Alfred University

AA UU CC II

CCAALLCCUULLUUSS II

NINTH EDITION

BBOOOOKK 22

APPENDICES: QUICK REFERENCE

INTRODUCTION TO THE TI-84

ACTIVITY HINTS AND ANSWERS

SUPPLEMENTAL EXERCISES

QUIZ AND HOMEWORK SETS

ALFRED UNIVERSITY CALCULUS INITIATIVE

CALCULUS I Ninth Edition

BOOK 2

Joseph Petrillo

Division of Mathematics

Alfred University

© 2015

Text, graphs, and figures typeset in Microsoft Word.

Equations created in Microsoft Equation 3.0.

This material is based upon work supported by the National Science Foundation

under Grant No. 1140437.

Any opinions, findings and conclusions or recommendations expressed in this

material are those of the author and do not necessarily reflect the views of

the National Science Foundation.

ALFRED UNIVERSITY CALCULUS INITIATIVE

A U C I

CONTENTS

APPENDIX A QUICK REFERENCE 1

Geometry Formulas 3

Graphs of Common Functions 4

Graphs of Trigonometric Functions and Their Inverses 5

Precalculus Concepts and Formulas 6

Derivative and Integral Formulas 7

APPENDIX B INTRODUCTION TO THE TI-84 9

APPENDIX C ACTIVITY HINTS AND ANSWERS 15

Chapter 1 17

Chapter 2 22

Chapter 3 28

Chapter 4 34

Chapter 5 37

Chapter 6 43

Chapter 7 47

Chapter 8 52

APPENDIX D SUPPLEMENTAL EXERCISES 59

Chapter 1 61

Chapter 2 62

Chapter 3 63

Chapter 4 64

Chapter 5 65

Chapter 6 66

Chapter 7 67

Chapter 8 68

APPENDIX E QUIZ AND HOMEWORK SETS 69

Quiz Sets 71

Homework Sets 113

APPENDIX A

QUICK REFERENCE

2 APPENDIX A: QUICK REFERENCE

APPENDIX A: QUICK REFERENCE 3

GEOMETRY FORMULAS

Triangle Right Triangle

Area = Area =

Law of Cosines: Pythagorean Thm:

c2 = a

2 + b

2 – 2ab cos θ c

2 = a

2 + b

2

Height = h = a sin θ

Parallelogram Trapezoid

Area = bh Area =

Circle Sector of a Circle

Area = πr2 Area =

Circumference = 2πr Arc = s = rθ

Cone Right Circular Cylinder

Volume = Volume = πr2h

Lateral Surface = 2πrh

Rectangular Solid Sphere

Volume = lwh Volume =

Surface Area = Surface Area = 4πr2

2lw + 2lh + 2wh

bh2

1ba

2

1

hba )(2

1

2 2

1r

Ah3

1

3 3

4r

a

b

h

r

a h

b

c

θ b

a c

h

b

θ r

s

h

A

r

r

l w

h


GRAPHS OF COMMON FUNCTIONS

Linear y = x

x

y

Absolute Value y = |x|

y = x y = –x

x

y

Quadratic y = x2

x

y

x

Square Root y = x

y

x

y Cubic y = x

3

x

y Cube Root y = 3 x Reciprocal y =

x1

x

y

x

y

Reciprocal Squared y = 2

1

x

Constant y = c

(0, c)

x

y

Exponential y =

xe

x

y

(0, 1)

Logarithm y = ln x

x

y

(1, 0)

x

y

Semi-Circle y =

21 x

(0, 1)

–1

1


GRAPHS OF TRIGONOMETRIC FUNCTIONS AND THEIR INVERSES

–2π –π π 2π

–3π/2 –π/2 π/2 3π/2

Tangent y = tan x

y

x

Sine y = sin x

1

–2π –π π 2π

–3π/2 –π/2 π/2 3π/2

y

x

–1

1

–2π –π π 2π

–3π/2 –π/2 π/2 3π/2

Cosine y = cos x

y

x

–1

π

π/2

y

–1 1 x

Arccosine y = cos–1

x

y

Arcsine y = sin–1

x

–1 1

π/2

–π/2

x

Arctangent y = tan–1

x

π/2

–π/2

y

x


PRECALCULUS CONCEPTS AND FORMULAS

Average rate of change in y on the interval 10 , xx :

Slope of the line between any two points and :

Slope-intercept form: , where m is the slope and b is the y-intercept

Point-slope form: , where m is the slope and is a point on the line

Quadratic formula: If , then .

Special forms: Difference of squares:

Sum of squares: 22 ax Never factors over the reals.

Difference of cubes:

Sum of cubes:

Properties of exponents:

nmnm xxx mnnm xx n

n

xx

1

mnn mnm xxx /

Inverse and algebraic properties:

xbxb )(log and xb

xb log

xex )ln( and xe x ln

yxyx bbb loglog)(log yxyx lnln)ln(

yxyx bbb loglog)/(log yxyx lnln)/ln(

xyx by

b log)(log xyx y ln)ln(

y

xyb

bxlog

)(log yxy

x ln)ln(

Exponential change of base: xbx aab

)(log xbx eb )(ln

Logarithmic change of base: b

xb

a

axlog

loglog

bx

b xlnlnlog

Right triangle trigonometry: Pythagorean theorem:

h

ax cos

h

ox sin

x

x

a

ox

cos

sintan

xa

hx

cos

1sec

xo

hx

sin

1csc

x

x

xo

ax

sin

cos

tan

1cot

01

01 )()(

xx

xfxf

x

y

),( 11 yx ),( 22 yx12

12

xx

yy

x

ym

bmxy

)( 00 xxmyy ),( 00 yx

02 cbxaxa

acbbx2

42

))((22 axaxax

))(( 2233 aaxxaxax

))(( 2233 aaxxaxax

222 hoa

h o

a

x


DERIVATIVE AND INTEGRAL FORMULAS

Assume a, b, c, k, m and n are constants and f and g are differentiable/integrable functions.

Differentiation Rules/Formulas/Properties:

1. Limit Definition: 12. Sine:

2. Constant: 13. Cosine:

3. Linear: 14. Tangent:

4. Power: 15. Cotangent:

5. Reciprocal: 16. Secant:

6. Square Root: 17. Cosecant:

7. Constant Multiple: 18. Logarithm:

8. Sum/Difference: 19. Exponential:

9. Product: 20. Arcsine:

10. Quotient: 21. Arccosine:

11. Chain: 22. Arctangent:

x

xfxxf

xxf

)()(

0lim)( xx

dx

dcossin

0cdx

d xx

dx

dsincos

mbmxdx

d xx

dx

d 2sectan

1 nn nxxdx

d xx

dx

d 2csccot

2

11

xxdx

d

xxxdx

dtansecsec

x

xdx

d

2

1 xxx

dx

dcotcsccsc

)()( xfkxfkdx

d

xx

dx

d 1ln

)()()()( xgxfxgxfdx

d xx ee

dx

d

)()()()()()( xgxfxgxfxgxfdx

d

2

1

1

1sin

xx

dx

d

2)(

)()()()(

)(

)(

xg

xgxfxgxf

xg

xf

dx

d

2

1

1

1cos

xx

dx

d

)())(())(( xgxgfxgfdx

d

2

1

1

1tan

xx

dx

d


Indefinite Integration Rules/Formulas/Properties:

1. 9.

2. , if n ≠ –1 10.

3. 11.

4. 12.

5. 13. Cxdxx

1

2tan

1

1

6. 14.

7. 15.

8.

Definite Integration Rules/Formulas/Properties:

1. Limit Definition: , where

2. Same Limits of Integration:

3. Switching Limits of Integration:

4. Combining Intervals of Integration:

5. Fundamental Theorem of Calculus (Part 1):

6. Fundamental Theorem of Calculus (Part 2):

Cmxdum Cedxe xx

Cxdxx n

n

n

1

11 Cxdx

x ||ln

1

Cxdxx cos sin Cxdxx

1

2sin

1

1

Cxdxx sin cos Cxdxx

1

2cos

1

1

Cxdxx tan sec2

Cxdxx cot csc2 dxxfkdxxfk )( )(

Cxdxxx sec tansec dxxgdxxfduxgxf )()()()(

Cxdxxx csc cotcsc

b

adxxf )(

n

k

kn

xxf1

)(limn

abx

0 )( a

adxxf

)( )( b

a

a

bdxxfdxxf

dxxfdxxfdxxfb

a

b

c

c

a )()()(

)()( )( aFbFdxxFb

a

)( )( xfdttfdx

d x

a

APPENDIX B

INTRODUCTION TO THE TI-84

10 APPENDIX B: INTRODUCTION TO THE TI-84

APPENDIX B: INTRODUCTION TO THE TI-84 11

Section 1: Calculating

The calculator is programmed to follow the “order of operations”: Parentheses, Exponents,

Multiplication/Division, and Addition/Subtraction. Be sure to use parentheses whenever

necessary and know the difference between subtraction [ – ] and the negative sign [ (-) ].

Press the carat key [ ^ ] to raise values to exponents. Press [2ND] [ x2 ] to obtain a square root.

Other roots can be found in the [MATH] menu or you can convert them to exponents first.

The FRAC command (option 1) in the [MATH] menu will express answers as fractions.

Simply type in the expression to be evaluated followed by the FRAC command.

The following exercises demonstrate common errors. Be sure to focus on the differences

between each pair. Answers are given to three decimal places.

EXERCISES

Answers: Answers:

1. (a) (–27.858) (b) (27.858)

2. (a) (5) (b) (6.952)

3. (a) (28.09) (b) (14.65)

4. (a) (11.312) (b) (18.922)

5. (a) (–11.2) (b) (–2)

6. (a) (197.857) (b)

(137.857)

7. (a) (0.25) (b) (–0.125)

A summation is entered as “sum( seq( term, index, lower limit, upper limit, increment ) ).” In

particular, is entered as “sum(seq( , k, 1, n, 1)).” To select the summation

command, press [2ND] [STAT] [ ] [ ] and choose 5:sum(. To select the sequence command,

press [2ND] [STAT] [ ] and choose 5:seq( .

4.24 4.2)4(

817 817

2)2.31.2( 22 2.31.2

414.123 414.1 32

)6.47.6(1.0 6.47.61.0

7

895145

7

895145

48

1

4

1

8

1

n

k

kxf1

)( )( kxf


Section 2: Graphing

Press the [ Y= ] key and enter the equation(s) you want to graph. Use the [X,T,θ,n] key to type

the variable x. Only the equations which have a black box over the = sign will be graphed.

Move the cursor over the equal sign and press [ENTER] to turn an equation on or off.

Press the [GRAPH] key to view the graph using the current viewing settings. If necessary,

change the viewing window in one of the following ways:

Press the [WINDOW] key to manually set the viewing window.

Press the [ZOOM] key and choose one of the following zoom options:

ZStandard Automatically sets the window from -10 to 10 along the x- and y-

axes.

ZBox Position the cursor at one corner of the region you wish to view

and press [ENTER] . Position the cursor at the opposite corner

and press [ENTER].

ZoomIn/Out Position the cursor at the point where you wish to center the new

window and press [ENTER].

Press [TRACE] to trace the graph(s). An equation will appear at the top of the screen and the x-

and y-coordinates of the cursor position will appear at the bottom. Switch from one graph to the

other by using the up and down arrow keys, and move along a graph by using the right and left

arrow keys. Tracing can be used to get a rough approximation of points, including points of

intersection.

