As part of the advanced placement program, there are two calculus courses offered. Calculus Ab and Calculus Bc. The ap calculus ab book contains study material which is equivalent to the college calculus course in the first semester. Whereas, on the other hand, the ap calculus bc is equivalent to both, the first as well as the second-semester calculus course.
- AP Calculus is structured around three foundational concepts: limits, derivatives, and integrals and the Fundamental Theorem of Calculus. Throughout the course, students should have opportunities to engage with these concepts so that by the end of the course, students can demonstrate proficiency in understandings related to each concept.
- Calculus is a method for studying rates of change of functions. Because we often represent functions by their graphs, you could say that calculus is all about the analysis of graphs. We will review the main topics that you’ll need to know for the AP Calculus exams.
- Download: AP Calculus AB 2020 (5 Steps to a 5 AP Calculus AB/BC) PDF -Read Online: AP Calculus AB 2020 (5 Steps to a 5 AP Calculus AB/BC) PDF AP Calculus AB 2020 (5 Steps to a 5 AP Calculus AB/BC) Review This AP Calculus AB 2020 (5 Steps to a 5 AP Calculus AB/BC) book is not really ordinary book, you have it then the world is in your hands.
AP Calculus AC Review Week 5 Applications of Integrals Review the topics before you being the problems. Solutions are posted on my teacher page. If you have difficulty with a question look at the detailed answer postings BEFORE you ask your teacher for help. Get a hint.DON’T COPY THE ANSWER!!! THAT IS NOT HELPFUL!! AP Calculus BC Chapter 5 – AP Exam Problems 1 Riemann Sums – Rectangle Approximation Method 1. The data for acceleration at of a car from 0 to 6 seconds are given in the table above.
But for a complete study, I would recommend you to first go for the ap calculus ab books. As they are the best to get the proper knowledge from the base. But which book to study?
Now, this is a tough choice because one relevant book could help you as much as one irrelevant book could ruin you. But you don’t have to worry as today I am going to provide you with the Best ap Calculus ab Review Book.
To hone yourself in all to facets, You only need the right ap calculus book. Here is the complete list which will help you to decide the right ap calculus ab study guide among all the Best ap Calculus ab Review Book. So here, let’s check out the complete ap calculus review.
Top 5 Best ap Calculus Book In 2020
1. Barron’s Ap Calculus: By Dennis Donovan M.S
Barron’s has long excelled in making AP exam review books. This material is not at all any different. No wonder, it is termed as one of the best study material available in the market for the ap calculus ab exam.
All the topics which are covered in the exam are mentioned thoroughly and with complete details in this Best ap Calculus ab Review Book. It also includes practice tests as well as questions for proper practice and a better understanding.
The best part is that each practice test includes appropriate answers and solutions. It gives you an idea of how you will get the solution to the respective questions next time.
The most important part of the ap calc ab exam is the graphing calculator. No one tells you how to use the graphing calculator. This is where this book will help you out in the best manner. Because this review book is the best to teach you how you can utilize a graphing calculator in the most efficient way possible. So that you save lots of time for you while you are attempting your exam.
In a nutshell, this book contains all the important material you require for your exam. But the only thing in which it lacks behind is that it is not as updated as other books. So, if you are thinking to use it as your main study material then it could create a little problem for you. Otherwise, it is a great book for a great practice and learning perspective.
Why You Should Buy This Book
It is a well-balanced book. Along with the excellent study material, it also helps you to learn many other things that are also important for passing the exam. It even tells you how you can use your graphing calculator efficiently during the exam. This could be a good plus point for you to save lots of time. No doubt, it is a great book that covers the practice paper for both ap calculus ab as well as ap calculus bc exam.
- All-round material
- Practice tests and questions for both ap calculus ab and ap calculus bc exam
- Teaches you to use the graphing calculator efficiently
2. Cracking the AP Calculus AB Exam 2020: By Princeton Review
If you are looking for the best ap calculus ab review book, the best you could get for you is the book by Princeton Review. Cracking is one of the must use review books.
With this book, you will surely get a score of 5 or at least a passing score in your ap calculus ab exam. It contains a complete step by step review of all the topics included in the college board. In this ap calc ab review book, you will find an in-depth and thorough review of all the concepts you need to master or recall for your exam.
This book also comprises of the latest information related to the upcoming exam which could also be helpful for you. The best part is that it consists of all the tips and tactics which you need to adapt to improve the way you answer all the questions in your exam. It helps to avoid any silly mistakes and waste of time.
To solve a calculus exam, there are different approaches available. This book provides you with the easiest to understand and efficient to use techniques. All the tricks and techniques are personally tried and tested before specifying in this book. It helps you to get a better understanding of every point.
After all, math is all about practice and repetition. Therefore, to improve your retention power this book consists of drills at the end of each chapter. At the end of all the chapters, you will get three full-length practice tests. These tests match the difficulty level of the original ap calculus ab exam. So, it will help you improve your skill and knowledge about every little thing which you need to remember while doing the actual exam.
Why You Should Choose This Book
This book contains good quality study material to help you study like a pro. It is easy to understand and includes the complete guide to help you prepare at its best for the exam. If you are looking to recall or even master calculus for your exam. Then nothing is a better option than cracking the ap calculus ab exam 2020 by Princeton Review. This book is an all-rounder to help you get the best results.
