first term calculus {...

$: First Term Calculus { 2016-2017rhs-richards.weebly.com/uploads/5/9/8/9/59892339/calculus_2016-2017.pdf · AP/CC Calculus M-220 Syllabus Joe Richards jrichards@roseburg.k12.or.us (541)$
First Term Calculus – 2016-2017

Monday Tuesday Wednesday Thursday Friday

Sep 5th

No School

Sep 6th

No School

Sep 7th

Assignment 1

Sep 8th

Assignment 2

Sep 9th

Assignment 3

Sep 12th

Assignment 3

Sep 13th

Assignment 4

Sep 14th

Assignment 5

Sep 15th

Assignment 5

Sep 16th

Assignment 6

Sep 19th

Assignment 7

Sep 20th

Quiz 1

Sep 21st

Assignment 8

Sep 22nd

Assignment 9

Sep 23rd

Assignment 10

Sep 26th

Assignment 11

Sep 27th

Assignment 12

Sep 28th

Quiz 2

Sep 29th

Assignment 13

Sep 30th

Assignment 14

Oct 3rd

Assignment 15

Oct 4th

Midterm 1 -Review

Oct 5th

Midterm 1

Oct 6th

Midterm 1 -Corrections

Oct 7th

Assignment 16

Oct 10th

Assignment 16

Oct 11th

Assignment 17

Oct 12th

No School

Oct 13th

No School

Oct 14th

No School

Monday Tuesday Wednesday Thursday FridayOct 17th

Assignment 18

Oct 18th

Assignment 19

Oct 19th

Day of Testing

Oct 20th

Assignment 20

Oct 21st

Assignment 21

Oct 24th

Max/Min Practice

Oct 25th

Assignment 22

Oct 26th

Assignment 23

Oct 27th

Quiz 3

Oct 28th

No School

Oct 31st

Assignment 24

Nov 1st

Assignment 24

Nov 2nd

Assignment 25

Nov 3rd

Assignment 25

Nov 4th

Assignment 26

Nov 7th

Assignment 27

Nov 8th

Assignment 28

Nov 9th

Assignment 29

Nov 10th

Assignment 30

Nov 11th

No School

Nov 14th

Midterm 2 -Review

Nov 15th

Midterm 2 -Review

Nov 16th

Midterm 2

Nov 17th

Midterm 2 -Corrections

Nov 18th

Assignment 31

Nov 21st

Assignment 32

Nov 22nd

Assignment 33

Nov 23rd

Assignment 34

Nov 24th

No School

Nov 25th

No School

Nov 28th Nov 29th Nov 30th Dec 1st Dec 2nd

Updated November 11, 2016.

Contents

Handouts 1Calendar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1High School Syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5UCC First Term Syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

September 8Assignment 1 – Precalculus Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Assignment 2 – Precalculus Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Assignment 3 – Derivatives of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Assignment 4 – More Derivatives of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18Assignment 5 – Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Assignment 6 – Trigonometric Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Assignment 7 – Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Quiz 1 – Assignments 1 - 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Quiz 1 – Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Assignment 8 – Beginning Trigonometric Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 30Assignment 9 – Inverse Function Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Assignment 10 – Reciprocals and Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Assignment 11 – Quotients and Trig Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36Assignment 12 – All of the Trigonometric Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . 38

Quiz 2 – Assignments 8 - 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40Quiz 2 – Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Assignment 13 – Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Assignment 14 – Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

October 46Assignment 15 – Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Midterm 1 – Assignments 1 - 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48Midterm 1 – Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Assignment 16 – Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56Assignment 17 – More Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58Assignment 18 – Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60Assignment 19 – More Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66Assignment 20 – Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71Assignment 21 – More Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Max/Min Practice – Four Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83Assignment 22 – Mean Value Theorm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85Assignment 23 – Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Quiz 3 – Assignments 16 - 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89Quiz 3 – Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3

November 95Assignment 24 – Pre-Calculus Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95Assignment 25 – Areas Under Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98Assignment 26 – More Areas Under Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100Assignment 27 – Net and Total Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Assignment 28 – Areas Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105Assignment 29 – Average Heights of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107Assignment 30 – Slope Fields and Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . 110

Midterm 2 – Assignments 16 - 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114Midterm 2 – Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

Assignment 31 – U-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116Assignment 32 – Trigonometric U-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118Assignment 33 – Volumes of Revolution (Disks) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121Assignment 34 – Volumes of Shapes with Similar Cross Sections . . . . . . . . . . . . . . . . . . . 124

AP/CC CalculusM-220Syllabus

Joe [email protected]

(541) 440-4142

Welcome to AP/CC Calculus!

1. Calculus basically has two chapters. Finding slopes of curves at a point, and findingareas of strange regions. You can go to the AP Central website to find out more of thespecifics of the curriculum:

https://apstudent.collegeboard.org/apcourse/ap-calculus-ab

2. This course is several courses in a single two-hour interval. You are taking an APcourse, which means that in early May you will be given a nationwide exam over theentire syllabus. You WILL pass this exam (RHS has a 96% success rate over the past20 years and I absolutely believe in your ability to improve that rate) and this creditis transferable to any college in the nation. You are also taking a UCC math course,which means that you will get community college credit, which is transferable to anypublic college in Oregon. You are also taking a high school math course, complete withall of the diversions, historical commentaries, and odd side routes that you have cometo expect from the RHS math program.

3. You will get a college grade, whose percent breaks occur at the traditional 90,80,70fences. You also get a high school grade, whose percent breaks occur at 85,75,65 fences.You can think about the incentive related implications of always being within 2.5% ofanother grade.

4. Your goal should be to pass the AP exam and get the college credit this year. Studieshave shown that students who go into high school Calculus, knowing ahead of timethat they are going to repeat it in college, don’t fare well. They don’t wrestle with thedifficult things this year because they know they’ll see it again, and they don’t pay closeattention to the tricky stuff the following year because they feel like “they’ve alreadyseen it”. This is the year . . . with only 27 students, and a teacher with multiple years ofteaching Calculus in his background . . . it seems like the odds of “getting it right” arebetter than next year, when you might have 200 peers in a lecture hall being taught bya foreign-born grad student who has a thick accent. Your call.

5. Cell phones are amazing tools. Having said that, I don’t want to hear or see yours atany point this year. This means they stay in your bag on silent mode. The potential forderailing your education is too great for me to have a more lenient policy.

6. You’ve been down the Ninth, Tenth, and Eleventh Honors roads, so all of the littlepaperwork details won’t be surprising to you. We’ll do (5 pt) assignments daily, take(20 pt) quizzes (more frequently than you did last year), take (50 pt) tests about thricea college term, and do other fun stuff. Falling behind is to be HIGHLY discouraged;this class moves far too fast for easy recovery from this error. I do, however, accept latework for full credit, provided it does not become a habit. We will not have extra creditopportunities beyond test corrections.

7. If you don’t already have one, I highly recommend getting a graphing calculator. Wehave a class set of TI-84’s but they are not borrowable outside of our regular meetingtime. You will need to be very familiar with a graphing calculator of some sort longbefore May, and it is a bad idea to wait too long to get one. If you absolutely can’tafford one, please see me. Incidentally, a math binder, a pencil, and a composition book(preferably with graph paper rather than lined paper) are the only other materials youneed to make sure to have.

8. Notes and assignments will be on our class webpage:

http://rhs-richards.weebly.com/

Therefore, there is no excuse for you to not have notes and/or homework any day, eventhose for which you were absent. On a related note, it would be wise of you to checkyour Synergy StudentVue account at least once a week or so to make sure that nothinghas slipped through the cracks of your, undoubtedly, very busy life. A link to log in toyour account can be found on our class webpage, as well as the high school’s webpage.

9. If you have ANY questions or concerns, you may contact me through any of the sourcesbelow:

• [email protected] (school email)

• [email protected] (home email)

• (541) 440-4142 (school phone)

10. Finally, a concluding phrase about behavior expectations. If you consistently do thesethings daily in class, youll be fine.

Be prompt, be kind, be perspicacious, and live PRIDE!

Course Title: MTH 251 Type of Program: AP/CC CalculusCRN: 23096 Instructor Name: Joe RichardsCredits: 5 Contact Info: (541) 440-4142Institution: Roseburg High School [email protected]/Time: M–F, 11:57–1:44 Office Hours: M–F, 2:40–3:30

Course Description:

This course is designed for accelerated high school students planning to take the AB Calculus exam in May, andintending to major in mathematics, science, and/or engineering. Topics include limits, continuity, differentiationformulas, chain rule, implicit differentiation, applications including rates of change and optimization, Newton’smethod, and the mean value theorem.

Course Objectives:

• We will develop the ability to make sense of problems involving limits and rates of change.

• We will develop an extensive set of skills and successfully analyze problems involving derivatives.

• We will develop a variety of strategies and solve problems involving applications of derivatives.

Length of Course: 9/7/15 – 11/31/15Grading Method: Daily Homework, Fortnightly Quizzes, Periodic Presentations, and Monthly ExamsGrading Scale: 90%–100% A 80%–89.9% B 70%–79.9% C 60%–69.9% D 0%–59.9% FRequired Text: Calculus with Analytic Geometry, by Edwards and Penney

Student Learning Outcomes:

• Students will be able to use both the limit definition and rules of differentiation to differentiate a wide varietyof functions.

• Students will be able to sketch the graph of a function using asymptotes, critical points, the derivative test forincreasing/decreasing functions, and concavity.

• Students will be able to apply differentiation to solve applied max/min problems.

• Students will be able to apply differentiation to solve related rates problems.

Major Topic Outline (by week):

1. Derivatives of Polynomials

2. Limits

3. Continuity

4. Derivatives of Trigonometric Functions

5. Derivatives of Products and Quotients

6. Position, Velocity, and Acceleration

7. Chain Rule

8. Implicit Differentiation

9. Related Rates

10. Optimization Problems

11. Mean Value Theorem

12. Finals Week

Course Policies:

Unless arrangements have been made ahead of time, late work will be accepted for only partial credit. Other thantest corrections, there will be no extra credit opportunities. Cell phones are to be neither seen nor heard. Classroombehavior is expected to conform to the same high level that our school’s PRIDE matrix has already outlined.

AP/CC CalculusAssignment 1

Name

1. What acute angle does the graph of g(x) =3

5x make with the x-axis?

1.

2. Suppose A = (10, 0), B = (0, 0), and C = (5, 5p

3). What is . . .

(a) . . . the length of segment BC?

(a)

(b) . . . the measure of angle CBA?

(b)

3. Find the sum of the infinite geometric series 1 +3

4+

9

16+

27

64+ . . .

3.

4. Sketch the graphs of each function below.

(a) y = cot x (b) y = 5Cx (c) y = log2 x (d) y = x2 +1

x + 1

x

y

x

y

x

y

x

y

5. If f(x) = 2x, find the value of f�1(8).

5.

6. Simplify the expression

p2

2+ i

p2

2

!6

.

6.

7. For which intervals below is sin x larger than cos x? (indicate all that apply)

(a) 0 < x <⇡

4

(b)⇡

4< x <

⇡

2

(c)⇡

2< x <

3⇡

4

(d)3⇡

4< x < ⇡

(e) ⇡ < x <5⇡

4

(f)5⇡

4< x <

3⇡

2

8. Write an equation for each graph shown below.

x

y

(a)

x

y

(b)

x

y

(c)

x

y

(d)

9. Solve the inequality x3 � 25x 0 and graph the solution on the provided number line.

9.

10. A population of bacteria is growing exponentially. If there are 8 million at noon and there are 10 millionat 1pm, at what time will there be 40 million?

10.

11. Find, roughly, the millionth term of each sequence below.

(a) tn =2n + 11

4n + 7

(a)

(b) tn = tan�1 n

(b)

(c) tn =�1 + 1

n

�n

(c)

12. Prove that log2 3 is an irrational number.


Name


5x make with the x-axis?

1.⇡ 30.96�

2. Suppose A = (10, 0), B = (0, 0), and C = (5, 5p

3). What is . . .

(a) . . . the length of segment BC?

(a)10

(b) . . . the measure of angle CBA?

(b)60�

3. Find the sum of the infinite geometric series 1 +3

4+

9

16+

27

64+ . . .

3.4


(a) y = cot x (b) y = 5Cx (c) y = log2 x (d) y = x2 +1

x + 1

x

y

x

y

x

y

x

y

5. If f(x) = 2x, find the value of f�1(8).

5.3


p2

2+ i

p2

2

!6

.

6.�i

7. For which intervals below is sin x larger than cos x? (indicate all that apply)

(a) 0 < x <⇡

4

(b)⇡

4< x <

⇡

2

(c)⇡

2< x <

3⇡

4

(d)3⇡

4< x < ⇡

(e) ⇡ < x <5⇡

4

(f)5⇡

4< x <

3⇡

2


x

y

(a)y = �2x

x

y

(b)y =

p9 � x2

x

y

(c)y = 2 sin(3x)

x

y

(d)y =

x2 + x � 2

x � 1

9. Solve the inequality x3 � 25x 0 and graph the solution on the provided number line.

9. -5 0 5


10.⇡ 7:13 pm

11. Find, roughly, the millionth term of each sequence below.

(a) tn =2n + 11

4n + 7

(a)

1

2

(b) tn = tan�1 n

(b)90�

(c) tn =�1 + 1

n

�n

(c)⇡ 2.718

12. Prove that log2 3 is an irrational number.

Let us suppose that the given statement is false. We can then be certain that there exists somepair of integers a and b such that log2 3 = a

b . It follows that 2ab = 3, hence 2a = 3b. But 2a is

certainly even while 3b is odd. Thus we have reached a contradiction, and must instead concludethat the original statement is true.


Name


4x make with the y-axis?

1.

2. Suppose A = (0, 8), B = (0, 0), and C = (8p

3, 0). What is . . .

(a) . . . the length of segment AC?

(a)

(b) . . . the measure of angle ACB?

(b)

3. Find the sum of the infinite geometric series1X

k=1

✓2

3

◆k

.

3.


(a) y = tan x

(b) y = x3 � x

(c) y =�

12

�x

(d) (x � 1)y = (x � 2)(x � 1)

(a)

x

y

(b)

x

y

(c)

x

y

(d)

x

y

5. If f�1(x) = log4 x, find the value of f(8).

5.


p3

2+ i

1

2

!6

.

6.

7. For which intervals below is tan x larger than sec x? (indicate all that apply)

(a) 0 < x <⇡

4

(b)⇡

4< x <

⇡

2

(c)⇡

2< x <

3⇡

4

(d)3⇡

4< x < ⇡

(e) ⇡ < x <5⇡

4

(f)5⇡

4< x <

3⇡

2


x

y

(a)

x

y

(b)

x

y

(c)

x

y

(d)

9. For what values of x will the expression x3 � 25x be positive?

9.


10.

11. Find the limit of each sequence given below.

(a) tn =11n � 2

9n + 4

(a)

(b) tn = sec�1 n

(b)

(c) tn =2n

n3

(c)

12. Prove that all odd squares are one more than a multiple of 8.


Name


4x make with the y-axis?

1.⇡ 75.96�

2. Suppose A = (0, 8), B = (0, 0), and C = (8p

3, 0). What is . . .

(a) . . . the length of segment AC?

(a)16

(b) . . . the measure of angle ACB?

(b)30�

3. Find the sum of the infinite geometric series1X

k=1

✓2

3

◆k

.

3.2


(a) y = tan x

(b) y = x3 � x

(c) y =�

12

�x

(d) (x � 1)y = (x � 2)(x � 1)

(a)

x

y

(b)

x

y

(c)

x

y

(d)

x

y

5. If f�1(x) = log4 x, find the value of f(8).

5.216


p3

2+ i

1

2

!6

.

6.�1

7. For which intervals below is tan x larger than sec x? (indicate all that apply)

(a) 0 < x <⇡

4

(b)⇡

4< x <

⇡

2

(c)⇡

2< x <

3⇡

4

(d)3⇡

4< x < ⇡

(e) ⇡ < x <5⇡

4

(f)5⇡

4< x <

3⇡

2


x

y

(a)y = � log6 x

x

y

(b)x =

p9 � y2

x

y

(c)y = 3 cos

⇣x

2

⌘

x

y

(d)y =

x2 + 1

x

9. For what values of x will the expression x3 � 25x be positive?

9. -5 0 5


10.6:00 pm

11. Find the limit of each sequence given below.

(a) tn =11n � 2

9n + 4

(a)

11

9

(b) tn = sec�1 n

(b)

⇡2

(c) tn =2n

n3

(c)Divergent

12. Prove that all odd squares are one more than a multiple of 8.

Let m be an arbitrary odd square. Then there must exist some natural number n such thatm = (2n � 1)

2. Note that this expands to m = 4n2 � 4n + 1 = 4n(n � 1) + 1. Since n and (n � 1)

are consecutive natural numbers, one of them is necessarily even. Consequently, the product of nand (n� 1) is a multiple of 2. Thus m = 4 · (a multiple of 2) + 1 = (a multiple of 8) + 1, as desired.


Name

1. Consider the parabola y = x2.

(a) What is the slope of the line tangent to the curve at x = 7?

(b) What is the x-intercept of the line tangent to the curve at x = 4?

(c) What is the height of the point at which the tangent line has a slope of 28?

(d) What is the slope of the curve at the point which has a height of 529?

(e) What is the y-intercept of the tangent line to the curve at the point where the height is 73?

(f) When the tangent line at (2, 4) hits the x-axis, what is the measure of the acute angle formed?

(g) When the tangent lines at x = 2 and x = 6 are both drawn, where do they cross and what acuteangle do they form?

(h) Find an equation for the line perpendicular to the curve at (3, 9).

2. Find the derivative (slope rule) of each function below, as well as the value of the derivative at x = 2.

(a) f(x) = 124x

12 (b) y = 2x4 − 5x3 − 4x + 7 (c) y = 9x + 73

3. Find the slope of the line tangent to f(x) = (x− 2)2 at the point (3, 1).

4. For f(x) = 2x6 − 7x4 − 2x, find the slopes at both x = 0 and x = 1.

5. For f(x) = 2x3 − 5x + 8, write an equation of the tangent line at the point (2, 14).

6. For f(x) = x3 + 9x2 + 24x− 13, find the coordinates of all points on the graph at which the tangent lineis horizontal.

7. The graph of f(x) = x3−9x2−16x+1 has a slope equal to 5 at exactly two points. Find the coordinatesof those points.

8. Find an equation for the line perpendicular to f(x) = 5x3 − 7x2 + 3x + 8 at the point where x = 2.

9. There’s nothing we’ve done with arithmetic yesterday and today that is beyond what you could havedone back in your 9th Honors days. What if you could be time-travelled back to yourself, back in thoseyounger years, and teach this stuff to younger-you? Alas, you can’t, but . . . what if you could sendyourself a letter? Your last task for this assignment is to write a brief letter to yourself (the 9th Honorsversion) that explains how to find the slope of f(x) = x4. No fair just telling them the formula we distilledtoday; explain how to create that formula using symbols like (x + h)4, and phrases like “infinitesimal”,or “really, really small”, or “limit”. We want correct terminology, but remember your audience. Don’tuse phrases or words that you wouldn’t have understood. If you think that your younger-you wouldjust toss this letter, then please make the effort to make it interesting reading, so that you know thatyou would (will?) read it. Who knows? Maybe you can change the past, and tomorrow you will havediscovered that you already took Calculus last year.

If you think that younger-you was especially precocious (and perspicacious) and want to teach themhow to find the slope of y = xn instead, go ahead. (Just remember to tell them that n must be a wholenumber!)

Oh, and if younger-you was just shy of genius, and you think you can explain to them how to get theslope for a POLY-nomial, instead of just a monomial, you have my blessing. Just don’t push younger-you so hard that they despair of ever taking Calculus and tomorrow you discover that you weren’t everenrolled!

Answer Key

1. You will make repeated use of the fact that the slope rule for y = x2 is 2x . . .

(a) 14

(b) 2

(c) 196

(d) 46

(e) -73

(f) tan−1 4 ≈ 76◦

(g) They intersect at (4, 12), and form an acute angle of tan−1 12 − tan−1 4 ≈ 9.3◦.

(h) y − 9 = − 16 (x− 3)

2.

(a) The slope rule = 12x

11, and plugging in 2 gives 210 = 1024.

(b) The slope rule = 8x3 − 15x2 − 4, and plugging in 2 gives 64 − 60 − 4 = 0.

(c) The slope rule = 9, and plugging in 2 gives, well, 9.

3. The slope is 2, just like the slope of y = x2 at (1, 1).

4. The slope rule is 12x5 − 28x3 − 2. Plugging in 0 and 1 gives hence −2 and −18, respectively.

5. y − 14 = 19(x− 2)

6. 3x2 + 18x + 24 = 0 has solutions x = −4 and x = −2. This gives us the points (−4,−29) and (−2,−33).

7. 3x2 − 18x− 16 = 5 has solutions x = −1 and x = 7. This gives us the points (−1, 7) and (7,−209).

8. y − 26 = − 135 (x− 2)

9. I am looking forward to reading these!


Name

1. Consider the function f(x) = x3.

(a) What is the slope of the line tangent to the curve at x = 4?

(b) What is the x-intercept of the line tangent to the curve at x = −1?