EXERCISES

1. Enter the expressions \Y1 = –x3 + 13x and \Y2 = x

2 – 9x – 11, and view the graphs on

the same screen in the standard window using 6:ZStandard.

2. The graphs of Y1 and Y2 intersect three times. Find a viewing window that includes all

three intersection points.

3. Use [TRACE] to approximate the points of intersection. To get better approximations

you can use 1:Zbox repeatedly for each point of intersection. Later we will learn how to

find the exact values.

APPENDIX B: INTRODUCTION TO THE TI-84 13

Section 3: Evaluating Functions [Find the output given an input.]

Given a function y = f (x) and a number a, we “evaluate f at x = a” by replacing all x’s in the

formula for f with the number a. For example, suppose we want to plug in x = –4.9766, x = 1.25,

and x = 11 into the function Y2 = x2 – 9x – 11. We could evaluate Y2 by using one of (at least)

three options: a trace, a table, or function notation.

Graph the function in a window that includes the given x-value(s). Press [TRACE].

Type –4.9766 and press [ENTER]. The approximate function value will show at the

bottom of the screen. Repeat for the other x-values.

Press [2ND] [WINDOW] to get the TABLE SETUP menu. Be sure Indpnt: is set to

Ask and Depend: is set to Auto. Go to the table by pressing [2ND

] [GRAPH]. Type

–4.9766 and then press [ENTER]. The function value appears in the Y2 column.

Repeat for the other x-values. Note that you could evaluate multiple functions at once.

From the home screen, press [VARS] [Y-VARS]. Choose 1:Function and then 2:Y2.

The expression “Y2” will appear on the home screen. To evaluate Y2 at x = –4.9766,

type Y2(–4.9766) and press [ENTER]. To evaluate Y2 at x = 1.25, press [2ND] then

[ENTER], move inside the parentheses to change the –4.9766 to 1.25, and press

[ENTER]. Repeat with x = 11.

Section 4: Solving Equations [Find the inputs that yield a given output.]

Given a function y = f (x) and a number b, we “solve f (x) = b for x” by finding all values that

yield b when they are plugged in for x. For example, suppose we want to solve the equation

x2 – 9x – 11 = –17. Note that the left side is Y2. Enter the right side as Y3. Graphically, solving

an equation is the same as finding the intersection points of the graphs of the separate sides of the

equation. First graph Y2 and Y3 on the same screen (you might want to temporarily turn off Y1)

making sure that the intersection points are visible. Press [2ND] [TRACE] and then choose

5:intersect. The upper left corner of the screen will show which equation the cursor is on. Place

the cursor on Y2 and press [ENTER], and then on Y3 and press [ENTER]. When prompted to

enter a “Guess,” move the cursor near the desired point of intersection and press [ENTER]. The

result will appear at the bottom of the screen.

EXERCISES

1. Use a trace, a table, and function notation to evaluate Y1 = –x3 + 13x at x = –4.9766,

1.25, and 11.

2. Solve the following equations: (a) –x3 + 13x = –17, (b) –x

3 + 13x = 0 (the x-intercepts

or roots of Y1), and (c) –x3 + 13x = x

2 – 9x – 11 (compare with Exercise 3 in Part B.)


Section 5: Working with Data

A set of discrete data points can be graphed as a “scatter plot” by typing the inputs and outputs

into lists. Press [STAT] and choose 1:Edit, which will bring you to the list menu. Clear old

data by moving the cursor to the heading of a column, press [CLEAR] [ENTER], and then

move the cursor back down to the top of the list. Type inputs into L1, one at a time, pressing

[ENTER] after each one. Repeat for the outputs into L2.

To view a scatter plot of the data, press [2ND] [ Y= ] to go to the STAT PLOTS menu. Choose

1:Plot1, turn the plot ON, and press [ENTER]. You might want to turn off or clear any other

irrelevant equations. Now press [ZOOM] and choose 9:ZoomStat.

EXERCISES

Input the following data into your calculator as two lists:

x 1995 1998 2001 2004 2007 2010

y 28 34 37 29 19 15

1. View a scatter plot of the data.

2. Sometimes inputs (such as years) are too large to use and we need to “align” inputs to

smaller values. Align the inputs so that x = 0 corresponds to 1995 by doing the

following:

Press [STAT] and choose 1:Edit, which will bring you to the list menu.

Move the cursor to the heading of the input column, say L1, and type L1 – 1995.

Press [ENTER].

View the “new” scatter plot. Why does it look the same?

Section 6: Writing a Regression Model

Study the scatter plot and decide on which function type would be the best fit. Possibilities

include linear, quadratic, cubic, natural logarithm, exponential, and sine. To write and store the

model, press [STAT] and then [ ] to the CALC menu. This will bring you to the list of

regression models. Depending on your calculator model, do one of the following:

Press [2ND] [ 1 ] [ , ] [2ND] [ 2 ] [ , ] [VARS] [ ] [ 1 ] [ 1 ] . You should

see QuadReg L1,L2,Y1 on the home screen; OR

Fill in Xlist (press [2ND] [ 1 ] [ , ]), Ylist (press [2ND] [ 2 ] [ , ]), and Store

RegEQ (press [VARS] [ ] [ 1 ] [ 1 ]).

Press [ENTER]. The model will be stored in the graph menu under Y1. Press [GRAPH] to

view the graph of the model over the scatter plot.

APPENDIX C

ACTIVITY HINTS AND ANSWERS

16 APPENDIX C: ACTIVITY HINTS AND ANSWERS

APPENDIX C : ACTIVITY HINTS AND ANSWERS 17

Activity 1.1 – Average Rate of Change

1. (a) Above zero on (–2.5, 3) and (3, 5). Equal to or below zero on (–3, –2.5] and at x = 3.

(b) Rising on (–3, –1) and (3, 5). The average rate of change is positive.

(c) Falling on (–1, 3). The average rate of change is negative.

(d) A peak at x = –1 means that on December 31, the temperature was higher than that of

surrounding days. A valley at x = 3 means that on January 4, the temperature was lower

than that of surrounding days.

(e) The x-intercepts are at x = –2.5 and at x = 3. These are the inputs at which the temp. was

zero. The y-intercept is is at y = 10. This means that the temp. on January 1 was 10°C.

(f) (i) 03

)0()3(

ff =

3

100 = –10/3;

Between January 1 and January 4, the daily high

temperature decreased by about 3.3°C per day.

(ii) )2()1(

)2()1(

ff =

1

612 = 6;

Between December 30 and December 31, the

daily high temperature increased by 6°C per day.

2. (a)

100

2.0

25125

50.530.5

25125

)25()125( TT –0.002 °C per meter

(b)

75

1.0

50125

20.530.5

50125

)50()125( TT

d

T 0.001 °C per meter

(c)

75

70.0

125200

30.500.6

125200

)125()200( TT 0.009 °C per meter

(d) The correct answers are (iii) and (v).

3. (a) The shape of the graph is linear.

(b)

(c) 70 mi/hr

(d) The vehicle had a constant velocity of 70 mi/hr.

Interval ΔD Δt ΔD/Δt

[1, 2.5] 105 1.5 70 [0, 2] 140 2 70

[0.5, 3] 175 2.5 70

–3 –2 –1 0 1 2 3 4 5

12

9

6

3

0

–3

–6

f (x)


Activity 1.2 – Linear Functions

1. (a) y – 5 = 2(x – 4) or f (x) – 5 = 2(x – 4)

(b) y = 2x – 3 or f (x) = 2x – 3

(c) x = 3/2

2. (a) Between 1915 and 1920, the population changed by 3100 – 3250 = –150 people, and

changed at a rate of 3019151920

32503100

people per year. The negative answers represent

a decrease in population.

(b) P(t) = –30t + 3250 people, where t is years after 1915.

(c) P(10) = –30(10) +3250 = 2950 people at the end of 1925.

3. (a)

(b) y = s(t) = 40t – 15 miles from Bill’s house.

(c) Set 40t – 15 = 0 to get 40t = 15, or t = 15/40 = 0.375. This is the time at which the

position from Bill’s house is zero. That is, they pass Bill’s house after 0.375 hours.

(d) Since s(0) = –15, we can conclude that the initial position was 15 miles west of Bill’s.

(e) s'(t) = 40 miles per hour (eastward)

4. (a) y = s(t) = 40t + C miles from Bill’s house 40

(b) s'(t) = 40 miles per hour (eastward)

(c) Infinitely many, since any line of the form 40t + C has a slope of 40. Examples include

40t – 10, 4t, and 4t + 3. The differences between these lines are their y-intercepts.

(d) Since the distance traveled at the start of the trip is zero, the constant C = 0. Therefore,

s(t) = 40t miles traveled.

Time t Position s

0 –15

1 25

2 65

3 105

4 145

0 1 2 3 4

180

150

120

90

60

30

0

t

y


Activity 1.3 – Derivatives of Linear Functions

1. (a) 0)( xf

(b) 0)( xg

(c) 0)( xh

(d) mxF )(

(e) 9)( xG

(f) 1)( xH

2. (a) 2)()( tstv ft/s

(b) 2)10( v ft/s

(c) 0)()( tvta ft/s2

3. (a) The given point is (1, 22) and the slope is –0.4. Therefore, )1(4.022 tH , so

4.224.0)( ttH ft3.

(b) 4.22)5(4.0)5( H = 20.4 ft3

4. (a) The given point is (50, 150) and the slope –0.1. Therefore, )50(1.0150 xP , so

P(x) = –0.1x +155 dollars, where 40 ≤ x ≤ 60 is the number of shirts sold.

(b) 151 P

149

40 60

(c) The net change in P is P(60) – P(40) = (149 dollars) – (151 dollars) = –2 dollars. The

negative shows a decrease in profit.

(d) P'(x) = –0.1 dollars per shirt.

(e) 40 60

–0.1

P’

(f) Net area bounded by P' = length × height = (20 shirts) × (–0.1 dollars/shirt) = –2 dollars

(g) The answers are the same! This result is called the Fundamental Theorem of Calculus…


Activity 1.4 – Integrals of Constant Functions

1. (a) dx 3 3x + C

(b) y = 3x + 6

y = 3x

3 y = 3x – 3

1 2

(c) 15)1(3)4(33 3 4

1

4

1

xdx

(d) y = 3

–1 4

2. (a) Cxdx 1

(b) Ctdt 4.5 4.5

(c) Cudu 62 62

(d) 3)2()5()( 1 5

2

5

2 xdx

(e) 96.12)0(32.4)3(32.4)32.4( 32.43

0

3

0 xdx

(f) 1.0))3(02.0())2(02.0()02.0( 02.02

3

2

3

vdv

3. (a) dttv )( = dt 45 = –45t + C miles from Bill’s at t hours

(b) 90)0(45)2(45)45( 45 )(2

0

2

0

2

0 tdtdttv

Over the first 2 hours, Bill and Sally traveled a net distance of 90 miles westward.

(c) No, we need to know the distance from Bill’s at the start of the trip.

(d) Since s(t) = –45t + C and s(0) = 200, s(0) = C = 200. Therefore, the distance function is

s(t) = –45t + 200 and s(2) = –45∙2 + 200 = 110 miles east of Bill’s house.

(e) Part (b) is the net distance traveled, but Part (d) is the distance from Bill’s.