- Practice Test
- Tips And Strategies To Solve The Answers Efficiently
- Drills After Every Chapter To Improve Retention
- Full-Length Practice Test
3. AP Calculus AB Lecture Notes: By Rita Korsunsky
This Best ap Calculus ab Review Book is written by a well-renowned teacher the Rita Korsunsky. This book even includes the lecture notes which she uses in her ap calculus ab classes. The author is so prided at herself. As all her students pass the exam and around 90% of those who did the course under her guidance were able to get a grade of 5.
Looking at her track record. It is clear that there is something very special about the lecture notes this book contains. The lecture notes in this book are very helpful for every calculus student. Because every topic contains a corresponding example that helps you develop proper skills.
The problems mentioned in this Best ap Calculus ab Review Book also guide you through accurate step by step procedures. This is the best and helps you not to get confused and lost whenever you decide to give a try to solve these problems.
This book’s approach may seem like a very simple one. But it is the best seller on Amazon and has a rating of 4.7. The reason is quite simple. It excels at one thing which does the best. That is to teach you the basic fundamentals of calculus in a very effective and efficient manner. It is a highly recommended book among most of the satisfied customers. And it has also received many positive feedbacks which you can also find on the Amazon.
Why You Should Go For This Book
This book is a complete study material in itself. Burn mac os x dmg to dvd on windows10. Once you study this complete book you need nothing extra. The lecture notes by Rita Korsunsky are the best to prepare for the ap calculus ab exam. Also, these notes are very easy to understand and retain for a long time.
If you are looking for a highly detailed and thorough review experience to understand everything in a complete manner. Then this ap calculus ab study guide should be your main material. It focuses on all the important topics to help you get the best.
- The lecture notes of the Teacher itself
- Every Topic Has Corresponding Examples For Better Understanding
- Best Seller On Amazon
- Unlike Other Review books, It Teaches Calculus Using Slideshow Lectures And Not In The Conventional Way
- Some Problems Are Of Higher Level Which Is Difficult For Some People To Understand
Also Read: Best Calculator For Sat
4. 320 AP Calculus AB Problems: By Steve Warner
The name of this book completely specifies the details. In this book, you will get great benefits of 320 ap calculus ab problems. All these problems are arranged according to the topics covered and the difficulties of the problems.
The main focus of this book is to focus on improving your knowledge as well as skills in calculus. The questions in this book have their own answers and solutions which help you get to know how to get the right answer for every type of problem. It is the best as well as the most efficient way to learn how to achieve your goals.
This book might be a little expensive as compared to other books available here. But no other book focuses on practice as much as this ap calc ab review book does. And I don’t need to tell you how important practice is not only in calculus but in the complete maths.
This book is all about providing you so many problems along with their solutions. But after completing this whole book. You will feel like an expert because you will encounter amazing practice sessions while solving this book. But unfortunately, this doesn’t consist of in-depth details of every topic. So, I will recommend you not to consider it as your main study material. But this Best ap Calculus ab Review Book is the right one to help you recall your fundamentals of calculus.
Why To Choose This Book
This ap calculus ab book covers all the topics you need to practice for your exam. It is the best suitable as a practice book. The main focus is on practical questions to help you improve your knowledge and skills. It gives you hundreds of problems to help you get used to all those problems and understand them in a better way.
Also, you will get all the problems arranged based on the topics as well as their difficulty level. So, you won’t get many difficulties while solving them. Rather you will experience a better understanding of solving them and great learning.
- Hundreds of problems to help you understand everything in a better way
- Covers all the topic you need to learn for your exam
- Different solutions for every question to make you understand better
- No in-depth detailed study, more focus on the practical questions
5. 5 Steps to a 5: AP Calculus AB 2020 Elite Student Edition: By William Ma
This review book follows the usual 5 step plan which even made the McGraw- Hill’s line of Ap review book so successful. This material is so useful to help you gain a 5 in your ap calculus ab exam. This book is so amazing because of its elite student edition. It improves the normal edition by providing bonus content which you will not find anywhere else.
In the 5 steps, it covers all the topics that come in the ap calculus ab exam. The review part of the book helps you to understand how you can simply solve the problem by using an easy step by step explanation which is designed to make you understand and master the subject.
As an excellent all-rounder review book. This material includes an appendix of the formulas which you need to learn by heart for your exam. This book includes a section that tells you how to answer free-response questions and also tells you how the ap scores all your answers.
After all, for any mathematical subject, practice is a must. Therefore, this Best ap Calculus ab Review Book includes 4 practice tests that are formed like the original exam. And nearly 200 questions and activities that keep your mind working and engaged. As you build the necessary skills to get a score of 5 in your upcoming ap Calculus ab exam.
Why You Should Choose This Book
It is a great all-round study book. It not only covers all the topics essential for your exam. But it has complete details to help you have enhanced learning.
It covers even every little detail which helps you get the learning to the edge. It even comes with lots of practice questions and mock test papers that help you know the exam pattern, difficulty and how to attempt the exam in the best way to get a score of 5. For a detailed study and good marks, this book is surely a great choice to get.