(c) What are the heights of the points at which the tangent lines has slopes of 12?

(d) What is the slope of the curve at the point which has a height of 343?

(e) What is the y-intercept of the tangent line to the curve at the point where the height is -8?

(f) When the tangent line at ( 13 ,

127 ) hits the x-axis, what is the measure of the acute angle formed?

(g) Where do the tangent lines at x = −1 and x = 2 cross each other?

(h) Find an equation for the line perpendicular to the curve at (3, 27).

(i) Give the coordinates of two points on the curve at which the tangent lines are parallel.

2. Sketch the graph of a curve, some of whose values are given in the table below. (There are MANYpossible sketches . . . )

x f(x) f ′(x)5 2 03 4 -27 4 1

28 0 -20 6 -1

3. Now sketch the derivative of the graph you made in the previous problem.

4. Find f ′(−4) for f(x) = 2|x|.

5. Find f ′(π2 ) for f(x) = sinx.

6. Find the coordinates of the valley floor (local minimum point) on the graph of y = x3 − 6x2.

7. For what values of x will the SLOPE of the graph of y = x4 − 18x2 be positive?

8. How much higher is the local maximum point of the graph of y = x3 − 6x2 + 2 than the local minimumpoint?

Answer Key

1. You will make repeated use of the fact that f ′(x) = 3x2 . . .

(a) 48

(b) − 23

(c) 8 and -8

(d) 147

(e) 16

(f) tan−1( 13 ) ≈ 18.4·

(g) (2, 8)

(h) y − 27 = − 127 (x− 3)

(i) There are MANY possible answers. For example, the slopes at (−5,−125) and (5, 125) are equal . . .

2. Answers may vary.

x

y

f(x)

f ′(x)

3. Answers may vary.

4. -2. Notice that there are only two possible slopes at any point on f(x) = 2|x|.

5. 0. There are very few fair slope questions involving trig functions . . . for now.

6. There are two possible candidates: x = 0 or x = 4. Checking both, we find (4,−32) to be the minimum.

7. We want 4x3 − 36x > 0. Note that the zeros of 4x3 − 36x are {−3, 0, 3}. Checking values between these“boundary points” results in positive values in exactly two regions: −3 < x < 0 and x > 3.

8. 32


Name

1. For the function f(x), whose graph is given, state the value of the given quantities, if they exist.

x

y

(a) f(3)

(b) limx→3−

f(x)

(c) limx→3+

f(x)

(d) limx→3

f(x)

(e) limx→−2+

f(x)

(f) limx→−2−

f(x)

(g) f(−2)

(h) limx→−2

f(x)

2. Sketch the graph of the following function: g(x) =

1− x if x < 0

3 if x = 0

2 + x if x > 0

3. Use the graph you sketched in the previous question to state the value of the following limits, if theyexist.

(a) limx→0−

g(x) (b) limx→0+

g(x) (c) limx→0

g(x)

4. Estimate the slope of h(x) = 5x at the point (2, 25) to the nearest tenth.

5. Find the limit of the expression8− 9n

2n− 6as . . .

(a) n→ 0 (b) n→∞ (c) n→ 3

6. Find the value of each limit below.

(a) limx→−1

x2 − x− 3

x+ 5

(b) limx→∞

5 + 2n

13− 3n

(c) limx→0

5 + 2n

13− 3n

(d) limx→∞

5n+ 3

13n2 − 4

(e) limx→∞

5n2 + 8

13n− 7

(f) limx→3

x3 − 9x

x− 3

(g) limh→0

(h+ 5)2 − 25

h

(h) limx→2

x3 − 8

x− 2

(i) limh→0

(2 + h)3 − 8

h

(j) limx→7

1x − 1

7

x− 7

(k) limx→9

x− 9√x− 3

(l) limx→4

(x3

x− 4− 4x2

x− 4

)

(m) limx→−5

x−1 + 0.2

x+ 5

(n) limx→4

x2 − 16√x− 2

(o) limt→0

√2− t−

√2

t

(p) limh→0

(3 + h)−1 − 3−1

h

7. For f(x) =x3

3− 9x and g(x) = 3.5x2 − 21x, evaluate the expression lim

x→3

f ′(x)g′(x)

.

Answer Key

1. (a) 1

(b) 2

(c) −2(d) DNE

(e) −1(f) −1

(g) −3(h) −1

2.

x

y

3. (a) 1 (b) 2 (c) DNE

4. 40.2

5. (a) − 43 (b) − 9

2 (c) DNE

6. (a) − 14

(b) 513

(c) − 23

(d) 0

(e) DNE

(f) 18

(g) 10

(h) 12

(i) 12

(j) − 149

(k) 6

(l) 16

(m) − 125

(n) 32

(o) − 12√2= −

√24

(p) − 19

7. limx→3

x2 − 9

7x− 21= lim

x→3

x+ 3

7=

6

7


Name

1. Graph (at least) one full period of each of the trig functions below.

(a) y = sinx

(b) y = cosx

(c) y = tanx

(d) y = cscx

(e) y = secx

(f) y = cotx

(g) y = sin−1 x

(h) y = cos−1 x

(i) y = tan−1 x

2. How does the graph of the second function listed (for each pair below) differ from the graph of the firstfunction listed?

(a) y = sinx and y = 2 sinx

(b) y = cosx and y = cos 2x

(c) y = sinx and y = sin (x− 30◦)

(d) y = cosx and y = cos (−x)

3. Find the coordinates of the y-intercept of each graph below.

(a) y = cosx (b) y = 7 cosx (c) y = sin (x− 30◦) (d) y = cos (x− 30◦)

4. Find the value of each limit below.

(a) limx→0

cosx

(b) limx→π

sinx

(c) limx→π

2

sinx

(d) limx→π

2

tanx

(e) limx→ 3π

2

x sinx

(f) limx→π

2

sinx

x

(g) limx→0

sinx

x

(h) limx→0

cosx

x

(i) limx→0

1− cosx

x

(j) limx→0

sin 3x

x

(k) limx→0

csc 2x

cotx(l) lim

x→π6

tan 4x

(m) limx→∞

tan−1 x

(n) limx→−

√2

2

sin−1 x

(o) limx→ 1

2

cos−1 x

(p) limx→∞

sec−1 x

5. For each of the functions y = f(x) graphed below, make a sketch of the graph of its derivative y = f ′(x).

(a)

x

y

(b)

x

y

(c)

x

y

(d)

x

y

Answer Key

1. (a)

x

y

(b)

x

y

(c)

x

y

(d)

x

y

(e)

x

y

(f)

x

y

(g)

x

y

(h)

x

y

(i)

x

y

2. (a) Every point is twice as far from the x-axis.

(b) Every point is half as far from the y-axis.

(c) The graph is shifted to the right 30◦.

(d) It doesn’t.

3. (a) (0, 1) (b) (0, 7) (c) (0,− 12 ) (d) (0,

√32 )

4. (a) 1

(b) 0

(c) 1

(d) DNE

(e) − 3π2

(f) 2π

(g) 1

(h) DNE

(i) 0

(j) 3

(k) 12

(l) −√

3

(m) π2

(n) −π4(o) π

3

(p) π2

5. (a)

x

y

(b)

x

y

(c)

x

y

(d)

x

y


Name

1. For each function listed below, tell whether or not the function is continuous at the indicated point. Ifit is discontinuous, state whether the discontinuity is essential or removable.

(a) f(x) = x3 − 5x+ 1, at x = 1

(b) f(x) =x2 − 9

x− 3, at x = 3

(c) f(x) =x2 − 9

x− 3, at x = −3

(d) f(x) =x2 − 9

x− 2, at x = 2

(e) f(x) =x2 − 4

x− 2, at x = 7

(f) f(x) =

{x for x ≥ 0

−x for x < 0, at x = 0

(g) f(x) =

{x2 + 4 for x < 2

x3 for x ≥ 2, at x = 2

(h) f(x) =

x2 − 1

x− 1for x 6= 1

2 for x = 1, at x = 1

(i) f(x) =

{−x2 for x < 0

1−√x for x ≥ 0, at x = 0

(j) f(x) =

{√9− x2 for |x| ≤ 3

3− |x| for |x| > 3, at x = 3

2. Sketch the graph of each below, locating and classifying any discontinuities.

(a) f(x) =x2 + 9

x+ 3

(b) f(x) =

{x for x ≥ x2x2 for x < x2

(c) f(x) =

x− 1 for x < 1

0 for x = 1

x2 for x > 1

(d) f(x) =

sinx

xfor x 6= 0

1 for x = 0, at x = 0

3. Let f(x) =

{x2 for x < 1

Ax− 3 for x ≥ 1. Find A to make the function continuous at x = 1.

4. Let f(x) =

{A2x2 for x ≤ 2

(1−A)x for x > 2. Find A to make the function continuous at x = 2.

5. Define f(5) so that the function f(x) =

√x+ 4− 3

x− 5becomes continuous at x = 5.

6. Define f(4) so that the function f(x) =

1

x− 1

4x− 4

becomes continuous at x = 4.

7. Let f(x) =

{Ax+B for x < 3

x2 + 1 for x ≥ 3. Find A and B so that the function is differentiable at x = 3.

8. Let f(x) =

{Ax+B for x < π

5 sinx for x ≥ π . Find A and B so that the function is differentiable at (π, f(π)).

Answer Key

1. (a) Continuous

(b) Discontinuous (Removable)

(c) Continuous

(d) Discontinuous (Essential)

(e) Continuous

(f) Continuous

(g) Continuous

(h) Continuous

(i) Discontinuous (Essential)

(j) Continuous

2. Sketch the graph of each below, locating and classifying any discontinuities.

(a)

x

y

(essential discontinuity at x = −3)

(b)

x

y

(continuous)

(c)

x

y

(essential discontinuity at x = 1)

(d)

x

y

(continuous)

3. A = 4

4. A =1

2or −1

5. f(5) =1

6

6. f(4) = − 1

16

7. A = 6 and B = −8

8. A = −5 and B = 5π

AP/CC CalculusQuiz 1

Name

1. The graph of f(x) =

{5x− 3 for x ≤ 1

x3 + 1 for x > 1has which below?

1.

A. It has a removable discontinuity at x = 1.

B. It has an essential discontinuity at x = 1.

C. It is both continuous and differentiable at x = 1.

D. It is continuous but not differentiable at x = 1.

2. For what values of x between 0 and π is the expression cosx− sinx positive?

2.

3. What are the two slopes of y = x2 + 3x at the two points where the height is 4?

3.

4. Find the coordinates of the local maximum (mountain top) of y = 2x3 − 24x+ 25.

4.

5. Evaluate each limit below.

(a) limt→0

3 + 7t

17− 8t

(a)

(b) limx→3−

f(x), where f(x) =

{7 + x for x < 3

3− 4x for x ≥ 3

(b)

(c) limx→2

sin−1(1x

)

x+ 1

(c)

(d) limx→11

1x − 1

11

x− 11

(d)

(e) limx→0

√x+ 64− 8

x

(e)

(f) limx→0

tanx

x secx

(f)

6. What is the x-intercept of the tangent line to y = x4 at x = 2?

6.

7. For the graph of y = x2, what angle does the tangent line at x = 2 make with the y-axis?

7.

8. The graph of y = x3 − 9x2 − 16x+ 1 has a slope equal to 5 at exactly 2 points.Give the coordinates. 8.

9. For the graph of y = x3, what is the slope of the curve at the point where the height is 64?

9.

10. Find values A and B so that f(x) =

{3x2 + 2 for x < 2

Ax−B for x ≥ 2is continuous and differentiable at x = 2.

10.

11. For what values of x will the expression x3 + 4x2 be positive?

11.

12. For the function f(x) =√x, what is the y-intercept of the tangent line at (64, 8)?

12.

13. Find equations for all three of the lines tangent to y = (x− 3)(x− 6) at one of its intercepts.

13.


Answer Key

1. The graph of f(x) =

{5x− 3 for x ≤ 1

x3 + 1 for x > 1has which below?

1.D

A. It has a removable discontinuity at x = 1.

B. It has an essential discontinuity at x = 1.

C. It is both continuous and differentiable at x = 1.

D. It is continuous but not differentiable at x = 1.

2. For what values of x between 0 and π is the expression cosx− sinx positive?

2.0 < x < π

4

3. What are the two slopes of y = x2 + 3x at the two points where the height is 4?

3.5 and −5

4. Find the coordinates of the local maximum (mountain top) of y = 2x3 − 24x+ 25.

4.(−2, 57)


(a) limt→0

3 + 7t

17− 8t

(a)− 7

8

(b) limx→3−

f(x), where f(x) =

{7 + x for x < 3

3− 4x for x ≥ 3

(b)10

(c) limx→2

sin−1(1x

)

x+ 1

(c)π18

(d) limx→11

1x − 1

11

x− 11

(d)− 1

121

(e) limx→0

√x+ 64− 8

x

(e)116

(f) limx→0

tanx

x secx

(f)1

6. What is the x-intercept of the tangent line to y = x4 at x = 2?

6.1.5

7. For the graph of y = x2, what angle does the tangent line at x = 2 make with the y-axis?

7.≈ 14◦

8. The graph of y = x3 − 9x2 − 16x+ 1 has a slope equal to 5 at exactly 2 points.Give the coordinates. 8.

(−1, 7) (7,−209)

9. For the graph of y = x3, what is the slope of the curve at the point where the height is 64?

9.48

10. Find values A and B so that f(x) =

{3x2 + 2 for x < 2

Ax−B for x ≥ 2is continuous and differentiable at x = 2.

10.A = 12, B = 10

11. For what values of x will the expression x3 + 4x2 be positive?

11.−4 < x 6= 0

12. For the function f(x) =√x, what is the y-intercept of the tangent line at (64, 8)?

12.(0, 4)

13. Find equations for all three of the lines tangent to y = (x− 3)(x− 6) at one of its intercepts.

13.y = 3x− 18 y = −3x+ 9 y = −9x+ 18


Name

1. For the function f(x) = sinx, find the value of each expression below.

(a) f(0)

(b) f ′(0)

(c) f(π3

)

(d) f ′(2π3

)(e) f ′ (π)

(f) f ′(7π6

)(g) f (30◦)

(h) f ′ (30◦)

2. For the function y = cosx, find the value(s) of each expression below.

(a) the y-intercept

(b) the slope at the y-intercept

(c) the slope at x =π

3

(d)dy

dxat x =

5π

3

(e)dy

dxat the positive x-intercept closest to (0, 0)

(f) y′ when x = −11π

6

(g) y′ when y = 0

(h) y′ when y = 1

3. At x =4π

3, evaluate each expression below.

(a)dy

dxfor y = sinx

(b)d (cosx)

dx

(c)d (− cosx)

dx

(d) y′′ for y = cosx

4. Find an equation for the line tangent to y = sinx at each of the following points.

(a) (0, 0) (b) (π, 0) (c) (π2 , 1) (d) when x = π

4 (e) at x = 1

5. Find an equation for the line tangent to y = cosx at x = π3 .

6. Find an equation for the line perpendicular to the curve y = sinx at its first positive x-intercept.

7. Determine three points on the graph of y = sinx at which the slope of the curve will be 12 .

8. Determine the slope of the curve y = 4 cosx at x = 5π6 .

9. Sketch the graph of y = sinx+ cosx and the line tangent to that curve at the point where x = π4 .

10. Write an equation for the line sketched in the previous problem.

11. Sketch the graph of y = x+ sinx and the line tangent to that curve at the point where x = π2 .

12. Write an equation for the line sketched in the previous problem.

13. Find the coordinates of a point on the graph of y = x− sinx at which the curve has a slope of 12 .

14. Find the coordinates of two different points on f(x) = 12x+ cosx which each have a horizontal tangent.

15. What is the slope of y = sin−1(x) at the point(√

32 ,

π3

)?

Answer Key

1. (a) 0

(b) 1

(c)√32

(d) − 12

(e) −1

(f) −√32

(g) 12

(h) You don’t know. . .

2. (a) (0, 1)

(b) 0

(c) −√32

(d)√32

(e) −1

(f) − 12

(g) 1 or −1

(h) 0

3. (a) − 12 (b)

√32 (c) −

√32

(d) 12

4. (a) y = x

(b) y = −1 (x− π)

(c) y = 1

(d) y −√22 =

√22

(x− π

4

)(e) y − sin 1 = cos 1 (x− 1)

5. y − 12 = −

√32

(x− π

3

).

6. y = x− π

7.(

π3 ,√32

),(

5π3 ,−

√32

),(

7π3 ,√32

),(−π

3 ,−√32

),(−5π3 ,

√32

)

8. −2

9.

x

y

10. y =√

2

11.

x

y

12. y = x+ 1

13.(

π3 ,

π3 −

√32

)or(−π

3 ,−π3 +

√32

)

14.(

π6 ,

π12 +

√32

)and

(5π6 ,

5π12 −

√32

)

15. 2


Name

1. Consider the function f(x) =√x.

(a) What is the slope of the line tangent to f(x) at x = 25?

(b) What is the x-intercept of the line tangent to f(x) at x = 9?


(d) What is the slope of f(x) at the point which has a height of 7?

(e) What is the y-intercept of the tangent line to f(x) at the point where the height is 8?

(f) When the tangent line at (4, 2) hits the x-axis, what is the measure of the acute angle formed?

(g) Find an equation for the line perpendicular to f(x) at (1, 1).

(h) Where do the tangent lines to f(x) at x = 4 and x = 9 intersect?

2. There is a function F such that F (5) = 8 and F ′(5) = 23 . What is the slope of F−1 at (8, 5)?

3. For the function H(x) = 3 + 2x, find the value of each expression below.

(a) H ′(17) (b) H−1(17) (c) the slope of H−1 at (12, 4.5)

4. For the function g(x) = x3, find the value of each expression below.

(a) g′(2)

(b)d(g−1(x)

)

dx

(c) the slope of g−1 at x = 8

(d)d(g−1(8)

)

dx

5. For the function h(x) = sinx, find the value of each expression below.

(a)d(h−1(x)

)

dxat x = −

√2

2(b) the slope of h−1 at x =

1

2(c) h′

(π3

)

6. Find the slope of the curve f(x) = 20 3√x+ 16 6

√x at x = 64.

7. Find the slope of the curve g(x) =

√x

3√x

at x = 64.

8. Name the coordinates of a point on the inverse cosine curve at which the slope is −2.

9. Find the measure of the acute angle formed by the y-axis and the tangent line to the curve h(x) = sin−1 x

at y =π

3.

Answer Key

1. You will make repeated use of the fact that f ′(x) = 12√x

.

(a)1

10(b) −9 (Not coincidentally, just as far left of the origin as the point of tangency is to the right.)

(c) 9

(d)1

14

(e) 4 (Not coincidentally, half as far from the x-axis as the point of tangency.)

(f) tan−11

4≈ 14.04◦

(g) y − 1 = −2(x− 1), or more simply, y = −2x+ 3

(h) (6, 2.5)

2.3

2

3. (a) 2 (b) 7 (c) 0.5

4. (a) 12

(b)1

3x−

23

(c)1

12(d) 0

5. (a)2√2

=√

2 (b)2√3

(c)1

2

6. 20 · 1

3· 64−

23 + 16 · 1

6· 64−

56 =

20

48+

16

192=

1

2

7.1

192

8. Either

(√3

2,π

6

)or

(−√

3

2,

5π

6

)

9. ≈ 26.57◦


Name

1. Consider the function f(x) =1

x.


(b) What is the x-intercept of the line tangent to f(x) at x = 3?

(c) What is the height of the point at which the tangent line has a slope of − 125?



(f) When the tangent line at (2, 12 ) hits the x-axis, what is the measure of the acute angle formed?

(g) Find an equation for the line perpendicular to f(x) at (1, 1).

(h) Consider the point P = (2, 12 ) and the tangent line to f(x) at that point. Which point is closer toP , the x-intercept of the tangent line or the y-intercept of the tangent line?

(i) For what other points on f(x) will the x-intercept of the tangent line and the y-intercept of thetangent line be equally far from the point of tangency?

(j) Let g(x) be the inverse function of f(x) and let h(x) equal the reciprocal function of f(x). Findthe derivatives of both g(x) and h(x).

2. Find an equation for the line tangent to the graph of xy = 12 at the point (4, 3).

3. Find an equation for the line perpendicular to the graph of f(x) = 2x− 1

x2+ 8x−3 at the point (2, 194 ).

4. Let h(x) = x3 · sinx. Find the slope of the curve at x = π.

5. Suppose that f(5) = 2, f ′(5) = 6, g(5) = −3, and g′(5) = 4. If the function h(x) = f(x) · g(x), find . . .

(a) h(5) (b) h′(5) (c) an equation of the tangentline to h(x) at x = 5

6. The graphs of functions f and g are given below. Let v(x) = f(x) · g(x) and find . . .

(a) both v(1) and v′(1).

(b) an equation for the tangent line at (1, v(1)).

(c) the measure of the angle formed by that tan-gent line from the previous question and they-axis.

x

y

f

g

7. At the point (2, 5), the slope of function f is 3. At the point (2, 7), the slope of function g is −4. Ifh = fg, what is the value of . . .

(a) h(2)?

(b) the slope of h at x = 2?

