factor the numerator in your above expression. Cancel. When wouldcanceling not be allowed? What happens to the remaining expressionas x gets closer and closer to 3?[Ans: g(3.1)−g(3)

3.1−3 = 5.1, g(3.01)−g(3)3.01−3 = 5.01, g(x)−g(3)

x−3 , x2−x−6x−3 , x+ 2, goes to 5

](e) Write a formula for the difference quotient for g(x) from 3 to 3 + h.

(f) Multiply the numerator out and combine like terms. Cancel. Whenwould canceling not be allowed? What happens to the remainingexpression as h get closer and closer to 0?[

Ans: ((3+h)2−(3+h))−6

(3+h)−3 = 5h+h2

h ; goes to 5]

1.E-3 Refer to q(x) = x2.

(a) Write a formula for the difference quotient for q(x) from 1 to x.Factor the numerator in your difference quotient and cancel. Whathappens to the remaining expression as x gets closer and closer to 1?[Ans: x1−1

x−1 = x+ 1, goes to 2]

(b) Repeat part (a), replacing 1 with 2, then 3, and then −1. Look for apattern, and guess a formula for q′(x).

1.E-4 Let r(x) =√x. [17]

(a) Graph r(x) on your calculator. Zoom in to the point (1, 1) repeatedly,until the graph looks like a straight line.

(b) Use that point, i.e., (1, 1), and one other nearby point on the ‘line’ tofind the slope of the ‘line’, which is an estimate of r′(1). [Ans: 0.5]

(c) Use zooming to estimate r′(1/4), r′(9/4), r′(4), r′(25/4), and r′(9).[Ans: 1; 1

3 ; 14 ; 1

5 ; 16

](d) Use these results to sketch a graph of r′(x) over the interval [−1, 2].

(e) Use your results and your graph to guess a formula for r′(x).[Ans: 1

2√x

]1.E-5 Let f(x) = lnx. [17]

(a) Use zooming to estimate f ′(1/5), f ′(1/2), f ′(1), f ′(2), and f ′(5).[Ans: 5, 2, 1, 0.5, 0.2]

(b) Use your results to sketch a graph of f ′ over the interval [0, 5].

(c) Use your results to guess a formula for f ′(x).[Ans: 1

x

]1.E-6 Repeat the zooming process for a function of your choosing.



1.E-7 (AB ’01) The temperature, in degrees Celsius (◦C), of the water in apond is a differentiable function W of time t. The table shows the watertemperature as recorded every 3 days over a 15-day period.

t W (t)(days) (◦C)

0 203 316 289 2412 2215 21

(a) Use data from the table to find an approximation for W ′(12). Showthe computations that lead to your answer. Indicate units of measure.

(b) A student proposes the function P , given by P (t) = 20 + 10te(−t/3),as a model for the temperature of the water in the pond at time t,where t is measured in days and P (t) is measured in degrees Celsius.Estimate P ′(12). Using appropriate units, explain the meaning ofyour answer in terms of water temperature.[

Ans: 21−24◦C15−9 days = − 1

2

◦C/day; −0.549◦C/day: water in pond decreasing at a rate of 0.549◦C/day]




AP Unit 1, Day 6: Slope and Area: Pulling It Together 53

1.6 Slope and Area: Pulling It Together

Do the following problems neatly on graph paper.

1.F-1 Suppose Mr. Budd is driving to the Utah Shakespearean Festival in CedarCity, UT. Once he gets on the road, he sets his cruise control for 55 mph.Let t be the number of hours since he started driving on cruise control.

(a) How far has he gone during the first half hour on cruise control? thefirst hour? the first two hours?

(b) Write an equation for the velocity, i.e., v(t) =(something).

(c) Graph the velocity versus time.

(d) Find the area under the curve of v(t) from t = 0 to t = 0.5. Also,find the area from t = 0 to t = 1 and also to t = 2.

(e) What shape are these areas in? If I look at the area from t = 0 tot = tstop, what is the width of the figure (as an expression with tstop

in it)? the height? the area (as an expression of tstop)? Call yourexpression for area A(tstop).

(f) Plot a graph of distance traveled versus time. Use the points (0.5,distance for 0.5),(1,distance for 1), and (2,distance for 2). Look for a pattern, anduse your result for A(tstop) to connect the dots.

(g) On your graph of distance versus time, what is the slope at t = 0.5?at t = 1? at t = 2? Indicate units.

1.F-2 Let f(t) = 2t

(a) Find f(0), f(0.5), f(1), f(1.5), and f(2).

(b) Graph f(t).

(c) Using the graph of f(t), find∫ 0

0f(t) dt,

∫ 0.5

0f(t) dt,

∫ 1

0f(t) dt,

∫ 1.5

0f(t) dt,∫ 2

0f(t) dt. Do you see a pattern?

(d) Think of a generic area under f(t) that starts at 0 and ends at x, i.e.,∫ x0f(t) dt. What shape is it in? Write an expression (with x in it)

for the height of the shape. What is the width of the shape? Writean expression in terms of x for the area from t = 0 to t = x. Callthis expression A(x) (A for area).

(e) On a separate graph, plot(

0,∫ 0

0f(t) dt

),(

0,∫ 0.5

0f(t) dt

),(

1,∫ 1

0f(t) dt

),(

1.5,∫ 1.5

0f(t) dt

),(

2,∫ 2

0f(t) dt

). Connect the dots using A(x).

1.F-3 Let G(t) = t2. While doing this problem, keep the previous problem inthe back of your mind.



(a) Write the expression for the Average Rate of Change of G(t) from 1to 1 + h. If this slope is to be a better and better approximation ofG′(1), what should be happening to h? Is this a forward, backward,or symmetric difference quotient?

(b) In your difference quotient to estimate G′(1), replace G(1 + h) with(1 + h)2, and G(1) with the actual value of G(1). Multiply out andcombine like terms. Factor and cancel.

(c) If you let h → 0, what happens to 1 + h? What happens to yourremaining expression for G′(1)?

(d) Repeat the above process to approximate G′(2) and G′(3).

1.F-4 Let f(x) = 3x2.

(a) Find f(1), f(2), and f(3).

(b) Use the TRAP program with more and more subintervals to makeconjectures for the following definite integrals:

∫ 1

0f(x) dx,

∫ 2

0f(x) dx,

and∫ 3

0f(x) dx. Do you see a pattern?

1.F-5 Let G(x) = x3. While doing this problem, keep the previous problem inthe back of your mind.

(a) Use symmetric difference quotients with h = 0.1, h = 0.01, andh = 0.001 to approximate G′(1). Continue using smaller h’s untilthere is no change in the hundredths place in your estimates of G′(1).

(b) Repeat for G′(2) and G′(3). What do you notice?


Unit 2

Limits and the Definition ofthe Derivative

1. Limits for Continuous Functions and Removable Discontinuities

2. Limit Definition of the Derivative (at x = c form)

3. Limit Definition of the Derivative (h or ∆x form)

4. One-Sided Limits and Infinite Limits

5. Limits at Infinity

Advanced Placement

Limits of functions (including one-sided limits).

• An intuitive understanding of the limiting process.

• Calculating limits using algebra.

• Estimating limits from graphs or tables of data.

Asymptotic and unbounded behavior.

• Understanding asymptotes in terms of graphical behavior.

• Describing asymptotic behavior in terms of limits involving infinity.

55

56 AP Unit 2 (Limits)

• Comparing relative magnitudes of functions and their rates of change. (Forexample, contrasting exponential growth, polynomial growth, and logarithmicgrowth.)

Continuity as a property of functions.

• An intuitive understanding of continuity. (Close values of the domain lead toclose values of the range.)

• Understanding continuity in terms of limits.

• Geometric understanding of graphs of continuous functions.


• Derivative presented graphically, numerically, and analytically.


• Derivative defined as the limit of the difference quotient.

Derivative at a point.

• Slope of a curve at a point.

• Tangent line to a curve at a point.

• Instantaneous rate of change as the limit of average rate of change.

Derivative as a function.


AP Unit 2, Day 1: Introduction to Limits 57

2.1 Introduction to Limits

Advanced Placement








Textbook §1.3 The Concept of Limit and §1.4 Computation of Limits [16]

Resources §1-5 Limit of a Function in Foerster [10]. Exploration 1-5: “Intro-duction to Limits” in [9]. §2.4 Introduction to Limits in Varberg, et al. [21]

2.1.1 Graphic Introduction to Limits

Understanding limits is a fairly intuitive process, and usually the easiest way tounderstand is to study examples and counterexamples.

The graph in Figure 2.1 is given by the following function:

f(x) =

x+ 1 −2 < x < 02 x = 0−x 0 < x < 20 x = 2x− 4 2 < x ≤ 4

Remember This? Recall the different types of discontinuities.

One thing to notice is that this is one function, not several. It is what is calleda piecewise function, because it is defined in terms of several pieces, rather than



Figure 2.1: A piecewise function [14]

one smooth, nice neat curve. One thing to realize is that functions don’t haveto look pretty, they just have to pass the vertical line test.

Make sure you understand what is going on with the function. Notice thatf(1) = −1, f(0) = 2 and f(2) = 0. The domain includes 4, but not −2. Thefunction is continuous over (−2, 0), which basically means that I can draw thatportion of the graph without lifting my pencil. At x = 0, the function undergoesa step (or jump) discontinuity, because it goes from 1 to 2 all the way down to0. The function is again continuous over (0, 2) and over (2, 4). At x = 2 there isa removable discontinuity. Removable discontinuities are recognized by a holein an otherwise continuous portion of the graph. A function has a removablediscontinuity at a point if changing that one point would make the functioncontinuous.

As x→ −1, f(x)→?

I want to look at the limits of f(x), but first lets go over some notation. Whenwe write lim

x→1f(x) = −1, we mean that as x get closer and closer to 1, but not

equal to 1, f(x) (or y) gets closer and closer to −1.

Simple Case: Continuity

To find the limit of f(x) as x goes to −1, trace your right forefinger along thefunction to the right of x = −1, moving left towards the point (−1, 0). Traceyour left forefinger along the function, moving right towards the same point(−1, 0). You will see that as the x value gets closer to −1, the y value getscloser to 0.



Example 2.1.1 From the graph in Figure 2.1, find the limit of f(x)as x approaches −1.

As x→ −1, f(x)→ 0.

Notationlimx→−1

f(x) = 0

This is read: “the limit of f(x) as x approaches −1 is zero”.

The Important Case: Removable Discontinuity

An important point should be made, however, and that is that if I’m looking atthe limit as x approaches −1, I don’t actually care what happens at −1, onlynear −1. So let’s take a look at the limit of f(x) as x approaches 2 (Figure 2.1still). If I let my right finger approach x = 2 along the function from the right,and let my left finger approach x = 2 from the left, I notice that as the x valuegets closer and closer to 2, the f(x) value gets closer and closer to −2. Hence

limx→2

f(x) = −2

Notice that limx→2

f(x) 6= f(2). In other words, the limit is not the same as thefunctional value. Remember: I don’t care what’s happening at 2 if I’m takingthe limit near 2.

Here are two authors’ informal definitions of limits:

Definition 2.1 (Intuitive Meaning of Limit). To say that limx→c

f(x) = L means

that when x is near but different from c then f(x) is near L. [21]

Definition 2.2 (Limit). L is the limit of f(x) as x approaches cif and only ifL is the one number you can keep f(x) arbitrarily close tojust by keeping x close enough to c, but not equal to c.[10]

Remember This? Remember the difference between if and if and only if.

The formal definition of limit is quite a bit more complicated. The trick is: howdo you objectively, logically, and mathematically reason what close and nearmean.

Now You Quickly look at numbers 1-4, 7, and 8 in Foerster §1-5. Find limx→c

f(x).Why didn’t I ask you to find the limits for numbers 5,6 and 9, 10. What types of



discontinuities are represented in the problems that do have limits. For problems5 and 6, what types of discontinuities are represented? For problems 9 & 10?

2.1.2 Step Discontinuities & One-Sided Limits

Notation

1. limx→2−

h(x) means the limit as x approaches 2 from the left, i.e., the negative

side of the axis.

2. limx→2+

h(x) means the limit as x approaches 2 from the right, i.e., the

positive side of the axis.

Graphically

Example 2.1.2 For the function f(x) graphed in Figure 2.1, findlimx→0

f(x).

Now let’s take a look at the limit as x goes to 0 of the function graphed in Figure2.1. As x gets closer and closer to 0 from the right, my y values get closer andcloser to 0. As x gets closer and closer to 0 from the left, however, f(x) or y isgetting closer and closer to 1. Which is the limit? Neither. The limit does notexist as x approaches 0 because the one-sided limits don’t match.

limx→0

f(x) does not exist

becauselimx→0−

f(x) = 1 6= limx→0+

f(x) = 0.

limx→0−

f(x) is the one-sided limit from the left, or negative side. The one-sided

limit from the right, or positive side, is given by using a “+” superscript.

Theorem 2.1. limx→c

f(x) = L if and only if limx→c−

f(x) = L and limx→c+

f(x) = L.

If we reexamine limx→2

f(x), we see that it exists (and is equal to −2) because

limx→2−

f(x) = limx→2+

f(x) = −2. So the overall limit exists because the two one-

sided limits match, and the value of that overall limit is the same as that of theone-sided limits.



2.1.3 Limits from a Table

Continuity: The Simple Case

Example 2.1.3 Use your calculator to make a table of values forf(x) = x2 + 2x + 4 for values of x near 2. Find lim

x→2f(x). Repeat

for limx→3

f(x).

[Ans: 12; 19]

Removable Discontinuity: The Important Case

Example 2.1.4 Make a table of values for g(x) =x3 − 8x− 2

for values

of x near 3. Find limx→3

g(x). Repeat for limx→2

.

[Ans: 19; 12]

Example 2.1.5 What is the difference between the graph of f(x)and the graph of g(x)?

[Ans: g(x) has a hole at (2, 12), representing the removable discontinuity]

Remember: in evaluating limx→c

f(x), imagine yourself blind to what is happeningto f at x = c. From all the evidence near x = c, what is your best guess as towhat f(c) is, or should be?

Another way to think about finding limits is: what would f(c) have to be inorder to make f continuous at c? That is the limit of f as x approaches c. Infact the definition of continuity is that the limit of f matches the value of f .

One-Sided Limits, Numerically

Example 2.1.6 On your graphing calculator, make a table ofh(x) = x3−8

|x−2| near x = 2. Find

(a) the limit of h(x) as x→ 2 from the left, aka from below.(b) the limit of h(x) as x→ 2 from the right, aka from above.(c) the limit of h(x) as x→ 2.



2.1.4 Limits from an Expression

If a function is continuous at x = c, then, by definition limx→c

f(x) = f(c). There-fore, if I know ahead of time that a function is continuous, then I simply plug inthe x value that I’m approaching. Polynomial functions and functions involvingsine and cosine are continuous, so long as there aren’t any variable expressionsin any denominators.

Over the next few class periods, we will learn several different methods fordealing with discontinuities.

I. Continuous Functions

A. The Simple Case

1. Be on the look out for:a. on a graph: graph you can draw without lifting your pencilb. on a table: no singular points that stick out or are undefinedc. with an expression: adding or multiplying polynomials, sine, co-

sine2. How we deal with it: plug it in, plug it in

II. Discontinuities

A. Removable Discontinuities These are the only types of disconti-nuities for which limits exist. Limits only exist where a function iscontinuous, or where there is a removable discontinuity.

1. Be on the look out for:a. on a graph: holesb. on a table: a single point that sticks outc. with an expression:• piecewise functions• Rational Functions that give you 0

0 . 00 is called an indeter-

minate form, and could be many things, but are typical can-didates for removable discontinuities. Be careful, though, in-determinate forms may be vertical asymptotes, or even stepdiscontinuities.

2. How we deal with it:a. Use Algebra to cancel factors, after• factoring• rationalizing with conjugates• de-denominatorizing with LCDs



b. Comparing to known limits, such as the grand-daddy of all known

limits, limθ→0

sin θθ

B. Step Discontinuities

1. Be on the look out for:a. on a graph: y-values that all of the sudden jump to a different

value.b. on a table: y-values that all of the sudden jump to a different

valuec. with an expression:• piecewise functions

• variations on limx→0

|x|x

2. How we deal with it: one-sided limits

C. Vertical Asymptotes

1. Be on the look out for:a. on a graph: Curves that go up or down off the graph.b. on a table: Numbers that get hugely positive or hugely negative

as x approaces a specific value.

c. with an expression: rational expressions where you havenonzero

02. How we deal with it:

a. one-sided limitsb. infinity, ∞

D. Infinitesimal Oscillations This is a relatively minor and rare type ofdiscontinuity. Be on the look out for:

1 on a graph: oscillations with periods that get smaller and smaller,eventually getting infinitesimally small near a specific x-value.

2 with an expression: variations on limx→0

sin(

1x

)

2.1.5 Substitution and Properties of Limits

These are the theorems that allow me to simply plug in values if I know thatmy function is continuous.

Theorem 2.2 (Main Limit Theorem). Let n be a positive integer, k be aconstant, and f and g be functions that have limits at c. [21] Then

1. limx→c

k = k

2. limx→c

x = c



3. limx→c

kf(x) = k limx→c

f(x)

4. limx→c

[f(x) + g(x)] = limx→c

f(x) + limx→c

g(x)

5. limx→c

[f(x)− g(x)] = limx→c

f(x)− limx→c

g(x)

6. limx→c

[f(x) · g(x)] = limx→c

f(x) · limx→c

g(x)

7. limx→c

f(x)g(x) =

limx→c

f(x)

limx→c

g(x) , provided limx→c

g(x) 6= 0

8. limx→c

[f(x)]n =[

limx→c

f(x)]n

9. limx→c

n√f(x) = n

√limx→c

f(x), provided limx→c

f(x) > 0 when n is even.

An example of the first limit property is that limx→−3

7 = 7. Think about why

this makes sense numerically and graphically.

The second limit property might be exemplified by limx→π

x = π.

The third property can be seen in action by limx→π

2x = 2 limx→π

x, which can thenbe simplified with the second property to 2π.

Example 2.1.7 As an exercise to familiarize yourself with the

properties, evaluate limx→2

x3 − 42x

, naming the property used at eachstep.

Theorem 2.3 (Limits of Trigonometric Functions). For every real number c inthe function’s domain [21]

1. limx→c

sin t = sin c

2. limx→c

cos t = cos c

3. limx→c

tan t = tan c

4. limx→c

cot t = cot c

5. limx→c

sec t = sec c

6. limx→c

sec t = sec c



Example 2.1.8 Findlimx→π

sinx

[Ans: 0]

Theorem 2.4 (Limit of a Composite Function). If f is continuous at b andlimx→a

g(x) = b, then limx→a

f (g(x)) = f(b). In other words, [20]

limx→a

f (g(x)) = f(

limx→a

g(x))

Example 2.1.9 Find

limθ→π

3

sin(

2θ +π

3

)

[Ans: 0]

Problems

big giant blue-green Calculus book p. 85: # 1,2, 15, 17, 21, 23, 25, 29, 31

2.A-1 Go to http://www.calculus-help.com/funstuff/phobe.html

Watch:

(a) Chapter 1, Lesson 1: What is a Limit?

(b) Chapter 1, Lesson 2: When Does a Limit Exist?

(c) Chapter 1, Lesson 3: How do you evaluate a limit?

(d) Chapter 2, Lesson 1: The Difference Quotient.

What differences in terminology do you notice? What did the tutorialhelp to clarify?

2.A-2 For the function f graphed in Figure 2.2, find the indicated limit or func-tion value, or state that it does not exist. [21]

(a) limx→−3

f(x)

(b) f(−3)

(c) f(−1)

(d) limx→−1

f(x)


http://www.calculus-help.com/funstuff/phobe.html


Figure 2.2: [21]

(e) f(1)

(f) limx→1

f(x)

(g) limx→1−

f(x)

(h) limx→1+

f(x)

[Ans: 2; 1; d.n.e.; 2.5; 2; d.n.e.; 2; 1]

Figure 2.3: [21]

2.A-3 Follow the directions of problem 2 for the function graphed in Figure 2.3.[21] [Ans: d.n.e.; 1; 1; 2; 1; d.n.e.; 1; d.n.e.]


AP Unit 2, Day 2: Limits at Cancelable Discontinuities 67

2.2 Limits at Cancelable Discontinuities

Advanced Placement









Textbook §1.3 Computation of Limits and §2.1 Tangent Lines and Velocity[16]

Resources §1-5 Limit of a Function in Foerster [10]. Exploration 1-5: “Intro-duction to Limits” in [9]. §2.4 Introduction to Limits in Varberg, et al. [21]

2.2.1 Limits at Cancelable Discontinuities

Theorem 2.5 (Functions that Agree at All But One Point). Let c be a realnumber and let f(x) = g(x) for all x 6= c in an open interval containing c. Ifthe limit of g(x) as x approaches c exists, then the limit of f(x) also exists and

limx→c

f(x) = limx→c

g(x).

In order to find the limx→c

f(x), the value of f(c) is irrelevant. Only the valuesinfinitesimally close to c matter.

Example 2.2.1 Find limx→2

x3−8x−2 by comparing it to lim

x→2x2 + 2x+ 4.

Example 2.2.2 Find

limx→3

3− xx2 − 9



Why are limits important in calculus?

Recall that f ′(2), the derivative of f(x) at x = 2, may be considered two ways,both of which are average rates of change over an interval that always includesx = 2, but where the interval gets smaller and smaller:

• the average rate of change of f form 2 to 2 +h, where h is getting smaller

and smaller, i.e.,f(2 + h)− f(2)

(2 + h)− 2as h→ 0;

• the average rate of change of f from 2 to x, where x is getting closer and

closer to 2, i.e.,f(x)− f(2)

x− 2as x→ 2.

Remember that if h > 0, i.e., h is positive, thenf(2 + h)− f(2)

(2 + h)− 2is a forward

difference quotient, but if h < 0, i.e., h is negative, then you have a backwarddifference quotient. Also recall that for the limit as h → 0, you would includeboth positive and negative values of h.

Example 2.2.3 For f(x) = x3, find lim∆x→0

f(2 + ∆x)− f(2)(2 + ∆x)− 2

Example 2.2.4 In your mighty, mighty groups of four: Pick aquadratic function f(x). Find

(a) f(2)

(b) limx→2

f(x)

(c) limh→0

f(2 + h)− f(2)(2 + h)− 2

;

(d) limx→2

f(x)− f(2)x− 2

.

Talk about what each of these things means, in terms of the graph of

f . Put your answers on the board. When you finish, find limh→0

f(x+ h)− f(x)h

2.2.2 De-rationalizing with Conjugates

Sometimes we have to squeeze the factors out of the expression. One techniquefor doing that when we have radicals is to use conjugates.



Example 2.2.5 [20] Find

limt→0

t2√t2 + 9− 3

Example 2.2.6 [16] Find

limx→16

4−√x

x− 16

Example 2.2.7 In your mighty, mighty groups of four: pick a linearfunction mx+ b. Let f(x) =

√mx+ b. Make sure that f(2) exists.

(a) f(2)

(b) limx→2

f(x)

(c) limx→2

f(x)− f(2)x− 2

.

(d) limh→0

f(2 + h)− f(2)(2 + h)− 2

;

Talk about what each of these things means, in terms of the graphof f . When you finish, put your answers on the board. When you

finish, find limh→0

f(x+ h)− f(x)h

Example 2.2.8 Find

lim∆x→0

√x+ ∆x−

√x

∆x

Note: compare this problem to lim∆x→0

f(x+∆x)−f(x)∆x , which is the for-

mula for the instantaneous rate of change, or... derivative. Thisproblem is finding the derivative for f(x) =

√x, which means that

f(x+ ∆x) =√x+ ∆x.

2.2.3 Derivative at a Point

Definition 2.3 (Derivative (at x = c form)). [16] The derivative of f(x), withrespect to x, at the point x = c is given by

f ′(c) = limx→c

f(x)− f(c)x− c



e.g.,

f ′(2) = limx→2

f(x)− f(2)x− 2

Meaning: The instantaneous rate of change of f with respect to x at x = c.Graphically: The slope of the line tangent to the graph of f at the point x = c.

Example 2.2.9 Let f(x) = x4. Find f ′(2). In your mighty, mightygroups of four: Find

(a) f ′(−1);

(b) f ′(3);

(c) f ′(−2).

What is the pattern?

Example 2.2.10 The graph of the function f shown in Figure 2.4consists of a semicircle and three line segments. Find the following

Figure 2.4: Graph of f

limits of difference quotients:

(a) limx→−2

f(x)− f(−2)x+ 2

[Ans: − 1

3

](b) lim

x→2.3

f(x)− f(2.3)x− 2.3

[Ans: −1]

(c) limx→−3−

f(x)− f(−3)x+ 3

[Ans: 2]

(d) limx→−3+

f(x)− f(−3)x+ 3

[Ans: − 1

3

](e) lim

x→−3

f(x)− f(−3)x+ 3

[Ans: d.n.e.]

Example 2.2.11 Each of the following limits are derivatives. Tellfor which function, and for what point



(a) limx→−3

x4 − 81x+ 3

(b) limt→25

√t− 5

t− 25

(c) limx→2

x2 + x− 6x− 2

Using conjugates to fetch factors

Example 2.2.12 Write an equation of the line tangent to the graphof y =

√2x+ 4 at the point

(−1,√

2)

[Ans: y −

√2 = 1√

2(x+ 1)

]

2.2.4 De-denominatorizing with LCDs

Example 2.2.13 Find

limx→3

1x+ 1

− 14

x− 3

Example 2.2.14 In your mighty, mighty groups of four: Let f(x) =1

3x+ 1. Pick a value of c 6= − 1

3

(a) f(c)

(b) limx→c

f(x)

(c) limx→c

f(x)− f(c)x− c

.

(d) limh→0

f(c+ h)− f(c)(c+ h)− c

;

Talk about what each of these things means, in terms of the graphof f . When you finish, put your answers on the board. After youput your answers on the board: How would you find the point at

which the derivative is − 43? Also, find lim

h→0

f(x+ h)− f(x)h

.



Example 2.2.15 Find

limh→0

1x+ h

− 1x

h

Note: compare this problem to limh→0

f(x+h)−f(x)h , which is the formula

for the instantaneous rate of change, or... derivative. This problemis finding the derivative for f(x) = 1

x , which means that f(x+ h) =1

x+h .

Problems

big giant blue-green Calculus book p. 95: Writing Exercises # 2; #3, 11, 19p. 156: # 23

2.B-1 (adapted from ?) If a 6= 0, then limx→a

x2 − a2

x4 − a4is

[Ans: 1

2a2

]

2.B-2 f(x) =

x2 − 64x− 8

x 6= 8

k x = 8What value of k will make f continuous at x = 8? [Ans: 16]

2.B-3 [16] Evaluate analytically, then check your answer by making a table onyour handy-dandy calculator.

(a) limx→−3

x2 + x− 6x2 − 9

(b) limh→0

((x+ h)2 − 2 (x+ h) + 1

)−(x2 − 2x+ 1

)h

[Ans: 2x− 2]

2.B-4 Evaluate limv→3

3− v√v −√

3.

[Ans: −2

√3]

2.B-5 [16] Evaluate analytically, then check your answer by making a table onyour handy-dandy calculator, if possible.

(a) limx→16

4−√x

x− 16[Ans: − 1

8

](b) lim

x→4

√x+ 5− 3x− 4

[Ans: 1

6

](c) lim

h→0

√3 (x+ h)−

√3x

h

[Ans: 3

2√

3x



2.B-6 Find limh→0

12− h

− 12

h.

[Ans: 1

4

]

2.B-7 Find limx→2

22x− 1

− 23

x− 2. This limit happens to be the derivative of what

function, at what point?[Ans: − 4

9 ; 22x−1 at x = 2

]2.B-8 Find lim

x→3

x− 32

5x− 8− 2

5

[Ans: − 49

10

]

2.B-9 Find lim∆x→0

13 (x+ ∆x) + 1

− 13x+ 1

∆x

2.B-10 limx→2

x√x−√

8x− 2

is the derivative of what function for what value of x?

Evaluate the limit, and check your answer by making a table on yourcalculator.

[Ans: f ′(2) for f(x) = x

√x; 3

√2

2

]2.B-11 Use the limit definition of the derivative to write an equation of the line

tangent to f(x) = x2 at the point (−4, f(−4)). [Ans: y − 16 = −8 (x+ 4)]

2.B-12 Each of the following limits is a derivative. For each limit,

• state for what function it is a derivative, and at what x-value;

• approximate the limit numerically, using a table on your calculator;

• confirm your answer using algebraic techniques, without using a cal-culator.

(a) limx→−1

x2 − 1x+ 1

[Ans: x2 at −1

](b) lim

x→2

x3 − 8x− 2

[Ans: x3 at 2

](c) lim

x→0

12 + x

− 12

x

[Ans: 1

2+x at 0]

(d) limx→0

√x+ 2−

√2

x

[Ans:

√x+ 2 at 0

]




AP Unit 2, Day 3: Limit Definition of Derivative as a Function 75

2.3 Limit Definition of Derivative as a Function

Advanced Placement

Limits of functions.









Textbook §2.1 Tangent Lines and Velocity and §2.2 The Derivative [16]

Resources §3-2 Difference Quotients and One Definition of Derivative in Fo-erster [10]. Exploration 3-2: “Exact Value of a Derivative” and Exploration 3-3“Numerical Derivative by Grapher” in [9].

2.3.1 Derivative as a Function

Example 2.3.1 For f(x) =1

3x+ 1, find f ′(−1), f ′(0), f ′(1), f ′(2),

etc.

Instead of using the formula for the derivative at a point formula several times,we will find an expression for f ′(x), and then simply plug in −1, 0, 1, and 2.

Definition 2.4 (Derivative (∆x or h form)). [16] The derivative function off(x) is given by

f ′(x) = lim∆x→0

∆y∆x

= lim∆x→0

f(x+ ∆x)− f(x)∆x

= limh→0

f(x+ h)− f(x)h



This gives me a formula for finding the derivative at any generic point, x, insteadof only at a specific point like 2 or −1.

Note that this formula can be adapted to find the derivative at a specific pointby replacing x with a specific value, like 2 or −1.

Example 2.3.2 Find the point on the graph of y =1

3x+ 1where

the tangent line is parallel to y = −34x+

54

.

First, let’s graph the function on our calculator, with the line − 34x + 5

4 . See ifyou can spot the point where the tangent line is parallel to the line that we’vegraphed. As it turns out, there are two points that will work for this problem.

Example 2.3.3 Find the point on the graph of y =√

7x+ 4 wherethe tangent line is perpendicular to the line 10x+7y = 5, i.e., wherethe normal line is parallel to 10x+ 7y = 5.

Example 2.3.4 (adapted from [2]) A function g is defined for allreal numbers and has the following property: g(a + b) − g(a) =6a2b+ 6ab2 + 2b3 − 3b. Find g′(x).

[Ans: 6x2 − 3

]Example 2.3.5 (BC89) Let f be a function that is everywheredifferentiable and that has the following properties.

(i) f(x+ h) =f(x) + f(h)

f(−x) + f(−h)for all real numbers h and x.

(ii) f(x) > 0 for all real numbers x.

(iii) f ′(0) = −1.

(a) Find the value of f(0). [Ans: 1]

(b) Show that f(−x) =1

f(x)for all real numbers x.

(c) Using part (b), show that f(x + h) = f(x)f(h) for all realnumbers h and x.

(d) Use the definition of the derivative to find f ′(x) in terms off(x). [Ans: f ′(x) = −f(x)]



Notation

The derivative of a function f(x) might be written as f ′(x) ordy

dx.dy

dxis called

Liebniz notation, and is technically the derivative of y. Other notations include

Dx f(x) andd

dxf(x).

Example 2.3.6 Finddy

dxwhen y =

1x

2.3.2 Tangent Lines

Example 2.3.7 (adapted from BC97) Refer to the graph in Figure2.5. The function f is defined on the closed interval [0, 8]. The graphof its derivative f ′ is shown. Think about what the graph is tellingyou. An equation of a line tangent to the graph of f is 3x− y = −1.

Figure 2.5:

(a) What is the x-coordinate of the point of tangency?

(b) What is the y-coordinate of the point of tangency?

[Ans: (1, 4)]



Example 2.3.8 (adapted from [2]) If p(x) = (x+ 2) (x+ k) and ifthe line tangent to the graph of p at the point (4, p(4)) is perpen-dicular to the line 2x+ 4y + 5 = 0, then k =

[Ans: −8]

Example 2.3.9 (adapted from [2]) If the line 3x − y + 5 = 0 istangent in the second quadrant to the curve y = x3 + k, then k =

[Ans: 3]

Problems

2.C-1 Find g′(x) if g(a+ b)− g(a) is defined as follows:

(a) 3a2b+ 3ab2 + b3[Ans: 3x2

](b) 2ab+ b2 − 2b [Ans: 2x− 2]

2.C-2 Suppose E(x) is a function for which E(x+ h)−E(x) = E(x) (E(h)− 1)

and limh→0

E(h)− 1h

= 1. Show that E′(x) = E(x), i.e., show that, if these

criteria are met for a function, then that function is its own derivative.

2.C-3 Use the limit definition of the derivative to find:

(a) f ′(x) if f(x) = 12− x2;

(b) dydx if y = 1

4x3;

(c) ddx

(3x2 − 4x+ 1

);

(d) dydx if y =

√x;

(e) f ′(x) if f(x) = 2x2 + x+ k, where k is a constant;

(f)d

dx

(1x

)[Ans: −2x; 3

4x2; 6x− 4; 1

2√x

; 4x+ 1; − 1x

2]

2.C-4 The following limit is the derivative of what function:

lim∆x→0

3√

2 (x+ ∆x)2 − 1− 3√

2x2 − 1

∆x [Ans: 3

√2x2 − 1



2.C-5 (BC90) Let f(x) = 12 − x2 for x ≥ 0 and f(x) ≥ 0. The line tangent tothe graph of f at the point (k, f(k)) intercepts the x-axis at x = 4. Whatis the value of k? [Ans: k = 2]

2.C-6 (adapted from AB97) At what point on the graph of y =14x3 is the

tangent line parallel to the line 3x− 4y = 7?[Ans:

(1, 1

4

)]2.C-7 (adapted from [2]) Let f(x) = 3x3−4x+1. An equation of the line tangent

to y = f(x) at x = 2 is [Ans: y = 32x− 47]

2.C-8 (adapted from [2]) Find the point on the graph of y =√x between (4, 2)

and (9, 3) at which the normal to the graph is perpendicular to the linethrough (4, 2) and (9, 3).

[Ans:

(254 ,

52

)]2.C-9 (adapted from [3]) If the graph of the parabola y = 2x2 +x+ k is tangent

to the line 3x+ y = 3, then k = [Ans: 5]




AP Unit 2, Day 4: Basic Calculus of Polynomials 81

2.4 Basic Calculus of Polynomials

Advanced Placement

Applications of derivatives

• Interpretation of the derivative as a rate of change in varied applied contexts,including velocity, speed, and accleeration.

Computation of derivatives

• Knowledge of derivatives of basic functions, including power.

• Basic rules for the derivative of sums of functions.

Textbook §2.3 Computation of Derivatives: The Power Rule [16]

Resources Explorations 3–4a: “Algebraic Derivative of a Power Function”,3–5a: “Velocity and Acceleration Reading”, and Exploration 3–5b: “ DerivingVelocity and Acceleration Data from Displacement Data” in [9].

2.4.1 Notation

Notation

The derivative of a function f(x) might be written as f ′(x) ordy

dx.dy

dxis called

Liebniz notation, and is technically the derivative of y. Other notations include

Dx f(x) andd

dxf(x).

Example 2.4.1 Find the f ′(x) for f(x) = x4

Plug in 2, −1, and 3. Compare your answers to the answers you got by usingthe at x = c form of the derivative for each point.

Example 2.4.2 Find the derivative of x3

Example 2.4.3 Find the derivative of x3 + 1



Example 2.4.4 Find f ′(x) for f(x) =√x and for f(x) =

√x3 + 1


dxwhen y =

1x

and when y =1

x3 + 1

2.4.2 Basic Properties of Derivatives

Derivative of a Constant

If f(x) = k, then f ′(x) = 0, i.e.,d

dxk = 0

Derivative of a Line

If f(x) = mx+ b, then f ′(x) = m, i.e.,d

dx(mx+ b) = m

Derivative of a Sum (or Difference)

If f(x) = g(x) + h(x), then f ′(x) = g′(x) + h′(x), i.e.,d

dx(u+ v) =

du

dx+dv

dx.

Derivative of a Scalar Multiple

If f(x) = k g(x), where k is some constant, then f ′(x) = k g′(x), i.e.,d

dxky =

kdy

dx.

Theorem 2.6. Linearity of Differentiation

d

dx[a f(x) + b g(x)] = a f ′(x) + b g′(x)

Differentiation will be intuitive as long as you are adding, or multiplying by aconstant. When things get more complicated than that, things will get morecomplicated than that.



2.4.3 Power Rule

Power Rule

d

dxxn = nxn−1

To differentiate a power function y = axb:

• Multiply the coefficient by the old exponent;

• Lower the exponent by one.

Example 2.4.6 (adapted from AB97) If f(x) = x3 + x − 1x

, then

f ′(−1) =

[Ans: 5]

Example 2.4.7 (adapted from [2]) For f(x) =√x, find the point on

the graph, between (1, 1) and (4, 2), where the slope of the tangentline is equal to the slope of the line between (1, 1) and (4, 2).

[Ans:

(94,

32

)]

Example 2.4.8 (adapted from AB97) Let f(x) = x 3√x. If the rate

of change of f at x = c is thrice its rate of change at x = 1, then c =

[Ans: 27]

Example 2.4.9 (AB86) Let f be the function defined by f(x) =7− 15x+ 9x2 − x3 for all real numbers x. Write an equation of theline tangent to the graph of f at x = 2.

[Ans: y − 5 = 9 (x− 2)]



Example 2.4.10 (adapted from AB98) Write an equation (in slope-intercept form) for the line tangent to the graph of f(x) = x4 − x2

at the point where f ′(x) = 1

[Ans: y = x− 1.055]

Example 2.4.11 Use the power rule to find limh→0

3√

8 + h− 2h

[Ans:

112

]

2.4.4 Higher Order Derivatives

f ′′(x) means the derivative of f ′(x), or the second derivative of f(x). f ′′′(x)would be the third derivative of f(x).

Example 2.4.12 Find f ′′′(x) if f(x) = x3 + 1.

[Ans: 6]

In Liebniz notation, the second derivative isd

dx

(dy

dx

). To abbreviate this,

we pretend like thed

dxis multiplied:

d2y

dx2. In reality,

d

dxis very much not

being multiplied, but it is a useful notation. (Think of the dx2 on the bottom

as (dx)2.) The sixth derivative would bed6y

dx6.

Example 2.4.13 Findd2

dt2(2t3 − 6t2 + 5

)[Ans: 12t− 12]

2.4.5 Kinematics

Velocity is the derivative of displacement:

v(t) = d′(t)



Acceleration is the derivative of the velocity:

a(t) = v′(t)

Acceleration is the derivative of the derivative of displacement. Acceleration isthe second derivative of displacement:

a(t) = v′(t) = s′′(t)

Also: speed is the magnitude of velocity. In one dimension, this means thatspeed is the absolute value of velocity.

speed(t) = |v(t)|

Example 2.4.14 A particle moves along the x-axis so that itsposition at time t, where t is in seconds, is given by d(t) = 2t3 −6t2 + 5, where d(t) is given in feet. What is the velocity when theacceleration is zero?

[Ans: −6 feet/sec]

Problems

Power Rule

2.D-1 (AB89) Let f be the function given by f(x) = x3 − 7x+ 6.

(a) Find the zeros of f .[Ans: 1, 2, −3]

(b) Write an equation of the line tangent to the graph of f at x = −1.[Ans: y − 12 = −4 (x+ 1)]

2.D-2 limx→5

x4 − 625x− 5

is the derivative of what function, at what point? Use the

power rule to evaluate the limit, without going through all that factor andcancel shtuff. [Ans: 500]

2.D-3 As with the problem before, limh→0

√4 + h− 2

his a derivative at a point.

Evaluate the limit, by identifying at which point and for which functionthis is the derivative, then use the power rule instead of squeezing out anh using conjugates.

[Ans: 1

4



2.D-4 [10] Misconception Problem Mae Danerror needs to find f ′(3), wheref(x) = x4. She substitutes 3 for x, gets f(3) = 81, differentiates 81,and gets zero for the answer. Explain why she also gets zero for her grade.

2.D-5 [10] For f(x) =x3

3− x2 − 3x + 5, plot the graphs of f and f ′ on the

same screen. Show that each place where the f ′ graph crosses the x-axiscorresponds to a high or low point on the f graph.

2.D-6 [15] The graphs of a function f and its derivative f ′ are given on the samecoordinate axes in Figure 2.6. Label the graphs as f or f ′ and state the

Figure 2.6: [15]

reasons for your choice.

2.D-7 [15] The graphs of a function f and its derivative f ′ are given on the samecoordinate axes in Figure 2.7. Label the graphs as f or f ′ and state the

Figure 2.7: [15]

reasons for your choice.



2.D-8 (adapted from [2])d

dx

(ln e2x3

)=

[Ans: 6x2

]2.D-9 [17] When an oil tank is drained for cleaning, there are V (t) = 100, 000−

4000t+ 40t2 gallons of oil left in the tank t minutes after the drain valveis opened.

(a) At what average rate does oil drain from the tank during the first 20minutes? [Ans: 3200 gal/min]

(b) At what rate does oil drain out of the tank 20 minutes after the drainvalve is opened? [Ans: 2400 gal/min]

(c) Explain what V ′′(t) says about the rate at which oil is draining fromthe tank.

[Ans: rate of change is increasing, i.e., rate is getting less negative, i.e., rate of drainage is decreasing]

2.D-10 Find the derivative of x3 +x2− 2, x3 +x2− 1, x3 +x2, and x3 +x2 + 58.7[Ans: 3x2 + 2x

]2.D-11 Work backwards: find the functions, f(x), for the following derivatives:

(a) f ′(x) = 6x5

(b) f ′(x) = 13x12

(c) f ′(x) = 10x

(d) f ′(x) = 3x2 + 2x

(e) f ′(x) = m

Make up a word: if f ′ is the derivative of f , then f is the (blank) of f ′.[Ans: x6; x13; 5x2; x3 + x2; mx+ b;

]Kinematics

2.D-12 (MM99(2)) A ball is thrown vertically upwards into the air. The height,h metres, of the ball above the ground after t seconds is given by

h = 2 + 20t− 5t2, t ≥ 0.

(a) Find the initial height above the ground of the ball (that is, itsheight at the instant when it is released.) [Ans: 2 m]

(b) Show that the height of the ball after one second is 17 metres.

(c) At a later time the ball is again at a height of 17 metres.

i. Write down an equation that t must satisfy when the ball is ata height of 17 metres.

ii. Solve the equation algebraically. [Ans: t = 1 s, t = 3 s]

(d) i. Finddh

dt.



ii. Find the initial velocity of the ball (that is, velocity at the in-stant when it is released). [Ans: v(0) = 20 m/s]

iii. Find when the ball reaches its maximum height. [Ans: t = 2 s]iv. Find the maximum height of the ball. [Ans: 22 m]

2.D-13 [10] Find equations for the velocity, v, and the acceleration, a, of a movingobject if y = 5t4−3t2.4+7t is its displacement.

[Ans: v = 20t3 − 7.2t1.4 + 7, a = 60t2 − 10.08t0.4

]2.D-14 [10] Car Problem Calvin’s car runs out of gas as it is going up a hill.

The car rolls to a stop, then starts rolling backward. As it rolls, itsdisplacement, d(t) feet, from the bottom of the hill at t seconds sinceCalvin’s car ran out of gas is given by d(t) = 99 + 30t− t2.

(a) Plot graphs of d and d′ on the same screen. Use a window largeenough to include the point where the d graph crosses the positivet-axis. Sketch the result.

(b) For what range of times is the velocity positive? How do you interpretthis answer in terms of Calvin’s motion up the hill? [Ans: Velocity is positive for 0 ≤ t < 15. Calvin is going uphill for the first 15 sec]

(c) At what time did Calvin’s car stop rolling up and start rolling back?How far was it from the bottom of the hill at this time? [Ans: At 15 seconds, his car stopped. d(15) = 324, so distance is 324 feet]

(d) If Calvin doesn’t put on the brakes, when will he be back down atthe bottom of the hill? [Ans: t = 33 sec]

(e) How far was Calvin from the bottom of the hill when the car ran outof gas? [Ans: 99 feet from the bottom]


Unit 3

Basic Differentiation

1. Finding Derivatives Using Limits

2. Power Rule - Derivatives

3. Antiderivatives of Polynomials

4. Product and Quotient Rules

5. Chain Rule

6. Tangent Lines

AB

I. DerivativesConcept of the derivative.





• Slope of a curve at a point. Examples are emphasized, including pointsat which there are vertical tangents and points at which there are notangents.

• Tangent line to a curve at a point and local linear approximation.


89

90 AP Unit 3 (Basic Differentiation)

• Approximate rate of change from graphs and tables of values.


• Corresponding characteristics of graphs of f and f ′.

• Relationship between the increasing and decreasing behavior of f and thesign of f ′.

Applications of derivatives.

• Interpretation of the derivative as a rate of change in varied applied con-texts, including velocity, speed, and acceleration.

Computation of derivatives.

• Knowledge of derivatives of basic functions, including power functions.

• Basic rules for the derivative of sums, products, and quotients of func-tions.

• Chain rule.

II. IntegralsTechniques of antidifferentiation.

• Antiderivatives following directly from derivatives of basic functions.


AP Unit 3, Day 1: Antidifferentiation of Polynomials 91

3.1 Antidifferentiation of Polynomials

Advanced Placement

Techniques of antidifferentiation.


Applications of antidifferentiation.

• Finding specific antiderivatives using initial conditions, including applicationsto motion along a line.

Textbook §4.1 Antiderivatives [16]

Resources Exploration 5–2b: “A Motion Antiderivative Problem” in [9].

3.1.1 Notation of Antiderivatives

We need a notation for antiderivative that is as confusing as humanly possible,so we will use a notation already available to us for a completely unrelatedconcept: the definite integral.

The antiderivative of f ′(x) is written as∫f ′(x) dx

Notice that the function, f ′(x), that you are antidifferentiating is surroundedby two things. On the right side is the integration sign,

∫. The integral sign

is used because antiderivatives are also known as indefinite integrals. On theright, is dx, which tells us what the variable is.

How will you be able to tell the difference between a definite integral, and anantiderivative?

3.1.2 Anti-Power Rule

Revisit problem 11 on page 87.

To anti-differentiate, we need to do the opposite of differentiation, in the oppo-site order.



Theorem 3.1. Anti-Power Rule∫xn dx =

xn+1

n+ 1+ C

To antidifferentiate a power function y = axb:

• Raise the exponent by one;

• Divide by the new exponent;

• Add an arbitrary constant C (the antiderivative of 0).

The antiderivative of a scalar multiple is the scalar multiple of the antiderivative.∫k f(x) dx = k

∫f(x) dx

As long as you are multiplying a function by a constant, you can pull the constantout. For your own safety, never try to pull a variable out of the antiderivative:63% of the time when you pull a variable out of the antiderivative, your pencilwill explode. If it doesn’t happen the first time, know that you got lucky, andyou are tempting the fates by trying it again.

The antiderivative of a sum is the sum of the antiderivatives:∫(f(x) + g(x)) dx =

∫f(x) dx+

∫g(x) dx

Theorem 3.2. Linearity of Antidifferentiation∫[a f(x) + b g(x)] dx = a

∫f(x) dx+ b

∫g(x) dx

Antidifferentiation is intuitive, so long as you are adding(/subtracting) func-tions, or multiplying by a constant. It will not be intuitive for multiplication ordivision.

Example 3.1.1 (adapted from AB93)∫ (x3 + x

)2dx =

[Ans:

x7

7+

2x5

5+x3

3+ c



Notice that we cannot get the correct answer by squaring the antiderivative of

x3 + x, i.e., the answer is not(x4

4 + x2

2

)2

+ C, or even

(x4

4 + x2

2

)3

3+ C. All

we know how to antidifferentiate right now is power functions (axn) and sumsof power functions.

Example 3.1.2 (adapted from [3]) If limh→0

g(x+ h)− g(x)h

=x3 − 2x3

,

then g(x) could be equal to

(A) −2x−4

(B) 6x−4

(C)x3 + 1x2

(D) x+ x2

(E) 1− 2x−3

[Ans:

x3 + 1x2

]Notice that we cannot find the antiderivative of a quotient by taking the quotient

of the antiderivatives: i.e., the answer is notx4

4 − 2xx4

4

+ C

Example 3.1.3 (BC86) For all real numbers x and y, let f bea function such that f(x + y) = f(x) + f(y) + 2xy and such that

limh→0

f(h)h

= 7.

(a) Find f(0). Justify your answer. [Ans: 0]

(b) Use the definition of the derivative to find f ′(x). [Ans: 7 + 2x]

(c) Find f(x).[Ans: 7x+ x2

]

3.1.3 Kinematics

Velocity is the derivative of displacement (v(t) =dx

dt), so that displacement is

the antiderivative of velocity:

x(t) =∫v(t) dt



orx(t) =

∫x′(t) dt

orx(t) =

∫dx

dtdt

Acceleration is the derivative of the velocity (a(t) =dv

dt=d2x

dt2), so that velocity

is the antiderivative of acceleration:

v(t) =∫a(t) dt

orv(t) =

∫v′(t) dt

orv(t) =

∫dv

dtdt

Example 3.1.4 (adapted from AB93) The acceleration of a particlemoving along the x-axis at time t is given by a(t) = 6t − 4. If thevelocity is 18 when the t = 3 and the position is 11 when t = 1, thenthe position x(t) =

[Ans: t3 − 2t2 + 3t+ 9

]Example 3.1.5 Exploration 3-9: “Displacement and Accelerationfrom Velocity”

3.1.4 General vs. Particular Solutions

In antidifferentiating v(t) = 50+6t0.6 [9], the displacement is d(t) = 50t+6t1.6

1.6+

C. Notice the +C, which is adding some arbitrary constant, the antiderivativeof +0.

This d(t) with the +C is a family of parallel curves. There are an infinite numberof curves in this family, corresponding to the infinite number of possibilities ofC. Note that these curves are considered parallel because their slopes are thesame for every x value. If your antiderivative includes +C, it is called a generalsolution.