- 4 Mock test paper for complete in-depth practice
- More than 200 questions to help you enhance your skills
- Bonus Content That You Can’t Find Anywhere Else
Ap Calculus Ab Study Tips
How should you study for your ap calculus exam? Now we are going to help you with some of the ap calculus ab study tips. Also, we will point out a few resources that will help you achieve a 5 in your ap exam.
AP Calculus AB Exam Study Tips
These shot tips will surely help you out in your exam. Some of the tips are general test-taking strategies. While others are specific for the ap calculus ab exam.
Know What’s On The Exam
If you expect to do well. Then it is essential to know all the topics which will be covered in the ap calculus ab exam. Basically, the material could be segregated into major three categories.
- Limits (including continuity, asymptotes, etc)
- Derivative (including applications)
- Integrals (Including applications)
Know The Format Of The Exam
The ap calculus ab is a standardized test that takes 3 hours and 15 minutes in total. There are majorly two sections in the test, multiple-choice and free-response. Also, these two sections are divided further into two parts, calculator and non-calculator. So, you need to properly manage all the questions within the time limit. In order to score good marks, you need to complete the whole paper within time.
Familiarize Yourself With Your Calculator
The worst thing which you could do is to run out and buy the most expensive graphing calculator for your exam. But it will be kept untouched in your exam. Because you don’t know how to use it. Therefore, it is way better to get a good graphing calculator a little earlier than the exam. And then try to know about all its functions.
Eventually, you might learn all the advanced features including numerical integration and symbolic derivatives. No wonder all these functions will help you save lots of time and effort.
Set Up And Follow A Study Plan
You require weeks and months of hard work for preparing for an ap calculus ab exam. Cramming a night before the exam is simply not going to help.
Plan on purchasing the ap calculus ab study guide and start studying. And work out at least 3-5 practice tests. The more practice you will do, the better will be your performance in the actual test. So, to study accurately, all you need is a proper plan to evaluate for your study.
Get A Tutor
Many students get excel independently. But other gets benefits from some quality one on one time with an expert. If you feel like there are some topics that you can’t understand no matter how many times you to try to read it in your textbook. Then maybe contacting a tutor could help you out in this situation.
But let me tell you that a private tutor will not come cheap. So, either you can refer to a good book. Or you should be willing to spend $50-100 for every hour of teaching. But according to me, referring to a good book is more than enough to understand all the important points.
You can go to any of the books I suggested above. All these Best ap Calculus ab Review Books are personally tried and tested. You tend to understand every single problem through those books for sure.
The Final Thoughts On Ap Calculus Ab Study Tips
Well, these are some of the tips to help you prepare for your ap calculus ab exam study. The exam is not going to be easy at all. But it doesn’t mean that it is impossible to pass it. You can do it easily. All it takes is consistency and hard work.
But most importantly you need to be focused in the right direction. All the above tips will help you to draw your focus in the right direction to help you score 5 or at least pass your ap calculus ab exam.
With the help of the above books and all these tips, nothing could stop you to pass your ap calculus ab exam. Ultimately it is you who is going to score in the exam. It all depends on you. But with good books and our study tips, you tend to move towards the edge you need for your success.
That’s all from my side. Now it is time for you to start preparing for your exam with the right study plan. You have very little time left so just stop everything and start preparing to the best extent. Just be calm, stay focus and be confident about your exam.
And nothing will be able to stop you from passing your exam if you are willing to do the hard work. This is all from my side, all the best for your exam. I hope you do well on your exam.
5 Steps to a 5: AP Calculus AB 2017 (2016)
STEP 4
Review the Knowledge You Need to Score High
CHAPTER 9
Applications of Derivatives
IN THIS CHAPTER
Summary: Two of the most common applications of derivatives involve solving related rate problems and applied maximum and minimum problems. In this chapter, you will learn the general procedures for solving these two types of problems and to apply these procedures to examples. Both related rate and applied maximum and minimum problems appear often on the AP Calculus AB exam.
Key Ideas
General Procedure for Solving Related Rate Problems
Common Related Rate Problems
Inverted Cone, Shadow, and Angle of Elevation Problems
General Procedure for Solving Applied Maximum and Minimum Problems
Distance, Area, Volume, and Business Problems
9.1 Related Rate
Main Concepts: General Procedure for Solving Related Rate Problems, Common Related Rate Problems, Inverted Cone (Water Tank) Problem, Shadow Problem, Angle of Elevation Problem
General Procedure for Solving Related Rate Problems
1. Read the problem and, if appropriate, draw a diagram.
2. Represent the given information and the unknowns by mathematical symbols.
3. Write an equation involving the rate of change to be determined. (If the equation contains more than one variable, it may be necessary to reduce the equation to one variable.)
4. Differentiate each term of the equation with respect to time.
5. Substitute all known values and known rates of change into the resulting equation.
6. Solve the resulting equation for the desired rate of change.
7. Write the answer and indicate the units of measure.
Common Related Rate Problems
Example 1
When the area of a square is increasing twice as fast as its diagonals, what is the length of a side of the square?
Let z represent the diagonal of the square. The area of a square is .
.
Since .
Let s be a side of the square. Since the diagonal z = 2, then s2 + s2 = z2 .
⇒ 2s2 = 4 ⇒ s2 = 4 ⇒ s2 = 2 or .