(c) the y-intercept of the tangent line to h(x) at the point whose y-coordinate is 35?

(d) the x-intercept of the perpendicular to h(x) at the point where x is 2?

Answer Key

1. You will make repeated use of the fact that f ′(x) = − 1

x2.

(a) −1

9

(b) 6

(c) There are two answers,1

5or −1

5.

(d) −49

(e) 16

(f) The slope of the line is −1

4, so the angle is ≈ 14.04◦.

(g) y = x

(h) It is a tie.

(i) All of them, and here is why. Consider some arbitrary point (c, 1c ) on the graph of f(x). Theequation of the tangent line there is y − 1

c = − 1c2 (x − c), which has an x-intercept at (2c, 0) and

y-intercept at (0, 2c ). Our arbitrary point is exactly the midpoint of these two intercepts!

(j) g′(x) = − 1x2 and h′(x) = 1

2. y − 3 = − 34 (x− 4)

3. y − 194 = − 4

3 (x− 2), because f ′(x) = 2 + 2x3 − 24

x4 , hence f ′(2) = 34 .

4. h′(x) = x3 · cosx+ 3x2 · sinx, so h′(π) = −π3.

5. (a) −6 (b) h′(5) = 2 · 4 − 3 · 6 = −10 (c) y + 6 = −10(x− 5)

6. (a) v(1) = 2 and v′(1) = 0

(b) y = 2

(c) 90◦

7. (a) h(2) = 35

(b) h′(2) = 5 · −4 + 7 · 3 = 1

(c) The y-intercept of y − 35 = 1(x− 2) is (0, 33).

(d) The x-intercept of y − 35 = −1(x− 2) is (37, 0).


Name

1. Consider the function f(x) = tanx.


(b) What is the x-intercept of the line tangent to f(x) at x = π4 ?




(f) When the tangent line at (π4 , 1) hits the x-axis, what is the measure of the acute angle formed?

(g) How many degrees steeper is the tangent line (to the tangent curve ,) at x = π3 than at x = π

4 ?

(h) How long is the quadrant 1 segment of the line perpendicular to the curve at (π6 ,1√3)?

2. Find the derivative of each product or quotient given below.

(a) f(x) =2x + 3

5x + 1

(b) f(x) = sinx cosx

(c) f(x) =sinx

x

(d) f(x) = x cosx

(e) f(x) =cosx

sinx

(f) f(x) = sin2 x

(g) f(x) =1

x5

(h) f(x) =8x2 + 5x

3

3. Find an equation for the tangent line to the graph of f(x) =3x + 4

4x + 1at the point where x = 3.

4. (Multiple Choice) The derivative of y = secx is which of the following?

A. sinx cosx B. secx tanx C. cscx cotx D. sinx secx

5. Suppose that f(5) = 2, f ′(5) = 6, g(5) = −3, and g′(5) = 4. If h(x) = f(x)g(x) and v(x) = g(x)

f(x) , find. . .

(a) h(5) (b) h′(5) (c) v(5) (d) v′(5)

6. The graphs of functions f and g are given below. Let u(x) = f(x)g(x) and v(x) = g(x)

f(x) and find . . .

(a) both u(5) and u′(5).

(b) both v(1) and v′(1).

(c) an equation of the tangent to v(x) at x = 1.x

y

f

g

7. At the point (2, 35), the slope of function f is 3. At the point (2, 7), the slope of function g is −4. Ifh = f

g , what is the value of . . .

(a) h(2)?

(b) the slope of h at x = 2?

(c) the y-intercept of the tangent line to h(x) at the point whose y-coordinate is 5?

(d) the x-intercept of the perpendicular to h(x) at the point where x is 2?

Answer Key

1. You will make repeated use of the fact that f ′(x) = sec2(x) = 1 + tan2(x).

(a) 1.

(b) π4 − 1

2 . . . which follows from the equation of the line being y − 1 = 2(x− π4 ).

(c) The height equals tan(x) = ±√

3. . . which follows from tan2(x) = 3.

(d) 50. . . which follows from the second version of the slope rule.

(e) 1− π2 . . . which follows from your solution to part (b).

(f) tan−1(2) ≈ 63.43◦.

(g) tan−1(4)− tan−1(2) ≈ 75.96◦ − 63.43◦ = 12.53◦.

(h) 53 (π8 + 1√

3). . . which follows from the line y − 1√

3= − 3

4 (x− π6 ) having a y-intercept of 1√

3+ π

8 and

from the fact that the segment in question is the hypotenuse of a shrunk 3− 4− 5 triangle.

2. (a) f ′(x) = − 13

(5x + 1)2

(b) f ′(x) = cos 2x

(c) f ′(x) =x cosx− sinx

x2

(d) f ′(x) = cosx− x sinx

(e) f ′(x) = − 1

sin2 x= − csc2 x

(f) f ′(x) = sin 2x

(g) f ′(x) = − 5

x6

(h) f ′(x) =16

3x +

5

3

3. y − 1 = − 113 (x− 3). . . because f ′(x) = − 13

(4x+1)2 , so f ′(3) = − 113 , while f(3) = 13

13 = 1.

4. B. secx tanx. . . because y′ = sin xcos2 x = sin x

cos x · 1cos x .

5. (a) − 23

(b) (−3)(6)−(2)(4)(−3)2 = − 26

9

(c) − 32

(d) (2)(4)−(−3)(6)(2)2 = 13

2

6. (a) u(5) = 32 and u′(5) =

(2)(− 13 )−(3)( 2

3 )

(2)2 = − 23

(b) v(1) = 12 and v′(1) = (2)(−1)−(1)(2)

(2)2 = −1

(c) y = −x + 32

7. (a) h(2) = 357 = 5

(b) h′(2) = (7)(3)−(35)(−4)(7)2 = 23

7

(c) (0,− 117 ). . . because the tangent line has the equation y − 5 = 23

7 (x− 2).

(d) ( 1297 , 0). . . because the perpendicular line has the equation y − 5 = − 7

23 (x− 2).


Name

1. Consider the function f(x) = secx.

(a) Carefully draw both f(x) = secx and g(x) = cosx on the same graph.

(b) What is the slope of the line tangent to f(x) at x = π3 ?

(c) What is the x-intercept of the line tangent to f(x) at x = π4 ?

(d) What are the heights of the points at which the tangent line has a slope of 0?

(e) What are the slopes of f(x) at the points which have a height of −2?

(f) What is the y-intercept of the tangent line to f(x) at the point where x = π3 ?

(g) When a tangent line to f(x) at a height of 3 hits the x-axis, what is the measure of the acute angleformed?

2. At x = π6 , which of the six trig functions is the steepest?

3. Cotangent is defined by two different looking quotients, cos xsin x and 1

tan x . Demonstrate that differentiatingeach yields the same slope formula.

4. Derive both f(x) = sec2 x and g(x) = tan2 x, and then explain the difference in the answers.

5. For the function g(x) = sin 2x, find . . .

(a) g(π2 )

(b) g′(π2 )

(c) an equation of the tangent to g(x) at its y-intercept.

6. For the function f(x) = 7 sin 5x, what is an equation of its tangent line at the origin?

7. For the function h(x) = tan−1 x, find . . .

(a) an equation of its tangent line at x = −1.

(b) h′(√

3)

(c) the slope of h−1(x) at x = π3

8. At which of the following values is f(x) = cotx steepest?

A. x = π6 B. x = π

4 C. x = π3 D. x = π

2 E. x = 2π3

9. Find the measure of the acute angle formed by the x-axis and the tangent line to h(x) = cscx at x = 3π4 .

Answer Key

1. You will make repeated use of the fact that f ′(x) = secx tanx.

(a)

x

y

(b) 2√

3

(c) π4 − 1. . . which follows from the equation of the tangent line, y −

√2 =√

2(x− π4 ).

(d) ±1. . . which can be readily be seen on the graph above.

(e) ±2√

3. . . note that because of the identity 1 + tan2 x = sec2 x, tanx = ±√

3.

(f) 2− 2√3π3 . . . . . . which follows from the equation of the tangent line, y − 2 = 2

√3(x− π

3 ).

(g) tan−1(3√

8) ≈ 83.28◦

2. cot(x) is steepest at x = π6 . All six slopes are computed below. . .

(a) sin′(π6 ) =√32

(b) cos′(π6 ) = − 12

(c) tan′(π6 ) = 43

(d) sec′(π6 ) = 23

(e) csc′(π6 ) = −2√

3

(f) cot′(π6 ) = −4

3. (a) ddx

(cos xsin x

)= (sin x)(− sin x)−(cos x)(cos x)

sin2 x= − 1

sin2 x= − csc2 x

(b) ddx

(1

tan x

)= (tan x)(0)−(1)(sec2 x)

tan2 x = − cos2 xsin2 x cos2 x

= − 1sin2 x

= − csc2 x

4. (a) f ′(x) = (secx)(secx tanx) + (secx)(secx tanx) = 2 sec2 x tanx

(b) g′(x) = (tanx)(sec2 x) + (tanx)(sec2 x) = 2 sec2 x tanx

(c) Take the derivative of both sides of the identity 1 + tan2 x = sec2 x.

5. (a) g(π2 ) = 0

(b) Since g(x) = 2 sinx cosx, it follows that g′(x) = (2 sinx)(− sinx) + (cosx)(2 cosx), which is equalto 2(cos2 x− sin2 x) = 2 cos 2x. Consequently, g′(π2 ) = −2.

(c) y = 2x

6. y = 35x. . . since f(x) is vertically stretched by a factor of 7 and horizontally squished by a factor of 5.

7. (a) y + π4 = 1

2 (x + 1). . . because tanx has a slope of 2 at (−π4 ,−1), and it follows that tan−1 x has aslope of 1

2 at (−1,−π4 ).

(b) Because tanx has a slope of 4 at (π3 ,√

3), it follows that tan−1 x has a slope of 14 at (

√3, π3 ).

(c) Since h−1(x) = tanx, we observe from the previous problem that the slope is 4.

8. A. x = π6 (see problem 2)

9. tan−1(√

2) ≈ 54.74◦


Name

1. The slope of y = 1 + sinx at

(2π

3,

2 +√

3

2

)is which of the following? 1.

A. −1 B. −0.5 C. 0 D. 0.5 E. 1

2. What is the equation of the tangent line to the curve y = 1x at x = 3? 2.

3. What are the heights of the points at which the slope of y = tanx is equal to 4? 3.

4. What is the x-intercept of the tangent line to y = secx at x = π4 ? 4.

5. For y =√x at (25, 5), what is the. . .

(a) . . . angle the tangent line makes with the x-axis? (a)

(b) . . . y-intercept of that tangent line? (b)

6. If h(x) = f(x)g(x) , f(3) = 8, f ′(3) = 4, g(3) = 2, and g′(3) = 3, find each below.

(a) h(3) (a)

(b) h′(3) (b)

7. Find an equation for the line perpendicular to y = tanx at x = π4 . 7.

8. If y = x−2 + 10√x+ 6x

13 , find dy

dx at x = 1. 8.

9. Find the coordinates of both points on y = 1x at which the slope is −4. 9.

10. Find the slope of the curve y = sin−1 x at y = π6 . 10.

11. What is the measure of the smaller angle between the hands of a clock at 2 : 45? 11.

12. Let f(x) =3x+ 1

2x+ 3. Find the x-intercept of the line tangent to the curve at x = 2. 12.

13. For the function f(x) = 13x+ 2, find the slope of f−1(x) at the point (41, 3). 13.

14. Let f(x) = g(x) · tanx. If g(π3 ) = 2 and g′(π3 ) =√

3, find the value of f ′(π3 ). 14.


Answer Key

1. The slope of y = 1 + sinx at

(2π

3,

2 +√

3

2

)is which of the following? 1.

B. −0.5

A. −1 B. −0.5 C. 0 D. 0.5 E. 1

2. What is the equation of the tangent line to the curve y = 1x at x = 3? 2.

y − 13 = − 1

9 (x− 3)

3. What are the heights of the points at which the slope of y = tanx is equal to 4? 3.±√

3

4. What is the x-intercept of the tangent line to y = secx at x = π4 ? 4.

π − 4

4

5. For y =√x at (25, 5), what is the. . .

(a) . . . angle the tangent line makes with the x-axis? (a)tan−1 1

10 ≈ 5.71◦

(b) . . . y-intercept of that tangent line? (b)(0, 2.5)

6. If h(x) = f(x)g(x) , f(3) = 8, f ′(3) = 4, g(3) = 2, and g′(3) = 3, find each below.

(a) h(3) (a)4

(b) h′(3) (b)−4

7. Find an equation for the line perpendicular to y = tanx at x = π4 . 7.

y − 1 = − 12 (x− π

4 )

8. If y = x−2 + 10√x+ 6x

13 , find dy

dx at x = 1. 8.5

9. Find the coordinates of both points on y = 1x at which the slope is −4. 9.

(± 1

2 ,±2)

10. Find the slope of the curve y = sin−1 x at y = π6 . 10.

2√3

11. What is the measure of the smaller angle between the hands of a clock at 2 : 45? 11.172.5◦

12. Let f(x) =3x+ 1

2x+ 3. Find the x-intercept of the line tangent to the curve at x = 2. 12.

(−5, 0)

13. For the function f(x) = 13x+ 2, find the slope of f−1(x) at the point (41, 3). 13.

1

13

14. Let f(x) = g(x) · tanx. If g(π3 ) = 2 and g′(π3 ) =√

3, find the value of f ′(π3 ). 14.11


Name

1. Given dxdy = 5, dydv = 2

7 , dxda = 0.33, and dy

dk = sin π6 , find each of the derivatives below.

(a) dxdv (b) da

dy (c) dadv (d) dx

dk (e) dkdv (f) da

dk

2. If y = t5 and t = 3x+ 7, find each of the derivatives below.

(a) dydt in terms of t

(b) dtdy in terms of x

(c) dxdt in terms of x

(d) dydx in terms of t

(e) dxdy in terms of x

(f) dxdy in terms of y

3. Evaluate each derivative below.

(a)dy3

dy

(b)dy3

dx

(c)duv

dx

(d)duv

du

(e)d(uv )

dt

(f)d(uv )

du

(g)d sin y

dt

(h)d tan y

dx

(i)d sec r

dt

(j)d sin(t2)

dt

(k)d sin(x2)

dt

(l)dx sin y

dt

4. Find the derivative of each function below.

(a) f(x) = (2x+ 1)5

(b) f(x) = 7 (8x− 3)−3

(c) f(x) = 4 · 5√x2 + 7x

(d) f(x) = tan2 x

(e) f(x) =√

cosx

(f) f(x) = sin2(3x)

5. If y = u2, u = sin t, and t = 6x+ 3 write each below.

(a) an expression for y in terms of t

(b) an expression for y in terms of x

(c) an expression for dudx in terms of x

(d) an expression for u in terms of x

(e) an expression for dydu in terms of t

(f) an expression for dydx in terms of x

6. Find an equation of the tangent line to f(x) = (2x− 1)10

at its y-intercept.

7. Find the y-intercept of the tangent line to the curve f(x) =3

(2x− 1)4at x = 1.

8. Find the slope of the tangent line to f(x) = 3 tan 5x at the origin.

9. Find the measure of the angle formed by the x-axis and the tangent to y = sec3(4x) at x = π12 .

10. Find the length of the quadrant one segment of the tangent line to f(x) =(sec2 x− tan2 x

)9at x = π.

Answer Key

1. (a)dx

dv=dx

dy· dydv

=10

7

(b)da

dy=da

dx· dxdy

= 15

(c)da

dv=da

dx· dxdy· dydv

=30

7

(d)dx

dk=dx

dy· dydk

=5

2

(e)dk

dv=dk

dy· dydv

=4

7

(f)da

dk=da

dx· dxdy· dydk

=15

2

2. (a) 5t4

(b)1

5(3x+ 7)4

(c)1

3

(d) 15t4

(e)1

15(3x+ 7)4

(f)1

15y45

3. (a) 3y2

(b) 3y2dy

dx

(c) udv

dx+ v

du

dx

(d) udv

du+ v

(e)v dudt − udvdt

v2

(f)v − u dvduv2

(g) cos ydy

dt

(h) sec2 ydy

dx

(i) sec r tan rdr

dt

(j) 2t cos(t2)

(k) 2x cos(x2)dx

dt

(l) x cos ydy

dt+ sin y

dx

dt

4. (a) f ′(x) = 10 (2x+ 1)4

(b) f ′(x) = −168 (8x− 3)−4

(c) f ′(x) =8x+ 28

5(x2 + 7x)45

(d) f ′(x) = 2 tanx sec2 x

(e) f ′(x) = − sinx

2√

cosx

(f) f ′(x) = 3 sin(6x)

5. (a) y = sin2 t

(b) y = sin2(6x+ 3)

(c) dudx = 6 cos(6x+ 3)

(d) u = sin(6x+ 3)

(e) dydu = 2 sin t

(f) dydx = 6 sin(12x+ 6)

6. First of all, we know that f(0) = 1, and f ′(x) = 20 (2x− 1)9

so f ′(0) = −20. Thus, an equation for thetangent line to f(x) at x = 0 is y = −20x+ 1.

7. First, we know that f(1) = 3, and f ′(x) = − 24

(2x− 1)5, hence f ′(1) = −24. Thus the tangent line has

an equation y − 3 = −24(x− 1), and a y-intercept at (0, 27).

8. 15

9. Since y′ = 12 sec3(4x) tan(4x), plugging in x = π12 gives y′ = 96

√3 for the slope in question. Thus

tan−1(96√

3) ≈ 89.66◦

10. The length is infinite. Did you recall that 1 + tan2 x = sec2 x?


Name

1. Consider the position function x(t) = t2 − 6t + 2, of a particle moving along the x-axis at time t ≥ 0.

(a) What is the location of the particle at time t = 5?

(b) At what other time(s) is it at that same position on the x-axis?

(c) What is its velocity at time t = 4?

(d) At what other time(s) is it traveling with that same velocity?

(e) At what other time(s) is it traveling with that same speed?

(f) What is its average “net” velocity over the interval 1 ≤ t ≤ 8?

(g) What is its average “total” velocity over the same interval?

(h) At time t = 2, is the particle traveling to the left or to the right?

(i) At time t = 4, is the particle’s position right or left of its position at time t = 1?

(j) At time t = 4, is the particle’s velocity greater or less than its velocity at time t = 1?

(k) At time t = 4, is the particle’s speed greater or less than its speed at time t = 1?

2. Position functions are given below for various particles. Find the position and velocity of each particleat the indicated time.

(a) x(t) = t3 − 6t, at t = 2

(b) x(t) = 18tt+2 , at t = 1

(c) x(t) = t3 − 3t2 + 3t, at t = 4

(d) x(t) = sin t + cos t, at t = π3

(e) x(t) = 3√t + 1

6t2 , at t = 1

(f) x(t) = 3 csc t, at t = 2π3

3. For each of the six particle paths described in the previous problem, determine when, if ever, the particlechanges direction.

4. For the particle whose path is given by x(t) = t4 − 12t3 + 28t2, over the time interval 0 ≤ t ≤ 10, whatpercent of the time is it moving to the right?

5. For the two particles whose paths are x1(t) = t4− 7t3 + 11t2− 4t+ 8 and x2(t) = t4− 6t3 + 5t2 + 5t+ 8,find each below. Assume t ≥ 0.

(a) At what time(s) are the particles at the same location?

(b) At what time(s) are the two particles traveling with the same velocity?

(c) At what time(s) is one particle traveling 3 speed units faster than the other?

6. Over the interval 0 ≤ t ≤ 5, where is the particle whose equation is x(t) = t3 − 9t2 + 24t + 6 going thefastest?

Answer Key

1. You will make repeated use of the fact that x′(t) = v(t) = 2t− 6.

(a) x(5) = −3 (b) t = 1 (c) v(4) = 2 (d) Never (e) t = 2

(f) 3 speed units, since the particle moved 21 distance units, from x(1) = −3 to x(8) = 18, in 7 timeunits.

(g) 297 speed units, since the particle traveled strictly to the left over the first two time units (from −3

to −7), then traveled strictly to the right over the next five time units (from −7 to 18), for a totalof 29 distance units in 7 time units.

(h) Left, since v(2) = −2.

(i) Left, since x(4) = −6 and x(1) = −3.

(j) Greater, since v(4) = 2 and v(1) = −4.

(k) Less, because |2| < |−4|.

2. (a) x(2) = −4 and v(2) = 6

(b) x(1) = 6 and v(1) = 4

(c) x(4) = 28 and v(4) = 27

(d) x(π3 ) = 1+√3

2 and v(π3 ) = 1−√3

2

(e) x(1) = 76 and v(1) = 0

(f) x( 2π3 ) = 2

√3 and v( 2π

3 ) = 2

3. (a) v(t) = 3t2 − 6 = 0 at ±√

2, and changes directions at both times.

(b) v(t) = 36(t+2)2 = 0 for no value of t, hence does not change direction.

(c) v(t) = 3t2 − 6t + 3 = 3(t− 1)2 = 0 at t = 1, but doesn’t turn around there, only pauses.

(d) v(t) = cos t− sin t = 0 at t = π4 , 5π

4 , 9π4 , . . . , and changes directions at each of these times.