If you want one function d(t) instead of a family of functions, what you wantis a particular solution, instead of the general solution. In order to narrowthe general solution to a particular solution, you need to incorporate some moreinformation into your answer. This is done by solving for C, after plugging insome known point (t, d(t)) into the formula for d(t).

For example, if v(t) = 50+6t0.6, then we can antidifferentiate to get the generalsolution for the displacement, d(t) = 50t+ 3.75t1.6 +C. If we want a particularsolution, we will have to use some more information. In this particular example,for instance, we may know that the displacement at time 0 is 100. Then we solve100 = 50 (0)+3.75 (0)1.6+C, to find that C = 100. Thus, our particular solutionwould be d(t) = 50t+ 3.75t1.6 + 100.

If the point used is (0, d(0)), this is said to be an initial value. Antidifferenti-ation problems that ask you to find a particular solution using an initial valueare called IVP’s, or Initial Value Problems.

Problems

3.A-1 [10] Find a function whose derivative is given. That is, write the generalequation for the antiderivative.

(a) f ′(x) = 7x6

(b) f ′(x) = x5

(c) f ′(x) = x−9

(d) f ′(x) = 36x72 [

Ans: x7 + C,16x6 + C, −1

8x−8 + C, 8x

92 + C

]3.A-2 [10] Find the particular function f(x) that has the given function f ′(x)

for its derivative and contains the given point.

(a) f ′(x) = x4 and f(1) = 10

(b) f ′(x) = x2 − 8x+ 3 and f(−2) = 13[Ans: 1

5x5 + 9.8, 1

3x3 − 4x2 + 3x+

1133

]3.A-3 [10] Derivative and Antiderivative Problem Let g′(x) = 0.6x.

(a) Find the general equation for the antiderivative, g(x).[Ans: g(x) = 0.3x2 + C

](b) Find the particular equation for g(x) in each case.

i. g(0) = 0[Ans: 0.3x2



ii. g(0) = 3[Ans: 0.3x2 + 3

]iii. g(0) = 5

[Ans: 0.3x2 + 5

](c) Plot the graph of g′(x) and the three graphs for g(x) on the same

screen, then sketch the results. Why are the three graphs of g(x)called a family of functions?

3.A-4 [2] If functions f and g are defined so that f ′(x) = g′(x) for all realnumbers x with f(1) = 2 and g(1) = 3, then the graph of f and the graphof g

(A) intersect exactly once;

(B) intersect no more than once;

(C) do not intersect;

(D) could intersect more than once;

(E) have a common tangent at each point of tangency.

[Ans: C]

3.A-5 (adapted from AB93)∫ (

x2 + 2)2dx =[

Ans:x5

5+

43x3 + 4x+ C

]

3.A-6∫

3 (ice)2d (ice) [Ans: iceberg]

3.A-7 Ifd2y

dx2is the second derivative, what do you suppose would be the meaning

ofd−1y

dx−1?

3.A-8 (BC86) For all real numbers x and y, let f be a function such that f(x+

y) = f(x) + f(y) + 2xy and such that limh→0

f(h)h

= 7.

(a) Find f(0). Justify your answer.

(b) Use the definition of the derivative to find f ′(x).

(c) Find f(x).

3.A-9 [10] Displacement Problem Ann Archer shoots an arrow into the air. Letd(t) be its displacement above the ground at time t seconds after she shootsit. From physics she knows that the velocity is given by d′(t) = 70− 9.8t.

(a) Write the general equation for d(t).[Ans: 70t− 4.9t2 + C



(b) Write the particular equation for d(t), using the fact that Ann isstanding on a platform that puts the bow 6 m above the groundwhen she shoots the arrow.[Ans: 70t− 4.9t2 + 6

](c) How far is the arrow above the ground when t = 5? When t = 6?

When t = 9? How do you explain the relationship among the threeanswers?[Ans: 233.5 m, 249.6 m, 239.1 m]

(d) When is the arrow at its highest? How high is it at that time?[Ans: 256 m, at about t = 7.1 sec]

3.A-10 (adapted from [2]) At t = 0, a particle starts at the origin with a velocityof 6 feet per second and moves along the x-axis in such a way that at timet its acceleration is 24t2 feet per second per second. Through how manyfeet does the particle move during the first 2 seconds? [Ans: 44 feet]

3.A-11 (adapted from [2]) The acceleration, a(t), of a body moving in a straightline is given in terms of time t by a(t) = 4−6t. If the velocity of the bodyis 20 at t = 0 and if s(t) is the distance of the body from the origin attime t, what is s(2)− s(1)?[Ans: 19]




AP Unit 3, Day 2: Product and Quotient Rules 99

3.2 Product and Quotient Rules

Advanced Placement

Concept of the derivative


Applications of derivatives

• Interpretation of the derivative as a rate of change in varied applied contexts,including velocity, speed, and acceleration.

• Geometric interpretation of differential equations via slope fields and the re-lationship between slope fields and solution curves for differential equations.


• Basic rules for the derivative of products and quotients of functions.

Textbook §2.4 The Product and Quotient Rules [16]

Resources Exploration 4-2: “Derivative of a Product” in [9]. §3.3 The Productand Quotient Rules in [20].

3.2.1 Product Rule

Example 3.2.1 Consider f(x) = x7 + 1, g(x) = x5− 4, and p(x) =(x7 + 1

) (x5 − 4

)= x12 − 4x7 + x5 − 4.

(a) Write an equation of the line tangent to f(x) at x = 1. Writeit in the form y = k + m (x− 1), i.e., treat (x− 1) as a singleentity, which you will leave alone.

(b) Write an equation of the line tangent to g(x) at x = 1. Writeit in the form y = a0 + a1 (x− 1).

(c) Write an equation of the line tangent to p(x) at x = 1. Asbefore, treat (x− 1) as a single entity.

(d) Multiply the tangent lines at x = 1 for f(x) and g(x), keep-ing (x− 1) as a single variable unto itself: don’t worry aboutdistributing it. The only like terms you need to combine arethe (x− 1) terms. Your answer should look like y = a0 +a1 (x− 1) + a2 (x− 1)2. What do you see?



Example 3.2.2

(a) Show thatd

dxx11 6= d

dxx7 · d

dxx4

(b) How could we squeeze the real derivative out of x4 and x7.Think about: how can we get the correct exponents, and thecorrect coefficients.

Product Rule

d

dx(u · v) =

du

dxv + u

dv

dx

If h(x) = f(x)g(x), then

h′(x) = f ′(x)g(x) + f(x)g′(x)

Derivative of First · Second + First · Derivative of Second.

Note that the order can be rearranged:

Derivative of First · Second + Derivative of Second · First.First · Derivative of Second + Derivative of First · Second.First · Derivative of Second + Second · Derivative of First.

Do whichever way you remember best.

Example 3.2.3 Use linear approximations of f(x) and g(x) atx = c to demonstrate the product rule.

Example 3.2.4 Use the limit definition of the derivative to showthat, if p(x) = f(x)g(x), then p′(x) = f ′(x)g(x) + f(x)g′(x).

Example 3.2.5 Suppose that f is the function shown in Figure 3.1and that h(x) = x2f(x) [17]. Evaluate h′(2) and h′(4)

[Ans: −16, 64]




Example 3.2.6 If f(x) =(x3 + 1

)2, find f ′(x)

Example 3.2.7 Findd

dx

[(x3 + 1

)3].This can be done two ways, one using the derivative of

(x3 + 1

)2, and splitting(x3 + 1

)3 into(x3 + 1

)2 (x3 + 1

).

Another way is to use a variation of the product rule for differentiating theproduct of three functions. For the product of three functions, u, v, and w:

d

dx(u · v · w) =

du

dx· v · w + u · dv

dx· w + u · v · dw

dx

For e(x) = f(x)g(x)h(x),

e′(x) = f ′(x)g(x)h(x) + f(x)g′(x)h(x) + f(x)g(x)h′(x)

Examine this formula and see if you can notice how the pattern for two functionshas been expanded for three functions.

Once you get your answer, simplify and factor it. In the back of your mind, belooking for a method to go straight to the answer very quickly.




[Ans: −8]

Example 3.2.9 Use the product rule to prove the power rule (forn = 1, 2, . . . using mathematical induction.

3.2.2 Quotient Rule

Quotient Rule

d

dx

(uv

)=vdu

dx− udv

dxv2

(v 6= 0)

If h(x) =f(x)g(x)

, then

h′(x) =g(x)f ′(x)− f(x)g′(x)

[g(x)]2

orLoDeHi−HiDeLo

LoLo

Because of the subtraction, it is important that you keep things in this order(unlike the product rule, where order is not really important).

Example 3.2.10 Use the limit definition of the derivative to provethe quotient rule.

Example 3.2.11 Use linear approximations of f(x) and g(x) atx = c to demonstrate the product rule.

Example 3.2.12 If y =

(x3 + 1

)2x3 + 1

, finddy

dx.

Notice that this problem has a quick and simple answer, if you cancel x3 + 1,and take the derivative of that to be 3x2. Confirm that the quotient rule works.

Example 3.2.13 Findd

dx

(1

x3 + 1

).



Figure 3.2:

Compare your answer with the derivative you get for(x3 + 1

)2 and(x3 + 1

)3and be looking for the pattern, while thinking of the derivatives of x2, x3, and1x .

Example 3.2.14 from 2000 AP Calculus AB-2. Two runners, Aand B, run on a straight racetrack for 0 ≤ t ≤ 10 seconds. Thegraph in Figure 3.2, which consists of two line segments, shows thevelocity, in meters per second, of Runner A. The velocity, in metersper second, of Runner B is given by the function v defined by v(t) =

24t2t+ 3

. Find the acceleration of Runner B at time t = 2 seconds.

Indicate units of measure.

Runner A: acceleration =103

= 3.333 meters/sec2

Runner B: a(2) = v′(2) =72

(2t+ 3)2

∣∣∣∣∣t=2

=7249

= 1.469 meters/sec2

Tangent Lines

When writing the equation of a line, use the point-slope form:

y − y1 = m (x− x1)

When talking about m, the slope of the tangent line, you should immediatelybe thinking about the derivative.

m = f ′(x1)

y − y1 = y′(x1) (x− x1)



Example 3.2.15 (adapted from AB93) An equation of the line

tangent to the graph of y =3x− 22x+ 3

at the point (−1,−5) is

[Ans: y + 5 = 13 (x+ 1)]

Problems

3.B-1 (adapted from AB98) Let f and g be differentiable functions with thefollowing properties:

(a) g(x) < 0 for all x(b) f(1) = π − 1

If h(x) = f(x)g(x) and h′(x) = f(x)g′(x), then f(x) = [Ans: π − 1]

3.B-2 [20]

(a) If F (x) = f(x)g(x), where f and g have derivatives of all orders,show that

F ′′ = f ′′g + 2f ′g′ + fg′′

(b) Find a similar formulas for F ′′′.

3.B-3 Use the Product Rule to show thatd

dx[f(x)]2 =

d

dx[f(x)f(x)] = 2f(x)f ′(x)

3.B-4 [20]

(a) Use the Product Rule twice to prove that if f , g, and h are differen-tiable, then

d

dx(fgh) = f ′gh+ fg′h+ fgh′

(b) Taking f = g = h in part 4a, show that

d

dx[f(x)]3 = 3 [f(x)]2 f ′(x)

3.B-5 [20] In this exercise we estimate the rate at which the total personal in-come is rising in the Miami-Ft. Lauderdale metropolitan area. In July,1993, the population of this area was 3,354,000 and the population wasincreasing at roughly 45,000 people per year. The average annual in-come was $21,107 per capita, and this average was increasing at about$1900 per year (well above the national average of about $660 yearly).Use the Product Rule and these figures to estimate the rate at which to-tal personal income was rising in Miami-Ft. Lauderdale in July, 1993.[Ans: $7.322 billion per year]



3.B-6 [17] Suppose that f(1) = 2 and f ′ is the function shown in Figure 3.3. Letm(x) = x3f(x).


(a) Evaluate m′(1). [Ans: 10]

(b) Show that m is increasing at 2.

(c) Estimate m′′(1). [Ans: 34]

3.B-7 (AB85) Let f be the function given by f(x) =2x− 5x2 − 4

.

(a) Find the domain of f . [Ans: {x|x ∈ R ∩ x 6= ±2}](b) Write an equation for each vertical and each horizontal asymptote

for the graph of f . [Ans: y = 0, x = 2, x = −2]

(c) Find f ′(x).[Ans: −2(x−4)(x−1)

(x2−4)2

](d) Write an equation for the line tangent to the graph of f at the point

(0, f(0)).[Ans: y − 5

4 = − 12x]

3.B-8 Differentiable functions f and g have the values shown in the table. [14]

x f f ′ g g′

0 2 1 5 −41 3 2 3 −32 5 3 1 −23 10 4 0 −1

(a) If A(x) = f(x) + 2g(x), then A′(3) = [Ans: 2]

(b) If B(x) = f(x) · g(x), then B′(2) = [Ans: −7]

(c) If K(x) =f(x)g(x)

, then K ′(0) =[Ans: 13

25



(d) If D(x) =1

g(x), then D′(1) =

[Ans: 1

3

]3.B-9 (adapted from [2]) If g(x) =

x+ 2x− 2

, then g′(−2) =[Ans: − 1

4

]3.B-10 (adapted from [2]) Consider the function f(x) =

6xa+ x3

for which f ′(0) =

0.75. The value of a is [Ans: 8]

3.B-11 (adapted from [2]) If y =4

3 + x2, then

dy

dx=

[Ans: −8x

(3+x2)2

]3.B-12 [20] The curve y =

11 + x2

is called a witch of Maria Agnesi. Find

an equation of the tangent line to this curve at the point(−1,

12

). On

your calculator, graph the curve and the tangent line on the same screen.[Ans: y = 1

2x+ 1]

3.B-13 [20] If f and g are the functions whose graphs are shown in Figure 3.4, let

u(x) = f(x)g(x) and v(x) =f(x)g(x)

.

Figure 3.4:

(a) Find u′(1). [Ans: 0]

(b) Find v′(5).[Ans: − 2

3

]3.B-14 [17] Suppose that f is the function shown in Figure 3.1. Let m(x) =

f(x)x2 + 1

. Evaluate m′(0). [Ans: −4]


AP Unit 3, Day 3: Chain Rule 107

3.3 Chain Rule

Advanced Placement


• Chain rule


• Slope of a curve at a point. Examples are emphasized, including points atwhich there are vertical tangents and points at which there are no tangents.


Textbook §2.5 The Chain Rule [16]

Resources §3-7 Derivatives of Composite Functions- The Chain Rule in Foer-ster [10]. Exploration 3-7:“Rubber-Band Chain Rule Problem” in [9]. §3-5 TheChain Rule in Stewart [20]. §3.6 New Derivatives from Old: The Chain Rule inOstebee & Zorn [17].

3.3.1 Chain Rule

Example 3.3.1 Findd

dx

((x3 + 1

)2),d

dx

((x3 + 1

)3),d

dx

(1

x3 + 1

),

d

dx

√x3 + 1

These are some of the results we found last class:

Chain Rule

dy

dx=dy

du· dudx

For h(x) = f (g(x)),h′(x) = f ′ (g(x)) · g′(x)

Derivative of the Outside · Derivative of the Inside



Power Chain Rule

d

dxun = nun−1 du

dx

Example 3.3.2 Use the product rule, together with the chain rule,to prove the quotient rule, by finding

d

dx

(f(x) · 1

g(x)

)

Example 3.3.3 Given that f ′(x) =x

x2 + 1and g(x) =

√3x− 1,

find F ′(x) if F (x) = f (g(x)) [19].

[Ans: 1

2x

]Example 3.3.4 Let h(x) = f (g(x)). Use the information aboutf and g given in the table below to fill in the missing informationabout h and h′ [17].

x f(x) f ′(x) g(x) g′(x) h(x) h′(x)1 1 2 4 32 2 1 3 43 4 3 1 24 3 4 2 1

Example 3.3.5 The function F is defined by F (x) = G (x−G(x))where the graph of the function G is shown in Figure 3.5. Find F ′(7)[2]

[Ans: − 2

3

]

Product and Quotient Rules did not vanish

Example 3.3.6 [18] Differentiate y = (2x+ 1)5 (x3 − x+ 1

)4[Ans: 2 (2x+ 1)4 (

x2 − x+ 1)3 (17x3 + 6x2 − 9x+ 3

)]Mr. Budd, compiled September 29, 2010


Figure 3.5: from Best & Lux [2]

Example 3.3.7 [18] Find the derivative of the function

g(t) =(t− 22t+ 1

)9

[Ans: 45(t−2)8

(2t+1)10

]

Problems

3.C-1 [14] Differentiable functions f and g have the values shown in the table.

x f f ′ g g′

0 2 1 5 −41 3 2 3 −32 5 3 1 −23 10 4 0 −1

(a) If H(x) =√f(x), then H ′(3) =

[Ans: 2√

10

](b) If M(x) = f (g(x)), then M ′(1) = [Ans: −12]

(c) If P (x) = f(x3), then P ′(1) = [Ans: 6]

3.C-2 [3] Let the function f be differentiable on the interval [0, 2.5] and define gby g(x) = f (f(x)). Use the table to estimate g′(1).

x 0.0 0.5 1.0 1.5 2.0 2.5f(x) 1.7 1.8 2.0 2.4 3.1 4.4

[Ans: 1.2]



3.C-3 [20] If f and g are the functions whose graphs are shown in Figure 3.6,let u(x) = f(g(x)), v(x) = g(f(x)), and w(x) = g(g(x)). Find eachderivative, if it exists. If it does not, explain why.

Figure 3.6: [20]

(a) u′(1)[Ans: 3

4

](b) v′(1) [Ans: nonexistent, g′(2) DNE]

(c) w′(1) [Ans: −2]

3.C-4 [17] Suppose that f is the function shown in Figure 3.7 and that g(x) =f(x2).


(a) For which values of x is g′(x) = 0?[Ans:

√2, 0

](b) Is g increasing or decreasing at −1? [Ans: increasing]

(c) Is g′ positive or negative over the interval(√

2,√

5)? [Ans: positive]

3.C-5 [3] The graphs of functions f and g are shown in Figure 3.8. If h(x) =f (g(x)), which of the following statements are true about the function h?

I. h(2) = 5.

II. h is increasing at x = 4.




III. The graph of h has a horizontal tangent at x = 1.

[Ans: II and III only]

3.C-6 (BC98) If f and g are twice differentiable and if h(x) = f(g(x)), thenh′′(x) =

[Ans: f ′′(g(x)) [g′(x)]2 + f ′(g(x))g′′(x)

]3.C-7 [2] A particle moves along the x-axis so that at time t, t >= 0, its position

is given by x(t) = (t+ 1) (t− 3)3. Find a formula for the velocity, v(t), andthe acceleration, a(t), of the particle.

[Ans: 4t (t− 3)2; 12 (t− 3) (t− 1)

]3.C-8 (adapted from AB93) Find the derivative of f(x) = (2x− 3)5 (7x+ 2)4.

At how many different values of x will f ′(x) be 0?[Ans: 2 (2x− 3)4 (7x+ 2)3 (63x− 32); Three

]3.C-9 (adapted from AB97) If f(x) = x

√6x− 2, then f ′(x) =

[Ans:

9x− 2√6x− 2

]3.C-10 [17] Suppose that f(1) = 2, that f ′ is the function shown in Figure 3.9,

and that k(x) = f(x3). Evaluate k′(−1). [Ans: 12]




3.C-11 [2] Suppose that g is a function with the following two properties: g(−x) =g(x) for all x and g′(a) exists. [I.e., g is an even, differentiable function.]Find g′(−x). [Ans: −g′(x)]

3.C-12 [20] Use the table to estimate the value of h′(0.5), where h(x) = f(g(x)).[Ans: −17.4]

x 0 0.1 0.2 0.3 0.4 0.5 0.6f(x) 12.6 14.8 18.4 23.0 25.9 27.5 29.1g(x) 0.58 0.40 0.37 0.26 0.17 0.10 0.05

3.C-13 [20] If f is the function whose graph is shown, let h(x) = f(f(x)) andg(x) = f(x2). Use the graph of f to estimate the value of each derivative.

Figure 3.10: [20]

(a) h′(2) [Ans: 0.64]

(b) g′(2) [Ans: 9]

3.C-14 Find∫f ′ (g(x)) g′(x) dx [Ans: f (g(x)) + C]

3.C-15 Find∫

13(x3 + 1

)12 (3x2)dx

[Ans:

(x3 + 1

)13 + C]


AP Unit 3, Day 4: Tangent Lines 113

3.4 Tangent Lines

Advanced Placement


• Slope of a curve at a point. Examples are emphasized, including points atwhich there are vertical tangents and points at which there are no tangents.


Textbook §3.1 Linear Approximations and Newton’s Method: “Linear Ap-proximations” [16]

3.4.1 Tangent Lines

Lines with known y-intercept:

y = mx+ b

Otherwise, in writing equations for lines, it is best to use the point-slope formula,in which you know a point and a slope:

y − y1 = m (x− x1)

If we are talking about writing the equation of a tangent line, remember thatm, the slope of the tangent line is the same as the derivative.

Example 3.4.1 (adapted from AB93) An equation of the line tan-

gent to the graph of y =3x− 22x+ 3

at the point (−1,−5) is

[Ans: y + 5 = 13 (x+ 1)]

Note: if you are simply asked to write an equation for the line, this would bean acceptable answer. If you’re asked to put the equation in standard form, itwould be 13x − y = −8. If you’re asked to put the equation in slope-interceptform, the answer would be y = 13x + 8. You should be aware of each of theseforms because you may need to spot them on a multiple choice exam.



Example 3.4.2 (adapted from AB98) Write an equation (in slope-intercept form) for the line tangent to the graph of f(x) = x4 − x2

at the point where f ′(x) = 1

[Ans: y = x− 1.055]


[Ans: −8]

Example 3.4.4 (adapted from [2]) If the line 3x − y + 5 = 0 istangent in the second quadrant to the curve y = x3 + k, then k =

[Ans: 3]

3.4.2 Horizontal Tangents

f(x) has a horizontal tangent at (c, f(c)) if the slope of the tangent line is zero,i.e., f ′(c) = 0. The equation of the tangent line will be y = f(c).

Example 3.4.5 The composite function h is defined by h(x) =f (g(x)), where f and g are functions whose graphs are shown inFigure 3.11. Find the number of horizontal tangent lines to thegraph of h. [2]

[Ans: 6]

Example 3.4.6 (AB92) Let f be the function defined by f(x) =3x5 − 5x3 + 2. Write the equation of each horizontal tangent line tothe graph of f .

[Ans: y = 0, y = 2, y = 4]




3.4.3 Vertical Tangents

Definition 3.1 (Vertical Tangent). f(x) has a vertical tangent at the point(c, f(c)) if the following conditions are met:

1. There is point of tangency, i.e., f(c) exists.

2. f ′(c) is infinite.

The equation of the line of tangency is x = c.

Recall that the usual way to get an infinite answer is to have nonzero over zero.Note that it is not sufficient to state that the denominator is zero.

Notice: both conditions must be met. Recall the definition of limit and realizethat the derivative cannot exist unless the point exists. Frequently we may findf ′ to be nonzero over zero when we plug the number into the formula, but f isas well. It is not sufficient to find an infinite derivative; the point must exist aswell.

Also, we cannot use the point-form of a line if the slope does not exist. Simplywrite the equation as x = c.

Example 3.4.7 Write the equations of all the vertical (and hori-

zontal tangents) to f(x) = 3√

(x2 − 4)2.

[Ans: x = 2, x = −2, (y = 3

√16]



Example 3.4.8 Write the equations of all the vertical tangents to

f(x) =1

x− 2.

[Ans: none]

Example 3.4.9 (adapted from [2]) Which of the following is afunction with a vertical tangent at x = 0?

(A) f(x) = x2

(B) f(x) = 3√x

(C) f(x) = 1x

(D) f(x) = |x|

[Ans: f(x) = 3√x]

Note that the derivative is infinite at x = 0 ford

dx

√x and for

d

dx

1x

, but the

function1x

is not defined at x = 0. For f(x) =1x

, there is a vertical asymptoteat x = 0, which is different from a vertical tangent.

3.4.4 Normal Lines

Normal lines are perpendicular to tangent lines. The slope of the normal linewill be the negative reciprocal of the slope of the tangent line. (Recall that theslope of the tangent line is the . . .)

Example 3.4.10 The line that is normal to the curve y = x2+2x−3at (1, 0) intersects the curve at what other point? [19]

[Ans:

(−13

4,

1716

)]

3.4.5 Tangent Line Approximations

Tangent Lines to Approximate Values of f

Example 3.4.11 You are stranded on a deserted island with yourbest friend, Wilson. He building some right triangles and wants toknow the square root of 9.3. Without a calculator, estimate

√9.3.



Example 3.4.12 Let f(x) =1

x+ 2. Use the tangent line to the

graph of f at x = −32

to find approximations for:

(a) f(−1.4)

(b) f(−1.45)

(c) f(−1.501)

[Ans: 1.6, 1.8, 2.004]

Compare these tangent-line approximations to the actual functional values.

Example 3.4.13 (adapted from BC98) Let f be the function givenby x2 − 2x+ 3. The tangent line to the graph of f at x = 3 is usedto approximate values of f(x). For what range of values is the errorresulting from this tangent line approximation less than 0.1? Whichof the following is the greatest value of x for which the error is lessthan 0.1?

(A) 3.1

(B) 3.2

(C) 3.3

(D) 3.4

(E) 3.5

[Ans: 3.3]Note that the linear approximation has an error below 0.1 from 2.684 . . . to3.316 . . .

Tangent Lines to Approximate Zeros of f

Example 3.4.14 (adapted from AB97) Let f be a differentiablefunction such that f(2) = 1 and f ′(2) = −4. If the tangent line tothe graph of f at x = 2 is used to find an approximation to a zeroof f , that approximation is

[Ans: 2.25]



3.4.6 Introduction to Slope Fields

Slope fields give you a sense of the direction or current of the general solutions.To find the particular solution, start at the known point, and draw a curve tothe left and right that is always parallel to the slopes as seen in the slope field.

Example 3.4.15 Draw a slope field fordy

dx= −2x + 2 at twenty-

five points using x-coordinates of 0, 12 , 1,

32 , 2, and y-coordinates of

−1,− 12 , 0,

12 , 1. Sketch an antiderivative within your slope field.

Example 3.4.16 With a window of [0, 4.7] × [−3.1, 3.1], use the

BIGSLOPE program to draw a slope field ofdy

dx= −4x + 8. Then

use the 5. Draw Solution option to graph∫

(−4x+ 8) dx for severalvalues of C (−10, −8, −6, −4).

Problems

3.D-1 99 MM2 The function f is given by

f(x) =2x+ 1x− 3

, x ∈ R , x 6= 3.

(a) i. Show that y = 2 is an asymptote of the graph of y = f(x).ii. Find the vertical asymptote of the graph. [Ans: x = 3]

iii. Write down the coordinates of the point P at which the asymp-totes intersect. [Ans: (3, 2)]

(b) Find the points of intersection of the graph and the axes.[Ans:

(− 1

2 , 0),(0,− 1

3

)](c) Hence sketch the graph of y = f(x), showing the asymptotes by

dotted lines.

(d) Show that f ′(x) =−7

(x− 3)2 and hence find the equation of the tan-

gent at the point S where x = 4. [Ans: y − 9 = −7 (x− 4)]

(e) The tangent at the point T on the graph is parallel to the tangent atS. Find the coordinates of T . [Ans: (2,−5)]

(f) Show that P is the midpoint of ST .

3.D-2 (adapted from BC97) Refer to the graph in Figure 3.12. The function fis defined on the closed interval [0, 8]. The graph of its derivative f ′ isshown. The point (1, 4) is on the graph of y = f(x). An equation of theline tangent to the graph of f at (1, 4) is [Ans: y − 4 = 3 (x− 1)]



Figure 3.12:

3.D-3 (BC90) Let f(x) = 12 − x2 for x ≥ 0 and f(x) ≥ 0. The line tangent tothe graph of f at the point (k, f(k)) intercepts the x-axis at x = 4. Whatis the value of k? [Ans: k = 2]

3.D-4 (adapted from AB97) At what point on the graph of y =14x3 is the

tangent line parallel to the line 3x− 4y = 7?[Ans:

(1, 1

4

)]3.D-5 (adapted from [2]) Let f(x) = 3x3−4x+1. An equation of the line tangent

to y = f(x) at x = 2 is [Ans: y = 32x− 47]

3.D-6 (adapted from [2]) Find the point on the graph of y =√x between (4, 2)

and (9, 3) at which the normal to the graph has the same slope as the linethrough (4, 2) and (9, 3).

[Ans:

(254 ,

52

)]3.D-7 (adapted from [3]) If the graph of the parabola y = 2x2 +x+ k is tangent

to the line 3x+ y = 3, then k = [Ans: 5]

3.D-8 (adapted from [3]) A tangent line drawn to the graph of y =8x

1 + x3at

the point (1, 4) forms a right triangle with the coordinate axes. The areaof the triangle is [Ans: 9]

3.D-9 (adapted from AB93) Find the derivative of f(x) = (x− 3)5 (x+ 2)4.At how many different values of x will the tangent line be horizontal?[Ans: Three]

3.D-10 (adapted from [2]) Which of the following is a function with a verticaltangent at x = 0?



(A) f(x) = x2

(B) f(x) = 3√x

(C) f(x) = 1x

(D) f(x) = |x|

[Ans: f(x) = 3√x]

3.D-11 (adapted from [2]) An equation of the normal to the graph of f(x) =x

3x− 2at (1, f(1)) is [Ans: x− 2y = −1]

3.D-12 (adapted from [3]) An equation of the line normal to the graph of f(x) =x

x− 2at (3, 3) is

[Ans: x− 2y + 3 = 0]

3.D-13 (AB88) Let f be the function given by f(x) =√x4 − 16x2.

(a) Find the domain of f . [Ans: Df : x ≥ 4, x ≤ −4, x = 0](b) Describe the symmetry, if any, of the graph of f . [Ans: Graph of f is symmetric w.r.t. the y-axis]

(c) Find f ′(x).[Ans: 2x3−16x√

x2(x2−16)

](d) Find the slope of the line normal to the graph of f at x = 5[

Ans: − 334

]3.D-14 (adapted from AB97) No calculator! Let f be a differentiable function

such that f(4) = 1.5 and f ′(4) = 3. If the tangent line to the graph of f atx = 4 is used to find an approximation to a zero of f , that approximationis[Ans: 3.5]

3.D-15 (adapted from [3]) No calculator! Let f be a function with f(2) = 6 andderivative f ′(x) =

√x3 + 1. Using a tangent line approximation to the

graph of f at x = 2, estimate f(2.05)[Ans: 6.15]

3.D-16 (adapted from [3]) No calculator! The approximate value of y =√x2 + 3

at x = 1.08, obtained from the tangent to the graph at x = 1, is[Ans: 2.04]

3.D-17 With a window of [0, 4.7] × [−3.1, 3.1], use the BIGSLOPE program to

draw a slope field ofdy

dx=

1√x

, i.e., put1√x

into Y1. After the slope

field is drawn, then graph 2√x+C for C = −3,−2,−1 using the 5. Draw

Solution option.

3.D-18 With a window of [0, 2.35] × [−3.1, 3.1], use the BIGSLOPE program to

draw a slope field ofdy

dx= 4 (x− 1). Then draw solution to 2

(x2 − 2x

)+C

for C = −1, 1, 3.



3.D-19 With a window of [−2.35, 2.35]× [−3.1, 3.1], use the BIGSLOPE program

to draw a slope field ofdy

dx= x2−1. Then draw solutions to

∫ (x2 − 1

)dx

for C = −2, 0, 2.

3.D-20 Draw by hand a slope field fordy

dx= 2x − 2 at twenty-five points using

x-coordinates of 0, 12 , 1,

32 , 2, and y-coordinates of −1,− 1

2 , 0,12 , 1. Sketch

an antiderivative within your slope field.

Basic Differentiation Review Problems

3.D-21 (AB97) If f(x) = −x3 + x+1x

, then f ′(−1) = [Ans: −3]

3.D-22 (AB97) A particle moves along the x-axis so that its velocity at any timet ≥ 0 is given by v(t) = 3t2 − 2t− 1. The position x(t) is 5 for t = 2.

(a) Write a polynomial expression for the position of the particle at anytime t ≥ 0.

(b) For what values of t, 0 ≤ t ≤ 3, is the particle’s instantaneous velocitythe same as its average velocity on the closed interval [0, 3]

3.D-23 [2] What is limx→1

√x− 1x− 1

?[Ans:

12

]3.D-24 [2] If h(x) =

(x2 − 4

)3/4 + 1, then the value of h′(2) is [Ans: nonexistent]

3.D-25 [2] Which of the following is a function with a vertical tangent at x = 0?

(A) f(x) = x3

(B) f(x) = 3√x

(C) f(x) =1x

[Ans: f(x) = 3√x]

3.D-26 [2] The derivative of√x− 1

x 3√x

is[Ans:

12x−1/2 +

43x−7/3

]3.D-27 (from [2]) A function f is defined for all real numbers and has the following

property: f(a+ b)− f(a) = 3a2b+ 2b2. Find f ′(x).[Ans: 3x2

]3.D-28 [2] If

d

dx(f(x)) = g(x) and

d

dx(g(x)) = f(3x), then

d2

dx2

(f(x2))

is

(A) 4x2f(3x2) + 2g(x2)

(B) f(3x2)



(C) f(x4)

(D) 2xf(3x2) + 2g(x2)

(E) 2xf(2x2) [Ans: 4x2f(3x2) + 2g(x2)

]3.D-29 [2] If y =

34 + x2

, thendy

dx=

[Ans:

−6x(4 + x2)2

]

3.D-30 [2] limh→0

3(

12

+ h

)5

− 3(

12

)5

h=

[Ans:

1516

]3.D-31 [2] An object moves along the x-axis so that at time t, t > 0, its position

is given by x(t) = t4 +t3−30t2 +88t. At the instant when the accelerationbecomes zero, the velocity of the object is [Ans: 12]

3.D-32 Using the limit definition of derivative, prove the Product Rule.

3.D-33 Use the Chain and Product Rule to prove the Quotient Rule.


Unit 4

Curve Sketching

1. Curve Sketching

2. Second Derivative Sketching

Advanced Placement

Derivative as a function


• Relationship between the increasing and decreasing behavior of f and the signof f ′.

Second derivatives.

• Corresponding characteristics of the graphs of f , f ′, and f ′′.

• Relationship between the concavity of f and the sign of f ′′.

• Points of inflection as places where the concavity changes.

123

124 AP Unit 4 (Curve Sketching)


AP Unit 4, Day 1: Relating Graphs of f and f ′ 125

4.1 Relating Graphs of f and f ′

Advanced Placement




Resources §4.3 Connecting f ′ and f ′′ with the Graph of f in [8]

4.1.1 Relative Extrema

Definition 4.1 (Critical Point). A critical point is a point on the graph of fwhere f ′ is either 0 or undefined.

Definition 4.2 (Critical Number). A critical number is the x-coordinate of thecritical point.

Note: the critical number is sometimes and confusingly referred to as the criticalpoint. Judge from context whether the critical point really means the criticalpoint or actually means the critical number.

Example 4.1.1 How many critical points does the function f(x) =(x+ 2)5 (

x2 − 1)4 have? [2]

Example 4.1.2 How many critical points does the function f(x) =3√

(x2 − 4)2 + 1 have? How about f(x) =1

x− 2?

A point on the graph of f where f ′ is 0 has a horizontal tangent line. A pointon the graph of f where f ′ is undefined is either a cusp or a vertical tangent.

An extremum is either a maximum or a minimum.

Definition 4.3 (Relative Extrema). If c is in the domain of a function f , thenf(c) is a

1. relative maximum of f at c if and only if f(c) ≥ f(x) for all x in nearc.



2. relative minimum of f at c if and only if f(c) ≤ f(x) for all x in near c.

Relative extrema are also called local extrema.

4.1.2 First Derivative Test

Critical points are important because they tell me where relative minima ormaxima might occur. Just because the derivative is zero or undefined does notmean that I have a local extrema. But finding the critical points allows me totest only a few points for extremity, as opposed to an infinite number of pointsin a typical domain.

To find out whether the critical points actually are relative extrema, we testthem. There are two tests, the First Derivative Test, and the Second DerivativeTest.

The First Derivative Test checks the sign of the first derivative before and afterthe critical point. If the sign of the first derivative is changing, then the originalfunction is changing direction, and there is a local extremum.

In order to perform the First Derivative Test, I need to cut the number line ateach critical number, so that I know how close I need to be when I check oneither side of the critical number.

Theorem 4.1 (First Derivative Test for Relative Extrema). (Notes stolen from[8]) At a critical point c:

1. If f ′ changes sign from positive to negative at c (f ′ > 0 for x < c andf ′ < 0 for x > c), then f has a relative maximum value at c.

2. If f ′ changes sign from negative to positive at c (f ′ < 0 for x < c andf ′ > 0 for x > c), then f has a relative minimum value at c.

3. If f ′ does not change sign at c (f ′ has the same sign on both sides of c),then f has no relative extreme value at c.



Example 4.1.3 (AB98) The graph of f ′, the derivative of f , isshown in Figure 4.1. Which of the following describes all relativeextrema of f on the open interval (a, b)?

(A) One relative maximum and two relative minima.(B) Two relative maxima and one relative minimum.(C) Three relative maxima and one relative minimum.(D) One relative maximum and three relative minima.(E) Three relative maxima and two relative minima.

[Ans: A]

Example 4.1.4 [2]The function f(x) = x4 − 18x2 has a relativeminimum at x =

[Ans: −3 and 3]

Example 4.1.5 Determine at what values of x the function f(x)has a relative minimum for:

• f ′(x) = x2(x2 − 9

)• f ′(x) = x2 (x− 9)

• f ′(x) = x (x− 3)2 (x+ 3)2

• f ′(x) = x2 (x− 3)2 (x+ 3)

Example 4.1.6 If the derivative of f is given by f ′(x) = sin (lnx),at which value of x, x ∈ [0, 1.07], does f have a relative maximumvalue? a relative minimum?

[Ans: 0.0432, 1]



Figure 4.1: Graph of f ′(x)

Connecting f and f ′


Example 4.1.7 (AB98) The graph of f is shown in Figure 4.2.Sketch the derivative of f .

Example 4.1.8 The graph of f ′, the derivative of f , is shown inFigure 4.1. Sketch a graph of f .

Example 4.1.9 (AB96) Figure 4.4 shows the graph of f ′, thederivative of a function f . The domain of f is the set of all realnumbers such that −3 < x < 5.



Figure 4.3: Derivative of f from Figure 4.2

Figure 4.4: AB ’98 Note: This is the graph of the derivative of f , not the graphof f .

(a) For what values of x does f have a relative maximum? Why?

(b) For what values of x does f have a relative minimum? Why?

(c) On what intervals is the graph of f concave upward? Use f ′ tojustify your answer.

(d) Suppose that f(1) = 0. Draw a sketch that shows the generalshape of the graph of the function f on the open interval 0 <x < 2.

[Ans: x = −2,f ′(x) changes from positive to negative at x = −2][Ans: x = 4, f ′(x) changes from negative to positive at x = 4][Ans: (−1, 1) and (3, 5), f ′ is increasing on these intervals][Ans: Figure 4.5]



Figure 4.5: AB ’96



Problems

4.A-1 (AB00) Figure 4.6 shows the graph of f ′, the derivative of the function f ,or −7 ≤ x ≤ 7. The graph of f ′ has horizontal tangent lines at x = −3,x = 2, and x = 5, and a vertical tangent line at x = 3.

Figure 4.6: AB ’00

(a) Find all values of x, for −7 < x < 7, at which f attains a relative min-imum. Justify your answer. [Ans: x = −1, f ′(x) changes from negative to positive at x = −1]

(b) Find all values of x, for−7 < x < 7, at which f attains a relative max-imum. Justify your answer. [Ans: x = −5, f ′(x) changes from positive to negative at x = −5]

(c) Find all values of x, for −7 < x < 7, at which f ′′(x) < 0.

[Ans: on the intervals (−7,−3), (2, 3), and (3, 5),f ′′(x) exists and f ′ is decreasing ]

4.A-2 (AB98) The graphs of the derivatives of the functions f , g, and h areshown in Figure 4.7. Which of the functions f , g, or h have a relativemaximum on the open interval a < x < b? [Ans: f only]

Figure 4.7: AB98

4.A-3 (adapted from [3]) Which of the following are (is) true about a particlethat starts at t = 0 and moves along a number line if its position at timet is given by s(t) = (t− 1)3 (t− 5)?

I. The particle is moving to the right for t > 4.

II. The particle is at rest at t = 1 and t = 4.

III. The particle changes direction at t = 1.



[Ans: I and II only]

4.A-4 (adapted from [3]) A particle starts at time t = 0 and moves along anumber line so that its position, at time t ≥ 0, is given by x(t) =(t− 3) (t− 7)3. The particle is moving to the left for what values of t?[Ans: t < 4]

4.A-5 (adapted from [3]) If the derivative of the function f is f ′(x) = −2 (x− 3) (x+ 1)2 (x+ 2)3,then f has a local minimum at x = [Ans: −2]

Figure 4.8: Problem 6: Graph of the derivative of f

4.A-6 (adapted from AB97) The graph of the derivative of f is shown in Figure4.8. Graph the function f . [Ans: Figure 4.9]

4.A-7 (adapted from AB93) For what value of x does the function f(x) =(x− 3) (x− 2)2 have a relative minimum?

[Ans: 8

3

]4.A-8 (adapted from AB93) How many critical points does the function f(x) =

(x− 3)5 (x+ 2)4 have? [Ans: Three]

4.A-9 (adapted from AB93) The function f given by f(x) = x3 − 12x + 24 isincreasing for what values of x? [Ans: x < −2, x > 2]

Figure 4.9: Answer to problem 6: Graph of f



Figure 4.10: Problem 10: Graph of h(x)

Figure 4.11: Answer to problem 10: Graph of h′(x)

4.A-10 (BC98) The graph of y = h(x) is shown in Figure 4.10. Sketch the graphof y = h′(x) [Ans: Figure 4.11]

4.A-11 [2] A particle moves along the x-axis so that at time t, t >= 0, its positionis given by x(t) = (t+ 1) (t− 3)3. For what values of t is the velocity ofthe particle increasing? [Ans: t < 1 or t > 3]

4.A-12 [2] Consider the function f(x) =x4

2− x5

10. The derivative of f attains it

maximum value at x = [Ans: 3]

4.A-13 The derivative of f is given by f ′(x) = ex2/5 cos

(x2

5

). Find the x–

coordinates of all relative minima of f on the interval x ∈ [0, 7]. [Ans: 4.854]

4.A-14 (adapted from AB98) The first derivative of the function f is given by

f ′(x) =cos2 x

x− 3

20. How many critical values does f have on the open

interval (0, 10)? [Ans: Five]

4.A-15 (adapted from AB97) If the derivative of f is given by f ′(x) = 2x2 − ex,at which value of x does f have a relative minimum value? [Ans: 1.488]




AP Unit 4, Day 2: Second Derivative Sketching 135

4.2 Second Derivative Sketching

Advanced Placement




Second derivatives.




4.2.1 Concavity

There are four basic curve types... (Under Construction)

The graph of f is concave upward if f ′ is increasing, i.e., f ′′ is positive. Thegraph of f is concave downward if f ′ is decreasing, i.e., f ′′ is negative.

If the function is concave down, then the curve is below the tangent line. If thefunction is concave up, then the curve is above the tangent line. If the functionis concave up, then the tangent line underestimates the actual function; if thefunction is concave down, the tangent line overestimates the actual function.

Figure 4.12:



Example 4.2.1 The graph of a twice-differentiable function f isshown in Figure 4.12. Which of the following is true?

(A) f(1) < f ′(1) < f ′′(1)

(B) f(1) < f ′′(1) < f ′(1)

(C) f ′(1) < f(1) < f ′′(1)

(D) f ′′(1) < f(1) < f ′(1)

(E) f ′′(1) < f ′(1) < f(1)

[Ans: D]

Second Derivative Test

Recall the First Derivative Test. The First Derivative Test tells me that:

• there is a local maximum if the first derivative changes from positive tonegative;

• there is a local maximum if the first derivative changes from negative topositive.

Note that, if the first derivative is changing from positive to negative throughzero, then it is decreasing. So, for a local maximum, if the second derivativeexists, it is negative. This makes sense, because if the second derivative isnegative, the function is concave down. If the function is concave down, it hasa local maximum at the place where f ′ is zero. Similar arguments can be madefor local minima.

Theorem 4.2 (Second Derivative Test, Graphically). Based on f ′:

• If f ′ is decreasing through zero at x = c, then f(c) is a local maximum,since the graph of f is concave up with a horizontal tangent.

• If f ′ is increasing through zero at x = c, then f(c) is a local minimum,since the graph of f is concave down with a horizontal tangent.

Or, based on f ′′:

• If f ′(c) = 0 and if f ′′(c) < 0, then f(c) is a local maximum;

• If f ′(c) = 0 and if f ′′(c) > 0, then f(c) is a local maximum;



Figure 4.13: BC Acorn ’02

• If f ′(c) = 0 and if f ′′(c) is 0 or undefined, then the Second Derivative Testfails to determine whether f(c) is a local extremum.

Note: we only use the Second Derivative Test for f ′(c) = 0. Why would we notuse the Second Derivative Test on those critical points where f ′(c) is undefined?

4.2.2 Points of Inflection

An inflection point is a point on the graph of f where the concavity changes.At inflection points on the graph of f ,

• f changes concavity;

• f ′ changes direction, i.e., extrema of f ′ yield points of inflection of f ;

• f ′′ changes sign, i.e., if f ′′ crosses the x-axis, then f has a point of inflec-tion.

For inflection points, the tangent line will overestimate the curve on one side ofthe point, and underestimate it on the other side.

Be careful: for there to be an inflection point, it is not enough for the secondderivative to be zero. The second derivative must change sign. This is similarto finding relative extrema: it is not enough for the first derivative to be 0, thefirst derivative must change sign for there to be a relative extremum.

Example 4.2.2 (BC Acorn ’02) Let f be a function whose domainis the open interval (1, 5). Figure 4.13 shows the graph of f ′′. Countthe number of extrema of f ′ and points of inflection of the graph off ′.



[Ans: one rel. min, two p.i.]

Example 4.2.3 (adapted from AB98) If f ′′(x) = x (x− 2) (x+ 1)2,then the graph of f has inflection points at what value(s) of x?

[Ans: 0 and 2 only]

Example 4.2.4 (adapted from BC98) If f is the function definedby f(x) = 3x5 +10x4 +10x3−60x+7, what are all the x-coordinatesof points of inflection for the graph of f?

[Ans: 0]

Example 4.2.5 [3] Which of the following statements are trueabout the function f if its derivative f ′ is defined by f ′(x) = x (x− a)3,a > 0.

I. The graph of f is increasing at x = 2a.

II. The function f has a local maximum at x = 0.

III. The graph of f has an inflection point at x = a.


Problems

4.B-1 (BC97) The function f is defined on the closed interval [0, 8]. The graphof its derivative f ′ is shown in Figure 4.14. How many points of inflectiondoes the graph of f have? [Ans: Six]

4.B-2 (adapted from AB98) What is the x-coordinate of the point of inflection

on the graph of y =13x3 − 5x2 − 13? [Ans: 5]

4.B-3 (adapted from AB98) If f ′′(x) = x (x+ 2) (x− 1)2, then the graph of fhas inflection points when x = [Ans: 0 and −2 only]

4.B-4 [3] Suppose a function f is defined so that it has derivatives f ′(x) =x2 (1− x) and f ′′(x) = x (2− 3x). Over which interval is the graph of fboth increasing and concave up?

[Ans: 0 < x < 2

3



Figure 4.14: BC ’97 The function f is defined on the closed interval [0, 8]. Thegraph of its derivative f ′ is shown.

4.B-5 [3] Which of the following are true about the function f if its derivative isdefined by

f ′(x) = (x− 1)2 (4− x) ?

I. f is decreasing for all x < 4.

II. f has a local maximum at x = 1.

III. f is concave up for all 1 < x < 3.

[Ans: III only]

4.B-6 (adapted from AB97) The graph of y = 3x4 − 8x3 − 24x2 + 16 is concavedown for what values of x?

[Ans: − 2

3 < x < 2]




Unit 5

Calculus of TrigonometricFunctions

1. Implicit Differentiation and Related Rates

2. Differentiation of Sine, Cosine

3. Antidifferentiation of Trigonometrics

4. Derivatives of Inverse Functions

Advanced Placement

I. Derivatives


• Equations involving derivatives. Verbal descriptions are translated intoequations involving derivatives and vice versa.


• Geometric interpretation of differential equations via slope fields and therelationship between slope fields and derivatives of implicitly defined func-tions.


• Knowledge of derivatives of basic functions, including exponential, loga-rithmic, and trigonometric functions.

141

142 AP Unit 5 (Trigonometrics)

II. Integrals



• Antiderivatives by substitution of variables.


• Finding specific antiderivatives using initial conditions, including applica-tions to motion along a line.

• Solving separable differential equations and using them in modeling. Inparticular, studying the equation y′ = ky and exponential growth.