Example 2
Find the surface area of a sphere at the instant when the rate of increase of the volume of the sphere is nine times the rate of increase of the radius.
Volume of a sphere: Surface area of a sphere: S = 4πr2 .
.
Since , you have or 9 = 4πr2 .
Since S = 4πr2 , the surface area is S = 9 square units.
Note: At 9 = 4πr2 , you could solve for r and obtain or . You could then substitute into the formula for surface area S = 4πr2 and obtain 9. These steps are of course correct but not necessary.
Example 3
The height of a right circular cone is always three times the radius. Find the volume of the cone at the instant when the rate of increase of the volume is twelve times the rate of increase of the radius.
Let r, h be the radius and height of the cone, respectively.
Since h = 3r , the volume of the cone .
When
Thus,
• Go with your first instinct if you are unsure. Usually that is the correct one.
Inverted Cone (Water Tank) Problem
A water tank is in the shape of an inverted cone. The height of the cone is 10 meters and the diameter of the base is 8 meters as shown in Figure 9.1-1 . Water is being pumped into the tank at the rate of 2 m3 /min. How fast is the water level rising when the water is 5 meters deep? (See Figure 9.1-1 .)
Figure 9.1-1
Solution:
Step 1: Define the variables. Let V be the volume of water in the tank; h be the height of the water level at t minutes; r be the radius of the surface of the water at t minutes; and t be the time in minutes.
Step 2: Given: . Height = 10 m, diameter = 8 m.
Find: at h =5.
Step 3: Set up an equation: .
Using similar triangles, you have ; or . (See Figure 9.1-2 .)
Figure 9.1-2
Thus, you can reduce the equation to one variable:
.
Step 4: Differentiate both sides of the equation with respect to t .
Step 5: Substitute known values.
Step 6: Thus, the water level is rising at when the water is 5 m high.
Shadow Problem
A light on the ground 100 feet from a building is shining at a 6-foot-tall man walking away from the light and toward the building at the rate of 4 ft/sec. How fast is his shadow on the building becoming shorter when he is 40 feet from the building? (See Figure 9.1-3 .)
Figure 9.1-3
Solution:
Step 1: Let s be the height of the man’s shadow; x be the distance between the man and the light; and t be the time in seconds.
Step 2: Given: the man is 6 ft tall; distance between light and building = 100 ft. Find at x = 60.
Step 3: (See Figure 9.1-4 .) Writing an equation using similar triangles, you have:
Figure 9.1-4
Step 4: Differentiate both sides of the equation with respect to t .
Step 5: Evaluate at x = 60.
Note: When the man is 40 ft from the building, x (distance from the light) is 60 ft.
Step 6: The height of the man’s shadow on the building is changing at ft/sec.
• Indicate units of measure, e.g., the velocity is 5 m/sec or the volume is 25 in3 .
Angle of Elevation Problem
A camera on the ground 200 meters away from a hot air balloon (also on the ground) records the balloon rising into the sky at a constant rate of 10 m/sec. How fast is the camera’s angle of elevation changing when the balloon is 150 m in the air? (See Figure 9.1-5 .)
Figure 9.1-5
Step 1: Let x be the distance between the balloon and the ground; θ be the camera’s angle of elevation; and t be the time in seconds.
Step 2: Given: distance between camera and the point on the ground where the balloon took off is 200 m, .
Step 3: Find at x = 150 m.
Step 4: Differentiate both sides with respect to t .
.
Step 5: and at x = 150.
Since y > 0, then y = 250. Thus, sec .
Step 6: The camera’s angle of elevation changes at approximately 1.833 deg/sec when the balloon is 150 m in the air.
9.2 Applied Maximum and Minimum Problems
Main Concepts: General Procedure for Solving Applied Maximum and Minimum Problems, Distance Problem, Area and Volume Problems, Business Problems
General Procedure for Solving Applied Maximum and Minimum Problems
Steps:
1. Read the problem carefully and if appropriate, draw a diagram.
2. Determine what is given and what is to be found, and represent these quantities by mathematical symbols.
3. Write an equation that is a function of the variable representing the quantity to be maximized or minimized.
4. If the equation involves other variables, reduce the equation to a single variable that represents the quantity to be maximized or minimized.
5. Determine the appropriate interval for the equation (i.e., the appropriate domain for the function) based on the information given in the problem.
6. Differentiate to obtain the first derivative and to find critical numbers.
7. Apply the First Derivative Test or the Second Derivative Test by finding the second derivative.
8. Check the function values at the endpoints of the interval.
9. Write the answer(s) to the problem and, if given, indicate the units of measure.
Distance Problem
Find the shortest distance between the point A (19, 0) and the parabola y = x2 – 2x + 1.
Solution:
Step 1: Draw a diagram. (See Figure 9.2-1 .)
Figure 9.2-1
Step 2: Let P (x , y ) be the point on the parabola and let Z represent the distance between points P (x , y ) and A (19, 0).
Step 3: Using the distance formula,
(Special case: In distance problems, the distance and the square of the distance have the same maximum and minimum points.) Thus, to simplify computations, let L = Z2 = (x – 19)2 + (x – 1)4 . The domain of L is (–∞, ∞).
Step 4: Differentiate:
is defined for all real numbers.
