An Unofficial Guide to Math Spring Ellen Weld

Transcription

1 An Unofficial Guide to Math 6 Spring 8 Ellen Weld

2

3 Contents Course Information vii. Course Calendar: Spring 8 vii. Basic Information viii 3. Course Policies ix How to Pass This Class xi Lesson R: Review of Basic Integration. Differentiation versus Integration. Indefinite and Definite Integration 3. Practice Quiz Questions 7 Lesson : Integration by Substitution (I) 9. Motivation 9. Integration by Substitution 9 3. Additional Examples 6 4. Practice Quiz Questions 8 Lesson : Integration by Substitution (II). u-substitution. Average Value of a Function 4 3. Additional Examples 6 4. Practice Quiz Questions 9 Lesson 3: Natural Log Function: Integration 3. Review of Natural Log 3. Examples of u-substitution with Natural Log 3 3. Practice Quiz Questions 38 Lesson 4: Integration by Parts (I) 39. Integration by Parts 39. Why Integration by Parts Works Additional Examples Practice Quiz Questions 47 Lesson 5: Integration by Parts (II) 49. Solutions to In-Class Examples 49. Possible Quiz Problems 54 Lesson 6: Differential Equations: Solutions, Growth, and Decay 55. Separable Differential Equations 55. Proportionality Constants Basic Examples Additional Examples 6 i

4 ii CONTENTS 5. Practice Quiz Questions 64 Lesson 7: Differential Equations: Separation of Variables (I) 67. Separable Differential Equations Examples 67. Practice Quiz Questions 7 Lesson 8: Differential Equations: Separation of Variables (II) 73. Solutions to In-Class Examples 73. Practice Quiz Questions 79 Lesson 9: First Order Linear Differential Equations (I) 8. First Order Linear Differential Equations 8. Practice Quiz Questions 86 Lesson : First-Order Linear Differential Equations (II) 87. Solutions to In-Class Examples 87. Additional Examples 9 3. Practice Quiz Questions 94 Lesson : Area of a Region between Two Curves 95. Area between Two Curves 95. Additional Examples 3. Practice Quiz Questions 3 Lesson : Volume of Solids of Revolution (I) Disk Method 5. Disk Method 5. Practice Quiz Questions Lesson 3: Volume of Solids of Revolution (II) Washer Method 3. Washer Method 3. Additional Examples 3. Practice Quiz Questions 3 Lesson 4: Volume of Solids of Revolution (III) 5. Revolving about Horizontal and Vertical Lines 5. Additional Examples Practice Quiz Problems 36 Lesson 5: Improper Integrals 39. Motivation 39. Review of Basic Limits 4 3. Improper Integrals 4 4. Practice Quiz Questions 46 Lesson 6: Geometric Series and Convergence (I) 47. Introduction to Series 47. Geometric Series Practice Quiz Problems 5 Lesson 7: Geometric Series and Convergence (II) 53. Solutions to In-Class Examples 53. Additional Examples Practice Quiz Problems 6

5 CONTENTS iii Lesson 8: Functions of Several Variables Intro 65. Functions of Several Variables 65. Level Curves Practice Quiz Questions 7 Lesson 9: Partial Derivatives (I) 7. Partial Derivatives 7. Practice Quiz Questions 75 Lesson : Partial Derivatives (II) 77. Second Order Partial Derivatives 77. Practice Quiz Questions 8 Lesson : Differentials of Multivariable Functions Quick Review of Differentials Differentials of Multivariable Functions Solution to In-Class Examples Practice Quiz Questions 88 Lesson : Chain Rule for Functions of Several Variables 89. Chain Rule for Multivariable Functions 89. Practice Quiz Questions 9 Lesson 3: Extrema of Functions of Two Variables (I) 93. Extrema of Multivariable Functions 93. Practice Quiz Questions 99 Lesson 4: Extrema of Functions of Two Variables (II). Solutions to In-Class Examples. Practice Quiz Questions 7 Lesson 5: Lagrange Multipliers Constrained Min/Max (I) 9. LaGrange Multipliers 9. Practice Quiz Questions 6 Lesson 6: LaGrange Multipliers Constrained Min/Max (II) 9. Solutions to In-Class Examples 9. Practice Quiz Questions 3 Lesson 7: Double Integrals, Volume, Applications (I) 5. Integrating Functions of Several Variables 5. Practice Quiz Questions 3 Lesson 8: Double Integrals, Volume, and Applications (II) 33. Double Integrals 33. Practice Quiz Questions 37 Lesson 9: Double Integrals, Volume, and Applications (III) 39. Double Integral Examples 39. Average Value of Functions of Several Variables 4 3. Practice Quiz Questions 43 Lesson 3: Systems of Equations, Matrices, Gaussian Elimination 45. Solutions to In-Class Examples 45

6 iv CONTENTS. Additional Examples Practice Quiz Questions 5 Lesson 3: Gauss-Jordan Elimination 53. Solutions to In-Class Examples 53. Practice Quiz Questions 56 Lesson 3: Matrix Operations 57. Solutions to In-Class Examples 57. Practice Quiz Questions 6 Lesson 33: Inverses and Determinants of Matrices (I) 6. Solutions to In-Class Examples 6. Practice Quiz Questions 64 Lesson 34: Inverses and Determinants of Matrices (II) 67. Solutions to In-Class Examples 67. Practice Quiz Questions 7 Lesson 35: Eigenvalues and Eigenvectors (I) 73. Solutions to In-Class Examples 73. Practice Quiz Questions 78 Lesson 36: Eigenvalues and Eigenvectors (II) 79. Solutions to In-Class Examples 79. Practice Quiz Questions 85 Appendix A. Functions 87. General Theory 87. Overview of Types of Functions Exponential and Logarithmic Basic Trigonometry 89 Appendix B. Basic Differentiation Table 93 Appendix C. Basic Integration Table 95 Appendix D. Exact VS. Decimal Answers 97 Appendix E. Set Builder Notation 99 Appendix F. Answers to Practice Quiz Questions 3 Lesson R: Review of Basic Integration 3 Lesson : Integration by Substitution (I) 3 Lesson : Integration by Substitution (II) 3 Lesson 3: Natural Log Function 3 Lesson 4: Integration by Parts (I) 3 Lesson 5: Integration by Parts (II) 33 Lesson 6: Differential Equations: Solutions, Growth, and Decay 33 Lesson 7: Differential Equations: Separation of Variables (I) 33 Lesson 8: Differential Equations: Separation of Variables (II) 34 Lesson 9: First Order Linear Differential Equations (I) 34 Lesson : First Order Linear Differential Equations (II) 34 Lesson : Area of a Region Between Two Curves 34

7 CONTENTS v Lesson : Volume of Solids of Revolution (I) 35 Lesson 3: Volume of Solids of Revolution (II) 35 Lesson 4: Volume of Solids of Revolution (III) 35 Lesson 5: Improper Integrals 36 Lesson 6: Geometric Series and Convergence (I) 36 Lesson 7: Geometric Series and Convergence (II) 36 Lesson 8: Functions of Several Variables Intro 37 Lesson 9: Partial Derivatives (I) 37 Lesson : Partial Derivatives (II) 37 Lesson : Differentials of Multivariable Functions 38 Lesson : Chain Rule for Functions of Several Variables 38 Lesson 3: Extrema of Functions of Two Variables (I) 38 Lesson 4: Extrema of Functions of Two Variables (II) 38 Lesson 5: Lagrange Multipliers (I) 39 Lesson 6: Lagrange Multipliers (II) 39 Lesson 7: Double Integrals, Volume, and Applications (I) 39 Lesson 8: Double Integrals, Volume, and Applications (II) 39 Lesson 9: Double Integrals, Volume, and Applications (III) 3 Lesson 3: System of Equations, Matrices, Gaussian Elimination 3 Lesson 3: Gauss-Jordan Elimination 3 Lesson 3: Matrix Operations 3 Lesson 33: Inverses and Determinants of Matrices (I) 3 Lesson 34: Inverses and Determinants of Matrices (II) 3 Lesson 35: Eigenvalues and Eigenvectors (I) 3 Lesson 36: Eigenvalues and Eigenvectors (II) 3 Index 33