(e) v(t) = 13 t− 2

3 − 13t3 = 0 when 1

3 t− 2

3 = 13t3 , hence t

73 = 1, or t = 1. Testing values on both sides of

this value of t, we find it did, in fact, change directions there.

(f) v(t) = −3 csc t cot t = 0 at t = π2 , 3π

2 , 5π2 , . . . , and changes directions at each of these times.

4. Since v(t) = 4t3 − 36t2 + 56t = 4t(t− 2)(t− 7), the particle has positive velocity for t larger than 7, aswell as between 0 and 2. This is exactly half of the values between 0 and 10, hence 50% of the time, theparticle is moving to the right.

5. For the two particles whose paths are x1(t) = t4− 7t3 + 11t2− 4t+ 8 and x2(t) = t4− 6t3 + 5t2 + 5t+ 8,find each below. Assume t ≥ 0.

(a) If x1(t) = x2(t), then t3 − 6t2 + 9t = t(t− 3)2 = 0, hence t = 0 or t = 3.

(b) If v1(t) = v2(t), then 3t2 − 12t + 9 = 3(t− 1)(t− 3) = 0, hence t = 1 or t = 3.

(c) First, let us suppose that v1 − v2 = 3. Then −3t2 + 12t − 9 = 3 or (t − 2)2 = 0, which only holdsfor t = 2. If we suppose instead that v2 − v1 = 3, then 3t2 − 12t + 9 = 3 or t2 − 4t + 2 = 0, whichholds only for t = 2±

√2. Hence, one particle is exactly 3 speed units faster than the other at the

three times 2−√

2, 2, and 2 +√

2, and never again.

6. Note first that v(t) = 3t2 − 18t+ 24 = 3(t− 2)(t− 4). The lowest value this function attains is at t = 3,which gives a velocity of -3, hence speed of 3. The highest value this function attains (in the domaingiven) is at the time furthest from 3, namely t = 0. Since v(0) = 24, we find that the greatest speed isattained at t = 0, and our solution is x(0) = 6.


Name

1. Consider the position function x(t) = t3 − 9t2 + 15t + 4 of a particle moving along the x-axis at timet ≥ 0.

(a) What is the location of the particle at time t = 5?

(b) What is the velocity of the particle at time t = 5?

(c) At what other time(s) does it have that same velocity on the x-axis?

(d) What is its acceleration at time t = 4?

(e) At what other time(s) is it accelerating that same amount in that same direction?

(f) At what other time(s) is it accelerating that same amount but in the other direction?

(g) At time t = 2, is the particle being tugged to the left or to the right?

(h) At time t = 3, is the particle being tugged to the left or to the right?

(i) At the time that its acceleration is 12, what is its velocity?

(j) At the times its speed is 15, what are its accelerations?

(k) At the times its speed is 15, what are its locations?

(l) At time t = 2, is it speeding up or slowing down?

(m) Over what time intervals is the particle speeding up?

2. Position functions are given below for various particles. Find the acceleration of each particle at theindicated time.

(a) x(t) = t3 − 6t, at t = 2

(b) x(t) = (2t− 3)5, at t = 1

(c) x(t) = t3 − 3t2 + 3t, at t = 4

(d) x(t) = sin2 t, at t = 2π3

(e) x(t) = 3√t + 1

6t2 , at t = 1

(f) x(t) = 9 sec t, at t = π3

3. For the particle whose path is given by x(t) = t3 − 5t2 + 8t + 2, over what time intervals is it. . .

(a) . . . moving to the left?

(b) . . . being tugged to the left?

(c) . . . speeding up?

4. Over the interval 0 ≤ t ≤ 8, where is the particle whose equation is x(t) = t3 − 10t2 going the fastest?

Answer Key

1. You will make repeated use of the fact that v(t) = 3(t− 1)(t− 5) and a(t) = 6(t− 3).

(a) x(5) = −21

(b) v(5) = 0

(c) v(t) = 0 only for t = 5 and t = 1.

(d) a(4) = 6

(e) Never

(f) a(t) = −6 only for t = 2.

(g) Left, because a(2) < 0.

(h) Neither, becuase a(3) = 0.

(i) 0, since a(t) = 12 only for t = 5, and v(5) = 0.

(j) First, v(t) = 15 only at times t = 0 and t = 6, with corresponding accelerations a(0) = −18 anda(6) = 18. Since speed doesn’t care about direction, we also consider v(t) = −15, but this equationdoes not have solutions that are real numbers.

(k) x(0) = 4, and x(6) = −14.

(l) v(2) and a(2) both have the same sign (negative), so the particle is speeding up

(m) The particle is speeding up when velocity and acceleration have the same sign. Below is a numberline which shows when v(t) is positive and when a(t) is positive.

0 1 3 5

By inspection, we see that v(t) and a(t) have the same sign for 1 < t < 3 and for t > 5.

2. (a) a(2) = 12 because a(t) = 6t

(b) a(1) = −80 because a(t) = 80(2t− 3)3

(c) a(4) = 18 because a(t) = 6t− 6

(d) a( 2π3 ) = −1 because a(t) = 2 cos(2t)

(e) a(1) = 79 because a(t) = − 2

9 t− 5

3 + t−4

(f) a(π3 ) = 126 since a(t) = 9(sec3 t+ tan2 x secx)

3. You will make use of the fact that v(t) = 3t2− 10t+ 8 = (3t− 4)(t− 2) and a(t) = 6t− 10. The numberline below shows both where v(t) is positive and where a(t) is positive.

43

53

2

(a) The particle is moving left at times 43 < t < 2.

(b) The particle is being tugged left at times t < 53 .

(c) The particle is speeding up at times 43 < t < 5

3 and t > 2.

4. Since x(t) is cubic, it follows that v(t) is a quadratic (parabolic) function. Computing v(t) = 3t2 − 20t,we easily find the lowest value of v(t) will occur at t = 10

3 . Since the number furthest away from 103 in

the given domain is 8, it follows that the greatest value of v(t) will occur at t = 8. Since v( 103 ) = −33 1

3while v(8) = 32, we find that speed is at a maximum at t = 10

3 . Hence, our solution is x( 103 ) = − 2000

27 .

AP/CC CalculusMidterm 1

Name

1. Consider the parabola f(x) = x2.

(a) What is the y-intercept of the tangent line to f(x) at the point where theheight of f(x) is 36? (a)

(b) What is the height of the point at which the tangent line has a slope of14? (b)

(c) When the tangent line at (1.5, 2.25) hits the x-axis, what is the measureof the acute angle formed? (c)

2. Consider the function g(x) =√x.

(a) What is the y-intercept of the tangent line to g(x) at (49, 7)? (a)

(b) What is the slope of g(x) at (9, 3)? (b)

(c) What is the height of the point at which the tangent line to g(x) has aslope of 1

8? (c)

3. Find the slope of f(x) = (x+ 2)4

at its y-intercept. 3.

4. Find f ′(−4) for f(x) = 3|x− 4|. 4.

5. Evaluate each expression below at x =π

6.

(a)dy

dxfor y = cosx (a)

(b)d (cscx)

dx (b)

(c) y′ for y = tanx (c)

(d) y′′ for y = 3 cos 2x (d)

6. If f−1(x) = log2 x, find the value of f(−2). 6.

7. Find the slope of f(x) = 2√x− 1

2x2+ 1 at x = 4. 7.

8. What is the slope of the curve g(x) = x3 at the point where the height is 64? 8.

9. Find the slope of the tangent line to f(x) =2x+ 3

5x+ 1at x = 1. 9.

10. Find the slope of the line tangent to the curve y = (2x− 5)3

at x = 2. 10.

11. Consider the function f(x) = x3 − 7x2 + 10x.

(a) Graph, on a number line, the values of x for which f(x) < 0. (a)

(b) Find the x-coordinates of the two points on f(x) at which the slope is 15. (b)

(c) Find an equation for the perpendicular to f(x) at the point where x = 3. (c)

(d) For what interval is the graph of f(x) concave up? (d)

12. Given that h(x) = x2 · f(x), f(2) = −3, and f ′(2) = 5, what is h′(2)? 12.

13. Consider the function g(x) = x3 − 3x2 − 24x+ 25.

(a) Find the x-coordinates of the points where the slope is zero. (a)

(b) Find the coordinates of the point of inflection. (b)

14. For the function f(x) = 5− 3x, find the value of each expression below.

(a) f ′(−16) (a)

(b) f−1(−16) (b)

(c) The slope of f−1(x) at the point (−1, 2) (c)

15. Find the slope of g(x) =

√x

5√x

at x = 1. 15.

16. For the function g(x) = −6 tan(3x), what is an equation for the tangent line tog(x) at the origin? 16.

17. For the function f(x) =2

cosx, what is the value of f ′

(π6

)? 17.

18. For f(x) = x−2 · cosx, what is the value of f ′(π)? 18.


(a) limn→∞

9 + 7n

11− 13n

(a)

(b) limh→0

(3 + h)−1 − 3−1

h (b)

(c) limt→0

√64 + t− 8

t (c)

(d) limx→3−

|x− 3|2x− 6 (d)

(e) limx→0

1− cos2 x

6x2 (e)

20. Consider the function f(x) =

9x− 2 for x < 4

x+ 30 for 4 ≤ x ≤ 6

x2 − 10 for x ≥ 6

(a) Find limx→6+

f(x). (a)

(b) Is f(x) continuous at x = 6? (b)

(c) Is f(x) continuous at x = 4? (c)

(d) Is f(x) differentiable at x = 4? (d)

21. Find the y-intercept of the tangent line to y = 2x+ sinx at x = π. 21.

22. What is the slope of y = sin−1 x at the point

(√3

2,π

3

)?

22.

23. A particle is moving along the x-axis according to the equation x(t) = t3 − 6t2 + 40.Assume that t ≥ 0.

(a) Find the value of v(3). (a)

(b) At time t = 1, is the particle speeding up, slowing down, or coasting? (b)

(c) When the velocity is zero, what are the particle’s locations? (c)

(d) What is the acceleration when the particle has a velocity of 15? (d)

24. If a particle moves on the x-axis according to x(t) = t2 − 12t + 21, how far tothe left will it ever get? 24.

25. Finddy

dx(in terms of x) for each function below.

(a) y =√x2 + 6 (a)

(b) y = (5x+ 1)3 (b)

(c) y = sin4 x (c)

EC 1.The perpendicular to y = 1

x at the point(12 , 2)

intersects the other branch ofthe curve as well. Where? 1.

EC 2.Two lines are drawn tangent to y = x2, one at (a, a2) and the other at (b, b2).Where do these two tangent lines cross, in terms of a and b? 2.

AP/CC CalculusMidterm 1

Answer Key

1. Consider the parabola f(x) = x2.

(a) What is the y-intercept of the tangent line to f(x) at the point where theheight of f(x) is 36? (a)

-36

(b) What is the height of the point at which the tangent line has a slope of14? (b)

49

(c) When the tangent line at (1.5, 2.25) hits the x-axis, what is the measureof the acute angle formed?

(c)tan−1 3 ≈ 71.57◦

2. Consider the function g(x) =√x.

(a) What is the y-intercept of the tangent line to g(x) at (49, 7)?(a)

(0, 3.5)

(b) What is the slope of g(x) at (9, 3)?(b)

16

(c) What is the height of the point at which the tangent line to g(x) has aslope of 1

8?(c)

4

3. Find the slope of f(x) = (x+ 2)4

at its y-intercept. 3.32

4. Find f ′(−4) for f(x) = 3|x− 4|. 4.−3

5. Evaluate each expression below at x =π

6.

(a)dy

dxfor y = cosx

(a)− 1

2

(b)d (cscx)

dx (b)−2√

3

(c) y′ for y = tanx(c)

43

(d) y′′ for y = 3 cos 2x (d)−6

6. If f−1(x) = log2 x, find the value of f(−2).6.

14

7. Find the slope of f(x) = 2√x− 1

2x2+ 1 at x = 4.

7.3364

8. What is the slope of the curve g(x) = x3 at the point where the height is 64? 8.48

9. Find the slope of the tangent line to f(x) =2x+ 3

5x+ 1at x = 1.

9.− 13

36

10. Find the slope of the line tangent to the curve y = (2x− 5)3

at x = 2. 10.6

11. Consider the function f(x) = x3 − 7x2 + 10x.

(a) Graph, on a number line, the values of x for which f(x) < 0. (a) 20 5

(b) Find the x-coordinates of the two points on f(x) at which the slope is 15.(b)

5 and − 13

(c) Find an equation for the perpendicular to f(x) at the point where x = 3.(c)

y + 6 = 15 (x− 3)

(d) For what interval is the graph of f(x) concave up?(d)

x > 73

12. Given that h(x) = x2 · f(x), f(2) = −3, and f ′(2) = 5, what is h′(2)? 12.8

13. Consider the function g(x) = x3 − 3x2 − 24x+ 25.

(a) Find the x-coordinates of the points where the slope is zero. (a)4 and −2

(b) Find the coordinates of the point of inflection.(b)

(1,−1)

14. For the function f(x) = 5− 3x, find the value of each expression below.

(a) f ′(−16) (a)−3

(b) f−1(−16) (b)7

(c) The slope of f−1(x) at the point (−1, 2)(c)

− 13

15. Find the slope of g(x) =

√x

5√x

at x = 1.

15.310

16. For the function g(x) = −6 tan(3x), what is an equation for the tangent line tog(x) at the origin? 16.

y = −18x

17. For the function f(x) =2

cosx, what is the value of f ′

(π6

)?

17.43

18. For f(x) = x−2 · cosx, what is the value of f ′(π)?18.

2π3


(a) limn→∞

9 + 7n

11− 13n (a)

911

(b) limh→0

(3 + h)−1 − 3−1

h (b)− 1

9

(c) limt→0

√64 + t− 8

t (c)116

(d) limx→3−

|x− 3|2x− 6

(d)− 1

2

(e) limx→0

1− cos2 x

6x2 (e)16

20. Consider the function f(x) =

9x− 2 for x < 4

x+ 30 for 4 ≤ x ≤ 6

x2 − 10 for x ≥ 6

(a) Find limx→6+

f(x).(a)

26

(b) Is f(x) continuous at x = 6? (b)no

(c) Is f(x) continuous at x = 4? (c)yes

(d) Is f(x) differentiable at x = 4? (d)no

21. Find the y-intercept of the tangent line to y = 2x+ sinx at x = π. 21.(0, π)

22. What is the slope of y = sin−1 x at the point

(√3

2,π

3

)?

22.2

23. A particle is moving along the x-axis according to the equation x(t) = t3 − 6t2 + 40.Assume that t ≥ 0.

(a) Find the value of v(3). (a)−9

(b) At time t = 1, is the particle speeding up, slowing down, or coasting? (b)speeding up

(c) When the velocity is zero, what are the particle’s locations? (c)x = 40 and x = 8

(d) What is the acceleration when the particle has a velocity of 15? (d)18

24. If a particle moves on the x-axis according to x(t) = t2 − 12t + 21, how far tothe left will it ever get? 24.

−15

25. Finddy

dx(in terms of x) for each function below.

(a) y =√x2 + 6 (a)

x√x2+6

(b) y = (5x+ 1)3(b)

15(5x+ 1)2

(c) y = sin4 x(c)

4 sin3 x cosx

EC 1.The perpendicular to y = 1

x at the point(12 , 2)

intersects the other branch ofthe curve as well. Where? 1.

(−8,− 18 )

EC 2.Two lines are drawn tangent to y = x2, one at (a, a2) and the other at (b, b2).Where do these two tangent lines cross, in terms of a and b?

2.(a+b2 , ab)


Name

1. Consider the relation x2 + xy + 2y2 = 28.

(a) What is the slope at the point (2, 3).

(b) What is an equation for the tangent line at its rightmost x-intercept?

(c) What angle does the tangent line at a y-intercept make with the x-axis?

(d) What is the slope of the line that passes through the highest and lowest points on the graph?

(e) What are the coordinates of the highest point on the graph?

(f) What function would you type into your calculator to graph the top “half”?

(g) What are the coordinates of the leftmost point on the graph?

2. Find the slope (derivative) of each relation at the indicated point.

(a) x5 = y3 at (8, 32)

(b)√x+√y = 3, at (4, 1)

(c) x2 + xy = sinx at (π,−π)

(d) x2 − 2xy + y2 = 4 at (3, 16)

3. Find the angle of any intersection of a branch of xy = 60 with the circle x2 + y2 = 169.

4. Find an equation of the normal (perpendicular) to y3 + 2x2 + 3xy = 4x+ 8 at its y-intercept.

5. Find dydx and d2y

dx2 for each relation below at the indicated point.

(a) 2x2 − y2 = 17 at (3, 1) (b) cos(x+ 2y) = 0, at (π6 ,π6 )

6. For each point and relation pair below, tell whether the curve is increasing or decreasing at that point,concave up or concave down at that point, and getting steeper or less steep at that point.

(a) x3 + y3 = 9 . . . at the point (2, 1)

(b) y4 + sin y = x2 − 1 . . . at the point (1, 0)

(c) dydx = x− y . . . at the point (9, 5)

Answer Key

1. You will make repeated use of the fact that dydx = − 2x+y

x+4y .

(a) − 4+32+12 = − 1

2

(b) The relevant point is (√

28, 0), and the slope there is −2. Hence y = −2(x−√

28).

(c) The slope is − 14 at both y-intercepts. Thus the angle is tan−1(− 1

4 ) ≈ 14.04◦.

(d) The slope of the relation equals zero when y = −2x. Since zero slopes correspond to highest/lowestpoints, the answer is −2.

(e) Substitute y = −2x (see previous problem) into the original relation to get x2− 2x2 + 2(4x2) = 28.Solving gets us x = ±2. Substituting these values back into y = −2x gives us the points (2,−4)and (−2, 4). Of these two points, (−2, 4) is clearly the highest.

(f) We can re-write the relation as a standard quadratic in terms of y: (2)y2 + (x)y + (x2 − 28) = 0.

According to the quadratic formula, we have y =−x±√x2−4(2)(x2−28)

4 . The top half is the one

involving the positive root. Simplified, this boils down to y = −x+√224−7x2

4 .

(g) The slope of the relation is undefined (vertical) when x = −4y. Substituting this into the originalrelation gives us 16y2 − 4y2 + 2y2 = 28, or y = ±

√2. Plugging these values of y into x = −4y give

us the points (−4√

2,√

2), and (4√

2,−√

2). Of these, (−4√

2,√

2), is clearly furthest to the left.

2. (a) dydx = 5x4

3y2

∣∣∣(8,32)

= 5·2123·210 = 20

3

(b) dydx = −

√y√x

∣∣∣(4,1)

= − 12

(c) dydx = cos x−2x−y

x

∣∣∣(π,−π)

= − 1+ππ

(d) dydx = x−y

x−y

∣∣∣(3,16)

= 1

3. The curves intersect in four places (a rough graph will reveal this). We can either happily noticethat (5, 12) is one of these points, or (unhappily) solve the system resulting from plugging y = 60

x inx2 + y2 = 169. Either way, the hyperbola’s slope there is − 12

5 while the circle’s slope is − 512 . Thus the

angle we are hunting for is tan−1(− 125 )− tan−1(− 5

12 ) ≈ 44.76◦.

4. Since the y-intercept is (0, 2) and dydx = 4−4x−3y

3y2+3x

∣∣(0,2)

= − 16 ⊥ 6, we have y = 6x+ 2.

5. (a) dydx = 2x

y

∣∣(3,1)

= 6 and d2ydx2 =

2y−2x dydx

y2

∣∣(3,1)

= 2−6·61 = −34.

(b) x+ 2y = ±(any odd number) · π2 . Thus dydx = − 1

2 and d2ydx2 = 0.

6. (a) Since dydx = −x2

y2

∣∣(2,1)

= −4 and d2ydx2 = − 2xy2−2yx2 dy

dx

y4

∣∣(2,1)

= −36, the graph is decreasing, concave

down, and getting steeper.

(b) Since dydx = 2x

4y3+cos y

∣∣(1,0)

= 2 and d2ydx2 = − 2(4y3+cos y)−2x(12y2−sin y) dy

dx

(4y3+cos y)2

∣∣(1,0)

= 2, the graph is

increasing, concave up, and getting steeper.

(c) Since dydx = x− y

∣∣(9,5)

= 4 and d2ydx2 = 1− dy

dx

∣∣(9,5)

= −3, the graph is increasing, concave down, and

getting less steep.


Name

1. Solve each of the equations for A (mentally would be preferred).

(a) 3 + 2A + 6 = 8 (b) 5− 3A + 6 + 7A = 8 (c) 3A− 6 + 6A + 9 = 8 + 4A

2. Solve each of the equations for dydx (mentally would be preferred).

(a) x + ydy

dx+ 6y = 8x

(b) 3x2 − 2ydy

dx+ 6y + 3x

dy

dx= 8

(c) x3 + ydx

dy+ 6y2 − sinx

dx

dy= 17x4

(d) 3xdy

dx− y3 + 7y

dy

dx+ 9x4 = 8xy + 5x

dy

dx

3. Implicitly differentiate each relation below.

(a) x2 + y7 = x + 5x3

(b) xy3 + sin y = 56.3

(c) y3 + sin y =√y + 3x2y

(d) x2 + 8xy = tanx

(e) y5 − 4x4 + 7y3 − 8x2 + 3y = 3√x + 1

y2

(f) sin(x + y) = cos(x− y)

4. Find the slope of the hyperbola x2 − 4y2 = 9 at the point (5, 2).

5. Find an equation for the tangent line to the ellipse x2 + 4y2 = 8 at the point (2, 1).

6. In class today, we found a curve whose derivative was xy , one whose derivative was −x

y , and one whose

derivative was − yx . Find an equation for a curve whose derivative is y

x .