Set = 0; 2x3 – 6x2 + 7x – 21 = 0. The factors of 21 are ±1, ±3, ±7, and ± 21.
Using Synthetic Division, 2x3 – 6x2 + 7x – 21 = (x – 3)(2x2 + 7) = 0 ⇒ x = 3.
Thus the only critical number is x = 3.
(Note: Step 4 could have been done using a graphing calculator.)
Step 5: Apply the First Derivative Test.
Step 6: Since x = 3 is the only relative minimum point in the interval, it is the absolute minimum.
Step 7: At x = 3, . Thus, the shortest distance is .
• Simplify numeric or algebraic expressions only if the question asks you to do so.
Area and Volume Problems
Example Area Problem
The graph of encloses a region with the x -axis and y -axis in the first quadrant. A rectangle in the enclosed region has a vertex at the origin and the opposite vertex on the graph of . Find the dimensions of the rectangle so that its area is a maximum.
Solution:
Step 1: Draw a diagram. (See Figure 9.2-2 .)
Figure 9.2-2
Step 2: Let P (x , y ) be the vertex of the rectangle on the graph of .
Step 3: Thus, the area of the rectangle is:
The domain of A is [0, 4].
Step 4: Differentiate:
Step 5: is defined for all real numbers.
Set = 0 ⇒ –x + 2 = 0; x = 2.
A (x ) has one critical number x = 2.
Step 6: Apply the Second Derivative Test:
⇒ A (x ) has a relative maximum point at x = 2; A (2) = 2.
Since x = 2 is the only relative maximum, it is the absolute maximum. (Note that at the endpoints: A (0) = 0 and A (4) = 0.)
Step 7: At x = 2, .
Therefore, the length of the rectangle is 2, and its width is 1.
Example Volume Problem (with calculator)
If an open box is to be made using a square sheet of tin, 20 inches by 20 inches, by cutting a square from each corner and folding the sides up, find the length of a side of the square being cut so that the box will have a maximum volume.
Solution:
Step 1: Draw a diagram. (See Figure 9.2-3 .)
Figure 9.2-3
Step 2: Let x be the length of a side of the square to be cut from each corner.
Step 3: The volume of the box is V (x ) = x (20 – 2x )(20 – 2x ).
The domain of V is [0, 10].
Step 4: Differentiate V (x ).
Enter d (x ∗ (20 – 2x ) ∗ (20 – 2x ), x ) and we have 4(x – 10)(3x – 10).
Step 5: V ′(x ) is defined for all real numbers:
Set V ′(x ) = 0 by entering: [Solve ] (4(x – 10)(3x – 10) = 0, x ), and obtain x = 10 or . The critical numbers of V (x ) are x = 10 and . V (10) = 0 and . Since V (10) = 0, you need to test only .
Step 6: Using the Second Derivative Test, enter d (x ∗ (20 – 2x ) ∗ (20 – 2x ), x ,2)| and obtain –80. Thus, is a relative maximum. Since it is the only relative maximum on the interval, it is the absolute maximum. (Note at the other endpoint x = 0, V (0) = 0.)
Step 7: Therefore, the length of a side of the square to be cut is .
• The formula for the average value of a function f from x = a to x = b is .
Business Problems
Summary of Formulas
1. P = R – C : Profit = Revenue – Cost
2. R = xp : Revenue = (Units Sold)(Price Per Unit)
3. Average
4. Marginal Revenue ≈ Revenue from selling one more unit
5. Marginal Profit ≈ Profit from selling one more unit
6. Marginal Cost ≈ Cost of producing one more unit
Example 1
Given the cost function C (x ) = 100 + 8x + 0.1x2 , (a) find the marginal cost when x = 50; and (b) find the marginal profit at x = 50, if the price per unit is $20.
Solution:
(a) Marginal cost is C ′(x ). Enter d (100 + 8x + 0.1x2 , x )|x = 50 and obtain $18.
(b) Marginal profit is P ′(x )
P = R – C
P = 20x – (100 + 8x + 0.1x2 ). Enter d (20x – (100 + 8x + 0.1x∧ 2, x )|x = 50 and obtain 2.
• Carry all decimal places and round only at the final answer. Round to 3 decimal places unless the question indicates otherwise.
Example 2
Given the cost function C (x ) = 500 + 3x + 0.01x2 and the demand function (the price function) p (x ) = 10, find the number of units produced in order to have maximum profit.
Solution:
Step 1: Write an equation.
Profit = Revenue – Cost
P = R – C
Revenue = (Units Sold)(Price Per Unit)
R = xp (x ) = x (10) = 10x
P = 10x – (500 + 3x + 0.01x2 )
Step 2: Differentiate.
Enter d (10x – (500 + 3x + 0.01x∧ 2, x )) and obtain 7 – 0.02x .
Step 3: Find critical numbers.
Set 7 – 0.02x = 0 ⇒ x = 350.
Critical number is x = 350.
Step 4: Apply Second Derivative Test.
Enter d (10x – (500 + 3x + 0.01x∧ 2), x , 2)|x = 350 and obtain –0.02.
Since x = 350 is the only relative maximum, it is the absolute maximum.
Step 5: Write a solution.
Thus, producing 350 units will lead to maximum profit.