8

9 Course Information. Course Calendar: Spring 8 Date Lesson Assignment/Topic /8 M R Review of Basic Integration / W Integration by Substitution (I) / F Integration by Substitution (II) No Class /5 M Martin Luther King Jr. Day /7 W 3 The Natural Logarithm Function: Integration /9 F 4 Integration by Parts (I) / M 5 Integration by Parts (II) /4 W 6 Differential Equations: Solutions, Growth, and Decay /6 F 7 Differential Equations: Separation of Variables (I) Exam /9 M R-6 Exam ; Lessons R-6 /3 W 8 Differential Equations: Separation of Variables (II) / F 9 First-Order Linear Differential Equations (I) /5 M First-Order Linear Differential Equations (II) /7 W Area of a Region between Two Curves /9 F Volume of Solids of Revolution (I) / M 3 Volume of Solids of Revolution (II) /4 W 4 Volume of Solids of Revolution (III) /6 F 5 Improper Integrals Exam /9 M 7-3 Exam ; Lessons 7-3 / W 6 Geometric Series and Convergence (I) /3 F 7 Geometric Series and Convergence (II) /6 M 8 Functions of Several Variables Intro /8 W 9 Partial Derivatives (I) 3/ F Partial Derivatives (II) Exam 3 3/5 M 3-9 Exam 3; Lesson 3-9 3/7 W Differentials of Multivariable Functions 3/9 F Chain Rule, Functions of Several Variables vii

10 viii COURSE INFORMATION No Class 3/ M Spring Break No Class 3/4 W Spring Break No Class 3/6 F Spring Break 3/9 M 3 Extrema of Functions of Two Variables (I) 3/ W 4 Extrema of Functions of Two Variables (II) 3/3 F 5 LaGrange Multipliers - Constrained Min/Max (I) Exam 4 3/6 M 8-4 Exam 4; Lessons 8-4 3/8 W 6 LaGrange Multipliers - Constrained Min/Max (II) 3/3 F 7 Double Integrals, Volume, Applications (I) 4/ M 8 Double Integrals, Volume, Applications (II) 4/4 W 9 Double Integrals, Volume, Applications (III) 4/6 F 3 Systems of Equations, Matrices, Gaussian Elimination 4/9 M 3 Gauss-Jordan Elimination 4/ W 3 Matrix Operations 4/3 F 33 Inverses and Determinants of Matrices (I) Exam 5 4/6 M 5-3 Exam 5; Lessons 5-3 4/8 W 34 Inverses and Determinants of Matrices (II) 4/ F 35 Eigenvalues and Eigenvectors (I) 4/3 M 36 Eigenvalues and Eigenvectors (II) 4/5 W Review Review for Final Exam 4/7 F Review Review for Final Exam Finals 4/3-5/5 Finals Week. Basic Information Webpage Grades General Information On Blackboard Course Coordinator: Owen Davis davisok@purdue.edu Instructor: Ellen Weld weld@purdue.edu Webpage weld/ Office Math 639

11 3. COURSE POLICIES ix 3. Course Policies Please be sure to read the syllabus on the course webpage for more general course policies. Calculators: You are allowed a -line scientific calculator on all quizzes and exams. Check the course webpage for a graphic detailing what a -line scientific calculator is. Homework: Homework on each lesson is due the morning before the next lesson. For example, Lesson is on Monday, February 5 and so its corresponding homework is due Wednesday morning, February 7. But the homework for Lesson 5 (Friday, February 6) is due Wednesday, February because Exam is Monday, February 9. All homework is done on Loncapa, an online homework system developed by Michigan State University with problems programmed by Purdue graduate students. You will have tries on each homework problem unless otherwise noted. All homework questions must be directed to Piazza (linked through Blackboard). I will not answer homework questions in my . Exams: About 95% of exam problems come directly from the homework. Up to the numbers in the problems, your exam experience will literally be homework questions from the appropriate sections. The key to succeeding on the exams is to do the homework with the approved calculator and learn how to correctly input answers into Loncapa. You will have 99 tries on each problem. Further, on all open-ended problems, you will receive correct/incorrect feedback; that is, if you answer correctly, you will be told your answer is correct. Exceptions include a few multiple choice problems wherein feedback is not given. Exams are taken during your usual class period but will all be held in a Stanley Coulter computer lab. You will choose your exam room when you sign up for the exam. Quizzes: Quizzes are graded out of points where turning in anything with your name on it will garner point. The only way to get points is to not turn in anything. I will typically give minutes for a quiz that has -3 questions. All potential quiz questions are included in this document in the appropriately titled section. The only difference between the potential quiz questions here and questions on the actual quiz will be the numbers in the problem. Quizzes will only cover past lessons. For example, if I give a quiz on Wednesday, February 8, then it will not cover material from Lesson 9 which was taught that day. Textbook: The official course textbook is found on Loncapa between the homework questions. It includes videos and written examples. Take time to watch the videos and read through the examples because many times they will directly address homework problems. This document is not the course textbook.

12

13 How to Pass This Class Learn to do all the homework questions using only an approved calculator. The typical exam lasts 5 minutes and has questions (the final will be minutes with questions). Because the exam questions come directly from the homework, it is imperative to be able to solve each homework question in an average of 5 minutes and be able to correctly input the answer into Loncapa. Not every problem can be completed in 5 minutes, but many problems should take no more than minutes. Before the exam, look at the homework from the appropriate sections and strategize which problems to focus on first to maximize your efficiency. After each homework is due, the course coordinator will open those homework problems in a practice mode that allows you to try a problem and then randomize the question for more practice. xi

14

15 Lesson R: Review of Basic Integration. Differentiation versus Integration Recall, derivatives tell you how a function is changing integrals tell you how a function is accumulating Theorem (Fundamental Theorem of Calculus (FTC)). (a) (b) b a d dx f (x) dx f(b) f(a) ( x ) f(t) dt f(x) a Fact. The Fundamental Theorem of Calculus (FTC) tells us that derivatives and integrals are opposites, i.e., differentiation undoes integration and integration undoes differentiation.. Indefinite and Definite Integration Definition 3. An antiderivative of a function f(x) is a function F (x) such that F (x) f(x). Ex. If f(x) x, then both 3 x3 + and 3 x3 7

16 LESSON R: REVIEW OF BASIC INTEGRATION are antiderivatives because d dx ( 3 x3 + ) x d ( ) dx 3 x3 7. In fact, we can list all the antiderivatives of x by writing 3 x3 + C where C is any constant. This is called an indefinite integral which we denote x dx 3 x3 + C. Integrals are useful because they describe the area under a curve. This is done using definite integrals. A definite integral will always have a specific interval in mind. Ex. The definite integral of x from to is represented by this picture Figure. This picture is the geometric representation of x dx. For a definite integral, you need bounds. Here, the bounds are x and x (which corresponds to the interval, ). Ex 3 (Definite VS. Indefinite Integrals). The indefinite integral of x is x dx 3 x3 + C which is just a list of all the antiderivatives of x. The definite integral of x from x to x is x dx 3 x3 F T C 3 ()3 3 ( )3 3

17 . INDEFINITE AND DEFINITE INTEGRATION 3 Definite integrals use FTC (a). Some Integral Rule Examples: See the table in the appendix or online textbook for a more comprehensive list. 3 dx 3x + C x dx + x+ + C 3 x3 + C e x dx e x + C sin x dx cos x + C cos x dx sin x + C sec x dx tan x + C sec x tan x dx sec x + C Note 4. Take time to memorize the integral lists provided. Examples.. (6e x + x / ) dx (6e x + x / ) dx 6 6e x dx + e x dx }{{} e x + x / dx x / dx 6e x + + x/+ + C 6e x + 3 x 3/ + C 6e x + 3 x3/ + C Note 5. You only add a single C at the very end of the integral.. What definite integral is described by the following:

18 4 LESSON R: REVIEW OF BASIC INTEGRATION We need two things: () the function and () the bounds. () This line passes through (3, ) and (6, ). Thus, its slope is given by 6 3. Further, the line passes through the point (, ). Recall 3 that point slope form is given by y y m(x x ) which means by function is given by y 3 (x ) y 3 x. () The bounds are x and x 6 because the region lies between these x-values. Thus, the definite integral that is represented by the above graph is 6 3 x dx. 3. What is the value of the definite integral from #? We write x dx 3 x dx ( 3 + x+ 3 x 6 6 x 6 ) 6 6 (6) 6 () x + x 3 x dx This looks tricky, but all you need to do is divide the top through by the bottom. x + x 3 ( ) x x3 dx + dx x x/ x /

19 . INDEFINITE AND DEFINITE INTEGRATION 5 (x 3/ + x 5/ ) dx 5 x5/ + 7 x7/ Find the area of the region bounded by First, draw the graph. y x + 3, y, x, x. Second, set up the definite integral. We want to find the area below x + 3 between x and x. Thus, (x + 3) dx. Finally, compute the definite integral. (x + 3) dx x dx + x dx x+ x + 3x 3 dx dx + 3x

20 6 LESSON R: REVIEW OF BASIC INTEGRATION ( ) () + 3() () + 3() (6 3) Suppose you start driving on the highway at : AM at a speed given by s(t) 5 t + 45 mile/hour. (a) How far have you gone by :3PM? Round your answer to the nearest hundredth. Observe that this problem asks for a rounded answer, not an exact answer. Check Appendix D for an example of the difference between rounded and exact answers. We need to find the distance you have gone, that is, the number of miles your car has accumulated, after.5 hours. Our definite integral is given by.5 ( ) 5 t + 45 dt. So, we need only integrate. Write.5 ( ) 5 t + 45 dt 5 ( ) 5 4 t + 45t t + 45t (.5) + 45(.5) miles. (b) After how many hours will you have gone miles? Round your answer to the nearest hundredth. In the previous question, we were asked about the distance traveled after a certain time. Here, we are given the distance and asked about the time.

21 3. PRACTICE QUIZ QUESTIONS 7 By our work in the previous problem, we have distance traveled time traveled ( ) 5 t + 45 dt. We are given distance traveled. So, if we let x time traveled, then x ( ) 5 t + 45 dt. Thus, our goal is to find x. Write x ( ) 5 t + 45 dt 5 4 x + 45x. Subtracting from both sides, we see we need to solve 5 4 x + 45x. By the quadratic formula, we get x.9 hours.. Evaluate (x + x) dx. 3. Practice Quiz Questions π/4. Evaluate 3 sec (x) dx. π/ 3. Evaluate (e x + 7 cos(x)) dx. x + x 3 4. Evaluate dx. x /3 5. Evaluate sec(x)(sec(x) + tan(x)) dx. 6. Evaluate sin(x) dx. 7. Consider the following region:

22 8 LESSON R: REVIEW OF BASIC INTEGRATION (a) Write down a definite integral that describes the region above. (b) Compute the definite integral from part (a). 8. Consider the following region: (a) Write down a definite integral that describes the region above. (b) Compute the definite integral from part (a).

23 Lesson : Integration by Substitution (I). Motivation As a general rule, differentiation is easy but integration is hard. We have fixed rules to differentiate essentially any function which can be differentiated. However, integration doesn t have a set of rules so much as a collection of methods which work in very specific circumstances. Integration by substitution (more commonly referred to as u-substitution) is the easiest of these methods. This method tackles integrals that are the product of differentiation via the chain rule. Ex. Suppose we want to find. Integration by Substitution x dx. Our list of integral rules didn t include something that looked exactly like this. But x this looks really similar to dx, which we know how to integrate. So let s try to x make it look more like dx. The idea is this: suppose we create a new variable u x. But if we replace immediately, u dx still doesn t look like x dx because the dx doesn t match the new variable u. We need to find a way to get rid of the dx. The key to this is differentiating. Write u x d dx u d (x ) dx du dx du dx. Hence, we have u x and du dx, so our integral becomes x dx u dx u du. 9

24 LESSON : INTEGRATION BY SUBSTITUTION (I) We know how to integrate the right hand side (RHS). Write u du u / du u/+ + + C 3 u3/ + C. But because the problem was originally in terms of x, we must return to x else we have not answered the question that was asked. So we use u x again to get x dx 3 u3/ + C 3 (x }{{ } ) 3/ + C. CHECK: ( ) d dx 3 (x )3/ + C x Note 6. The above check is important because it verifies that differentiation undoes integration. Not sure if you integrated correctly? Differentiate and check if what you get is exactly the function you started with. Note 7. For u-substitution to work, you must completely eliminate the original variable. So we are not allowed to write x dx u dx u because this doesn t help us integrate. This is why it is important to always figure out what the dx is. Ex. Find xe x dx. Again, this doesn t look exactly like something from our list, so we need to rewrite it in a better form. Here, we want to take u x and then determine how dx relates to u (because we need to completely remove x to use u-sub). Write u x d dx u d dx x du dx x du x dx. So, we are ready to make our substitution: xe x dx e x (x }{{ dx} ) du e u du e u + C.

25 . INTEGRATION BY SUBSTITUTION Again, since we started in one variable, we must finish in that variable. So our final answer is xe x dx e u + C e x + C. Examples.. 3x (x 3 + 6) 6 dx One way to do this problem is to multiply the function out and then apply the polynomial techniques we have. But this is time-consuming and leaves a lot of room for error. Another way is to make the following observation: d dx (x3 + 6) 3x. So suppose we take u x 3 + 6, then du dx 3x du 3x dx. Hence 3x (x 3 + 6) 6 dx (x } 3 {{ + 6 } ) 6 (3x }{{ dx} ) u du u 6 du 7 u7 + C. Find 5x e x3 dx 7 (x3 + 6) 7 + C. Let u x 3, then du dx 3x du 3x dx du 3x dx. However, we don t have a factor of 3x dx in our integral we have a factor of 5x dx. How do we resolve this situation? Consider Then it is clear we can write 5x e x3 dx du 3x dx du 3 x dx 5du 3 } 5x {{ dx} What we need to replace. e x3 (5x dx) }{{} replace ( ) 5du e u 3

26 LESSON : INTEGRATION BY SUBSTITUTION (I) 5 3 eu du 5 3 eu + C 5 3 ex3 + C. Alternatively, we could write 5x e x3 dx 5 u x 3 as before, we get du 3x dx du 3 x dx and so 5x e x3 dx x e x3 dx e x3 ( x dx ) e u ( du 3 e u du ) x e x3 dx. Then, letting, 5 3 eu + C 3. Evaluate + 4t dt 5 3 ex3 + C. Take u + 4t, then du dt dt, not 4 dt so we write 4 du 4 dt. Now, we have a factor of instead. Hence, du 4dt du }{{} dt du dt 4 dt + 4t + 4t ( dt) u ( du ) du u/ u / du

27 . INTEGRATION BY SUBSTITUTION 3 ( ) + + C u /+ ( ) u / + C ()u/ + C u / + C + 4t + C. 4. Find a function f such that f (x) 6x(5x ) 5 and f() 3. What is this question asking us to do? We are given a derivative and asked to find the original function. That is, we need to undo differentiation, which means we need to integrate. But a derivative alone is not enough since when we integrate we find a list of antiderivatives. This is why f() 3 is important this lets us figure out the correct C value to find the particular antiderivative. First, we find all possible antiderivatives of f, which means we compute 6x(5x ) 5 dx. This is a u-substitution problem. Let u 5x, then du x du x dx. So, 6du 6x dx and dx we substitute: 6x(5x ) 5 dx (5x ) 5 (6x dx) u 5 (6du) 6u 5 du u 6 + C (5x ) 6 + C.