Answer Key

1. (a) A = −1

2(b) A = −3

4(c) A =

5

5= 1

2. (a)dy

dx=

7x− 6y

y

(b)dy

dx=

8− 6y − 3x2

3x− 2y

(c)dy

dx=

y − sinx

17x4 − x3 − 6y2

(d)dy

dx=

y3 − 9x4 + 8xy

7y − 2x

3. (a)dy

dx=

15x2 − 2x + 1

7y6

(b)dy

dx= − y3

3xy2 + cos y

(c)dy

dx=

6xy

3xy2 + cos y − 3x2 − 12√y

(d)dy

dx=

sec2 x− 2x− 8y

8x

(e)dy

dx=

16x3 + 16x + 13x− 2

3

5y4 + 21y2 + 3 + 2y−3

(f)dy

dx=

cos(x + y) + sin(x− y)

sin(x− y)− cos(x + y)

4.dy

dx=

x

4y

∣∣∣(5,2)

=5

8

5.dy

dx= − x

4y

∣∣∣(2,1)

= −1

2, so an equation that works is y − 1 = − 1

2 (x− 2).

6. Any line that passes through the origin will do.


Name

1. A heap of rubbish in the shape of a cube is being compacted. The volume is decreasing at a rate of 2cubic meters every minute. What is the rate of change of an edge of the cube when the volume is exactly125 cubic meters?

2. A baseball player attempts to steal second base, 90 feet away from first base. At the time he is exactlymidway between the bases, his speed is 25 feet/second. At that time, how fast is his distance from homeplate changing?

3. A circle is inscribed in a square. The radius of the circle is expanding at a constant rate of 6 inches/second.The square is also expanding, so as to maintain tangency. What is the rate of increase in the area betweenthe circle and square at the time the radius is 5 inches?

4. A tightrope is stretched 30 feet above the ground between the Ford Building and the Honda Building,which are 50 feet apart. A tightrope walker, walking at a constant rate of 2 feet per second across thetightrope, starting at the Ford Building, is illuminated by a spotlight 70 feet above her on the corner ofthat building. How fast is the shadow of the tightrope walker’s feet moving along the ground when sheis exactly midway between the buildings?

5. Romeo is climbing up to Juliet’s balcony on a 25 foot ladder which leans against a vertical wall of hercastle. Juliet wants to play “hard to get” and pushes the top of the ladder toward the ground. If thebottom of the ladder slides away from the castle at a rate of 4 feet/second, find each below.

(a) How fast are Romeo and his ladder sliding down the wall of the castle when the top of the ladderis 15 feet above the ground?

(b) How fast is the angle between the ground and the ladder changing at that same instant?

6. My son Simon spent all day last Saturday loading sand into his fireman bag, which subsequently gothung on a hook. Unfortunately, a hole occurred in the bottom of the bag, and sand leaked out forminga perfectly conical pile whose altitude and diameter remained exactly equal. If the radius of the pilechanged at a rate of 2 inches/minute, what was the rate of change of the volume of the pile when theradius was 3

4 of an inch?

Answer Key

1. Picture:

e

ee

V = cube volume

Known:

dV

dt= −2

Desired:

de

dt. . . when V = 125 (or e = 5)

Equation(s):

V = e3

Calculations:

V = e3

dV

dt= 3e2

de

dt

−2 = 3(5)2de

dtde

dt= − 2

75

Solution:

The edges are shrinking at a rate of 275

meters per minute.

2. Picture:

h

90 feet

x

Known:

dx

dt= 25. . . when x = 45

Desired:

dh

dt. . . when x = 45 (or h = 45

√5)

Equation(s):

x2 + 902 = h2

Calculations:

x2 + 902 = h2

2xdx

dt+ 0 = 2h

dh

dt

2 · 45 · 25 = 2 · 45√

5 · dhdt

dh

dt=

25√5

Solution:

His distance from home plate is increasingat a rate of 5

√5 feet per second.

3. Picture:

r2r

S = shaded region

Known:

dr

dt= 6

Desired:

dS

dt. . . when r = 5

Equation(s):

Area of 2= 4r2, Area of #= πr2

Calculations:

S = 4r2 − πr2DS

dt= 8r

dr

dt− 2πr

dr

dtDS

dt= 8 · 5 · 6− 2π · 5 · 6

DS

dt≈ 51.50

Solution:

The shaded region is increasing byapproximately 51.5 square inches persecond.

4. Picture:

70

30x

y

light

Known:

dx

dt= 2

Desired:

dy

dt. . . when x = 25

Equation(s):

x

70=

y

100

Calculations:

x

70=

y

1001

70

dx

dt=

1

100

dy

dt1

70· 2 =

1

100

dy

dtdy

dt=

20

7

Solution:

The shadow is moving toward the HondaBuilding at a rate of 20

7 feet per second.

5. (a) Picture:

wall

ground

25 feety

xθ

Known:

dx

dt= 4

Desired:

dy

dt. . . when y = 15, x = 20, sin θ =

15

25

Equation(s):

x2 + y2 = 252

Calculations:

x2 + y2 = 252

2xdx

dt+ 2y

dy

dt= 0

160 + 30dy

dt= 0

dy

dt= −16

3

Solution:

Romeo (at the top of the ladder) is fallingat 16

3 feet per second.

(b) Picture:

wall

ground

25 feety

xθ

Known:

dx

dt= 4

Desired:

dθ

dt. . . when y = 15, x = 20, sin θ =

15

25

Equation(s):

cos θ =x

25

Calculations:

cos θ =x

25

− sin θdθ

dt=

1

25

dx

dt

−15

25

dθ

dt=

1

25· 4

dθ

dt= − 4

15

Solution:

The angle between the ladder and theground is dropping at 4

15 radians persecond.

6. Picture:

r

2r

V = cone volume

Known:

dr

dt= 2

Desired:

dV

dt. . . when r = 3

4

Equation(s):

V =1

3πr2h , h = 2r

Calculations:

V =2

3πr3

dV

dt= 2πr2

dr

dtdV

dt= 2π

9

16· 2

dV

dt=

9π

4

Solution:

The volume of the pile was growing by 9π4

cubic inches per minute.


Name

1. Bertha and Bertram, a newlywed couple, go out for a romantic dinner at McDairy King. They order abottle of sparkling water, and the waiter brings them conical glasses (which are magically able to balanceon their tips) of altitude 4 inches and diameter 3 inches. If the sparkling water is poured into a glass ata rate of 15 cubic inches per minute (the waiters at McDairy King are very well trained), what is therate at which the water is rising when it is halfway up to the top of the glass?

2. Oh no! Winnie the Pooh has lost his honey in the Hundred Acre Woods. Eeyore volunteers to helpPooh (though I’m sure it’s lost forever. . . ) in his search for the missing honey. They start at the bridge.Pooh stumbles south at a rate of 2 ft/sec while Eeyore waddles west at a rate of 3 ft/sec. At what rateis the distance between Pooh and Eeyore changing after they have been searching for 5 minutes?

3. A man (who is 6 feet tall) is walking toward a lamppost at a rate of 5 feet per second. The light atthe top of the lamppost (20 feet above the ground) is casting a shadow of the man. At what rate is thelength of his shadow changing when the man is 10 feet from the base of the lamppost?

4. Find the rate of change of the volume of a cylinder when its radius is 6 feet, if its height is always 50%longer than its radius and its radius is increasing at the rate of 2 feet per minute.

5. A kite 100 feet above the ground is being blown away from the person holding its string in a directionparallel to the ground at a steady rate of 10 feet/second. At what rate must the string be let out whenthe length of the string already let out is 200 feet?

6. A camera is located 50 feet from a straight road along which a car is traveling at a steady 100 feet persecond. The camera turns so that it is always pointed directly at the car at all times. In radians persecond, how fast is the camera turning as the car passes closest to the camera?

Answer Key

1. Picture:

h

r

V = water volume

4

1.5

Known:

dV

dt= 15

Desired:

dh

dt. . . when h = 2

Equation(s):

V =1

3πr2h ,

h

4=

r

1.5

Calculations:

V =3π

64· h3

dV

dt=

9π

64· h2 · dh

dt

15 =9π

64· 4 · dh

dt80

3π=dh

dt

Solution:

The sparkling water is rising at a rate of803π inches per minute.

2. Picture:

bridgea

bc

Known:

da

dt= 3 ,

db

dt= 2

Desired:

dc

dt. . . when a = 900, b = 600,c = 300

√13

Equation(s):

a2 + b2 = c2

Calculations:

a2 + b2 = c2

2a · dadt

+ 2b · dbdt

= 2c · dcdt

2 · 900 · 3 + 2 · 600 · 2 = 2 · 300√13 · dc

dt√13 =

dc

dt

Solution:

The distance from Pooh to Eeyore isincreasing at a rate of

√13 ft/sec.

3. Picture:

x y

20

6

Known:

dx

dt= −5

Desired:

dy

dt. . . when x = 10

Equation(s):

x+ y

20=y

6

Calculations:

x+ y

20=y

61

20·(dx

dt+dy

dt

)=

1

6· dydt

−1

4=

(1

6− 1

20

)· dydt

−15

7=dy

dt

Solution:

The shadow’s length is decreasing at arate of 15

7 feet per second.

4. Picture:

V = cylinder volume

h

r

Known:

dr

dt= 2

Desired:

dV

dt. . . when r = 6

Equation(s):

V = πr2h , h = 1.5r

Calculations:

V =3π

2· r3

dV

dt=

9π

2· r2 · dr

dtdV

dt=

9π

2· 36 · 2

dV

dt= 324π

Solution:

The volume of the cylinder is growing ata rate of 324π cubic feet per minute.

5. Picture:

ground

y

kite

100

x

Known:

dx

dt= 10

Desired:

dy

dt. . . when y = 200, x = 100

√3

Equation(s):

1002 + x2 = y2

Calculations:

10000 + x2 = y2

2x · dxdt

= 2y · dydt

2 · 100√3 · 10 = 2 · 200 · dy

dt

5√3 =

dy

dt

Solution:

The kite’s string must be let out at a rateof 5√3 feet per second.

6. Picture:

road

camera

θ

x

50

car

Known:

dx

dt= 100

Desired:

dθ

dt. . . when x = 0, θ = 0

Equation(s):

tan θ =x

50

Calculations:

tan θ =x

50

sec2 θ · dθdt

=1

50· dxdt

1 · dθdt

=1

50· 100

dθ

dt= 2

Solution:

The camera is turning at a rate of 2radians (≈ 114.59◦) per second.


Name

1. What positive number exceeds its cube by the greatest amount?

2. A farmer has 120 feet of fencing, with which to build a rectangular yard for his pigs. He cleverly buildsit against one wall of a HUGE barn, so that he only has to fence the remaining three sides of the yard.What is the maximum area he can make the yard?

3. Racquet-World charges its members $250 for a year’s membership. They currently have 500 members.They want to increase their fees, but estimate they will lose 10 members for each $25 increase. Whatyearly fee will give them the most income?

4. A sheet of cardboard, 18 inches by 30 inches, is to have a square cut out of each corner, and then theremainder folded up into an open-top box. What size squares should be cut out to maximize the volumeof the box?

5. Bertha’s Ice Cream Parlor sells 800 ice cream cones per week when the price is only 80 cents per cone.For each 5 cent increase in price, 16 fewer cones are sold per week. What price gives the maximumweekly revenue?

6. A rectangular playground is to be fenced off and divided in two by another fence parallel to one side ofthe playground. A total of 600 feet of fencing is used. Find the dimensions of the playground that willenclose the greatest total area.

7. A rectangular box has a top that is twice as long as it is wide. What would be the smallest possiblesurface area if the volume of the box was 72 cubic meters?

8. Another rectangular box has a top that is three times as long as it is wide. What would be the largestpossible volume if the surface area was 72 square meters?

1. Picture:

x

a

Relevant Equation(s):

a(x) = x− x3

Function whose extrema is desired:

a(x) = x− x3, for x > 0

Find Derivative(s):

a(x) = x− x3

a′(x) = 1− 3x2

Identify Critical Point(s):

1− 3x2 = 0

x = ±√

3

3

Justification:

The function a(x) has a single critical

point in its domain at x =√33 . Since

a′(x) changes from positive to negative atthis point, the critical point correspondsto the maximum value of a(x).

Solution:

The positive number which exceeds its cube by the greatest amount is√33 .

2. Picture:

barn wall

x x

y

A = area of yard

Relevant Relations:

120 = 2x + y , A = xy , 0 < x < 60


A(x) = x(120− 2x), for 0 < x < 60.

Find Derivative(s):

A(x) = 120x− 2x2

A′(x) = 120− 4x


120− 4x = 0

x = 30

Justification:

The function A(x) has a single criticalpoint in its domain at x = 30. SinceA′(x) changes from positive to negativeat this point, the critical pointcorresponds to the maximum value ofA(x), namely 1800.

Solution:

The maximum area the farmer can make his yard is 1800 square feet (30 feet by 60 feet).

3. Picture:

xmembers

fee

y

(500, $250)

Relevant Relations:

P = xy , y − 250 = −25

10(x− 500) , x > 0


P (x) = x

(−5

2x + 1500

), x > 0

Find Derivative(s):

P (x) = 1500x− 2.5x2

P ′(x) = 1500− 5x


1500− 5x = 0

x = 300

Justification:

The function P (x) has a single criticalpoint in its domain at x = 300. SinceP ′(x) changes from positive to negativeat this critical point, it corresponds to themaximum value of P (x). Note that whenx = 300, y = $750, and P = $225000.

Solution:

Racquet-World will maximize their income if they set their yearly fee at $750.

4. Picture:

30

18x x

x x

x

x

x

x

Relevant Relations:

V = (30− 2x)(18− 2x)(x) , 0 < x < 9


V (x) = 540x− 96x2 + 4x3, 0 < x < 9

Find Derivative(s):

V (x) = 540x− 96x2 + 4x3

V ′(x) = 540− 192x + 12x2


540− 192x + 12x2 = 0

x2 − 16x + 45 = 0

x = 8±√

19

Justification:

The function V (x) has a single criticalpoint in its domain at x = 8−

√19. Since

V ′(x) changes from positive to negativeat this critical point, it corresponds tothe maximum value of V (x).

Solution:

To maximize the box volume, (8−√

19) inch by (8−√

19) inch squares should be cut out.

5. Picture:

xcones sold per week

price

y

(800, 80¢)

Relevant Relations:

P = xy , y − 80 = − 5

16(x− 800) , x > 0


P (x) = x

(− 5

16x + 330

), x > 0

Find Derivative(s):

P (x) = − 5

16x2 + 330x

P ′(x) = −5

8x + 330


−5

8x + 330 = 0

x = 528

Justification:

The function P (x) has a single criticalpoint in its domain at x = 528. SinceP ′(x) changes from positive to negativeat this critical point, it corresponds tothe maximum value of P (x). Note thatwhen x = 528, y = 165, and P = 87120.

Solution:

In order to maximize the weekly revenue, the price per cone should be set at $1.65.

6. Picture:

x

y

x x

y

A = area of playground

Relevant Relations:

A = xy , 3x + 2y = 600, 0 < x < 200


A(x) = x

(−3

2x + 300

), 0 < x < 200

Find Derivative(s):

A(x) = −3

2x2 + 300x

A′(x) = −3x + 300


−3x + 300 = 0

x = 100

Justification:

The function A(x) has a single criticalpoint in its domain at x = 100. SinceA′(x) changes from positive to negativeat this critical point, it corresponds tothe maximum value of A(x). Note thatwhen x = 100, y = 150, and A = 15000.

Solution:

To maximize the playground’s area, it should be made 100 feet by 150 feet.

7. Picture:

y

x2x

S = surface area

Relevant Relations:

S = 6xy + 4x2 , 2x2y = 72, 0 < x < 6


S(x) = 6x · 36

x2+ 4x2, 0 < x < 6

Find Derivative(s):

S(x) = 216x−1 + 4x2

S′(x) = −216x−2 + 8x


−216x−2 + 8x = 0

8x3 = 216

x = 3

Justification:

The function S(x) has a single criticalpoint in its domain at x = 3. Since S′(x)changes from negative to positive at thiscritical point, it corresponds to theminimum value of S(x). Note that whenx = 3, y = 4, and S = 108.

Solution:

The smallest possible surface area for the box is 108 square meters.

8. Picture:

y

x3x

V = volume

Relevant Relations:

72 = 8xy + 6x2 , V = 3x2y, 0 < x < 3


V (x) = 3x2 · 72− 6x2

8x, 0 < x < 3

Find Derivative(s):

V (x) = −9

4x3 + 27x

V ′(x) = −27

4x2 + 27


−27

4x2 + 27 = 0

x2 = 4

x = ±2

Justification:

The function V (x) has a single criticalpoint in its domain at x = 2. Since V ′(x)changes from positive to negative at thiscritical point, it corresponds to themaximum value of V (x). Note that whenx = 2, y = 3, and V = 36.

Solution:

The largest possible volume for the box is 36 cubic meters.


Name

1. Paint cans are made in the shape of a right circular cylinder. What dimensions of the can will requirethe least amount of material? (1 gallon = 231 cubic inches)

2. A lighthouse lies three miles offshore, directly across from a point P on a straight shoreline. Five milesdown the coast from P there is a store. The lighthouse keeper can row his boat at a rate of 4 mph andhe can walk 6 mph. To what point on the shore should he row, so as to reach the store as quickly aspossible?

3. A steel storage tank for propane gas is to be constructed in the shape of a right circular cylinder witha hemisphere at each end. The construction cost per square foot for the hemispheres is twice the costof each square foot of the lateral area of the cylinder. If the desired capacity of the storage tank is 10πcubic feet, what dimensions will minimize the cost of construction?

4. What coke can dimensions would use the least amount of aluminum while still holding 375 ml?

5. A closed box with a square base is to contain 252 cubic feet. The bottom costs $5 per square foot,the top costs $2 per square foot, and the sides cost $3 per square foot. Find the dimensions that willminimize the cost.

6. The ABC Cable TV Company intends to run a cable from its central station A, on one side of a one milewide river, to a substation C, which is eight miles downriver on the other side of the (straight) river. Itcosts $10 per foot to run the cable underwater, and only $6 per foot to run the cable overland. Find themost economical path to run the TV cable from point A to point C.

7. What is the greatest possible percent of a sphere’s volume that can be taken up by a cylinder that fitsinside it?

8. Which point on the graph of y =√x is closest to the point (5, 0)?

1. Picture:

r

h

Relevant Equation(s):

231 = πr2h , S = 2πr2 + 2πrh , 0 < r


S(r) = 2πr2 +462

r, for r > 0

Find Derivative(s):

S(r) = 2πr2 + 462r−1

S′(r) = 4πr − 462r−2


4πr − 462r−2 = 0

2πr3 = 231

r =3

√231

2π

Justification:

The function S(r) has a single criticalpoint in its domain at r ≈ 3.325. SinceS′(r) changes from negative to positive atthis point, the critical point correspondsto the minimum value of S(r).

Solution:

A radius of ≈ 3.325 inches and height of ≈ 6.650 inches minimizes surface area.

2. Picture:

lighthouse

3

xP store

5− x

√9 + x2

Relevant Relations:

distancerate = time , 0 < x


T (x) =

√9 + x2

4+

5− x6

, for 0 < x.

Find Derivative(s):

T (x) =

√9 + x2

4+

5− x6

T ′(x) =x

4√

9 + x2− 1

6


x

4√

9 + x2− 1

6= 0

x = ±√

36

5

Justification:

The function T (x) has a single criticalpoint in its domain at x ≈ 2.683. SinceT ′(x) changes from negative to positive atthis point, the critical point correspondsto the minimum value of T (x).

Solution:

He should row to the point ≈ 2.683 miles toward the store from P .

3. Picture:

rr h

Relevant Relations:

10π = πr2h+ 43πr

3

c1 = $1 · 2πrh , c2 = $2 · 4πr2


C(r) = 8πr2 + 2πr

(10π − 4

3πr3

πr2

)

Find Derivative(s):

C(r) =16

3πr2 + 20πr−1

C ′(r) =32

3πr − 20πr−2


32

3πr − 20πr−2 = 0

r =3

√15

8

Justification:

The function C(r) has a single criticalpoint in its domain at x ≈ 1.233. SinceC ′(r) changes from negative to positiveat this critical point, it corresponds tothe minimum value of C(r).

Solution:

Setting the cylinder radius at ≈ 1.233 feet and height at ≈ 4.932 feet minimizes cost.