9.3 Rapid Review
1. Find the instantaneous rate of change at x = 5 of the function .
Answer:
2. If h is the diameter of a circle and h is increasing at a constant rate of 0.1 cm/sec, find the rate of change of the area of the circle when the diameter is 4 cm.
Answer:
3. The radius of a sphere is increasing at a constant rate of 2 inches per minute. In terms of the surface area, what is the rate of change of the volume of the sphere?
Answer: since
4. Using your calculator, find the shortest distance between the point (4, 0) and the line y = x . (See Figure 9.3-1 .)
Figure 9.3-1
Answer:
Enter
Use the [Zero ] function for y2 and obtain x = 2. Note that when x < 2, y2< 0 which means y1 is decreasing and when x > 2, y2> 0 which means y1 is increasing and thus at x = 2, y1 is a minimum. Use the [Value ] function for y1 at x = 2 and obtain y1 = 2.82843. Thus, the shortest distance is approximately 2.828.
9.4 Practice Problems
Part A—The use of a calculator is not allowed.
1 . A spherical balloon is being inflated. Find the volume of the balloon at the instant when the rate of increase of the surface area is eight times the rate of increase of the radius of the sphere.
2 . A 13-foot ladder is leaning against a wall. If the top of the ladder is sliding down the wall at 2 ft/sec, how fast is the bottom of the ladder moving away from the wall when the top of the ladder is 5 feet from the ground? (See Figure 9.4-1 .)
Figure 9.4-1
3 . Air is being pumped into a spherical balloon at the rate of 100 cm3 /sec. How fast is the diameter increasing when the radius is 5 cm?
4 . A woman 5 feet tall is walking away from a streetlight hung 20 feet from the ground at the rate of 6 ft/sec. How fast is her shadow lengthening?
5 . A water tank in the shape of an inverted cone has a height of 18 feet and a base radius of 12 feet. If the tank is full and the water is drained at the rate of 4 ft3 /min, how fast is the water level dropping when the water level is 6 feet high?
6 . Two cars leave an intersection at the same time. The first car is going due east at the rate of 40 mph and the second is going due south at the rate of 30 mph. How fast is the distance between the two cars increasing when the first car is 120 miles from the intersection?
7 . If the perimeter of an isosceles triangle is 18 cm, find the maximum area of the triangle.
8 . Find a number in the interval (0, 2) such that the sum of the number and its reciprocal is the absolute minimum.
9 . An open box is to be made using a piece of cardboard 8 cm by 15 cm by cutting a square from each corner and folding the sides up. Find the length of a side of the square being cut so that the box will have a maximum volume.
10 . What is the shortest distance between the point and the parabola y = –x2 ?
11 . If the cost function is C (x ) = 3x2 + 5x + 12, find the value of x such that the average cost is a minimum.
12 . A man with 200 meters of fence plans to enclose a rectangular piece of land using a river on one side and a fence on the other three sides. Find the maximum area that the man can obtain.
Part B—Calculators are allowed.
13 . A trough is 10 meters long and 4 meters wide. (See Figure 9.4-2 .) The two sides of the trough are equilateral triangles. Water is pumped into the trough at 1 m3 /min. How fast is the water level rising when the water is 2 meters high?
Figure 9.4-2
14 . A rocket is sent vertically up in the air with the position function s = 100t2 where s is measured in meters and t in seconds. A camera 3000 m away is recording the rocket. Find the rate of change of the angle of elevation of the camera 5 sec after the rocket went up.
Review 5ap Calculus 2
15 . A plane lifts off from a runway at an angle of 20°. If the speed of the plane is 300 mph, how fast is the plane gaining altitude?
16 . Two water containers are being used. (See Figure 9.4-3 .)
Figure 9.4-3
One container is in the form of an inverted right circular cone with a height of 10 feet and a radius at the base of 4 feet. The other container is a right circular cylinder with a radius of 6 feet and a height of 8 feet. If water is being drained from the conical container into the cylindrical container at the rate of 15 ft3 /min, how fast is the water level falling in the conical tank when the water level in the conical tank is 5 feet high? How fast is the water level rising in the cylindrical container?
17 . The wall of a building has a parallel fence that is 6 feet high and 8 feet from the wall. What is the length of the shortest ladder that passes over the fence and leans on the wall? (See Figure 9.4-4 .)
18 . Given the cost function C(x ) = 2500 + 0.02x + 0.004x2 , find the product level such that the average cost per unit is a minimum.
19 . Find the maximum area of a rectangle inscribed in an ellipse whose equation is 4x2 + 25y2 = 100.
20 . A right triangle is in the first quadrant with a vertex at the origin and the other two vertices on the x – and y -axes. If the hypotenuse passes through the point (0.5, 4), find the vertices of the triangle so that the length of the hypotenuse is the shortest possible length.
Figure 9.4-4
9.5 Cumulative Review Problems
(Calculator) indicates that calculators are permitted.
21 . If y = sin2 (cos(6x – 1)), find .
22 . Evaluate .
23 . The graph of f ′ is shown in Figure 9.5-1 . Find where the function f : (a) has its relative extrema or absolute extrema; (b) is increasing or decreasing; (c) has its point(s) of inflection; (d) is concave upward or downward; and (e) if f (3) = –2, draw a possible sketch of f .