AP Unit 5, Day 1: Differentiation of Trigonometric Functions 143

5.1 Differentiation of Trigonometric Functions

Advanced Placement


• Knowledge of derivatives of basic functions, including trigonometric functions.

5.1.1 Special Limits

Theorem 5.1 (Squeeze Theorem). Let f , g, and h be functions satisfyingf(x) ≤ g(x) ≤ h(x) for all x near c, except possibly at c. If lim

x→cf(x) =

limx→c

h(x) = L, then limx→c

g(x) = L.

Example 5.1.1 Show that limx→0

x2 sin1x

= 0. [20]

Example 5.1.2 Find f ′(0) for f(x) = sinx using the definition ofthe derivative.

(a) Show that f ′(0) can be written as limh→0

sinhh

.

(b) Prove limh→0+

sinhh

= 1 for positive angles h along the unit circle.

(c) Find the limit of the average rate of change using a table.

(d) Use a symmetric difference quotient using ∆x = 0.001 on bothsides of x = 0 to find f ′(0).

(e) Use nDeriv to show find f ′(0).

Example 5.1.3 Find g′(0) for g(x) = cosx using the definition ofthe derivative.

(a) Show that g′(0) can be written as limh→0

cosh− 1h

.

(b) Prove limh→0

cosh− 1h

= 0 using limh→0

sinhh

= 1.

(c) Find the limit of the average rate of change using a table.

(d) Use a symmetric difference quotient using ∆x = 0.001 on bothsides of x = 0 to find g′(0).

(e) Use nDeriv to show find g′(0).



5.1.2 Trigonometric Derivatives

Derivative of Sine

Example 5.1.4 Find the derivative of f(x) = sinx:

(a) If f(x) = sinx, sketch f ′(x).

(b) Make a conjecture about whatd

dxsinx is.

(c) Prove or disprove your conjecture:

(a) Start with the definition of the derivative as the limit of anaverage rate of change.

(b) Use the sum of angle formula for sine: sin (α+ β) = sinα cosβ+cosα sinβ.

(c) Use the known limits for sinhh and 1−cosh

h as h approaches0.

(d) Give graphic evidence that the derivative is the same as yourconjecture by comparing the graphs of nDeriv(Y1, X, X) andyour proposed derivative using the happy bouncing ball.

d

dxsinx = cosx

Using the Chain Rule

d

dxsin (g(x)) = cos (g(x)) g′(x)

Derivative of Cosine

Example 5.1.5 Graph the derivative of cosx. Make and prove aconjecture for the derivative of cosx.

d

dxcosx = − sinx

Using the Chain Rule

d

dxcos (g(x)) = − sin (g(x)) g′(x)



Derivative of Tangent

Example 5.1.6 Findd

dxtanx.

d

dxtanx = sec2 x

Using the Chain Rule:

d

dxtan (g(x)) = sec2 (g(x)) g′(x)

Derivative of Secant

Example 5.1.7 Findd

dxsecx.

Use the Quotient Rule:

d

dxsecx = secx tanx

Using the Chain Rule:

d

dxsec (g(x)) = sec (g(x)) tan (g(x)) g′(x)

Example 5.1.8 (adapted from AB93) A particle moves along aline so that at time t, where 0 ≤ t ≤ π, its position is given by

s(t) = −3 cos t− t2

2+ 10. What is the velocity when its acceleration

is zero?

[Ans: 1.597]

Example 5.1.9 Let f(x) = sin(2x). Find the intersection of thetwo lines tangent to the graph of f at x =

π

6and at x =

π

3.



[Ans:

(π4 ,

π12 +

√3

2

)≈ (0.785, 1.128)

]

Example 5.1.10 (adapted from AB97)d

dxcos3(x2) =

[Ans: −6x cos2(x2) sin(x2)

]Example 5.1.11 (BC97) The position of an object attached to a

spring is given by y(t) =16

cos(5t) − 14

sin(5t), where t is time inseconds. In the first 3 seconds, how many times is the velocity ofthe object equal to 0? [You may use a calculator.]

[Ans: Five]

Example 5.1.12 (adapted from [2]) Administrators at MassachusettsGeneral Hospital believe that the hospital’s expenditures E(B), mea-sured in dollars, are a function of how many beds B are in use with

E(B) = 7000 + (B + 1)2.

On the other hand, the number of beds B is a function of time, t,measured in days, and it is estimated that

B(t) = 25 sin(t

10

)+ 50.

At what rate are the expenditures decreasing when t = 100?[Ans: $157/day]

Checking Antiderivatives

Example 5.1.13 (adapted from AB97) Which of the following areantiderivatives of f(x) = sinx cosx?

I. F (x) = − sin2 x

2

II. F (x) = −cos2 x

2

III. F (x) = −cos(2x)4




Problems

5.A-1 (adapted from AB acorn ’02) What is limh→0

cos(

3π2

+ h

)− cos

(3π2

)h

?

[Ans: 1]

5.A-2 (adapted from AB acorn ’02) If f(x) = sin2(π

4− x)

, then f ′(0) =

[Ans: −1]

5.A-3 (adapted from AB98) Find equation of the line tangent to the graph ofy = 2x + cos(πx) at the point (1, 1). Graph the function and its tangentline on your calculator. Find the equation of the line tangent to the graphof y = 2x+ cos(πx) at the point (−1,−3). [Ans: y = 2x− 1]

5.A-4 (adapted from AB93) If f(x) = (x− 1)2 cosx, then f ′(0) = [Ans: −2]

5.A-5 (adapted from AB98) If f(x) = tan(2x), then f ′(π

8

)= [Ans: 4]

5.A-6 (AB93) The fundamental period of 2 cos(3x) is[Ans: 2π

3

]5.A-7 (adapted from [2]) For x 6= 0, the slope of the tangent to y = x sinx equals

zero whenever

(A) tanx = −x

(B) tanx =1x

(C) tanx = x

(D) sinx = x

(E) cosx = x

[Ans: tanx = −x]

5.A-8 (adapted from [2]) If y = 2 sinx cosx =, then y′ = [Ans: 2 cos(2x)]

5.A-9 (adapted from [2]) For f(x) = cos2 x and g(x) = −0.5x2 on the interval[−π

2,π

2

], the instantaneous rate of change of f is greater than the instan-

taneous rate of change of g for which value of x? [Note: you should use acalculator.]

(A) −1.5

(B) −1.2

(C) −0.8

(D) 0

(E) 0.9



[Ans: −0.8]

5.A-10 (adapted from [2]) limh→0

(sin (x+ h)− sinx

h

)= [Ans: cosx]

5.A-11 (adapted from [2]) If g(x) = x2 − sinx, then limh→0

g(x+ h)− g(x)h

=

[Ans: 2x− cosx]

5.A-12 (adapted from [2]) At how many points on the interval −2π ≤ x ≤ 2πdoes the tangent to the graph of the curve y = x sinx have slope

π

2?

[Ans: Three]

5.A-13 (adapted from [2]) If y = cos3(3x), thendy

dx=[Ans: −9 cos2(3x) sin(3x)

]5.A-14 (from Stewart [20]) Let f(x) = sinx.

(a) Find values of the constants a and b so that the linear function L(x) =a+ bx has the properties L(0) = f(0) and L′(0) = f ′(0).

(b) Find the value of the constant c for which the quadratic functionQ(x) = L(x) + cx2 has the properties Q(0) = f(0), Q′(0) = f ′(0),and Q′′(0) = f ′′(0).

(c) Find the value of the constant d for which the cubic function C(x) =Q(x) + dx3 has the properties C(0) = f(0), C ′(0) = f ′(0), C ′′(0) =f ′′(0), and C ′′′(0) = f ′′′(0).

(d) Plot f , L, Q, and C on the same axes over the interval [−4, 4].[Ans: L(x) = x = Q(x); C(x) = x− 1

6x3]

5.A-15 Repeat the previous problem for f(x) = cosx[Ans: L(x): Q(x) = 1− 1

2x2 = C(x)

]


AP Unit 5, Day 2: Implicit Differentiation 149

5.2 Implicit Differentiation

Advanced Placement


• Chain rule and implicit differentiation.

Textbook §2.8 Implicit Differentiation and Inverse Trigonometric Functions[16]

Resources Exploration 4-8: “Implicit Relation Derivatives” in [9]

5.2.1 Implicit Differentiation

Example 5.2.1 For x = tan (f(x)), find f ′(x) in terms of x only.Note that your answer is d

dx arctanx.

Example 5.2.2 Write an equation of the line tangent to the graphof x2 + y2 = 16 at the point

(2√

3, 2)

[Ans: y − 2 = −

√3(x− 2

√3)]

Implicit Differentiation is Latin for “Remember the Freakin’ Chain Rule”.

Remember: anytime you differentiate a y-term with respect to x, you must

multiply bydy

dx.

d

dxO(y) = O′(y) · dy

dx

Example 5.2.3 Identify the variable, then Remember the Freakin’Chain Rule:

(a)d

dxx2

(b)d

dyy2

(c)d

dx[f(x)]2



(d)d

dxy2

(e)d

dtx2

Example 5.2.4 Remember the Product Rule:

(a)d

dx(xy)

(b)d

dx(−4xy)

Example 5.2.5 For x2 + y2 = 16:

(a) For what coordinate pairs is the graph increasing?

(b) For what coordinate pairs does the curve have a vertical tan-gent?

(c) Remember the Quotient Rule: For what values of x or y isd2ydx2 > 0 ?

Example 5.2.6 Do Foerster’s Exploration 4-8 in your mighty,mighty groups of four.

Example 5.2.7 Second Derivative Problems (adapted fromBC98)

(a) Ifdy

dx=√

1− y2, what isd2y

dx2? [Ans: −y]

(b) Ifdy

dx= −

√1− y2, what is

d2y

dx2? [Ans: −y]

(c) Name two functions for which f ′′(x) = −f(x). Is it true for

those functions thatdy

dx=√

1− y2 anddy

dx= −

√1− y2?


dxfor y = cos (xy)

The derivative at a specific point

You may plug in the values for x and y at any point after differentiating. If you

only need a slope at a specific value, and don’t need a formula fordy

dx, it may



behoove you to plug in values right after differentiation. It’s a lot easier to dealwith 12, 6, and −24 than with 8y, 3x2y, and 4xy.

Note that sometimes, you must find one of the coordinate pairs from the other.

The explicit option

Some equations can be solved explicitly. Just solve for y, and take the derivativeper usual.

Example 5.2.9 Solve x2 + y2 = 16 explicitly for y, then finddy

dx, and show that the answer is the same as we get using implicit

differentiation.

Problems

5.B-1 (from [2]) Consider the curve x + xy + 2y2 = 6. The slope of the linetangent to the curve at the point (2, 1) is

[Ans: − 1

3

]5.B-2 (from [2]) If x2 +2xy−3y = 3, then the value of

dy

dxat x = 2 is [Ans: −2]

5.B-3 (from [2]) If y2 − 3x = 7, thend2y

dx2=

[Ans: − 9

4y3

]5.B-4 (from [2]) If tan (x+ y) = x, then

dy

dx=

[Ans: − sin2 (x+ y)

]5.B-5 (from [2]) If y is a differentiable function of x, then the slope of the tangent

to the curve xy − 2y + 4y2 = 6 at the point where y = 1 is[Ans: − 1

10

]5.B-6 (adapted from AB ’93) If 2x3 + 2xy+ 4y3 = 17, then in terms of x and y,

dy

dx=

[Ans: − 3x2+y

x+6y2

]5.B-7 (adapted from AB ’97) If x2 + y2 = 10, what is the value of

d2y

dx2at the

point (1,−3)?[Ans: 10

27

]5.B-8 If x2 − y2 = 25, for what coordinate pairs will the curve have vertical

tangents? [Ans: (5, 0), (−5, 0)]

5.B-9 If y2 − x2 = 25, for what coordinate pairs will the curve have verticaltangents? [Ans: none]



5.B-10 (adapted from AB ’98) If x2+xy = −10, then when x = 2,dy

dx=[Ans: 3

2

]5.B-11 (adapted from BC ’97) If 2y = xy + x2 + 1, then when x = 1,

dy

dx=

[Ans: 4]

5.B-12 (adapted from BC ’98) The slope of the line tangent to the curve y4 +(xy + 1)3 = 0 at (2,−1) is

[Ans: 3

2

]5.B-13 (adapted from AB Acorn ’00) What is the slope of the tangent to the

curve y3x2 + y2x = 6 at (2, 1)?[Ans: − 5

16

]5.B-14 (adapted from BC Acorn ’00) If x = y+cos (xy), what is

dy

dx?[Ans: 1+y sin(xy)

1−x sin(xy)

]5.B-15 (HL 5/02) A curve has equation xy3 + 2x2y = 3. Find the equation of the

tangent to this curve at the point (1, 1). [Ans: y − 1 = − (x− 1)]

5.B-16 (HL 5/03) A curve has equation x3y2 = 8. Find the equation of the normalto the curve at the point (2, 1). [Note: the normal sticks out of the curvelike hairs, and is perpendicular to the tangent.]

[Ans: y − 1 = 4

3 (x− 2)]


AP Unit 5, Day 3: Inverse Functions 153

5.3 Inverse Functions

Advanced Placement


• Use of implicit differentiation to find the derivative of an inverse function.


• Knowledge of derivative of basic functions, including inverse trigonometricfunctions.

• Chain rule and implicit differentiation.

5.3.1 Inverse Functions

To get an inverse relation, switch the x’s and y’s, then solve for y.

Example 5.3.1 (adapted from D&S) Let f(x) = x3−9x2+31x−39and let g be the inverse of f . What is the value of g′(0)?

[Ans: 1

4

]If you like, you can use

d

dxf−1(x) =

1f ′ (f−1(x))

Example 5.3.2 (adapted from AB ’07) The functions f and g aredifferentiable for all real numbers, and g is strictly increasing. Table5.1 gives values of the functions and their first derivatives at selectedvalues of x.

(a) If g−1 is the inverse function of g, write an equation for the linetangent to the graph of y = g−1(x) at

(a) x = 3(b) x = 4(c) x = 5



Table 5.1: From AB 2007 Exam

x f(x) g(x) f ′(x) g′(x)1 6 4 2 52 9 2 3 13 10 −4 4 24 −1 3 6 7

[Ans: y − 2 = 1 (x− 3); y − 3 = 1

2 (x− 4);]

(b) What is the problem with asking for an equation for the linetangent to the graph of y = f−1(x) at x = 9?

[Ans: the inverse of f ain’t a function]

In order for a function f to have an inverse function f−1, i.e., an inverse relationthat is also a function, then f−1 must pass the vertical line test, and f mustpass a horizontal line test. One way to ensure that f passes a horizontal linetest is if f is monotonically increasing or monotonically decreasing.

The inverse relation won’t always be a function. Sometimes we must take se-lective pieces of the graph to make sure that the inverse is a function.

• For example, with√x, we use only the top half of the parabola.

• For arcsinx, we use only a portion of the graph that will give a full spec-trum of sines, i.e., from −1 to 1. We use that portion of the graph in thefourth and first quadrants where the cosine is always positive. The rangeof arcsinx is

[−π

2,π

2

].

• For arccosx, we use a portion of the graph that gives a full spectrum ofcosine values, so the domain is x ∈ [−1, 1]. The range is the angles inquadrants I and II where the sine is positive.

• The domain of arctanx is all real numbers, but the range is limited to theopen interval

(−π

2,π

2

).

• If f(x) passes a horizontal line test, then its inverse relation passes avertical line test, and is therefore a function.

Example 5.3.3 Find√

(−3)2


AP Unit 5, Day 3: Inverse Functions 155

Example 5.3.4

(a) Find arcsin√

32

(b) Find arccos 1√2

(c) Find arctan−1

(d) Find arcsin(sin 4π

3

)(e) Graph y = arcsin (sinx)

Example 5.3.5 Use a right triangle to find cos(arcsin 3

5

)Example 5.3.6 Use a right triangle to find sec (arctan 2x)

Example 5.3.7 For x = cos y, finddy

dxin terms of x only. Note

that your answer is ddx arccosx.

5.3.2 Differentiating Inverse Functions

Example 5.3.8 Find an inverse function, g(x) = f−1(x), for f(x) =x3, and find g′(x) two ways.

Example 5.3.9 Find the derivatives of

(a) arcsinx

(b) arccosx

(c) arctanx

d

dxarcsinx =

1√1− x2

d

dxarccosx = − 1√

1− x2

d

dxarctanx =

11 + x2


dxarcsin x

5




dx

(x√

1− x2 + arcsinx)

[Ans: 2

√1− x2

]

Problems

5.C-1 For x = tan f(x), find f ′(x) in terms of x only. Use right triangles, withOH = x

1 .[Ans: 1

1+x2

]5.C-2 For x = sin y, find

dy

dxin terms of x only. Remember that sin2 y+cos2 y =

1, or use right triangles, with OA = x

1 . Repeat for x = cos y, so AH = x

1 .[Ans: 1√

1−x2 ;− 1√1−x2

]5.C-3 (adapted from [2]) An equation for a tangent to the graph of y = arctan

x

3at the point

(3,π

4

)is:

[Ans: y − π

4 = 16 (x− 3)

]5.C-4 [2] If g(x) = 3

√x− 1 and f is the inverse function of g, then f ′(x) =[

Ans: 3x2]

5.C-5 (adapted from [2])d

dx[arctan 2x] =

[Ans: 2

1+4x2

]5.C-6 (adapted from AB ’07) The functions f and g are differentiable for all

real numbers, and g is strictly increasing. Table 5.1 gives values of thefunctions and their first derivatives at selected values of x. If g−1 is theinverse function of g, write an equation for the line tangent to the graphof y = g−1(x) at x = 6.

[Ans: y − 4 = 1

7 (x− 6)]

5.C-7 (adapted from D&S)d

dxarcsin

4x3

=[Ans: 4

9−16x2

]5.C-8 (adapted from D&S) If f(x) = x−

14 , what is the derivative of the inverse

of f(x)?[Ans: − 4

x5

]


AP Unit 5, Day 4: Related Rates (Triangles) 157

5.4 Related Rates (Triangles)

Advanced Placement


• Equations involving derivatives. Verbal descriptions are translated into equa-tions involving derivatives and vice versa.


• Modeling rates of change, including related rates problems.

Textbook §3.8 Related Rates [16]

5.4.1 Introduction to Related Rates

Example 5.4.1 (adapted from Acorn BC ’00) A point (x, y) ismoving along a curve y = f(x). At the instant when the slope of

the curve is −15

, the x-coordinate of the point is increasing at therate of 3 units per second. The rate of change, in units per second,of the y-coordinate of the point is

[Ans: − 3

5

]Example 5.4.2 (BC ’98) When x = 8, the rate at which 3

√x is

increasing is1k

times the rate at which x is increasing. What is thevalue of k?

[Ans: 12]

Example 5.4.3 [2] If xy2 = 20, and x is decreasing at the rate of3 units per second, the rate at which y is changing when y = 2 isapproximately

[Ans: 0.6 units/sec]



5.4.2 Related Rates w/ Triangles

Example 5.4.4 (adapted from AB ’93) The top of a 13-foot ladderis sliding down a vertical wall at a constant rate of 2 feet per minute.When the top of the ladder is 5 feet from the ground, what is therate of change of the distance between the bottom of the ladder andthe wall?

[Ans: 5

6 feet per minute]

Example 5.4.5 (AB 2002 Form B) Ship A is traveling due west to-ward Lighthouse Rock at a speed of 15 kilometers per hour (km/hr).Ship B is traveling due north away from Lighthouse Rock at a speedof 10 km/hr. Let x be the distance between Ship A and LighthouseRock at time t, and let y be the distance between Ship B and Light-house Rock at time t, as shown in Figure 5.1.

Figure 5.1: From 2002 AP Calculus AB Exam

(a) Find the distance, in kilometers, between Ship A and Ship Bwhen x = 4 km and y = 3 km.

(b) Find the rate of change, in km/hr, of the distance between thetwo ships when x = 4 km and y = 3 km.

(c) Let θ be the angle shown in Figure 5.1. Find the rate of changeof θ, in radians per hour, when x = 4 km and y = 3 km.


AP Unit 5, Day 4: Related Rates (Triangles) 159

Problems

5.D-1 [2] One ship traveling west is W (t) nautical miles west of a lighthouse and

Figure 5.2: From [2]

a second ship traveling south is S(t) nautical miles south of the lighthouseat time t (hours). The graphs of W and S are shown in Figure 5.2. Atwhat approximate time is the distance between the ships increasing att = 1? (nautical miles per hour = knots) [Ans: 4 knots]

5.D-2 (adapted from AB ’97) A railroad track and a road cross at right angles.An observer stands on the road 70 meters south of the crossing and watchesan eastbound train traveling at 80 meters per second. At how many metersper second is the train moving away from the observer 3 seconds after itpasses through the intersection? [Ans: 76.8]

5.D-3 (from HL 11/03) An airplane is flying at a constant speed at a constantaltitude of 3 km in a straight line that will take it directly over an observerat ground level. At at given instant the observer notes that the angle θ is13π radians and is increasing at

160

radians per second. Find the speed, inkilometers per hour, at which the airplane is moving towards the observer.[

Ans: 115 km/s = 240 km/hr

]Figure 5.3: Airplane flying towards an observer.



5.D-4 Figure 5.4 shows an isosceles triangle ABC with AB = 10 cm and AC =BC. The vertex C is moving in a direction perpendicular to (AB) with

Figure 5.4: Growing isosceles triangle.

speed 2 cm per second. Calculate the rate of increase of the angle CABat the moment the triangle is equilateral.

[Ans: 1

10 radians per second]

5.D-5 (adapted from D&S) Two cars start at the same place and at the sametime. One car travels east at a constant velocity of 40 miles per hour anda second car travels north at a constant velocity of 48 miles per hour.Approximately how fast is the distance between them changing after halfan hour? Round your answer to the nearest mile per hour. [Ans: 62 mph]

5.D-6 (adapted from D&S) A missile rises vertically from a point on the ground65, 000 feet from a radar station. If the missile is rising at the rate of17, 500 feet per minute at the instant when it is 38, 000 feet high, what isthe rate of change, in radians per minute, of the missile’s angle of elevationfrom the radar station at this instant? [Ans: .201 radians per minute]


AP Unit 5, Day 5: Antidifferentiating Trig 161

5.5 Antidifferentiating Trig

Advanced Placement

Techniques of antidifferentiation



Textbook §2.8 Implicit Differentiation and Inverse Trigonometric Functions[16]

5.5.1 Antidifferentiating to Inverse Functions

∫1√

1− x2dx = arcsinx+ C

Example 5.5.1 Antidifferentiate:∫1√

1− 9x2dx

[Ans: 1

3 arcsin 3x+ C]

∫1

1 + x2dx = arctanx+ C

Example 5.5.2 Antidifferentiate:∫1√

x (1 + x)dx

[Ans: 2 arctan√x+ C]



5.5.2 Antidifferentiation of Trigonometric Functions

Antiderivative of Sine

Example 5.5.3 Use a program to draw a slope field ofdy

dx= sinx.

Conjecture the antiderivative of sine. Prove your conjecture usingu-substitution.

∫sinx dx = − cosx+ C

u-substitution gives:∫sin (g(x)) g′(x) dx = − cos (g(x)) + C

Antiderivative of Cosine

Example 5.5.4 Use a program to draw a slope field ofdy

dx= cosx.

Conjecture the antiderivative of cosine.

∫cosx dx = sinx+ C

u-substitution gives: ∫cos (g(x)) g′(x) dx = sin (g(x)) + C

Other Trigonometric Antiderivatives

∫sec2 x dx = tanx+ C

∫secx tanx dx = secx+ C


AP Unit 5, Day 5: Antidifferentiating Trig 163

Example 5.5.5 (adapted from AB97) At time t ≥ 0, the accel-eration of a particle moving on the x-axis is a(t) = t + sin t. Att = 0, the velocity of the particle is −3. For what value of t will thevelocity of the particle be zero?

[Ans: 1.855]

Be careful! C 6= −3.

Problems

5.E-1 Find∫

3x2

√1− x6

dx using the u-substitution u = x3.[Ans: arcsinx3 + C

]5.E-2 (AB93) If the second derivative of f is given by f ′′(x) = 2x− cosx, which

of the following could be f(x)?

(A)x3

3+ cosx− x+ 1

(B)x3

3− cosx− x+ 1

(C) x3 + cosx− x+ 1

(D) x2 − sinx+ 1

(E) x2 + sinx+ 1 [Ans: x3

3 + cosx− x+ 1]

5.E-3 (adapted from [2]) A particle moves along the x-axis with velocity at timet given by: v(t) = t+ 2 sin t. If the particle is at the origin when t = 0, itsposition at the time when v = 5 is x = [Ans: 17.277]




Unit 6

Calculus of ExponentialFunctions

1. Derivatives Involving e

2. Anti-derivatives Involving e

3. Separable Differential Equations

Advanced Placement

I. Derivatives


• Equations involving derivatives. Verbal descriptions are translated intoequations involving derivatives and vice versa.


• Knowledge of derivatives of basic functions, including exponential andlogarithmic, and trigonometric functions.

II. Integrals




165

166 AP Unit 6 (Exponentials)


• Finding specific antiderivatives using initial conditions, including applica-tions to motion along a line.

• Solving separable differential equations and using them in modeling. Inparticular, studying the equation y′ = ky and exponential growth.


AP Unit 6, Day 1: Antidifferentiation by Simplification 167

6.1 Antidifferentiation by Simplification

Advanced Placement




Textbook §4.6 Integration by Substitution [16]

6.1.1 u-Simplification

Example 6.1.1 ∫6x2

(x3 + 1

)dx

[Ans:

(x3 + 1

)2 + C]

Example 6.1.2 ∫3x2

2√x3 + 1

dx

[Ans:

√x3 + 1 + C

]Example 6.1.3 (adapted from AB93)∫

4x3

√x4 − 3

dx =

[Ans: 2

√x4 − 3 + C

]Completing the Derivative of the Inside

Example 6.1.4 ∫x2(x3 + 1

)2dx =



[Ans: 1

9

(x3 + 1

)3 + C]

Example 6.1.5 ∫x2

(x3 + 1)2 dx

[Ans: − 1

3(x3+1)

]

Example 6.1.6 (adapted from [2])∫x(x2 − 1

)4dx =

[Ans: 1

15

(x3 + 1

)5 + C]

Example 6.1.7 [11] ∫r2 − 1

(r3 − 3r + 3)2 dr

Example 6.1.8 [11] ∫1x2

5

√1x

+ 5 dx

6.1.2 Simplification with Trigonometrics Inside

Example 6.1.9 Solve∫

cosx sinx dx two different ways, andthen show that the solutions are really the same. Do the same for∫

sec2 x tanx dx.

Example 6.1.10 (adapted from Stewart [20])∫

sec4 x tanx dx

[Ans: sec4 x

4 + C]



Example 6.1.11 [18] ∫sinx√

1 + cosxdx

[Ans: −2

√1 + cosx+ C

]

6.1.3 Simplification with Trigonometrics Outside

Example 6.1.12 (adapted from [20]) Find∫x2 cos

(2x3)dx.

[Ans: 1

6 sin(2x3)

+ C]

Example 6.1.13 (adapted from [18]) Calculate∫

sec√x tan

√x√

xdx

[Ans: 2 sec√x+ C]

Example 6.1.14 Calculate∫

x

cos2 x2dx

[Ans: 1

2 tanx2 + C]

Motivation for finding the happy function

Example 6.1.15 What happens when we try to find∫

tanx dx?

I need a function whose derivative is 1u .

Suppose I had a function h(x) (the happy function), for which h′(x) = h(x).

Then suppose I had a function I(x) = h−1(x), which was the inverse functionof h(x). Remember, to get an inverse function, I switch the x’s and y’s



So:

y = I(x)

x = h(y) I = h−1

d

dxx =

d

dxh(y) Differentiate both sides

1 = h′(y)dy

dxChain Rule, inside is y

1h′(y)

=dy

dx

1h(y)

=dy

dxh′ = h

1x

=dy

dxh(y) = x from before

So, if I can find a happy function, h(x), such that h′(x) = h(x), then the inverseof the happy function has a derivative 1

x , and I can finally find the antiderivativeof tanx. Now, if I only had a happy function...

Problems

6.A-1∫x

3√

1− x2 dx =[Ans: − 3

8

(1− x2

)4/3 + C]

6.A-2∫G′ (v(x)) v′(x) dx = [Ans: G (v(x)) + C]

6.A-3∫H ′(ax+ b) dx =

[Ans: 1

aH(ax+ b) + C]

6.A-4 [11]∫

sin7 (θ) cos (θ) dθ[Ans: sin8 θ

8 + C]

6.A-5 (adapted from [2]) Ifdy

dx= cos3 x sinx, then y =

[Ans: − 1

4 cos4 x+ C]

6.A-6 (adapted from [2])∫

6 cosx sin2 x dx =[Ans: 2 sin3 x+ C

]6.A-7 (adapted from [20])

∫tan3 θ sec2 θ dθ

[Ans: tan4 θ

4 +C]

6.A-8 [18] Find∫

cos4 x sinx dx.[Ans: − 1

5 cos5 x+ C]



6.A-9 [18] Find∫ √

1 + sinx cosx dx.[Ans: 2

3 (1 + sinx)3/2 + C]

6.A-10 [18] Find∫x sin3 x2 cosx2 dx.

[Ans: 1

8 sin4 x2 + C]

6.A-11 [18] Find∫

sec2 x√1 + tanx

dx.[Ans: 2

√1 + tanx+ C

]6.A-12 [18] Find

∫ (1 + tan2 x

)sec2 x dx.

[Ans: tanx+ 1

3 tan3 x+ C]

6.A-13 (adapted from [2])∫

cos (3x+ 2) dx =[Ans: 1

3 sin (3x+ 2) + C]

6.A-14 [20] Find∫x sin

(1− x2

)dx

[Ans: 1

2 cos(1− x2

)+ C

]6.A-15 [18] Find

∫ sin√x√

xdx [Ans: −2 cos

√x+ C]

6.A-16 [18] Find∫x sec2 x2 dx.

[Ans: 1

2 tanx2 + C]

6.A-17 [18] Find∫

cos (3x+ 1) dx.[Ans: 1

3 sin (3x+ 1) + C]

6.A-18 [18] Find∫

sin (3− 2x) dx.[Ans: 1

2 cos (3− 2x) + C]




AP Unit 6, Day 2: The Happy Function 173

6.2 The Happy Function

Advanced Placement


• Knowledge of derivatives of basic functions, including exponential and loga-rithmic functions.




Textbook §2.7 Derivatives of Exponential and Logarithmic Functions: “Deriva-tives of the Exponential Functions”; §4.1 Antiderivatives; §4.6 Integration bySubstitution [16]

6.2.1 Differentiating the Exponential Function

d

dxex = ex

Example 6.2.1 Findd

dxbx

d

dxbx =

d

dx

(eln b

)x=

d

dxex ln b

= ex ln b ln b

=(eln b

)xln b

= bx ln b

Applying the Chain Rule:

d

dxeg(x) = eg(x)g′(x)



Example 6.2.2 (adapted from AB acorn ’02) Two particles start atthe origin and move along the x-axis. For 0 ≤ t ≤ 10, their respectiveposition functions are given by x1 = cos t and x2 = e−2t − 1. Forhow many values of t do the particles have the same velocity?

[Ans: Four]

Example 6.2.3 (adapted from BC97) limh→0

eh − 13h

is

[Ans: 1

3

]Example 6.2.4 (adapted from AB97) Let f be the function givenby f(x) = 2e4x2

. For what value of x is the slope of the line tangentto the graph of f at (x, f(x)) equal to 5?

[Ans: 0.246]

Example 6.2.5 (adapted from BC93) If f(x) = etan3 x, then f ′(x) =

[Ans: 3 tan2 x sec2 x etan3 x

]Example 6.2.6 For ef(x) = x, find f ′(x) in terms of x only. Note

that your answer isd

dxlnx.

u-Substitution with exponentials

Example 6.2.7 Ifdy

dx=

e−x

(3 + 2e−x)2 , then y =

[Ans: 1

2(3+2e−x) + C]

Example 6.2.8 [18]∫e−x

[1 + cos

(e−x

)]dx

[Ans: −e−x − sin (e−x) + C]



6.2.2 Antidifferentiating the Exponential Function

Since the derivative of ex is ex the antiderivative of ex is ex.

∫ex dx = ex + C

Example 6.2.9 Find the antiderivative of∫

2x dx.

Change 2x to(eln 2

)x to ex ln 2. Let u = x ln 2. If the antiderivative of yourinside function is just a constant number (such as ln 2), you can skip the wholeprocess of u-substitution and just divide by that constant number.

∫bx dx =

bx

ln b+ C

Applying u-substitution to the general antiderivative of ex:∫eg(x)g′(x) dx = eg(x) + C

Example 6.2.10 (BC93) A particle moves along the x-axis so thatat any time t ≥ 0 the acceleration of the particle is a(t) = e−2t. If

at t = 0 the velocity of the particle is52

and its position is174

, then

its position at any time t > 0 is x(t) =

[Ans: e−2t

4 + 3t+ 4]

Example 6.2.11 (adapted from [2])∫ex

3 − x2

ex3 dx =

[Ans: x+

e−x3

3+ C

]



6.2.3 Skippable u-Simplification

If the inside function is linear, i.e.

• u = mx+ b;

• du

dx= m or k;

• the derivative of the inside is constant;

Then all that u-substitution will accomplish is to divide by the derivative of theinside.

Do not try this unless the derivative of the inside is a constant.

∫f ′(mx+ b) dx =

f(mx+ b)m

+ C

Example 6.2.12 Prove the above equation.

Example 6.2.13 Find∫

(4− 7x)3

2dx.

[Ans: − (4−7x)4

56 + C]


cos (3x+ 2) dx.

[Ans: sin(3x+2)

3 + C]


sec (πx) tan (πx) dx.

[Ans: 1

π sec (πx) + C]



Problems

6.B-1 (adapted from AB98) Let f be the function given by f(x) = 2e3x and letg be the function given by g(x) = 6x3. At what value of x do the graphsof f and g have parallel tangent lines? [Ans: −0.344]

6.B-2 (adapted from AB98) If f(x) = cos (e−x), then f ′(x) = [Ans: e−x sin (e−x)]

6.B-3 (adapted from BC97) If f(x) = (x− 1)32 +

ex−5

2, then f ′(5) =

[Ans: 7

2

]6.B-4 (adapted from AB97) If f(x) =

ex2

x2, then f ′(x) =

[Ans:

2ex2(x2−1)x3

]6.B-5 (adapted from AB93)

d

dx(3x) = [Ans: (3x) ln 3]

6.B-6 (adapted from [2]) The approximate value of y =√

3 + ex at x = 0.04,obtained from the tangent to the graph at x = 0, is [Ans: 2.01]

6.B-7 (adapted from [2]) Two particles move along the x-axis and their positionsat time 0 ≤ t ≤ 2π are given by x1 = sin 3t and x2 = e(t−3)/2 − 0.75.For how many values of t do the two particles have the same velocity?[Ans: Six]

6.B-8 (adapted from [2]) If y = ekx, thend6y

dx6=

[Ans: k6ekx

]6.B-9 [11]

∫ex cos (πex) dx

[Ans: 1

π sin (πex) + C]

6.B-10 [18] ∫ex√ex + 1

dx [Ans: 2

√ex + 1 + C

]6.B-11 (from [2]) Find

dy

dxfor ey = xy.

[Ans: y

xy−x

]6.B-12 (from [2]) If cosx = ey, 0 < x <

π

2, what is

dy

dxin terms of x?

[Ans: − tanx]

6.B-13 (from Stewart [20]) Let f(x) = ex.

(a) Find the value of the constants a, b, c, and d cubic function C(x) =a + bx + cx2 + dx3 has the properties C(0) = f(0), C ′(0) = f ′(0),C ′′(0) = f ′′(0), and C ′′′(0) = f ′′′(0).

(b) Plot f and C on the same axes over the interval [−4, 4].[Ans: C(x) = 1 + x+ 1

2x2 + 1

6x3]



6.B-14 For ey = x, finddy

dxin terms of x only.

[Ans: 1

x

]6.B-15 (adapted from AB97) 1

π

∫etπ dt =

[Ans: e

tπ + C

]6.B-16 (adapted from AB97) ∫

etan x

cos2 xdx =

[Ans: etan x + C]

6.B-17 (adapted from [2]) The acceleration of a particle at time t moving alongthe x-axis is given by: a = 9e3t. At the instant when t = 0, the particleis at the point x = 2 moving with velocity v = −2. The position of the

particle at t =13

is[Ans: e− 2

3

]6.B-18 (adapted from [2]) ∫

e3√x

3 3√x2

dx [Ans: e

3√x + C]

6.B-19 (adapted from [2]) The number of bacteria in a culture is growing at a rateof 1000e2t/3 per unit of time t. At t = 0, the number of bacteria presentwas 1500. Find the number present at t = 3.

[Ans: 1500e2

]6.B-20 [11]

∫ex cos (πex) dx

[Ans: 1

π sin (πex) + C]

6.B-21 (adapted from [18])∫

sinx ecos x dx [Ans: −ecos x + C]

6.B-22 [18] ∫e1/x

x2dx [

Ans: −e1/x + C]

6.B-23 Find∫

ex√1− e2x

dx using the u-substitution u = ex. [Ans: arcsin ex + C]


AP Unit 6, Day 3: Inverse of the Happy Function 179

6.3 Inverse of the Happy Function

Advanced Placement


• Knowledge of derivatives of basic functions, including exponential and loga-rithmic functions.




Textbook §2.7 Derivatives of Exponential and Logarithmic Functions: “Deriva-tive of the Natural Logarithm”; §4.1 Antiderivatives; §4.6 Integration by Sub-stitution [16]

6.3.1 Inverse of the Exponential Function

y = lnxey = x

eydy

dx= 1

dy

dx=

1ey

dy

dx=

1x

d

dxlnx =

1x

x > 0

Using the Chain Rule gives:

d

dxln (g(x)) =

1g(x)

g′(x)

d

dxln (g(x)) =

g′(x)g(x)

g(x) > 0



Example 6.3.1 Find the derivative:

d

dxln (−x) x < 0

Combiningd

dxlnx =

1x

x > 0

andd

dxln (−x) =

1x

x < 0

into one statement gives

d

dxln |x| = 1

xx 6= 0

.

Example 6.3.2 (adapted from AB97) If f(x) = ln∣∣3x2 − 1

∣∣, thenf ′(x) =

(A)∣∣∣∣ 6x3x2 − 1

∣∣∣∣(B)

6x|3x2 − 1|

(C)6 |x|

3x2 − 1

(D)6x

3x2 − 1

(E)1

3x2 − 1

[Ans: 6x

3x2−1

]

Example 6.3.3 Show that

d

dxlnxa =

d

dxa lnx


d

dx(ln a+ lnx) =

d

dxln (ax)




d

dxln ex =

d

dxx


d

dxeln x =

d

dxx

Example 6.3.7 (adapted from [2]) If y = cosu, u = 2v− 1v

+π

2−1,

and v = lnx, then the value ofdy

dxat x = e is

[Ans: − 3

e

]

Example 6.3.8 Show that y = ln(x2 + e

)is a solution to the

differential equationdy

dx= 2xe−y.

6.3.2 Implicit Differentiation with ln

The derivative at a specific point

Example 6.3.9 [3] The slope of the tangent to the graph of ln (x+ y) =x2 at the point where x = 1 is

[Ans: 2e− 1]

Note that sometimes, you must find one of the coordinate pairs from the other.

The explicit option

Example 6.3.10 Solve ln (x+ y) = x2 explicitly for y, then finddy

dxwhen x = 1.



6.3.3 Antidifferentiating Reciprocals

Why does the Anti-Power Rule not work for∫x−1 dx?

But, we recently found a function whose derivative is 1x .

∫1xdx = ln |x|+ C

Why is the antiderivative ln |x| and not just lnx?

The domain of1x

is x 6= 0, but the domain of lnx is only the real numbers

greater than 0. So,∫

1x dx = lnx only for x > 0. The problem is that x in 1

x canbe anything but zero, but lnx is impossible for those x’s below zero. I didn’thave this problem before, when I was differentiating lnx, because the domainof lnx is just those numbers above 0, and my derivative, 1

x can handle thosenumbers.

I need an antiderivative of1x

for x < 0, since lnx only works for x > 0.

So let’s take a look at the real numbers less than 0, For x > 0, we’ve already

seen thatd

dxlnx =

1x

. Now let’s look at x < 0:

= ln (−x)

dy

dx=

1−x

d

dx(−x)

dy

dx= − 1

x(−1)

dy

dx=

1x

So what we see is that1x

is the derivative of both lnx (for x > 0) and ln (−x)

(for x < 0).



d

dxlnx =

1x

(x > 0)

d

dxln (−x) =

1x

(x < 0)

Combining these into one statement for all x 6= 0 gives me

d

dxln |x| = 1

x

The typical expression that I’m use to seeing, ddx lnx = 1

x , is basically the samething, only your restricting yourself to positive x’s, i.e., those numbers for whichx = |x|

Since1x

is the derivative of ln |x| (for all x 6= 0, the antiderivative of1x

is ln |x|(plus some arbitrary constant).

∫dx

x= ln |x|+ C

u-substitution gives: ∫g′(x)g(x)

= ln |g(x)|+ C

Example 6.3.11 Find ∫tanx dx

[Ans: − ln |cosx|+ C]

Example 6.3.12 Use the substitution u = secx+ tanx to find∫secx dx

[Ans: ln |secx+ tanx|+ C]



Example 6.3.13 (adapted from [2]) Find a family of curves thatintersect at right angles every curve of the family y = 0.5x2 + k forevery real value of k?

[Ans: y = − ln |x|+ C]

6.3.4 Antidifferentiating Fractions

If the order on top is the same or higher than that on bottom, use division(long or synthetic) to reduce the problem to a polynomial + some remainderover the original denominator. The polynomial is easy to antidifferentiate usingthe anti-power rule, and the remainder over the original denominator should beeasier to antidifferentiate than the original fraction.

Example 6.3.14 (adapted from [2])∫x− 1x− 2

dx =

[Ans: x+ ln |x− 2|+ C]

Example 6.3.15 If the velocity of Runner B, in meters per second,

is given by v(t) =24t

2t+ 3, find an expression for the runner’s position

if her position at time t = 0 seconds is 0 meters.

[Ans: 12t− 18 ln |2t+ 3|+ 18 ln 3]

Problems

6.C-1 (BC97) limh→0

ln(e+ h)− 1h

is[Ans:

1e

]6.C-2 (adapted from BC93) If f(x) = ln

(e3x2

), then f ′(x) = [Ans: 6x]

6.C-3 (adapted from AB93) The slope of the line normal to the graph of y =3 ln (secx) at x =

π

4is [Note: normal lines are perpendicular to tangent

lines.][Ans: − 1

3



6.C-4 (adapted from [2]) If y = 2u+ 3eu and u = 1 + lnx, finddy

dxwhen x =

1e

.

[Ans: 5e]

6.C-5 (adapted from [2]) The formula x(t) = 2 ln t +t2

9+ 1 gives the position

of an object moving along the x-axis during the time interval 1 ≤ t ≤ 5.At the instant when the acceleration of the object is zero, what is thevelocity?

[Ans: 4

3

]6.C-6 (adapted from [2]) What is the slope of the line tangent to the graph of

y = ln 3√x at

(e3, 1

)?

[Ans: 1

3e3

]6.C-7 [3] There is a point between P (1, 0) and Q (e, 1) on the graph of y =

lnx such that the tangent to the graph at that point is parallel to theline through points P and Q. What is the x-coordinate of this point?[Ans: e− 1]

6.C-8 (adapted from AB acorn ’02) Which of the following are antiderivatives

ofln3 x

x?

I.ln4 x

4

II.ln4 x

4+ 6

III.3 lnx− ln3 x

x2

(A) I only

(B) III only

(C) I and II only

(D) I and III only

(E) II and III only


6.C-9 (adapted from AB98) Let F (x) be an antiderivative of(lnx)3

x. If F (1) = 2,

then F (9) = [Ans: 7.827]

6.C-10 (adapted from [3]) If eg(x) = 3x− 1, then g′(x) =[Ans: 3

3x−1

]6.C-11 (MM spec) The function f is given by

f(x) = 1− 2x1 + x2

By using an appropriate substitution, find∫f(x) dx.

[Ans: x− ln

(1 + x2

)+ C



6.C-12 (adapted from [2]) Find ∫sec2 x

tanxdx

[Ans: ln |tanx|+ C]

6.C-13 (adapted from [2]) A particle starts at (3, 0) when t = 0 and moves alongthe x-axis in such a way that at time t > 0 its velocity is given by v(t) =

11 + t

. Determine the position of the particle at t = 5. [Ans: 3 + ln 6]

6.C-14 The antiderivative of1

cabin[Ans: Evan’s ark]

6.C-15 (adapted from [2]) ∫12x2

1 + x3dx = [

Ans: 4 ln∣∣1 + x3

∣∣+ C]

6.C-16 (adapted from AB98) ∫x2 − 1x

dx = [Ans: x2

2 − ln |x|+ C]


AP Unit 6, Day 4: Separable Differential Equations 187

6.4 Separable Differential Equations

Advanced Placement





• Solving separable differential equations and using them in modeling.

Textbook §7.1 Modeling with Differential Equations; §7.2 Separable DifferentialEquations [16]

6.4.1 Separable Differential Equations

Definition 6.1. Differential Equation A differential equation is an equationthat contains the derivative of a function. [10]

Example 6.4.1 The solution to the differential equationdy

dx=

2xe−y, where y(0) = 1, is

[Ans: y = ln

(x2 + e

)]Example 6.4.2 (adapted from AB acorn ’02) The solution to the


dx=x2

y3where y(3) = 2, is

[Ans: y = 4

√43x

3 − 20]

Example 6.4.3 (adapted from [2]) If the graph of y = f(x) contains

the point (0, 1) and ifdy

dx=

2x sin(x2)

y, then f(x) =



[Ans:

√3− cos (x2)

]

6.4.2 Separable Differential Equations with Logs

Example 6.4.4 (AB97) Let v(t) be the velocity, in feet per second,of a skydiver at time t seconds, t ≥ 0. After her parachute opens,

her velocity satisfies the differential equationdv

dt= −2v − 32, with

initial condition v(0) = −50.

(a) Use separation of variables to find an expression for v in termsof t, where t is measured in seconds.

(b) Terminal velocity is defined as limt→∞

v(t). Find the terminalvelocity of the skydiver to the nearest foot per second.

(c) It is safe to land when her speed is 20 feet per second. At whattime t does she reach this speed?

[Ans: v = −34e−2t − 16, −16, 1.070

]Example 6.4.5 A turkey is cooking in the oven at 300 degreesFahrenheit. It starts out at room temperature (70 degrees). After1 hour, it is ? degrees. How long before it reaches 170 degrees, atwhich point it will be done. The rate of change in the temperature ofthe turkey is proportional to the difference between the temperaturesof the environment and the turkey.

Problems

6.D-1 (adapted from AB93) Ifdy

dx= 2y2 and if y = 1 when x = 2, then when

x = 3, y = [Ans: −1]

6.D-2 (adapted from AB acorn ’02) The solution to the differential equationdy

dx=x2


[Ans: y = 4

√43x

3 − 20]

6.D-3 (AB05B) Consider the differential equation given bydy

dx=−xy2

2. Let

y = f(x) be the particular solution to this differential equation with theinitial condition f(−1) = 2.

(a) On the axes provided (Figure 6.1), sketch a slope field for the givendifferential equation at the twelve points indicated.



Figure 6.1: 2005B AB Exam

(b) Write an equation for the line tangent to the graph of f at x = −1.

(c) Find the solution y = f(x) to the given differential equation with theinitial condition f(−1) = 2. [

Ans: y − 2 = 2 (x+ 1);y = 4x2+1

]6.D-4 (adapted from [2]) The point (1, 4) lies on the graph of an equation y =

f(x) for whichdy

dx= 6x2√y where x ≥ 0 and y ≥ 0. When x = 0 the

value of y is [Ans: 1]

6.D-5 (adapted from [2]) If the graph of y = f(x) is defined for all x ≥ 0,

contains the point (0, 4), hasdy

dx= 3√xy and f(x) > 0 for all x, then

f(x) =[Ans:

(x3/2 + 2

)2]6.D-6 (AB98) Let f be a function with f(1) = 4 such that for all points (x, y)

on the graph of f the slope is given by3x2 + 1

2y.

(a) Find the slope of the graph of f at the point where x = 1.[Ans: 1

2

](b) Write an equation for the line tangent to the graph of f at x = 1 and

use it to approximate f(1.2).[Ans: y − 4 = 1

2 (x− 1), 4.1]

(c) Find f(x) by solving the separable differential equationdy

dx=

3x2 + 12y

with the initial condition f(1) = 4.[Ans:

√x3 + x+ 14

](d) Use your solution to find f(1.2). [Ans: 4.114]



6.D-7 (adapted from [2]) Ifdy

dx= 2xy and if y = 3 when x = 0, then y =[

Ans: 3ex2̀]

6.D-8 (adapted from [2]) A solution of the equationdy

dx+ 2xy = 0 that contains

the point(0, e2

)is

[Ans: y = e2−x2

]


AP Unit 6, Day 5: Exponential Growth and Decay 191

6.5 Exponential Growth and Decay

Advanced Placement





• Solving separable differential equations and using them in modeling. In par-ticular, studying the equation y′ = ky and exponential growth.


Resources §7.2 Exponential Growth and Decay and §7.3 Other DifferentialEquations for Real-World Applications in [10]

6.5.1 Proportional Growth

Example 6.5.1 The rate growth of the population of Escherichiacoli is proportional to the number of E. coli. Find a general expres-sion for the population as a function of time if the initial populationis P0.