28 4 LESSON : INTEGRATION BY SUBSTITUTION (I) But we aren t done yet since we are looking for a particular antiderivative. We know f() 3, which means Therefore, our solution is (5() ) 6 + C 3 ( ) 6 +C 3 }{{} C 5. Evaluate f(x) (5x ) sec 9 (5x) tan(5x) dx This one is a bit tricky. Before you choose a u, you need to rewrite the function slightly. Consider sec 9 (5x) tan(5x) (sec 8 (5x))sec(5x) tan(5x) sec(5x) 8 sec(5x) tan(5x). Now, if we take u sec(5x), then du 5 sec(5x) tan(5x) dx. substitute, 5 sec 9 (5x) tan(5x) dx (sec(5x)) 8 (5 sec(5x) tan(5x) ) dx }{{} u u 8 du } {{ } du When we 9 u9 + C 9 (sec(5x))9 + C 6. Suppose an oven is currently at 75 F and is heating up at a rate of t t + deg/min. When will the oven be 5 F? Round your answer to the nearest hundredth. Let T (t) be the function that describes the temperature of the oven at time t in minutes. We are told that T (t) t t + (because it is how fast the temperature of the oven is changing). We can use this to find T (t). Note that t t + dt is a u-substitution problem. Let u t +, then du t dt du t dt. We write t t t + dt + (t dt)

29 . INTEGRATION BY SUBSTITUTION 5 ( ) u du u du ( ) u 3/ + C 3 3 u3/ + C 3 (t + ) 3/ + C. Our temperature function is a particular antideriative of T (t), specifically the one that satisfies T () 75. So, Hence, 75 3 (() + ) 3/ + C 3 + C C 4 3. T (t) 3 (t + ) 3/ Now, we are asked to find when the oven will be 5. This means we are asked to find the t such that Subtracting 4 3 We solve for t: 3 (t + ) 3/ from both sides, we see 3 (t + ) 3/ 6 3 (t + ) 3/ 6 3 (t + ) 3/ 6 3. t + (6) /3 t (6) /3 t ((6) /3 ) / 6. min Note 8. u-sub is a skill that you get better at in time. At first, it is going to be very difficult to spot the appropriate choice of u, but after enough practice it becomes second-nature.

30 6 LESSON : INTEGRATION BY SUBSTITUTION (I) Examples. 3. Additional Examples. The key to u-sub is learning how to choose what to take as u. Suppose we are asked to integrate (cos(x)) sin(x) dx. The correct u is u cos(x). But how do we know that u works? Remember that the goal of u-sub is to completely eliminate the original variable. So, we write d dx u d dx (cos(x)) du dx sin(x) du sin(x) dx du sin(x) dx. Now, we substitute: write cos(x) u and dx du sin(x) to get (cos(x)) sin(x) dx ( (u) sin(x) du ) sin(x) }{{} dx u ( du) u du. From here, we integrate as usual. We write u du 3 u3 + C 3 (cos(x))3 + C. Remember, you always return to the original variable. Suppose instead that you had chosen u (cos(x)). Why doesn t this work? Well, we have d dx u d dx (cos(x)) du sin(x) cos(x) by chain rule dx du sin(x) cos(x) dx

31 3. ADDITIONAL EXAMPLES 7 Then, if we substitute (cos(x)) du u and dx, our integral sin(x) cos(x) becomes ( ) (cos(x)) du sin(x) dx u sin(x) u sin(x) cos(x) cos(x) du. }{{} dx But this is WRONG because we have not completely eliminated the original variable. In summary, to determine if you have found a good choice for u, check that you have eliminated the original variable.. Find the function f(x) whose tangent line has the slope ( + x) / x for any x and whose graph passes through the point (9, 7). What is f(x)? Remember, the slope of the tangent lines of a function is the derivative of the function, that is to say, f (x) ( + x) / x. So, we need to integrate and then use f(9) 7 to find the particular antiderivative. We want f (x) dx ( + x) / x dx. This is a u-substitution problem. Take u + x, then u + x + x / du dx x/ du dx x / du dx x / du dx x du x dx du x dx

32 8 LESSON : INTEGRATION BY SUBSTITUTION (I) Thus, ( + x) / dx x ( + ( ) x) / x dx u / ( du) u / du + u/+ + C 3 u 3/ + C 3 u3/ + C 3 ( + x) 3/ + C Now that we have the list of all antiderivatives of f, we need to find the particular antiderivative specified by f(9) 7. So, 7 3 ( + 9) 3/ + C 3 ( + 3)3/ + C 3 43/ + C 3 (8) + C C C 3 3 C Putting this all together, we get f(x) 3 ( + x) 3/ Practice Quiz Questions. Evaluate 3xe x dx.

33 . Evaluate 3. Evaluate 4. Evaluate 5. Evaluate 6. Evaluate 7. Evaluate 8. Evaluate 4 x + dx. 4. PRACTICE QUIZ QUESTIONS 9 (cos (x)) sin(x) dx. x (x 3 + 4) 9 dx. x 3 x 4 + dx. sin(5x) dx. ( ) x cos dx. 3 x 9 3 9t dt.

34

35 Lesson : Integration by Substitution (II). u-substitution In lesson, we addressed how to evaluate indefinite integrals using u-sub, but how do we compute definite integrals? Fortunately, not much changes: we go about choosing our u in the same way, but this time we need to be careful with the bounds. Ex. Evaluate 5 x dx. Take u x, then du dx. So that s easy enough. But the bounds as given are relative to x. When we substitute, we need to change our bounds to be relative to u. 5 something x dx something u du We know that u x and the bounds of our integral in terms of x are x 5 and x, so our bounds in terms of u are just u evaluated at the bounds of x: Hence, 5 u(5) 5 4 and u() 9. u() x dx u(5) 9 u du 4 u du. Since everything is in terms of u and a definite integral gives a number, we don t need to go back to x. So 5 9 x dx u du du u u / du u / du ( + ) u /+ 9 4

36 LESSON : INTEGRATION BY SUBSTITUTION (II) Ex. Evaluate 5 3x x dx. u This looks very similar to the Ex. but the key difference is that there is another factor of x on the top. We might ignore this difference and use the same substitution as before: take u x and du dx. Then, when we substitute, we get () 5 3x x dx u() u(5) 3x u du. But this isn t quite right since we still have an extra x that we need to eliminate. Here s where we need to be a little clever. Observe that u x also means u + x. So, instead of equation (), write 5 3x x dx u() u(5) 3(u + ) u du 9 4 3u + 3 u This way, we have completely eliminated x and we can integrate as usual. Write 9 4 3u + 3 u du ( 3u + 3 ) du u u ( 3u u + 3 ) du / u / (3u / + 3u / ) du du. ( ) ( ) u / u /+ 4 ( ) 9 3 u 3/ + 3()u / 3 4 u 3/ + 6u / 9 4 (9) 3/ + 6(9) / ( (4) 3/ + 6(4) /) (7) + 6(3) (8) 6() Examples. 44.