Figure 9.5-1
24 . (Calculator) At what value(s) of x does the tangent to the curve x2 + y2 = 36 have a slope of –1?
25 . (Calculator) Find the shortest distance between the point (1, 0) and the curve y = x3 .
9.6 Solutions to Practice Problems
Part A—The use of a calculator is not allowed .
1 . Volume:
Surface Area: .
Since ,
or
At cubic units.
2 . Pythagorean Theorem yields x2 + y2 = (13)2 .
Differentiate:
.
At x = 5, (5)2 + y2 = 132 ⇒ y = ± 12, since y > 0, y = 12.
Therefore, . The ladder is moving away from the wall at when the top of the ladder is 5 feet from the ground.
3 . Volume of a sphere is
Differentiate:
.
Substitute: 100 = 4π(5)2 .
Let x be the diameter. Since .
Thus,
= . The diameter is increasing at when the radius is 5 cm.
4 . (See Figure 9.6-1 .) Using similar triangles, with y the length of the shadow you have:
Figure 9.6-1
Differentiate:
5 . (See Figure 9.6-2 .) Volume of a cone
Figure 9.6-2
.
Using similar triangles, you have
or , thus
reducing the equation to
Differentiate: .
Substituting known values:
or . The water level is dropping at when h = 6 ft.
6 . (See Figure 9.6-3 .)
Figure 9.6-3
Step 1: Using the Pythagorean Theorem, you have x2 + y2 = z2 . You also have and .
Step 2: Differentiate:
.
At x = 120, both cars have traveled 3 hours and thus, y = 3(30) = 90. By the Pythagorean Theorem, (120)2 + (90)2 = z2 ⇒ z = 150.
Step 3: Substitute all known values into the equation:
Step 4: The distance between the two cars is increasing at 50 mph at x = 120.
7 . (See Figure 9.6-4 .)
Figure 9.6-4
Step 1: Applying the Pythagorean
Theorem, you have x2 = y2 + (9 – x )2 ⇒ y2 = x2 – (9 – x )2 = x2 – 81 – 18x + x2 = 18x – 81 = 9(2x – 9), or since y< 0, .
The area of the triangle
Step 2:
Step 3: Set
is undefined at critical numbers are and 6.
Step 4: First Derivative Test:
Thus at x = 6, the area A is a relative maximum.
Step 5: Check endpoints. The domain of A is [9/2, 9]. A (9/2) = 0; and A (9) = 0. Therefore, the maximum area of an isosceles triangle with the perimeter of 18 cm is cm2 . (Note that at x = 6, the triangle is an equilateral triangle.)
8 . Step 1: Let x be the number and be its
Step 2: with 0 < x< 2.
Step 3:
Step 4: Set ⇒ x = ±1, since the domain is (0, 2), thus x = 1. is defined for all x in (0, 2). Critical number is x = 1.
Step 5: Second Derivative Test: and . Thus at x = 1, s is a relative minimum. Since it is the only relative extremum, at x = 1, it is the absolute minimum.
9 . (See Figure 9.6-5 .)
Figure 9.6-5
Step 1: Volume: V = x (8 – 2x )(15 – 2x ) with 0 ≤ x ≤ 4.
Step 2: Differentiate: Rewrite as V = 4x3 – 46x2 + 120x
= 12x2 – 92x + 120.
Step 3: Set V = 0 ⇒ 12x2 – 92x + 120 = 0 ⇒ 3x2 – 23x + 30 = 0. Using the quadratic formula, you have x = 6 or and is defined for all real numbers.
Step 4: Second Derivative Test:
.
Thus, is a relative maximum.
Step 5: Check endpoints.
At x = 0, V = 0 and at x = 4, V = 0. Therefore, at , V is the absolute maximum.
10 . (See Figure 9.6-6 .)
Figure 9.6-6
Step 1: Distance Formula:
Step 2: Let S = Z2 , since S and Z have the same maximums and minimums.
Step 3: Set x = 1 and is defined for all real numbers.
Step 4: Second Derivative Test:
Thus at x = 1, Z has a minimum, and since it is the only relative extremum, it is the absolute minimum.
Step 5: At x = 1,
Therefore, the shortest distance is
11 . Step 1: Average Cost:
Step 2:
Step 3: Set Since x > 0, x = 2 and is undefined at x = 0 which is not in the domain.
Step 4: Second Derivative Test:
and
Thus at x = 2, the average cost is a minimum.
12 . (See Figure 9.6-7 .)
Figure 9.6-7
Step 1: Area:
A = x (200 – 2x ) = 200x – 2x2 with 0 ≤ x ≤ 100.
Step 2: A ′(x ) = 200 – 4x
Step 3: Set A ′(x ) = 0 ⇒ 200 – 4x = 0; x = 50.
Step 4: Second Derivative Test:
A ″(x ) = –4; thus at x = 50, the area is a relative maximum. A (50) = 5000 m2 .
Step 5: Check endpoints.
A (0) = 0 and A (100) = 0; therefore at x = 50, the area is the absolute maximum and 5000 m2 is the maximum area.
Part B—Calculators are allowed.
13 .
Step 1: Let h be the height of the trough and 4 be a side of one of the two equilateral triangles. Thus, in a 30–60 right triangle, .