[Ans: P = P0e

kt]

Example 6.5.2 (adapted from AB98) Population y grows according

to the equationdy

dt= ky, where k is a constant and t is measured in

years. If the population doubles every 8 years, then the value of k is

[Ans: 0.087]



Example 6.5.3 [10] Chemical Reaction Problem Calculus buddite(a rare substance) is converted chemically into Glamis thanus. Bud-dite reacts in such a way that the rate of change in the amount leftunreacted is directly proportional to that amount.

(a) Write a differential equation that expresses this relationship.Solve it to find an equation that expresses amount in terms oftime. Use the initial conditions that the amount is 50 mg whent = 0 min and 30 mg when t = 20 min.

[Ans: dB

dt = kB, B = 50 (0.6)t/20 = 50e−0.025541...t]

(b) Sketch the graph of amount versus time.

(c) How much buddite remains an hour after the reaction starts?[Ans: 10.8 mg]

(d) When will the amount of buddite equal 0.007 mg? [Ans: 5 hr 47 min]

6.5.2 Other Applications of Differential Equations

Example 6.5.4 Tin Can Leakage Problem [10] Suppose you fill atall (topless) tin can with water, then punch a hole near the bottomwith an ice pick. The water leaks quickly at first, then more slowlyas the depth of the water increases. In engineering or physics, youwill learn that the rate at which the water leaks out is directly pro-portional to the square root of its depth. Suppose that at time t = 0min, the depth is 12 cm and dy

dt is −3 cm/min.

(a) Write a differential equation stating that the instantaneous rateof change of y with respect to t is directly proportional to thesquare root of y. Find the proportionality constant.

(b) Solve the differential equation to find y as a function of t. Usethe given information to find the particular solution. Whatkind of function is this?

(c) Plot the graph of y as a function of t. Sketch the graph. Con-sider the domain of t in which the function gives reasonableanswers.

(d) Solve algebraically for the time at which the can becomes empty.Compare your answer with the time it would take at the initialrate of −3 cm/min.

[Ans: k = −3

(12−1/2

); y = 3

16 t2 − 3t+ 12; ;8 (twice as long)

]Example 6.5.5 Dam Leakage Problem [10] A new dam is con-structed across Scorpion Gulch. Engineers want to predict the amount



of water in the lake behind the dam as a function of time. At t = 0days the water starts flowing in at a fixed rate F ft3/hr. Unfortu-nately, as the water level rises, some leaks out. The leakage rate,L, is directly proportional to the amount of water, W ft3, presentin the lake. Thus the instantaneous rate of change of W is equal toF − L.

(a) What does L equal in terms of W? Write a differential equationthat expresses dW/dt in terms of F , W , and t.

[Ans: dW

dt = F − kW]

(b) Solve for W in terms of t, using the initial condition W = 0when t = 0.

[Ans: W = F

k

(1− e−0.04t

)](c) Water is known to be flowing in at F = 5000 ft3/hr. Based on

geological considerations, the proportionality constant in theleakage equation is assumed to be 0.04/hr. Write the equationforW , substituting these quantities.

[Ans: W = 125000

(1− e−0.004t

)](d) Predict the amount of water after 10 hr, 20 hr, and 30 hr. After

these numbers of hours, how much water has flowed in and howmuch has leaked out? [Ans: L : 8790, 31166, 72649]

(e) When will the lake have 100000 ft3 of water? [Ans: bit more than 40 hr]

(f) Find the limit of W as t approaches infinity. State the realworld meaning of this number.

[Ans: 125000 ft3

](g) Draw the graph of W versus t. Clearly show an asymptote.

(h) The lake starts filling with water. The actual amount of waterat time t = 10 is exactly 40000 ft3. The flow rate is still 5000ft3/hr, as predicted. Use this information to find a more precisevalue of the leakage constant k. [Ans: k = 0.0464...]

Problems

6.E-1 (AB98) Ifdy

dt= ky and k is a nonzero constant, then y could be

(A) 2ekty

(B) 2ekt

(C) ekt + 3

(D) kty + 5

(E)12ky2 +

12 [

Ans: 2ekt]



6.E-2 (adapted from AB93) A puppy weighs 2.1 pounds at birth and 3.5 poundstwo months later. If the weight of the puppy during its first 6 monthsis increasing at a rate proportional to its weight, then how much willthe puppy weigh when it is 4 months old (to the nearest 0.1 pound)?[Ans: 5.8 pounds]

6.E-3 (adapted from [2]) If g′(x) = 3g(x) and g(−1) = 1, then g(x) is[Ans: e3x+3

]6.E-4 (adapted from [2]) The change in N , the number of bacteria in a culture

dish at time t is given by:dN

dt= 3N . If N = 4, when t = 0, the

approximate value of t when N = 1614 is [Ans: 2]

6.E-5 (AB ’96) The rate of consumption of cola in the United States is givenby S(t) = Cekt, where S is measured in billions of gallons per year and tis measured in years from the beginning of 1980. The consumption ratedoubles every 5 years and the consumption rate at the beginning of 1980was 6 billion gallons per year. Find C and k.

[Ans: C = 6, k = 1

5 ln 2]

6.E-6 [10] You run over a nail. As the air leaks out of your tire, the rate of changeof air pressure inside the tire is directly proportional to that pressure.

(a) Write a differential equation that states this fact. Evaluate the pro-portionality constant if the pressure was 35 psi and decreasing at 0.28psi/min at time zero.

[Ans: dP

dt = −0.008P]

(b) Solve the differential equation subject to the initial condition impliedin step (a).

[Ans: P = 35e−0.008t

](c) Sketch the graph of the function. Show its behavior a long time after

the tire is punctured.

(d) What will be the pressure at 10 min after the tire was punctured?[Ans: about 32.3 psi]

(e) The car is safe to drive as long as the tire pressure is 12 psi orgreater. For how long after the puncture will the car be safe todrive? [Ans: about 134 min]

6.E-7 (AB93) Let P (t) represent the number of wolves in a population at timet years, when t ≥ 0. The population P (t) is increasing at a rate directlyproportional to 800− P (t), where the constant of proportionality is k.

(a) If P (0) = 500, find P (t) in terms of t and k.[Ans: P (t) = 800− 300e−kt

](b) If P (2) = 700, find k.

[Ans: k = ln 3

2 ≈ 0.549]

(c) Find limt→∞

P (t) [Ans: 800]


Unit 7

Existence Theorems

1. Continuity

2. Extreme and Intermediate Value Theorems

3. Differentiability and Rolle’s Theorem

4. Average Rate of Change and Mean Value Theorem

5. Riemann Sums: Evaluating Definite Integrals

6. Fundamental Theorem of Calculus

7. Area

Advanced Placement

1. DerivativesConcept of the derivative.

• Relationship between differentiability and continuity.


• The Mean Value Theorem and its geometric consequences.

2. IntegralsFundamental Theorem of Calculus.

• Use of the Fundamental Theorem to evaluate definite integrals.

• Use of the Fundamental Theorem to represent a particular antiderivative,and the analytical and graphical analysis of functions so defined.

195

196 AP Unit 7 (Existence Theorems)


AP Unit 7, Day 1: Quasi-Limits: One-Sided and Infinite 197

7.1 Quasi-Limits: One-Sided and Infinite

Advanced Placement












Textbook §1.3 Computation of Limits and §1.5 Limits Involving Infinity; Asymp-totes [16]

Resources §2-5 Limits Involving Infinity in Foerster [10].

7.1.1 Step Discontinuities & One-Sided Limits

Piecewise Functions

Example 7.1.1 For the function f(x) =

−x3 − 8x− 2

x < 2

0.2 x = 2

x3 − 8x− 2

x > 2

Is f(x) continuous? Where do you expect to have problems? If



you are looking at values near 2, but different from 2, what formulashould you use?

Determine:

(a) f(2)

(b) limx→2−

f(x)

(c) limx→2+

f(x)

(d) limx→2

f(x)

(e) limx→0+

f(x)

(f) limx→3

f(x)

Example 7.1.2 In your mighty, mighty groups of four: [18] Let

g(x) =

{2x− 1, x ≤ 2x2 − x x > 2

Is g(x) continuous? Where do you expect to have problems? Deter-mine:

(a) g(2)

(b) limx→2−

g(x)

(c) limx→2+

g(x)

(d) limx→2

g(x)

(e) limx→0+

g(x)

(f) limx→3

g(x)

Absolute Value

Example 7.1.3 Examine limx→2

x3 − 8|x− 2|

analytically. Check your an-

swer with a table. Predict the behavior of the graph near x = 2.

7.1.2 One-Sided Derivatives

Example 7.1.4 Find

(a) lim∆x→0+

|2 + ∆x| − |2|∆x



(b) limh→0+

|−3 + h| − |−3|h

(c) lim∆x→0+

|0 + ∆x| − |0|∆x

In your mighty, mighty groups of four: Find

(a) lim∆x→0−

|2 + ∆x| − |2|∆x

(b) lim∆x→0−

|−3 + ∆x| − |−3|∆x

(c) limh→0−

|0 + h| − |0|h

(d) lim♥→0

|2 +♥| − |2|♥

(e) lim∆x→0

|−3 + ∆x| − |−3|∆x

(f) lim∆x→0

|0 + ∆x| − |0|∆x

What do each one of these limits represent, in terms of a graph of

y = |x|? Use your answers to sketch a graph ofd

dx|x|.

Example 7.1.5 The graph of the function f shown in Figure 7.1


consists of a semicircle and three line segments. Find the followinglimits of difference quotients:

(a) limx→−2

f(x)− f(−2)x+ 2

[Ans: − 1

3

](b) lim

h→0

f(2.3 + h)− f(2.3)h

[Ans: −1]

(c) limx→−3−

f(x)− f(−3)x+ 3

[Ans: 2]



(d) limx→−3+

f(x)− f(−3)x+ 3

[Ans: − 1

3

](e) lim

x→−3

f(x)− f(−3)x+ 3

[Ans: d.n.e.]

(f) limh→0+

f(2 + h)− f(2)h

[Ans: −1]

Example 7.1.6 In your mighty, mighty groups of four: Write alinear function, mx + b, such that if you plug 2 in, you get 7 out.For example, I will use 3x + 1, which works because 3 (2) + 1 = 7.Let

R(x) =

√

3x+ 1, x < 2√7, x = 2√mx+ b, x > 2

Why is it important that m (2) + b is 7?

(a) Find limx→2−

R(x)−R(2)x− 2

(b) On the board, claim your functionmx+b. Find limx→2+

R(x)−R(2)x− 2

,

and put your answer in a table next to your function mx+ b.

(c) Find limx→2

R(x)−R(2)x− 2

(d) Find limh→0+

R(2 + h)−R(2)h

(e) Pick a value c, c > −13

. Find limx→c

R(x)−R(c)x− c

7.1.3 Vertical Asymptotes: Infinite Limits

Nonexistent, Infinite Limits

Example 7.1.7 How many vertical asymptotes does the function

f(x) =2x2 − x− 6x2 − x− 2

have? What are they? Check your answers on

your grapher.

Remember This? How do you distinguish between vertical asymptotes andremovable discontinuities?



Example 7.1.8 Using a table or graph, examine

limx→2

∣∣∣∣ 1x− 2

∣∣∣∣Technically, as x goes to 2,

∣∣∣ 1x−2

∣∣∣ does not go to one particular number (infinity

is not a number). Therefore, we say that limx→2

∣∣∣ 1x−2

∣∣∣ does not exist, or that it isnonexistent. Sometimes, however, we like to treat infinity like a number, so wewrite lim

x→2

∣∣∣ 1x−2

∣∣∣ =∞, and we describe the limit as being infinite. It is true thatthe limit is both nonexistent and infinite.

Positive and Negative Zero

On a case by case basis, you need to decide the sign (positive or negative) ofany zeros in the denominator which yield infinite limits.

Example 7.1.9 Find limθ→π

2−

tan θ and limθ→π

2+

tan θ. Confirm your

answer with a graph and with a table.

Derivatives at a Step Discontinuity

Example 7.1.10 (adapted from [18]) Let

g(x) =

{2x− 1, x ≤ 2x2 − x x > 2

Is g(x) continuous? Where do you expect to have problems? Deter-mine:

(a) g(2)

(b) limx→2−

g(x)

(c) limx→2+

g(x)

(d) limx→2

g(x)

(e) limx→2−

g(x)− g(2)x− 2

(f) limx→2+

g(x)− g(2)x− 2



(g) limx→2

g(x)− g(2)x− 2

(h) limx→3

g(x)− g(3)x− 3

Problems

7.A-1 [18] Let f(x) =

{x2, x ≤ 15x x > 1

Determine:

(a) f(2)

(b) limx→2−

f(x)

(c) limx→2+

f(x)

(d) limx→2

f(x)

(e) limx→1+

f(x)

(f) limx→1−

f(x)

(g) limx→1

f(x)

(h) limx→0+

f(x)

(i) limx→3

f(x)

7.A-2 [18] Let g(x) =

x2, x < 37, x = 32x+ 3, x > 3

Find limx→3

f(x) [Ans: 9]

7.A-3 For the function f(x) =

x− 4√x− 2

0 < x ≤ 4

|4− 2x| 4 < x ≤ 4.27determine whether lim

x→4f(x) exists, and if so, what the limit is. [Ans: 4]

7.A-4 For the function g(x) =

3− x1

x−2 − 10 < x ≤ 4

√x+ 5 4 < x ≤ 4.27

determine whether limx→3

g(x) exists, and if so, what the limit is. [Ans: 1]



7.A-5 For the function h(x) =

4x− 16|x− 4|

0 < x ≤ 4

√x+ 12 4 < x ≤ 4.27

determine whether limx→4

h(x) exists, and if so, what the limit is.[Ans: d.n.e. b/c lim

x→4−h(x) = −4 6= lim

x→4+h(x) = 4

]7.A-6 lim

x→π−cotx is [Ans: −∞]

7.A-7 f(x) =(x− 1)2

3x2 − 5x+ 2

(a) Is f continuous at x = 1?

(b) Give the equation(s) of all vertical asymptotes. [Ans: no; x = 2

3

]7.A-8 Let W be the function given by W (x) =

(x+ 2)(x2 − 1

)x2 − c

.

(a) Will f be continuous if c = 1? Explain.

(b) Will f be continuous if c = 4? Explain.

(c) For what positive values of c is f continuous for all real numbers x?

[Ans: no ; no; none]

7.A-9 Find limt→1−

t

ln t, being as specific as possible. [Ans: −∞]

7.A-10 Do you know your asymptote from a hole in the graph?




AP Unit 7, Day 2: Limits at Infinity and Horizontal Asymptotes 205

7.2 Limits at Infinity and Horizontal Asymp-totes

Advanced Placement




• Comparing relative magnitudes of functions and their rates of change. (Forexample, contrasting exponential growth, polynomial growth, and logarithmicgrowth.)

Textbook §1.5 Limits Involving Infinity; Asymptotes [16]

Resources §2-5 Limits Involving Infinity in Foerster [10].

7.2.1 Limits at Infinity

If limx→c

f(x) = 0 then limx→c

1f(x)

is infinite. In other words, 0 in the denominator

wants an infinite limit.

Similarly, if limx→c

f(x) is infinite, then limx→c

1f(x)

= 0, i.e., ∞ in the denominator

wants the limit to be zero.

Note also that 0 in the numerator wants a zero limit, and ∞ in the numeratorwants an infinite limit.

00

= indeterminatenonzero

0= ±∞ ±∞

0= ±∞

0nonzero

= 0nonzerononzero

= answer±∞

nonzero= ±∞

0±∞

= 0nonzero±∞

= 0±∞±∞

= inderterminate

As x→∞ or x→ −∞, f(x) behaves like its highest order term.

Example 7.2.1 Find the horizontal asymptotes of y =2x2 − x− 6x2 − x− 2



(a) limx→−∞

2x2 − x− 6x2 − x− 2

(b) limx→−∞

2x6x2 − x− 2

(c) limx→∞

2x2 − x− 6x3

x2 − x− 2

Confirm your answers on your handy, dandy calculator with a graph,and with a table.

Example 7.2.2 Find

(a) limx→∞

12 + 7x− 5x3

3x2 − 12x+ 9

(b) limx→−∞

12 + 7x− 5x2

3x2 − 12x+ 9

(c) limx→∞

12 + 7x− 5x2

3x2 − 12x3 + 9

Do it the textbook way, and then using highest order terms. Whathappens if instead of taking the limit to ∞, you take x→ −∞?

In using the highest order term, I’m saying that 3x2 − 12x + 9 behaves as 3x2

for larger and larger values of x. Essentially, I’m ignoring the −12x. But ifx → ∞, −12x should be extremely massive, not negligible. The differenceis in the relative values. Think of how many molecules are in a drop of water.Conceptually, we will say an infinite amount. Add a drop of water to the AtlanticOcean. Has the amount of water in the ocean really changed significantly? Eventhough you’ve added an infinite number of water molecules? So, now imaginethat you have three oceans, take out 12 drops, and add nine molecules of water.You still have three oceans. Which is why we say that 3x2 − 12x + 9 behavesas 3x2 as x→∞.

Note that if the degree is higher on top, the limit at infinity is nonexistent(infinite). If the degree on top is lower, then the limit at infinity is zero. If thedegree is the same, then the limit at infinity is the ratio of the coefficients.

In determining degrees, ex >> · · · >> xn >> · · · >> x3 >> x2 >> x >>lnx >> sinx.

Example 7.2.3 Find

(a) limx→∞

3x2 − 13x+ 7 sinx5x2 + 4x− 9


AP Unit 7, Day 2: Limits at Infinity and Horizontal Asymptotes 207

(b) limx→∞

3x2 − 13ex + 7 sinx5x2 + 4 lnx− 9ex

(c) limx→∞

3x2 − 13x183 + 7 sinx5x2 + 4 lnx− 9ex

You could also see that the sinx becomes irrelevant by using a variation on thesqueeze theorem.

Example 7.2.4 [18] Find limh→0

1− 1h2

1− 1h

[Ans: does not exist]

7.2.2 Horizontal Asymptotes

There is a horizontal asymptote at y = L if (and only if) at least one of thefollowing conditions is met:

• limx→∞

f(x) = L

• limx→−∞

f(x) = L

Note that there are two possibilities for the horizontal asymptote. There mightbe two horizontal asymptotes, or one, or none. In typical examples of ratio-nal expressions, where we just have polynomials over polynomials, the limits atpositive and negative infinity are the same. So, to make sure that you know totake the limits at both positive and negative infinity, test-makers devise prob-lems where the limits are different, yielding two horizontal asymptotes insteadof one. Typically, this is done with radicals, or with absolute values.

Remember This? When is√

9x2 6= 3x? When is√

9x4 6= 3x2?

Example 7.2.5 Find all horizontal asymptotes of−7x+ 12√

9x2 + 3x+ 12

[Ans: y = 7

3 , y = − 73



Problems

big giant blue-green Calculus book p. 96: # 33; p. 118: #25, 27, 29, 39, 43,57, 59

7.B-1 Find all the horizontal asymptotes of h(x) =x− 1√

−3− x+ 4x2.[Ans: y = 1

2 , y = − 12

]7.B-2 (a) What is lim

x→−∞

x− 1−3− x+ 4x2

? Show your process.

(b) Make a single change to the problem so that the answer would be14[

Ans: y = 0; x2 − 1 in numerator]

7.B-3 Let f(x) =x− 1

−3− x+ 4x2.

(a) Create a new function g(x) by making a single change to f(x) so that

limx→−∞

g(x) = −12

.

(b) Show the analysis necessary to evaluate the limit at negative infinityof your new function. A table is not sufficient analysis.

(c) Use your calculator to find limx→−∞

g(x). Does this confirm your pre-

vious answer?

(d) Use your calculator to write a table to find limx→+∞

g(x).[Ans: 1

2

]

7.B-4 [18] Find limh→0

1− 1h

1 +1h

[Ans: −1]

7.B-5 [18] Find limt→0

t+a

t

t+b

t

[Ans: a

b

]


AP Unit 7, Day 3: Continuity and Differentiability 209

7.3 Continuity and Differentiability

Advanced Placement




• Geometric understanding of graphs of continuous functions (IntermediateValue Theorem and Extreme Value Theorem).




Definition 7.1 (Continuity at a point). A function f is continuous at x = c ifand only if

• limx→c

f(x) exists

• f(c) exists

• limx→c

f(x) = f(c)

Ways for a function to be not continuous:

• removable discontinuity

• step discontinuity

• vertical asymptote

Example 7.3.1 (adapted from AB 1969) If

f(x) =√

4x+ 9−√

2x+ 5x+ 2

, for x 6= −2

f(−2) = kand if f is continuous at x = −2, then k =

[Ans: 1]



Definition 7.2 (Differentiability at a point). A function f is differentiable at

x = c if and only if limx→c

f(x)− f(c)x− c

exists

Ways for a function to be not continuous:

• cusp

• vertical tangent

• discontinuity

When checking differentiability, make sure

• the y-values on the left and right match;

• the slopes on the left and right match.

Example 7.3.2 Determine whether f(x) = |x− 2| is differentiableat x = 2. Write f ′(x).

Example 7.3.3 Let g(x) =

{2x x ≤ 112x

2 + x x > 1

Determine whether the function g(x) is differentiable at x = 1. Writeg′(x).

Example 7.3.4 How could you change g(x) to make it differen-tiable?

Differentiability Implies Continuity

Differentiability is a higher standard than continuity.

If f is differentiable, then f is continuous.

If f is continuous, then f may or may not be differentiable.

If f is not continuous, then f cannot be differentiable.

Definition 7.3 (Continuity over an open interval).

Definition 7.4 (Continuity over a closed interval).

Definition 7.5 (Differentiability over an open interval).


AP Unit 7, Day 3: Continuity and Differentiability 211

Problems

7.C-1 (adapted from Acorn ’05) Let f be defined as follows, where a 6= 0,

f(x) =

x2 − a2

x− afor x 6= a

0 for x = a

Describe each of the following as true or false:

(a) limx→a

f(x) exists

(b) f(a) exists

(c) f(x) is continuous at x = a

7.C-2 (adapted from Acorn ’05) Let f be defined as follows, where a 6= 0,

f(x) =

x3 − a3

x− afor x 6= a

3a2 for x = a

Describe each of the following as true or false:

(a) limx→a

f(x) exists

(b) f(a) exists

(c) f(x) is continuous at x = a

7.C-3 (adapted from [2]) The function f is continuous at x = 1. If f(x) =√x+ 7−

√5x− 1

x− 1for x 6= 2

k for x = 2then k =

[Ans: − 2

3

]

7.C-4 (adapted from [2]) Given f(x) =

x2 + 4x+ 3

x+ 1for x 6= −1

k for x = −1

Determine the value of k for which f is continuous for all real x. [Ans: 2]

7.C-5 (adapted from [2]) The function f is defined for all real numbers such that

f(x) =

{x2 + kx− 3 for x ≤ 15x+ b for x > 1

If f is both continuous and differentiable,

what are the values of k and b? What changes if you are only told that fneeds to be differentiable? [Ans: k = 3, b = −4]

7.C-6 (adapted from AB ’97) Let f be a function such that limh→0

f(2 + h)− f(2)h

=

5. Why must f be continuous at x = 2?



7.C-7 (adapted from [2]) Let m and b be real numbers and let the function f bedefined by

f(x) =

{1 + 3bx+ 2x2 for x ≤ 2mx+ b for x > 2

Find m and b if f is differentiable at x = 2. [Ans: m = −13, b = −7]


AP Unit 7, Day 4: Some Basic Calculus Theorems 213

7.4 Some Basic Calculus Theorems

Advanced Placement


• Geometric understanding of graphs of continuous functions (IntermediateValue Theorem and Extreme Value Theorem).



Resources Workshop Calculus [12]

Notation

⇒ impliesbecausetherefore

∀ for all, for any∃ there exists∈ is a member of3 such that

7.4.1 Intermediate Value Theorem

Example 7.4.1 Connect the dots.

Theorem 7.1 (Intermediate Value Theorem). If f is a function that is contin-uous over the closed interval [a, b], and k is a y-value between f(a) and f(b),then there exists an x-value, c, which is between a and b, such that f(c) = k.

In other words, if f is a nice function (i.e., continuous function) on some interval,then it goes through every y-value between the two endpoints.

7.4.2 Extreme Value Theorem

Theorem 7.2 (Extreme Value Theorem). If f is a function that is continuousover the closed interval [a, b], then there exists



• at least one x-value cM , where a ≤ cM ≤ b, such that f(cM ) ≥ f(x) forall x, a ≤ x ≤ b.

• at least one x-value cm, where a ≤ cm ≤ b, such that f(cM ) ≤ f(x) for allx, a ≤ x ≤ b.

In other words, if f is nice on some interval, then there is a maximum value of fsomewhere on that interval, and a minimum value somewhere on that interval.

7.4.3 Rolle’s Theorem

Theorem 7.3 (Rolle’s Theorem). If f is a function that is continuous overthe closed interval [a, b], and differentiable over the open interval (a, b), and iff(a) = f(b) [= 0], then there exists at least one x-value c, where a < c < b, suchthat f ′(c) = 0.

If f is a nice, smooth function (i.e., a continuous and differentiable function)which happens to start and stop on a horizontal line, then somewhere in betweenf will have a horizontal tangent.

Problems

7.D-1 Memorize the conditions and conclusions of the Intermediate Value The-orem.

7.D-2 [12] Apply the Intermediate value Theorem for Continuous Functions toa specific example. Consider f(x) = 2x+ 1, where −2 ≤ x ≤ 4.

(a) Sketch the graph of f for −2 ≤ x ≤ 4. Label −2 and 4 on thehorizontal axis and f(−2) and f(4) on the vertical axis.

(b) In order to apply the theorem, f must be continuous over the givenclosed interval. Explain why f is continuous on the closed interval[−2, 4].

(c) The theorem claims that if y is any value between f(−2) and f(4),there exists an input value c between −2 and 4 such that f(c) = y.Apply the theorem for y = 6; that is, find a value for c between −2and 4 such that f(c) = 6. Label c and f(c) on the graph you sketchedin part 2a and indicate the relationship between c and f(c).

(d) Apply the theorem for y = −2; that is, find a value for c between−2 and 4 such that f(c) = −2. Label c and f(c) on the graph yousketched in part 2a and indicate the relationship between c and f(c).



7.D-3 [12] Explain why the Intermediate Value Theorem for Continuous Func-tions makes sense. Support your explanation with an appropriate diagram;that is, on a pair of axes:

(a) Label a and b on the horizontal axis, where a < b.

(b) Sketch the graph of a squiggly function f that is continuous over theclosed interval [a, b].

(c) Label f(a) and f(b) on the vertical axis.

(d) Pick a y-value between f(a) and f(b) and label it k

(e) Show that the conclusion of the Intermediate Value Theorem forContinuous Functions holds; that is, show that there exists an x-valuec between a and b such that f(c) = k. Label c on the horizontal axis.

7.D-4 [12] For a given value of y between f(a) and f(b), is it possible for thereto be more than one choice of c between a and b such that f(c) = y? Ifso, draw a diagram supporting your conclusion. If not, explain why not.

7.D-5 [12] Apply the Intermediate Value Theorem to some real-life situations.For each of the following situations:

i Model the scenario with a graph.

ii Illustrate the conclusion on your graph.

(a) Scenario: The light turns green, and you step on the gas pedal inyour car. Fifteen seconds later, you level off your speed at 60 mph.Conclusion: According to the Intermediate Value Theorem, thereexists a time in the 15-second time interval when you are going 28mph.

(b) Scenario: You walk back and forth in front of a motion detector for20 seconds. You vary your velocity. Your minimum distance is 0.5meters from the detector and your maximum distance is 8.5 meters.Conclusion: According to the Intermediate Value Theorem, thereexists a time in the 20-second time interval when you are 5.25 metersfrom the detector .

(c) Scenario: You have the flu. Over a three-day period, your tempera-ture fluctuates between 99.8◦ and 103.2◦.Conclusion: According to the Intermediate Value Theorem, thereexists a time in the three-day period when your temperature is 100◦.

7.D-6 [12] Use the Intermediate Value Theorem to show that the equation x5 −3x4 − 2x3 − x+ 1 = 0 has a solution between 0 and 1.

7.D-7 Sketch a continuous function f for which does not meet the conclusions ofthe Intermediate Value Theorem, (i.e., there is no c ∈ (a, b) 3 f(c) = k)because k is not between f(a) and f(b).



7.D-8 Sketch a function f which does not meet the conclusions of the Interme-diate Value Theorem because f is not continuous.

7.D-9 Sketch a function f which does meet the conclusions of the IntermediateValue Theorem (i.e., there is c ∈ (a, b) 3 f(c) = k) despite the fact that kis not between f(a) and f(b).

7.D-10 Sketch a function f which does meet the conclusions of the IntermediateValue Theorem (i.e., there is c ∈ (a, b) 3 f(c) = k) despite the fact that fis not continuous.

7.D-11 Memorize the conditions and conclusions of the Extreme Value Theorem.

7.D-12 [12] Apply the Max-Min Theorem (aka the Extreme Value Theorem) forContinuous Functions to a specific example. Consider f(x) = x2 − 1,where −2 ≤ x ≤ 3.

(a) Sketch the graph of f for −2 ≤ x ≤ 3. Label −2 and 3 on thehorizontal axis and f(−2) and f(3) on the vertical axis.

(b) In order to apply the theorem, f must be continuous over the givenclosed interval. Explain why f is continuous on the closed interval[−2, 3].

(c) The theorem claims that f has an absolute minimum on [−2, 3] – thatis, there exists a value cm between −2 and 3 such that f(cm) ≤ f(x)for all x between −2 and 3. Find cm. Label cm and f(cm) on thegraph you sketched in part 12a.

(d) The theorem claims that f has an absolute maximum on [−2, 3] – thatis, there exists a value cM between −2 and 3 such that f(x) ≤ f(cM )for all x between −2 and 3. Find cM . Label cM and f(cM ) on thegraph you sketched in part 12a.

7.D-13 [12] Explain why the Max-Min Theorem for Continuous Functions makessense. Support your explanation with an appropriate diagram. On a pairof axes:

(a) Label a and b on the horizontal axis, where a < b.

(b) Sketch the graph of a (squiggly) function f which is continuous overthe closed interval [a, b].

(c) Show that there exists an input value cm between a and b such thatf(m) ≤ f(x) for all x between a and b. Label cm and f(cm) on yourdiagram.

(d) Show that there exists an input value cM between a and b such thatf(x) ≤ f(cM ) for all x between a and b. Label cM and f(cM ) onyour diagram.



7.D-14 [12] The Max-Min Theorem for Continuous Functions states that everyfunction f which is continuous over a closed interval [a, b] takes on bothan absolute minimum and an absolute maximum on [a, b]. If you wantto determine where the absolute extrema exist, what values would youexamine? In other words, which values in the interval [a, b] would becandidates for the absolute maximum and absolute minimum?

7.D-15 [12] Is it possible for there to be more than one choice of cm between aand b such that f(cm) ≤ f(x) for all x between a and b? If so, draw adiagram supporting your conclusion. If not, explain why not. Similarly,is it possible for there to be more than one choice of cM between a and bsuch that f(x) ≤ f(cM ) for all x between a and b? If so, draw a diagramsupporting your conclusion. If not, explain why not.

7.D-16 [12] Apply the Max-Min Theorem to some real-life situations. For each ofthe following situations:

i Model the scenario with a graph.ii Label the absolute extrema for your model.

(a) You work an 8-hour shift at a pretzel factory. At the start of yourshift, your production rate is low, but it continues to increase as yousettle into a routine. Two hours before the end of the shift, youstart thinking about what you are going to do after work, and yourproduction rate decreases until it’s time to quit.

(b) You’re home alone watching a scary movie, on a dreary night. Eachtime a scary part comes on, your heart rate increases dramaticallyand then returns to normal when the scary part is over. The movieis 117 minutes long, and there are 7 scenes that frighten you.

(c) You create a distance-versus-time graph using a motion detector.Starting 7.75 meters from the detector, you walk toward the detectorfor 6.5 seconds. You walk faster and faster for the first 4 seconds,and then slower and slower for the next 2.5 seconds. You stop at thehalf–meter mark.

7.D-17 Sketch a function that has no maximum because it has a vertical asymp-tote.

7.D-18 Sketch a function that has no minimum because it has a removable dis-continuity.

7.D-19 Sketch a function that has a maximum despite the fact that it is notcontinuous over [a, b].

7.D-20 Memorize the conditions and conclusions of Rolle’s Theorem.

7.D-21 Sketch a continuous and differentiable function which does not meet theconclusions of Rolle’s Theorem (i.e., the function does not have a horizon-tal tangent).



7.D-22 Sketch a function which does not meet the conclusions of Rolle’s Theorem(i.e., the function does not have a horizontal tangent) even though it startsand stops on the same horizontal line.

7.D-23 Sketch a function that does not start and stop on the same horizontal line,but still has a horizontal tangent.


AP Unit 7, Day 5: Mean Value Theorem 219

7.5 Average Rate of Change and the Mean ValueTheorem

Advanced Placement





• The Mean Value Theorem and its geometric consequences.

Textbook §2.9 The Mean Value Theorem [16]

Resources Workshop Calculus

7.5.1 Average Rate of Change

7.5.2 Mean Value Theorem

Theorem 7.4 (Mean Value Theorem). If f(x) is a function that is continuousover [a, b] and differentiable over (a, b), then ∃ c ∈ (a, b) 3

f ′(c) =f(b)− f(a)

b− a

A nice, smooth function has a spot where the tangent line is parallel to thesecant line, i.e., where the instantaneous rate of change matches the averagerate of change.

Example 7.5.1 The function L(t) = 10 000(e−0.2 − e−0.2t

)gives

the number of gallons that have leaked out of a tanker, where t thetime in hours after noon.

(a) FindL(3)− L(1)

3− 1. Explain the meaning of this value. [Ans: 1349.596]

(b) Find the average rate at which oil leaked out of the tanker from3 p.m. to 9 p.m. Indicate units. [Ans: 639.188 gal/hr]



(c) Find L′(t). Using correct units, explain the meaning of L′(t).

(d) At what time between 1 p.m. and 3 p.m. is the instantaneousrate of leakage the same as the average rate of leakage over thatsame time interval?

(e) At what time between 3 p.m. and 9 p.m. is the instantaneousrate of leakage the same as the average rate of leakage over thatsame time interval?

[Ans: 2 000e−0.2t, rate of leakage; t = 1.96671 (1:58); t = 5.70352 (5:42)

]Example 7.5.2 (adapted slightly from AB ’02) Let f be a functionthat is differentiable for all real numbers. Table 7.1 gives the values

Table 7.1: AB ’02

x −1.5 −1.0 −0.5 0 0.5 1.0 1.5f(x) −1 −4 −6 −7 −6 −4 −1f ′(x) −7 −5 −3 0 3 5 7

of f and its derivative f ′ for selected points x in the closed interval−1.5 ≤ x ≤ 1.5. The second derivative of f has the property thatf ′′(x) > 0 for −1.5 ≤ x ≤ 1.5.

(a) Find a positive real number r having the property that theremust exist a value c with 0 < c < 0.5 and f ′′(c) = r. Give areason for your answer.

(b) Let g be the function given by g(x) =

{2x2 − x− 7 for x < 02x2 + x− 7 for x ≥ 0

The graph of g passes through each of the points (x, f(x)) givenin the table above. Is it possible that f and g are the samefunction? Give a reason for your answer.

(c) Write an equation of the line tangent to the graph of f at thepoint where x = 1. Use this line to approximate the valueof f(1.2). Is this approximation greater than or less than theactual value of f(1.2)? Give a reason for your answer.

(d) Write an equation of the line tangent to the graph of f at thepoint where x = 1.5. Use this line to approximate the value off(1.2). Which of these approximations do you suppose is moreaccurate, and why?

[Ans: 6; no; −3 < f(1.2); −3.1 < −3 < f(1.2)]



Example 7.5.3 (from Acorn ’02) Let f be a function such thatf ′′(x) < 0 for all x in the closed interval [1, 2]. Selected values of fare shown in Table 7.2. Which of the following must be true about

Table 7.2: Acorn ’02 # 18

x 1.1 1.2 1.3 1.4f(x) 4.18 4.38 4.56 4.73

f ′(1.2)?

(a) f ′(1.2) < 0(b) 0 < f ′(1.2) < 1.6(c) 1.6 < f ′(1.2) < 1.8(d) 1.8 < f ′(1.2) < 2.0(e) f ′(1.2) > 2.0

Example 7.5.4 (adapted slightly from AB ’06B) A car travels ona straight track. During the time interval 0 ≤ t ≤ 60 seconds, thecar’s velocity v, measured in feet per second, and acceleration a,measured in feet per second per second, are continuous functions.Table 7.3 shows selected values of these functions.

Table 7.3: AB ’06B

t(sec)

0 15 25 30 35 50 60

v(t)(ft/sec)

−20 −30 −20 −14 −10 0 10

a(t)(ft/sec2)

1 5 2 1 2 4 2

(a) Using appropriate units, explain the meaning of∫ 60

30

v(t) dt in

terms of the car’s motion. Approximate∫ 60

30

v(t) dt using a

trapezoidal approximation with the three subintervals deter-mined by the table.

(b) Using appropriate units, explain the meaning of∫ 60

30

|v(t)| dt

in terms of the car’s motion. Approximate∫ 60

30

|v(t)| dt using



a trapezoidal approximation with the three subintervals deter-mined by the table.

(c) For 0 < t < 60, must there be a time t when v(t) = −5? Justifyyour answer.

(d) For 0 < t < 60, must there be a time t when a(t) = 0? Justifyyour answer.

(e) For 0 < t < 60, must there be a time t when a′(t) = 0? Justifyyour answer.

Example 7.5.5 (adapted slightly from AB ’03) A blood vessel is360 millimeters (mm) long with circular cross sections of varyingdiameter. Table 7.4 gives the measurements of the diameter of the


Distancex (mm)

0 60 120 180 240 300 360

DiameterB(x) (mm)

24 30 28 30 26 24 26

blood vessel at selected points along the length of the blood vessel,where x represents the distance from one end of the blood vessel andB(x) is a twice–differentiable function that represents the diameterat that point.

(a) Using correct units, explain the meaning of1

360

∫ 360

0

B(x)2

dx

in terms of the blood vessel. Approximate the value of1

360

∫ 360

0

B(x)2

dx

using the data from the table and a midpoint Riemann sumwith three subintervals of equal length. Show the computa-tions that lead to your answer. Is your answer reasonable?

(b) Using correct units, explain the meaning of π∫ 360

0

(B(x)

2

)2

dx

in terms of the blood vessel.

(c) Explain why there must be at least one value x, for 0 < x < 360,such that B′′(x) = 0.

[Ans: 14 mm; volume of the b.v. from x = 0 to x = 360; B′(c1) = B′(c2) = 0, . . . ]



Problems

7.E-1 (from [2]) The graph of a function f whose domain is the interval [−4, 4]is shown in the Figure 7.2. Which of the following statements are true?

Figure 7.2: Venture III IB #11

(a) The average rate of change of f over the interval from x = −2 to

x = 0 is12

(b) The slope of the tangent line at the point where x = 2 is 0.

(c) The left–sum approximation of∫ 3

−1

f(t) dt with four equal subdivi-

sions is 4.

[Ans: T, F, T]

7.E-2 (from [2]) The line x− 3y + 7 = 0 is tangent to the graph of y = f(x) at(2, f(2)) and is also parallel to the line through (1, f(1)) and (7, f(7)). Iff is differentiable on the closed interval [1, 7] and f(1) = 2, then find

(a) f(7)

(b) f(2)

[Ans: 4, 3]

7.E-3 (AB ’93) If f(x) = sin(x

2

), then there exists a number c in the interval

π

2< x <

3π2

that satisfies the conclusion of the Mean Value Theorem.

What is c? [Ans: π]

7.E-4 (AB ’97) Let f be the function given by f(x) = 3 cosx. As shown inFigure 7.3, the graph f crosses the y–axis at point P and the x–axis atpoint Q.

(a) Write an equation for the line passing through points P and Q.

(b) Write an equation for the line tangent to the graph at point Q.



Figure 7.3: AB ’97 Section II, # 2

(c) Find the x–coordinate of the point on the graph of f , between pointsP and Q, at which the line tangent to the graph of f is parallel tothe line PQ.[

Ans: y − 3 = − 6π (x− 0); y − 0 = −3 (x− π/2); 0.690

]7.E-5 (AB ’89) Let f be the function given by f(x) = x3 − 7x+ 6.

(a) Find the zeros of f .

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

(c) Find the number c that satisfies the conclusion of the Mean ValueTheorem for f on the closed interval [1, 3][

Ans: 1, 2, −3; y − 12 = −4 (x+ 1);√

133

]7.E-6 (adapted from [2]) Find the point on the graph of y = 3

√x between (0, 0)

and (1, 1) at which the line tangent to the graph has the same slope asthe line through (0, 0) and (1, 1).

[Ans: 1

3√

3

]7.E-7 (AB ’04B) A test plane flies in a straight line with positive velocity v(t),

in miles per minute at time t minutes, where v is a differentiable functionof t. Selected values of v(t) for 0 ≤ t ≤ 40 are shown in the Table 7.5.

(a) Use a midpoint Riemann sum with four subintervals of equal length

and values from the table to approximate∫ 40

0

v(t) dt. Show the




t(min)

0 5 10 15 20 25 30 35 40

v(t)(mpm)

7.0 9.2 9.5 7.0 4.5 2.4 2.4 4.3 7.3

computations that lead to your answer. Using correct units, explain

the meaning of∫ 40

0

v(t) dt in terms of the plane’s flight.

(b) Based on the values in the table, what is the smallest number ofinstances at which the acceleration of the plane could equal zero onthe open interval 0 < t < 40? Justify your answer.

[Ans: 229 miles flown during the forty minutes; Twice]

7.E-8 (AB ’05) A metal wire of length 8 centimeters (cm) is heated at one end.Table 7.6 gives selected values of the temperature T (x), in degrees Celsius

Table 7.6: AB ’05

Distancex (cm)

0 1 5 6 8

TemperatureT (x) (◦C)

100 93 70 62 55

(◦C), of the wire x cm from the heated end. The function T is decreasingand twice differentiable.

(a) Estimate T ′(7). Show the work that leads to your answer. Indicateunits of measure.

(b) Are the data in the table consistent with the assertion that T ′′(x) > 0for every x in the interval 0 < x < 8? Explain your answer.[

Ans: − 72

◦C/cm; No, T ′(c1) = −5.75, T ′(c2) = −8, . . .]

7.E-9 The function L(t) = 10 000(e−0.2 − e−0.2t

)gives the number of gallons

that have leaked out of a tanker, where t the time in hours after noon.

(a) FindL(5)− L(1)

5− 1. Explain the meaning of this value. [Ans: 1127.128]

(b) Find the average rate at which oil leaked out of the tanker from 5p.m. to 9 p.m. Indicate units. [Ans: 506.451 gal/hr]



(c) Find the average rate at which oil leaked out of the tanker from 1p.m. to 9 p.m. Indicate units. [Ans: 816.790 gal/hr]

(d) Find L′(t). Using correct units, explain the meaning of L′(t).

(e) At what time between 1 p.m. and 5 p.m. is the instantaneous rate ofleakage the same as the average rate of leakage over that same timeinterval?

(f) At what time between 5 p.m. and 9 p.m. is the instantaneous rate ofleakage the same as the average rate of leakage over that same timeinterval?

(g) At what time between 1 p.m. and 9 p.m. is the instantaneous rate ofleakage the same as the average rate of leakage over that same timeinterval?[

Ans: 2 000e−0.2t, rate of leakage; t = 2.86737 (2:52); t = 6.86737 (6:52); t = 4.477603 (4:29)]


AP Unit 7, Day 6: Riemann Sums 227

7.6 Riemann Sums: Evaluating Definite Inte-grals

Advanced Placement



• Definite integral as a limit of Riemann sums over equal subdivisions.

7.6.1 Sigma Notation

Example 7.6.1 Find5∑i=1

i2

[Ans: 55]

Example 7.6.2 Write in sigma notation:

(a)1

1 + 1+

21 + 2

+3

1 + 3+ · · ·+ 15

1 + 15

[Ans:

∑15i=1

i1+i

]

7.6.2 Riemann Sums

Example 7.6.3 (adapted from Acorn ’05) Oil is leaking from atanker at the rate of R(t) = 2 000e−0.2t gallons per hour, where t ismeasured in hours after noon.

(a) Using correct units, explain the meaning of∫ 9

1R(t) dt.

(b) Approximate∫ 9

1R(t) dt using

(a) M4, a midpoint Riemann sum with 4 equal subintervals;(b) T4, a trapezoidal approximation with 4 equal subintervals;

(c) S8 =2M4 + T + 4

3(d) A Riemann sum, using the partition a = x0 = 1 < x1 =

3 < x2 = 9, and evaluation points c1 = 1.96671 and c2 =5.70352

[Ans: 6490.959; 6621.211; 6534.376; 6534.319]



Example 7.6.4 (AB ’97) The expression

150

(√150

+

√250

+

√350

+ . . .+

√5050

)is a Riemann sum approximation for

(a)∫ 1

0

√x

50dx

(b)∫ 1

0

√x dx

(c)150∫ 1

0

√x

50dx

(d)150∫ 1

0

√x dx

(e)150∫ 50

0

√x dx

[Ans:

∫ 1

0

√x dx

]

7.6.3 Evaluating Definite Integrals Exactly

Example 7.6.5

(a) I will give you a table of values for certain definite integrals.Find a pattern, in terms of a and b, for:

(a)∫ bax dx

(b)∫ bax2 dx

(b) Use geometric reasoning to justify the pattern for∫ bax dx

(c) Pick a function, f(x), and deduce a pattern for∫ baf(x) dx.

(d) Deduce a pattern for:

(a)∫ baex dx

(b)∫ ba

cosx dx

(c)∫ ba

1xdx

(d)∫ ba

11 + x2

dx

(e)∫ ba

sec2 x dx

Example 7.6.6 Write the Mean Value Theorem, as it applies toG(x).



7.6.4 Proof

Given: Let f(x) be the derivative of some function G(x) over the interval [a, b].

G(x) is differentiable over [a, b] and over any subinterval inside [a, b]. Why?

G(x) is continuous over [a, b] and any subinterval therein. Why? If G(x) iscontinuous and differentiable, what conclusions can we draw?

Recall:

• We partition the interval [a, b] into n subintervals: a = x0 < x1 < x2 <· · · < xn−1 < xn

• The Riemann sum is given by∑ni=1 f(ci)4xi, where 4xi = xi−xi−1, and

ci ∈ [xi−1, xi]. ci represents the sample points, or evaluation points withineach and every subinterval, where the height is taken for the rectangularapproximation of the funky strip.

• There are multiple rules for how to choose the sample points, e.g., left,right, midpoint, Monte Carlo, etc.

The proof involves setting up a Riemann sum such that the sample points, ci,for each subinterval were chosen as the points guaranteed by the Mean ValueTheorem for the antiderivative G(x) =

∫f(x) dx for each subinterval [xi−1, xi].

That is,

G′(ci) =G(xi)−G(xi−1)

xi − xi−1

Recognizing that G′(ci) = f(ci) and xi − xi−1 = 4xi, the conclusion of theMVT becomes

f(ci) =G(xi)−G(xi−1)

4xi



Replacing this in the formula for the Riemann sum, we have

Rn =n∑i=1

f(ci)4xi

=n∑i=1

G(xi)−G(xi−1)4xi

4xi

=n∑i=1

G(xi)−G(xi−1)

= G(x1)−G(x0)+G(x2)−G(x1)+G(x3)−G(x2)...+G(xn)−G(xn−1)

= G(xn)−G(x0)Rn = G(b)−G(a)

Note that if we choose our sample points this way, then the value we get forRn, G(b) − G(a), is independent of n, the number of subintervals. So that wecan have three, six, one hundred, one million, or even one subinterval, and weget the same value for the Riemann sum, which is therefore the same value forthe definite integral. ∫ b

a

f(x) dx = limn→∞

Rn

= limn→∞

G(b)−G(a)

= G(b)−G(a)

Example 7.6.7 (adapted from BC ’97) The closed interval [a, b]is partitioned into n equal subintervals, each of width ∆x, by thenumbers x0, x1, . . . , xn where a = x0 < x1 < x2 < · · · < xn−1 <

xn = b. What is limn→∞

n∑i=1

3√xi∆x?

[Ans:

34(b4/3 − a4/3

)]Mr. Budd, compiled September 29, 2010


7.6.5 Evaluating Definite Integrals

Example 7.6.8 (adapted from AB ’97)

(a) Find

i∫ 2

1

(4x3 − 6x2

)dx

ii∫ 3

1

(4x3 − 6x2

)dx

iii∫ b

1

(4x3 − 6x2

)dx

iv∫ x

1

(4t3 − 6t2

)dt

(b) What do you suppose is the meaning of

i1

2− 1

∫ 2

1

(4x3 − 6x2

)dx

ii1

3− 1

∫ 3

1

(4x3 − 6x2

)dx

iii1

b− 1

∫ b

1

(4x3 − 6x2

)dx

(c) Let A(x) =∫ x

1

(4t3 − 6t2

)dt. Find A′(x).

[Ans: 1, 28, b4 − 2b3 + 1, ; ; 4x3 − 6x2

]Example 7.6.9 (AB97) If

∫ baf(x) dx = a+2b, then

∫ ba

(f(x) + 5) dx =

[Ans: 7b− 4a]

Example 7.6.10 (adapted from Acorn ’05) Oil is leaking from atanker at the rate of R(t) = 2 000e−0.2t gallons per hour, where t ismeasured in hours after noon.


1R(t) dt.


1R(t) dt using

(a) M4, a midpoint Riemann sum with 4 equal subintervals;(b) T4, a trapezoidal approximation with 4 equal subintervals;

(c) S8 =2M4 + T + 4

3[Ans: 6490.959; 6621.211; 6534.376]



(c) Find the exact amount of oil that leaked out between 1 p.m.and 9 p.m.