4. Picture:

r

h

Relevant Relations:

375 = πr2h , S = 2πr2 + 2πrh , 0 < r


S(r) = 2πr2 +750

r, for r > 0

Find Derivative(s):

S(r) = 2πr2 + 750r−1

S′(r) = 4πr − 750r−2


4πr − 750r−2 = 0

2πr3 = 375

r =3

√375

2π

Justification:

The function S(r) has a single criticalpoint in its domain at r ≈ 3.908. SinceS′(r) changes from negative to positive atthis point, the critical point correspondsto the minimum value of S(r).

Solution:

A radius of ≈ 3.908 cm and height of ≈ 7.816 cm minimizes surface area.

5. Picture:

y

xx

Relevant Relations:

252 = x2y , cb = 5x2, ct = 2x2, cs = 12xy


c(x) = 7x2 + 12x

(252

x2

)

Find Derivative(s):

c(x) = 7x2 + 3024x−1

c′(x) = 14x− 3024x−2


14x− 3024x−2 = 0

x = 6

Justification:

The function c(x) has a single criticalpoint in its domain at x = 6. Since c′(x)changes from negative to positive at thiscritical point, it corresponds to theminimum value of c(x).

Solution:

The cost is minimized when the box is 6 feet by 6 feet by 7 feet.

6. Picture:

A

1

xB C

8− x

√1 + x2

Relevant Relations:

c1 = 10√

1 + x2 , c2 = 6(8− x), 0 < x < 8


c(x) = 10√

1 + x2 + 6(8− x), 0 < x < 8

Find Derivative(s):

c(x) = 10√

1 + x2 + 48− 6x

c′(x) =10x√1 + x2

− 6


10x√1 + x2

− 6 = 0

x = ±3

4

Justification:

The function c(x) has a single criticalpoint in its domain at x = 0.75. Sincec′(x) changes from negative to positive atthis critical point, it corresponds to theminimum value of c(x).

Solution:

The cable should go underwater to the point (see picture) when x = 0.75 miles.

7. Picture:

r

h1

Relevant Relations:

h = 2√

1− r2 , 0 < r < 1 ,Vcylinder = πr2h , Vsphere = 4

3π


P (r) =2πr2

√1− r2

43π

, 0 < r < 1

Find Derivative(s):

P (r) =3

2

√r4 − r6

P ′(r) =3(4r3 − 6r5)

4√r4 − r6


3(4r3 − 6r5) = 0

r = ±√

2

3

Justification:

The function P (r) has a single criticalpoint in its domain at r ≈ 0.816. SinceP ′(r) changes from positive to negativeat this critical point, it corresponds tothe maximum value of P (r). Note that

when r =√

23 , P (r) =

√13 .

Solution:

The greatest percentage of the sphere’s volume taken up by the cylinder is ≈ 57.7%.

8. Picture:

x

y

5

l(x, y)

Relevant Relations:

y =√x , y2 + (5− x)2 = l2, 0 < x


l(x) =√x2 − 9x+ 25 , 0 < x

Find Derivative(s):

l(x) =√x2 − 9x+ 25

l′(x) =2x− 9

2√x2 − 9x+ 25


2x− 9 = 0

x = 4.5

Justification:

The function l(x) has a single criticalpoint in its domain at l = 4.5. Since l′(x)changes from negative to positive at thiscritical point, it corresponds to theminimum value of l(x).

Solution:

The point on y =√x closest to (5, 0) is (4.5,

√4.5).

AP/CCCalculus Name_________________________Max/MinandRelatedRatesPractice

1. Atroughis27feetlongand8feetacrossthetop.Itssidesareisoscelestriangleswithaltitudesof5feet.Thewaterisbeingpumpedintothetankatarateof8cubicfeetpersecond.

a) Findtherateatwhichthewaterlevelischanging.

b) Findtherateatwhichthesurfaceareaofthewater(thepartexposedtotheair)ischanging.

2. Twoparallelsidesofarectanglearebeinglengthenedattherateof2inchespersecond,whilethe

othertwosidesareshortenedinsuchawaythatthefigureremainsarectanglewithaconstantareaof50squareinches.a) Whatistherateofchangeoftheperimeterwhenthelengthofanincreasingsideis5inches?

b) Istheperimeterincreasingordecreasingatthattime?

3. Theradiusofarightcircularcylinderisincreasingatarateof2feetperminuteandtheheightis

decreasingatarateof3feetperminute.

a) Atwhatrateisthevolumechangingwhentheradiusis8inchesandtheheightis12inches?

b) Isthevolumeincreasingordecreasingatthattime?4. Findtheequationofthelinethrough(2,4)thatcutsofftheleastareafromthefirstquadrant.


Name

1. Find the admissible values of c for which the Mean Value Theorem is satisfied for each function below.

(a) f(x) = x2 for 1 ≤ x ≤ 2

(b) f(x) =√

1− x2 for 0 ≤ x ≤ 1

(c) g(x) = x3 − 12x for −1 ≤ x ≤ 1

(d) h(x) = 1x for −1 ≤ x ≤ 1

2. Sketch the graph of f(x) =

{x2 for x < 3

−2x2 + 18x− 27 for x ≥ 3, and determine whether or not the function

satisfies the conditions of the mean value theorem over [−1, 4]. If so, find the admissible value(s) of c.

3. A continuous function has the following values: f(3) = −6 and f(5) = 2. Which statements below mustbe true?

(a) The point (4, 0) must be on the graph of f .

(b) There must be an x-intercept on the right half of the x-axis.

(c) There must be an odd number of x-intercepts.

(d) f(4) is negative.

(e) There must be at least one value of x between 3 and 5 where f ′(x) > 0.

4. A particle is moving along the x-axis, and its position (in feet) at time t (in seconds) is given by x(t)where x is a differentiable function. Furthermore, x(3) = 0, x(4) = 40, and x(6) = 0.

(a) What does Rolle’s Theorem say about the motion of the particle?

(b) What does the Mean Value Theorem say about the motion of the particle?

(c) What does the Intermediate Value Theorem say about the motion of the particle?

5. Use the graph of f(x), shown below, to estimate the value(s) of c that satisfy the Mean Value Theoremfor the interval [0, 5].

x

y

Answer Key

1. (a) Since f(1) = 1 and f(2) = 4, the average slope of f over the interval [1, 2] is equal to 4−12−1 = 3.

Since f is differentiable over this entire interval, by the MVT there must be some value c in theinterval such that f ′(c) = 3. Finally, since f ′(c) = 2c, we solve 2c = 3 to find c = 1.5 .

(b) Note that f(0) = 1 and f(1) = 0, hence the average slope over the interval [0, 1] is equal to−1. Furthermore, f is differentiable over that entire interval, which means the MVT is applicable:there must exist some c in the interval for which f ′(c) = −1. Now f ′(x) = −x(1 − x2)−

12 , thus

−1 = −c(1 − c2)−12 , which leads to c = ±

√22 . Of these two values of c, only one is in the domain

in question. Thus c =√22 .

(c) The average slope of g over the interval [−1, 1] is equal to −11. Again, g is differentiable over thisentire interval, thus the MVT applies and we are guaranteed at least one solution to the equation

−11 = 3c2 − 12 in the interval [−1, 1]. In fact, there are two such values of c. Thus, c = ±√33 .

(d) There are no admissible values of c for which the MVT is satisfied. This is not a problem, as h isnot a differentiable function over the entire interval, hence the MVT does not apply.

2. Since 32 = −2 ·32+18 ·3−27, the function f is continuous over its entire domain. Since 2 ·3 = −4 ·3+18,the function f is differentiable over its entire domain. Consequently, the MVT applies.

• Note that f(−1) = 1 and f(4) = 13. Thuswe are guaranteed at least one solution tof ′(c) = 12

5 .

• One possibility is that −1 ≤ c < 3, giving us2c = 12

5 or c = 65 .

• Another possibility is that 3 ≤ c ≤ 4, givingus −4c + 18 = 12

5 or c = 3910 .

• Both c values satisfy the conditions set. Thusc = 1.2 or 3.9 .

x

y

3. Only (b) (because of the IVT) and (e) (because of the MVT) must be true.

4. (a) There was a time between 3 and 6 seconds at which the particle was stopped.

(b) There was a time between 3 and 4 seconds at which the particle was traveling exactly 40 ft/sec.

(c) There was a time between 4 and 6 seconds at which the particle was exactly at the 17.2 ft mark.

5. The MVT guarantees us that the average slope, found by connecting (0, 0) to (5, 4), must be equal tothe instantaneous slope at some point(s) between x = 0 and x = 5. My best guesses are drawn in below.It appears that c ≈ 0.7, c ≈ 3, and c ≈ 4.8 each satisfy the MVT.

x

y


Name

1. Suppose you want to find a solution to the equation f(x) = 0 over the interval [0, 3] for the functionf(x) = x3 + 3x2 + 3x− 6.

(a) Find f(0), f(1), f(2), and f(3) either with direct substitution or synthetic substitution.

(b) Make a sketch of the graph based on the points you have found.

(c) Find the derivative of the function. When will the derivative be negative?

(d) Since the graph never turns around, how many roots will there be over the interval from 0 to 3?

(e) What positive integer is closest to that positive root?

(f) Find the root to the nearest thousandth with Newton’s Method.

2. The problem of calculating a square root,√a, can be handled using Newton’s Method on the funtion

f(x) = x2−a. Use this idea to find an approximation for√

10, accurate to at least four decimal places.

3. Find the smallest positive root for the function f(x) = sinx by computing two approximations withNewton’s Method, starting with the closest integer to that zero, and comparing the result with what youknow the exact value of that zero to be. Why does this method seem to work so well in this example?

4. A student tries Newton’s Method on a function and uses a first guess of x = 2. However, f(2) = 5 andf ′(2) = −3. What will the student’s next guess be?

5. A student tries Newton’s Method on a function and uses a first guess of x = 5. However, f(5) = 12, sothe student’s next guess was x = 2. What was the slope of the curve at the point (5, f(5))?

6. Show how finding the x-intercept of the line that passes through the point (x0, f(x0)) with a slope off ′x0 leads to Newton’s Formula.

Answer Key

1. Your arithmetic might go more smoothly if you notice f(x) = (x + 1)3 − 7. Either way, you will makeuse of the facts that f ′(x) = 3x2 + 6x+ 3 = 3(x+ 1)2 and Newton’s Method,

xn+1 = xn −f(xn)

f ′(xn).

(a) f(0) = −6, f(1) = 1, f(2) = 20, and f(3) = 57.

(b)

x

y

(c) f ′(x) = 3x2 + 6x+ 3 = 3(x+ 1)2, which is never negative.

(d) One root (because of the intermediate value theorem, which you will hear about tomorrow).

(e) By inspection, 1 is the closest integer.

(f) x0 = 1, x1 = 0.916, and x2 = 0.91293845 . . . , thus the root is approximately 0.913.

2. Since xn+1 = xn − x2n−102xn

, and 3 is the integer closest to√

10, we have x0 = 3, x1 = 3.16, and

x2 = 3.162280701 . . . , thus√

10 ≈ 3.16228.

3. The smallest positive root of f(x) = sinx is exactly π, hence approximately 3. Since xn+1 = xn−tan(xn),we have x0 = 3, x1 = 3.142546543 . . . , and x2 = 3.141592653300476 . . . , which is really close to the exactvalue of π because f(x) = sinx is locally linear there.

4. xn+1 = 2− 5−3 = 3.6

5. Since 2 = 5− 12f ′(5) , we have f ′(5) = 4.

6. The point-slope form of the tangent line at (x0, f(x0)) is y − f(x0) = f ′x0(x − x0). Setting y equal tozero (in order to find the line’s x-intercept) and solving for x gives Newton’s Formula.


Name

1. The slope of 3x2 + y2 = xy + 5 at (1, 2) is which of the following? 1.

A. 34 B. − 3

4 C. − 43 D. 4

3

2. Write an equation of the tangent line to the curve x2 − 4y2 = 5 at (3, 1). 2.

3. Given that 3x dydx + y2 − 2y dy

dx − 5x = 8, find the value of dydx at the point (5, 6). 3.

4. Implicitly differentiate tan y = y3 + 5x4. 4.

5. What are the coordinates of the highest point on the graph of x2+xy+3y2 = 44? 5.

6. A student uses Newton’s Method to approximate an x-intercept of a function,and his first guess was x = 3. Since f(3) was 10, he made a second guess ofx = 5. What is the slope of the line tangent to the function at x = 3? 6.

7. Find the value of c that Rolle’s Theorem claims exists between 2 and 6 for thefunction f(x) = x(x − 2)(x − 6). 7.

8. Find the value of c that the Mean Value Theorem claims exists between 4 and9 for the function f(x) = 6

√x − 4x. 8.

9. For the curve x3 + y3 = 19 at the point (3,−2), is the curve concave up ordown? 9.

10. Choose one of the two related rates problems below. Cross out the one you are NOT doing, Youwill be awarded points based on your picture and labeling, your writing of the relevant equations andinformation, doing the appropriate differentiation, and solving for the requested item.

Ms. Stanton is blowing up a balloon in the shapeof a sphere. She blows in air at a constant rate of8cc/sec. At what rate is the radius increasing atthe instant that the radius equals 2 cm?

A rocket leaves the ground 200 feet away from anobserver, and rises vertically at an increasing speed.The angle of inclination of the observer’s line ofsight at the instant when the rocket is exactly 400feet above the ground is changing at 0.5 radians perminute. How fast is the rocket rising at that time?

11. Choose one of the two optimization problems below. Cross out the one you are NOT doing. Youwill be awarded points based on your picture and labeling, your writing of the relevant equations andinformation, doing the appropriate differentiation, and solving for the requested item.

An OPEN TOPPED box with a square base is tocontain 432 cubic feet. The bottom costs $12 persquare foot, and each wall costs $3 per square foot.What dimensions of the box will minimize the costof building the box?

Farmer Simmons wants to fence in a rectangularplot of area 600 square feet. He also wants to useadditional fencing to build ONE internal dividerfence, parallel to the length of the fence. Whatis the minimum total length of fencing the projectrequires?


Answer Key

1. The slope of 3x2 + y2 = xy + 5 at (1, 2) is which of the following?1.

C. − 43

A. 34 B. − 3

4 C. − 43 D. 4

3

2. Write an equation of the tangent line to the curve x2 − 4y2 = 5 at (3, 1).2.

y − 1 = 34 (x − 3)


dx − 5x = 8, find the value of dydx at the point (5, 6).

3.−1

4. Implicitly differentiate tan y = y3 + 5x4.4.

dydx = 20x3

sec2 y−3y2

5. What are the coordinates of the highest point on the graph of x2+xy+3y2 = 44?5.

(−2, 4)

6. A student uses Newton’s Method to approximate an x-intercept of a function,and his first guess was x = 3. Since f(3) was 10, he made a second guess ofx = 5. What is the slope of the line tangent to the function at x = 3?

6.−5

7. Find the value of c that Rolle’s Theorem claims exists between 2 and 6 for thefunction f(x) = x(x − 2)(x − 6).

7.

8+√

283 ≈ 4.4305


√x − 4x.

8.

254

9. For the curve x3 + y3 = 19 at the point (3,−2), is the curve concave up ordown?

9.Concave Up




r

V = volume

Known:

dV

dt= 8

Desired:

dr

dt. . . when r = 2

V =4

3πr3

dV

dt= 4πr2 dr

dtdr

dt=

1

2πdr

dt≈ 0.159

Solution:

Radius grows at≈ 0.159 cm/sec.

y

θ200

Known:

dθ

dt=

1

2

Desired:

dy

dt. . . at y = 400

tan θ =y

200

sec2 θdθ

dt=

1

200

dy

dt

5 · 1

2=

1

200

dy

dtdy

dt= 500

Solution:

Rocket rises at 500ft/min.




y

xx

Known:

y =432

x2

C = 12x2 + 12xy

Need Crit Pt of:

C(x) = 12x2 + 5184x

C �(x) = 24x − 5184

x2

0 = 24x − 5184

x2

x = 6 & y = 12

The lone crit ptis a min since C �

changes from neg.to pos. there.

Solution:

6ft by 6ft by 12ft

y

x

y y

x

Known:

y =600

xL = 2x + 3y

Need Crit Pt of:

L(x) = 2x + 1800x

L�(x) = 2 − 1800

x2

0 = 2 − 1800

x2

x = 30 & y = 20

The lone crit ptis a min since L�


Solution:

120 feet of fencing


Answer Key

1. The slope of 3x2 + y2 = xy + 5 at (1, 2) is which of the following?1.

C. − 43

A. 34 B. − 3

4 C. − 43 D. 4

3

2. Write an equation of the tangent line to the curve x2 − 4y2 = 5 at (3, 1).2.y − 1 = 3

4 (x− 3)


dx − 5x = 8, find the value of dydx at the point (5, 6).

3.−1

4. Implicitly differentiate tan y = y3 + 5x4.4.

dydx = 20x3

sec2 y−3y2

5. What are the coordinates of the highest point on the graph of x2+xy+3y2 = 44?5.

(−2, 4)

6. A student uses Newton’s Method to approximate an x-intercept of a function,and his first guess was x = 3. Since f(3) was 10, he made a second guess ofx = 5. What is the slope of the line tangent to the function at x = 3?

6.−5

7. Find the value of c that Rolle’s Theorem claims exists between 2 and 6 for thefunction f(x) = x(x− 2)(x− 6).

7.

8+√28

3 ≈ 4.4305


√x− 4x.

8.

254

9. For the curve x3 + y3 = 19 at the point (3,−2), is the curve concave up ordown?

9.Concave Up




r

V = volume

Known:

dV

dt= 8

Desired:

dr

dt. . . when r = 2

V =4

3πr3

dV

dt= 4πr2

dr

dtdr

dt=

1

2πdr

dt≈ 0.159

Solution:

Radius grows at≈ 0.159 cm/sec.

y

θ200

Known:

dθ

dt=

1

2

Desired:

dy

dt. . . at y = 400

tan θ =y

200

sec2 θdθ

dt=

1

200

dy

dt

5 · 1

2=

1

200

dy

dtdy

dt= 500

Solution:

Rocket rises at 500ft/min.




y

xx

Known:

y =432

x2

C = 12x2 + 12xy

Need Crit Pt of:

C(x) = 12x2 + 5184x

C ′(x) = 24x− 5184

x2

0 = 24x− 5184

x2

x = 6 & y = 12

The lone crit ptis a min since C ′


Solution:

6ft by 6ft by 12ft

y

x

y y

x

Known:

y =600

xL = 2x+ 3y

Need Crit Pt of:

L(x) = 2x+ 1800x

L′(x) = 2− 1800

x2

0 = 2− 1800

x2

x = 30 & y = 20

The lone crit ptis a min since L′


Solution:

120 feet of fencing


Name

1. When a circle is inscribed in a square, what percent of the area of the square is also inside the circle?

2. Which is a better fit (percent-wise), a square peg in a round hole, or a round peg in a square hole?

3. A 4.5 by 2√

2 rectangle has the same center as a circle of radius 2. Is there a higher percentage ofrectangle inside the circle, or circle inside the rectangle?

4. A triangle has sides of 8 inches, 15 inches, and 17 inches.

(a) What is the area of the inscribed circle?

(b) What is the circumference of the circumscribed circle?

5. A circle with an area of π and a circle with an area of 16π are externally tangent. A third circle istangent to the first two and also to one of their common external tangents (as shown below). What isthe area of the third circle?

6. An equilateral triangle and a square have equal areas. What is the ratio of their perimeters?

7. Suppose OPQ is a quadrant of a circle, and semicircles are drawn on OP and OQ, forming regions Aand B (shown below). Find the ratio of the area of A to the area of B.

P

O Q

A

B

8. A regular octagon ABCDEFGH is drawn below. If the area of rectangle ABEF = 16, what is the areaof the octagon?

A

B

C D

E

F

GH

Answer Key

1. Declare the radius of the circle to be r. Then the sides of the square are each 2r. Consequently, the areaof the square is 4r2 while the area of the circle is πr2. Thus, the percent of the area of the square that

is also inside the circle is πr2

4r2 = π4 ≈ 78.54%.

r

2. We already know that a round peg fills approximately 78% of the area of a square hole (see previousproblem). For the other scenario, let the circle’s radius once again be labelled r. Then the side of theinscribed square equals r

√2. Consequently the area of the square is 2r2, and the percent of the circle

filled by the square is 2r2

πr2 = 2π ≈ 63.66%. As a result, a round peg in a square hole is a better fit.

r

3. Notice that the area common to both the circle and the rectangle (shaded below) is composed of tworight triangles (each with area 2) and two quarter circles (each with area π). Therefore, the fraction ofthe rectangle that is inside the circle is 4+2π

4π , while the fraction of the circle that is inside the rectangleis 4+2π

9√2

. Since the first fraction is larger, there is a higher percentage of the rectangle in the circle.

2

4. (a) Since 8− r + 15− r = 17, the radius of the inscribed circle is 3, making its area 9π.

(b) Since the hypotenuse is equal to the circumscribed circle’s diameter, its circumference is 17π.

5. Consider a right triangle drawn with its legs horizontal and vertical and with its hypotenuse connectingthe centers of the largest two circles (shown in orange below). The hypotenuse has length 4 + 1 = 5 andthe vertical leg has length 4− 1 = 3. Consequently, the horizontal leg has length

√52 − 32 = 4.