37 . Evaluate. u-substitution 3 ln ln xe x dx. Let u x, then du dx bounds for x, x du x dx. Further, evaluating u at the u( ln ) ( ln ) ln and u( ln ) ( ln ) ln. So, ln ln xe x dx ln ln ln ln e u ln ln e x (x dx) }{{} du e u du e ln e ln. If 9. 6(x 4 + x ) f(x) dx (x 4 + x ) 3 + C, find f. The point of this question is to emphasize the connection between integration and differentiation; in particular, that they undo one another. If we differentiate an integral, we get the integrand and if we integrate a derivative, we get an antideriative (this is what FTC states). How do we apply this to the example? By FTC: () 6(x 4 + x ) f(x) d ( dx ) 6(x 4 + x ) f(x) dx But, we are told that 6(x 4 + x ) f(x) dx (x 4 + x ) 3 + C, which means that equation () becomes (3) 6(x 4 + x ) f(x) d dx ( (x 4 + x ) 3 + C ). }{{} 6(x 4 + x ) f(x) dx

38 4 LESSON : INTEGRATION BY SUBSTITUTION (II) So, d ( (x 4 + x ) 3 + C ) 3(x 4 + x ) (4x 3 + x) by chain rule dx 3(x 4 + x ) ((x + x)) 6(x 4 + x ) (x 3 + x). By equation (3), 6(x 4 + x ) f(x) d ( (x 4 + x ) 3 + C ) 6(x 4 + x ) (x 3 + x) dx 6(x 4 + x ) f(x) 6(x 4 + x ) (x 3 + x) Dividing both sides by 6(x 4 + x ), we conclude f(x) x 3 + x.. Average Value of a Function We now address finding the average value of a function over a specified interval. Ex 3. Suppose we wanted to find the average value of the function f(x) x + over the interval x 4. We first want to draw a picture. To find the average of a bunch of numbers we add them up and then divide by how many there are. We use the same idea here: we add up all the functions values (which is the integral) and then divide by the length of the interval (which is sort of

39 like the number of things). So Average Value 4. AVERAGE VALUE OF A FUNCTION 5 4 (x + ) dx ( ) 4 x + x ( ( )) (4) + 4 () + (8 + 4 ) 4. Definition 9 (Average Value of a Function over an Interval). If f(x) is defined on an interval a, b, then the average value of f(x) over a, b is Examples. b a b a f(x) dx. 3. Find the average value of f(x) x + over the interval x. Here, a and b, so by our formula, b a b a f(x) dx (x + ) dx. Thus Average Value (x + ) dx (x + x) (4 + ) Suppose during a hot dog eating contest you eat 3 t + hot dogs/minute. What is the average number of hot dogs you eat in the first 4 minutes of the contest? We are asked to find the average value of the function 3 t + over the interval t 4. By our formula, the average value is given by t + dt. Observe this is a u-sub problem with bounds. Take u t +, then du dx du dt du dt. Moreover, u() () + and u(4) (4) + 9.

40 6 LESSON : INTEGRATION BY SUBSTITUTION (II) Hence, t + dt u(4) u() 9 ( 3 3 u du 3 u du ) 9 u 3/ 4 (9)3/ 4 ()3/ hotdogs Examples. 3. Additional Examples. Evaluate 7t 3t + 4 dt. Solution: This is a u-substitution problem, although in a more subtle way. Let u 3t + 4, then du 3dt du dt. But we re not quite done yet 3 since we cannot write 7t 3t + 4 dt 7t ( ) du u 3 because we must completely eliminate the original variable. Instead, we observe that u 3t + 4 t u 4 3 and write 7t ( u 4 3t + 4 dt 7 3 ( (u 4) u du. ) ( ) du u 3 ) (u 4) u du

41 3. ADDITIONAL EXAMPLES 7 So now we can solve the integral as usual, even if it requires more computation than we would like. (u 4) u du (u 8u + 6) u du (u 4/ 8u / + 6)u / du (u 5/ 8u 3/ + 6u / ) du 7 u7/ 8 ( ) ( ) u 5/ + 6 u 3/ + C u7/ 6 5 u5/ u3/ + C 7 (3t + 4)7/ 6 5 (3t + 4)5/ (3t + 4)3/ + C. You would get full credit on quiz for leaving your answer in the above form.. Consider the graph of f(x) x + area under the curve is, what is a? over the interval x a. If the Solution: If a question asks for the area under a curve, that means you are going to need to compute a definite integral (since a definite integral is the area under a curve). So the question comes down to whether you can set up the appropriate integral. Because the interval is x a (which is the same as, a) and we know the area under the curve is, we have a f(x) dx a x + dx. Let s compute the RHS. Take u x +, du dx. Then We know that So this becomes a x + dx u(a) u() u du. u() + and u(a) a +. u(a) u() a+ u du u du.

42 8 LESSON : INTEGRATION BY SUBSTITUTION (II) Now we continue on as we would with any other definite integral: a+ a+ u du du u/ a+ u / du ( ) a+ + u /+ u / a+ u a+ a + a +. The question asks for the value of a such that the area under the curve is. But this means that Since we know a a x + dx. x + dx a +, to solve for a we just need to figure out Adding to both sides, we get a +. a + 4 a + 3 a. 3. Suppose a population is changing at a rate of P (t) 3t such that its initial population is,. What is the average size of the population in the first years? Solution: For this problem, you need to be very clear on what everything represents. Firstly, 3t dt is not correct? Why not? Well, the question doesn t ask for the average in the change of the population (that s P (t)), but the average in the actual population (which is P (t)). Thus, they want you to find the area under the curve of P (t), not P (t).

43 4. PRACTICE QUIZ QUESTIONS 9 So we need to find P (t) and the C value that the initial population specifies. We see that P (t) P (t) dt 3t dt t 3 + C. Since P (), we see that C. Our population function is then P (t) t 3 +. Next, the average of the population over the first years is (t 3 + ) dt ( ) 4 t4 + t 5 members. 4. A certain plant grows at a rate of H (t) te t inches/day, t days after it is planted. By how many inches will the height of the plant change on the fourth day? Solution: This problem comes down to really understanding the wording. If H(t) is the height of the plant t days after it is planted, then t <, }{{} t <, }{{} t < 3, }{{} 3 t < 4. }{{} Day Day Day 3 Day 4 Moreover, the question asks only for how the plant changes on the fourth day. Not any other day. Thus, the integral that we need to compute is: 4 3 te t dt. Let u t, then du t dt and u(3) 3, u(4) 4. Write 4 3 te t dt 4 3 eu du eu 6 9 e6 e9. Evaluate 4. Practice Quiz Questions x x dx.. Evaluate 7 x dx.

44 3 LESSON : INTEGRATION BY SUBSTITUTION (II) 3. Find the average value of the function f(x) 9x + 3 over the interval x Evaluate 9(x 3)(x 3 9x) 5 dx. 5. Find the average of the function x e x3 over the interval x 4.

45 Lesson 3: Natural Log Function: Integration. Review of Natural Log Consult Appendix A.?? for more details. Natural log (denoted ln x) is the function inverse of e x, that is, ln e x x and e ln x x. Ex. ln e and e ln(37t+4) 37t + 4. Ex. Observing that e e and e, show that ln e and ln. Because e x >, ln x is only defined for x in the interval (, ). This means that ln and ln( 7) are not valid but ln 3 makes perfect sense. This does not mean that ( ) lnx can t be negative. In fact, ln.693. This statement means that the input of ln x cannot be negative (but this has nothing to do with its output). Figure. Graph of ln x () a ln b ln b a Properties of ln x () ln(ab) ln a + ln b ( a ) (3) ln ln a ln b b Ex 3. 3

46 3 LESSON 3: NATURAL LOG FUNCTION: INTEGRATION is, () 3 ln ln 3 ln 8 () ln 6 ln( 3) ln + ln 3 (3) ln 3 ln ( ) 6 ln 6 ln Ex 4. The rules above do not apply in the case of addition or subtraction, that ln(3 + ) ln 3 + ln nor ln(3 ) ln 3 ln. Moreover, (ln(x)) 6 ln(x 6 ). You can check these using your calculator. Recall, dx ln x + C when x. x Note that x is the only exception to the polynomial integration rule. Further, x notice the absolute value symbols ( ). This is important because ln x only takes positive values and the absolute values allow us to compute integrals like 4 4 ( ) ( ) dx ln x 4 6 x ln 4 ln 6 ln 4 ln 6 ln ln }{{} by (3) without incident. We may drop the absolute values if we know the input is always non-negative. That is, ln x + ln(x + ) because x + > for all x. However, ln 3x + ln(3x + ) because 3x + < whenever x < 3. Ex 5. Observe that d dx (ln(x)) x. But even more is true: for f(x) a function, More concretely, d dx (ln(f(x))) f (x) f(x). d dx (ln(x + x + )) x + x + x +. Remark. The lesson today is using u-sub with ln x. Examples.. Examples of u-substitution with Natural Log

47 . EXAMPLES OF u-substitution WITH NATURAL LOG 33. Evaluate x (ln(x)) 6 dx Since x x, x (ln(x)) 6 dx ( ) (ln(x)) 6 dx. x So, if we take u ln x, then du dx and we may write x ( ) ( ) (ln(x)) 6 dx (ln x) 6 x x dx u 6 du 7 u7 + C 7 (ln(x))7 + C.. Evaluate tan x dx. Here, we need to observe that tan sin x and so cos x sin x tan x dx cos x dx. Let u cos x, then du sin x dx du sin x dx. Therefore, sin x tan x dx cos x dx (sin x dx) cos x }{{} du u du since (cos x) cos x ln u + C ln cos x + C ln cos x + C ln (cos x) + C by () ln sec x + C sec x.