[Ans: 10 000

(e−0.2 − e−1.8

)= 6534.319 gallons

](d) Find 1

9−1

∫ 9

1R(t) dt. Using correct units, explain the meaning

of this value. You can think of this as (signed) area divided bywidth. [Ans: 816.790 gal/hr; average of the rate of leakage from 1 p.m. to 9 p.m.]

(e) Let L(T ) be the amount of oil that has leaked out from 1 p.m.to time t = T . Write an integral expression for L(T ). Find aformula for L(T ) that does not require an integral, then findL′(T )

[Ans:

∫ T1R(t) dt = 10 000

(e−0.2 − e−0.2T

); 2 000e−0.2T

]

Problems

In your big giant calculus book, do problems 1-19 odd on page 390.

7.F-1 (adapted from Acorn ’05) Oil is leaking from a tanker at the rate of R(t) =2 000e−0.2t gallons per hour, where t is measured in hours after noon.


1R(t) dt.


1R(t) dt using a Riemann sum, using the partition

a = x0 = 1 < x1 = 5 < x2 = 9, and sample points c1 = 3 and c2 = 7.How were these sample points chosen?

(c) Approximate∫ 9

1R(t) dt using a Riemann sum, using the partition

a = x0 = 1 < x1 = 5 < x2 = 9, and evaluation points c1 = 2.86737and c2 = 6.86737. Refer to your homework on the Mean ValueTheorem to determine how these sample points were chosen.

(d) Approximate∫ 9

1R(t) dt using a Riemann sum with only one rectan-

gle, using the evaluation point c1 = 4.477603. What is the height ofthis rectangle?

[Ans: 6363.269 gal; 6534.319 gal; 6534.319 gal, 816.790 gal/hr]

7.F-2 (from [2]) Use a right-hand Riemann sum with four equal subdivisions to

approximate the integral∫ 3

−1

|2x− 3| dx [Ans: 8]

7.F-3 (from [2]) When∫ 5

−1

√x3 − x+ 1 dx is approximated by using the mid-

points of three rectangles of equal width, then the approximation is nearestto [Ans: 22.9]

7.F-4 [2]∫ 1

0

sin (πx) dx =[Ans:

2π



7.F-5 What is the area of the region above the x-axis, below the graph of

y = sec2 (πx), between the lines x = 0 and x =13

? What do you

suppose is the significance of dividing this area by the width,(

13 − 0

)?[

Ans:√

3π ; average height

]




Unit 8

Integral Theorems

1. Average Value and Mean Value Theorem for Integrals

2. Accumulation Functions and Fundamental Theorem of Calculus part II

3. Curve Sketching with Accumulation Functions

Advanced Placement



• Definite integral as the rate of change of a quantity over an interval interpretedas the change of the quantity over the interval:∫ b

a

f ′(x) dx = f(b)− f(a)

• Basic properties of definite integrals. (Examples include additivity and linear-ity.)

Applications of integrals. Appropriate integrals are used in a variety of applica-tions to model physical, social, or economic situations. Although only a samplingof applications can be included in any specific course, students should be able toadapt their knowledge and techniques to solve other similar application problems.Whatever applications are chosen, the emphasis is on using the integral of a rate ofchange to give accumulated change or using the method of setting up an approxi-mating Riemann sum and representing its limit as a definite integral. To provide acommon foundation, specific applications should include finding the area of a region,

235

236 AP Unit 8 (Integral Theorems)

the volume of a solid with known cross sections, the average value of a function,and the distance traveled by a particle along a line.

Fundamental Theorem of Calculus.


• Use of the Fundamental Theorem to represent a particular antiderivative, andthe analytical and graphical analysis of functions so defined.

Numerical approximations to definite integrals. Use of Riemann and trapezoidalsums to approximate definite integrals of functions represented algebraically, geo-metrically, and by tables of values.


AP Unit 8, Day 1: MVT for Integrals 237

8.1 Average Value and Mean Value Theorem forIntegrals

Advanced Placement

Applications of Integrals Appropriate integrals are used in a variety of applicationsto model physical, biological, or economic situations. The emphasis is on using theintegral of a rate of change to give accumulated change or using the method ofsetting up an approximating Riemann sum and representing the limit as a definiteintegral. To provide a common foundation, specific applications should include theaverage value of a function.



• Definite integral of the rate of change of a quantity over an interval interpretedas the change of the quantity over the interval:∫ b

a

f ′(x) dx = f(b)− f(a)

• Basic properties of definite integrals. (Examples include additivity and linear-ity.)

Fundamental Theorem of Calculus

• Use of Fundamental Theorem to evaluate definite integrals.



• Antiderivatives by substitution of variables (including change of limits fordefinite integrals).

8.1.1 Substitution of Variables

Example 8.1.1 (adapted from Acorn ’05) The area of the regionin the first quadrant between the graph of y = x

√9− x2 and the

x-axis is



[Ans: 9]

Example 8.1.2 (AB ’97)∫ π

4

0 cos2 xdx is

8.1.2 Properties of Definite Integrals

8.1.3 Average Value

Suppose we were to average the following numbers: 17, 7, 5, 11. We would add17 + 7 + 5 + 11, and then divide by 4. Note that 4 is the same as 1 + 1 + 1 + 1.So, what we have is essentially

17 + 7 + 5 + 111 + 1 + 1 + 1

= 10

Also note that 17 is 7 more than the average, 7 is 3 less, 5 is 5 less, and 11 is 1more. Overall there is a total of 8 above the average, and 8 below the average.

In general, to find the average, we take

n∑i=1

yn

n∑i=1

1

.

Now consider averaging a continuous function f(x) over an interval [a, b]. Thereare an infinite number of points to average, so how do we add an infinite numberof numbers? ∫ b

af(x) dx∫ ba

1 dx

Definition 8.1 (Average Value). The average value of a function f over theinterval [a, b] is given by∫ b

af(x) dxb− a

=1

b− a

∫ b

a

f(x) dx

The average value is basically the average height of the function over an interval.We get this by taking the “area” (the definite integral), and dividing by thewidth (b− a).



Example 8.1.3 (adapted from AB97) What is the average valueof cosx on the interval [−2, 6]?

[Ans: sin 6+sin 2

8

]Example 8.1.4 Graphically estimate the average value of sec2 (πx)

over the interval[0,

13

]. Confirm your answer analytically.

Example 8.1.5 (adated from BC Acorn ’02) If f is a continuousfunction for all real x, then what is lim

h→0

1h

∫ a+h

af(x) dx?

[Ans: f(a)]

Substitution of Variables

Example 8.1.6 (adapted from AB ’98) Graphically estimate themean value of y = x2

√x3 + 1 on the interval [−1, 2]. Confirm or

reject analytically.

Average Rate of Change given the derivative

If you are given f ′(x), the derivative of f with respect to x, then the averagerate of change of f is the same as the average value of f ′. The average rate ofchange is the average value of the rate of change, i.e., the average value of thederivative.

Example 8.1.7 (adapted from BC93) Ifdy

dx=

1x

, what is theaverage rate of change of y with respect to x on the closed interval[1, e]?

[Ans: 1

e−1

]Average Value of f : ∫ b

af(x) dxb− a



Average Rate of Change of f :

f(b)− f(a)b− a

=

∫ baf ′(x) dxb− a

Connection between the average value of a function and the Funda-mental Theorem of Calculus

Recall the proof of the Fundamental Theorem of Calculus that shows that∫ baG′(x) dx = G(b) − G(a). The proof involved setting up a Riemann sum

such that the sample points, ci, for each subinterval were chosen as the pointsthat satisfied the Mean Value Theorem for the antiderivative G(x) =

∫G′(x) dx

for each subinterval [xi−1, xi]. That is,


xi − xi−1

Recognizing that xi − xi−1 = 4xi, the MVT becomes


4xiReplacing this in the formula for the Riemann sum, we have

Rn =n∑i=1

G′(ci)4xi

=n∑i=1

G(xi)−G(xi−1)4xi

4xi

=n∑i=1

G(xi)−G(xi−1)

= G(x1)−G(x0)+G(x2)−G(x1)+G(x3)−G(x2)...+G(xn)−G(xn−1)

= G(xn)−G(x0)Rn = G(b)−G(a)

Note that if we choose our sample points this way, then the value we get forRn, G(b) − G(a), is independent of n, the number of subintervals. So that we



can have three, six, one hundred, one million, or even one subinterval, and weget the same value for the Riemann sum, which is therefore the same value forthe definite integral.

Suppose that we choose one rectangle instead of three, or six, or one hundred.If we choose our one sample point as above, so that the Mean Value Theorem issatisfied for the antiderivative, then our one rectangle will have exactly the samearea as the definite integral. Both shapes have the same “area” and width, butthe rectangle has a constant height, and the definite integral has a (possibly)variable height. Since the one constant height gives the same signed area as thevariable height, that one height behaves similarly to all the combined variableheights. This one constant height that mimics our changing heights is said tobe the average or mean height.

The proof of the Fundamental Theorem basically involves taking sample pointssuch that the height of each rectangle is equal to the average value of the functionover the corresponding subinterval.

Finding the average value in some ways involves replacing the funky shape underf(x) with a rectangle of exactly the same “area” and width. The rectangle hasa height equal to the average value of the function over the interval. When com-paring the funky shape to the rectangle, parts sticking out above the rectangleexactly match the parts drooping below the rectangle in size.

8.1.4 Mean Value Theorem for Integrals

Theorem 8.1. Mean Value Theorem for Integrals If f(x) is continuous overthe interval [a, b], then there exists a number c ∈ [a, b] such that

f(c) (b− a) =∫ b

a

f(x) dx

or

f(c) =1

b− a

∫ b

a

f(x) dx

Example 8.1.8 (BC93) If f is continuous on the closed interval[a, b], then there exists c such that a < c < b and

∫ baf(x) dx =

(A)f(c)b− a

(B)f(b)− f(a)

b− a(C) f(b)− f(a)



(D) f ′(c) (b− a)

(E) f(c) (b− a)

[Ans: E]

Example 8.1.9 Oil is leaking from a tanker at the rate of R(t) =2 000e−0.2t gallons per hour, where t is measured in hours after noon.

(a) Find1

3− 1

∫ 3

1

R(t) dt. Explain the meaning of this value. [Ans: 1349.596]

(b) Find the average rate at which oil leaked out of the tanker from3 p.m. to 9 p.m. Indicate units. [Ans: 639.188 gal/hr]

(c) At what time between 1 p.m. and 3 p.m. is the instantaneousrate of leakage the same as the average rate of leakage over thatsame time interval?

(d) At what time between 3 p.m. and 9 p.m. is the instantaneousrate of leakage the same as the average rate of leakage over thatsame time interval?

[Ans: 2 000e−0.2t, rate of leakage; t = 1.96671 (1:58); t = 5.70352 (5:42)

]Connection between the Mean Value Theorem for Integrals and theaverage value of a function

Note that f(c) is the average (or mean) value of the function f(x) over theinterval [a, b].

The Mean Value Theorem for Integrals is little more than applying the MeanValue Theorem to the antiderivative.

If we look at the Mean Value Theorem for G(x), the antiderivative of G′(x),over [a, b]:

G′(c) =G(b)−G(a)

b− aThis is similar to the approach we take in the proof of the Fundamental The-orem, except that we only have one subinterval. Now we learned (from theFundamental Theorem) that G(b) − G(a) is the same as

∫ baG′(x) dx. Using

these gives us:

G′(c) =

∫ baG′(x) dxb− a

,



which is the signed area divided by the width, so that G′(c) is the average valueof G′(x) over the interval [a, b].

If we let f(x) = G′(x), i.e., G(x) is the antiderivative of f(x), then

f(c) =

∫ baf(x) dxb− a

at some point c ∈ (a, b).

Problems

8.A-1 [2]π/3∫π/4

sec2 x

tanxdx =

[Ans: ln

√3]

8.A-2 (Acorn ’05) If f is continuous for all x, which of the following integralsnecessarily have the same value as

∫ baf(x) dx?

I∫ b−a

0f(x+ a) dx

II∫ b+ca+c

f(x+ c) dx

III 12

∫ 2b

2af(x2

)dx

Use u-simplification, as well as geometric arguments. [Ans: I, III]

8.A-3 ([2]) The average value of sec2 x over the interval 0 ≤ x ≤ π

4is[Ans: 4

π

]8.A-4 ([2]) The average rate of change of the function f(x) =

∣∣x2 − 2 |x+ 2|∣∣

over the interval −3 < x < −1 is [Ans: −3]

8.A-5 ([2]) The average (mean) value of1x

over the interval 1 ≤ x ≤ e is[Ans: 1

e−1

]8.A-6 ([2]) The average value of f(x) = e2x + 1 on the interval 0 ≤ x ≤ 1

2is

[Ans: e]

8.A-7 (adapted from [2]) What is the average value of −2t3 + 6t2 + 4 over theinterval −1 ≤ t ≤ 1? [Ans: 6]

8.A-8 (adapted from BC97) Let f be a twice differentiable function such thatf(1) = −2 and f(4) = 7. Which of the following must be true for thefunction f on the interval 1 ≤ x ≤ 4?

I. The average rate of change of f is 3.



II. The average value of f is53

.

III. The average value of f ′ is 3.

[Ans: I and III only]

8.A-9 Draw a function that has an average value of 0 over the closed interval[−π, π].

8.A-10 (AB ’97) Let f be the function given by f(x) = x3 − 6x2 + p, where p isan arbitrary constant.

(a) Write an expression for f ′(x) and use it to find the relative maximumand minimum values of f in terms of p. Show the analysis that leadsto your conclusion. [Ans: max: f(0) = p, min: f(m = 4) = p− 32]

(b) For what values of the constant p does f have three distinct roots?[Ans: 0 < x < 32]

(c) Find the value of p such that the average value of f over the closedinterval [−1, 2] is 1.

[Ans: 23

4

]8.A-11 (AB ’96) The rate of consumption of cola in the United States is given by

S(t) = Cekt, where S is measured in billions of gallons per year and t ismeasured in years from the beginning of 1980.

(a) The consumption rate doubles every 5 years and the consumptionrate at the beginning of 1980 was 6 billion gallons per year. Find Cand k.

[Ans: C = 6, k = 1

5 ln 2]

(b) Find the average rate of consumption of cola over the 10-year timeperiod beginning January 1, 1983. Indicate units of measure.[Ans: 19.680 billion gallons/year]

(c) Use the trapezoidal rule with four equal subdivisions to estimate∫ 7

5S(t) dt. [Ans: 27.668]

(d) Using correct units, explain the meaning of∫ 7

5S(t) dt in terms of

cola consumption.

[Ans: amount, in billions of gallons, of cola consumed in the two year period from 1/1/85 to 1/1/87]

8.A-12 (AB) A particle moves along the x-axis so that its velocity at time t,0 ≤ t ≤ 5, is given by v(t) = 3 (t− 1) (t− 3). At time t = 2, the positionof the particle is x(2) = 0. Find the average velocity of the particle overthe interval 0 ≤ t ≤ 5. [Ans: 4]

8.A-13 (AB ’88) Without a calculator, find the average value of y =x

x2 + 2on

the interval[0,√

6].

[Ans: ln 2√

6

]8.A-14 Using a calculator only to graph the function, exactly find the average

value of y = −√

1− x2 + 1 over the interval [−1, 1].[Ans: 4−π

4



8.A-15 (AB ’85) Let f(x) = 14πx2 and g(x) = k2 sin(πx

2k

)for k > 0.

(a) Find the average value of f on [1, 4].

(b) For what value of k will the average value of g on [0, k] be equal tothe average value of f on [1, 4]?

[Ans: 98π; 7π]

8.A-16 [3] Which of the following statements is true?

I. If the graph of a function is always concave up, then the left-handRiemann sums with the same subdivisions over the same interval arealways less than the right-hand sums.

II. If the function f is continuous on the interval [a, b] and∫ baf(x) dx =

0, then f must have at least one zero between a and b.

III. If f ′(x) > 0 for all x in an interval, then the function f is concaveup in that interval.

[Ans: II only]

8.A-17 Oil is leaking from a tanker at the rate of R(t) = 2 000e−0.2t gallons perhour, where t is measured in hours after noon.

(a) Find1

5− 1

∫ 5

1

R(t) dt. Explain the meaning of this value. [Ans: 1127.128]

(b) Find the average rate at which oil leaked out of the tanker from 5p.m. to 9 p.m. Indicate units. [Ans: 506.451 gal/hr]

(c) Find the average rate at which oil leaked out of the tanker from 1p.m. to 9 p.m. Indicate units. [Ans: 816.790 gal/hr]

(d) At what time between 1 p.m. and 5 p.m. is the instantaneous rate ofleakage the same as the average rate of leakage over that same timeinterval?

(e) At what time between 5 p.m. and 9 p.m. is the instantaneous rate ofleakage the same as the average rate of leakage over that same timeinterval?

(f) At what time between 1 p.m. and 9 p.m. is the instantaneous rate ofleakage the same as the average rate of leakage over that same timeinterval?

[Ans: t = 2.86737 (2:52); t = 6.86737 (6:52); t = 4.477603 (4:29)]

8.A-18 Sometimes the Mean Value Theorem for Integrals is called the AverageValue Theorem. Explain why, in that case, the Mean Value Theoremproper could be called the Average Value Theorem for Derivatives.




AP Unit 8, Day 2: Accumulation Functions 247

8.2 Accumulation Functions

Advanced Placement



Textbook §4.4 The Fundamental Theorem of Calculus: “The Second Funda-mental Theorem of Calculus” [16]

8.2.1 Accumulation Functions

Example 8.2.1 (AB98)∫ x

0sin t dt =

[Ans: 1− cosx]

Example 8.2.2 (AB97) Let f(x) =∫ xah(t) dt, where h has the

Figure 8.1:

graph shown in Figure 8.1. Sketch the graph of f .

[Ans: Figure 8.2]

8.2.2 Fundamental Theorem of Calculus, part II

d

dx

∫ x

a

f(t) dt = f(x)



Figure 8.2:

As soon as you see an accumulation function, differentiate it. Stop immediately,do not continue reading the problem until you have taken the derivative of theaccumulation function. It is very easy to do, and the chances are extremelyhigh that you will need the derivative. You can start reading the problemagain, realizing that you know the derivative.

Example 8.2.3 (AB98) If F (x) =∫ x

0

√t3 + 1 dt, then what is

F ′(2)?

[Ans: 3]

Example 8.2.4 [20] The error function

erf(x) =2√π

∫ x

0

e−t2dt

is used in probability, statistics, and engineering. Show that thefunction y = ex

2erf(x) satisfies the differential equation y′ = 2xy +

2√π

.

Example 8.2.5 (BC93) If F and f are differentiable functionssuch that F (x) =

∫ x0f(t) dt, and if F (a) = −2 and F (b) = −2

where a < b, which of the following must be true?

(A) f(x) = 0 for some x such that a < x < b.

(B) f(x) > 0 for all x such that a < x < b.

(C) f(x) < 0 for all x such that a < x < b.

(D) F (x) ≤ 0 for all x such that a < x < b.



(E) F (x) = 0 for some x such that a < x < b.

[Ans: A]

Using the Chain Rule,

d

dx

∫ v(x)

a

f(t) dt = f (v(x)) v′(x)

Example 8.2.6 (BC97) Let f(x) =∫ x2

0sin t dt. At how many

points in the closed interval [0,√π] does the instantaneous rate of

change of f equal the average rate of change of f on that interval?

[Ans: Two]

Example 8.2.7 [18] Let

F (x) = 2x+∫ x2

0

sin 2t1 + t2

dt

Determine

(a) F (0)

(b) F ′(0)

(c) F ′′(0)

[Ans: 0;2;2]

To carry extend part II even further,

d

dx

∫ v(x)

u(x)

f(t) dt = f (v(x)) v′(x)− f (u(x))u′(x)

Example 8.2.8 Let F (x) =∫ 3x

x/2

1tdt. Find F ′(x) and explain your

answer by simplifying F (x).

[Ans: 0; ln 6]



Example 8.2.9 [20] Find the derivative of g(x) =∫ 3x

2x

u2 − 1u2 + 1

du.

[Ans:

3(9x2−1)9x2+1 − 2(4x2−1)

4x2+1

]

8.2.3 Curve Sketching with Accumulation Functions

Example 8.2.10 (BC97) The graph of f is shown in Figure 8.3. If

Figure 8.3:

g(x) =∫ xaf(t) dt, for what value of x does g(x) have a maximum?

[Ans: c]

Example 8.2.11 (BC97) Refer to the graph in Figure 8.4. At

Figure 8.4: The function f is defined on the closed interval [0, 8]. The graph ofits derivative f ′ is shown.

what value does the absolute minimum of f occur? The absolutemaximum?



[Ans: 0]

Example 8.2.12 (AB Acorn ’02) If the function g is defined byg(x) =

∫ x0

sin(t2) dt on the closed interval −1 ≤ x ≤ 3, then g has alocal minimum at x =

[Ans: 2.507]

Example 8.2.13 (AB 2002) The graph of function f shown in

Figure 8.5: From AP Calculus AB 2002 Exam

Figure 8.5. Let g be the function given by g(x) =∫ x

0f(t) dt.

(a) Find g(−1), g′(−1), and g′′(−1).[Ans: − 3

2 ; 0; 3]

(b) For what values of x in the open interval (−2, 2) is g increasing?Explain your reasoning. [Ans: −1 < x < 1]

(c) For what values of x in the open interval (−2, 2) is the graphof g concave down? Explain your reasoning. [Ans: 0 < x < 2]

(d) Sketch the graph of g on the closed interval [−2, 2].

Problems

8.B-1 (AB ’97) Let f be the function given by f(x) =√x− 3.

(a) Sketch the graph of f and shade the region R enclosed by the graphof f , the x-axis, and the vertical line x = 6.

(b) Find the area of the region R.(c) Rather than using the line x = 6, consider the line x = w, where

w can be any number greater than 3. Let A(w) be the area of theregion enclosed by the graph of f , the x-axis, and the vertical linex = w. Write an integral expression for A(w).



(d) Find the rate of change of A with respect to w when w = 6.[Ans: ; 2

√3;∫ w

3

√x− 3 dx;

√3]

8.B-2 (AB ’93) What isd

dx

∫ x0

cos(2πu) du? [Ans: cos(2πx)]

8.B-3 (adapted from [2]) Suppose F (x) =∫ x2

0

12 + t3

dt for all real x, then

F ′(−1) =[Ans: − 2

3

]8.B-4 (adapted from [2]) Consider the function F defined so that F (x) +

52

=∫ x2

cos(πt

3

)dt. The value of F (2) + F ′(2) is [Ans: −3]

8.B-5 (adapted from [2]) If the function G is defined for all real numbers byG(x) =

∫ 3x

0cos(t2)dt, then G′(

√π) = [Ans: −3]

8.B-6 (adapted from [2]) Suppose F (x) =∫ 2 sin x

0

√9 + t3 dt for all real x, then

F ′(π) = [Ans: −6]

8.B-7 (adapted from [2]) If for all x > 0, G(x) =∫ x

1cos(π

2ln t)dt, then the

value of G′′(e) is[Ans: − π

2e

]8.B-8 [2] Which of the following are true about the function F (x) =

∫ x1

ln (2t− 1) dt?

I. F (1) = 0II. F ′(1) = 0

III. F ′′(1) = 1


8.B-9 (AB ’94) Let F (x) =∫ x

0sin(t2)dt for 0 ≤ x ≤ 3.

(a) Use the trapezoidal rule with four equal subdivisions of the closedinterval [0, 1] to approximate F (1). [Note: you may use a calculator,but don’t use a program.]

(b) On what intervals is F increasing?(c) If the average rate of change of F on the closed interval [1, 3] is k,

find∫ 3

1sin(t2)dt in terms of k. [

Ans: 0.316; (0,√π),

(√2π, 3

); 2k

]8.B-10 [20] The sine integral function

Si (x) =∫ x

0

sin tt

dt

is important in electrical engineering. Using your calculator to help you:



(a) Draw the graph of Si .

(b) At what values of x does this function have local maximum values?

(c) Find the coordinates of the first inflection point to the right of theorigin.

(d) Does this function have horizontal asymptotes?

(e) Solve the following equation:∫ x

0

sin tt

dt = 1

8.B-11 (AB 2002B) The graph of a differentiable function f on the closed interval

Figure 8.6: From AP Calculus AB 2002 Exam

[−3, 15] is shown in Figure 8.6. The graph of f has a horizontal tangentline at x = 6. Let g(x) = 5 +

∫ x6f(t) dt for −3 ≤ x ≤ 15.

(a) Find g(6), g′(6), and g′′(6). [Ans: 5; 3; 6]

(b) On what intervals is g decreasing? Justify your answer. [Ans: [−3, 0] and [12, 15]]

(c) On what intervals is the graph of g concave down? Justify youranswer. [Ans: (6, 15)]

(d) Find a trapezoidal approximation of∫ 15

−3f(t) dt using six subintervals

of length ∆t = 3. [Ans: 12]

8.B-12 [2] The graph of the function f is shown in Figure 8.7. If the function Gis defined by G(x) =

∫ x−4f(t) dt, for −4 ≤ x ≤ 4, which of the following

statements about G are true?

I. G is increasing on (1, 2)

II. G is decreasing on (−4,−3)

III. G(0) < 0


8.B-13 (BC Acorn ’02) Let g be the function given by g(x) =∫ x

1100

(t2 − 3t+ 2

)e−t

2dt.

Which of the following statements about g must be true?




I. g is increasing on (1, 2).

II. g is increasing on (2, 3).

III. g(3) > 0.

[Ans: II only]

8.B-14 (BC Acorn ’02) The graph of f in Figure 8.8 consists of four semicircles.


If g(x) =∫ x

0f(t) dt, where is g(x) nonnegative? [Ans: [−3, 3]]

8.B-15 (BC ’98) Let g(x) =∫ xaf(t) dt, where a ≤ x ≤ b. Figure 8.9 shows the

Figure 8.9:



graph of g on [a, b]. Sketch the graph of f on [a, b].

Ans:




AP Unit 8, Day 3: Quick, Cheap Antiderivatives 257

8.3 Quick, Cheap Antiderivatives

Advanced Placement



8.3.1 Creating Quick, Cheap Antiderivatives

Suppose I have an initial value problem, such as finding velocity given acceler-ation. Remember that velocity is the antiderivative of acceleration. In doingthese problems, we antidifferentiate, and then we plug in a value so that we canfind the value of the arbitrary constant C. The antiderivative of f(x), whichwe will call G(x), given the point G(c) = k can be written in one step as anaccumulation function:

G(x) =∫ x

c

f(t) dt+ k

Note that this function satisfies our two conditions, namely:

• G′(x) = f(x) by the Fundamental Theorem of Calculus part II

• G(c) = k since∫ ccf(t) dt = 0

Example 8.3.1 (BC93) If p is a polynomial of degree n, n > 0,what is the degree of the polynomial Q(x) =

∫ x0p(t) dt

[Ans: n+ 1]

Example 8.3.2 (AB97) At time t ≥ 0, the acceleration of a particlemoving on the x-axis is a(t) = t+ sin t. At t = 0, the velocity of theparticle is −2. For what value of t will the velocity be zero?

[Ans: 1.48]

Example 8.3.3 (adapted from AB ’03) A particle moves alongthe x–axis so that at any time t > 0, its acceleration is given bya(t) = ln (1 + 2t). If the velocity of the particle is 2 ln 3− 1 at timet = 1, then the velocity of the particle at time t = 2 is



[Ans: 2.544]

Example 8.3.4 (adapted from BC ’97) If f is the antiderivative ofx3

2 + x5such that f(1) = 0, then f(4) is what?

[Ans: 0.577]

Example 8.3.5 Let A(x) =∫ x−1

√t3 + 1 dx.

(a) Find the average rate of change of A over [0, 2].

(b) Find the value of c guaranteed by the Mean Value Theorem forA(x) over [0, 2].

[Ans: 1.621; 1.176]

Example 8.3.6 The graph of f ′, the derivative of the function f ,

Figure 8.10: From AP Calculus BC 2003 Exam

is shown in Figure 8.10. If f(0) = 0, which of the following must betrue?

(a) f(0) > f(1)

(b) f(2) > f(3)

(c) f(1) > f(3)


AP Unit 8, Day 3: Quick, Cheap Antiderivatives 259

Problems

8.C-1 (BC ’97) If f is the antiderivative ofx2

1 + x5such that f(1) = 0, then f(4)

is what?

[Ans: 0.376]

8.C-2 (BC Acorn ’04-05) If the function f is defined by f(x) =√x3 + 2 and g is

an antiderivative of f such that g(3) = 5, then what is g(1)? [Ans: −1.585]

8.C-3 (adapted from BC ’03) A particle moves along the x–axis so that at anytime t ≥ 0, its velocity is given by v(t) = cos

(2− t2

). The position of

the particle is 3 at time t = 0. What is the position of the particle at thesecond time when its velocity is equal to zero? [Ans: 3.563]

8.C-4 (adapted from [2]) The rate at which ice is melting in a pond is given bydV

dt=√

1 + 2t, where V is the volume of ice in cubic feet, and t is thetime in minutes. What amount of ice has melted in the first 4 minutes?[Ans: 9.645 ft3

]8.C-5 (adapted from [2]) Oil is leaking from a tanker at the rate of R(t) =

500e−0.2t gallons per hour, where t is measured in hours. The amount ofoil that has leaked out, starting at the end of the second hour, until theend of the tenth hour is

8.C-6 [2] A particle moves along the x-axis with velocity at time t given byv(t) = t+ 2 sin t. If the particle is at the origin when t = 0, its position atthe time when v = 5 is x = [Ans: 17.277]

8.C-7 (adapted from AB ’98) Let F (x) be an antiderivative of(lnx)2

x. If F (2) =

10, then what is F (9)? [Ans: 13.425]

8.C-8 (adapted from AB Acorn ’04-05) If f ′(x) = cos(πex

2

)and f(0) = 2, then

f(1) = [Ans: 1.351]

8.C-9 Let C(x) be the antiderivative of n(x) =1√2πe−x

2/2 such that C(0) =12

.

(a) Find C(2).

(b) Find limx→∞ C(x)

[Ans: 0.97725; 1]

8.C-10 (adapted from BC ’03) A particle starts at point A on the positive x-axis attime t = 0 and travels along the x-axis. The particle’s position x(t) is a dif-

ferentiable function of t, where x′(t) =dx

dt= −9 cos

(πt

6

)sin(π√t+ 12

).



At time t = 9, the particle reaches its final destination at point D on thepositive x-axis. How far apart are points A and D, the initial and finalpositions, respectively, of the particle? [Ans: 39.255 apart]

8.C-11 (adapted slightly from AB ’07) The amount of water in a storage tank, ingallons, is modeled by a continuous function on the time interval 0 ≤ t ≤ 7,where t is measured in hours. In this model, rates are given as follows:

(a) The rate at which water enters the tank is f(t) = 100t2 sin√t gallons

per hour for all seven hours.

(b) The rate at which water leaves the tank is: 250 gallons per hour forthe first three hours, and 2000 gallons per hour for the next fourhours.

At t = 0, the amount of water in the tank is 5000 gallons.

(a) On the same graph, sketch the rate at which water enters the tank,together with the rate at which water leaves the tank. At what timesare the two rates the same?

(b) Using correct units, explain the meaning of∫ 7

0

f(t) dt in terms of

water in the tank.

(c) How many gallons of water are there in the tank at t = 0? t = 3?t = 7?

[Ans: ; ; 5000, 5126.591, 4513.807]

8.C-12 (adapted from AB ’07B) A particle moves along the x–axis so that itsvelocity v at time t ≥ 0 is given by v(t) = sin

(t2). The position of the

particle at time t is x(t) and its position at time t = 0 is x(0) = 5.

(a) Find the position of the particle at t = 0 and at t = 3. [Ans: 5, 5.774]

(b) Find the first time at which the particle changes direction. Does itchange from right to left, or left to right? Find the position of theparticle at that time.

[Ans: 1.772, 5.895]


Unit 9

Area and Volume

1. Area

2. Slicing with Discs

3. Slicing with Washers

4. Non-circular Slicing

5. Volume-Oriented Related Rates

Advanced Placement

Interpretations and properties of definite integrals



• Basic properties of definite integrals.

Applications of integrals Appropriate integrals are used in a variety of applica-tions to model physical, biological, economic situations. Although only a samplingof applications can be included in any specific course, students should be able toadapt their knowledge and techniques to solve other similar application problems.Whatever applications are chosen, the emphasis is on using the integral of a rate ofchange to give accumulated change or using the method of setting up an approxi-mating Riemann sum and representing its limit as a definite integral. To provide acommon foundation, specific applications should include finding the area of a region,the volume of a solid with known cross sections, the average value of a function,

261

262 AP Unit 9 (Area and Volume)

and the distance traveled by a particle along a line. Fundamental Theorem ofCalculus




• Antiderivatives by substitution of variables (including change of limits fordefinite integrals)

Numerical approximations to definite integrals. Use of Riemann and trape-zoidal sums to approximate definite integrals of functions represented algebraically,graphically, and by tables of values.


AP Unit 9, Day 1: More Definite Integrals 263

9.1 More Definite Integrals

Advanced Placement

Applications of Integrals Appropriate integrals are used in a variety of applicationsto model physical, biological, or economic situations. The emphasis is on using theintegral of a rate of change to give accumulated change or using the method ofsetting up an approximating Riemann sum and representing the limit as a definiteintegral. To provide a common foundation, specific applications should include find-ing the area of a region, the average value of a function, and the distance traveledby a particle along a line.

9.1.1 Definite Integral

The definite integral can be seen as several things:

1. An infinite sum. Area is an infinite sum of infinitesimally thin rectan-gles.

2. A product. Displacement is velocity times time. How is a product reallya sum?

3. An accumulated change. If oil is leaking out of a tank at certain rateR(t), then

∫ baR(t) dt represents how much oil has leaked out from t = a to

t = b. This is the accumulation of all the oil that has leaked out at a rateR(t). Change in velocity is the accumulation of accleration over a certaintime period.

9.1.2 Area: Slicing dx

The area between two curves is a sum of an infinite number of infinitesimallythin rectangles.

If you slice the area into vertical strips, then the area of each infinitesimally thinrectangle is given by dA = (highy − lowy) dx, and the area is given by

Area =∫ b

a

(highy − lowy) dx

Example 9.1.1 (adapted from Ostebee & Zorn [17]) Find the areaof the region R bounded by the curves x = 0, y = 2, and y = ex.



(a) Find the area of R, with and without a calculator.

(b) If the line x = h divides the region R into two regions of equalarea, what is the value of h?

[Ans: 0.386; 0.219 ; 1.683]

Example 9.1.2 (adapted from AB ’00) Let R be the region in thefirst quadrant enclosed by the graphs of y = e−x

2, y = 1−cosx, and

the y-axis.

(a) Find the area of R.

(b) If the line x = k divides the region R into two regions of equalarea, what is the value of k?

[Ans: 0.591; 0.310]

Example 9.1.3 Find the area of the region bounded by the graphs

of y = ex/2, y =1x2

, x = 2, and x = 3. Try this with, and withouta calculator.

[Ans: 2

(e3/2 − e

)− 1

6 = 3.360]

Example 9.1.4 (adapted from AB ’98) If 0 ≤ k ≤ π

2and the area

under the curve y = cosx from x = k to x =π

2is 0.2, then k =

[Ans: 0.927]

9.1.3 Total Distance

Distance : Displacement :: Area : Definite Integral

There is a difference between the total distance traveled and the displacement.When you go backward, distance is counted positively, but displacement iscounted negatively. How is that similar to the relationship between area andthe definite integral?



Speed is the magnitude (absolute value of velocity), and total distance trav-eled is the accumulation (i.e., definite integral) of speed. Displacement is theaccumulation of velocity.

D =∫ ba|v(t)| dt

Just like area, total distance traveled must be positive.

In a graph of velocity, the definite integral yields the displacement, i.e., changein position. The area yields the total distance traveled.

Example 9.1.5 (adapted from AB 1997) A bug begins to crawl upa vertical wire at time t = 0. The velocity v of the bug at time t,0 ≤ t ≤ 8, is given by the function whose graph is shown in Figure9.1

Figure 9.1: Vertical velocity of a bug

(a) At value of t does the bug change direction?(b) What is the total distance the bug traveled from t = 0 to t = 8?(c) What is the net displacement of the bug between t = 0 and

t = 8?(d) What is the total distance traveled downward by the bug? up-

ward?(e) What is the bug’s velocity at t = 5? The acceleration at t = 5?(f) What other questions could we ask about the bug?

Example 9.1.6 (AB ’03) A particle moves along the x-axis so thatits velocity at time t is given by

v(t) = − (t+ 1) sin(t2

2

)At time t = 0, the particle is at position x = 1.



(a) Find the acceleration of the particle at time t = 2. Is the speedof the particle increasing at t = 2? Why or why not?

(b) Find all times t in the open interval 0 < t < 3 when the particlechanges direction. Justify your answer.

(c) Find the total distance traveled by the particle from time t = 0until time t = 3.

(d) During the time interval 0 ≤ t ≤ 3, what is the greatest distancebetween the particle and the origin? Show the work that leadsto your answer.

[Ans: 1.588, yes;

√2π; 4.334; 2.265

]Example 9.1.7 (adapted slightly from AB ’83) A particle movesalong the x-axis so that at time t its position is given by x(t) =t3 − 6t2 + 9t+ 11.

(a) What is the velocity of the particle at time t?

(b) During what time intervals is the particle moving to the left?

(c) What is the total distance traveled by the particle from t = 0to t = 2? Do this two ways.

(d) What does∫ 2

0v(t) dt represent?

[Ans: 3t2 − 12t+ 9; 1 < t < 3; 6; x(2)− x(0)

]

9.1.4 Other Applications

Example 9.1.8 (adapted minimally from AB ’06) At an inter-section in Thomasville, Oregon, cars turn left at the rate L(t) =

60√t sin2

(t

3

)cars per hour over the time interval 0 ≤ t ≤ 9 hours.

(a) To the nearest whole number, find the total number of carsturning left at the intersection over the time interval 0 ≤ t ≤ 9hours.

(b) Traffic engineers will consider turn restrictions when L(t) ≥ 100cars per hour. Find all values of t for which L(t) ≥ 100 andcompute the average value of L over this time interval. Indicateunits of measure.



Example 9.1.9 (adapted minimally from AB ’05) The tide removessand from Sandy Point Beach at a rate modeled by the function R,given by

R(t) = 2 + 4 cos(

4πt25

)A pumping station adds sand to the beach at a rate modeled by thefunction S, given by

S(t) =12t

1 + 3t

Both R(t) and S(t) have units of cubic yards per hour and t ismeasured in hours for 0 ≤ t ≤ 6. At time t = 0, the beach contains2000 cubic yards of sand.

(a) Find the rate at which sand is being removed by the tide attime t = 4. Find the rate at which sand is being added by thepumping station at time t = 4.

(b) How much sand will the tide removed from the beach duringthis 6–hour period? What is the average rate at which the tideis removing sand from the beach during this period? Indicateunits of measure.

(c) Write an expression for the amount of sand removed by the tideduring the first t hours.

(d) How much sand will the pumping station add to the beachduring this 6–hour period? What is the average rate at whichthe pumping station is adding sand to the beach during thisperiod? Indicate units of measure.

(e) Write an expression for the amount of sand added by the pump-ing station during the first t hours.

(f) Write an expression for Y (t), the total number of cubic yardsof sand on the beach at time t. Find Y ′(t).

(g) Find the rate at which the total amount of sand on the beachis changing at time t = 4.

(h) For 0 ≤ t ≤ 6, what times are the best candidates for amountof sand on the beach to be an (absolute) maximum? minimum?How would you choose amongst the candidates?

Problems

9.A-1 (adapted from AB ’98) If 0 ≤ k ≤ π

2and the area under the curve y = cosx

from x = k to x =π

2is 0.2, then k = [Ans: 0.927]



9.A-2 (AB ’96) Let R be the region in the first quadrant under the graph of

y =1√x

for 4 ≤ x ≤ 9.[Ans: 2; 25

4

]9.A-3 (AB ’02) Let f and g be the functions given by f(x) = ex and g(x) = lnx.

Find the area of the region enclosed by the graphs of f and g between

x =12

and x = 1. [Ans: 1.223]

9.A-4 (AB acorn ’02) What is the average value of the function f(x) = e−x2

onthe closed interval [−1, 1]? [Ans: 0.747]

9.A-5 [2] The average rate of change of the function f(x) =∫ x

0

√1 + cos (t2) dt

over the interval [1, 3] to three decimal places is: [Ans: 0.858]

9.A-6 (adapted from [2]) The approximate average rate of change of the functionf(x) =

∫ x0

sin(t2)dt over the interval [2, 5] is [Ans: −0.092]

9.A-7 (BC97) If 0 ≤ x ≤ 4, of the following, which is the greatest value of x suchthat

∫ x0

(t2 − 2t

)dt ≥

∫ x2t dt?

(A) 1.35

(B) 1.38

(C) 1.41

(D) 1.48

(E) 1.59

[Ans: B]

9.A-8 (adapted from AB ’04) Traffic flow is defined as the rate at which carspass through an intersection, measured in cars per minute. The trafficflow at a particular intersection is modeled by the function F defined by

F (t) = 80 + 6 cos(t

2

)for 0 ≤ t ≤ 30.

where F (t) is measured in cars per minute and t is measured in minutes.

(a) To the nearest whole number, how many cars pass through the in-tersection during the middle twenty minutes, i.e., for 5 ≤ t ≤ 25?

(b) What is the average traffic flow during the middle twenty minutes?Indicate units of measure.

(c) What is the average rate of change of traffic flow during the middletwenty minutes? Indicate units of measure.

[Ans: 1592 cars; 79.601 cars per minute; 0.540 cars per minute per minute]



9.A-9 (AB ’05B) A water tank at Camp Newton holds 960 gallons of water attime t = 0. During the time interval 0 ≤ t ≤ 18 hours, water is pumpedinto the tank at the rate of

W (t) = 76√t cos2

(t

6

)gallons per hour.

During the same time interval, water is removed from the tank at the rate

R(t) = 220 cos2

(t

3

)gallons per hour.

(a) At what rate is water being removed from the tank at t = 15? Indi-cate units of measure.

(b) Is the amount of water in the tank increasing at t = 15? Why or whynot? [Ans: Y;]

(c) How much water has been pumped into the tank during the firstfifteen hours? What is the average rate at which water is pumped intothe tank during that time period? [Ans: 933.042 gal; 62.203 gph]

(d) How much water has been removed from the tank during the firstfifteen hours? What is the average rate at which water is removedfrom the tank during that time period?

(e) To the nearest gallon, how much water is in the tank at t = 15?t = 18? [Ans: 333; ]

(f) Write an expression for the gallons of water in the tank at time t.

(g) Write an expression for the rate at which the total gallons of waterin the tank is changing at time t.

[Ans: 960 +

∫ t0W (z)−R(z) dz

](h) Find the best candidates for the time, 0 ≤ t ≤ 18 at which the amount

of water in the tank is an absolute minimum. [Ans: 2.487, 12.450]

(i) Find the best candidates for the time, 0 ≤ t ≤ 18 at which the amountof water in the tank is an absolute maximum. [Ans: 0, 6.198, 18]

(j) How would you choose between the best candidates to determine theactual time in which the water is at an absolute extremum?

9.A-10 (BC Acorn 2000) A particle moves along the x-axis so that at any timet ≥ 0 its velocity is given by v(t) = ln (t+ 1)− 2t2 + 4t− 1.

(a) What is the total distance traveled by the particle from t = 0 tot = 2? [Ans: 2.178]

(b) What is the net displacement of the particle between t = 0 and t = 2?[Ans: 1.963]

9.A-11 (AB ’87) A particle moves along the x-axis so that its acceleration at anytime t is given by a(t) = 6t−18. At time t = 0 the velocity of the particleis v(0) = 24, and at time t = 1 its position is x(1) = 20.



(a) Write an expression for the velocity v(t) of the particle at any timet.

(b) For what values of t is the particle at rest?

(c) Write an expression for the position of the particle at any time t.

(d) Find the total distance traveled by the particle from t = 1 to t = 3.[Ans: 3t2 − 18t+ 24; 2, 4; t3 − 9t2 + 24t+ 4; 6

]9.A-12 (AB ’93) A particle moves on the x-axis so that its position at any time

t ≥ 0 is given by x(t) = 2te−t. Find the total distance traveled by theparticle from t = 0 to t = 5. Try this using fnInt, then try this on yourcalculator, without calculating a definite integral. [Ans: 1.404]

9.A-13 (AB 1997) A particle moves along the x-axis so that its velocity at anytime t ≥ 0 is given by v(t) = 3t2− 2t− 1. The position x(t) is 5 for t = 2.


(b) For what values of t, 0 ≤ t ≤ 3, is the particle’s instantaneous velocitythe same as its average velocity on the closed interval [0, 3]?

(c) Find the total distance traveled by the particle from t = 0 until timet = 3. Try it with a calculator, then without a calculator.[

Ans: t3 − t2 − t+ 3; t = 1.786; 17]


AP Unit 9, Day 2: Area 271

9.2 Area

Advanced Placement

Applications of IntegralsAppropriate integrals are used in a variety of applications to model physical, bio-logical, or economic situations. The emphasis is on using the integral of a rate ofchange to give accumulated change or using the method of setting up an approxi-mating Riemann sum and representing the limit as a definite integral. To provide acommon foundation, specific applications should include finding the area of a region,the volume of a solid with known cross sections, the average value of a function,and the distance traveled by a particle along a line.

Fundamental Theorem of Calculus

• Use of Fundamental Theorem to evaluate definite integrals.

9.2.1 High and Low y Switch

At times, the graphs may intersect between the bounds of integration, and thehigh and low y functions might switch. In this case, you should divide theintegral up.

When you get to use the calculator, feel free to use abs

When you must use the fundamental theorem instead of a calculator to evaluatethe definite integral, the absolute value will be worthless. Why? In that case,you need to separate


of y = x, y =4x

, x = 1, and x = 4. Try this with

(a) without a calculator;

(b) with a calculator, but without using absolute value;

(c) with a calculator, using absolute value.

[Ans: 9

2 + 4 ln 4− 8 ln 2 = 92 = 4.500

]Example 9.2.2 (adapted from AB ’83) Do the following problemwithout a calculator, then with a calculator. Find the area bounded



by the curve f(x) = 3x2− 12x+ 9 and the x-axis, between the linesx = 0 and x = 2.

[Ans: 6]


of y =1x2

, y = x, and y = 2.

[Ans: 7

2 − 2√

2]

9.2.2 Area: Slicing dy

If you slice the area into vertical strips, then the area of each infinitesimally thinrectangle is given by dA = (rightx− leftx) dy, and the area is given by

Area =∫ b

a

(rightx− leftx) dy

Example 9.2.4 (adapted from Ostebee & Zorn [17]) Find the areaof the region R bounded by the curves x = 0, y = 2, and y = ex.

(a) Find the area of R, with and without a calculator.(b) If the line y = k divides the region R into two regions of equal

area, what is the value of k?

[Ans: 0.386; 0.219 ; 1.683]


of y =1x2

, y = x, and y = 2. Try this two ways.

[Ans: 7

2 − 2√

2]

9.2.3 Total Distance

Remember that

Distance : Displacement :: Area : Definite Integral



Example 9.2.6 (AB ’02B) A particle moves along the x-axis sothat its velocity v at any time t, for 0 ≤ t ≤ 16, is given by v(t) =e2 sin t − 1. At time t = 0, the particle is at the origin.

(a) Sketch the graph of v(t) for 0 ≤ t ≤ 16.

(b) During what intervals of time is the particle moving to the left?Give a reason for your answer.

(c) Find the total distance traveled by the particle from t = 0 tot = 4. Do this using absolute value, and without using absolutevalue.

(d) Is there any time t, 0 < t ≤ 16, at which the particle returnsto the origin? Justify your answer.

[Ans: ; (π, 2π), (3π, 4π), (5π, 16]; 10.542, no]

Problems

9.B-1 (AB ’93) Set up a definite integral to find the area of the shaded regionin Figure 9.2.

[Ans:

∫ ba

(d− f(x)) dx]

Figure 9.2: AP Calculus AB (1993)

9.B-2 (AB ’96) [NO CALCULATOR] Let R be the region in the first quadrant

under the graph of y =1√x

for 4 ≤ x ≤ 9.


(b) If the line x = k divides the region R into two regions of equal area,what is the value of k? [

Ans: 2; 254



9.B-3 (adapted from AB ’97) [NO CALCULATOR] The area of the region en-closed by the graph of y = x2 − 2x+ 2 and the line y = 10 is [Ans: 36]

9.B-4 (adapted from AB ’98) [NO CALCULATOR] What is the area of theregion between the graphs of y = x3 and y = −x from x = 0 to x = 3?[Ans: 99

4

]9.B-5 (AB ’02B) Let R be the region bounded by the y–axis and the graphs of

y =x3

1 + x2and y = 4− 2x. Find the area of R. [Ans: 3.215]

9.B-6 (adapted from BC Acorn) Find the area of the region R bounded by thecurves x = 1, y = 1, and y = e3x.

(a) Find the area of R, by slicing dx, then again by slicing dy.

(b) If the vertical line x = h divides the region R into two regions ofequal area, what is the value of h?

(c) If the horizontal line y = k divides the region R into two regions ofequal area, what is the value of k?

[Ans: 5.362; 0.814 ; 5.065]

9.B-7 Find the area enclosed by the ellipsex2

16+y2

9= 1. What is area in the

form kπ? What do you suppose is the area enclosed by any generic ellipsex2

a2+y2

b2= 1? [Ans: 12π]

9.B-8 (AB ’02) Let f and g be the functions given by f(x) = ex and g(x) = lnx.Find the area of the region enclosed by the graphs of f and g between

x =12

and x = 1. [Ans: 1.223]

9.B-9 (AB ’03) Let R be the region bounded by the graphs of y =√x and

y = e−3x and the vertical line x = 1. Find the area of R. [Ans: 0.443]

9.B-10 (adapted from AB ’03B) Let f be the function given by f(x) = 4x2 − x3,and let ` be the line y = 18 − 3x, where ` is tangent to the graph of f .Let R be the region bounded by the graph of f and the x-axis, and let Sbe the region bounded by the graph of f , the line `, and the x-axis.