4. 2)1(2)0(2)2( 20

1

0

1

xdx

5. dxx )82(1

1 = – [area of trapezoid, or area of rectangle and triangle] = –16

6. If 2)( xf , then Cxxf 2)( and 72)1( Cf . Hence, C = 5 and 52)( xxf .


Activity 1.5 – Rectilinear Motion

1. Position s(t) = 35 – t feet Position s(5) =30 ft

Velocity v(t) = –1 ft/s Velocity v(5) = –1 ft/s

Speed |v(t)| = 1 ft/s Speed |v(5)| = 1 ft/s

Acceleration a(t) = 0 ft/s2 Acceleration a(5) = 0 ft/s

2

2. (a) s(t) = –20t + 100 m

(b) Setting –20t + 100 = 0 yields t = 5 s.

(c) 60))5.9(20())5.12(20()20(205.12

5.9

5.12

5.9 tdt m

(d) 60)5.9(20)5.12(20)20(205.12

5.9

5.12

5.9 tdt m

3. (a)

(b) Setting 10 – 4t = 0 yields t = 10/4 = 2.5 s.

(c) 8)6)(5.1()10)(5.2()410(21

21

4

0 dtt m

(d) 17)6)(5.1()10)(5.2(|410|21

21

4

0 dtt m

4. (a) Linear because for each unit change in t, there is a constant change of –30 in r. The

slope is –30 and the initial value (y-int.) is 110. Therefore, r(t) = –30t + 110 gal/min.

(b) Setting –30t + 110 = 0 yields t = 11/3 min.

(c)

(e) Quadratic

(f) V(t) = –15t2 + 110t + 500 gallons, where t is minutes after the malfunction

t

(min)

r(t)

(gal/min)

V(t)

(gal)

0 110 500

1 80 595

2 50 660

3 20 695

4 –10 700

5 –40 675

0 1 2 3 4 5

Rate is 0 at

t = 11/3 min

t

r

100

80

60

40

20

0

–20

(1/2)(110 + 80) = 95


Activity 2.1 – Derivatives of Quadratic Functions

1. (a) 19.4 ft/s

(b) 20.84 ft/s

(c) 20.984 ft/s

(d) 21 ft/s

(e) v(t) = 85 – 32t; v(2) = 21 ft/s.

(f) a(t) = –32 ft/s2

2. (a) 104)( xxf ; 4)( xf

(b) wwy 1)( ; 1)( wy

(c) 2432)( rrh ; 32)( rh

3. 348)( ttP bacteria per hour; 50)2( P bacteria per hour

4. (a) f (4) = 5

(b) f’(4) = –2

(c) y – 5 = –2(x – 4)


Activity 2.2 – Analyzing Quadratic Functions

1. (a) x = –4, 4

(b) x = –1, 0, 1

(c) x = 0 (Factor as 022

212 xx , and note that 22

21 x is never zero.)

(d) x = 4/5, 1

(e) x = –3, –1, 1, 3

2. (a) 5°F

(b) 1:00 a.m. and 2:30 a.m.

(c) 74)( ttT ; rising by 5°F per hour

(d) The low temperature occurred at 1:45 a.m. and was approximately –6°F.

(e) Falling on (0, 7/4); rising on (7/4, 4) – + T’

0 7/4 4

3. (a)

(b) If Vʹ(t) = –30t + 110 = 0, then t = 11/3. The maximum volume is V(11/3) = 701.67 gal.

(c) V(t) = –15t2 + 110t + 500 = 0 when t =

)15(2

500)15(42110110

≈ –3.17 or 10.5. We cast

out the negative solution and find that it takes 10.5 minutes for the tank to empty.

4. Since C'(x) = 0.0004x + 7, we have C'(1000) = 7.4. The marginal cost is $7.40.

0 1 2 3 4 5

710

680

650

620

590

560

530

500


)(

)()(lim

)()(lim)(

0

0

xfk

x

xfxxfk

x

xfkxxfkxfk

x

x

)()(

)()(lim

)()(lim

)()()()(lim

)()()()(lim

)()()()(lim)()(

00

0

0

0

xgxf

x

xgxxg

x

xfxxf

x

xgxxg

x

xfxxf

x

xgxxgxfxxf

x

xgxfxxgxxfxgxf

xx

x

x

x

Activity 2.3 – Definition and Properties of the Derivative

9

9lim

9lim

29299lim

29299lim

292)(9lim)( (a) .1

0

0

0

0

0

x

x

x

x

x

x

x

x

xxx

x

xxx

x

xxxxf

(b) xxf 4)(

(c) 26)( xxf

2

22

0

322

0

33223

0

33

0

3

)33(lim

33lim

33lim

)(lim .2

x

xxxx

x

xxxxx

x

xxxxxxx

x

xxxy

x

x

x

x

3. (a) cbxaxdcxbxaxdcxbxaxxf

23)( 22323

(b) (i) xxxf 26)( 2 ; 212)( xxf

(ii) 6.06.46.3)( 2 xxxg ; 6.42.7)( xxg

4.


Activity 2.4 – Analyzing Cubic Functions

1. (a) The graph is…………………….. Increasing Decreasing

The derivative (slope) is………… Positive Negative

The derivative (slope) is………… Increasing Decreasing

(b) The graph is…………………….. Increasing Decreasing

The derivative is………….…….. Positive Negative

The derivative is………………… Increasing Decreasing

2. (a) The graph is…………………….. Increasing Decreasing

The derivative is………………… Positive Negative


(b) The graph is…………………….. Increasing Decreasing

The derivative is………….…….. Positive Negative


3. 543 2 xxy ; 046 xy yields x = 2/3. A sign test shows that y is concave

up on (2/3, ∞) and concave down on (–∞, 2/3). The inflection point is at x = 2/3 and the

coordinates are (2/3, 56/27).

4. (a) 027183)( 2 ttts yields t = 3. + + s'

3

0186)( tts yields t = 3. – + s''

3

(b) Speeding up on (3, ∞); slowing down on (–∞, 3).

5. (a) (x – 5)(x2 + 5x + 25); x-intercept at x = 5.

(b) (x + 4)(x2 – 4x + 16); x-intercept at x = –4.

6. x3 + 2x

2 – 5x – 6 = (x + 1)(x

2 + x – 6) = (x + 1)(x + 3)(x – 2); the solutions are x = –1, –3, 2.

7. 1212)5()12)(5()5(12)5(60125 2223 xxxxxxxxxxx ;

the solutions are x = 5, 12 , 12 .


Activity 2.5 – Linear Approximation

1. (a) xxdx

dy26 2

2122

2

xdx

yd

(b) 8)1(2)1(626 2

1

2

1

xxdx

dyxx

142)1(122121

12

2

x

xdx

ydx

2. (a) Δy = y(2.03) – y(2) = 2(2.03)3 – 2(2)

3 = 0.731

(b) dy = 6x2dx = 6(2)

2(.03) = 0.720

3. (a) f (2) = 4; f '(2) = 12

(b) y – 4 = 12(x – 2); y = 12x – 20

(c) f(2.01) ≈ 12(2.01) – 20 = 4.12

(d) f(2.01) = 2(2.01)3 – 3(2.01)

2 = 4.120902

4. (a) dA = 10xdx = 10(36)(±0.125) = ±45 in2

(b) %69.00069.0)36(5

)125.0)(36(10

5

1022

x

xdx

A

dA

5. dV = 4πr2dr = 4π(19)

2(±0.5) ≈ ±2268.23 cm

3

%89.70789.0)19()3/4(

(±0.5)(19)4

)3/4(

43

2

3

2

r

drr

V

dV

6. dA = 2πrdr = 2π(24)(±0.1) ≈±15.08 cm2

%83.00083.0)24(

(24)(±0.1)2222

r

rdr

A

dA

7. dV = 74πrdr = 74π(12)(±0.15) ≈ ±418.46 m3

%5.2025.0)12(37

)(12)(±0.1547

37

7422

r

rdr

V

dV


Activity 2.6 – Integrals of Linear and Quadratic Functions

1. (a) Cx 25

(b) Ctt 42

23

(c) Cuu 2

21

2. (a) 15)1(5)2(55 222

1

2 x

(b) 2112

232

23

1

0

2

23 )0(4)0()1(4)1(4 tt

(c) 252

212

21

3

2

2

21 )2()2()3()3(

uu

3. (a) Cxxxdxxx 772 2

213

322

(b) Cwwwdwww 3

3122 332632

4. 186)5(4)5(10)8(4)8(10410)810( 228

5

28

5 ttdtt cows

5. (a) 1632)( ttv ft/s

(b) 1751616)( 2 ttts ft

(c) Set 01751616 2 tt and solve for t (using the quadratic formula) to get t = 3.84 s.

(d) v(3.84) ≈ –107 ft/s

6. 7.

8. 2222

4 75.65.05.225.11 L ; 75.105.035.225.1 22224 R .

That is, 6.75 ≤ dxx3

1

2 ≤ 10.75.

9. Δx = (3 – 1)/8 = 0.25.

9

8

7

6

5

4

3

2

1

0

f (x) = x2

0 1 2 3

9

8

7

6

5

4

3

2

1

0

f (x) = x2

0 1 2 3


x

xxx

xxxx

x

xxx

xxx

x

xxxx

dx

d

x

x

x

2

1

1lim

lim

lim

0

0

0

2

0

0

11

0

1

)(

1lim

)(lim

)(

)(lim

1

x

xxx

xxxx

x

xxx

xxx

xxdx

d

x

x

xxx

x

Activity 3.1 – Power Functions

1. (a) 0y

(b) 1y

(c) 31y

(d) xy

(e) 47

6

xdx

dy

(f) 930x

dx

dy

(g) 55

8

xdx

dy

(h) 3 53

8

xdx

dy

2. (a) x

y2

1

(b) 2

1

xy

3. (a) 23 2

1

3

8)(rr

rf ; 2411

)2(2

1

)2(3

823

)2( f

(b) 3

22

3

3

1225)(

t

ttg ; 673

31

225)1( g

4. (a) Set 03)(2

12 x

xf to get x = 2, –2. + – – +

–2 0 2

(b) local minimum at (2, 12)

(c) local maximum at (–2, –12)

5. (a) The domain of 211

xxdxd

is x ≠ 0, so xy is not differentiable at x = 0.

(b) The domain of xdx

d x2

1

is x > 0, so xy is not differentiable for x ≤ 0.

6. (a) (b)


0)6)(3(183 23 xxxxxx

Activity 3.2 – Polynomial Functions

1. (a) 136108)( 47 xxxxf

(b) 364056)( 36 xxxf

(c) 25 120336)( xxxf

2. (a) 203)( 2 ttv m/s

(b) tta 6)( m/s2

(c) 8)2( v m/s; 12)2( a m/s2

(d) Moving left (v < 0); slowing down (v and a have opposite signs)

3. (a)

62 123lim xxx

;

62 123lim xxx

(b)

25 513lim xxx

;

25 513lim xxx

4. (a) Setting yields x = –3, 0, 6.

(b) Setting 01863)( 2 xxxf yields 71x (using the quadratic formula).

Relative minimum at x = 71 ; relative maximum at x = 71 .

– + –

71 71

(c) Setting 066)( xxf yields x = 1. Inflection point at x = 1.

+ –

1

(d)

)(lim)183(lim 323 xxxxxx

;

)(lim)183(lim 323 xxxxxx

5. (b) yes;

(c) f(x) = c;

(d) continuity

(e) Intermediate Value Theorem:

If f is a continuous function defined on the

closed interval [a, b], and c is any number between

f (a) and f (b) , then there exists a number x

between a and b such that f (x) = c .