Next, consider a right triangle drawn with its legs horizontal and vertical and with its hypotenuseconnecting the centers of the leftmost and middle circles (shown in blue below). Let us say that themiddle circle has radius r. Then the hypotenuse of this triangle has length 1 + r and the vertical leghas length 1− r. Therefore, the horizontal leg has length

√(1 + r)2 − (1− r)2 = 2

√r. Finally, consider

a right triangle with its legs horizontal and vertical with its hypotenuse connecting the centers of therightmost and middle circles (shown in red below). This hypotenuse has length 4 + r, while its verticalleg has length 4− r. Consequently, its horizontal leg has length

√(4 + r)2 − (4− r)2 = 4

√r.

Finally, since the horizontal legs of the two smaller right triangles have a sum of 4, we conclude that6√r = 4 or r = 4

9 . Thus, the area of the third circle is 1681π.

6. Let’s say that the side of our triangle is equal to x. Then its area is equal to x2 · x2√

3 = x2·√3

4 . Since the

square must have this same area, its sides must have length x· 4√3

2 . Thus the ratio of the perimeters of

our shapes is equal to 3·x2·x· 4√3

or 33/4

2 ≈ 1.14

7. Assume that segment OQ has length 2r. Then each of the half circles has an area of 12 · πr2, while the

quarter circle has an area of 14 · 4πr2. Now, notice that adding the two half circles to region B gives

the area of the quarter circle plus region A. Using the areas we have already found, this means that12 · πr2 + 1

2 · πr2 +B = 14 · 4πr2 +A which boils down to just B = A. The ratio of areas is just 1.

8. Rectangle ABEF is made up of 4 triangles with equal areas. (All four of them have a base equalto the length of segment BO and a corresponding height equal to the distance from point A to linesegment BO.) The octagon is made up of 8 of these same triangles (each congruent to triangle ABO).Consequently, the octagon’s area is exactly twice the area of the rectangle.

A

B E

FO


Name

1. Calculate the area of each shaded region below.

(a) x

y

7

y = x2

(b) x

y

2 5

y = x2

(c) x

y

1 3

y = x3

(d) x

y

3 9

y = 2x

(e) x

y

π2

y = sinx

(f) x

y

y = sin 7x

(g) x

y

y = 3 sin 5x

(h) x

y

π4

y = sec2 x

(i) x

y

π2

y = x cosx+ sinx

(j) x

y

3

y = 2x3 + x2 + 4x

(k) x

y

-5 -2

y = 2|x+ 2|

(l) x

y

2

y = |x2 − 1|

2. Evaluate each integral below. None of these need technology, just perspicacity. Think about the graph.

(a)

∫ 5

0

10x dx

(b)

∫ 5

−2

|x| dx

(c)

∫ 5

0

√25− x2 dx

(d)

∫ 5

−5

x3 dx

(e)

∫ π2

0

sin 2x dx

(f)

∫ 9

0

√x dx

(g)

∫ 5

0

8 dx

(h)

∫ 5

−4

x+ |x| dx

3. If

∫ 1

0

f(x) dx = 6 and

∫ 3

1

f(x) dx = 4, find the value of each integral described below.

(a)

∫ 3

0

f(x) dx (b)

∫ 1

0

5f(x) dx (c)

∫ 2

1

f(x− 1) dx (d)

∫ 3

0

f(x) + 2 dx

4. A rectangular “dartboard” is made on the coordinate plane. Its boundaries are the x-axis, the y-axis,the line y = 10, and the line x = 6. The parabola y = 6x−x2 is drawn on the plane. If the points in thedartboard are chosen at random, what is the probability that they will be underneath the parabola?

Answer Key

1. (a)1

3x3∣∣∣7

0=

343

3

(b)1

3x3∣∣∣5

2= 39

(c)1

4x4∣∣∣3

1= 20

(d) x2∣∣∣9

3= 72

(e) − cosx∣∣∣π

−π2

= 1

(f) −1

7cos 7x

∣∣∣π7

0=

2

7

(g) −3

5cos 5x

∣∣∣π5

0=

6

5

(h) tanx∣∣∣π4

0= 1

(i) x sinx∣∣∣π2

0=π

2

(j)1

2x4 +

1

3x3 + 2x2

∣∣∣3

0=

135

2

(k) 9 + 4 = 13

(l) x− 1

3x3∣∣∣1

0+

1

3x3 − x

∣∣∣2

1= 2

2. (a) x

y

5

50

125

(b) x

y

-2 5

2 + 252 = 29

2

(c) x

y

5

25π4

(d) x

y

-5

5

0

(e) x

y

π2

1

(f) x

y

9

18

(g) x

y

5

8

40

(h) x

y

5-4

10

25

3. (a) x

y

6 4

1 3

4 + 6 = 10

(b) x

y

30 20

1 3

30

(c) x

y

6 4

1 2 4

6

(d) x

y

6 4

62

3

16

4. The area of the rectangle is 60, while the area of the parabolic sector is

∫ 6

0

6x−x2 dx = 3x2− 1

3x3∣∣∣6

0= 36.

Thus, the probability is36

60=

3

5.

x

y

10

6


Name

1. Draw a picture and shade the relevant area for each integral written below.

(a)

∫ 1

0

√x dx

(b)

∫ 17

−17

3√x dx

(c)

∫ 13

3

5 dx

(d)

∫ π3

π4

3 sin 12x dx

(e)

∫ 3

3

2x dx

(f)

∫ 1

−2

|x| dx

(g)

∫ 3

0

√9− x2 dx

(h)

∫ 1

0

sin−1 x dx

(i)

∫ π3

−π4

3 sin 12x dx

(j)

∫ 2

1

1

xdx

(k)

∫ 12

0

x(12− x) dx

(l)

∫ 2

−1

x3 − x dx

2. Determine the value of each definite integral above. You may use technology for AT MOST two of them.

3. Find an answer for each indefinite integral below. You may use technology for AT MOST none of them.

(a)

∫sinx dx

(b)

∫csc2 x dx

(c)

∫1 dx

(d)

∫3

4x2dx

(e)

∫2x cosx + x2 sinx

cos2 xdx

(f)

∫3 sin2 x cosx dx

4. Given that f is an odd function, the derivative of f is g, f(2) = 8, and f(4) = 10, evaluate each below.

(a)

∫ 4

2

g(x) dx (b)

∫ 9

−9

f(x) dx (c)

∫ 2

0

g(x + 2) dx (d)

∫ 7

0

f(4)−f(−2) dx

5. A particle moves on the x-axis according to the velocity equation v(t) = 3t2− 12t+ 12, with the particlestarting (when t = 0) at position x = 5. Where is the particle at time t = 3?

6. The graph of f(x) is shown below. Let A(x) =

∫ x

0

f(t) dt.

(a) Find A(0), A(1), A(4), A(6), and A(−2).

(b) Make a rough sketch of y = A(x).x

y

1. (a) x

y

1

1

(b) x

y

-17

17

(c) x

y

5

3 13

(d) x

y

3

(e) x

y

3

(f) x

y

-2 1

(g) x

y

3

3

(h) x

y

1

π2

(i) x

y

3

(j) x

y

1 2

(k) x

y

(l) x

y

2. (a) 23

(b) 0

(c) 50

(d) − 12

(e) 0

(f) 52

(g) 9π4

(h) π2 − 1

(i) − 12

(j) ≈ 0.693

(k) 288

(l) 94

3. (a) − cosx + C

(b) − cotx + C

(c) x + C

(d) − 34x + C

(e) x2

cos2 x + C

(f) sin3 x + C

4. (a)

∫ 4

2

g(x) dx = f(4)− f(2) = 2

(b)

∫ 9

−9

f(x) dx = 0 (odd function)

(c)

∫ 2

0

g(x + 2) dx =

∫ 4

2

g(x) dx = 2

(d)

∫ 7

0

f(4)−f(−2) dx =

∫ 7

0

18 dx = 7 ·18 = 126

5. Since x(t) = t3 − 6t2 + 12t + 5, from finding the antiderivative of v(t), it follows that x(3) = 14.

6. A(0) = 0, A(1) = 3, A(4) = 7.5, A(6) = 3.5, and A(−2) = −6.

x

y


Name

1. The velocity functions (in m/s) for three different particles traveling along the x-axis are given below.For each, find

i. the NET distance traveled between t = 0 and t = 5, and

ii. the TOTAL distance traveled between t = 0 and t = 5.

(a) v(t) = 8t− 4 (b) v(t) = 3t2 − 12t+ 9 (c) v(t) = sin(π2 t)

2. The acceleration functions (in m/s2) and initial velocities are given below for two different particlesmoving along the x-axis. For each, find

i. the velocity function and

ii. the TOTAL distance traveled during the given time interval.

(a) a(t) = t+ 4, v(0) = 5, 0 ≤ t ≤ 10 (b) a(t) = 2t+ 3, v(0) = −4, 0 ≤ t ≤ 3

3. For the trip described by x(t) = cos t over the interval [0, 3π/2], find. . .

(a) its location at time t = π/2.

(b) its velocity at t = π/2.

(c) its direction at t = π/2.

(d) its acceleration at t = π/2.

(e) its net distance traveled over the entire interval.

(f) its total distance traveled over the entire interval.

4. For the trip described by v(t) = sec2 t over the interval [−π/4, π/4], with x(0) = 1, find. . .

(a) its location at time t = π/6.

(b) its velocity at t = π/6.

(c) its direction at t = π/6.

(d) its acceleration at t = π/6.

(e) its net distance traveled over the entire interval.

(f) its total distance traveled over the entire interval.

5. In a river estuary, customs officers boarded and searched a ship suspected of drug smuggling, but foundnothing. Later they were told that a floating package had been dropped overboard just before theyboarded the ship. In addition to the current in the river, there is a strong tidal flow. The two combineto produce a current with an estimated speed of v(t) = 2 − 4 sin(πt/6) nautical miles per hour, where tis the time in hours since the ship was intercepted.

(a) If the customs officers return 4 hours later to begin to search for the package, where would yourecommend them to begin looking?

(b) When should an observer at a post 4 nautical miles further downstream be alerted to look out forthe package?

Answer Key

1. (a) Since v(t) = 8t− 4, t = 1/2 is a critical point. Also, x(t) = 4t2 − 4t+C, though for just computingdistance traveled, the value of C is irrelevant and can be made to be anything we might findconvenient (I like C = 0 this time, giving us x(t) = 4t2 − 4t).

i. The NET distance traveled equals x(5)− x(0) = 80 meters.

ii. The important positions are the beginning, the ending, and any potential “turn arounds”.These positions are x(0) = 0, x(1/2) = −1, and x(5) = 80. The TOTAL distance traveled of82 meters can be seen relatively easily by plotting the important positions and sketching theparticle’s trip.

0-1 80

(b) First 3t2− 12t+ 9 = 0 at t = 1 and t = 3, giving us two critical points in the domain [0, 5]. Second,using the same logic as above, we have x(t) = t3 − 6t2 + 9t.

i. The NET distance traveled equals x(5)− x(0) = 20 meters.

ii. The critical positions are x(0) = 0, x(1) = 4, x(3) = 0, and x(5) = 20. The TOTAL distancetraveled is 28 meters.

40 20

(c) Notice that sin(π2 t)

= 0 whenever t is an even integer, giving us t = 2 and t = 4 for critical points.

Also, x(t) = − 2π cos

(π2 t).

i. The NET distance traveled equals x(5)− x(0) = 2/π.

ii. The critical positions are x(0) = −2/π, x(2) = 2/π, x(4) = −2/π, and x(5) = 0. The TOTALdistance traveled is 10/π.

− 2π

0 2π

2. (a) Integrating and choosing our “C” to fit the given data, we have v(t) = 12 t

2 + 4t + 5. Either byusing some clever factoring or perhaps the quadratic formula, we discover that this function’s rootsare both negative (t = −4 ±

√6). Hence there are no critical points in the domain [0, 10]. Anti-

differentiating one more time gives us x(t) = 16 t

3+2t2+5t (plus an irrelevant constant I am choosingto let equal 0).

i. From the thinking above, v(t) = 12 t

2 + 4t+ 5.

ii. The NET distance traveled equals x(10) − x(0) = 12503 = 416.6 meters. Since there were no

“turn arounds” in the domain given, this is also the TOTAL distance traveled.

(b) Integrating, just as above, we get v(t) = t2 +3t−4. Furthermore, this function has two roots: t = 1and t = −4. However, only one of these two roots is in the domain [0, 3], giving us a single criticalpoint at t = 1. Anti-differentiating again gives us x(t) = 1

3 t3 + 3

2 t2 − 4t.

i. As stated above, v(t) = t2 + 3t− 4.

ii. The critical positions are x(0) = 0, x(1) = − 136 , and x(3) = 63

6 . The TOTAL distance traveledis 89

6 = 14.83 meters.

− 136

0 636

3. Since x(t) = cos t, it follows that v(t) = − sin t and a(t) = − cos t. Furthermore, v(t) = 0 whenever t isan integer multiple of π, giving us a single critical point of t = π in the domain [0, 3π/2].

(a) x(π2 ) = 0.

(b) v(π2 ) = −1.

(c) v(π2 ) < 0, so the particle is heading left.

(d) a(π2 ) = 0.

(e) The NET distance traveled equals x( 3π2 ) − x(0) = −1. In other words, the NET distance traveled

is one unit left.

(f) The critical positions are x(0) = 1, x(π) = −1, and x( 3π2 ) = 0. The TOTAL distance traveled is 3.

-1 0 1

4. Using anti-differentiation and the initial condition x(0) = 1, we have x(t) = tanx + 1. Furthermore,since v(t) = sec2 t is strictly positive over the domain [−π/4, π/4], we do not have any critical points.Finally, a(t) = 2 sec2 t tan t.

(a) x(π6 ) =√33 + 1.

(b) v(π6 ) = 43 .

(c) v(π6 ) > 0, so the particle is heading right.

(d) a(π6 ) = 2 ·(

2√3

)2· 1√

3= 8

3√3.

(e) The NET distance traveled equals x(π4 )− x(−π4 ) = 2− 0 = 2.

(f) Since there are no critical points, the TOTAL distance equals the NET distance of 2.

5. Anti-differentiating the velocity rule gives us x(t) = 2t + 24π cos(πt6 ). I would rather not throw an

additional constant into that rule, and will instead be content that the intercepted boat is at positionx(0) = 24

π . Also, it is worth noticing that our position rule is capable of both positive and negativevalues. It is same to assume that the river is flowing, ultimately, to the sea. Since large values of t makelarge positive values of x(t), we can safely assume that moving right is moving downstream (seaward)and that moving left is moving upstream.

(a) Calculating x(4) we get 8 − 12π . However, x(4) isn’t as important as how far away (and in what

direction) x(4) is from x(0). We can handle that, though: x(4) − x(0) = 8 − 36π ≈ −3.459.

Consequently, they should look for the package about 3.459 (nautical) miles upstream (furtheraway from the ocean).

(b) Solving x(t) − x(0) = 4 is impossible (for me at least) without the aid of technology. Sincex(t) = 2t + 24

π cos(πt6 ) and x(0) = 24π , this means solving 2t + 24

π cos(πt6 ) − 24π = 4. I graphed

both the left and right-hand sides of this equation and used the “intersect” button on my TI to gett ≈ 7.9 hours after intercepting the ship.

Just for the record, and because I didn’t want my curiosity to go undocumented, I worked out thepath of the package according to the equations we were given. Here it is:

• x(0) = 24π ≈ 7.6

• x(1) = 2 + 12√3

π ≈ 8.6

• x(5) = 10− 12√3

π ≈ 3.4

• x(13) = 26 + 12√3

π ≈ 32.6

• x(17) = 34− 12√3

π ≈ 27.4

7.6 8.63.4 32.627.4


Name

1. Using the graph of intersecting functions below, write an integral which would evaluate each region.

x

yf(x)

g(x)

j k l m

(a)

(b)

(c)

(d)

(e)

(f) (g)

(h)

(i)

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

2. Find the area of the region bounded by each pair of curves below.

(a) y = x2 and y =√x

(b) y = x2 and y = 8− x2

(c) y = sinx and y = sin 2x over [0, π/3]

(d) y = x2 + 3 and y = x2 + 8 over [2, 7]

3. Find the approximate area of the quadrant one region bounded by y = x, y = cosx, and x = 0.

4. Find the area trapped between y = x2 + 2 and y = −x2 + 6x+ 2.

5. Find the area trapped between y = x2 + 3 and the line segment joining (−2, 7) and (3, 12).

Answer Key

1. (a)

∫ 0

j

f(x)− g(x) dx

(b)

∫ 0

j

g(x) dx

(c)

∫ k

0

f(x)− g(x) dx

(d)

∫ k

0

g(x) dx

(e)

∫ l

k

g(x)− f(x) dx

(f)

∫ l

k

f(x) dx

(g)

∫ m

l

g(x) dx

(h)

∫ m

l

f(x)− g(x) dx

(i) (m− k) · f(m)−∫ l

k

g(x) dx−∫ m

l

f(x) dx

2. (a)

x

y

1

∫ 1

0

√x− x2 dx =

2

3x

3/2 − 1

3x3∣∣∣1

0=

1

3

(b)

x

y ∫ 2

−2

8− x2 − x2 dx = 2

∫ 2

0

8− 2x2 dx = 2

(8x− 2

3x3

) ∣∣∣2

0=

64

3

(c) x

y∫ π

3

0

sin 2x− sinx dx = −1

2cos 2x+ cosx

∣∣∣π3

0=

(1

4+

1

2

)−(−1

2+ 1

)=

1

4

(d)

x

y

2 7

∫ 7

2

(x2 + 8)− (x2 + 3) dx =

∫ 7

2

5 dx = 5x∣∣∣7

2= 25

3. x

y∫ .739

0

cosx− x dx = sinx− 1

2x2∣∣∣.739

0≈ 0.400

4.

3x

y ∫ 3

0

(−x2 + 6x+ 2

)−(x2 + 2

)dx =

∫ 3

0

−2x2 + 6x dx = −2

3x3 + 3x2

∣∣∣3

0= 9

5. (−2, 7)

(3, 12)∫ 3

−2

(x+ 9)−(x2 + 3

)dx =

∫ 3

−2

−x2 + x+ 6 dx = −1

3x3 +

1

2x2 + 6x

∣∣∣3

−2=

125

6


Name

1. Find the average value (height) of each function below over the indicated interval.

(a) f(x) = 2x + 5, over [3, 7]

(b) f(x) = x2, over [3, 7]

(c) f(x) = sin x, over [0,⇡]

(d) f(x) = sin x, over [0, 3⇡]

(e) f(x) =p

9 � x2, over [�3, 3]

(f) f(x) = (x + 2) (x) (x � 2), over [0, 2]

(g) f(x) =1px

, over [1, 16]

(h) f(x) = sin2 x cos x, over [0, ⇡/2]

(i) f(x) =p

4x + 1, over [0, 2]

(j) f(x) =3

(1 + 4x)2 , over [1, 6]

2. For questions (b), (e), (g), and (j) above, find a value of c within the interval for which f(c) is the averagevalue (height).

3. If f(x) is a continually decreasing function over the interval [a, b], which of the following will be more:

Z b

a

f(x) dx or (b � a) · f(a)?

4. If f(x) is an even function, over which interval is the average value of f(x) the greatest?

A. [�a, 0] B. [0, a] C. [�a, a] D. (�a, a)

5. A function f(x) has the property that f 0(x) = 6x + 8 and f(0) = 3. What is the average value of thefunction over the interval [0, 2]?

6. A circle is growing from an infinitesimal speck. If dr/dt is a constant 3 feet per minute, what is theaverage area of the circle in its first 10 minutes of existence?

7. If, in the previous problem, it had been dA/dt that grew at a constant 3 cubic feet per minute (hence dr/dt

is no longer constant), what would the average value of the area of the circle be in its first 10 minutesof existence?

Answer Key

1. (a)1

4

Z 7

3

2x + 5 dx =1

4

�x2 + 5x

� ��7

3=

1

4(84 � 24) = 15

(b)1

4

Z 7

3

x2 dx =1

4

✓1

3x3

◆ ��7

3=

1

4

✓343

3� 27

3

◆=

79

3

(c)1

⇡

Z ⇡

0

sin x dx =1

⇡(� cos x)

��⇡

0=

2

⇡

(d)1

3⇡

Z 3⇡

0

sin x dx =1

3⇡(� cos x)

��3⇡

0=

2

3⇡

(e)1

6

Z 3

�3

p9 � x2 dx =

1

6· 9⇡

2=

3⇡

4

(f)1

2

Z 2

0

x3 � 4x dx =1

2

✓1

4x4 � 2x2

◆ ��2

0=

1

2(4 � 8) = �2

(g)1

15

Z 16

1

1px

dx =1

15

�2p

x� ��

16

1=

1

15(8 � 2) =

2

5

(h)1

⇡/2

Z ⇡/2

0

sin2 x cos x dx =2

⇡

✓1

3sin3 x

◆ ��⇡/2

0=

2

3⇡

(i)1

2

Z 2

0

p4x + 1 dx =

1

2

✓1

4· 2

3(4x + 1)

3/2

◆ ��2

0=

1

12(27 � 1) =

13

6

(j)1

5

Z 6

1

3

(1 + 4x)2 dx =

1

5

✓�3

4(1 + 4x)�1

◆ ��6

1= � 3

20

✓1

25� 1

5

◆=

3

125

2.