(a) At what point is ` tangent to f(x)?

(b) Find the area of R.

(c) Find the area of S. [Ans: (3, 0); 64

3 ;7.917]

9.B-11 (AB ’05)



Let f and g be the functions given by f(x) =14

+sin (πx) and g(x) = 4−x.Let R be the shaded region in the first quadrant enclosed by the y–axisand the graphs of f and g, and let S be the shaded region in the firstquadrant enclosed by the graphs of f and g, as shown in the figure above.


(b) Find the area of S.

[Ans: 0.0648; 0.410]

9.B-12 (AB ’05B) Let f and g be the functions given by f(x) = 1 + sin (2x) andg(x) = ex/2. Let R be the region in the first quadrant enclosed by thegraphs of f and g. Find the area of R. [Ans: 0.429]

9.B-13 (adapted from AB ’86) A particle moves along the x-axis so that at any

time t ≥ 1 its acceleration is given by a(t) =1t. At time t = 1, the velocity

of the particle is v(1) = −2 and its position is x(1) = 4.

(a) Find the velocity at time t = 9.

(b) What is the position at time t = 9?

(c) What is the total distance traveled from t = 1 to t = 9?

(d) How far did the particle travel backwards, starting from t = 1?

[Ans: ln 9− 2; −0.225; 4.553; 4.389]

9.B-14 (AB ’00) Two runners, A and B, run on a straight racetrack for 0 ≤ t ≤10 seconds. Figure 9.3, which consists of two line segments, shows the



Figure 9.3: from AP Calculus AB 2000 exam

velocity, in meters per second, of Runner A. The velocity, in meters per

second, of Runner B is given by the function v defined by v(t) =24t

2t+ 3.

Find the total distance run by Runner A and the total distance run byRunner B over the time interval 0 ≤ t ≤ 10 seconds. Indicate unitsof measure. Then find Runner B’s distance without using a calculator.Recall what to do when antidifferentiating and the degree on top is thesame or higher as on bottom.

[Ans: 85 m; 83.336 m; 120− ln

(233

)]


AP Unit 9, Day 3: Volume 277

9.3 Volume: Sweet, Sweet Loaves of Calculus

Advanced Placement

Applications of integrals Appropriate integrals are used in a variety of applica-tions to model physical, biological, economic situations. Although only a samplingof applications can be included in any specific course, students should be able toadapt their knowledge and techniques to solve other similar application problems.Whatever applications are chosen, the emphasis is on using the integral of a rate ofchange to give accumulated change or using the method of setting up an approxi-mating Riemann sum and representing its limit as a definite integral. To provide acommon foundation, specific applications should include finding the area of a region,the volume of a solid with known cross sections, the average value of a function,and the distance traveled by a particle along a line.

Numerical approximations to definite integrals. Use of Riemann and trape-zoidal sums to approximate definite integrals of functions represented algebraically,graphically, and by tables of values.

Resources §8.2 Finding Volumes by Integration in Ostebee and Zorn [17]. §6.2Volumes in Stewart [20]. §7.4 Volumes in Finney [8].

9.3.1 Volumes of Rotation

In general,

V =∫ b

a

A(x) dx

or

V =∫ d

c

A(y) dy

where A(x) or (A(y)) represents the cross-sectional area of the solid at a par-ticular value of x (or y)

For volumes of rotation where cross sections are discs, then A(x) = π [r(x)]2 orA(y) = π [r(y)]2.

Example 9.3.1 (adapted from AB 1997) Let R be the regionbounded by the y-axis, the line y = 2, and the curve y =

√x.

(a) Find the area of region R.

(b) Find the volume of the solid generated when region R is rotatedabout the y-axis.



[Ans: 8

3 ; 32π5

]Things to keep in mind for volume of rotation problems:

• If the axis of rotation is the x-axis, or parallel to the x-axis, slice dx. Ifthe axis of rotation is the y-axis, or parallel to the y-axis, slice dy.

• V =∫ baπr2d

• If slicing dy, the radius will be a high x minus a low x. If slicing dx, theradius will be a high y minus a low y. If the axis of rotation is either thex- or y-axis, one of these values will be zero.

Example 9.3.2 [3] A region in the plane is bounded by y =1√x

,

the x-axis, the line x = m, and the line x = 2m where m > 0. Asolid is formed by revolving the region about the x-axis. The volumeof this solid

(A) is independent of m

(B) increases as m increases

(C) decreases as m decreases

(D) increases until m =12

, then decreases

(E) is none of the above

[Ans: A]

Example 9.3.3 Derive, from scratch, the formula for the volumeof a sphere of radius r.

Example 9.3.4 Derive, from scratch, the formula for the volumeof a cone with height h and base radius r.

Example 9.3.5 [17] The following table gives the circumference (ininches) of a pole at several heights (in feet).

Height 0 10 20 30 40 50 60Circumference 16 14 10 5 3 2 1

Assuming that cross sections of the pole taken parallel to the groundare circles, estimate the volume of the pole using:


AP Unit 9, Day 3: Volume 279

(a) T3

(b) M3

(c) T6

(d) S2·3 =M3 + 2T3

3, a weighted average of M3 and T3.

Nonstandard axes of rotation

Remember that each radius is a high y minus a low y (or high x minus low x).

Example 9.3.6 Let R be the region bounded by the graphs ofy = ex/2, y = 1, and x = ln 2.

(a) Set up, but do not solve, a definite integral that could be usedto find the area of R.

(b) Set up, but do not solve, a definite integral that could be usedto find the volume of the region obtained by rotating R aboutthe line y = 1.

(c) Set up, but do not solve, a definite integral that could be usedto find the volume of the region obtained by rotating R aboutthe line x = ln 2.

(d) Preview: what changes if the axes of rotation are y = −1 orx = 0?

Problems

9.C-1 (adapted from AB ’03B) Let R be the region bounded by the graph off(x) = 4x2 − x3 and the x-axis. Find the volume of the solid generatedwhen R is revolved about the x-axis. [Ans: 490.208]

9.C-2 [20] A log 10 m long is cut at 1-meter intervals and its cross-sectional areasA (at a distance x from the end of the log) are listed in the table.

x (m) A (m2) x (m) A (m2)0 0.68 6 0.531 0.65 7 0.552 0.64 8 0.523 0.61 9 0.504 0.58 10 0.485 0.59

Estimate the volume of the log using:



(a) the Midpoint Rule with n = 5.[Ans: 5.8 m3

](b) the Trapezoid Rule with n = 5.

[Ans: 5.7 m3

](c) the Trapezoid Rule with n = 10.

[Ans: 5.75 m3

](d) S2·5 =

M5 + 2T10

3, a weighted average ofM5 and T10.

[Ans: 5.767 m3

]9.C-3 [8] We wish to estimate the volume of a flower vase using only a calculator,

a string, and a ruler. We measure the height of the vase to be 6 inches.We then use the string and the ruler to find the circumference of the vase(in inches) at half-inch intervals. (We list them from starting at the top

left, moving down)

Circumferences5.4 10.84.5 11.64.4 11.65.1 10.86.3 9.07.8 6.39.4

(a) Sketch the vase.

(b) Find the areas of the cross sections that correspond to the givencircumferences.

(c) Express the volume of the vase as an integral with respect to y overthe interval [0, 6].

(d) Approximate the integral using the Trapezoidal Rule with n = 12.[Ans: 2.3, 1.6, 1.5, . . .; 1

4π

∫ 6

0[C(y)]2 dy; 34.7 in3

]9.C-4 (AB 1993) [No calculator! ]The region enclosed by the x-axis, the line

x = 3, and the curve y =√x is rotated about the x-axis. What is the

volume of the solid generated?[Ans: 9

2π]


AP Unit 9, Day 4: Volume: Slicing with Washers 281

9.4 Volume: Slicing with Washers

Advanced Placement

Applications of integrals Appropriate integrals are used in a variety of applica-tions to model physical, biological, economic situations. Although only a samplingof applications can be included in any specific course, students should be able toadapt their knowledge and techniques to solve other similar application problems.Whatever applications are chosen, the emphasis is on using the integral of a rate ofchange to give accumulated change or using the method of setting up an approxi-mating Riemann sum and representing its limit as a definite integral. To provide acommon foundation, specific applications should include finding the area of a region,the volume of a solid with known cross sections, the average value of a function,and the distance traveled by a particle along a line.

9.4.1 Slicing with Washers

For washers,

A(x) =[π (R(x))2 − π (r(x))2

]= π

[(R(x))2 − (r(x))2

]or

A(y) =[π (R(y))2 − π (r(y))2

]= π

[(R(y))2 − (r(y))2

]Remember that each radius is [high y - low y].

Example 9.4.1 Let R be the region bounded by the graphs ofy = ex/2, y = 1, and x = ln 2. Set up, but do not solve, a definiteintegral that could be used to find the volume of the region obtainedby rotating R about

(a) the y-axis;(b) the line y = −1;(c) the line x = −1;(d) the line x = 1;(e) the line y =

√2

Example 9.4.2 ([3]) Let R be the region in the first quadrantbounded above by the graph of f(x) = 2 arctanx and below by thegraph of y = x. What is the volume of the solid generated when Ris rotated about the x-axis?



[Ans: 7.151]

Example 9.4.3 (AB ’02) Let f and g be the functions given byf(x) = ex and g(x) = lnx. Find the volume of the solid generated

when the region enclosed by the graphs of f and g between x =12

and x = 1 is revolved about the line y = 4.

[Ans: 23.609]

Example 9.4.4 (AB ’97) Let f be the function given by f(x) =3 cosx. As shown in Figure 9.4, the graph of f crosses the y-axis atpoint P and the x-axis at point Q.

Figure 9.4: AB Exam 1997

(a) Write an equation for the line passing through points P and Q.

(b) Write an equation for the line tangent to the graph of f at pointQ. Show the analysis that leads to your equation.

(c) Find the x-coordinate of the point on the graph of f , betweenpoints P and Q, at which the line tangent to the graph of f isparallel to line PQ.

(d) Let R be the region in the first quadrant bounded by the graphof f and the line segment PQ. Write an integral expressionfor the volume of the solid generated by revolving the region Rabout the x-axis.


AP Unit 9, Day 4: Volume: Slicing with Washers 283

[Ans: y − 3 = − 6

π (x− 0); y − 0 = −3 (x− π/2); 0.690; π∫ π/2

0

[(3 cosx)2 −

(− 6πx+ 3

)2]dx]

Homework

For Monday, February 22, 2010, put # 1 on the front and #2,3 on the back.For Tuesday, February 23, 2010, put #4 on the front and #5,6 on the back.Each is due before the final bell at 3:42 p.m.

9.D-1 (adapted from AB ’00) Let R be the region in the first quadrant enclosedby the graphs of y = e−x

2, y = 1− cosx, and the y-axis.

(a) Find the volume of the solid generated when the region R is revolvedabout the x–axis.

(b) Find the volume of the solid generated when the region R is revolvedabout the line y = −1.

(c) Suppose R is revolved around the line y = k, where k > 1 so that theline is above the region. Find k if the volume of this solid is the sameas the volume of the solid in the previous part, where R is revolvedaround the line y = −1.

[Ans: 1.747; 5.460; 1.941]

9.D-2 (AB ’03) Let R be the region bounded by the graphs of y =√x and

y = e−3x and the vertical line x = 1. Find the volume of the solidgenerated when R is revolved about the horizontal line y = 1. [Ans: 1.424]

9.D-3 (AB Acorn ’04-05) Let S be the region enclosed by the graphs of y = 2xand y = 2x2 for 0 ≤ x ≤ 1. Write a definite integral for the vol-ume of the solid generated when S is revolved about the line y = 3.[Ans: π

∫ 1

0

((3− 2x2

)2 − (3− 2x)2)dx]

9.D-4 (adapted from AB ’02B) Let R be the region bounded by the y–axis and

the graphs of y =x3

1 + x2and y = 4 − 2x. Find the volume of the solid

generated when R is revolved about

(a) the x–axis.

(b) the line y = 4.

[Ans: 31.885;48.906]

9.D-5 (AB ’05B) Let f and g be the functions given by f(x) = 1 + sin (2x) andg(x) = ex/2. Let R be the region in the first quadrant enclosed by thegraphs of f and g. Find the volume generated when R is revolved aboutthe x–axis. [Ans: 4.267]



9.D-6 (AB ’05)

Let f and g be the functions given by f(x) =14

+sin (πx) and g(x) = 4−x.Let S be the shaded region in the first quadrant enclosed by the graphs of fand g, as shown in the figure above. Find the volume of the solid generatedwhen S is revolved about the horizontal line y = −1. [Ans: 4.558]


AP Unit 9, Day 5: Non-Circular Slicing 285

9.5 Flat-Bottomed Volume: Non-Circular Slic-ing

Advanced Placement

Interpretations and properties of definite integrals



Applications of integrals Appropriate integrals are used in a variety of applica-tions to model physical, biological, economic situations. Although only a samplingof applications can be included in any specific course, students should be able toadapt their knowledge and techniques to solve other similar application problems.Whatever applications are chosen, the emphasis is on using the integral of a rate ofchange to give accumulated change or using the method of setting up an approxi-mating Riemann sum and representing its limit as a definite integral. To provide acommon foundation, specific applications should include finding the area of a region,the volume of a solid with known cross sections, the average value of a function,and the distance traveled by a particle along a line.Numerical approximations to definite integrals. Use of Riemann and trape-zoidal sums to approximate definite integrals of functions represented algebraically,graphically, and by tables of values.

Figure 9.5: AB 1998

Example 9.5.1 (AB 1998) The base of a solid is a region in the firstquadrant bounded by the x-axis, the y-axis, and the line x+2y = 8,as shown in the Figure 9.5. If cross sections of the solid perpendicularto the x-axis are semicircles, what is the volume of the solid?

[Ans: 16.755]

Example 9.5.2 (AB 1997) The base of a solid S is the regionenclosed by the graph of y =

√lnx, the line x = e, and the x-axis.



If the cross sections of S perpendicular to the x-axis are squares,then the volume of S is

[Ans: 1]

Example 9.5.3 (BC 1997) The base of a solid is the region in thefirst quadrant enclosed by the graph of y = 2−x2 and the coordinateaxes. If every cross section of the solid perpendicular to the y-axisis a square, the volume of the solid is given by

[Ans:

∫ 2

0(2− y) dy

]

Example 9.5.4 Repeat for equilateral triangles, isosceles right tri-angles, etc.

Problems

For Friday, February 26, 2010, put # 1 on the front and #2,3 on the back.For Monday, March 1, 2010, put #4,5 on the front and #6 on the back.Each is due before the final bell at 3:42 p.m.

9.E-1 (AP ’96) Let R be the region in the first quadrant under the graph of

y =1√x

for 4 ≤ x ≤ 9. Find the volume of the solid whose base is

the region R and whose cross-sections cut by planes perpendicular to thex-axis are squares.

[Ans: ln 9

4

]9.E-2 (AB ’00) Let R be the region in the first quadrant enclosed by the graphs

of y = e−x2, y = 1 − cosx, and the y-axis. The region R is the base of

a solid. For this solid, each cross section perpendicular to the x–axis is asquare. Find the volume of the solid. [Ans: 0.461]

9.E-3 (adapted from AB ’02B) Let R be the region bounded by the y–axis and

the graphs of y =x3

1 + x2and y = 4 − 2x. The region R is the base of

a solid. For this solid, each cross section perpendicular to the x–axis is asemicircle. Find the volume of this solid. [Ans: 3.533]

9.E-4 (AB ’03)


AP Unit 9, Day 5: Non-Circular Slicing 287

Let R be the region bounded by the graphs of y =√x and y = e−3x and

the vertical line x = 1. The region R is the base of a solid. For this solid,each cross section perpendicular to the x–axis is a rectangle whose heightis 5 times the length of its base in region R. Find the volume of this solid.[Ans: 1.554]

9.E-5 (AB ’05B) Let f and g be the functions given by f(x) = 1 + sin (2x) andg(x) = ex/2. Let R be the region in the first quadrant enclosed by thegraphs of f and g. The region R is the base of a solid. For this solid, thecross sections perpendicular to the x–axis are semicircles with diametersextending from y = f(x) to y = g(x). Find the volume of this solid.[Ans: 0.078]

9.E-6 [20] A CAT scan produces equally spaced cross-sectional views of a humanorgan that provide information about the organ otherwise obtained onlyby surgery. Suppose that a CAT scan of a human liver shows cross-sectionsspaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas,in square centimeters, are 0, 18, 58, 79, 94, 106, 117, 128, 63, 39, and 0.

(a) Use the Midpoint Rule to estimate the volume of the liver.[Ans: 1110 cm3

](b) Use the Trapezoidal Rule to estimate the volume of the liver.[

Ans: 1053 cm3]




AP Unit 9, Day 6: Area and Volume 289

9.6 Area and Volume

Problems

For Tuesday, March 2, 2010, put #1 on the front and #2 on the back.For Thursday, March 4, 2010, put #3 on the front and #4 on the back.For Monday, March 8, 2010, put #5 on the front.For Tuesday, March 9, 2010, put #6 on the front.Each is due before the final bell at 3:42 p.m.

9.F-1 (2010–4) [NO CALCULATOR] Let R be the region in the first quadrantbounded by the graph of y = 2

√x, the horizontal line y = 6, and the

y–axis, as shown in the figure below.


(b) Write, but do not evaluate, an integral expression that gives thevolume of the solid generated when R is rotated about the horizontalline y = 7.

(c) Region R is the base of a solid. For each y, where 0 ≤ y ≤ 6, the crosssection of the solid taken perpendicular to the y–axis is a rectanglewhose height is 3 times the length of its base in region R. Write, butdo not evaluate, an integral expression that gives the volume of thesolid.



9.F-2 (2010B–1) In the figure above, R is the shaded region in the first quadrantbounded by the graph of y = 4 ln (3− x), the horizontal line y = 6, andthe vertical line x = 2.


(b) Find the volume of the solid generated when R is revolved about thehorizontal line y = 8.

(c) The region R is the base of a solid. For this solid, each cross sectionperpendicular to the x–axis is a square. Find the volume of the solid.



9.F-3 (2004B AB-1) Let R be the region enclosed by the graph of y =√x− 1,

the vertical line x = 10, and the x–axis.



(c) Find the volume of the solid generated when R is revolved about thevertical line x = 10.

9.F-4 (2005B AB-1)

Let f and g be the functions given by f(x) = 1 + sin(2x) and g(x) = ex/2.Let R be the shaded region in the first quadrant enclosed by the graphsof f and g as shown in the figure above.


(b) Find the volume of the solid generated when R is revolved about thex–axis.

(c) The region R is the base of a solid. For this solid, the cross sectionsperpendicular to the x–axis are semicircles with diameters extendingfrom y = f(x) to y = g(x). Find the volume of this solid.

9.F-5 (2008P AB–2)



Let R and S in the figure above be defined as follows: R is the region inthe first and second quadrants bounded by the graphs or y = 3− x2 andy = 2x. S is the shaded region in the first quadrant bounded by the twographs, the x–axis, and the y–axis.

(a) Find the area of S.

(b) Find the volume of the solid generated when R is rotated about thehorizontal line y = −1.

(c) The region R is the base of a solid. For this solid, each cross sectionperpendicular to the x–axis is an isosceles right triangle with one legacross the base of the solid. Write, but do not evaluate, an integralexpression that gives the volume of the solid.



9.F-6 (2007B AB–1)

Let R be the region bounded by the graph of y = e2x−x2and the horizontal

line y = 2, and let S be the region bounded by the graph of y = e2x−x2

and the horizontal lines y = 1 and y = 2, as shown above.



(c) Write, but do not evaluate, an integral expression that gives thevolume of the solid generated when R is rotated about the horizontalline y = 1.



9.F-7 (1999 AB-2)

The shaded region, R, is bounded by the graph of y = x2 and the liney = 4, as shown in the figure above.


(b) Find the volume of the solid generated by revolving R about thex–axis.

(c) There exists a number k, k > 4, such that when R is revolved aboutthe line y = k, the resulting solid has the same volume as the solid inpart (b). Write, but do not solve, an equation involving an integralexpression that can be used to find the value of k.

9.F-8 (2007 AB–1) LetR be the region in the first and second quadrants bounded

above by the graph of y =20

1 + x2and below by the horizontal line y = 2.


(b) Find the volume of the solid generated when R is rotated about thex–axis.

(c) The region R is the base of a solid. For this solid, the cross sectionsperpendicular to the x–axis are semicircles. Find the volume of thissolid.


AP Unit 9, Day 7: Related Rates with Volume 295

9.7 Related Rates with Volume

Advanced Placement




• Modeling rates of change, including related rates problems.

Textbook §2.6 Related Rates [16]

Resources §10-4 Related Rates in Foerster; Exploration 10-4

9.7.1 Volume problems

Example 9.7.1 (adapted from AB ’98) The radius of a circle isdecreasing at a constant rate of 0.2 centimeter per second. In termsof the circumference C, what is the rate of change of the area of thecircle, in square centimeters per second?

[Ans: −0.2C]

Example 9.7.2 (AB 2002) A container has the shape of an openright circular cone, as shown in Figure 9.6. The height of the con-tainer is 10 cm and the diameter of the opening is 10 cm. Water inthe container is evaporating so that its depth h is changing at theconstant rate of − 3

10 cm/hr.(The volume of a cone of height h and radius r is given by V =13πr2h.)

(a) Find the volume V of water in the container when h = 5 cm.Indicate units of measure.

[Ans: 125

12 π cm3]

(b) Find the rate of change of the volume of water in the container,with respect to time, when h = 5 cm. Indicate units of measure.[Ans: − 15

8 π cm3/hr]



Figure 9.6: From 2002 AP Calculus AB Exam

(c) Show that the rate of change of the volume of water in the con-tainer due to evaporation is directly proportional to the exposedsurface area of the water. What is the constant of proportion-ality?

[Ans: − 3

10

]Example 9.7.3 [2] The function V whose graph is sketched in Fig-ure 9.7 gives the volume of air, V (t), (measured in cubic inches) thata man has blown into a balloon after t seconds.(V =

43πr3

)The rate at which the radius is changing after 6 seconds is approxi-mately what?



AP Unit 9, Day 7: Related Rates with Volume 297

[Ans: 0.1 in/sec]

Problems

9.G-1 (adapted from Acorn AB ’00) If r is positive and increasing, for what valueof r is the rate of increase of r3 forty-eight times that of r? [Ans: 4]

9.G-2 [2] Let y = 2ecos x. Both x and y vary with time in such a way that yincreases at the constant rate of 5 units per second. The rate at which x

is changing when x =π

2is [Ans: −2.5 units/sec]

9.G-3 [2] When the area of an expanding square, in square units, is increasingthree times as fast as its side is increasing, in linear units, the side is[Ans: 3

2

]9.G-4 [2] Water is flowing into a spherical tank with 6 foot radius at the constant

rate of 30π cubic ft per hour. When the water is h feet deep, the volumeof water in the tank is given by

V =πh2

3(18− h)

What is the rate at which the depth of the water in the tank is increasingat the moment when the water is 2 feet deep? [Ans: 1.5 ft per hr]

9.G-5 [2] The edge of a cube is increasing at the uniform rate of 0.2 inches persecond. At the instant when the total surface area becomes 150 squareinches, what is the rate of increase, in cubic inches per second, of thevolume of the cube?

[Ans: 15 in3/sec

]9.G-6 (adapted from AB ’98) If the base b of a triangle is increasing at a rate of

3 inches per minute while its height h is decreasing at a rate of 6 inchesper minute, what relationship must exist between b and h for the area Aof the triangle to be decreasing?

[Ans: b > 1

2h]

9.G-7 [2] Sand is being dumped on a pile in such a way that it always forms acone whose base radius is always 3 times its height. The function V whosegraph is sketched in Figure 9.8 gives the volume of the conical sand pile,V (t), measured in cubic feet, after t minutes. At what approximate rateis the radius of the base changing after 6 minutes. [Ans: 0.22 ft/min]





Unit 10

Extrema and Optimization

1. Solving Inequalites

2. First Derivative Test

3. Points of Inflection

4. Second Derivative Test

5. Absolute Extrema

6. Curve Sketching

7. Optimization

Advanced Placement




Second derivatives.




299

300 AP Unit 10 (Extrema and Optimization)


• Optimization, both absolute (global) and relative (local) extrema.


AP Unit 10, Day 1: Absolute Extrema 301

10.1 Absolute Extrema

Advanced Placement



Textbook §3.7 Optimization Problems and §3.10 Business and Economic Ap-plications [16]

Resources §8-3 Maxima and Minima in Plane and Solid Figures in [10]

10.1.1 Absolute Extrema

First of all know the difference between absolute extrema and relative extrema

• absolute = global = higher (or lower) than everything

• relative = local = higher (or lower) than everything nearby

The absolute extremum (of a continuous function over a closed in-terval) will occur either at an endpoint or at a critical point.

To find absolute, aka global, extrema of f over [a, b]. [16]

1. Find the critical numbers of f in (a, b).

2. Evaluate f at each critical number in (a, b).

3. Evaluate f at the endpoints, i.e., at a and at b.

4. The lowest of these values is the absolute minimum, the highest is theabsolute maximum.

5. Answer the question that’s asked. Do you need an x-value, or a y-value?

Example 10.1.1 Find the absolute minimum and the absolutemaximum values of the following functions, over the following inter-vals:

(a) f(x) = πx2 (12− 3x) over [0, 3]

(b) f(x) = x

(200− 3

4x

)over [5, 146.25]



10.1.2 Absolute Extrema from the Derivative

Example 10.1.2 (adapted from AB ’02) Let H(t) be a functionfor 9 ≤ t ≤ 23 whose derivative is given by

H ′(t) =15600

t2 − 24t+ 160− 9890t2 − 38t+ 370

such that H(9) = 0.

(a) For what values of t, t ∈ [9, 23], is H(t) increasing? Justifyyour answer.

(b) Based on a graph of H ′(t), why are the endpoints not goodcandidates for the absolute maximum?

(c) What is the maximum value (i.e., the absolute maximum) ofH(t)? At what value of t does the maximum occur?

(d) Based on a graph of H ′(t), why is the critical point not a goodcandidate for the absolute minimum? How would you choosefrom the remaining candidates?

(e) At what time does the absolute minimum occur? What is theminimum value of H(t) over [9, 23].

(f) How many points of inflection will the graph of H(t) have?Justify your answer.

Example 10.1.3 (adapted from AB ’02B) The number of gal-lons, P (t), of a pollutant in a lake changes at the rate P ′(t) =

2 cos(

2πt24

)−3e−0.2

√t gallons per day, where t is measured in days.

There are 100 gallons of the pollutant in the lake at time t = 0. Thelake is considered to be safe when it contains 40 gallons or less ofpollutant.

(a) Find the critical values of P (t) over (0, 50).

(b) Find all values of x, x ∈ [0, 50], for which P (t) is increasing.

(c) Based on knowledge of the intervals for which P (t) is increas-ing or decreasing, which are the two best candidates for theabsolute minimum? Explain your reasoning.

(d) How would you decide which is smaller, P (20.489 . . .) or P (43.574 . . .)?

(e) Is the lake safe when the number of gallons of pollutant is atits absolute minimum? Justify your answer.

(f) Is the lake safe for both relative minima of P (t)?

(g) Based on knowledge of the intervals for which P (t) is increas-ing or decreasing, which are the three best candidates for theabsolute minimum? Explain your reasoning.



(h) What is the maximum number of gallons of pollutant over[0, 50]? At what time does it occur?

[Ans: 20.490 = c, 27.904 = d, 43.574 = e; [c, d], [e, 50]][Ans: lake is safe at t = e, but not at c]

Example 10.1.4 (adapted from AB ’00) Figure 10.1 shows the

Figure 10.1:

graph of f ′, the derivative of the function f , for −7 ≤ x ≤ 7. Thegraph of f ′ has horizontal tangent lines at x = −3, x = 2, and x = 5,and a vertical tangent line at x = 3.

(a) What are the critical values of f over (−7, 7)?

(b) Of the endpoints and critical points, which are good candidatesfor the absolute maximum? For the absolute minimum? Ex-plain your reasoning.

(c) Without being able to exactly find f(−5) or f(7), explain howyou can still use the fundamental theorem to decide which islarger.

(d) Which is smaller, f(−7) of f(−1)? Explain your reasoning.

(e) Find all values of x, x ∈ [−7, 7], for which f

(i) has relative maxima;(ii) has relative minima;(iii) has absolute maxima;(iv) has absolute minima;(v) is increasing;

(vi) is decreasing;(vii) is concave down;

(viii) is concave up;(ix) has a point of inflection.



10.1.3 Optimization

Example 10.1.5 Old Mac-Donald had a farm. And on that farm,he had some cows, which fought incessantly. In order to separatethe Cowpulets from the Montamoos, he built two identical, adjacentrectangular pens which share a side. Old Mac-Donald hires you tofind out the dimensions of the pen which give you the greatest areaif Old MacDonald has, say, 600 yards of fencing.

[Ans: 75 yards by 100 yards adjoining]

• Draw a (big) picture.

• Label variables.

• Write an expression for what you’re maximizing or minimizing.

• Write an equation for the constraints.

• Use the constraint to reduce your maximizing/ minimizing expression toone variable.

• Use the physical reality of the problem to determine end values.

• Differentiate to find critical values.

• Plug end and critical values into your maximizing/ minimizing expression.

• Answer the question that’s asked, not the question that you want to an-swer.

Technique: Analysis of Maximum-Minimum Problems [10]

1. Make a sketch if one isn’t already drawn.

2. Write an equation for the variable you are trying to maximize or minimize.

3. Get the equation in terms of one variable and specify a domain.

4. Find an approximate maximum or minimum by grapher.

5. Find the exact maximum or minimum by seeing where the derivative iszero or infinite. Check any endpoints of the domain.

6. Answer the question by writing what was asked for in the problem state-ment.



College Board does not recognize maxima or minima calculated on the graphingcalculator. Justifications require mathematical (i.e., noncalculator) reasoning.

Example 10.1.6 A 400 meter track is to be built around a fieldthat consists of a rectangle with two semicircles at either end. (Thebase of each semicircle spans the entire width of the rectangle.) Howshould the track be built in order to maximize

(a) the area of the rectangle;

(b) the total enclosed area.

[Ans: length 100 m, radius 100

π m ; circle of radius 200π m

]Example 10.1.7 [from Acorn ’02] Consider one arch of cosx abovethe x-axis. Draw a rectangle that lies on the x-axis so that its toptwo vertices lie on the curve y = cosx. Shade the area betweeny = cosx and the x-axis that is not in the rectangle. Find theminimum area of the shaded region.

[Ans: 0.878]

Example 10.1.8 [10] Barb Dwyer must build a rectangular corralalong the river bank. Three sides of the corral will be fenced witha barbed wire. The river forms the fourth side of the corral (Figure10.2). The total length of fencing available is 1000 feet. What is themaximum area the corral could have? How should the fence be builtto enclose this maximum area? Justify your answers.

[Ans: 125,000 square feet, 250 feet ⊥ to river by 500 feet parallel]

Example 10.1.9 [10] The part of the parabola y = 4 − x2 fromx = 0 to x = 2 is rotated about the y-axis to form a surface. Acone is inscribed in the resulting paraboloid with its vertex at theorigin and its base touching the parabola (Figure 10.3). At whichradius and altitude does the maximum volume occur? What is themaximum volume? Justify your answer.

[Ans: x =

√2, y = 2, V = 4π

3



Figure 10.2:

Figure 10.3:

Example 10.1.10 (adapted from BC93) Consider all right circularcylinders for which the sum of the height and the circumferenceis 30π centimeters. What is the radius of the one with maximumvolume?

[Ans: 10 cm]



Curve Sketching

Advanced Placement




Second derivatives.




Resources §4.3 Connecting f ′ and f ′′ with the Graph of f in [8]




AP Unit 10, Day 2: First Derivative 309

10.2 First Derivative

Advanced Placement




10.2.1 First Derivative Test

Example 10.2.1 Find all local extrema of f(x) =12x+ sinx over

the interval (0, 2π).

Example 10.2.2 Find all local extrema of g(x) = 3√

(x2 − 2x− 8)2.

Example 10.2.3 Show that the maximum value of h(x) = bxe−bx,b > 0 is independent of b.

Example 10.2.4 (MM 2004) Consider the function f(x) = 2 +1

x− 1.

(a) Sketch f(x)

(b) Write down the x-intercepts and y-intercepts of f(x).

(c) Write down the equations of the asymptotes of f(x).

(d) Find f ′(x)

(e) There are no maximum or minimum points on the graph off(x). Use your equation for f ′(x) to explain why.

Example 10.2.5 (MM 2003) Figure blah-blah-blah shows the graphof y = ex (cosx+ sinx), −2 ≤ x ≤ 3. The graph has a maximumturning point at C (a, b).

(a) Finddy

dx.

(b) Find the exact value of a and b.



Example 10.2.6 (adapted from MM 2002) Let the function f(x) =2

1 + x3, x 6= 1.

(a) Write down the equation of the vertical asymptote of the graphof f .

(b) Write down the equation of the horizontal asymptote of thegraph of f .

(c) Show that f ′(x) =−6x2

(1 + x3)2 .

(d) Determine the x-coordinates for which f

(a) is increasing;(b) is decreasing;(c) has a relative maximum;(d) has a relative minimum.

Problems

10.B-1 For the function h(x) = xe−x

(a) Find the critical values of h(x);

(b) Find all x-coordinates for which h is increasing;

(c) Find the x-coordinates of any relative maxima on the graph of h;

(d) Find the x-coordinates of any relative minima on the graph of h.

Justify all answers. [Ans: 1; x < 1; x = 1; ∅]

10.B-2 Exactly find all critical values of f(x) = x2 lnx and classify each as arelative minimum, a relative maximum, or neither. Justify your answer.[Ans: 1√

e, relative min

]


AP Unit 10, Day 3: Second Derivative 311

10.3 Second Derivative

Advanced Placement



Second derivatives.




Textbook §3.4 Concavity and the Second Derivative Test: [16]

10.3.1 Concavity

Recall concavity

If f is concave up, then:

• f is like a cup;

• f ′ is increasing (if f ′ exists);

• f ′′ is positive (if f ′′ exists);

• the curve will lie above the tangent line.

Example 10.3.1 (AB97) For what interval is the graph of y =x4 − 9x3 + 27x2 − 45x+ 36 is concave up?

[Ans: x < 3

2 or x > 3]

Example 10.3.2 For x2 + y2 = 16: Remember the Quotient Rule:For what values of x or y is the graph of x2 + y2 = 16 concave up?



Example 10.3.3 (MM 2003) Figure blah-blah-blah shows the graphof y = ex (cosx+ sinx), −2 ≤ x ≤ 3. The graph has a maximumturning point at C (a, b) and a point of inflection at D.

(a) Finddy

dx.

(b) Find the exact value of a and b.

(c) Show that at D, y =√

2eπ4 .

Example 10.3.4 (adapted from MM 2002) Let the function f(x) =2

1 + x3, x 6= 1.

(a) Write down the equation of the vertical asymptote of the graphof f .

(b) Write down the equation of the horizontal asymptote of thegraph of f .

(c) Using the fact that f ′(x) =−6x2

(1 + x3)2 , show that the second

derivative f ′′(x) =12x

(2x3 − 1

)(1 + x3)3 .

(d) Find the x-coordinates of the points of inflection of the graphof f .

(e) Use the trapezium rule with five subintervals to approximate

the integral3∫1

f(x) dx.

(f) Given that3∫1

f(x) dx = 0.637599, use a diagram to explain why

your answer is greater than this.

Points of Inflection

An inflection point is a point on the graph of f where the concavity changes.At inflection points on the graph of f ,

• f changes concavity;

• f ′ changes direction (points of inflection of f occur at extrema of f ′);

• f ′′ changes sign (points of inflection of f occur when f ′′ crosses the x-axis).



Be careful: for there to be an inflection point, it is not enough for the secondderivative to be zero. The second derivative must change sign. This is similarto finding relative extrema: it is not enough for the first derivative to be 0, thefirst derivative must change sign for there to be a relative extremum.

10.3.2 Second Derivative Test

Theorem 10.1 (Second Derivative Test for Local Extrema). Let c be a criticalvalue of f such that f ′(c) = 0 and f ′′(c) exists.

1. If f ′′(c) > 0, then f(c) is a relative minimum.

2. If f ′′(c) < 0, then f(c) is a relative maximum.

3. If f ′′(c) = 0, then the test fails. In such cases, you can use the FirstDerivative Test.

Example 10.3.5 Find the critical values of y = 5x3−3x5, and testeach value to decide whether it corresponds to a relative maximum,a relative minimum, or neither. Use the second derivative test.

[Ans: −1 (rel min); 0 (neither); 1 (rel max)]

For this problem, there is one point for which the second derivative test fails.Can you determine whether it is a local extremum merely from the informationyou’ve received from the second derivative test?

Example 10.3.6 Let f(x) = x+b

x, where b is a positive number.

Use the second derivative test to find and distinguish any relativeextrema.

[Ans: −

√b (rel max);

√b (rel min)

]

Example 10.3.7 (BC97) Where are the relative extrema of thefunction f given by f(x) = 3x5 − 4x3 − 3x.

[Ans: rel max at x = −1, rel min at x = 1]



Example 10.3.8 [3] Which of the following are true about thefunction f if its derivative is defined by

f ′(x) = (x− 1)2 (4− x)

I. f is decreasing for all x < 4.

II. f has a local maximum at x = 1.

III. f is concave up for all 1 < x < 3.

[Ans: III only]

Problems

In textbook, §3.4: # 11, 17, 19; 27-35 odd; §5.1:#71; §5.4:#53, 57

10.C-1 [3] At x = 0, which of the following is true of the function f(x) = sinx+e−x

(a) f is increasing

(b) f is decreasing

(c) f is discontinuous

(d) f is concave up

(e) f is concave down

[Ans: concave up]

10.C-2 (adapted from AB93) At what value of x does the graph of y =1x2

+1x3

have a point of inflection? [Ans: −2]

10.C-3 (adapted from [2]) Consider the function f(x) =(x2 − 5

)3 for all realnumbers x. At what x-coordinates are the inflection points for the graphof f?

[Ans: −

√5, −1, 1,

√5]

10.C-4 (adapted from [3]) The slope of the curve y = 32x

2 − e−x at its point ofinflection is [Ans: 3− ln 27]

10.C-5 (BC98) Let f be a function defined and continuous on the closed interval[a, b]. If f has a relative maximum at c and a < c < b, which of thefollowing statements must be true?

I. f ′(c) exists.

II. If f ′(c) exists, then f ′(c) = 0.

III. If f ′′(c) exists, then f ′′(c) ≤ 0.




10.C-6 (AB99) Suppose that the function f has a continuous second derivativefor all x, and that f(0) = 2, f ′(0) = −3, and f ′′(0) = 0. Let g be afunction whose derivative is given by g′(x) = e−2x (3f(x) + 2f ′(x)) for allx.

(a) Write an equation of the line tangent to the graph of f at the pointwhere x = 0. [Ans: y − 2 = −3 (x− 0)]

(b) Is there sufficient information to determine whether or not the graphof f has a point of inflection when x = 0? Explain your answer.[Ans: No]

(c) Given that g(0) = 4, write an equation of the line tangent to thegraph of g at the point where x = 0. [Ans: y = 4]

(d) Show that g′′(x) = e−2x (−6f(x)− f ′(x) + 2f ′′(x)). Does g have alocal maximum at x = 0? Justify your answer. [Ans: Yes; g′(x) = 0, g′′(x) < 0]





Second Derivative Test

Advanced Placement

Second derivatives.




Textbook §3.4 Concavity and the Second Derivative Test: “Concavity” and“The Second Derivative Test” [16]




Unit 11

Review

1. Solving Inequalites

2. First Derivative Test

3. Points of Inflection

4. Second Derivative Test

5. Absolute Extrema

6. Curve Sketching

7. Optimization

Advanced Placement




Second derivatives.




319

320 AP Unit 11 (Review)





11.1 Separable Differential Equations

Advanced Placement





• Solving separable differential equations and using them in modeling.


11.1.1 Separable Differential Equations

Definition 11.1. Differential Equation A differential equation is an equationthat contains the derivative of a function. [10]

Example 11.1.1 (2005B AB-6) [NO CALCULATOR] Consider

the differential equationdy

dx=−xy2

2. Let y = f(x) be the partic-

ular solution to this differential equation with the initial conditionf(−1) = 2.

(a) On the axes provided, sketch a slope field for the given differ-ential equation at the twelve points indicated.



(b) Write an equation for the line tangent to the graph of f atx = −1.

(c) Find the solution y = f(x) to the given differential equationwith the initial condition f(−1) = 2.

Example 11.1.2 (2002 AB–5) Consider the differential equationdy

dx=

3− xy

.

(a) Let y = f(x) be the particular solution to the given differentialequation for 1 < x < 5 such that the line y = −2 is tangent tothe graph of f . Find the x–coordinate of the point of tangency,and determine whether f has a local maximum, local minimum,or neither at this point. Justify your answer.

(b) Let y = g(x) be the particular solution to the given differentialequation for −2 < x < 8 with the initial condition g(6) = −4.Find y = g(x)

Example 11.1.3 The solution to the differential equationdy

dx=

2xe−y, where y(0) = 1, is

[Ans: y = ln

(x2 + e

)]Example 11.1.4 (adapted from AB acorn ’02) The solution to the


dx=x2


[Ans: y = 4

√43x

3 − 20]



Example 11.1.5 (adapted from [2]) If the graph of y = f(x) con-

tains the point (0,−1) and ifdy

dx=

2x sin(x2)

y, then f(x) =

[Ans: −

√3− 2 cos (x2)

]

11.1.2 Separable Differential Equations with Logs

Example 11.1.6 (AB97) Let v(t) be the velocity, in feet per second,of a skydiver at time t seconds, t ≥ 0. After her parachute opens,

her velocity satisfies the differential equationdv

dt= −2v − 32, with

initial condition v(0) = −50.

(a) Use separation of variables to find an expression for v in termsof t, where t is measured in seconds.

(b) Terminal velocity is defined as limt→∞

v(t). Find the terminalvelocity of the skydiver to the nearest foot per second.

(c) It is safe to land when her speed is 20 feet per second. At whattime t does she reach this speed?

[Ans: v = −34e−2t − 16, −16, 1.070

]

Example 11.1.7 (2004 AB-5) Consider the differential equationdy

dx= x2 (y − 1).




(b) While the slope field in part (a) is drawn at only twelve points,it is defined at every point in the xy–plane. Describe all pointsin the xy–plane for which the slopes are positive.

(c) Find the particular solution y = f(x) to the given differentialequation with the initial condition f(0) = 3.

Example 11.1.8 A turkey is cooking in the oven at 300 degreesFahrenheit. It starts out at room temperature (70 degrees). After1 hour, it is ? degrees. How long before it reaches 170 degrees, atwhich point it will be done. The rate of change in the temperature ofthe turkey is proportional to the difference between the temperaturesof the environment and the turkey.

Homework

For Wednesday, March 10, put 107 on the front and 108 on the back.For Thursday, March 11, put 109 on the front and 110 on the back.



11.A-107 (2005 AB–6) [NO CALCULATOR] Consider the differential equationdy

dx=

−2xy

.

(a) On the axes provided, sketch a slope field for the given differentialequation at the twelve points indicated.

(b) Let y = f(x) be the particular solution to the differential equationwith the initial condition f(1) = −1. Write an equation for the linetangent to the graph of f at (1,−1) and use it to approximate f(1.1).

(c) Find the particular solution y = f(x) to the given differential equa-tion with the initial condition f(1) = −1.

11.A-108 (2004B AB-5) Consider the differential equationdy

dx= x4 (y − 2).


(b) While the slope field in part (a) is drawn at only twelve points, itis defined at every point in the xy–plane. Describe all points in thexy–plane for which the slopes are negative.

(c) Find the particular solution y = f(x) to the given differential equa-tion with the initial condition f(0) = 0.



11.A-109 (2008–5) [NO CALCULATOR] Consider the differential equationdy

dx=

y − 1x2

, where x 6= 0.

(a) On the axes provided, sketch a slope field for the given differentialequation at the nine points indicated.

(b) Find the particular solution y = f(x) to the differential equationwith the initial condition f(2) = 0.

(c) For the particular solution y = f(x) described in part (b), findlimx→∞

f(x).

11.A-110 (2006 AB-5) [NO CALCULATOR] Consider the differential equationdy

dx=

1 + y

x, where x 6= 0.

(a) On the axes provided, sketch a slope field for the given differentialequation at the eight points indicated.

(b) Find the particular solution y = f(x) to the differential equationwith the initial condition f(−1) = 1 and state its domain.



11.2 Exponential Growth and Decay

Advanced Placement





• Solving separable differential equations and using them in modeling. In par-ticular, studying the equation y′ = ky and exponential growth.


Resources §7.2 Exponential Growth and Decay and §7.3 Other DifferentialEquations for Real-World Applications in [10]

11.2.1 Proportional Growth

Example 11.2.1 The rate growth of the population of Escherichiacoli is proportional to the number of E. coli. Find a general expres-sion for the population as a function of time if the initial populationis P0.

[Ans: P = P0e

kt]

Example 11.2.2 (adapted from AB98) Population y grows accord-

ing to the equationdy

dt= ky, where k is a constant and t is measured

in years. If the population doubles every 8 years, then the value ofk is

[Ans: 0.087]



Example 11.2.3 [10] Chemical Reaction Problem Calculus bud-dite (a rare substance) is converted chemically into Glamis thanus.Buddite reacts in such a way that the rate of change in the amountleft unreacted is directly proportional to that amount.

(a) Write a differential equation that expresses this relationship.Solve it to find an equation that expresses amount in terms oftime. Use the initial conditions that the amount is 50 mg whent = 0 min and 30 mg when t = 20 min.

[Ans: dB

dt = kB, B = 50 (0.6)t/20 = 50e−0.025541...t]

(b) Sketch the graph of amount versus time.

(c) How much buddite remains an hour after the reaction starts?[Ans: 10.8 mg]

(d) When will the amount of buddite equal 0.007 mg? [Ans: 5 hr 47 min]

11.2.2 Other Applications of Differential Equations

Example 11.2.4 Tin Can Leakage Problem [10] Suppose you fill atall (topless) tin can with water, then punch a hole near the bottomwith an ice pick. The water leaks quickly at first, then more slowlyas the depth of the water increases. In engineering or physics, youwill learn that the rate at which the water leaks out is directly pro-portional to the square root of its depth. Suppose that at time t = 0min, the depth is 12 cm and dy

dt is −3 cm/min.

(a) Write a differential equation stating that the instantaneous rateof change of y with respect to t is directly proportional to thesquare root of y. Find the proportionality constant.

(b) Solve the differential equation to find y as a function of t. Usethe given information to find the particular solution. Whatkind of function is this?

(c) Plot the graph of y as a function of t. Sketch the graph. Con-sider the domain of t in which the function gives reasonableanswers.

(d) Solve algebraically for the time at which the can becomes empty.Compare your answer with the time it would take at the initialrate of −3 cm/min.

[Ans: k = −3

(12−1/2

); y = 3

16 t2 − 3t+ 12; ;8 (twice as long)

]Example 11.2.5 Dam Leakage Problem [10] A new dam is con-structed across Scorpion Gulch. Engineers want to predict the amount



of water in the lake behind the dam as a function of time. At t = 0days the water starts flowing in at a fixed rate F ft3/hr. Unfortu-nately, as the water level rises, some leaks out. The leakage rate,L, is directly proportional to the amount of water, W ft3, presentin the lake. Thus the instantaneous rate of change of W is equal toF − L.

(a) What does L equal in terms of W? Write a differential equationthat expresses dW/dt in terms of F , W , and t.

[Ans: dW

dt = F − kW]

(b) Solve for W in terms of t, using the initial condition W = 0when t = 0.

[Ans: W = F

k

(1− e−0.04t

)](c) Water is known to be flowing in at F = 5000 ft3/hr. Based on

geological considerations, the proportionality constant in theleakage equation is assumed to be 0.04/hr. Write the equationforW , substituting these quantities.

[Ans: W = 125000

(1− e−0.004t

)](d) Predict the amount of water after 10 hr, 20 hr, and 30 hr. After

these numbers of hours, how much water has flowed in and howmuch has leaked out? [Ans: L : 8790, 31166, 72649]

(e) When will the lake have 100000 ft3 of water? [Ans: bit more than 40 hr]

(f) Find the limit of W as t approaches infinity. State the realworld meaning of this number.

[Ans: 125000 ft3

](g) Draw the graph of W versus t. Clearly show an asymptote.

(h) The lake starts filling with water. The actual amount of waterat time t = 10 is exactly 40000 ft3. The flow rate is still 5000ft3/hr, as predicted. Use this information to find a more precisevalue of the leakage constant k. [Ans: k = 0.0464...]

Problems

For Friday, March 12, put 111 on the front and 112 on the back.It would be a good idea to turn the following in before spring break:For Monday, March 22, put 113 on the front, and be on the lookout for a quizon Moodle due Monday.

11.B-111 (a) (adapted from AB93) A puppy weighs 2.1 pounds at birth and 3.5pounds two months later. If the weight of the puppy during its first6 months is increasing at a rate proportional to its weight, then howmuch will the puppy weigh when it is 4 months old (to the nearest0.1 pound)?

(b) (adapted from [2]) If g′(x) = 3g(x) and g(−1) = 1, find g(x).



11.B-112 (a) (adapted from [2]) The change in N , the number of bacteria in a

culture dish at time t is given by:dN

dt= 3N . If N = 4, when t = 0,

the approximate value of t when N = 1614 is

(b) (AB ’96) The rate of consumption of cola in the United States isgiven by S(t) = Cekt, where S is measured in billions of gallons peryear and t is measured in years from the beginning of 1980. Theconsumption rate doubles every 5 years and the consumption rate atthe beginning of 1980 was 6 billion gallons per year. Find C and k.