(f) an x-intercept a x b

f (b)

c

f (a)


Activity 3.3 – Composite Functions

1. (a) 06.0dPdL ppm/h; (b) 4

dtdP h/mo; (c) 24.0)4)(06.0(

dtdP

dPdL

dtdL ppm/mo

2. (a) 910xy

(b) 712 xy

(c) 444 1215)3(153)3(5 xxxy

(d) xxxxy 14378 2323

(e) 324

3

96

12423324 124962

xx

xxxxxxy

(f) 5

45

4

525

251 252

xxy

(g) xxxy 623213

12

34

(h) 3101

1003)10(10110

xxy

(i) 5

62

56

745

242

53 874

x

xxxy

3. (a) xxdx

dy

52

5

52

1 5

(b) xx

x

xxdx

dyx

102

2510

1022

5

22)104(

4. (a) 22 )27(

7

)27(

1 7

xxdx

dy

(b) 25

4

25 )13(

1054

)13(

7 15

x

x

xdx

dyx

5. (a) 2.0)( tV , so the volume is decreasing by –0.2 liters per minute.

(b) tttV

nRTtP

2.06

61125.264

2.06

)215)(08205.0)(15(

)()(

atmospheres

(c) 22

2.06

92225.52)2.0(

2.06

61125.264)(

tttP

atmospheres per minute

(d) 6.4)7(2.06)7( V liters; 52418.57)7(2.06

61125.264)7(

P atmospheres

(e)

50105.2)7(2.06

92225.52)7(

2

P atmospheres per minute


Activity 3.4 – Products of Functions

1. (a) 2276226)( 2 xxxxxf

(b) 21032252)32(5)(524 xxxxxxg

(c)

8

282

1283)( 32

ttttts

2. (a) 32131092 xxxy

2131032)32(13909282

xxxxxxy

(b) xxy 4273

22

31

42744273

23

52

312

92

xxxxy

3.

Horizontal tangents at x = 9 and at x = 45/19. Vertical tangent at x = 0.

4. (a) )()()()()( tPtStPtStR dollars per day

(b) 405)2)(1290()199)(15()60()60()60()60()60( PSPSR dollars per day;

That is, the revenue was decreasing by $405 per day.

5. (a) )()()()())(( gfgfgfgfggff

(b) )()()()())(()( gfgfgfgfggffgf

(c) gx

f

x

gfg

x

f

x

gfgfgf

x

gf

)()()()(

(d) Since dx

df

exists (is finite), and Δg → 0 as Δx → 0, we have 00lim

0

dx

dfg

x

f

x.

(e) dx

dgfg

dx

dfg

x

f

x

gfg

x

f

x

gfgf

dx

d

xx

00lim

)(lim

72

745

719

757

2

75

272

75

)9(

2)9()9(

)9(29)(

x

xx

xxxx

xxxxxf


–3 –2 –1 0 1 2 3

6

–4

Activity 3.5 – Piecewise Functions

1. (a)

1 if,

1 if,62)(

xa

xxxg

(b) Set 2(–1) – 6 = a to get a = –8; set (–1)2 – 6(–1)+5 = (–8)(–1) + b to get b = 4.

2. a = 1/2; b = 8

3. (a) 6)3( f (b) 2)1( f (c) 1)0( f

(d) 2)1( f (e) 2)5( f

4. (a) 2)(lim1

xfx

(b) 0)(lim1

xfx

(c)

)(lim1

xfx

DNE

(d) 2)(lim1

xfx

(e) 2)(lim1

xfx

(f) 2)(lim1

xfx

(g) No (h) Yes

5.

1 if,1

11 if,1

1 if,2

)(

x

x

xx

xf

6. (a) 2)(lim1

xfx

(b) 1)(lim1

xfx

(c)

)(lim1

xfx

DNE

(d) 1)(lim1

xfx

(e) 1)(lim1

xfx

(f)

)(lim1

xfx

DNE

(g) No (h) No

7. (a) The function f is not continuous on [–3, 3].

(b)

(c) 3

2038

3

1

1

1

1

3

23

322)3()1()3()(

dxxdxxdxxdxxf


Activity 3.6 – Integrals of Polynomials

1. (a) Ctttt 42

213

354

21

(b) Cxx 7

59 3

5

(c) Cx 23

32

(d) Cxx 21

34

1043

(e) 144634129231

1

231

1

21

1

2

uuudxuudxu

(f) 29

4

1

84

1

164

1 2

322233

3

1

xxx

x xdxdx

2. (a) CdxxxxFxx 42 2

3

2

753 67)( ; set 3)1(23

27 CF to get C = 5;

5)(42 2

3

2

7 xx

xF

(b) Set 3241

323

87)2( CF to get C =

21 ;

21

2

3

2

742

)( xx

xF

3. Let xxf )( and let xxg )( .

The integral of the product is 3

312 )()( xdxxdxxgxf , but

the product of the integrals is 4

412

212

21 )( )( xxxxdxxdxdxxgdxxf .

4. (a) (i) Since )()( xfkxFkdx

d , it follows that )()( xFkdxxfk .

(ii) Since )()( xfxFdx

d , it follows that )()( xFdxxf .

(iii) From (ii), then (i), dxxfkxFkdxxfk )()()( .

(b) (i) Since )()()()( xgxfxGxFdx

d , we have )()())()(( xGxFdxxgxf .

(ii) Since )()( xfxFdx

d , we have )()( xFdxxf .

(iii) Since )()( xgxGdx

d , we have )()( xGdxxg .

(iv) From (ii) and (iii), then (i), dxxgxfxGxFdxxgdxxf ))()(()()()()( .


Activity 4.1 – Analyzing Rational Functions

1. (a) N(1) = (1)3 – 5(1)

2 + 7(1) – 3 = 0

(b) g(x) = x2 – 4x + 3

(c) N(x) = (x – 1)2(x – 3)

(d) )1)(1(

)3()1()(

2

xx

xxxf

(e) x = 1, –1

(f) x = 3

(g) y = 3

(h) x = –1

(i) x = 1

(j) 1

88)5()(

2

x

xxxf

3. (a) 22 )32(

3

)32(

)2)(()32)(1(

xx

xxy

(b) 2

2

2

2

)4(

128

)4(

)1)(3()4)(32(

x

xx

x

xxxxy

(c) 22

2

22

22

)693(

18249

)693(

)96)(2()693)(2(

xx

xx

xx

xxxxxy

(d)

22

2

22

2

22

1

)9(22

983

)9(

)2(2)9()1(

xx

xx

x

xxx

yx

4. Set 0)4(

)6)(2(

)4(

12822

2

x

xx

x

xxy to get x = –2, –6.

5. (a) C(3) = 0.032 mg/cm3

(b) 22

2

22

2

)6(

96.016.0

)6(

)2)(16.0()6)(16.0()(

t

t

t

ttttC mg/cm

3 per hour

(c) 002.0)3( C mg/cm3 per hour


Activity 4.2 – Horizontal and Vertical Asymptotes

1. Since 29

29

2

9

172

269 limlimlim4

4

4

34

xx

x

xxx

xxx

x, the line y =

29 is a horizontal asymptote.

Similarly for x .

2. (a) 41

41

446

5 limlimlim2

2

2

2

xx

x

xx

xx

x

(b) 0limlimlim31

313

106

5

46

25

xxx

x

xxx

xx

x

(c)

1812 limlimlim

22 x

xxx

xxxx

x

(d) 31

31

3343

1 limlimlimlim22

xxx

xx

x

xx

x

x

3. (a) Since f has a vertical asymptote at x = 2, the limit on either side will be +∞, or –∞.

We only need to check the signs:

If 2x , then 2x and 0)2( x . Therefore,

))((

))((

)3)(2(

)2()1(

22

2

limxx

xx

x

.

If 2x , then 2x and 0)2( x . Therefore,

))((

))((

)3)(2(

)2()1(

22

2

limxx

xx

x

.

Hence, )(lim2

xfx

DNE.

(b) If 3x , then 0)3( x . Therefore,

))((

))((

)3)(2(

)2()1(

32

2

limxx

xx

x

.

If 3x , then 0)3( x . Therefore,

))((

))((

)3)(2(

)2()1(

32

2

limxx

xx

x

.

Hence,

)(lim3

xfx

.

4. (a)

2222

22

100

3000

100

21510030)(

x

x

x

xxxxxf ; f is zero at x = 0 and undefined at x = 10, –10.

(b) Increasing on (–∞,–10) and (–10, 0); + + – –

decreasing on (0, 10) and (10, +∞). –10 0 10

(c) 100

15

102

2

limx

x

x

; 100

15

102

2

limx

x

x

; 100

15

102

2

limx

x

x

; 100

15

102

2

limx

x

x

5. (a) +∞; (b) –∞; (c) DNE; (d) +∞; (e) +∞; (f) +∞; (g) 1; (h) –2

6. (b) y(2) must exist

(d) existmust )(lim2

xyx

(f) (2) equalmust )(lim2

yxyx


Activity 4.3 – Continuity and L’Hôpital’s Rule

1. (a) 34

3

62

11

56

123

2

limlim

x

x

x

LR

x

xx

x

(0/0)

(b) 23

8666

2483

63

244

43

2limlimlim

2

2

23

23

xx

x

LR

xx

xx

x

LR

xxx

xx

x

(0/0) (0/0)

(c) 3

52

3

3

lim

xxx

x has the form –39/0, so L’Hôpital’s Rule does not apply here. The limit is

either +∞, –∞, or DNE (vert. asymp. at x = 3). We must check left- and right-hand limits:

)(

)(

352

3

3

limx

xx

x;

)(

)(

352

3

3

limx

xx

x; Therefore,

352

3

3

lim

xxx

xDNE.

(d) 0lim40

410

43

027

359

xx

xxx

x

(e) 41

463

5

443

205

4223

limlim

xxx

LR

xxx

x

x

(0/0)

(f) 0limlimlim4

02666

1123

363

11

133

12

2

23

23

xx

x

LR

xx

xx

x

LR

xxx

xxx

x

(0/0) (0/0)

2. (a) 1511

3022

30422

215

7411 limlimlim2

2

x

LR

xx

x

LR

x

xx

x

(+∞/+∞) (+∞/+∞)

(c) (i) 1511

15

11

262215

1041197

97

125497

23397

limlim

x

x

xxxx

xxx

x

(ii) 4limlim5

5

5

235 4

78

34

x

x

xxx

xxx

x

3. (a) (i) 0limlimlim3012

615

412

235

46223

2

xx

LR

xx

x

x

LR

xx

xx

x

(+∞/–∞) (–∞/+∞) (12/–∞)

(ii)

2854

323827

3

2349 limlimlim2

2

23 x

x

LR

xxx

x

LR

xx

xxx

x

(–∞/+∞) (+∞/–∞) (–∞/2)

(b) (i) 0limlimlim2171

50

123371

25950

5

6

5

6

166215

10376

xxx

x

xxxx

xxxx

x

(ii)

199

94021

4102109 6

39

45

8153039

2101745

limlimlim x

xx

x

xxxxxx

xxxx

x

4. Set c(2)2 + 7(2) = (2)

3 – c(2) to get 4c + 14 = 8 – 2c. Therefore, c = –1.


Activity 5.1 – Exponential Growth and Decay

1. (a) If M(t) = at + b, then M(0) = b = 500, M(2) = 2a + 500 = 245, and a = –127.5.

The model is M(t) = –127.5t + 500 mg, where t is hours after the injection.

(b) If M(t) = A(1 – r)t, then M(0) = A = 500, M(2) = 500(1 – r)

2 = 245, and 1 – r = 0.7.