48 34 LESSON 3: NATURAL LOG FUNCTION: INTEGRATION 3. Find x + x dx. Let u + x, then du 4x dx du dx du Moreover, u() + () and u() + () So we may write x + x dx (x dx) + x u() ( ) du u() u 9 ( ) du u ln u 9 x dx. 4. Find 7 x /3 ( + x /3 ) dx. ln(9) ln() ln(9) ln(9 / ) ln 3 We might be tempted here to try to rewrite this by x /3 ( + x /3 ) x /3 + x /3 x /3 x /3 + x /3+/3 x /3 + x ( ) and then taking u x /3 + x. But then du 3 x /3 + dx, which is not a factor up to a constant in this integral (try doing this substitution on your own to determine that you can t eliminate the original variable). Instead, leave the integral as it is and take u + x /3, then du 3 x /3 dx 3 du x /3 dx. Finally, notice that x /3. Thus, x/3 7 7 ( ) x /3 ( + x /3 ) dx dx + x /3 x/3 u(7) u() (3 du) u

49 . EXAMPLES OF u-substitution WITH NATURAL LOG 35 Since u(7) u() 3 u du u() + /3 + and u(7) + 7 / , we have 5. Evaluate 7 e 4 e u(7) x /3 ( + x /3 ) dx 3 u() u du x ln(x) dx. 4 3 u du 4 3 ln u 3 ln(4) 3 ln() 3(ln(4) ln()) ( ) 4 3 ln by (3) 3 ln() ln( 3 ) ln 8 by () Take u ln(x), then du dx. Further, x So u(e) ln(e) ln(e ) and u(e 4 ) ln(e 4 ) 4. e 4 e e 4 x ln(x) dx e u(e 4 ) u(e) 4 u du 4 ln u ( ) ln x x dx u du ln(4) ln() ln(4) ln(4)

50 36 LESSON 3: NATURAL LOG FUNCTION: INTEGRATION 6. Suppose a population of penguins changes at a rate of P (t) e t ln()( + e t ) penguins/year and that the current population is penguins. (a) What is the penguin population after years? Round your answer to the nearest penguin. Since P (t) is the change in population, we need to find P (t) (which is the number of penguins at year t) and then compute P (). Write P (t) dt e t ln()( + e t ) dt. So, let u + e t, then du e t dt which means e t ( ) ln()( + e t ) dt e t dt ln + e t ln ln() u du u du ln u + C ln() ln() ln + et + C. This lists all the possible P (t), but we need to find the specific P (t) that satisfies P (). Write P () ln() ln + e + C P () ln( + ) + C ln() ln() ln() + C + C C 98 Thus, P (t) ln() ln( + et ) + 98.

51 . EXAMPLES OF u-substitution WITH NATURAL LOG 37 Thus, to answer the question, we compute P () ln() ln( + e ) + 98,69 penguins. (b) What is the average change in the penguin population from now to years from now? Round your answer to the nearest penguin. We are asked to find the average change in the penguin population. Hence, by our formula for function averages, we need to compute By FTC, P (t) dt. P (t) dt P () P () ln( + e ) ln( + e ) ln() 8 penguins 7. Suppose a factory produces sponges at rate of s (t) 3t + t 3 + t + thousand sponges/day. Find the total amount of sponges created in the factory s first week (which is 7 days) of production. Round your answer to the nearest hundred. by The number of sponges (in thousands) created in the first week is given 7 3t + t 3 + t + dt. Let u t 3 + t +, then du (3t + ) dt. Write 7 3t + t 3 + t + dt 7 u(7) t 3 + t + (3t + ) dt u() u du u(7) ln u u() ln t 3 + t + 7 ln((7 3 ) + (7) + ) ln(( 3 ) + () + )

52 38 LESSON 3: NATURAL LOG FUNCTION: INTEGRATION ln((7 3 ) + 5) ln }{{} ln(358) 5.9 thousand sponges. Evaluate 3. Practice Quiz Questions x (ln x) 8 dx.. Evaluate 3. Evaluate x + x dx. x ln x dx. cos(ln(x)) 4. Evaluate dx. x 5. Evaluate x 3/4 ( + x /4 ) dx. 6. The slope f (x) at each point (x, y) on a curve y f(x) is given by f (x) 4x + x. Find f(x) if f(x) passes through the point (, 4). 7. Find the average value of the function over the interval, 9. f(x) et + e t

53 Lesson 4: Integration by Parts (I). Integration by Parts We introduce another method for evaluating integrals called integration by parts. The key is the following : (4) u dv uv v du MEMORIZE this formula. Ex. Suppose we want to evaluate xe x dx. Here, u-substitution fails to produce anything useful. integration by parts. Instead, we need to use To use integration by parts, we must apply equation (4) which means we must identify our u, du, v, dv. The integrand is u dv, which means once u is chosen, everything else is dv. Then, we find du by differentiating u and we find v by integrating dv. u du derivative of u dv everything leftover v dv For the example give above, we should take u x. This means that everything leftover in the integral (e x dx) is our dv. To get our du and v, we differentiate and integrate respectively: u x du dx dv e x dx v e} x {{ dx} e x dv Note. In this context, whenever we integrate dv, we always assume the constant is. Now that we have each part of equation (4), we just plug in our values: u dv uv v du }{{} x u e} x {{ dx} }{{} x dv u 39 e x }{{} v e x }{{} v }{{} dx du.

54 4 LESSON 4: INTEGRATION BY PARTS (I) Thus, our integral becomes xe x dx xe x } {{ } original question e x dx xe x e x + C Note. We add a single C at the end of the integration by parts process. CHECK: d dx (xex e x + C) xe x Integration by parts requires attention to detail and is a difficult method to apply, but it is very useful. The trickiest part of integration by parts is determining u. To choose an appropriate u, we use the following: L - Logarithms like ln x, ln(x 3 + ), etc I - Inverse trig functions (not for this class) A - Algebraic functions like x, x 3 + x + 7, polynomials (NO ROOTS) T - Trig functions like cos x, tan x, etc E - Exponential functions like e x, x, etc We use this list as follows: if ln x appears in the integral, then we take u ln x. If not, then we let u be whatever inverse trig function is present. If neither of these, we take u to be a polynomial, etc. Ex. To evaluate (3x + x )e x dx, we take u 3x + x and get the following table: u 3x + x du (6x + ) dx dv e x dx v e x dx e x Note 3. The above is not really a set of hard and fast rules but rather a rule of thumb for choosing u. However, for this class, LIATE should be sufficient. Examples.. Evaluate x ln x dx By LIATE, we take u ln x. Then we get the following table: u ln x du x dx dv x dx v x dx x

55 We apply equation (4), x ln x dx ln x. INTEGRATION BY PARTS 4 }{{} u ( x ) }{{} v x ln x x ln x ( x }{{} v ) x ( x x dx ) ( ) x dx }{{} du dx x ln x 4 x + C. Find π/3 x cos x dx This is a definite integral but we still apply the same method. By LIATE, u x. So u x dv cos x dx du dx v cos x dx sin x By equation (4), we write π/3 x cos x dx }{{} x sin }{{} x u v π/3 x sin x x sin x x sin x π/3 π/3 π/3 π/3 π/3 sin }{{} x }{{} dx v du sin x dx ( cos x) + cos x π/3 π/3 π/3 x sin x + cos x π ( π ) ( π ) 3 sin + cos }{{ 3 } 3 } sin {{ } + }{{} cos }{{} 3 ( ) π 3 + 3π 3 6