11.B-113 [10] You run over a nail. As the air leaks out of your tire, the rate of changeof air pressure inside the tire is directly proportional to that pressure.

(a) Write a differential equation that states this fact. Evaluate the pro-portionality constant if the pressure was 35 psi and decreasing at 0.28psi/min at time zero.

(b) Solve the differential equation subject to the initial condition impliedin step (a).

(c) Sketch the graph of the function. Show its behavior a long time afterthe tire is punctured.

(d) What will be the pressure at 10 min after the tire was punctured?

(e) The car is safe to drive as long as the tire pressure is 12 psi or greater.For how long after the puncture will the car be safe to drive?

11.B-114 Be on the lookout for a quiz on moodle due Monday, March 22. (Hereafter,moodle quizzes will be due on Sunday evening, but I give you some leewaybecause of spring break.)


AP Unit 11, Day 3: Related Rates 331

11.3 Related Rates

Example 11.3.15 (2002 AB-5) [NO CALCULATOR]

A container has the shape of an open right circular cone, as shownin the figure above. The height of the container is 10 cm and thediameter of the opening is 10 cm. Water in the container is evapo-

rating so that its depth h is changing at the constant rate of − 310

cm/hr.(Note: the volume of a cone of height h and radius r is given by

V =13πr2h.)

(a) Find the volume V of water in the container when h = 5 cm.Indicate units of measure.

(b) Find the rate of change of the volume of water in the container,with respect to time, when h = 5 cm. Indicate units of measure.

(c) Show that the rate of change of the volume of water in the con-tainer due to evaporation is directly proportional to the exposedsurface area of the water. What is the constant of proportion-ality?




The trough shown in the figure above is 5 feet long, and its verticalcross sections are inverted isosceles triangles with base 2 feet andheight 3 feet. Water is being siphoned out of the trough at the rateof 2 cubic feet per minute. At any time t, let h be the depth and Vbe the volume of water in the trough.

(a) Find the volume of water in the trough when it is full.

(b) What is the rate of change in h at the instant when the trough

is14

full by volume?

(c) What is the rate of change in the area of the surface of thewater (shaded in the figure) at the instant when the trough is14

full by volume?



Example 11.3.17 (2002B AB-6) [NO CALCULATOR]

Ship A is traveling due west toward Lighthouse Rock at a speed of15 kilometers per hour (km/hr). Ship B is traveling due north awayfrom Lighthouse Rock at a speed of 10 km/hr. Let x be the distancebetween Ship A and Lighthouse Rock at time t, and let y be thedistance between Ship B and Lighthouse Rock at time t, as shownin the figure above.

(a) Find the distance, in kilometers, between Ship A and Ship Bwhen x = 4 km and y = 3 km.

(b) Find the rate of change, in km/hr, of the distance between thetwo ships when x = 4 km and y = 3 km.

(c) Let θ be the angle shown in the figure. Find the rate of changeof θ, in radians per hour, when x = 4 km and y = 3 km.




The balloon shown above is in the shape of a cylinder with hemi-spherical ends of the same radius as that of the cylinder. The balloonis being inflated at the rate of 261π cubic centimeters per minute. Atthe instant the radius of the cylinder is 3 centimeters, the volume ofthe balloon is 144π cubic centimeters and the radius of the cylinderis increasing at the rate of 2 centimeters per minute. (The volumeof a cylinder with radius r and height h is πr2h, and the volume of

a sphere with radius r is43πr3.)

(a) At this instant, what is the height of the cylinder?

(b) At this instant, how fast is the height of the cylinder increasing?



Homework

11.C-115 (1988 BC-3) [NO CALCULATOR]

The figure above represents an observer at point A watching balloon B asit rises from point C. The balloon is rising at a constant rate of 3 metersper second and the observer is 100 meters from point C.

(a) Find the rate of change in x at the instant when y = 50.(b) Find the rate of change in the area of right triangle BCA at the

instant when y = 50.(c) Find the rate of change in θ at the instant when y = 50.

11.C-116 (1995 AB-5)

As shown in the figure above, water is draining from a conical tank withheight 12 feet and diameter 8 feet into a cylindrical tank that has a basewith area 400π square feet. The depth h, in feet, of the water in theconical tank is changing at the rate of (h− 12) feet per minute. (The

volume V of a cone with radius r and height h is V =13πr2h.)

(a) Write an expression for the volume of water in the conical tank as afunction of h.

(b) At what rate is the volume of water in the conical tank changingwhen h = 3? Indicate units of measure.

(c) Let y be the depth, in feet, of the water in the cylindrical tank. Atwhat rate is y changing when h = 3? Indicate units of measure.



11.C-117 (1992 AB-6) [NO CALCULATOR] At time t, t ≥ 0, the volume of a sphereis increasing at a rate proportional to the reciprocal of its radius. At t = 0,the radius of the sphere is 1 and at t = 15, the radius is 2. (The volume

V of a sphere with radius r is V =43πr3.)

(a) Find the radius of the sphere as a function of t.

(b) At what time t will the volume of the sphere be 27 times its volumeat t = 0?

11.C-118 (1984 AB-5) [NO CALCULATOR] The volume of a cone(V =

13πr2h

)is increasing at the rate of 28π cubic units per second. At the instantwhen the radius r of the cone is 3 units, its volume is 12π cubic units and

the radius is increasing at12

unit per second.

(a) At the instant when the radius of the cone is 3 units, what is the rateof change of the area of its base?

(b) At the instant when the radius of the cone is 3 units, what is the rateof change of its height h?

(c) At the instant when the radius of the cone is 3 units, what is theinstantaneous rate of change of the area of its base with respect toits height h?


AP Unit 11, Day 4: Graphs 337

11.4 Graphs


The graph of the function f shown above consists of a semicircleand three line segments. Let g be the function given by g(x) =∫ x

−3

f(t) dt.

(a) Find g(0) and g′(0).(b) Addendum: Find g′′(0)(c) Find all values of x in the open interval (−5, 4) at which g

attains a relative maximum. Justify your answer.(d) Find the absolute minimum value of g on the closed interval

[−5, 4]. Justify your answer.(e) Find all values of x in the open interval (−5, 4) at which the

graph of g has a point of inflection.(f) Addendum: Find the average rate of change of g on the interval−3 ≤ x ≤ 0; on the interval −5 ≤ x ≤ 0.

(g) Addendum: For how many values c, where −3 < c < 0, if any,is g′(c) equal to the average rate of change of g on the interval−3 ≤ x < 0?

(h) Addendum: For how many values c, where −5 < c < 0, if any,is g′(c) equal to the average rate of change of g on the interval−5 ≤ x < 0?

(i) Addendum: For how many values c, where −5 < c < 0, if any,is g′(c) equal to the average rate of change of g on the interval−5 ≤ x < 0?

(j) Addendum: Write an equation of the line tangent to the graphof g at x = −3. Write an equation of the line tangent to thegraph of g at x = 0. Do the tangent lines lie above or belowthe graph of g?



Example 11.4.22 (1999 AB-5)

The graph of the function f , consisting of three line segments, is

given above. Let g(x) =∫ x

1

f(t) dt.

(a) Compute g(4) and g(−2).

(b) Find the instantaneous rate of change of g, with respect to x,at x = 1.

(c) Find the absolute minimum value of g on the closed interval[−2, 4]. Justify your answer.

(d) The second derivative of g is not defined at x = 1 and x =2. How many of these values are x-coordinates of points ofinflection of the graph of g? Justify your answer.




x 0 0 < x < 1 1 1 < x < 2 2 2 < x < 3 3 3 < x < 4f(x) −1 Negative 0 Positive 2 Positive 0 Negativef ′(x) 4 Positive 0 Positive DNE Negative −3 Negativef ′′(x) −2 Negative 0 Positive DNE Negative 0 Positive

Let f be a function that is continuous on the interval [0, 4). Thefunction f is twice differentiable except at x = 2. The function fand its derivatives have the properties indicated in the table above,where DNE indicates that the derivatives of f do not exist at x = 2.

(a) For 0 < x < 4, find all values of x at which f has a relativeextremum. Determine whether f has a relative maximum or arelative minimum at each of these values. Justify your answer.

(b) On the axes provided, sketch the graph of a function that hasall the characteristics of f .

(c) Let g be the function defined by g(x) =∫ x

1

f(t) dt on the open

interval (0, 4). For 0 < x < 4, find all values of x at which ghas a relative extremum. Determine whether g has a relativemaximum or a relative minimum at each of these values. Justifyyour answer.

(d) For the function g defined in part (c), find all values of x, for0 < x < 4, at which the graph of g has a point of inflection.Justify your answer.



Example 11.4.24 (2009 AB–6) [NO CALCULATOR]

The derivative of a function f is defined by f ′(x) =

{g(x) for − 4 ≤ x ≤ 05e−x/3 − 3 for 0 < x ≤ 4

The graph of the continuous function f ′, shown in the figure above,

has x–intercepts at x = −2 and x = 3 ln(

53

). The graph of g on

−4 ≤ x ≤ 0 is a semicircle, and f(0) = 5.

(a) For −4 < x < 4, find all values of x at which the graph of fhas a point of inflection. Justify your answer.

(b) Find f(−4) and f(4).

(c) For −4 ≤ x ≤ 4, find the value of x at which f has an absolutemaximum. Justify your answer.



Homework

11.D-121 [NO CALCULATOR]

Let f be a function defined on the closed interval −3 ≤ x ≤ 4 withf(0) = 3. The graph of f ′, the derivative of f , consists of one line segmentand a semicircle, as shown above.

(a) On what intervals, if any, is f increasing? Justify your answer.

(b) Find the x-coordinate of each point of inflection of the graph of f onthe open interval −3 < x < 4. Justify your answer.

(c) Find an equation for the line tangent to the graph of f at the point(0, 3).

(d) Find f(−3) and f(4). Show the work that leads to your answers.




The graph of the function f shown above consists of two line segments.

Let g be the function given by g(x) =∫ x

0

f(t) dt.

(a) Find g(−1), g′(−1), and g′′(−1).

(b) For what values of x in the open interval (−2, 2) is g increasing?Explain your reasoning.

(c) For what values of x in the open interval (−2, 2) is the graph of gconcave down? Explain your reasoning.

(d) On the axes provided, sketch the graph of g on the closed interval[−2, 2].




The graph of the function f above consists of three line segments.

(a) Let g be the function given by g(x) =∫ x

−4

f(t) dt. For each of g(−1),

g′(−1), and g′′(−1), find the value or state that it does not exist.

(b) For the function g defined in part (a), find the x-coordinate of eachpoint of inflection of the graph of g on the open interval −4 < x < 3.Explain your reasoning.

(c) Let h be the function given by h(x) =∫ 3

x

f(t) dt. Find all values of

x in the closed interval −4 ≤ x ≤ 3 for which h(x) = 0.

(d) For the function h defined in part (c), find all intervals on which h isdecreasing. Explain your reasoning.


Let f be a function defined on the closed interval [0, 7]. The graph of f ,consisting of four line segments, is shown above. Let g be the function

given by g(x) =∫ x

2

f(t) dt.

(a) Find g(3), g′(3), and g′′(3).

(b) Find the average rate of change of g on the interval 0 ≤ x ≤ 3.

(c) For how many values c, where 0 < c < 3, is g′(c) equal to the averagerate found in part (b)? Explain your reasoning.

(d) Find the x-coordinate of each point of inflection of the graph of g onthe interval 0 < x < 7. Justify your answer.




AP Unit 11, Day 5: Integral as Accumulator 345

11.5 Integral as Accumulator

Example 11.5.25 (2007 AB–2)

The amount of water in a storage tank, in gallons, is modeled bya continuous function on the time interval 0 ≤ t ≤ 7, where t ismeasured in hours. In this model, rates are given as follows:

(i) The rate at which water enters the tank is f(t) = 100t2 sin(√t)

gal-lons per hour for 0 ≤ t ≤ 7.

(ii) The rate at which water leaves the tank is

g(t) =

{250 for 0 ≤ t < 32000 for 2 < t ≤ 7

gallons per hour

The graphs of f and g, which intersect at t = 1.617 and t = 5.076,are shown in the figure above At time t = 0, the amount of water inthe tank is 5000 ga1lons.

(a) How many gallons of water enter the tank during the time in-terval 0 ≤ t ≤ 7? Round your answer to the nearest gallon.

(b) For 0 ≤ t ≤ 7, find the time intervals during which the amountof water in the tank is decreasing. Give a reason for each an-swer.

(c) For 0 ≤ t ≤ 7, at what time t is the amount of water in thetank greatest? To the nearest gallon, compute the amount ofwater at this time. Justify your answer.

(d) Addendum: For how many times t, 0 ≤ t ≤ 7, is the instan-taneous rate at which water enters the tank equal to the aver-age rate at which water enters the tank over the time interval0 ≤ t ≤ 7?



(e) Addendum: For how many times t, 0 ≤ t ≤ 7, is the instan-taneous rate at which water leaves the tank equal to the aver-age rate at which water leaves the tank over the time interval0 ≤ t ≤ 7?

(f) Addendum: What is the average rate of change of the rate atwhich the water leaves the tank over the time interval 0 ≤ t ≤7?

Example 11.5.26 (2002 AB-2) The rate at which people enter anamusement park on a given day is modeled by the function E definedby

E(t) =15600

(t2 − 24t+ 160)

The rate at which people leave the same amusement park on thesame day is modeled by the function L defined by

L(t) =9890

(t2 − 38t+ 370)

Both E(t) and L(t) are measured in people per hour and time tis measured in hours after midnight. These functions are valid for9 ≤ t ≤ 23, the hours during which the park is open. At time t = 9,there are no people in the park.

(a) How many people have entered the park by 5:00 p.m. (t = 17)?Round your answer to the nearest whole number.

(b) The price of admission to the park is $15 until 5:00 p.m. (t =17). After 5:00 p.m., the price of admission to the park is $11.How many dollars are collected from admissions to the parkon the given day? Round your answer to the nearest wholenumber.

(c) Let H(t) =∫ t

9

(E(x)− L(x)) dx for 9 ≤ t ≤ 23. The value of

H(17) to the nearest whole number is 3725. Find the value ofH ′(17), and explain the meaning of H(17) and H ′(17) in thecontext of the amusement park.

(d) At what time t, for 9 ≤ t ≤ 23, does the model predict that thenumber of people in the park is a maximum?



Example 11.5.27 (2008P AB–1) The rate at which raw sewage

enters a treatment tank is given by E(t) = 850 + 715 cos(πt2

9

)gal-

lons per hour for 0 ≤ t ≤ 4 hours. Treated sewage is removed fromthe tank at the constant rate of 645 gallons per hour. The treatmenttank is empty at time t = 0.

(a) How many gallons of sewage enter the treatment tank duringthe time interval 0 ≤ t ≤ 4? Round your answer to the nearestgallon.

(b) For 0 ≤ t ≤ 4, at what time t is the amount of sewage in thetreatment tank greatest? To the nearest gallon, what is themaximum amount of sewage in the tank? Justify your answers.

(c) For 0 ≤ t ≤ 4, the cost of treating the raw sewage that entersthe tank at time t is (0.15 − 0.02t) dollars per gallon. To thenearest dollar, what is the total cost of treating all the sewagethat enters the tank during the time interval 0 ≤ t ≤ 4?

Example 11.5.28 (2005B AB-2) A water tank at Camp Newtonholds 1200 gallons of water at time t = 0. During the time interval0 ≤ t ≤ 18 hours, water is pumped into the tank at the rate

W (t) = 95√t sin2

(t

6

)gallons per hour.

During the same time interval, water is removed from the tank atthe rate

R(t) = 275 sin2

(t

3

)gallons per hour.

(a) Is the amount of water in the tank increasing at time t = 15?Why or why not?

(b) To the nearest whole number, how many gallons of water arein the tank at time t = 18?

(c) At what time t, for 0 ≤ t ≤ 18, is the amount of water in thetank at an absolute minimum? Show the work that leads toyour conclusion.

(d) For t > 18, no water is pumped into the tank, but water con-tinues to be removed at the rate R(t) until the tank becomesempty. Let k be the time at which the tank becomes empty.Write, but do not solve, an equation involving an integral ex-pression that can be used to find the value for k.



Homework

11.E-125 A tank contains 125 gallons of heating oil at time t = 0. During the timeinterval 0 ≤ t ≤ 12 hours, heating oil is pumped into the tank at the rate

H(t) = 2 +10

(1 + ln (t+ 1))gallons per hour.

During the same time interval, heating oil is removed from the tank at therate

R(t) = 12 sin(t2

47

)gallons per hour.

(a) How many gallons of heating oil are pumped into the tank duringthe time interval 0 ≤ t ≤ 12 hours?

(b) Is the level of heating oil in the tank rising or falling at time t = 6hours? Give a reason for your answer.

(c) How many gallons of heating oil are in the tank at time t = 12 hours?

(d) At what time t, for 0 ≤ t ≤ 12, is the volume of heating oil in thetank the least? Show the analysis that leads to your conclusion.

11.E-126 For 0 ≤ t ≤ 31, the rate of change of the number of mosquitoes on TropicalIsland at time t days is modeled by R(t) = 5

√t cos

(t5

)mosquitoes per

day. There are 1000 mosqitoes on Tropical Island at time t = 0.

(a) Show that the number of mosquitoes is increasing at time t = 6.

(b) At time t = 6, is the number of mosquitoes increasing at an increasingrate, or is the number of mosquitoes increasing at a decreasing rate?Give a reason for your anwer.

(c) According to the model, how many mosquitoes will be on the islandat time t = 31?

(d) To the nearest whole number, what is the maximum number ofmosquitoes for 0 ≤ t ≤ 31? Show the analysis that leads to yourconclusion.



11.E-127 The tide removes sand from Sandy Point Beach at a rate modeled by thefunction R, given by

R(t) = 2 + 5 sin(

4πt25

).

A pumping station adds sand to the beach at a rate modeled by thefunction S, given by

S(t) =15t

1 + 3t.

Both R(t) and S(t) have units of cubic yards per hour and t is measuredin hours for 0 ≤ t ≤ 6. At time t = 0, the beach contains 2500 cubic yardsof sand.

(a) How much sand will the tide remove from the beach during this 6–hour period? Indicate units of measure.

(b) Write an expression for Y (t), the total number of cubic yards of sandon the beach at time t.

(c) Find the rate at which the total amount of sand on the beach ischanging at time t = 4.

(d) For 0 ≤ t ≤ 6, at what time t is the amount of sand on the beach aminimum? What is the minimum value? Justify your answer.

11.E-128 [NO CALCULATOR] Water is pumped into an underground tank at aconstant rate of 8 gallons per minute. Water leaks out of the tank at therate of

√t+ 1 gallons per minute, for 0 ≤ t ≤ 120 minutes. At time t = 0,

the tank contains 30 gallons of water.

(a) How many gallons of water leak out of the tank from time t = 0 tot = 3 minutes?

(b) How many gallons of water are in the tank at time t = 3 minutes?

(c) Write an expression for A(t), the total number of gallons of water inthe tank at time t.

(d) At what time t, for 0 ≤ t ≤ 120, is the amount of water in the tanka maximum? Justify your answer.




AP Unit 11, Day 6: Particle Motion 351

11.6 Particle Motion

Example 11.6.29 (2003 AB-2) A particle moves along the x-axisso that its velocity at time t is given by

v(t) = − (t+ 1) sin(t2

2

)At time t = 0, the particle is at position x = 1.

(a) Find the acceleration of the particle at time t = 2. Is the speedof the particle increasing at t = 2? Why or why not?

(b) Find all times t in the open interval 0 < t < 3 when the particlechanges direction. Justify your answer.

(c) Find the total distance traveled by the particle from time t = 0until time t = 3.

(d) During the time interval 0 ≤ t ≤ 3, what is the greatest distancebetween the particle and the origin? Show the work that leadsto your answer.

(e) Addendum: For what values of t is the particle moving to theright? Justify your answer.

(f) Addendum: Find the position of the particle at time t = 2. Isthe particle moving toward the origin or away from the originat time t = 2? Justify your answer.

Example 11.6.30 (2004 AB-3) A particle moves along the y-axisso that its velocity v at time t ≥ 0 is given by v(t) = 1− tan−1(et).At time t = 0, the particle is at y = −1. (Note: tan−1 x = arctanx)

(a) Find the acceleration of the particle at time t = 2.

(b) Is the speed of the particle increasing or decreasing at timet = 2? Give a reason for your answer.

(c) Find the time t ≥ 0 at which the particle reaches its highestpoint. Justify your answer.

(d) Find the position of the particle at time t = 2. Is the particlemoving toward the origin or away from the origin at time t = 2?Justify your answer.




t(seconds)

0 10 20 30 40 50 60 70 80

v(t)(feet per second)

5 14 22 29 35 40 44 47 49

Rocket A has positive velocity v(t) after being launched upward froman initial height of 0 feet at time t = 0 seconds. The velocity of therocket is recorded for selected values of t over the interval 0 ≤ t ≤ 80seconds, as shown in the table above.

(a) Find the average acceleration of rocket A over the time interval0 ≤ t ≤ 80 seconds. Indicate units of measure.


10

v(t) dt in terms

of the rocket’s flight. Use a midpoint Riemann sum with 3

subintervals of equal length to approximate∫ 70

10

v(t) dt.

(c) Rocket B is launched upward with an acceleration of a(t) =3√t+ 1

feet per second per second. At time t = 0 seconds, the

initial height of the rocket is 0 feet, and the initial velocity is 2feet per second. Which of the two rockets is traveling faster attime t = 80 seconds? Explain your answer.

Example 11.6.32 (2007B AB–2)

A particle moves along the x–axis so that its velocity v at timet ≥ 0 is given by v(t) = sin

(t2). The graph of v is shown above for

0 ≤ t ≤√

5π. The position of the particle at time t is x(t) and itsposition at time t = 0 is x(0) = 5.




(b) Find the total distance traveled by the particle from time t = 0to t = 3.

(c) Find the position of the particle at time t = 3.

(d) For 0 ≤ t ≤√

5π, find the time t at which the particle is farthestto the right. Explain your answer.

Homework

11.F-129 [NO CALCULATOR]

A particle moves along the x–axis so that its velocity at time t, for 0 ≤t ≤ 6, is given by a differentiable function v whose graph is shown above.The velocity is 0 at t = 0, t = 3, and t = 5, and the graph has horizontaltangents at t = 1 and t = 4. The areas of the regions bounded by thet–axis and the graph of v on the intervals [0, 3], [3, 5], and [5, 6] are 8, 3,and 2, respectively. At time t = 0, the particle is at x = −2.

(a) For 0 ≤ t ≤ 6, find both the time and the position of the particlewhen the particle is farthest to the left. Justify your answer.

(b) For how many values of t, where 0 ≤ t ≤ 6, is the particle at x = −8?Explain your reasoning.

(c) On the interval 2 < t < 3, is the speed of the particle increasing ordecreasing? Give a reason for your answer.

(d) During what time intervals, if any, is the acceleration of the particlenegative? Justify your answer.

11.F-130 [NO CALCULATOR]



t(seconds)

0 8 20 25 32 40

v(t)(meters per second)

3 5 −10 −8 −4 −7

The velocity of a particle moving along the x–axis is modeled by a differ-entiable function v, where the position x is measured in meters, and timet is measured in seconds. Selected values of v(t) are given in the tableabove. The particle is at position x = 7 meters when t = 0 seconds.

(a) Estimate the acceleration of the particle at t = 36 seconds. Show thecomputations that lead to your answer. Indicate units of measure.


20

v(t) dt in the context

of this problem. Use a trapezoidal sum with the three subintervals

indicated by the data in the table to approximate∫ 40

20

v(t) dt

(c) For 0 ≤ t ≤ 40, must the particle change direction in any of thesubintervals indicated by the data in the table? If so, identify thesubintervals and explain your reasoning. If not, explain why not.

(d) Suppose that the acceleration of the particle is positive for 0 < t <8 seconds. Explain why the position of the particle at t = 8 secondsmust be greater than x = 30 meters.

11.F-131 Caren rides her bicycle along a straight road from home to school, startingat home at time t = 0 minutes and arriving at school at time t = 12 min-utes. During the time interval 0 ≤ t ≤ 12 minutes, her velocity v(t), inmiles per minute, is modeled by the piecewise-linear function whose graphis shown below.

(a) Find the acceleration of Carens bicycle at time t = 7.5 minutes.Indicate units of measure.

(b) Using correct units, explain the meaning of∫|v(t)| dt in terms of

Carens trip. Find the value of∫|v(t)| dt.



(c) Shortly after leaving home, Caren realizes she left her calculus home-work at home, and she returns to get it. At what time does she turnaround to go back home? Give a reason for your answer.

(d) Larry also rides his bicycle along a straight road from home to schoolin 12 minutes. His velocity is modeled by the function w given byw(t) =

π

15sin( π

12t)

, where w(t) is in miles per minute for 0 ≤ t ≤12 minutes. Who lives closer to school: Caren or Larry? Show thework that leads to your answer.

11.F-132 An object moves along the x-axis with initial position x(0) = 2. Thevelocity of the object at time t ≥ 0 is given by v(t) = sin

(π3t)

.

(a) What is the acceleration of the object at time t = 4?

(b) Consider the following two statements.Statement I: For 3 < t < 4.5, the velocity of the object is decreasing.Statement II: For 3 < t < 4.5, the speed of the object is increasing.Are either or both of these statements correct? For each statementprovide a reason why it is correct or not correct.

(c) What is the total distance traveled by the object over the time inter-val 0 ≤ t ≤ 4?

(d) What is the position of the object at time t = 4?




AP Unit 11, Day 7: Data 357

11.7 Data

Example 11.7.33 (2007 AB–3)

x f(x) f ′(x) g(x) g′(x)1 6 4 2 52 9 2 3 13 10 −4 4 24 −1 3 6 7

The functions f and g are differentiable for all real numbers, and gis strictly increasing. The table above gives values of the functionsand their first derivatives at selected values of x. The function h isgiven by h(x) = f(g(x))− 6.

(a) Explain why there must be a value r for 1 < r < 3 such thath(r) = −5.

(b) Explain why there must be a value c for 1 < c < 3 such thath′(c) = −5.

(c) Let w be the function given by w(x) =∫ g(x)

1

f(t) dt. Find the

value of w′(3).

(d) Addendum: Use either a trapezoidal approximation or a mid-point Riemann sum with three subdivisions of equal length toapproximate w(3).

(e) Addendum: Estimate w′′(3).

(f) Addendum: Find the exact value of∫ 4

1

(3f ′(x)− 2g′(x) + 3) dx.

(g) If g−1 is the inverse function of g, write an equation for the linetangent to the graph of y = g−1(x) at x = 2.


x 2 3 5 8 13f(x) 1 4 −2 3 6

Let f be a function that is twice differentiable for all real numbers.The table above gives values of f for selected points in the closedinterval 2 ≤ x ≤ 13.

(a) Estimate f ′(4). Show the work that leads to your answer.

(b) Evaluate∫ 13

2

(3− 5f ′(x)) dx. Show the work that leads toyour answer.



(c) Use a left Riemann sum with subintervals indicated by the data

in the table to approximate∫ 13

2

f(x) dx. Show the work that

leads to your answer.

(d) Suppose f ′(5) = 3 and f ′′(x) < 0 for all x in the closed interval5 ≤ x ≤ 8. Use the line tangent to the graph of f at x = 5 toshow that f(7) ≤ 4. Use the secant line for the graph of f on

5 ≤ x ≤ 8 to show that f(7) ≥ 43

.

Example 11.7.35 (2003B AB–3)

Distancex 0 60 120 180 240 300 360

(mm)

DiameterB(x) 24 30 28 30 26 24 26(mm)

A blood vessel is 360 millimeters (mm) long with circular cross sec-tions of varying diameter. The table above gives the measurementsof the diameter of the blood vessel at selected points along the lengthof the blood vessel, where x represents the distance from one endof the blood vessel and B(x) is a twice-differentiable function thatrepresents the diameter at that point.

(a) Write an integral expression in terms of B(x) that representsthe average radius, in mm, of the blood vessel between x = 0and x = 360.

(b) Approximate the value of your answer from part (a) using thedata from the table and a midpoint Riemann sum with threesubintervals of equal length. Show the computations that leadto your answer.

(c) Using correct units, explain the meaning of∫ 275

125

π

(B(x)

2

)2

dx

in terms of the blood vessel.

(d) Explain why there must be at least one value, x, for 0 < x <360, such that B′′(x) = 0.



Example 11.7.36 (2005 AB–3)

Distancex (cm)

0 1 5 6 8

TemperatureT (x) (◦C)

100 93 70 62 55

A metal wire of length 8 centimeters (cm) is heated at one end. Thetable above gives selected values of the temperature T (x), in degreesCelsius (◦C), of the wire x cm from the heated end. The function Tis decreasing and twice differentiable.

(a) Estimate T ′(7). Show the work that leads to your answer.Indicate units of measure.

(b) Write an integral expression in terms of T (x) for the averagetemperature of the wire. Estimate the average temperatureof the wire using a trapezoidal sum with the four subintervalsindicated by the data in the table. Indicate units of measure.

(c) Find∫ 8

0

T ′(x) dx, and indicate units of measure. Explain the

meaning of∫ 8

0

T ′(x) dx in terms of the temperature of the

wire.

(d) Are the data in the table consistent with the assertion thatT ′′(x) > 0 for every x in the interval 0 < x < 8? Explain youranswer.



Homework

11.G-133 The graph of the velocity v(t), in ft/sec, of a car traveling on a straightroad, for 0 ≤ t ≤ 50, is shown below. A table of values for v(t), at 5second intervals of time t, is shown to the right of the graph.

(a) During what intervals of time is the acceleration of the car positive?Give a reason for your answer.

(b) Find the average acceleration of the car, in ft/sec2, over the interval0 ≤ t ≤ 50.

(c) Find one approximation for the acceleration of the car, in ft/sec2, att = 40. Show the computations you used to arrive at your answer.

(d) Approximate∫ 50

0

v(t) dt with a Riemann sum, using midpoints of

five subintervals of equal length. Using correct units, explain themeaning of this integral.

11.G-134 [NO CALCULATOR]

t(sec)

0 15 25 30 35 50 60

v(t)(ft/sec)

−20 −30 −20 −14 −10 0 10

a(t)(ft/sec2)

1 5 2 1 2 4 2

A car travels on a straight track. During the time interval 0 ≤ t ≤ 60seconds, the car’s velocity v, measured in feet per second, and accelerationa, measured in feet per second per second, are continuous functions. Thetable above shows selected values of these functions.



(a) Using appropriate units, explain the meaning of∫ 60

30

|v(t)| dt in terms

of the car’s motion. Approximate∫ 60

30

|v(t)| dt using a trapezoidal

approximation with the three subintervals determined by the table.

(b) Using appropriate units, explain the meaning of∫ 30

0

a(t) dt in terms

of the car’s motion. Find the exact value of∫ 30

0

a(t) dt.

(c) For 0 < t < 60, must there be a time t when v(t) = −5? Justify youranswer.

(d) For 0 < t < 60, must there be a time t when a(t) = 0? Justify youranswer.

11.G-135 Concert tickets went on sale at noon (t = 0) and were sold out within9 hours. The number of people waiting in line to purchase tickets at timet is modeled by a twice-differentiable function L for 0 ≤ t ≤ 9. Values ofL(t) at various times t are shown in the table below.

t (hours) 0 1 3 4 7 8 9L(t) (people) 120 156 176 126 150 80 0

(a) Use the data in the table to estimate the rate at which the number ofpeople waiting in line was changing at 5:30 P.M. (t = 5.5). Show thecomputations that lead to your answer. Indicate units of measure.

(b) Use a trapezoidal sum with three subintervals to estimate the averagenumber of people waiting in line during the first 4 hours that ticketswere on sale.

(c) For 0 ≤ t ≤ 9, what is the fewest number of times at which L′(t)must equal 0 ? Give a reason for your answer.

(d) The rate at which tickets were sold for 0 ≤ t ≤ 9 is modeled byr(t) = 550te−t/2 tickets per hour. Based on the model, how manytickets were sold by 3 P.M. (t = 3), to the nearest whole number?

11.G-136 [NO CALCULATOR] Let f be a twice-differentiab1e function such thatf(2) = 5 and f(5) = 2. Let g be the function given by g(x) = f(f(x)).

(a) Explain why there must be a value c for 2 < c < 5 such that f ′(c) =−1.

(b) Show that g′(2) = g′(5). Use this result to explain why there mustbe a value k for 2 < k < 5 such that g′′(k) = 0.

(c) Show that if f ′′(x) = 0 for all x, then the graph of g does not have apoint of inflection.

(d) Let h(x) = f(x) − x. Exp1ain why there must be a value r for2 < r < 5 such that h(r) = 0.




AP Unit 11, Day 8: Extrema and Optimization 363

11.8 Extrema and Optimization

Example 11.8.37 (1996 AB–4)This problem deals with functions defined by f(x) = x + b sinx,where b is a positive constant and −2π ≤ x ≤ 2π.

(a) Sketch the graphs of two of these functions, y = x + sinx andy = x+ 3 sinx, as indicated below.

(b) Find the x-coordinate of all points, −2π ≤ x ≤ 2π, where theline y = x+ b is tangent to the graph of f(x) = x+ b sinx.

(c) Are the points of tangency described in part (b) relative max-imum points of f? Why?

(d) For all values of b > 0, show that all inflection points of thegraph of f lie on the line y = x.



Example 11.8.38 (2008B–4) [NO CALCULATOR] The functions

f and g are given by f(x) =∫ 3x

0

√4 + t2 dt and g(x) = f(sinx).

(a) Find f ′(x) and g′(x).

(b) Write an equation for the line tangent to the graph of y = g(x)at x = π.

(c) Write, but do not evaluate, an integral expression that rep-resents the maximum value of g on the interval 0 ≤ x ≤ π.Justify your answer.

Example 11.8.39 (1997 AB–4) Let f be the function given byf(x) = x3 − 6x2 + p, where p is an arbitrary constant.

(a) Write an expression for f ′(x) and use it to find the relativemaximum and minimum values of f in terms of p. Show theanalysis that leads to your conclusion.

(b) For what values of the constant p does f have 3 distinct realroots?

(c) Find the value of p such that the average value of f over theclosed interval [−1, 2] is 1.

Example 11.8.40 (2008P AB–6) [NO CALCULATOR] Let g(x) =xe−x + be−x, where b is a positive constant.

(a) Find limx→∞

g(x)

(b) For what positive value of b does g have an absolute maximum

at x =23

? Justify your answer.

(c) Find all values of b, if any, for which the graph of g has a pointof inflection on the interval 0 < x <∞. Justify your answer.


AP Unit 11, Day 8: Extrema and Optimization 365

Homework

11.H-137 [NO CALCULATOR] A cubic polynomial function f is defined by

f(x) = 4x3 + ax2 + bx+ k

where a, b, and k are constants. The function f has a local minimum atx = −1, and the graph of f has a point of inflection at x = −2.

(a) Find the values of a and b.

(b) If∫ 1

0

f(x) dx = 32, what is the value of k?

11.H-138 Let f be the function given by f(x) = 2xe2x.

(a) Find limx→−∞

f(x) and limx→∞

f(x).

(b) Find the absolute minimum value of f . Justify that your answer isan absolute minimum.

(c) What is the range of f?

(d) Consider the family of functions defined by y = bxebx, where b is anonzero constant. Show that the absolute minimum value of bxebx isthe same for all nonzero values of b.



11.H-139 The figure below shows the graph of f ′, the derivative of the function f ,for −7 ≤ x ≤ 7. The graph of f ′ has horizontal tangent lines at x = −3,x = 2, and x = 5, and a vertical tangent at x = 3.

(a) Find all values of x, for −7 < x < 7, at which f attains a relativeminimum. Justify your answer.

(b) Find all values of x, for −7 < x < 7, at which f attains a relativemaximum. Justify your answer.

(c) Find all values of x, for −7 < x < 7, at which f ′′(x) < 0.

(d) At what value of x, for −7 ≤ x ≤ 7, does f attain its absolutemaximum? Justify your answer.

11.H-140 [NO CALCULATOR] Let f be a function defined by f(x) = k√x − lnx

for x > 0, where k is a positive constant.

(a) Find f ′(x) and f ′′(x).

(b) For what value of the constant k does f have a critical point at x = 1?For this value of k, determine whether f has a relative minimum,relative maximum, or neither at x = 1. Justify your answer.

(c) For a certain value of the constant k, the graph of f has a point ofinflection on the x–axis. Find this value of k.



11.9 Implicit Differentiation

Example 11.9.41 (2005B AB–5) [NO CALCULATOR] Considerthe curve given by y2 = 2 + xy.

(a) Show thatdy

dx=

y

2y − x.

(b) Addendum: Findd2y

dx2.

(c) Find all points (x, y) on the curve where the line tangent to the

curve has slope12

.

(d) Addendum: At these points, do the lines tangent to the graphlie above or below the graph?

(e) Show that there are no points (x, y) on the curve where the linetangent to the curve is horizontal.

(f) Addendum: Show that there are no points (x, y) on the curvewhere the line tangent to the curve is vertical.

(g) Let x and y be functions of time t that are related by theequation y2 = 2 + xy. At time t = 5, the value of y is 3 anddy

dt= 6. Find the value of

dx

dtat time t = 5.



Example 11.9.42 (1995 AB–3) Consider the curve defined by−8x2 + 5xy + y3 = −149.

(a) Finddy

dx.

(b) Write an equation for the line tangent to the curve at the point(4,−1).

(c) There is a number k so that the point (4.2, k) is on the curve.Using the tangent line found in part (b), approximate the valueof k.

(d) Write an equation that can be solved to find the actual valueof k so that the point (4.2, k) is on the curve.

(e) Solve the equation in part (d) for the value of k.

Example 11.9.43 (1998 AB–6) Consider the curve defined by 2y3+6x2y − 12x2 + 6y = 1.

(a) Show thatdy

dx=

4x− 2xyx2 + y2 + 1

.

(b) Write an equation of each horizontal tangent line to the curve.

(c) The line through the origin with slope −1 is tangent to thecurve at point P . Find the x- and y-coordinates of point P .

Example 11.9.44 (1992 AB–4) [NO CALCULATOR] Consider thecurve defined by the equation y + cos y = x+ 1 for 0 ≤ y ≤ 2π.

(a) Finddy

dxin terms of y.

(b) Write an equation for each vertical tangent to the curve.

(c) Findd2y

dx2in terms of y.



Homework

11.I-141 [NO CALCULATOR] Consider the curve defined by x2 + xy + y2 = 27.

(a) Write an expression for the slope of the curve at any point (x, y).

(b) Determine whether the lines tangent to the curve at the x–interceptsof the curve are parallel. Show the analysis that leads to your con-clusion.

(c) Find the points on the curve where the lines tangent to the curve arevertical.

11.I-142 [NO CALCULATOR] Consider the curve given by xy2 − x3y = 6.

(a) Show thatdy

dx=

3x2y − y2

2xy − x3.

(b) Find all points on the curve whose x–coordinate is 1, and write anequation for the tangent line at each of these points.

(c) Find the x-coordinate of each point on the curve where the tangentline is vertical



11.I-143 [NO CALCULATOR] Consider the curve given by x2 + 4y2 = 7 + 3xy.

(a) Show thatdy

dx=

3y − 2x8y − 3x

.

(b) Show that there is a point P with x–coordinate 3 at which the linetangent to the curve at P is horizontal. Find the y-coordinate of P .

(c) Find the value ofd2y

dx2at the point P found in part (b). Does the

curve have a local maximum, a local minimum, or neither at thepoint P? Justify your answer.

11.I-144 [NO CALCULATOR] Consider the closed curve in the xy-plane given by

x2 + 2x+ y4 + 4y = 5.

(a) Show thatdy

dx=− (x+ 1)2 (y3 + 1)

.

(b) Write an equation for the line tangent to the curve at the point(−2, 1).

(c) Find the coordinates of the two points on the curve where the linetangent to the curve is vertical.

(d) Is it possible for this curve to have a horizontal tangent at pointswhere it intersects the x–axis? Explain your reasoning.


AP Unit 11, Day 10: Differential Equations Again 371

11.10 Differential Equations Again


A coffeepot has the shape of a cylinder with radius 5 inches, asshown in the figure above. Let h be the depth of the coffee in thepot, measured in inches, where h is a function of time t, measured inseconds. The volume V of coffee in the pot is changing at the rate of−5π√h cubic inches per second. (The volume V of a cylinder with

radius r and height h is V = πr2h.)

(a) Show thatdh

dt= −√h

5.

(b) Given that h = 17 at time t = 0, solve the differential equationdh

dt= −√h

5for h as a function of t.

(c) At what time t is the coffeepot empty?



Example 11.10.46 (2005B AB-6) [NO CALCULATOR] Consider


dx=−xy2

2. Let y = f(x) be the partic-

ular solution to this differential equation with the initial conditionf(−1) = 2.


(b) Write an equation for the line tangent to the graph of f atx = −1.

(c) Find the solution y = f(x) to the given differential equationwith the initial condition f(−1) = 2.

Example 11.10.47 (2004 AB-6) [NO CALCULATOR] Consider


dx= x2 (y − 1).




(b) While the slope field in part (a) is drawn at only twelve points,it is defined at every point in the xy-plane. Describe all pointsin the xy-plane for which the slopes are positive.

(c) Find the particular solution y = f(x) to the given differentialequation with the initial condition f(0) = 3.

Example 11.10.48 (1993 AB-6) [NO CALCULATOR] Let P (t)represent the number of wolves in a population at time t years,when t ≥ 0. The population P (t) is increasing at a rate directlyproportional to 800− P (t), where the constant of proportionality isk.

(a) If P (0) = 500, find P (t) in terms of t and k.

(b) If P (2) = 700, find k.

(c) Find limt→∞

P (t).

Homework

11.J-145 [NO CALCULATOR] Consider the differential equationdy

dx=

12x+y−1.


(b) Findd2y

dx2, in terms of x and y. Describe the region in the xy–plane

in which all solution curves to the differential equation are concaveup.

(c) Let y = f(x) be a particular solution to the differential equation withthe initial condition f(0) = 1. Does f have a relative minimum, arelative maximum, or neither at x = 0? Justify your answer.

(d) Find the values of the constants m and b, for which y = mx+ b is asolution to the differential equation.




dx=x

y, where

y 6= 0.

(a) The slope field for the given differential equation is shown below.Sketch the solution curve that passes through the point (3,−1), andsketch the solution curve that passes through the point (1, 2).(Note: The points (3,−1) and (1, 2) are indicated in the figure.)

(b) Write an equation for the line tangent to the solution curve thatpasses through the point (1, 2).

(c) Find the particular solution y = f(x) to the differential equationwith the initial condition f(3) = −1, and state its domain.

11.J-147 [NO CALCULATOR] Consider the differential equation given bydy

dx=

x (y − 1)2.

(a) On the axes provided, sketch a slope field for the given differentialequation at the eleven points indicated.

(b) Use the slope field for the given differential equation to explain whya solution could not have the graph shown below.




(d) Find the range of the solution found in part (c).


dx= (y − 1)2 cos (πx).


(b) There is a horizontal line with equation y = c that satisfies the dif-ferential equation. Find the value of c.

(c) Find the particular solution y = f(x) to the differential equationwith the initial condition f(1) = 0.




AP Unit 11, Day 11: Related Rates Again 377

11.11 Related Rates Again

Example 11.11.49 (1996 AB-5)

An oil storage tank has the shape shown above, obtained by revolv-

ing the curve y =9

625x4 from x = 0 to x = 5 about the y-axis,

where x and y are measured in feet. Oil flows into the tank at theconstant rate of 8 cubic feet per minute.

(a) Find the volume of the tank. Indicate units of measure.

(b) To the nearest minute, how long would it take to fill the tankif the tank was empty initially?

(c) Let h be the depth, in feet, of oil in the tank. How fast is thedepth of the oil in the tank increasing when h = 4? Indicateunits of measure.

Example 11.11.50 (1999 AB-6)



In the figure above, line ` is tangent to the graph of y =1x2

at point

P , with coordinates(w,

1w2

), where w > 0. Point Q has coordi-

nates (w, 0). Line ` crosses the x-axis at point R, with coordinates(k, 0).

(a) Edited Write an equation of the line tangent to the graph at(3,

19

).

(b) Edited Write an equation of line `, which is tangent to the graph

at point P(w,

1w2

), for w > 0. Hence show that, for w > 0,

k =32w.

(c) Suppose that w is increasing at the constant rate of 7 units persecond. When w = 5, what is the rate of change of k withrespect to time?

(d) Suppose that w is increasing at the constant rate of 7 units persecond. When w = 5, what is the rate of change of the area of4PQR with respect to time? Determine whether the area isincreasing or decreasing at this instant.

Example 11.11.51 (2008B–2) For time t ≥ 0 hours, let r(t) =120

(1− e−10t2

)represent the speed, in kilometers per hour, at

which a car travels along a straight road. The number of litersof gasoline used by the car to travel x kilometers is modeled byg(x) = 0.05x

(1− e−x/2

).

(a) How many kilometers does the car travel during the first 2 hours?

(b) Find the rate of change with respect to time of the number ofliters of gasoline used by the car when t = 2 hours. Indicateunits of measure.

(c) How many liters of gasoline have been used by the car when itreaches a speed of 80 kilometers per hour?


AP Unit 11, Day 11: Related Rates Again 379

Homework

11.K-149 At a certain height, a tree trunk has a circular cross section. The radiusR(t) of that cross section grows at a rate modeled by the function

dR

dt=

116(3 + sin

(t2))

centimeters per year

for 0 ≤ t ≤ 3, where time t is measured in years. At time t = 0, theradius is 6 centimeters. The area of the cross section at time t is denotedby A(t).

(a) Write an expression, involving an integral, for the radius R(t) for0 ≤ t ≤ 3. Use your expression to find R(3).

(b) Find the rate at which the cross-sectional area A(t) is increasing attime t = 3 years. Indicate units of measure.

(c) Evaluate∫ 3

0

A′(t) dt. Using appropriate units, interpret the meaning

of that integral in terms of cross-sectional area.

11.K-150 [NO CALCULATOR]

t(minutes)

0 2 5 7 11 12

r′(t)(feet per minute)

5.7 4.0 2.0 1.2 0.6 0.5

The volume of a spherical hot air balloon expands as the air inside theballoon is heated. The radius of the balloon, in feet, is modeled by atwice-differentiable function r of time t, where t is measured in minutes.For 0 < t < 12, the graph of r is concave down. The table above givesselected values of the rate of change, r′(t), of the radius of the ba1loonover the time interval 0 ≤ t ≤ 12. The radius of the balloon is 30 feetwhen t = 5.

(Note: The Volume of a sphere of radius r is given by V =43πr3.)

(a) Estimate the radius of the balloon when t = 5.4 using the tangentline approximation at t = 5. Is your estimate greater than or lessthan the true value? Give a reason for your answer.

(b) Find the rate of change of the volume of the balloon with respect totime when t = 5. Indicate units of measure.

(c) Use a right Riemann sum with the five subintervals indicated by thedata in the table to approximate

∫ 12

0r′(t) dt. Using correct units, ex-

plain the meaning of∫ 12

0r′(t) dt in terms of the radius of the balloon.

(d) Is your approximation in part (c) greater than or less than∫ 12

0r′(t) dt?

Give a reason for your answer.



11.K-151 The wind chill is the temperature, in degrees Fahrenheit (◦F), a humanfeels based on the air temperature, in degrees Fahrenheit, and the windvelocity v, in miles per hour (mph). If the air temperature is 32◦F, then thewind chill is given by w(v) = 55.6− 22.1v0.16 and is valid for 5 ≤ v ≤ 60.

(a) Find W ′(20). Using correct units, explain the meaning of W ′(20) interms of the wind chill.

(b) Find the average rate of change of W over the interval 5 ≤ v ≤ 60.Find the value of v at which the instantaneous rate of change of W isequal to the average rate of change of W over the interval 5 ≤ v ≤ 60.

(c) Over the time interval 0 ≤ t ≤ 4 hours, the air temperature is aconstant 32◦F. At time t = 0, the wind velocity is v = 20 mph. Ifthe wind velocity increases at a constant rate of 5 mph per hour,what is the rate of change of the wind chill with respect to time att = 3 hours? Indicate units of measure.

11.K-152 Oil is leaking from a pipeline on the surface of a lake and forms an oilslick whose volume increases at a constant rate of 2000 cubic centimetersper minute. The oil slick takes the form of a right circular cylinder withboth its radius and height changing with time. (Note: The volume V of aright circular cylinder with radius r and height h is given by V = πr2h.)

(a) At the instant when the radius of the oil slick is 100 centimeters andthe height is 0.5 centimeter, the radius is increasing at the rate of2.5 centimeters per minute. At this instant, what is the rate of changeof the height of the oil slick with respect to time, in centimeters perminute?

(b) A recovery device arrives on the scene and begins removing oil. Therate at which oil is removed is R(t) = 400

√t cubic centimeters per

minute, where t is the time in minutes since the device began work-ing. Oil continues to leak at the rate of 2000 cubic centimeters perminute. Find the time t when the oil slick reaches its maximumvolume. Justify your answer.

(c) By the time the recovery device began removing oil, 60, 000 cubiccentimeters of oil had already leaked. Write, but do not evaluate, anexpression involving an integral that gives the volume of oil at thetime found in part (b).



11.12 Area and Volume

Example 11.12.53 (2006 AB-1)

Let R be the shaded region bounded by the graph of y = lnx andthe line y = x− 2, as shown above.


(b) Find the volume of the solid generated when R is rotated aboutthe horizontal line y = −3.

(c) Write, but do not evaluate, an integral expression that can beused to find the volume of the solid generated when R is rotatedabout the y–axis.