The model is M(t) = 500(0.7)t mg, where t is hours after the injection.

2. (a) m(0) = 90 grams

(b) m(40) ≈ 27 grams

(c) decay rate of 0.03 = 3%

3. P(t) = 77.2e0.016t

million tons, where t is years since 2004

4.

5.

METHOD 1: 1)0(33

0

RUNRISE

x

x

dxd ef

METHOD 2: 11

lim)0(00

x

ee

dx

df

x

xx

x

.

6. r

m

mm

rm

mm

mr

mm

mr

mrr

mr

n

nr

n

111 1lim1lim1lim1lim1lim ;

The limit inside the outer parentheses is equivalent to en

nn 11lim , hence

rrr

m

mm

n

nr

nee

11lim1lim .

n compounding per yr nn11 dollars after 1 yr

1 (yearly) 00000000.2$11

11

12 (monthly) 61303529.2$112

121

365 (daily) 71456748.2$

525,600 (every minute) 71827922.2$

↓ ↓

+∞ $2.718281828459…

Δx –0.1 –0.01 –0.001 → 0 ← 0.001 0.01 0.1

x

e x

1 0.9516 0.9950 0.9995 → 1 ← 1.0005 1.0050 1.0517

3

3

y = ex

(0, 1)


Activity 5.2 – Derivative and Antiderivative of ex

1. (a) xxx eeedx

d 444 20455

(b) 222 3233

teedt

d tttt

(c) 43234242 624241244621621 uuuueuueuueuuedu

d uuuu

(d)

22

222

2

2

19

18192

19

x

xx

xe

exex

e

x

dx

d

(e)

222

222

224

8210410

4

10

x

xxxx

x

x

ex

exeexe

ex

e

dx

d

2. Set 022)2(2)( 36626 222

xxexexexxf xxx to get x = –1, 0, 1.

3. (a) Cexdxex xx 2

25 5

(b) Cedxe xx

2

2525

(c) 11011

0

1

0

e

tt eeedte

4. (a) +∞; (b) 0; (c) 0; (d) +∞; (e) 0; (f) –5;

(g)25

16

40

81

20

4

102

2

2

2

2

2

limlimlim

x

x

x

x

x

x

e

e

x

LR

e

e

x

LR

ex

e

x

5. Set 0102522)( 555 teetetD ttt to get t = 1/5 s.

6. (c) ttU )1289.1( 1496.0)( million users, where t is months since 12/31/2003

(d) 5.50)48( U million users

(e) tt eetU 1213.0 1213.0 0181.01213.01496.0)( million users per month, where t is

months since 12/31/2003

(f) 11.6)48( U million users per month


Activity 5.3 – Implicit Differentiation and Inverse Functions

1. (a) y

xy

(b) x

yy

2

3

(c) 42

3 22

xy

yxy or

xy

yx

24

3 22

(d) x

yy

(e) xy

xy

xe

yey

5

2. 82

22

y

xy ;

Horizontal tangents: Set 2 – 2x = 0 to get x = 1. Substitute x = 1 into the original equation to

get a quadratic equation in y with solutions y = 0 and y = 8. The points at which the circle

has horizontal tangents are (1, 0) and (1, 8).

Vertical tangents: Set 2y – 8 = 0 to get y = 4. Substitute y = 4 into the original equation to

get a quadratic equation in x with solutions x = –3 and x = 5. The points at which the circle

has vertical tangents are (–3, 4) and (5, 4).

3. (a) 6

1)(1 x

xf

(b) 9)( 21 xxg

(c) 2

4)(1

x

xxh

4. Since xxf 2)( , xxf

xdx

d

2

1

))((2

11

.


Activity 5.4 – Logarithmic Functions

1. (a) )10ln(21x

(b) 35 ex

2. (a) 4ln)2ln(4

)2(ln 2

31

21

3 2

xxx

x

xx

(b) )4(log)1(log 102

10 xx , so (x + 1)2 = 4x. Solve this quadratic equation to get x = 1.

3. (a) Set 36.5b7 = 351.8 to get 3822.1

5.36

8.351 71

b

(b) Set 73)3822.1(5.36 t to get 14.2)3822.1ln(

)2ln(t years

4. Set 400400211590 ke to get

1590

ln 21

k . Therefore,

t

etA

1590

ln 21

400)( , and so

167400)2000(

)2000(1590

ln 21

eA mg.

5. (a) +∞

(b) –∞

(c) 0

(d) 0

6. (a) If )60()5580(5575)60( keT , then 2025 60 ke , and 0.003719 ln2520

601 k .

(b) Since 6.98)5580(55 0.003719 te , we have 55.149ln25

6.430.003719

1 t . This is

149.55 minutes before 11:00 p.m. Therefore, the time of death was around 8:30 p.m.


J/mol 142.4526190.46

)(

1000

290

1008.102

21051.31

1000

290

53

2

2

1

T

T

cT

TP

TT

dTbTadTTCH

K-J/mol 940.741051.31||ln90.461000

2902

1008.103

1000

290

)(

2

5

3

2

1

T

T

cTaT

T T

TC

TT

dTbdTS P

Activity 5.5 – Derivatives and Antiderivatives of Exponentials and Logarithms

1. (a) xdx

d x 1)ln( ; (b) xbbdx

d x)(ln

1)(log ; (c) 5

1102

2)102ln(

xxdx

d x

(d) 793)10(ln

96210 2

793log

xx

xdxd xx ; (e)

12223

)12ln(3)12ln(

tdt

d tt

(f) udu

d uuuuu 122 4)ln(8)ln(4

2. (a) xey ; (b)

xbby )(ln ; (c) xxf 5)5(ln)(

(d) 12)03.1()03.1ln(1000)( 12 ttA ; (e)

213

3101310)10(ln)(

x

x xx

xg

3. (a) Cedxe xxx x

5

51

5ln555

(b) Cxxdxdxxxxx

xx 4||ln34 131431

22

2

4. (a)

128

28

3228

1428lim

4

24

87

7limlim

)24ln(

)87ln(lim

244

877

x

xx

xx

x

xxx

x

x

LR

x

(b)

61

lim3

)8ln(lim 8

2

3

2

3

2

x

x

x

LR

x x

x

5. (a) 22)( xk is a constant function. Its derivative is 0)( xk .

(b) 2)( xxg is a power function. Its derivative is xxg 2)( .

(c) xxh 2)( is an exponential function. Its derivative is xxh 22ln)( .

6. (a) If xxy , then xxxy x lnlnln , and xxydxd

dxd lnln . Therefore, 1ln

x

y

y,

and 1ln1ln xxxyy x .

(b) If nxy , then xnxy n lnlnln , and xnydxd

dxd lnln . Therefore,

xn

y

y

,

and 1 n

xnn

xn nxxyy .

(c) 2

21 1ln5114lnln3ln xxxy

dxd

dxd yields

252

3

1

10114

23

)1(

114

x

xxxx

xxy

.

7.


Activity 5.6 – Definite Integrals of Exponentials and Logarithms

1. (a) Δx = 0.4

(b) 390565.454.0333333 222222 )8.1()4.1()1()6.0()2.0(2

0

eeeeedxex

2. (a) 86.342.52.52.5

2.52.5

1

1

2.51

1

2.5

eeedxe

xx

(b) 91.310ln

1

10ln

10

10ln

1010

1

0

1

0

xxdx

3. (a) The initial rate is a = 0.5, so R(t) = 0.5bt. Also, R(120) = 0.5b

120 = 0.25. From this last

equation, b120

= 0.5, so 9942.05.0120 b . The formula for R(t) is R(t) = 0.5(0.9942)t

million gallons per minute.

(b) R(280) = 0.5(0.9942)280

≈ 0.0981 million gallons per minute.

(c) 0938.69)9942.0ln(

)9942.0(5.0)9942.0(5.0

280

0

280

0

tt dt million gallons

4. (a) xxtdtt

xxln|1|ln||ln||ln

1

11

(b) 1ln 1

1 xdtx

t implies that ex .


Activity 6.1 – The Cosine and Sine Functions

1. θ x = cos θ y = sin θ

00 1 0

30

6

2

3

21

45

4

2

2

2

2

60

3

21

2

3

90

2 0 1

2.

3. (a) 4; (b) 3

2 ; (c) 32 ; (d) 1y ;

(e)

4. (a) A = (3.96–1.60)/2 = 1.18; B = 2π/11; C = 0; D = (3.96+1.60)/2 = 2.78; the model is

78.2cos18.1)(112 tth meters, where t is hours after 9:00 a.m. on July 22.

(b) 01.2)18( h meters; 61.2)47( h meters

(c) C/B = –3, so C = –6π/11; the model is 78.2cos18.1)(116

112 ttH meters, where t is

hours after noon on July 22.

θ cos θ sin θ tan θ

00 1 0 0

306

23

21

3

3

454

2

2 2

2 1

603

21

23 3

902 0 1 undefined

180 1 0 0

2702

3 0 1 undefined

1506

5 2

3

21

3

3

2403

4 21

2

3 3

1

–1

y = sin x

–2π –π π 2π

–3π –π π 3π

2 2 2 2

x

1

–1

y = cos x

x –2π –π π 2π

–3π –π π 3π

2 2 2 2

–5

x

3

32

32


Activity 6.2 – Derivatives and Antiderivatives of Cosine and Sine

1. (a) xxdxd cossin

(b) xxdxd sincos

(c) xxxdxd 10cos101010cos10sin

(d) xxxdxd sinsincos

221

21

(e) 32233 cos33cossin xxxxxdxd

(f) xxxxxdxd 223 cossin3)sin(cos3cos

(g) 222222 5cos105sin105cos5sin15sin tttttttttdtd

(h)

)2(sin

)2cos(2)2sin(5

)2sin( 2

555

eee

d

d

(i)

u

uu

uu

du

d

3cos

3sin3)3sin(3

3cos

13cosln

2. (a) Cxdxx sin cos

(b) Cxdxx cos sin

(c) Ctdtt )3sin( )3cos(232

(d) Cdx xx 44sin2 cos5.0

(e)

42

022

2

0 2cossin xdxx

3. (a) 8

5

)8cos(8

)5cos(5lim

)8sin(

)5sin(lim

00

x

x

x

x

x

LR

x

(b)

4sin

)42cos(2lim

cos

)42sin(lim

212

22

x

x

x

x

x

LR

xx

4. (a) 77.378.2)23(cos18.1)23(112 h meters

(b) ttth112

1136.2

112

112 sinsin18.1)( meters per hour, where t is hours after

9:00 a.m. on July 22.

(c) 36.0)23(sin)23(112

1136.2 h meters per hour; negative rate means tide is falling.