56 4 LESSON 4: INTEGRATION BY PARTS (I) (5) 3. Evaluate For definite integrals, we have the following integration by parts formula: e x ln(x 9 ) dx b By LIATE, u ln(x 9 ). Then a u dv uv b a b a v du Thus, d dx ln(x9 ) u ln(x 9 ) du 9 x dx d dx (x9 ) x 9 9x8 x 9 dv x dx 9 x v x dx x We write e ( ) x ln(x 9 ) dx ln(x 9 ) }{{} x u }{{} e x ln(x 9 ) 9 e x ln(x 9 ) 9 e x ln(x 9 ) 9 v e e ( ) ( ) 9 x x dx }{{}}{{} v du ( ) x dx x e e x ln(x 9 ) 9 4 x e x dx ( x ) e e ln(e 9 ) 9 }{{} 4 e () ln( 9 ) 9 }{{} 4 () 9 9 e 9 4 e e Find x 3 + x dx Sometimes you need to be very clever in how you choose your u even when following LIATE. If you take u x 3, then dv dx. But we + x

57 . INTEGRATION BY PARTS 43 don t know how to integrate this dv. Instead, take u x which leaves x dv dx. Note that + x So we write u+x x dx dux dx + x ( u du u / du ( + ) u / u x dv u + x. ) u /+ x + x dx du x dx v + x Note that finding v is itself a u-substitution problem. Thus we can write x 3 dx + x (x ) ( + x }{{} ) ( + x }{{} ) (x dx) }{{}}{{} u v v du x + x x + x dx. } {{ } Where ( ) is, again, a u-substitution problem. Therefore, x 3 + x dx x + x x + x dx } {{ } x + x 3 ( + x ) 3/ + C Note 4. To use LIATE effectively, choose a u that gets simpler when you differentiate and a dv which you know how to integrate. 5. Suppose a certain plant is growing at a rate of te t inches per day t days after it is planted. What is the height of the plant at the beginning of the third day (assuming it is planted as a seed on the first day)? x u+x + x ux dx u dx du 3 u3/ 3 ( + x ) 3/

58 44 LESSON 4: INTEGRATION BY PARTS (I) Solution: Because of how we are measuring t, we make the observation t <, }{{} t <, }{{} t < 3. }{{} Day Day Day 3 Let H(t) be the height of the plant t days after it is planted. We know that H() (since it was planted as a seed) and we want to find H() because t corresponds to the beginning of day 3. Note that H (t) te t because this is the rate of change of the height of the plant. Now, the integral te t dt is an integration by parts problem. By LIATE, we choose u t dv e t dt du dt v e t dt e t. So by equation (4), te t dt (t) (e t ) }{{}}{{} u v (e t ) }{{} v (dt) te t e t + C (t )e t + C. }{{} du We need to find H(t) given our initial condition H(). ( )e + C + C C. Thus, Finally, H(t) (t )e t +. H() ( )e + e +. The plant is e + inches tall at the beginning of the third day.. Why Integration by Parts Works Suppose we have functions f(x) and g(x). Then, by the product rule, f(x)g(x) f (x)g(x) + f(x)g (x). If we integrate both sides, then f(x)g(x) dx f (x)g(x) dx + f(x)g (x) dx. But integration undoes differentiation by FTC, so f(x)g(x) dx f(x)g(x).

59 3. ADDITIONAL EXAMPLES 45 Therefore, f(x)g(x) dx } {{ } f(x)g(x) f(x)g(x) f (x)g(x) dx + f (x)g(x) dx + f(x)g (x) dx f(x)g (x) dx Now, subtract f (x)g(x) dx from both sides, f(x)g(x) f(x)g (x) dx f (x)g(x) dx. Finally, let u f(x) du f (x) dx dv g (x) dx v g (x) dx g(x) to get f(x) g (x) dx f(x) g(x) }{{}}{{}}{{}}{{} u dv u v g(x) f (x) dx. }{{}}{{} v du 3. Additional Examples Examples. z +. Evaluate dz. e z First, we rewrite the integral a little. Since e z e z, we can write z + dz (z + )e z dz e z (z e z + e z ) dz z e z dz } {{ } ( ) + e z dz. We see that e z dz can be done via u-substitution but it is not immediately clear what z e z dz is. We turn our attention to ( ). } {{ } ( )

60 46 LESSON 4: INTEGRATION BY PARTS (I) Solving ( ): This is an integration by parts problem. By LIATE, we take u z du z dz dv e z dz v e z Then we simply plug this into equation (4): ( ) (z ) (e z dz) (z ) ( e z ) }{{}}{{}}{{}}{{} u dv u v z e z ( ) ( e z ) (z dz) }{{}}{{} v du ze z dz z e z + ze z dz. } {{ } ( ) Because ze z dz does not have an obvious antiderivative and u-substitution fails, we run into another situation wherein we need to use integration by parts. We focus on finding ( ). Solving ( ): By LIATE, we take u z du dx dv e z dz v e z Then by equation (4): ( ) (z) (e z ) (z) ( e z ) }{{}}{{}}{{}}{{} u u dv v ze z e z dz ze z ( ) ze z + ze z e z. e z dz ( e z ) (dx) }{{} v e z dz }{{} du Finally, we put it all back together. z + Putting it all back together: We were originally given the integral dz. By our work, we can now write z + dz e z z e z dz } {{ } ( ) + e z dz e z

61 4. PRACTICE QUIZ QUESTIONS 47 z e z + ze z dz } {{ } ( ) + e z dz z e z + ( ze z e z ) + z e z ze z e z e z + C z e z ze z 3e z + C e z dz. Compute. Compute 3. Compute 4. Practice Quiz Questions xe x dx. x ln x dx. x cos x dx. 4. Compute 5. Compute 6. Compute e/4 /4 e (x )e x dx. x ln(4x) dx. ln x x 3 dx.

62

63 Lesson 5: Integration by Parts (II). Solutions to In-Class Examples The following are solutions to examples done in class. Example. Find the area under the curve of f(x) x(x 3) 6 over the interval x 3. Solution: The area under a curve is given by a definite integral. In this case, the definite integral is 3 x(x 3) 6 dx. This can be computed using integration by parts, but it is easier to use u-substitution. Take u x 3, then du dx. Note u x 3 u + 3 x and so we can write 3 x(x 3) 6 dx u(3) u() (u + 3)u 6 du. Since u x 3, evaluating at x and x 3, we find Hence, Finally, we evaluate: 3 u() 3 3 and u(3) 3 3. u(3) u() (u + 3)u 6 du (u + 3)u 6 du 3 3 (u 7 + 3u 6 ) du 8 u u7 3 (u + 3)u 6 du. 8 ()8 + 3 ( 7 ()7 8 ( 3)8 + 3 ) 7 ( 3)7 ( ) 8 7 ( ) ( )

64 5 LESSON 5: INTEGRATION BY PARTS (II) Example. Suppose a turtle is moving at a speed of 8(t + ) 3 ln(t + ) /9 miles/hour. How far does the turtle travel in half an hour? Round your answer to the nearest thousandth. Solution: Here, we want to integrate the function 8(t + ) 3 ln(t + ) /9 because we need to know the distance accumulated by the turtle between and hours. This is described by / 8(t + ) 3 ln(t + ) /9 dt. u-substitution will not work here and so we need to use integration by parts. However, we should first simplify our integral. By our rules about ln x, We write / 8(t+) 3 ln(t+) /9 dt ln(t + ) /9 ln(t + ). 9 / 8 9 (t+)3 ln(t+) dt / (t+) 3 ln(t+) dt. Next, by LIATE, we take u ln(t + ), which means dv (t + ) 3 dt. Observe that ut+ ( ) (t + ) 3 dudt dt u 3 dt 4 u4 4 (t + )4. Hence, integrating dv, ( ) (t + ) 3 dt (t + ) 3 dt ( ) (t + ) 4 4 (t + )4. Thus, our table becomes u ln(t + ) du t + dt dv (t + ) 3 dt v ( ) (t + )4 So we write / (t + ) 3 ln(t + ) dt ( ) / ln(t + ) (t + )4 }{{} u }{{} v / (t + )4 ln(t + ) / / ( ) (t + )4 dt } {{} t + }{{} v du (t + ) 3 dt