(b) c2 =79

3(e)

p9 � c2 =

3⇡

4

c =

r79

3c =

s9 �

✓3⇡

4

◆2

c ⇡ 5.132 c ⇡ 1.857

(g)1pc

=2

5(j)

3

(1 + 4c)2 =

3

125

c =25

4c =

1

4

⇣p125 � 1

⌘

c = 6.25 c ⇡ 2.545

3. Sketching an example of f(x) as described makes this a very easy question to answer. There is clearlymore area in the (dotted) rectangle whose base is (b � a) and whose height is f(a) than there is below

the curve f(x) between a and b. Thus (b � a) · f(a) >R b

af(x) dx.

x

y

f(x)

f(a)

a b

4. Since f(x) is even, the average value of f(x) over [�a, 0] equals the average value of f(x) over [0, a], hencethe first two choices are equal. The third choice has twice the area AND twice the width, consequentlythe same average value as the first two choices. Finally, the last choice is identical to the third choice,since leaving out the endpoints won’t have an impact on the average value at all. Therefore, the averagevalue of f(x) is the same for all four domains.

5. Note that f(x) = 3x2 + 8x + 3. Thus the average value of f(x) over [0, 2] will be

1

2

Z 2

0

3x2 + 8x + 3 dx =1

2

�x3 + 4x2 + 3x

� ��2

0=

1

2(8 + 16 + 6) = 15.

6. Since dr/dt = 3, taking an anti-derivative gives us r = 3t (make sure that soaks in before going on). Also,

we know A = ⇡r2, so the single rule that captures our situation is A(t) = ⇡ (3t)2

= 9⇡t2. The averageof A(t) over the interval [0, 10] is

1

10

Z 10

0

9⇡t2 dt =1

10

�3⇡t3

� ��10

0=

3000⇡

10= 300⇡ square feet.

7. Since dA/dt = 3, taking an anti-derivative give us A = 3t. The average of A(t) over the interval [0, 10] is

1

10

Z 10

0

3t dt =1

10

✓3

2t2◆ ��

10

0=

300

20= 15 square feet.


Name

1. Solve the following differential equations for y, given the accompanying initial values.

(a)dy

dx=√x and y(0) = 3

(b)dy

dx=√y and (2, 9)

(c)dy

dx=

5

y + 2and (1, 1)

(d)dy

dx= sinx and (π, 4)

(e)dy

dx=

2− xy − 3

and (2, 7)

(f)dy

dx= sec y and (1, 0)

(g)dy

dx= xy2 and (1,−1)

(h)dy

dx= −2x+ 3y

3x+ 2yand (0, 0)

(i)dy

dx=x

yand (3, 4)

(j)dy

dx= sec2 x+ 2x sinx sec3 x and (0, 0)

2. Consider the differential equation dy/dx = −2x/y. Sketch a slope field for the equation at the twelve pointsindicated on the graph below.

x

y (a) Let y = f(x) be the particular solution to thedifferential equation with the initial conditionf(1) = −1. Write an equation for the line tan-gent to the graph of f(x) at (1,−1) and use itto approximate f(1.1).

(b) Find the particular solution y = f(x) tothe differential equation with initial conditionf(1) = −1.

3. Which of the following is a slope field for the differential equation dydx = x

y ?

A.

x

y

B.

x

y

C.

x

y

D.

x

y

E.

x

y

Answer Key

1. (a) First, “separating the variables”, integrating, and solving for C using the initial conditions gives us

∫1 dy =

∫ √x dx −→ y =

2

3x

3/2 + C −→ 3 =2

3· 03/2 + C −→ C = 3

Substituting our value of C back into the equation relating y to x gives us

y =2

3x

3/2 + 3

(b) First, “separating the variables”, integrating, and solving for C using the initial conditions gives us

∫1√ydy =

∫1 dx −→ 2

√y = x+ C −→ 2

√9 = 2 + C −→ C = 4


2√y = x+ 4 −→ y =

(x2

+ 2)2

(c) First, “separating the variables”, integrating, and solving for C using the initial conditions gives us

∫y + 2 dy =

∫5 dx −→ 1

2y2 + 2y = 5x+C −→ 1

212 + 2 · 1 = 5 · 1 +C −→ C = −5

2

Next, we solve for y.

1

2y2+2y = 5x− 5

2−→ y2+4y = 10x−5 −→ (y + 2)

2= 10x−1 −→ y = −2±

√10x− 1

Finally, we plug in the initial condition to decide whether to keep the positive or negative “half” ofthe picture. It is clear that the positive half is desired. Thus, the particular solution y = f(x) tothe differential equation with initial condition (1,−1) is

y = −2 +√

10x− 1

(d) First, “separating the variables”, integrating, and solving for C using the initial conditions gives us

∫1 dy =

∫sinx dx −→ y = − cosx+ C −→ 4 = − cosπ + C −→ C = 3


y = − cosx+ 3

(e) First, “separating the variables”, integrating, and solving for C using the initial conditions gives us

∫y−3 dy =

∫2−x dx −→ 1

2y2−3y = 2x− 1

2x2 +C −→ 1

2·72−3 ·7 = 2 ·2− 1

2·22 +C −→ C =

3

2


1

2y2−3y = 2x− 1

2x2+

3

2→ y2−6y = 4x−x2+3→ (y − 3)

2= 12+4x−x2 → y = 3±

√12 + 4x− x2

Finally, we plug in the initial condition to decide whether to keep the positive or negative “half” ofthe picture. It is clear that the positive half is desired. Thus, the particular solution y = f(x) tothe differential equation with initial condition (1,−1) is

y = 3 +√

12 + 4x− x2

(f) First, “separating the variables”, integrating, and solving for C using the initial conditions gives us

∫cos y dy =

∫1 dx −→ sin y = x+ C −→ sin 0 = 1 + C −→ C = −1


sin y = x− 1 −→ y = sin−1 (x− 1)

(g) First, “separating the variables”, integrating, and solving for C using the initial conditions gives us

∫y−2 dy =

∫x dx −→ −y−1 =

1

2x2 + C −→ − (−1)

−1=

1

2· (1)

2+ C −→ C =

1

2


−y−1 =1

2x2 +

1

2−→ −y−1 =

x2 + 1

2−→ y = − 2

x2 + 1

(h) First, you can’t “separate the variables”! We will have to be more perspicacious. . .

dy

dx= −2x+ 3y

3x+ 2y−→ 3x

dy

dx+ 2y

dy

dx= −2x− 3y −→ 3x

dy

dx+ 3y + 2y

dy

dx+ 2x = 0

If you squint at that last bit just right, you can see ddx

(3xy + y2 + x2

)= d

dx (C). Furthermore,since (0, 0) is our initial condition, it must be that C = 0. Thus, the relationship between y and xcan be written

y2 + 3xy + x2 = 0 −→ y =−3x±

√5x2

2−→ y =

−3±√

5

2x

In this particular instance, it is impossible to decide between the positive or negative option, asplugging in our initial condition works in both cases. There are, therefore two correct answers:

y =−3±

√5

2x

(i) First, “separating the variables”, integrating, and solving for C using the initial conditions gives us

∫y dy =

∫x dx −→ 1

2y2 =

1

2x2 + C −→ 16

2=

9

2+ C −→ C =

7

2


1

2y2 =

1

2x2 +

7

2−→ y2 = x2 + 7 −→ y = ±

√x2 + 7

Finally, we plug in the initial condition to decide whether to keep the positive or negative “half” ofthe picture. It is clear that the positive half is desired. Thus, the particular solution y = f(x) tothe differential equation with initial condition (3, 4) is

y =√x2 + 7

(j) Perspicacity pays off on this problem as well. Notice that

dy

dx= sec2 x+2x sinx sec3 x −→ dy

dx=

cos2 x− x · 2 cosx · − sinx

cos4 x−→ dy

dx=

d

dx

( x

cos2 x+ C

)

“Seeing” this one is much more likely to happen when you have been working with derivatives fora while. Nevertheless, given the initial conditions in this instance, we once again have C = 0. Thisleads to

y =x

cos2 x

2. x

y

(a) The slope at (1,−1) equals 2. Therefore, anequation for the line tangent to the graph off(x) at (1,−1) is y + 1 = 2(x − 1). Plug-ging in x = 1.1 gives us y = −0.8. Hence,f(1.1) ≈ −0.8.

(b) First, “separating the variables”, integrating,and solving for C using the initial conditionsgives us

∫y dy =

∫−2x dx→ 1

2y2 = −x2 + C

→ 1

2(−1)

2= − (1)

2+ C

→ C =3

2


1

2y2 = −x2 +

3

2→ y2 = 3− 2x2

→ y = ±√

3− 2x2

Finally, we plug in the initial condition to de-cide whether to keep the positive or negative“half” of the picture. It is clear that the nega-tive half is desired. Thus, the particular solu-tion y = f(x) to the differential equation withinitial condition (1,−1) is

f(x) = −√

3− 2x2

3. E.

Notyet.


Name

Evaluate each indefinite integral below.

1.

∫4

(1 + 2x)3 dx 2.

∫ √3x− 1 dx 3.

∫2

(t + 1)6 dt

4.

∫(2− x)

6dx 5.

∫x√

x2 + 3 dx 6.

∫6x

3√

4x2 + 7dx

7.

∫6x2

(x3 + 5)4 dx 8.

∫4x + 6

(x2 + 3x + 1)6 dx 9.

∫x(x2 − 7

)5dx

10.

∫x (x− 7)

5dx 11.

∫x− 3

4

(x

14 + 1

)−2

dx 12.

∫sinx

(1 + cosx)4 dx

Answer Key

1. u = 1 + 2x −→ du = 2dx

∫2 · 2

(1 + 2x)3 dx =

∫2

u3du =

−1

u2+ C =

−1

(1 + 2x)2 + C

2. u = 3x− 1 −→ du = 3dx1

3

∫3√

3x− 1 dx =1

3

∫ √u du =

2

9u

32 + C =

2

9(3x− 1)

3/2+ C

3. u = t + 1 −→ du = dt

∫2

(t + 1)6 dt =

∫2

u6du =

−2

5u−5 + C =

−2

5 (t + 1)5 + C

4. u = 2−x −→ du = −1dx −∫

(2− x)6 ·−1 dx = −

∫u6 du = −1

7u7 +C =

− (2− x)7

7+ C

5. u = x2+3 −→ du = 2xdx1

2

∫2x√x2 + 3 dx =

1

2

∫ √u du =

1

3u

32 +C =

1

3

(x2 + 3

)3/2+ C

6. u = 4x2+7 −→ du = 8xdx6

8

∫8x

3√

4x2 + 7dx =

3

4

∫u− 1

3 du =9

8u

23 +C =

9

8

(4x2 + 7

)2/3+ C

7. u = x3 + 5 −→ du = 3x2dx

∫2 · 3x2

(x3 + 5)4 dx =

∫2

u4du =

−2

3u−3 + C =

−2

3 (x3 + 5)3 + C

8. u = x2 + 3x + 1 −→ du = (2x + 3)dx

∫2 · (2x + 3)

(x2 + 3x + 1)6 dx =

∫2

u6du =

−2

5u−5 + C

=−2

5 (x2 + 3x + 1)5 + C

9. u = x2− 7 −→ du = 2xdx1

2

∫2x(x2 − 7

)5dx =

1

2

∫u5 du =

1

12u6 +C =

(x2 − 7

)6

12+ C

10. u = x− 7 −→ du = dx

∫(x− 7 + 7) (x− 7)

5dx =

∫(u + 7) · u5 du =

∫ (u6 + 7u5

)du

=1

7u7 +

7

6u6 + C =

(x− 7)7

7+

7 (x− 7)6

6+ C

11. u = x14 + 1 −→ du = 1

4x− 3

4 dx 4

∫1

4x− 3

4

(x

14 + 1

)−2

dx = 4

∫u−2 du = −4u−1 + C

=−4

4√x + 1

+ C

12. u = 1 + cosx −→ du = − sinxdx −∫ − sinx

(1 + cosx)4 dx = −

∫u−4 du =

1

3u−3 + C

=1

3 (1 + cosx)3 + C


Name

Evaluate each indefinite integral below.

1.

∫sin9 x cosx dx 2.

∫ (sin3 x− sin2 x

)cosx dx 3.

∫ (1− sin2 x

)cosx dx

4.

∫tan6 x sec2 x dx 5.

∫sec6 x tanx dx 6.

∫ (1 + tan2 x

)3tanx dx

7.

∫cotx csc2 x dx 8.

∫8 cosx

(1 + sinx)2 dx 9.

∫(1 + sinx)

32 cosx dx

10.

∫5 cos 8x dx 11.

∫sin5 x

cos7 xdx 12.

∫(3 + cosx)

4

cscxdx

13.

∫sec2 x tanx dx 14.

∫cos3 x

sin5 xdx 15.

∫sin 8x cos 8x dx

Answer Key

1. u = sinx −→ du = cosx dx

∫(sinx)

9cosx dx =

∫u9 du =

1

10u10 + C

=sin10 x

10+ C


∫ ((sinx)

3 − (sinx)2)

cosx dx =

∫ (u3 − u2

)du =

1

4u4−1

3u3+C

=sin4 x

4− sin3 x

3+ C


∫ (1− (sinx)

2)

cosx dx =

∫ (1− u2

)du = u− 1

3u3 + C

= sinx− sin3 x

3+ C

4. u = tanx −→ du = sec2 x dx

∫(tanx)

6sec2 x dx =

∫u6 du =

1

7u7 + C

=tan7 x

7+ C

5. u = secx −→ du = secx tanx dx

∫(secx)

5secx tanx dx =

∫u5 du =

1

6u6 + C

=sec6 x

6+ C

6. (exactly the same as the previous problem)

7. u = cotx −→ du = − csc2 x dx −∫

cotx · − csc2 x dx = −∫

u du = −1

2u2 + C

= −cot2 x

2+ C

8. u = 1 + sinx −→ du = cosx dx

∫8 cosx

(1 + sinx)2 dx =

∫8

u2du =

−8

u+ C

=−8

1 + sinx+ C

9. u = 1 + sinx −→ du = cosx dx

∫(1 + sinx)

32 cosx dx =

∫u

32 du =

2

5u

52 + C

=2 (1 + sinx)

5/2

5+ C

10. u = 8x −→ du = 8 dx5

8

∫8 cos 8x dx =

5

8

∫cosu du =

5

8sinu + C

=5

8sin 8x + C


∫(tanx)

5sec2 x dx =

∫u5 du =

1

6u6 + C

=tan6 x

6+ C

12. u = 3 + cosx −→ du = − sinx dx −∫

(3 + cosx)4 · − sinxdx = −

∫u4 du =

−1

5u5 + C

=− (3 + cosx)

5

5+ C


∫sec2 x tanx dx =

∫u du =

1

2u2 + C

=tan2 x

2+ C

14. u = cotx −→ du = − csc2 xdx −∫

(cotx)3 · − csc2 xdx = −

∫u3 du =

−1

4u4 + C

= −cot4 x

4+ C

15. u = sin 8x −→ du = 8 cos 8x dx1

8

∫sin 8x · 8 cos 8x dx =

1

8

∫u du =

1

16u2 + C

=sin2 8x

16+ C


Name

1. For the problems below, find the volumes generated by revolving the regions described around the x-axis.

(a) The region bounded by x + y = 2,the x-axis,and the y-axis

(b) The region bounded by y = x2 − 2x and thex-axis

(c) The region bounded by y = x4, x = 1, andy = 0

(d) The region bounded by y = secx, |x| =π

4,

and y = 0

2. For the problems below, find the volumes generated by revolving the regions described around the y-axis.

(a) The region bounded by x = 1− y2 and x = 0 (b) The region bounded by xy = 1, x = 0, y = 1,and y = 2

3. Find the volume generated by revolving the region bounded below by the parabola y = 3x2 + 1 andabove by the line y = 4. . . around the line y = 4.

4. A hemispherical bowl of radius 5 feet contains water to a depth h, as shown below.

h

(a) Find an expression for the volume of water in the bowl in terms of h.

(b) Suppose that water runs into the bowl at the rate of .2 cubic feet per second. How fast is the waterlevel rising when the water is 4 feet deep?

Answer Key

1. (a)

Volume of a single disk Volume of all of the disks Final answer

= πy2 dx = π

∫ 2

0

(4− 4x+ x2

)dx

8π

3

= π (x− 2)2dx = π

(4x− 2x2 +

1

3x3) ∣∣∣

2

0

(b)


= πy2 dx = π

∫ 2

0

(x4 − 4x3 + 4x2

)dx

16π

15

= π(x2 − 2x

)2dx = π

(1

5x5 − x4 +

4

3x3) ∣∣∣

2

0

(c)


= πy2 dx = π

∫ 1

0

x8 dxπ

9

= π(x4)2

dx = π · 1

9x9∣∣∣1

0

(d)


= πy2 dx = π

∫ π/4

−π/4

sec2 x dx 2π

= π (secx)2dx = π · tanx

∣∣∣π/4

−π/4

2. (a)


= πx2 dy = π

∫ 1

−1

(y4 − 2y2 + 1

)dy

16π

15

= π(1− y2

)2dy = π

(1

5y5 − 2

3y3 + y

) ∣∣∣1

−1

(b)


= πx2 dy = π

∫ 2

1

y−2 dyπ

2

= π(y−1

)2dy = π · −1

y

∣∣∣2

1

3.


= π(y1 − y2)2 dx = π

∫ 1

−1

(9x4 − 18x2 + 9

)dx

48π

5

= π(3− 3x2

)2dx = π

(9

5x5 − 6x3 + 9x

) ∣∣∣1

−1

4. I’ll tell you tomorrow.

This is the kind of problem and solution that you will likely need to read slowly and think about morethan once. . .

r

a

(a) You will notice that I added some different labels to the diagram. If we draw (I didn’t, becauseit made the picture too cluttered) a radius of the hemisphere bowl out to the right endpoint ofthe segment labelled r above, we would have a right triangle whose sides are r, 5 − a, and 5.Consequently, r2 = 25− (5− a)2 = 10a− a2. Notice that for larger and larger choices of a (up to5), we get larger and larger values of r2 (up to 25), which makes our thinking so far at least plausible.

(If you prefer, there is a pretty quick “point to circle” plan that also gets the relationship betweena and r. Do you see why the product of r and r has to equal the product of a and (10− a)?)

Now, a very skinny disk a height a above the bottom of the bowl would have volume

πr2 da = π(10a− a2) da.

Consequently, the sum of all disks (total amount of water) from a = 0 to a = h would be

∫ h

0

π(10a− a2

)da = π

(5a2 − 1

3a3) ∣∣∣

h

0= π

(5h2 − 1

3h3).

(b) From the previous problem, we have V = π(5h2 − 1

3h3). Taking the derivative of both sides with

respect to t gives us

dV

dt= π

(10h

dh

dt− h2 dh

dt

)

.2 = π

(40dh

dt− 16

dh

dt

)

.2

24π=dh

dt

Therefore, the height of the water is increasing at a rate of 1/120π feet per second at the instant thewater is 4 feet deep.


Name

1. The base of the solid is the circle x2 +y2 = 16. Each cross section perpendicular to the x-axis is a squarewith one side on the base. Find the volume of the solid.

2. The base of the solid is the area bounded by y2 = 16x and x = 4. Each cross section perpendicular tothe x-axis is an equilateral triangle, with one side on the base. Find the volume of the solid.

3. The base of the solid is the area bounded by y = tanx, y = 0, and x = π/3. Each cross sectionperpendicular to the x-axis is a square with one side on the base. Find the volume of the solid.

4. The base of the solid is the area bounded by xy = 1, x = 1, x = 4, and y = 0. Each cross sectionperpendicular to the x-axis is an isosceles right triangle whose hypotenuse is on the base. Find thevolume of the solid.

5. A hemispheric basin of radius 10 feet is being used to store water. When the water level is 5 feet deep,what percent of the basin’s volume is filled?

1.

Picture Volume of a single slab Final Answer

x

y

= (2y)2dx

= 4(16− x2

)dx

1024

3

Volume of all of the slabs

=

∫ 4

−4

(64− 4x2

)dx

=

(64x− 4

3x3) ∣∣∣

4

−4

2.


x

y=(y ·√

3y)dx

= 16√

3x dx 128√

3


=

∫ 4

0

16√

3x dx

= 8√

3x2∣∣∣4

0

3.


x

y= y2 dx

= tan2 x dx√

3− π

3


=

∫ π/3

0

(sec2 x− 1

)dx

= (tanx− x)∣∣∣π/3

0

4.


x

y=(y

2

)2dx

=1

4x−2 dx

3

16


=

∫ 4

1

1

4x−2 dx

=−1

4x

∣∣∣4

1

5. It is exactly 5/16 full, or 32.15%. (Yesterday’s last problem will help here.)

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