5. yxy

sin4cos3


Activity 6.3 – Other Trigonometric Functions

1. (a) xxx

cossintan

(b) x

xcos

1sec

(c) xxx

sincoscot

(d) x

xsin

1csc

2. (a) xxdxd 2sectan

(b) xxxdxd tansecsec

(c) xxdxd 2csccot

(d) xxxdxd cotcsccsc

3. (a) xxee xx

dxd tansecsecsec

(b) tdt

d tt 12 )(lnsec)tan(ln

(c) 2)2(csc)2cot(1)2cot( 2 dd

(d) xxxdxd 22 sectan2tan

4. (a) Cxdxxx sec tansec

(b) Cxdxx tan sec2

(c) Cxdxx cot csc2

(d) Cxdxxx csc cotcsc

5. (a) Cxdx )23tan( )2(3sec312

(b) Cxxdxxdxxx

xxx ||lncossin 1sin

2

2

6. 57

1sec4

cos7

0tan4sin7

02

lim lim

LR

7. (a) xx

xx

xx

x

xxxx

xx

dxd

dxd 2

cos

1

cos

sincos

cos

sinsincoscos

cossin sectan

22

22

2

(b) xxxx

xx

x

xxxx

xx

dxd

dxd 2

sin

1

sin

cossin

sin

coscossinsin

sincos csccot

22

22

2

(c) xxxxxxxx

xx

xdxd

dxd tansec)sin())(cos1()(cossec

cossin

cos1

cos

sin212

(d) xxxxxxxx

xx

xdxd

dxd cotcsc)(cos))(sin1()(sincsc

sincos

sin1

sin

cos212


Activity 6.4 – Inverse Trigonometric Functions

1. (a) 21

1arcsinx

dxd x

(b) 21

11cosx

dxd x

(c) 21

1arctanxdx

d x

2. (a) 22

11

11sinx

x

x e

ex

e

x

dxd ee

(b)

10

6

25 1

554

1

12552 arccos25arccos2arccosx

x

xdxd xxxxxxxx

(c)

2

11

21

11

ln

tanln

ln

tan

x

xx

x

x

dxd xx

3. (a) Cxdxx

arcsin21

1

(b) Cxdxx

arccos21

1

(c) Cxdxx

arctan

21

1

4. (a)

Cxdxx

)3arccos( 31

31

1

2

(b)

3

2622

12

12

2

1 1

1 22arcsin21arcsin2arcsin22

2

xdxx

(c)

Cdxdx x

x x

441

1

1161

16

1 arctan 2

4

2

5. (a) 2

arctanlim

xx

(b) 2

1tanlim

x

x

6. (a) (b) 21 1cossin xx

1

(c) 21

1

1

cossin

11cosxxdx

d x

7. (a) (b) 21 1tansec xx .

X x (c) 212 1tansec xx

(d) 212 1

1

tansec

11tanxxdx

d x

xx cos 1

1 tan 1 x

21 x

21 x


Activity 7.1 – Related Rates

1. Related Variables Equation: 222 13 yx 13 y

Given: 1dtdx ; Want:

dt

dy when x = 12 and y = 5144169169 2 x x

Related Rates Equation: 4.2 0)5(2)1)(12(2 022 dt

dy

dt

dy

dt

dy

dtdx yx

Answer: When the bottom of the ladder is 12 feet from the wall, the top of the ladder is

falling at a rate of 2.4 ft/s.

2. Related Variables Equation: 2rA

Given: 4dt

dA ; Want: dt

dy when x = 12 and y = 5144169169 2 x r

Related Rates Equation: 36.0 24 2 10 dtdr

dtdr

dtdr

dtdA r

A

Answer: When the area 10 mi2, the radius of the spill is increasing at a rate of 0.36 mi/h.

3. Related Variables Equation: 5280

cot x 5280

Given: 400dtdx ; Want:

dtd when θ = 30° =

6 x

Related Rates Equation:

02.0 400 csc5280

1

62sin

15280

12 dtd

dtd

dtdx

dtd

Answer: When the angle of elevation is 30°, is decreasing at a rate of 0.02 rad/s.

4. Related Variables Equation: y

x 122

12 or

yx 122

6 y

Given: 2.1dtdx ; Want:

dt

dy when x = 6 and 4

61224

y 12 – x x

Related Rates Equation: 8.0 )2.1( 22 )4(

122112

21

dt

dy

dt

dy

dt

dy

ydtdx

Answer: When the woman is 6 meters from the building, her shadow is decreasing at a rate

of –0.8 m/s.

5. Related Variables Equation: 222 50 zx z 50

Given: 2dtdx

; Want: dtdz

when θ = 6 and

2

3

6cos

zx

x

Related Rates Equation: 73.132 222

3 dtdx

zx

dtdz

dtdz

dtdx zx

Answer: When the angle of elevation is π/6 radians, the string must be let out at a rate of

1.73 ft/s.

6. Related Variables Equation: CVP 2.1

Given: 10dtdP

; Want: dtdV

when V = 330 and P = 90

RRE: 0)330(2.1)90()330(10)( 02.1 2.02.12.02.1 dtdV

dtdV

dtdP VPV

Answer: At this instant, the volume is increasing by 30.56 cm3/min.

θ

2

θ


Activity 7.2 – Graph Analysis Using First and Second Derivatives

1. (a) xexxf )28()(

(i) x = 4; (ii) 42 ,4 e ; (iii) none; (iv) (–∞, 4); (v) (4, ∞)

(b) xexxf )26()(

(i) 34 ,3 e ; (ii) (–∞, 3); (iii) (3, ∞)

(c) xexxf )210()(

(i) x = 5; (ii) y = 10

(iii)

))(()210(lim x

xex ; 0limlim 2210

xx ex

LR

e

x

x

2. 3.

4. (a) Moving to left: (0, 1) U (5, 6) – + –

Moving to right: (1, 5)

At rest: t = 1, 5

+ –

(b) Speeding up: (1, 3) U (5, 6)

Slowing down: (0, 1) U (3, 5)

– +

(c) Left of origin: (0, 2)

Right of origin: (2, 6)

At origin: t = 0, 2

3 (a)

–1 0 1

–2 –1 0 1 2

3 (s)

3 (v)


Activity 7.3 Solutions – Graph Analysis with the TI-84

1. 370.1778.3571.1169.0)(1Y 23 tttt million barrels per day, where t is years after

2008, 0 ≤ t ≤ 6.

2. (a) t = 0.55 years after 2008 → in 2009

t = 3.10 years after 2008 → in 2012

t = 5.67 years after 2008 → in 2014

(b) 75.3)1(1Y million barrels per day

(c) 14.1)1(1Y ; This is an increase of 1.14 million barrels per day per year.

(d) At t = 1.63 (2010), the surplus was at a maximum of 08.41Y million barrels per day.

(e) At t = 4.59 (2013), the surplus was at a minimum of 91.11Y million barrels per day.

3. 778.3143.3506.0)(1Y 2 ttt million barrels per day per year, where t is years after

2008, 0 ≤ t ≤ 6.

(a) 0)(1Y t when t = 1.63 and changes from positive to negative there. This verifies the

local maximum at t = 1.63.

0)(1Y t when t = 4.58 and changes from negative to positive there. This verifies the

local minimum at t = 4.58.

(b) The point of most rapid decline corresponds to the minimum of the slope graph, which

occurs at t = 3.11. This corresponds to the year 2012. Algebraically,

0143.3012.1)(1Y tt when t = 3.11.

(c) 10.1)11.3(1Y ; This is a decrease of 1.10 million barrels per day per year.


Activity 7.4 – The Extreme Value Theorem and Optimization

1. Length: L = 2x + 5y (minimize)

Area: x

yxyA 3600 3600 y

xx

xxxL 180003600 252)( x

9000 9000 02)( 2180002

xxxLx

– + L'

Minimum length of fencing is 3799000 L ft. 0 9000

2. Area: A = xy (maximize)

Cost: 3

5430

10540 5403010 xxyy = x + C = y

2

31

354 18)( xxxxA x x

27 018)(32 xxxA + – A'

Maximum area is 243)27( A ft2. 0 27

3. Volume: V = x2y (maximize)

Area: x

xyxyxA4

14002 2

14004 y

3

41

414002 350)(

2

xxxxVx

x x x

3

14003

140022

43 0350)( xxxxV + – V'

Maximum volume is 50403

1400 V cm3. 0

31400

4. Cost: xyxC 3026 2 (minimize)

Volume: 2

6.22 2.52x

yyxV y

xx

xxxxC 7826.22 263026)(2

x 2x

3378 5.1 5.1 052)(2

xxxxCx

– + C'

Minimum cost is 50.110$5.13 C . 0 3 5.1

5. Volume: hrV 2 (maximize) r

Area: rrhrhrA

2862 2

862

3

22862 43)(

2

rrrrVrr

h + – V'

386

38622

23 043)( rrrrV 0 3

86

Height is 02.3)3/(862

))3/(86(86

h in. Maximum liquid is 6.86

3

86

V in

3.


Activity 7.5 – Differential Equations

1. (a) teCy 5

0

(b) teCy 3

0

(c) tttt eCeCeCeCy 5

25

125

225

1

(d) tCtCy 15sin 15cos 21

2. ktetT 11570)( ;

17011570)5( )5( keT implies 115100

51 lnk ;

14611570)15()15(ln

115100

51

eT °F.

3. tt eey 4

474

45

4. ttty21

21 sin10cos5.1)( in

5. kyeCkyeCy ktkt 00

6.

tktk eCkeCky 21

kyeCeCkeCkeCky tktktktk

2121

7. tkCktkCky cos sin 21

kytkCtkCktkCktkCky sin cos sin cos 2121

8. (a) Ckt|y| kdtdyy

ln 1

(b) CktCCkty eCeeyeyeCkt|y|

0||ln || ln


Activity 8.1 – Sigma Notation and Summations

1. (a) 16; (b) 13

2. (a) nnnnkknn

n

k

n

k

n

k

2)1(2412424 22

4

)1(

11

3

1

322

(b)

3. (a) 898,256,16632)163()63(24 2263

1

3 k

k

(b)

4. (a)

(b)

5. (a)

(b)

(c)

)12)(1()1(2

44441441

32

6

)12)(1(

2

)1(

1

2

111

2

nnnnnn

nkkkknnnnn

n

k

n

k

n

k

n

k

880,91)1402)(140(40)140(402402132

40

1

2

k

k

2

32

32

32

2

6

)12)(1()1(2

6

)12)(1(12

)1(44

1 1

21

1

44

1

2144

1

1214

1

121

4

1

1

42

n

nn

n

n

nnn

n

nn

nn

n

k

n

kn

n

knn

n

knnn

n

knnn

n

knn

n

kk

kk

kkk

2

32

32

)12)(1(4)1(12

6

)12)(1(242

)1(246

1

224246

1

222

6

113

n

nn

n

n

nnn

n

nn

nn

n

knnn

n

knn

n

kkk

)()(

)()()()()()()()(

03

231201

3

1

1

xFxF

xFxFxFxFxFxFxFxFk

kk

)()(

)()()()()()()()()()(

04

34231201

4

1

1

xFxF

xFxFxFxFxFxFxFxFxFxFk

kk

)()()()( 0

1

1 xFxFxFxF n

n

k

kk


Activity 8.2 – The Definition of Net Area

1. a = 1; b = 2; Δx = nn112 ; kx

nk 1* 1 ; 2

23621 313 kkkxf

nnnk ;

2

33

26312

2363 kkkkxxf

nnnnnnk

2

32

3232

2

)12)(1()1(3

6

)12)(1(32

)1(63

1

23

1

6

1

3

1

2363

1

3

1

n

nn

n

n

nnn

n

nn

nn

n

kn

n

kn

n

kn

n

knnn

n

k

k

n

kkkkxxf

71333limlim 3 22

)12)(1()1(3

1

2

1

2

n

nn

n

n

n

n

k

kn

xxfdxx

2. a = 0; b = 3; Δx = nn303 ; kx

nk 3*; kkkkxf

nnnnk 32183232

2 ;

kkkkxxfnnnnn

k 232

925433218

n

n

n

nn

nn

n

nnn

n

n

kn

n

kn

n

knn

n

k

k kkkkxxf

2

)1(9)12)(1(9

2

)1(96

)12)(1(54

1

9

1

254

1

9254

1

2

232323

227

29

2

)1(9)12)(1(9

1

2

1

2 18limlim 3 2

n

n

n

nn

n

n

k

kn

xxfdxx

3. 3

583

5434

5

2

22

0

25

0

2 2 2 2 dtttdtttdttt m

4. (a) 41)(1

2

dxxf ; (b) 6)(

6

2 dxxg ; (c) 183)( 43 )(4

5

3

5

3

5

3 dxdxxhdxxh

5. (a) 0)(lim)(1

n

kn

aak

n

a

axfdxxf

(b)

b

a

n

kn

abk

n

n

kn

bak

n

a

bdxxfxfxfdxxf )()(lim)(lim)(

11

(c)

b

a

n

kn

abk

n

n

kn

abk

n

b

adxxfkxfkxfkdxxfk )()(lim)(lim)(

11

(d)

b

a

b

a

n

kn

abk

n

n

kn

abk

n

n

kn

abkk

n

b

a

dxxgdxxf

xgxfxgxfdxxgxf

)()(

)(lim)(lim)()(lim)()(111


Activity 8.3 – Rolle’s Theorem and the Mean Value Theorem

1. (a) Set 313)(4

)4(8

)2(2

)2()2(2

ffccf to get

34c in the interval (–2, 2).