Step 2: Volume:
V = (area of the triangle) · 10
.
Step 3: Differentiate with respect to t .
Step 4: Substitute known values:
The water level is rising when the water level is 2 m high.
14 . (See Figure 9.6-8 .)
Figure 9.6-8
Step 1: tan θ = S /3000
Step 2: Differentiate with respect to t .
Step 3: At t = 5; S = 100(5)2 = 2500; Thus, Z2 = (3000)2 + (2500)2 = 15,250,000. Therefore, , since Z > 0, . Substitute known values into the equation:
radian/sec. The angle of elevation is changing at 0.197 radian/sec, 5 seconds after liftoff.
15 . (See Figure 9.6-9 .)
Figure 9.6-9
h = (sin 20°)(300)t ; ≈ 102.606 mph. The plane is gaining altitude at 102.606 mph.
16 .
Similar triangles: or
Substitute known values:
The water level in the cone is falling at
ft/min ≈ 1.19 ft/min when the water level is 5 feet high.
Vcylinder = π R2 H = π(6)2H = 36πH .
ft/min ≈ 0.1326 ft/min or 1.592 in/min.
The water level in the cylinder is rising at ft/min = 0.133 ft/min.
17 . Step 1: Let x be the distance of the foot of the ladder from the higher wall. Let y be the height of the point where the ladder touches the higher wall. The slope of the ladder is or .
Thus,
Step 2: Pythagorean Theorem:
Step 3: Enter
The graph of y1 is continuous on the interval x > 8. Use the [Minimum ] function of the calculator and obtain x = 14.604; y = 19.731. Thus the minimum value of l is 19.731 or the shortest ladder is approximately 19.731 feet.
18 . Step 1: Average thus,
Step 2: Enter: + .02 + .004 * x
Step 3: Use the [Minimum ] function in the calculator and obtain x = 790.6.
Step 4: Verify the result with the First Derivative Test. Enter y 2 = d (2500/x +.02 +.004x , x ); Use the [Zero ] function and obtain x = 790.6. Thus at x = 790.6. Apply the First Derivative Test:
Thus the minimum average cost per unit occurs at x = 790.6. (The graph of the average cost function is shown in Figure 9.6-10 .)
Figure 9.6-10
19 . (See Figure 9.6-11 .)
Figure 9.6-11
Step 1: Area A = (2x )(2y ); 0 ≤ x ≤ 5 and 0 ≤ y ≤ 2.
Step 2: 4x2 + 25y2 = 100;
25y2 = 100 – 4x2 .
Step 3:
Step 4: Enter
Use the [Maximum ] function and obtain x = 3.536 and y1 = 20.
Step 5: Verify the result with the First Derivative Test.
Enter
. Use the [Zero ] function and obtain x = 3.536.
Note that:
The function f has only one relative extremum. Thus, it is the absolute extremum. Therefore, at x = 3.536, the area is 20 and the area is the absolute maxima.
20 . (See Figure 9.6-12 .)
Figure 9.6-12
Step 1: Distance formula:
l2 = x2 + y2 ; x > 0.5 and y > 4.
Step 2: The slope of the hypotenuse:
Step 3: .
Step 4: Enter and use the [Minimum ] function of the calculator and obtain x = 2.5.
Step 5: Apply the First Derivative Test. Enter y 2 = d (y 1(x ), x ) and use the [Zero ] function and obtain x = 2.5.
Note that:
Since f has only one relative extremum, it is the absolute extremum. Thus, at x = 3, it is an absolute minimum.
Step 6: Thus, at x = 2.5, the length of the hypotenuse is the shortest. At x = 2.5, . The vertices of the triangle are (0, 0), (2.5, 0), and (0, 5).
9.7 Solutions to Cumulative Review Problems
21 .
22 . As x → ∞, the numerator approaches 0 and the denominator increases without bound (i.e., ∞). Thus, the .
23 . (a) Summarize the information of f ′ on a number line.
Since f has only one relative extremum, it is the absolute extremum. Thus, at x = 3, it is an absolute minimum.
(b) The function f is decreasing on the interval (–∞, 3) and increasing on (3, ∞).
(c)
No change of concavity ⇒ No point of inflection.
(d) The function f is concave upward for the entire domain (–∞, ∞).
(e) Possible sketch of the graph for f (x ). (See Figure 9.7-1 .)
Figure 9.7-1
24 . (Calculator) (See Figure 9.7-2 .)
Figure 9.7-2
Step 1: Differentiate:
Step 2: Set y = x
Step 3: Solve for y : x2 + y2 = 36 ⇒ y2 = 36 – x2 ;
.
Step 4: Thus, = x ⇒ ± x ⇒ 36 – x2 = x2 ⇒ 36 = 2x2 or .
25 . (Calculator) (See Figure 9.7-3 .)
Step 1: Distance formula:
Figure 9.7-3
Ap Calculus Chapter 5 Review Worksheet
Step 2: Enter:
Use the [Minimum ] function of the calculator and obtain x = .65052 and y 1 = .44488. Verify the result with the First Deriative Test. Enter y 2 = d (y 1(x ), x ) and use the [Zero ] function and obtain x = .65052.
Thus the shortest distance is approximately 0.445.
Review 5ap Calculus Practice