Example 11.12.54 (1998 AB-1) Let R be the region bounded bythe x–axis, the graph of y =

√x, and the line x = 4.

(a) Find the area of the region R.

(b) Find the value of h such that the vertical line x = h divides theregion R into two regions of equal area.

(c) Find the volume of the solid generated whenR is revolved aboutthe x-axis.

(d) The vertical line x = k divides the region R into two regionssuch that when these two regions are revolved about the x–axis,they generate solids with equal volumes. Find the value of k.

Example 11.12.55 (2001 AB-1)

Let R and S be the regions in the first quadrant shown in the figureabove. The region R is bounded by the x–axis and the graphs ofy = 2 − x3 and y = tanx. The region S is bounded by the y–axisand the graphs of y = 2− x3 and y = tanx.



(c) Find the volume of the solid generated when S is revolved aboutthe x–axis.



Example 11.12.56 (2004 AB-2)

Let f and g be the functions given by f(x) = 2x (1− x) and g(x) =3 (x− 1)

√x for 0 ≤ x ≤ 1. The graphs of f and g are shown in the

figure above.

(a) Find the area of the shaded region enclosed by the graphs of fand g.

(b) Find the volume of the solid generated when the shaded re-gion enclosed by the graphs of f and g is revolved about thehorizontal line y = 2.

(c) Let h be the function given by h(x) = kx (1− x) for 0 ≤ x ≤ 1.For each k > 0, the region (not shown) enclosed by the graphsof h and g is the base of a solid with square cross sectionsperpendicular to the x–axis. There is a value of k for which thevolume of this solid is equal to 15. Write, but do not solve, anequation involving an integral expression that could be used tofind the value of k.



Homework

11.L-153 Let R be the region bounded by the graphs of y = sin(πx) and y = x3−4x,as shown in the figure below.


(b) The horizontal line y = −2 splits the region R into two parts. Write,but do not evaluate, an integral expression for the area of the partof R that is below this horizontal line.

(c) The region R is the base of a solid. For this solid, each cross sectionperpendicular to the x–axis is a square. Find the volume of this solid.

(d) The region R models the surface of a small pond. At all points in Rat a distance x from the y–axis, the depth of the water is given byh(x) = 3− x. Find the volume of water in the pond.

11.L-154 [NO CALCULATOR] Let R be the region bounded by the graphs of y =√x and y =

x

2, as shown in the figure below.


(b) The region R is the base of a solid. For this solid, the cross sectionsperpendicular to the x-axis are squares. Find the volume of this solid.

(c) Write, but do not evaluate, an integral expression for the volume ofthe solid generated when R is rotated about the horizontal line y = 2.



11.L-155 Let R be the region in the first quadrant bounded by the graphs of y =√x

and y =x

3.


(b) Find the volume of the solid generated when R is rotated about thevertical line x = −1.

(c) The region R is the base of a solid. For this solid, the cross sectionsperpendicular to the y–axis are squares. Find the volume of thissolid.

11.L-156 [NO CALCULATOR] Let R be the region in the first quadrant enclosedby the graphs of y = 2x and y = x2, as shown in the figure below.


(b) The region R is the base of a solid. For this solid, at each x the crosssection perpendicular to the x–axis has area A(x) = sin

(π2x)

. Findthe volume of the solid.

(c) Another solid has the same base R. For this solid, the cross sectionsperpendicular to the y–axis are squares. Write, but do not evaluate,an integral expression for the volume of the solid.





11.13 Tangent Lines

Example 11.13.57 (1999 AB-4) Suppose that the function f has acontinuous second derivative for all x, and that f(0) = 2, f ′(0) = −3,and f ′′(0) = 0. Let g be a function whose derivative is given byg′(x) = e−2x (3f(x) + 2f ′(x)) for all x.

(a) Write an equation of the line tangent to the graph of f at thepoint where x = 0.

(b) Is there sufficient information to determine whether or not thegraph of f has a point of inflection when x = 0? Explain youranswer.

(c) Given that g(0) = 4, write an equation of the line tangent tothe graph of g at the point where x = 0.

(d) Show that g′′(x) = e−2x (−6f(x)− f ′(x) + 2f ′′(x)). Does ghave a local maximum at x = 0? Justify your answer.



Example 11.13.58 (2002B AB–2) The number of gallons, P (t), ofa pollutant in a lake changes at the rate P ′(t) = 1−3e−0.2

√t gallons

per day, where t is measured in days. There are 50 gallons of thepollutant in the lake at time t = 0. The lake is considered to be safewhen it contains 40 gallons or less of pollutant.

(a) Is the amount of pollutant increasing at time t = 9? Why orwhy not?

(b) For what value of t will the number of gallons of pollutant beat its minimum? Justify your answer.

(c) Is the lake safe when the number of gallons of pollutant is atits minimum? Justify your answer.

(d) An investigator uses the tangent line approximation to P (t) att = 0 as a model for the amount of pollutant in the lake. Atwhat time t does this model predict that the lake becomes safe?

Example 11.13.59 (1996 AB-6)

Line ` is tangent to the graph of y = x − x2

500at the point Q, as

shown in the figure above.

(a) Find the x-coordinate of point Q.

(b) Write an equation for line `.

(c) Suppose the graph of y = x− x2

500shown in the figure, where x

and y are measured in feet, represents a hill. There is a 50-foottree growing vertically at the top of the hill. Does a spotlightat point P directed along line ` shine on any part of the tree?Show the work that leads to your conclusion.



Example 11.13.60 (2001 AB–4) [NO CALCULATOR] Let h be afunction defined for all x 6= 0 such that h(4) = −3 and the derivative

of h is given by h′(x) =x2 − 2x

for all x 6= 0.

(a) Find all values of x for which the graph of h has a horizontaltangent, and determine whether h has a local maximum, a lo-cal minimum, or neither at each of these values. Justify youranswers.

(b) On what intervals, if any, is the graph of h concave up? Justifyyour answer.

(c) Write an equation for the line tangent to the graph of h atx = 4.

(d) Does the line tangent to the graph of h at x = 4 lie above orbelow the graph of h for x > 4 Why?



Homework

11.M-157 (Wed, 21-Apr, front) [NO CALCULATOR] Let f be the function given by

f(x) =lnxx

for all x > 0. The derivative of f is given by f ′(x) =1− lnxx2

.

(a) Write an equation for the line tangent to the graph of f at x = e2.

(b) Find the x–coordinate of the critical point of f . Determine whetherthis point is a relative minimum, a relative maximum, or neither forthe function f . Justify your answer.

(c) The graph of the function f has exactly one point of inflection. Findthe x–coordinate of this point.

(d) Find limx→0+

f(x).

11.M-158 (Wed, 21-Apr, back) [NO CALCULATOR] Let f be the function givenby f(x) = (lnx) (sinx). The figure below shows the graph of f for 0 <

x ≤ 2π. The function g is defined by g(x) =∫ x

1

f(t) dt for 0 < x ≤ 2π.

(a) Find g(1) and g′(1).

(b) On what intervals, if any, is g increasing? Justify your answer.

(c) For 0 < x ≤ 2π, find the value of x at which g has an absoluteminimum. Justify your answer.

(d) For 0 < x ≤ 2π, is there a value of x at which the graph of g istangent to the x–axis? Explain why or why not.

11.M-159 (Thu, 22-Apr) No problems due today. Test 10 is on Thursday, Apr 22.There will not be retests; make-ups will not be taken from classwork andhomework.

11.M-160 (Thu, 22-Apr) You may want to get a jump start on your homework thatis due Friday, 23-Apr.



11.M-161 (Fri, 23-Apr, front) [NO CALCULATOR] The figure below shows thegraph of f ′, the derivative of the function f , on the closed interval −1 ≤x ≤ 5. The graph of f ′ has horizontal tangent lines at x = 1 and x = 3.The function f is twice differentiable with f(2) = 6.

(a) Find the x-coordinate of each of the points of inflection of the graphof f . Give a reason for your answer.

(b) At what value of x does f attain its absolute minimum value on theclosed interval −1 ≤ x ≤ 5? At what value of x does f attain itsabsolute maximum value on the closed interval −1 ≤ x ≤ 5? Showthe analysis that leads to your answers.

(c) Let g be the function defined by g(x) = xf(x). Find an equation forthe line tangent to the graph of g at x = 2.



11.M-162 (Fri, 23-Apr, back) The temperature, in degrees Celsius (◦C), of the waterin a pond is a differentiable function W of time t. The table below showsthe water temperature as recorded every 3 days over a 15-day period.

t W (t)(days) (◦C)

0 203 316 289 2412 2215 21

(a) Use data from the table to find an approximation for W ′(12). Showthe computations that lead to your answer. Indicate units of measure.

(b) Approximate the average temperature, in degrees Celsius, of the wa-ter over the time interval 0 ≤ t ≤ 15 days by using a trapezoidalapproximation with subintervals of length 4t = 3 days.

(c) A student proposes that function P , given by P (t) = 20 + 10te(−t/3),as a model for the temperature of the water in the pond at time t,where t is measured in days and P (t) is measured in degrees Celsius.Find P ′(12). Using appropriate units, explain the meaning of youranswers in terms of water temperature.

(d) Use the function P defined in part (c) to find the average value, indegrees Celsius, of P (t) over the time interval 0 ≤ t ≤ 15 days.

11.M-163 (Mon, 26-Apr) No problems due today.

11.M-164 (Mon, 26-Apr) Get ready for the big push over the next week.


AP Unit 11, Day 14: Miscellany 393

11.14 Miscellany

Example 11.14.65 (2009 AB–2) The rate at which people enteran auditorium for a rock concert is modeled by the function R givenby R(t) = 1380t2 − 675t3 for 0 ≤ t ≤ 2 hours; R(t) is measured inpeople per hour. No one is in the auditorium at time t = 0, when thedoors open. The doors close and the concert begins at time t = 2.

(a) How many people are in the auditorium when the concert be-gins?

(b) Find the time when the rate at which people enter the audito-rium is a maximum. Justify your answer.

(c) The total wait time for all the people in the auditorium is foundby adding the time each person waits, starting at the timethe person enters the auditorium and ending when the concertbegins. The function w models the total wait time for all thepeople who enter the auditorium before time t. The derivativeof w is given by w′(t) = (2− t)R(t). Find w(2) − w(1), thetotal wait time for those who enter the auditorium after timet = 1.

(d) On average, how long does a person wait in the auditoriumfor the concert to begin? Consider all people who enter theauditorium after the doors open, and use the model for totalwait time from part (c).



Example 11.14.66 (1998 AB-5) The temperature outside a houseduring a 24-hour period is given by

F (t) = 80− 10 cos(πt

12

), 0 ≤ t ≤ 24,

where F (t) is measured in degrees Fahrenheit and t is measured inhours.

(a) Sketch the graph of F on the grid below

(b) Find the average temperature, to the nearest degree Fahrenheit,between t = 6 and t = 14.

(c) An air conditioner cooled the house whenever the outside tem-perature was at or above 78 degrees Fahrenheit. For what val-ues of t was the air conditioner cooling the house?

(d) The cost of cooling the house accumulates at the rate of $0.05per hour for each degree the outside temperature exceed 78degrees Fahrenheit. What was the total cost, to the nearestcent, to cool the house for this 24-hour period?

Example 11.14.67 (2009 AB–3) Mighty Cable Company manufac-tures cable that sells for $120 per meter. For a cable of fixed length,the cost of producing a portion of the cable varies with its distancefrom the beginning of the cable. Mighty reports that the cost toproduce a portion of a cable that is x meters from the beginning ofthe cable is 6

√x dollars per meter. (Note: Profit is defined to be the

difference between the amount of money received by the companyfor selling the cable and the companys cost of producing the cable.)

(a) Find Mightys profit on the sale of a 25–meter cable.




25

6√x dx in the

context of this problem.(c) Write an expression, involving an integral, that represents Mightys

profit on the sale of a cable that is k meters long.(d) Find the maximum profit that Mighty could earn on the sale

of one cable. Justify your answer.

Example 11.14.68 (2006 AB-2)

At an intersection in Thomasville, Oregon, cars turn left at the rate

L(t) = 60√t sin2

(t

3

)cars per hour over the time interval 0 ≤ t ≤ 18

hours. The graph of y = L(t) is shown above.

(a) To the nearest whole number, find the total number of carsturning left at the intersection over the time interval 0 ≤ t ≤ 18hours.

(b) Traffic engineers will consider turn restrictions when L(t) ≥ 150cars per hour. Find all values of t for which L(t) ≥ 150 andcompute the average value of L over this time interval. Indicateunits of measure.

(c) Traffic engineers will install a signal if there is any two-hourtime interval during which the product of the total number ofcars turning left and the total number of oncoming cars travel-ing straight through the intersection is greater than 200, 000. Inevery two–hour time interval, 500 oncoming cars travel straightthrough the intersection. Does this intersection require a trafficsignal? Explain the reasoning that leads to your conclusion.



Homework

11.N-165 (Tue, 27-Apr, front) Traffic flow is defined as the rate at which cars passthrough an intersection, measured in cars per minute. The traffic flow ata particular intersection is modeled by the function F defined by

F (t) = 82 + 4 sin(t

2

)for 0 ≤ t ≤ 30,

where F (t) is measured in cars per minute and t is measured in minutes.

(a) To the nearest whole number, how many cars pass through the in-tersection over the 30–minute period?

(b) Is the traffic flow increasing or decreasing at t = 7? Give a reasonfor your answer.

(c) What is the average value of the traffic flow over the time interval10 ≤ t ≤ 15? Indicate units of measure.

(d) What is the average rate of change of the traffic flow over the timeinterval 10 ≤ t ≤ 15? Indicate units of measure.

11.N-166 (Tue, 27-Apr, back) A storm washed away sand from a beach, causing theedge of the water to get closer to a nearby road. The rate at which thedistance between the road and the edge of the water was changing duringthe storm is modeled by f(t) =

√t + cos t − 3 meters per hour, t hours

after the storm began. The edge of the water was 35 meters from the roadwhen the storm began, and the storm lasted 5 hours. The derivative of

f(t) is f ′(t) =1

2√t− sin t.

(a) What was the distance between the road and the edge of the waterat the end of the storm?

(b) Using correct units, interpret the value f ′(4) = 1.007 in terms of thedistance between the road and the edge of the water.

(c) At what time during the 5 hours of the storm was the distance be-tween the road and the edge of the water decreasing most rapidly?Justify your answer.

(d) After the storm, a machine pumped sand back onto the beach sothat the distance between the road and the edge of the water wasgrowing at a rate of g(p) meters per day, where p is the number ofdays since pumping began. Write an equation involving an integralexpression whose solution would give the number of days that sandmust be pumped to restore the original distance between the roadand the edge of the water.



11.N-167 (Wed, 28-Apr, front) The figure below is the graph of a function of x,which models the height of a skateboard ramp. The function meets thefollowing requirements.

(i) At x = 0, the value of the function is 0, and the slope of the graphof the function is 0.

(ii) At x = 4, the value of the function is 1, and the slope of the graphof the function is 1.

(iii) Between x = 0 and x = 4, the function is increasing.

(a) Let f(x) = ax2, where a is a nonzero constant. Show that it is notpossible to find a value for a so that f meets requirement (ii) above.

(b) Let g(x) = cx3− x2

16, where c is a nonzero constant. Find the value of

c so that g meets requirement (ii) above. Show the work that leadsto your answer.

(c) Using the function g and your value of c from part (b), show that gdoes not meet requirement (iii) above.

(d) Let h(x) =xn

k, where k is a nonzero constant and n is a positive

integer. Find the values of k and n so that h meets requirement (ii)above. Show that h also meets requirements (i) and (iii) above.

11.N-168 (Wed, 28-Apr, back) The rate of fuel consumption, in gallons per minute,recorded during an airplane flight is given by a twice-differentiable andstrictly increasing function R of time t. The graph of R and a table ofselected values of R(t), for the time interval 0 ≤ t ≤ 90 minutes, are shownbelow.



(a) Use data from the table to find an approximation for R′(45). Showthe computations that lead to your answer. Indicate units of measure.

(b) The rate of fuel consumption is increasing fastest at time t = 45minutes. What is the value of R′′(45)? Explain your reasoning.

(c) Approximate the value of∫ 90

0

R(t) dt using a left Riemann sum with

the five subintervals indicated by the data in the table. Is this nu-

merical approximation less than the value of∫ 90

0

R(t) dt? Explain

your reasoning.

(d) For 0 < b ≤ 90 minutes, explain the meaning of∫ b

0

R(t) dt in terms of

fuel consumption for the plane. Explain the meaning of1b

∫ b

0

R(t) dt

in terms of fuel consumption for the plane. Indicate units of measurein both answers.

11.N-169 (Thu, 29-Apr, front) [NO CALCULATOR] Let f be a function that isdifferentiable for all real numbers. The table below gives the values of fand its derivative f ′ for selected points x in the closed interval −1.5 ≤x ≤ 1.5. The second derivative of f has the property that f ′′(x) > 0 for−1.5 ≤ x ≤ 1.5.

x −1.5 −1.0 −0.5 0 0.5 1.0 1.5f(x) −1 −4 −6 −7 −6 −4 −1f ′(x) −7 −5 −3 0 3 5 7

(a) Evaluate∫ 1.5

0

(3f ′(x) + 4) dx. Show the work that leads to youranswer.

(b) Write an equation of the line tangent to the graph of f at the pointwhere x = 1. Use this line to approximate the value of f(1.2). Is this



approximation greater than or less than the actual value of f(1.2)?Give a reason for your answer.

(c) Find a positive real number r having the property that there mustexist a value c with 0 < c < 0.5 and f ′′(c) = r. Give a reason foryour answer.

(d) Let g be the function given by g(x) =

{2x2 − x− 7 for x < 02x2 + x− 7 for x ≥ 0

. The

graph of g passes through each of the points (x, f(x)) given in thetable above. Is it possible that f and g are the same function? Givea reason for your answer.

11.N-170 (Thu, 29-Apr, back) [NO CALCULATOR] The temperature, in degreesCelsius (◦C), of an oven being heated is modeled by an increasing differ-entiable function H of time t, where t is measured in minutes. The tablebelow gives the temperature as recorded every 4 minutes over a 16–minuteperiod.

t (minutes) 0 4 8 12 16H(t) (◦C) 65 68 73 80 90

(a) Use the data in the table to estimate the instantaneous rate at whichthe temperature of the oven is changing at time t = 10. Show thecomputations that lead to your answer. Indicate units of measure.

(b) Write an integral expression in terms of H for the average temper-ature of the oven between time t = 0 and time t = 16. Estimatethe average temperature of the oven using a left Riemann sum withfour subintervals of equal length. Show the computations that leadto your answer.

(c) Is your approximation in part (b) an underestimate or an overesti-mate of the average temperature? Give a reason for your answer.

(d) Are the data in the table consistent with or do they contradict theclaim that the temperature of the oven is increasing at an increasingrate? Give a reason for your answer.

11.N-171 (Fri, 30-Apr, front) [NO CALCULATOR] The twice–differentiable func-tion f is defined for all real numbers and satisfies the following conditions:

f(0) = 2, f ′(0) = −4, and f ′′(0) = 3.

(a) The function g is given by g(x) = eax + f(x) for all real numbers,where a is a constant. Find g′(0) and g′′(0) in terms of a. Show thework that leads to your answers.

(b) The function h is given by h(x) = cos (kx) f(x) for all real numbers,where k is a constant. Find h′(x) and write an equation for the linetangent to the graph of h at x = 0.



11.N-172 (Fri, 30-Apr, back) The rate at which water flows out of a pipe, in gallonsper hour, is given by a differentiable function R of time t. The table belowshows the rate as measured every 3 hours for a 24-hour period.

t R(t)(hours) (gallons per hour)

0 9.63 10.46 10.89 11.212 11.415 11.318 10.721 10.224 9.6

(a) Use a midpoint Riemann sum with 4 subdivisions of equal length to

approximate∫ 24

0

R(t) dt. Using correct units, explain the meaning

of your answer in terms of water flow.

(b) Is there some time t, 0 < t < 24, such that R′(t) = 0? Justify youranswer.

(c) The rate of water flowR(t) can be approximated byQ(t) =179(768 + 23t− t2

).

Use Q(t) to approximate the average rate of water flow during the24-hour time period. Indicate units of measure.

11.N-173 (Mon, 3-May, front) Let f be the function given by f(x) = 4x2 − x3, andlet ` be the line y = 18 − 3x, where ` is tangent to the graph of f . LetR be the region bounded by the graph of f and the x-axis, and let S bethe region bounded by the graph of f , the line `, and the x-axis, as shownbelow.



(a) Show that ` is tangent to the graph of y = f(x) at the point x = 3.


(c) Find the volume of the solid generated when R is revolved about thex-axis.

11.N-174 (Mon, 3-May, back) Let f be the function given by f(x) =x3

4− x

2

3− x

2+

3 cosx. Let R be the shaded region in the second quadrant bounded bythe graph of f , and let S be the shaded region bounded by the graph off and line `, the line tangent to the graph of f at x = 0, as shown below.



(c) Write, but do not evaluate, an integral expression that can be usedto find the area of S.



11.N-175 (Tue, 4-May, front of the provided sheet) [NO CALCULATOR] Consider


dx= −2x

y.


(b) Let y = f(x) be the particular solution to the differential equationwith the initial condition f(1) = −1. Write an equation for the linetangent to the graph of f at (1,−1) and use it to approximate f(1.1).




11.N-176 (Tue, 4-May, back of the provided sheet) We will talk more about this asthe time approaches, but commit to the following:

For Tuesday:

(a) Turn in problems 175 and 176 on the provided sheet. Beyond that,restrict yourself to very limited, light reviewing. No cramming thenight before

(b) Take some time to reflect on how much you’ve learned this year.

(c) Get at least eight hours sleep before your exam. This is not asuggestion. It is an assignment, part of your final assignment for thisclass. Some of you have already heard me say this, but I will repeat itanyway. When neurologists study sleep deprivation, with PET-scansand the like, they notice that one of the quickest things to go is yourability to do mathematics. And the more complicated the math, thefaster it goes. So get a good night’s sleep.

For Wednesday:

(a) Eat a good breakfast. Your brain uses a lot of energy. The ancientGreeks used to think that the center of thinking was the stomach,because they noticed when they sat around thinking a lot, they gothungry. We may know better now, but the point is that your brainneeds fuel to keep going, so you need to give it some good fuel thatwill last you through the end of your exam. Eat a good breakfast,with some protein; I’ll even go so far as to invoke the f-word: fiber.

(b) No cramming just before the exam, but do wake your brain up. Gothrough your flash cards. Relax. The hard part of AP Calculus is thestudying and the preparation. By this point, all that is finished.

(c) When you get to the test site, please go out of your way to be sin-cerely nice to everyone: including Mr. Guinther, any other proc-tors, and your peers. I want people to to out of their way to tell mehow nice you were.

(d) During the exam, know that there will be some problems that youdon’t immediately know how to do. Don’t panic. Do breathe. Don’tstare. Do write something. Jot down formulas you think might beappropriate. A lot of times you may start writing, having no realsense of how you’re going to get to the answer, but once you startwriting, it may take only a couple steps to get to the answer.

(e) After the exam, take some time to acknowledge what you’ve accom-plished. Celebrate in a way that’s not going to interfere with yourother upcoming exams.




AP Unit 11, Day 15: More Miscellany 405

11.15 More Miscellany

Example 11.15.77 (2008B–3)

Distance from therivers edge (feet)

0 8 14 22 24

Depth of the water (feet) 0 7 8 2 0

A scientist measures the depth of the Doe River at Picnic Point. Theriver is 24 feet wide at this location. The measurements are takenin a straight line perpendicular to the edge of the river. The dataare shown in the table above. The velocity of the water at PicnicPoint, in feet per minute, is modeled by v(t) = 16 + 2 sin

(√t+ 10

)for 0 ≤ t ≤ 120 minutes.

(a) Use a trapezoidal sum with the four subintervals indicated bythe data in the table to approximate the area of the cross sec-tion of the river at Picnic Point, in square feet. Show thecomputations that lead to your answer.

(b) The volumetric flow at a location along the river is the productof the cross-sectional area and the velocity of the water at thatlocation. Use your approximation from part (a) to estimate theaverage value of the volumetric flow at Picnic Point, in cubicfeet per minute, from t = 0 to t = 120 minutes.

(c) The scientist proposes the function f , given by f(x) = 8 sin(πx

24

),

as a model for the depth of the water, in feet, at Picnic Pointx feet from the river’s edge. Find the area of the cross sectionof the river at Picnic Point based on this model.

(d) Recall that the volumetric flow is the product of the cross-sectional area and the velocity of the water at a location. Toprevent flooding, water must be diverted if the average valueof the volumetric flow at Picnic Point exceeds 2100 cubic feetper minute for a 20–minute period. Using your answer frompart (c), find the average value of the volumetric flow duringthe time interval 40 ≤ t ≤ 60 minutes. Does this value indicatethat the water must be diverted?

Example 11.15.78 (2006 AB–3) The graph of the function f shownbelow consists of six line segments. Let g be the function given by

g(x) =∫ x

0

f(t) dt.



(a) Find g(4), g′(4), and g′′(4).

(b) Does g have a relative minimum, a relative maximum, or nei-ther at x = 1? Justify your answer.

(c) Suppose that f is defined for all real numbers x and is periodicwith a period of length 5. The graph above shows two periodsof f . Given that g(5) = 2, find g(10) and write an equation forthe line tangent to the graph of g at x = 108.


Unit 12

Makeup

407

408 AP Unit 12 (Makeup)

12.1 MU: Differential Equations

Replaces problems 107-110, 145-148, 175

Homework

12.A-107 (2010–6) [NO CALCULATOR] Solutions to the differential equationdy

dx=

xy3 also satisfyd2y

dx2= y3

(1 + 3x2y2

). Let y = f(x) be a particular

solution to the differential equationdy

dx= xy3 with f(1) = 2.

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

(b) Use the tangent line equation from part (a) to approximate f(1.1).Given that f(x) > 0 for 1 < x < 1.1, is the approximation for f(1.1)greater than or less than f(1.1)? Explain your reasoning.

(c) Find the particular solution y = f(x) with initial condition f(1) = 2.

12.A-108 (2010B–5) [NO CALCULATOR] Consider the differential equationdy

dx=

x+ 1y

(a) On the axes provided, sketch a slope field for the given differentialequation at the twelve points indicated, and for −1 < x < 1, sketchthe solution curve that passes through the point (0,−1).

(b) While the slope field in part (a) is drawn at only twelve points, it isdefined at every point in the xy–plane for which y 6= 0. Describe all

points in the xy–plane, y 6= 0, for whichdy

dx= −1.



AP Unit 12, Day 2: MU: Related Rates 409

12.2 MU: Related Rates

Replaces problems 115-118, 149-152

Makeup

12.B-115 (2010B–3) The figure below shows an aboveground swimming pool in theshape of a cylinder with a radius of 12 feet and a height of 4 feet.

The pool contains 1000 cubic feet of water at time t = 0. During thetime interval 0 ≤ t ≤ 12 hours, water is pumped into the pool at the rateP (t) cubic feet per hour. The table below gives values of P (t) for selectedvalues of t.

t 0 2 4 6 8 10 12P (t) 0 46 53 57 60 62 63

During the same time interval, water is leaking from the pool at the rateR(t) cubic feet per hour, where R(t) = 25e−0.05t. (Note: The volume Vof a cylinder with radius r and height h is given by V = πr2h.)

(a) Use a midpoint Riemann sum with three subintervals of equal lengthto approximate the total amount of water that was pumped into thepool during the time interval 0 ≤ t ≤ 12 hours. Show the computa-tions that lead to your answer.

(b) Calculate the total amount of water that leaked out of the pool duringthe time interval 0 ≤ t ≤ 12 hours.

(c) Use the results from parts (a) and (b) to approximate the volume ofwater in the pool at time t = 12 hours. Round your answer to thenearest cubic foot.

(d) Find the rate at which the volume of water in the pool is increasingat time t = 8 hours. How fast is the water level in the pool rising att = 8 hours? Indicate units of measure in both answers.



12.3 MU: Graphs

Replaces problems 121-124, 158, 160

Makeup

12.C-121 (2010–5) [NO CALCULATOR] The function g is defined and differentiableon the closed interval [−7, 5] and satisfies g(0) = 5. The graph of y = g′(x),the derivative of g, consists of a semicircle and three line segments, asshown in the figure below.

(a) Find g(3) and g(−2).

(b) Find the x–coordinate of each point of inflection of the graph ofy = g(x) on the interval −7 < x < 5. Explain your reasoning.

(c) The function h is defined by h(x) = g(x) − 12x2. Find the x–

coordinate of each critical point of h, where −7 < x < 5, and classifyeach critical point as the location of a relative minimum, relativemaximum, or neither a minimum nor a maximum. Explain yourreasoning.


AP Unit 12, Day 3: MU: Graphs 411

12.C-122 (2009B AB–3)

A continuous function f is defined on the closed interval −4 ≤ x ≤ 6. Thegraph of f consists of a line segment and a curve that is tangent to thex–axis at x = 3, as shown in the figure above. On the interval 0 < x < 6,the function f is twice differentiable, with f ′′(x) > 0.

(a) Is f differentiable at x = 0? Use the definition of the derivative withone-sided limits to justify your answer.

(b) For how many values of a, −4 ≤ a < 6, is the average rate of changeof f on the interval [a, 6] equal to 0 ? Give a reason for your answer.

(c) Is there a value of a, −4 ≤ a < 6, for which the Mean Value Theorem,applied to the interval [a, 6], guarantees a value c, a < c < 6, at which

f ′(c) =13

? Justify your answer.

(d) The function g is defined by g(x) =∫ x

0

f(t) dt for −4 ≤ x ≤ 6.

On what intervals contained in [−4, 6] is the graph of g concave up?Explain your reasoning.



12.C-123 (2009B AB–5) [NO CALCULATOR]

Let f be a twice-differentiable function defined on the interval −1.2 <x < 3.2 with f(1) = 2. The graph of f ′, the derivative of f , is shownabove. The graph of f ′ crosses the x–axis at x = −1 and x = 3 and has ahorizontal tangent at x = 2. Let g be the function given by g(x) = ef(x).

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

(b) For −1.2 < x < 3.2, find all values of x at which g has a localmaximum. Justify your answer.

(c) The second derivative of g is g′′(x) = ef(x)[(f ′(x))2 + f ′′(x)

]. Is

g′′(−1) positive, negative, or zero? Justify your answer.

(d) Find the average rate of change of g′, the derivative of g, over theinterval [1, 3].



12.C-124 (2006B AB–2)

Let f be the function defined for x ≥ 0 with f(0) = 5 and f ′, the firstderivative of f , given by f ′(x) = e−x/4 sin

(x2). The graph of y = f ′(x) is

shown above.

(a) Use the graph of f ′ to determine whether the graph of f is concaveup, concave down, or neither on the interval 1.7 < x < 1.9. Explainyour reasoning.

(b) On the interal 0 ≤ x ≤ 3, find the value of x at which f has anabsolute maximum. Justify your answer.

(c) Write an equation for the line tangent to the graph of f at x = 2.

12.C-125 (2008B–5) [NO CALCULATOR]

Let g be a continuous function with g(2) = 5. The graph of the piecewise-linear function g′, the derivative of g, is shown above for −3 ≤ x ≤ 7.

(a) Find the x–coordinate of all points of inflection of the graph of y =g(x) for −3 < x < 7. Justify your answer.



(b) Find the absolute maximum value of g on the interval −3 ≤ x ≤ 7.Justify your answer.

(c) Find the average rate of change of g(x) on the interval −3 ≤ x ≤ 7.

(d) Find the average rate of change of g′(x) on the interval −3 ≤ x ≤ 7.Does the Mean Value Theorem applied on the interval −3 ≤ x ≤ 7guarantee a value of c, for −3 < c < 7, such that g′′(c) is equal tothis average rate of change? Why or why not?

12.C-126 (2002B AB-4) [NO CALCULATOR]

The graph of a differentiable function f on the closed interval [−3, 15] isshown in the figure above. The graph of f has a horizontal tangent line

at x = 6. Let g(x) = 5 +∫ x

6

f(t) dt for −3 ≤ x ≤ 15.

(a) Find g(6), g′(6), and g′′(6).

(b) On what intervals is g decreasing? Justify your answer.

(c) On what intervals is the graph of g concave down? Justify youranswer.

(d) Find a trapezoidal approximation of∫ 15

−3

f(t) dt using six subintervals

of length 4t = 3.

12.C-127 (2000 AB–2)



Two runners, A and B, run on a straight racetrack for 0 ≤ t ≤ 10 seconds.The graph above, which consists of two line segments, shows the velocity,in meters per second, of Runner A. The velocity, in meters per second, of

Runner B is given by the function v defined by v(t) =24t

2t+ 3.

(a) Find the velocity of Runner A and the velocity of Runner B at timet = 2 seconds. Indicate units of measure.

(b) Find the acceleration of Runner A and the acceleration of Runner Bat time t = 2 seconds. Indicate units of measure.

(c) Find the total distance run by Runner A and the total distance runby Runner B over the time interval 0 ≤ t ≤ 10 seconds. Indicateunits of measure.

12.C-128 (1996 AB-1)

The figure above shows the graph of f ′, the derivative of a function f .The domain of f is the set of all real numbers x such that −3 < x < 5.



(a) For what values of x does f have a relative maximum? Why?

(b) For what values of x does f have a relative minimum? Why?

(c) On what intervals is the graph of f concave upward? Use f ′ to justifyyour answer.

(d) Suppose that f(1) = 0. In the xy-plane provided, draw a sketch thatshows the general shape of the graph of the function f on the openinterval 0 < x < 2.


AP Unit 12, Day 4: MU: Integral as Accumulator 417

12.4 MU: Integral as Accumulator


Makeup

12.D-125 (2010–1) There is no snow on Janet’s driveway when snow begins to fall atmidnight. From midnight to 9 a.m., snow accumulates on the driveway ata rate modeled by f(t) = 7tecos t cubic feet per hour, where t is measuredin hours since midnight. Janet starts removing snow at 6 a.m. (t = 6).The rate g(t), in cubic feet per hour, at which Janet removes snow fromthe driveway at time t hours after midnight is modeled by

g(t) =

0 for 0 ≤ t < 6125 for 6 ≤ t < 7108 for 7 ≤ t ≤ 9

(a) How many cubic feet of snow have accumulated on the driveway by6 a.m.?

(b) Find the rate of change of the volume of snow on the driveway at 8a.m.

(c) Let h(t) represent the total amount of snow, in cubic feet, that Janethas removed from the driveway at time t hours after midnight. Ex-press h as a piecewise-defined function with domain 0 ≤ t ≤ 9.

(d) How many cubic feet of snow are on the driveway at 9 a.m.?



12.D-126 (2010–3) There are 700 people in line for a popular amusement-park ridewhen the ride begins operation in the morning. Once it begins operation,the ride accepts passengers until the park closes 8 hours later. Whilethere is a line, people move onto the ride at a rate of 800 people per hour.The graph below shows the rate, r(t), at which people arrive at the ridethroughout the day. Time t is measured in hours from the time the ridebegins operation.

(a) How many people arrive at the ride between t = 0 and t = 3 ? Showthe computations that lead to your answer.

(b) Is the number of people waiting in line to get on the ride increasingor decreasing between t = 2 and t = 3 ? Justify your answer.

(c) At what time t is the line for the ride the longest? How many peopleare in line at that time? Justify your answers.

(d) Write, but do not solve, an equation involving an integral expressionof r whose solution gives the earliest time t at which there is no longera line for the ride.


AP Unit 12, Day 5: MU: Linear Motion 419

12.5 MU: Linear Motion

Replaces problems 129-132

Makeup

12.E-129 (2010B–4) [NO CALCULATOR] A squirrel starts at building A at timet = 0 and travels along a straight, horizontal wire connected to building B.For 0 ≤ t ≤ 18, the squirrel’s velocity is modeled by the piecewise-linearfunction defined by the graph below.

(a) At what times in the interval 0 < t < 18, if any, does the squirrelchange direction? Give a reason for your answer.

(b) At what time in the interval 0 ≤ t ≤ 18 is the squirrel farthest frombuilding A? How far from building A is the squirrel at that time?

(c) Find the total distance the squirrel travels during the time interval0 ≤ t ≤ 18.

(d) Write expressions for the squirrel’s acceleration a(t), velocity v(t),and distance x(t) from building A that are valid for the time interval7 < t < 10.



12.E-130 (2010B–6) [NO CALCULATOR] Two particles move along the x–axis. For0 ≤ t ≤ 6, the position of particle P at time t is given by p(t) = 2 cos

(π4t)

,

while the position of particle R at time t is given by r(t) = t3−6t2 +9t+3.

(a) For 0 ≤ t ≤ 6, find all times t during which particle R is moving tothe right.

(b) For 0 ≤ t ≤ 6, find all times t during which the two particles travelin opposite directions.

(c) Find the acceleration of particle P at time t = 3. Is particle Pspeeding up, slowing down, or doing neither at time t = 3? Explainyour reasoning.

(d) Write, but do not evaluate, an expression for the average distancebetween the two particles on the interval 1 ≤ t ≤ 3.

12.E-131 (1993 AB-2) [NO CALCULATOR] A particle moves on the x–axis so thatits position at any time t ≥ 0 is given by x(t) = 2te−t.

(a) Find the acceleration of the particle at t = 0.

(b) Find the velocity of the particle when its acceleration is 0.

(c) Find the total distance traveled by the particle from t = 0 to t = 5.

12.E-132 (2007 AB–4) [NO CALCULATOR] A particle moves along the x–axis withposition at time t given by x(t) = e−t sin t for 0 ≤ t ≤ 2π.

(a) Find the time at which the particle is farthest to the left. Justifyyour answer.

(b) Find the value of the constant A for which x(t) satisfies the equationAx′′(t) + x′(t) + x(t) = 0 for 0 < t < 2π.

12.E-133 (2003B AB-4) [NO CALCULATOR] A particle moves along the x-axiswith velocity at time t ≥ 0 given by v(t) = −1 + e1−t.


(b) Is the speed of the particle increasing at time t = 3? Give a reasonfor your answer.

(c) Find all values of t at which the particle changes direction. Justifyyour answer.

(d) Find the total distance traveled by the particle over the time interval0 ≤ t ≤ 3.


AP Unit 12, Day 5: MU: Linear Motion 421

12.E-134 (2005 AB-5) [NO CALCULATOR]

A car is traveling on a straight road. For 0 ≤ t ≤ 24 seconds, the car’svelocity v(t), in meters per second, is modeled by the piecewise-linearfunction defined in the graph above.

(a) Find∫ 24

0

v(t) dt. Using correct units, explain the meaning of∫ 24

0

v(t) dt.

(b) For each of v′(4) and v′(20), find the value or explain why it doesnot exist. Indicate units of measure.

(c) Let a(t) be the car’s acceleration at time t, in meters per second persecond. For 0 < t < 24, write a piecewise-defined function for a(t).

(d) Find the average rate of change of v over the interval 8 ≤ t ≤ 20.Does the Mean Value Theorem guarantee a value of c, for 8 < c < 20,such that v′(c) is equal to this average rate of change? Why or whynot?



12.E-135 (2005B AB-3) A particle moves along the x-axis so that its velocity v attime t, for 0 ≤ t ≤ 5, is given by v(t) = ln

(t2 − 3t+ 3

). The particle is at

position x = 8 at time t = 0.


(b) Find all times t in the open interval 0 < t < 5 at which the particlechanges direction. During which time intervals, for 0 < t < 5, doesthe particle travel to the left?

(c) Find the position of the particle at time t = 2.

(d) Find the average speed of the particle over the interval 0 ≤ t ≤ 2.

12.E-136 (1999 AB–1) A particle moves along the y–axis with velocity given byv(t) = t sin

(t2)

for t ≥ 0.

(a) In which direction (up or down) is the particle moving at time t =1.5? Why?

(b) Find the acceleration of the particle at time t = 1.5. Is the velocityof the particle increasing at t = 1.5? Why or why not?

(c) Given that y(t) is the position of the particle at time t and thaty(0) = 3, find y(2).

(d) Find the total distance traveled by the particle from t = 0 to t = 2.

12.E-137 (1997 AB–1) A particle moves along the x–axis so that its velocity at anytime t ≥ 0 is given by v(t) = 3t2− 2t− 1. The position x(t) is 5 for t = 2.


(b) For what values of t, 0 ≤ t ≤ 3, is the particle’s instantaneous velocitythe same as its average velocity on the closed interval [0, 3]?

(c) Find the total distance traveled by the particle from time t = 0 untilt = 3.


AP Unit 12, Day 6: MU: Data 423

12.6 MU: Data

Replaces problems 133-136, 161, 168-170, 172

Makeup

12.F-133 (2010–2) A zoo sponsored a one-day contest to name a new baby elephant.Zoo visitors deposited entries in a special box between noon (t = 0) and8 p.m. (t = 8). The number of entries in the box t hours after noon ismodeled by a differentiable function E for 0 ≤ t ≤ 8. Values of E(t), inhundreds of entries, at various times t are shown in the table below.

t (hours) 0 2 5 7 8E(t)

(hundreds of 0 4 13 21 23entries)

(a) Use the data in the table to approximate the rate, in hundreds ofentries per hour, at which entries were being deposited at time t = 6.Show the computations that lead to your answer.

(b) Use a trapezoidal sum with the four subintervals given by the table

to approximate the value of18

∫ 8

0

E(t) dt Using correct units, explain

the meaning of18

∫ 8

0

E(t) dt in terms of the number of entries.

(c) At 8 p.m., volunteers began to process the entries. They processedthe entries at a rate modeled by the function P , where P (t) =t3−30t2 +298t−976 hundreds of entries per hour for 8 ≤ t ≤ 12. Ac-cording to the model, how many entries had not yet been processedby midnight (t = 12)?

(d) According to the model from part (c), at what time were the entriesbeing processed most quickly? Justify your answer.



12.7 MU: Extrema and Optimization

Replaces problems 137-140, 157

Makeup

12.G-137 (2010B–2) The function g is defined for x > 0 with g(1) = 2, g′(x) =

sin(x+

1x

), and g′′(x) =

(1− 1

x2

)cos(x+

1x

).

(a) Find all values of x in the interval 0.12 ≤ x ≤ 1 at which the graphof g has a horizontal tangent line.

(b) On what subintervals of (0.12, 1), if any, is the graph of g concavedown? Justify your answer.

(c) Write an equation for the line tangent to the graph of g at x = 0.3.

(d) Does the line tangent to the graph of g at x = 0.3 lie above or belowthe graph of g for 0.3 < x < 1? Why?


AP Unit 12, Day 7: MU: Extrema and Optimization 425

12.G-138 (2007B AB–4) [NO CALCULATOR]

Let f be a function defined on the closed interval −5 ≤ x ≤ 5 withf(1) = 3. The graph of f ′, the derivative of f , consists of two semicirclesand two line segments, as shown above.

(a) For −5 < x < 5, find all values of x at which f has a relativemaximum. Justify your answer.

(b) For −5 < x < 5, find all values of x at which the graph of f has apoint of inflection. Justify your answer.

(c) Find all intervals on which the graph of f is concave up and also haspositive slope. Explain your reasoning.

(d) Find the absolute minimum value of f(x) over the closed interval−5 ≤ x ≤ 5. Explain your reasoning.

12.G-139 (2001 AB–3) A car is traveling on a straight road with velocity 55 ft/secat time t = 0. For 0 ≤ t ≤ 18 seconds, the car’s acceleration a(t), inft/sec2, is the piecewise linear function defined by the graph below.

(a) Is the velocity of the car increasing at t = 2 seconds? Why or whynot?



(b) At what time in the interval 0 ≤ t ≤ 18, other than t = 0, is thevelocity of the car 55 ft/sec? Why?

(c) On the time interval 0 ≤ t ≤ 18, what is the car’s absolute maximumvelocity, in ft/sec, and at what time does it occur? Justify youranswer.

12.G-140 (1985 AB-2) [NO CALCULATOR] A particle moves along the x-axis withacceleration given a(t) = cos t for t ≥ 0. At t = 0 the velocity v(t) of theparticle is 2 and the position x(t) is 5.

(a) Write an expression for the velocity v(t) of the particle.

(b) Write an expression for the position x(t).

(c) For what values of t is the particle moving to the right? Justify youranswer.

(d) Find the total distance traveled by the particle from t = 0 to t =π

2.


AP Unit 12, Day 7: MU: Extrema and Optimization 427

12.G-141 (1992 AB-2) [NO CALCULATOR] A particle moves along the x-axis sothat its velocity at time t, 0 ≤ t ≤ 5, is given by v(t) = 3 (t− 1) (t− 3).At time t = 2, the position of the particle is x(2) = 0.

(a) Find the minimum accleration of the particle.

(b) Find the total distance traveled by the particle.

(c) Find the average velocity of the particle over the interval 0 ≤ t ≤ 5

12.G-142 (2006B AB-4) [NO CALCULATOR] The rate, in calories per minute, atwhich a person using an exercise machine burns calories is modeled by the

function f . In the figure above, f(t) = −14t3 +

32t2 + 1 for 0 ≤ t ≤ 4 and

f is piecewise linear for 4 ≤ t ≤ 24.

(a) Find f ′(22). Indicate units of measure.

(b) For the time interval 0 ≤ t ≤ 24, at what time t is f increasing at itsgreatest rate? Show the reasoning that supports your answer.

(c) Find the total number of calories burned over the time interval 6 ≤t ≤ 18 minutes.



12.8 MU: Implicit Differentiation

Replaces problems 141-144

Makeup

12.H-141 (2001 AB-6) [NO CALCULATOR] The function f is differentiable for all

real numbers. The point(

3,14

)is on the graph of y = f(x), and the

slope at each point (x, y) on the graph is given bydy

dx= y2 (6− 2x).

(a) Findd2y

dx2and evaluate it at the point

(3,

14

).

(b) Find y = f(x) by solving the differential equationdy

dx= y2 (6− 2x)

with the initial condition f(3) =14

.


AP Unit 12, Day 9: MU: Area and Volume 429

12.9 MU: Area and Volume


Makeup

12.I-153 (2010–4) [NO CALCULATOR] Let R be the region in the first quadrantbounded by the graph of y = 2

√x, the horizontal line y = 6, and the

y–axis, as shown in the figure below.


(b) Write, but do not evaluate, an integral expression that gives thevolume of the solid generated when R is rotated about the horizontalline y = 7.

(c) Region R is the base of a solid. For each y, where 0 ≤ y ≤ 6, the crosssection of the solid taken perpendicular to the y–axis is a rectanglewhose height is 3 times the length of its base in region R. Write, butdo not evaluate, an integral expression that gives the volume of thesolid.



12.I-154 (2010B–1) In the figure above, R is the shaded region in the first quadrantbounded by the graph of y = 4 ln (3− x), the horizontal line y = 6, andthe vertical line x = 2.



(c) The region R is the base of a solid. For this solid, each cross sectionperpendicular to the x–axis is a square. Find the volume of the solid.



12.I-155 (2004B AB-1) Let R be the region enclosed by the graph of y =√x− 1,

the vertical line x = 10, and the x–axis.



(c) Find the volume of the solid generated when R is revolved about thevertical line x = 10.

12.I-156 (2005B AB-1)

Let f and g be the functions given by f(x) = 1 + sin(2x) and g(x) = ex/2.Let R be the shaded region in the first quadrant enclosed by the graphsof f and g as shown in the figure above.


(b) Find the volume of the solid generated when R is revolved about thex–axis.

(c) The region R is the base of a solid. For this solid, the cross sectionsperpendicular to the x–axis are semicircles with diameters extendingfrom y = f(x) to y = g(x). Find the volume of this solid.

12.I-157 (2008P AB–2)



Let R and S in the figure above be defined as follows: R is the region inthe first and second quadrants bounded by the graphs or y = 3− x2 andy = 2x. S is the shaded region in the first quadrant bounded by the twographs, the x–axis, and the y–axis.

(a) Find the area of S.


(c) The region R is the base of a solid. For this solid, each cross sectionperpendicular to the x–axis is an isosceles right triangle with one legacross the base of the solid. Write, but do not evaluate, an integralexpression that gives the volume of the solid.



12.I-158 (2007B AB–1)

Let R be the region bounded by the graph of y = e2x−x2and the horizontal

line y = 2, and let S be the region bounded by the graph of y = e2x−x2

and the horizontal lines y = 1 and y = 2, as shown above.



(c) Write, but do not evaluate, an integral expression that gives thevolume of the solid generated when R is rotated about the horizontalline y = 1.



12.I-159 (1999 AB-2)

The shaded region, R, is bounded by the graph of y = x2 and the liney = 4, as shown in the figure above.


(b) Find the volume of the solid generated by revolving R about thex–axis.

(c) There exists a number k, k > 4, such that when R is revolved aboutthe line y = k, the resulting solid has the same volume as the solid inpart (b). Write, but do not solve, an equation involving an integralexpression that can be used to find the value of k.

12.I-160 (2007 AB–1) LetR be the region in the first and second quadrants bounded

above by the graph of y =20

1 + x2and below by the horizontal line y = 2.


(b) Find the volume of the solid generated when R is rotated about thex–axis.

(c) The region R is the base of a solid. For this solid, the cross sectionsperpendicular to the x–axis are semicircles. Find the volume of thissolid.


AP Unit 12, Day 10: MU: Tangent Lines 435

12.10 MU: Tangent Lines

Makeup

12.J-81 This is blank now




Bibliography

[1] Stephen Bernstein and Ruth Bernstein. Schaum’s Outline of Theory andProblems of Elements of Statistics I: Descriptive Statistics and Probability.McGraw–Hill, New York, 1999.

[2] George W. Best and J. Richard Lux. Preparing for the (AB) AP CalculusExamination. Venture Publishing, Andover, Massachusetts, 1998.

[3] George W. Best and J. Richard Lux. Preparing for the (BC) AP CalculusExamination. Venture Publishing, Andover, Massachussetts, 1998.

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