(b) Set 151

235

)3()5(1 31

51

2)(

gg

ccg to get 15c , but only 15 is in (3, 5).

2. (a) Since f has a vertical asymptote at x = 0, it is not continuous on [0,4]. Since it is not

continuous at x = 0, it is not differentiable at x = 0.

(b) Since g only has one discontinuity at x = 6, it is continuous on [0, 4]. Since g' is

undefined only at x = 6 (use the quotient rule), g is differentiable everywhere except at

x = 6. In particular, it is differentiable on [0, 4].

(c) Since h has a vertical asymptote at x = –3, it is not continuous on [–4 ,5]. Since it is not

continuous at x = –3, it is not differentiable at x = –3.

(d) Since

3 if,6

3 if,6)(

2

2

xxx

xxxxF , it follows that

3 if,12

3 if,12)(

xx

xxxF .

From the left of 3, F has a slope of –5, and from the right of 3, F has a slope of 5.

Therefore, F is not differentiable at x = 3 (graph has a corner), and hence it is not

differentiable on [1, 4]. It is continuous everywhere, however, as is seen by substituting

x = 3 into the piecewise formulas for F.

3. (There are many possible correct answers.) )(xf

0 c1 c2 c3 5

4. (There are many possible correct answers.) )(xg

0 c 5

5. Your average velocity between 8:00 and 8:03 is mi/h 80hour 1

minutes 60minutes 3

miles 4time

Δposition

.

By the MVT, there exists a time between 8:00 and 8:03 at which your instantaneous velocity

was equal to your average velocity. In other words, you are guilty of traveling at a speed of

80 mi/h during this time interval.


Activity 8.4 – The Fundamental Theorem of Calculus (Part 1)

1. (a) 3593.11arctan82arctan8arctan81

8 2

1

2

1 2

xdx

x

(b) 2

3

1

2

4

2242

12

2

1 3

xdx

x

(c) 2

0 3

4dx

x The FTC cannot be used since the integrand is not continuous at x = 0.

(d)

3

0 2

1dx

x The FTC cannot be used since the integrand is not continuous at x = 2.

(e) 7644946

6 23

232

3 9

42

3

9

4

xdxx

(f) 00sin3sin3sin3 cos300

xdxx

(g) 681.295

515

51

1

1

5

51

1

1

5

eeedxe xx

(h) 6

)0arcsin()5.0arcsin(arcsin1

1 5.0

0

5.0

0 2

xdx

x

2. (a) NOTE THAT THE FOLLOWING COMPUTATION IS INCORRECT!

2)1()1(11

1

1

1

1

21

1 2

xdxxdx

x

(b) The graph is above the x-axis on [–1, 1], so the net area is positive, not negative.

3. (a) Set )03(23

0

2 tdxx so that 23

0

3

31 3 tx .

Therefore, 239 t , and so 732.13 t .

(b) ))(,( tft

(c) b

adxxf )(

(d) ))(( abtf

9

f (x) = x2

6

3

0 1 2 3


Activity 8.5 – The Fundamental Theorem of Calculus (Part 2)

1. (a) 25 5)( xxxf

(b) dte

txy

x

t

1

3

4)( , so

4)(

3

xe

xxy

(c) )104arctan(22)10)2arctan(()( 22 xxxF

(d)

3 23 2 22

1)(

xx

xx

xx ee

ee

ee

xH

(e) dttxgx

25

1)ln(cos)( , so 22 5lncos10105lncos)( xxxxxg

2. (a) 01

1

2

31

dteG t

(b) 292)3( 33)( xx eexG

(c) 33)0(2)0(9 eG

3. (a) Set 021)(22 xexxF x to get x = –1, 0, 1. (Note that 0

2

xe .)

(b) Rewrite 2

22)( 3 xexxxF . Set

0

1122

122

22226)(

2

2

22

22

24

32

x

x

xx

exx

exx

xexxexxF

to get 012 2 x , or 21x (Note that 0

2

xe and 012 x .) A sign test shows

that these are the locations of the inflection points:

+ – + F''

21

21


Activity 8.6 – Integration by Substitution

1. (a) CeCedueduedxe xuuux

54

41

41

41

4154

u = 4x + 5; du = 4dx ; dx = 41 du

(b) CeCedueduedxex xuuux 33

31

31

31

312

u = x3; du = 3x

2dx ; x

2dx =

31 du

(c) CtCudududtuut

|49|ln||ln

91

911

91

911

491

u = 9t + 4; du = 9dt; dt = 91 du

(d) CxxCudududxuuxx

x

|1012|ln||ln 2

21

211

21

211

1012

62

u = x2 + 12x + 10; du = 2x + 12 = 2(x + 6); (x + 6)dx =

21 du

(e) CCuduud 6

616

6155 sin cos sin

u = sin θ; du = cos θ dθ

2. (a) 25

3280

321

3281

3

132

3

1

3

81

3

1 8131

0

343 4

21 uduuduudxxx

u = 1 + 2x4; du = 8x

3dx; x

3 dx =

81 du

x = 0 → u = 1 + 2(0)4 = 1; x = 1 → u = 1 + 2(1)

4 = 3

(b) 2

43

83

83

83

1

483

1

4

3

21

2

1

3 2 33

43

43

4

415 uduudxxx

u = 5 – x2; du = –2xdx; xdx = –

21 du

x = 1 → u = 5 – (1)2 = 4; x = 2 → u = 5 – (2)

2 = 1

(c) 21

2

0

211

21

1

021

1

021

0

)2sin(4 )2cos(

euux eeeduedxxe

u = sin(2x); du = 2cos(2x)dx; cos(2x)dx = 21 du

x = 0 → u = sin(0) = 0; x = π/4 → u = sin(π/2) = 1

3. (a)

CCdudxxuux

x

88

3831

83

1

38228

7

;

163

1

088

31

0 1

3828

7

xx

x dx

u = x8 + 1; du = 8x

7dx; 3x

7dx =

83

du

(b) CxCuduudxx

x

3

313

312)(ln

)(ln2

; 3

31

3

1

3

313

1

)(ln)3(ln)(ln

2

xdxx

x

u = ln x; du =

x1 dx


Activity 8.7 – Modeling Accumulated Change with the TI-84

1. 50793332)( 2 tttr ft3/min, where t is in minutes

2. t = 11.48 min

3. t = 14.58 min; r(14.58) = 7308 ft3/min

4. 41053 5079333210

0

2 dttt ft3

5. CtttdttttA 50750793332)( 2

29333

3322 ; since A(0) = 5000, C = 5000.

Therefore, 5000507)( 2

29333

332 ttttA ft

3, where t is minutes.

6. 1164075000)20(507)20()20()20( 2

29333

332 A ft

3

7. Set 15000050005072

29333

332 ttt to get t = 27.55 min.

APPENDIX D

SUPPLEMENTAL EXERCISES

60 APPENDIX D: SUPPLEMENTAL EXERCISES

APPENDIX D: SUPPLEMENTAL EXERCISES 61

Chapter 1 – Supplemental Exercises

Lesson 1.3

1. Find the derivative of each constant function.

(a) y = b

(b) y = 4

(c) y = –3.1

2. Find the derivative of each linear function.

(a) y = mx + b

(b) y = x – 4

(c) y = –2x

3. Let T(t) = –2.1t + 63 be the temperature in degrees Fahrenheit t hours after noon for 0 ≤ t ≤ 6.

(a) Find the temperature at noon.

(b) Find the temperature at 3:00 p.m.

(c) How fast was the temperature changing at 3:00 p.m.?

Lesson 1.4

4. Find the family of antiderivatives of each constant function.

(a) y = m

(b) y = –9

(c) y = 0.23

5. Use the Fundamental Theorem of Calculus to evaluate each definite integral.

(a) dxmb

a

(b) dx 5

2 6

(c) dx1

21

2.4

6. Stan saves money by stashing $50 per week under his mattress.

(a) Write a definite integral that represents Stan’s total accumulation of savings between the

second and tenth weeks.

(b) Use the Fundamental Theorem of Calculus to evaluate the integral from part (a), and

interpret your answer in context.

Lesson 1.5

7. An ant is walking along a wire with velocity v(t) = 2 in/s on the time interval [0, 2],

v(t) = –t + 4 in/s on the time interval [2, 5], and v(t) = –1 in/s on the time interval [5, 7].

(a) Sketch the graph of the ant’s velocity on the interval [0, 7].

(b) Use geometry to find the ant’s displacement on the interval [0, 7].

(c) Use geometry to find the total distance traveled on the interval [0, 7].



Lesson 2.1

1. (a) Carefully sketch the graph of f (x) = (x – 3)2 in the interval [0, 4] on the x-axis, and in

the interval [0, 9] on the y-axis. Then sketch the tangent line to the graph of f at x = 1.

(b) Estimate the slope of the tangent line using “rise-over-run.”

(c) Set up a table like Example 2.1.1. Use it to estimate the slope of the tangent line at x = 1.

(d) Expand (x – 3)2 by multiplying (x – 3)(x – 3).

(e) Use the formula for the derivative of a quadratic to find a formula for f ʹ (x).

(f) Use the formula from part (f) to find f ʹ (1). Were your estimates good ones?

Lesson 2.3

2. (a) Write down the limit definition of the derivative of the function f.

(b) Use the limit definition of the derivative to find the derivative of f (x) = 3x2 – 2x.

3. Find the derivative of each quadratic function.

(a) y = ax2 + bx + c

(b) y = 0.5x2 – 10x

(c) y = (2 – x)2

Lesson 2.4

4. Find the derivative of each cubic function.

(a) y = ax3 + bx

2 + cx + d

(b) y = 1 – x + x2 – x

3

(c) y = (x + 2)(x2 – 2x + 4)

Lesson 2.5

5. (a) Find the equation of the tangent line to the graph of f (x) = 2x3 – 5x

2 at x = 2.

(b) Use the tangent line to estimate f (2.1). How close are you to the actual value?

6. A beaker with radius r inches is filled with acid to a height of 4 inches. The radius of the

beaker is measured to be 2 inches with a possible error in measurement of ±0.08 inches.

Estimate the propagated and relative errors in the calculated volume of acid in the beaker.

(Hint: The volume of a right circular cylinder of height 4 is V = 4πr2.)

Lesson 2.6

7. Evaluate each integral.

(a) dtt 3

1 )12(

(b) duu 0

1 )1.22.3(

(c) dwww 1042



Lesson 3.1

1. Find the derivative of each power function.

(a) y = 2x11

(b) y = 25

7

x

(c) y = 3

1

xx

Lesson 3.2

2. Let f (x) = 24

41 2xx .

(a) Find the x-intercepts of f.

(b) Find the critical points, the intervals of increase/decrease, and the local extrema of f.

(c) Find the intervals of concavity and the inflection points of f.

(d) Determine the end behavior of f.

Lesson 3.3

3. Find the derivative of each composite function.

(a) y = (3x – 4)5

(b) y = 3 2 2xx

(c) y = 42 12

6

xx

(d) y = 109 5 x (use the square-root rule)

(e) y = 17

139 xx

(use the reciprocal rule)

Lesson 3.4

4. Find the derivative of each product function.

(a) y = 235 1044 xxx

(b) y = 35 1 xx

Lesson 3.6


(a) dxxxx 3

(b) 1

0

59 243 dxxx

6. Let f (x) = 26 32

xx

. Find an antiderivative F of f that passes through the point (2, –5).



Lesson 4.1

1. Use the quotient rule to find the derivative of each function.

(a) y = 23

xx

(b) y = 12

42

x

x

(c) y = 74

232

3

x

xx

Lesson 4.2

2. Evaluate the following limits.

(a) 742

10325

4

lim

xx

xx

x

(b) 3

3

102

471limxx

xx

x

(c) 4

52

821

739limxx

xx

x

(d) 25

892 3

lim

xxx

x

(e) 12

32

lim xx

x

x

3. Evaluate the following limits.

(a) 3

74

2

2

lim

xxx

x

(b) xx

x

x 2

1

02

3

lim

(c) xxx

xx

x

23

5

2

232

1lim

Lesson 4.3

4. Evaluate the following limits. Use L’Hôpital’s Rule if necessary.

(a) 1

32

03

2

lim

x

xx

x

(b) 352

65

32

23

lim

xx

xxx

x

(c) xx

x

x 104

4

23

3 2

lim

(d) 2

1lim

x

x

x



Lesson 5.2

1. Find the derivative of each function.

(a) y = xe

(b) y = xex 2

(c) y = 2

2

31

5

t

et


(a) dxex

(b) dxekx

(c) dte t510

Lesson 5.3

3. Use implicit differentiation to find yʹ = dx

dy.

(a) 23 24 yxyx

(b) 111 yx

(c) xyxe y 35

Lesson 5.5


(a) xy ln

(b) xy blog

(c) xxy 43ln 2

(d) tty ln

(e) 310log xy

(f) xby

(g) ty )03.1(1000

Lesson 5.6

5. Evaluate each definite integral.

(a) due u

1

0

22.1

(b) dttt

4

2

3102



Lesson 6.2


(a) xey cos

(b) 518 3sin5.16 ty


(a) dxx )sin(5

(b) 1

1 1)2cos( dtt

Lesson 6.3


(a) )ln(tan xy

(b) )3(sec3 y

(c) tty tan2


(a) dtt )14(sec2

(b) dxxx tansec4

0

5. Evaluate each limit.

(a) )5tan(

)10sin(

0lim

x

x

x

(b) )2sin(

)3cos(lim

x

xx

x

Lesson 6.4


(a) )arctan(ln xy

(b) xxy 1sin


(a) dxx

2)2(1

1

(b) dxx

29

2

(c) dxx

24

1



Lesson 7.1

1. A spherical balloon is inflated so that the volume is increasing by 8 in3/s. How fast is the

radius of the balloon changing when the volume is 512 in3?

2. A boat is pulled toward a dock by a rope attached to a pulley on the dock. The dock is 12

feet above the point on the bow of the boat at which the rope is attached. If the rope is pulled

through the pulley at a rate of 10 ft/min, how fast is the boat approaching the dock when 50

feet of rope remain?

3. A man 6 ft tall is walking at a rate of 2 ft/s toward a stationary spotlight 20 ft high. How fast

is his shadow length changing? How fast is the tip of his shadow moving?

Lesson 7.2

4. Determine each of the following for the function 16

822

2

x

xy .

(a) Domain, x-intercepts, y-intercept

(b) Vertical asymptotes and nearby behavior

(c) Horizontal asymptotes

(d) Critical points

(e) Intervals of increase/decrease, and local extrema

(f) Intervals of concavity and inflection points

(g) Sketch of the graph

Lesson 7.4

5. A cylindrical can with no top must be constructed using 300 in2 of material. Find the height

and radius of the can having the greatest volume.

6. A fenced-in garden is to be laid out in a rectangular area and partitioned into two sections

with fencing parallel to one side. The fencing for the perimeter costs $10 per foot, while the

fencing for the partition costs $5 per foot. Find the largest area that can be enclosed for a

cost of $500.

7. A window consisting of a rectangle topped by a semicircle is to have a perimeter of 20 feet.

Find the radius of the semicircle that maximizes the area of the window.

Lesson 7.5

8. A mass attached to a vertical spring has position y(t) meters after t seconds, where y satisfies

yʹʹ = –0.25y. Positions below equilibrium and downward motion are considered positive.

Find the position function y if the initial position is –0.3 m and the initial velocity is 2 m/s.

9. Find the general solution to the differential equation ydt

dy125.0 .

10. Find the general solution to the differential equation 0105 yy .



Lesson 8.2

1. Evaluate dxxx 23

1

2 using the limit definition of the definite integral.

2. Use the properties of the definite integral to evaluate dxxx 23 2

0

2 .

Lesson 8.4

3. If possible, use Part 1 of the Fundamental Theorem to evaluate each integral.

(a) dxx

2

0

1

(b) dxx

1

0 231

(c) dxx tan0

Lesson 8.5

4. Use Part 2 of the Fundamental Theorem to find the derivative of each function.

(a) dteyx t 0

2

(b) dtyx t

t 2

sin1cos1

(c) dttyx

)arctan(ln0

3 2

5. Let dttxFx

cos)(2

0 .

(a) Find F(0).

(b) Find Fʹ(x)

(c) Find Fʹ(π)

Lesson 8.6

6. Use the method of u-substitution to evaluate each integral.

(a) 1

0

32 13 dxxx

(b)

2

1 63

12

dxxx

x

(c) dxx

x

)(lnsec2

(d) dxx

x 22 )32(1

(e) dxxx cos sin3

APPENDIX E

QUIZ AND HOMEWORK SETS

70 APPENDIX E: QUIZ AND HOMEWORK SETS

APPENDIX E: QUIZ AND HOMEWORK SETS 71

Quiz 1.1 – Average Rates of Change


Quiz 1.2 – Linear Functions


Quiz 1.3 – Derivatives of Linear Functions


Quiz 1.4 – Integrals of Constant Functions


Quiz 1.5 – Rectilinear Motion


Quiz 2.1 – Derivatives of Quadratic Functions


Quiz 2.2 – Analyzing Quadratic Functions


Quiz 2.3 – Definition and properties of the derivative


Quiz 2.4 – Analyzing Cubic Functions


Quiz 2.5 – Linear Approximation


Quiz 2.6 – Integrals of Linear and Quadratic Functions


Quiz 3.1 – Power Functions


Quiz 3.2 – Polynomial Functions


Quiz 3.3 – Composite Functions


Quiz 3.4 – Products of Functions


Quiz 3.5 – Piecewise Functions


Quiz 3.6 – Integrals of Polynomials


Quiz 4.1 – Analyzing Rational Functions


Quiz 4.2 – Horizontal and Vertical Asymptotes


Quiz 4.3 – Continuity and L’Hôpital’s Rule


Quiz 5.1 – Exponential Growth and Decay


Quiz 5.2 – Derivative and Antiderivative of ex


Quiz 5.3 – Implicit Differentiation and Inverse Functions


Quiz 5.4 – Logarithmic Functions


Quiz 5.5 – Derivatives and Antiderivatives of Exponentials and Logarithms


Quiz 5.6 – Definite Integrals of Exponentials and Logarithms


Quiz 6.1 – The Cosine and Sine Functions


Quiz 6.2 – Derivatives and Antiderivatives of Cosine and Sine


Quiz 6.3 – Other Trigonometric Functions


Quiz 6.4 – Other Trigonometric Functions


Quiz 7.1 – Related Rates


Quiz 7.2 – Graph Analysis Using First and Second Derivatives


Quiz 7.3 – Graph Analysis with the TI-84


Quiz 7.4 – The Extreme Value Theorem and Optimization


Quiz 7.5 – Differential Equations


Quiz 8.1 – Sigma Notation and Summations


Quiz 8.2 – The Definition of Net Area


Quiz 8.3 – Rolle’s Theorem and the Mean Value Theorem


Quiz 8.4 – The Fundamental Theorem of Calculus (Part 1)


Quiz 8.5 – The Fundamental Theorem of Calculus (Part 2)


Quiz 8.6 – Integration by Substitution


Quiz 8.7 – Modeling Accumulated Change with the TI-84


Homework 1.1 – Average Rates of Change



Homework 1.2 – Linear Functions


Homework 1.3 – Derivatives of Linear Functions


Homework 1.4 – Integrals of Constant Functions



Homework 1.5 – Rectilinear Motion



Homework 2.1 – Derivatives of Quadratic Functions


Homework 2.2 – Analyzing Quadratic Functions


Homework 2.3 – Definition and properties of the derivative


Homework 2.4 – Analyzing Cubic Functions


Homework 2.5 – Linear Approximation


Homework 2.6 – Integrals of Linear and Quadratic Functions


Homework 3.1 – Power Functions


Homework 3.2 – Polynomial Functions


Homework 3.3 – Composite Functions


Homework 3.4 – Products of Functions


Homework 3.5 – Piecewise Functions


Homework 3.6 – Integrals of Polynomials


Homework 4.1 – Analyzing Rational Functions



Homework 4.2 – Horizontal and Vertical Asymptotes



Homework 4.3 – Continuity and L’Hôpital’s Rule


Homework 5.1 – Exponential Growth and Decay


Homework 5.2 – Derivative and Antiderivative of ex


Homework 5.3 – Implicit Differentiation and Inverse Functions


Homework 5.4 – Logarithmic Functions


Homework 5.5 – Derivatives and Antiderivatives of Exponentials and Logarithms


Homework 5.6 – Definite Integrals of Exponentials and Logarithms


Homework 6.1 – The Cosine and Sine Functions


Homework 6.2 – Derivatives and Antiderivatives of Cosine and Sine



Homework 6.3 – Other Trigonometric Functions


Homework 6.4 – Other Trigonometric Functions


Homework 7.1 – Related Rates


Homework 7.2 – Graph Analysis Using First and Second Derivatives


Homework 7.3 – Graph Analysis with the TI-84


Homework 7.4 – The Extreme Value Theorem and Optimization


Homework 7.5 – Differential Equations


Homework 8.1 – Sigma Notation and Summations


Homework 8.2 – The Definition of Net Area


Homework 8.3 – Rolle’s Theorem and the Mean Value Theorem


Homework 8.4 – The Fundamental Theorem of Calculus (Part 1)


Homework 8.5 – The Fundamental Theorem of Calculus (Part 2)


Homework 8.6 – Integration by Substitution


Homework 8.7 – Modeling Accumulated Change with the TI-84

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