65 . SOLUTIONS TO IN-CLASS EXAMPLES 5 (t + )4 ln(t + ) / ( ) (t + )4 4 }{{} by ( ) (t + )4 ln(t + ) 8 (t + )4 / ( + ( ) 4 3 ln ( 34 3 ln 5.59 miles ) 4 ( ln ( ) 3 8 ) / ) + ( ) 4 ( 8 + ( ) 4 ( 3 ln() ) 8 ( + )4 ln( + ) ( + )4 8 4 ln(7t + ) Example 3. A factory produces pollution at a rate of tons/week. (7t + ) 3 How much pollution does the factory produce in a day? Round your answer to the nearest hundredth. Solution: Our function measures output in terms of weeks but we are asked about the pollution produced in a day. Hence, the integral we must compute is /7 4 ln(7t + ) (7t + ) 3 dt. ) However, this form looks somewhat unwieldy, so we introduce a small cosmetic change to our integral. Set x 7t +, then dx 7dt dx 7 dt with x() 7() + and x(/7) 7(/7) +. Thus, /7 4 ln(7t + ) 4 ln(x) dt (7t + ) 3 x 3 ( ) dx ln(x) dx. 7 x 3 So now, we need only compute the far right integral and we have solved the problem. Our integral cannot be computed using u-substitution, so we apply integration by parts. By LIATE, we take u ln(x), and so dv dx. Since x3 v x dt 3 x 3 dt 3 + x 3+ x x x,

66 5 LESSON 5: INTEGRATION BY PARTS (II) our table becomes u ln(x) dv x 3 dx Therefore, we write ln(x) x 3 du x dx v x dx ln(x) ( x ) }{{} u ln(x) x ln(x) x } {{ } v ln(x) x ln(x) + x ln() + ( ) ln + ( ) + x dx 3 x x ( x ) ( ) }{{ x dx }}{{} v du ( ln() + ) ( ) + ln() + ( ) 8 ( ln() + ) ln(). tons Example 4. Suppose the probability of a gold necklace having a gold purity of x percent (so x ) is given by P (x) 9e3 e 3 4 xe 3x. Find the probability that a gold necklace has a purity of at least 75%. Round your answer to the nearest percent. Solution: We want our gold necklace to have a purity of at least 75%. Hence, 75 x.75 x. Thus, the question comes down to computing.75 9e 3 e 3 4 xe 3x dx 9e3 e xe 3x dx.

67 . SOLUTIONS TO IN-CLASS EXAMPLES 53 For the moment, let s focus on the integral and forget about the constant out front. We want to solve.75 xe 3x dx. This is an integration by parts problem, which means it comes down to choosing the correct u and dv. By LIATE, let u x and dv e 3x dx. Integrating dv, we get ( ) v dv e 3x dx u 3x du 3 dx 3 eu du 3 eu 3 e 3x. So our table becomes u x du dx dv e 3x dx v 3 e 3x We write.75 xe 3x dx }{{} x ( 3 ) e 3x u }{{} v xe 3x + e 3x dx }{{} 3 xe 3x.75 9 e 3x 3 xe 3x 9 e 3x.75 ( ).75 ( 3 e 3x ) } {{ } v dx }{{} du 3 e 3 9 e 3 3 (.75)e 3(.75) 9 e 3(.75) Hence, 9e 3 xe 3x dx e So the probability is 8%. 4 e 3(.75) + 9 e 3(.75) 3 e 3 9 e 3 ( 4 + ) ( e 3(.75) ) e 3 9 ( 9e3 e ) ( e 3(.75) ) e 3 9.8

68 54 LESSON 5: INTEGRATION BY PARTS (II). Evaluate. Evaluate. Possible Quiz Problems x(x + 3) 7 dx. xe 3x dx. 3. A certain chemical reaction produces a compound X at a rate of t (t + 8) 3 kg/hr, where t is time in hours from the start of the reaction. How much of the compound is produced during the first hours of the reaction? Round your answer to 3 decimal places. 4. Suppose a car has a velocity of 4te t/4 where t time in hours. How far has the car traveled in an hour? Round your answer to decimal places.

69 Lesson 6: Differential Equations: Solutions, Growth, and Decay. Separable Differential Equations For this lesson, be sure to read the online textbook because it has some great examples. In particular, be sure to read the tank example since this will become especially relevant on later homework. Definition 5. A differential equation is an equation that includes one or more derivatives of a function. Ex. dy dt 8y, y t cos y, y x 3 y + xy, and dy dt (cos t)y + t + y are all 3 examples of differential equations. Definition 6. A differential equation is called separable if it can be written in the form dy dx f(x)g(y). Ex. dy dt 8y, y tcos y are separable y x 3 y + xy, dy dt (cos t)y + t + y are NOT separable 3 Ex 3. We show that dy dx x e 3y x4 We need to rewrite this as dy dx is separable. (function of x) (function of y). Write x e 3y x4 x e 3y e x4 x e 3y So this differential equation is separable. e x4 ( ) x (e 3y ) Definition 7. A solution to a differential equation is a function that you can plug into the differential equation and have it be true. 55 e x4 only x only y

70 56 LESSON 6: DIFFERENTIAL EQUATIONS: SOLUTIONS, GROWTH, AND DECAY Ex 4. A solution to is y(x) x 7 because dy dx d dx (y(x)) d dx dy dx x y ( x 7) } {{ } y(x) d dx (x 7) / (x 7) / (x) by chain rule ( ) (x) x 7 x x 7 x y Definition 8. A particular solution is a solution without any unknowns. Think of this as solving for C after an indefinite integral when you have initial conditions. Examples.. Find the particular solution to the following: dy dx x ; if y when x. y This is a separable differential equation because ( ) x y x. y only x only y We like separable differential equations because we can do the following: dy dx x y y dy dx x y dy x dx y dy x dx

71 . PROPORTIONALITY CONSTANTS 57 3 y3 3 x + C Note 9. We only add one C, and it doesn t matter what side you put it on. I tend to put it on the RHS but there is nothing wrong with putting it on the left hand side (LHS). Now, the solution we are looking for is a function y(x). So we need to solve for y. Write 3 y3 3 x + C y 3 x 3 + 3C y 3 x 3 + 3C 3 x 3 + C We call y(x) 3 x 3 + C the general solution because there are unknowns. Our final step is to find the appropriate C. We were told that y when x, so 3 () 3 + C 3 C. Raising both sides to the third power, we get C. Going back to our general solution, y 3 x 3 + C 3 x 3 +. Therefore, the particular solution is y 3 x 3 +. Note. The does NOT go on the outside of the cube root.. Proportionality Constants Differential equations are useful in modeling a variety of situations like populations and radioactive decay. A particular type of differential equation that appears frequently in this class is one that involves proportionality constants. A common theme is a substance or population changing proportionally to a function of itself. We address what this means through a series of examples. Ex 5. Suppose a population of deer in Pulaski County, Arkansas changes proportionally to itself. Use a differential equation to describe this phenomenon. Let y(t) be the population of deer in Pulaski County at time t. The change in population is described by the derivative of the population function, which is y (t) dy. The phrase changes proportionally to itself is represented by dt y dy dt ky

72 58 LESSON 6: DIFFERENTIAL EQUATIONS: SOLUTIONS, GROWTH, AND DECAY where k is called the proportionality constant. This constant depends on the specifics of what is being modeled and is often something which must be found when solving for the particular solution. Ex 6. Suppose after a February snow storm in Madison, Wisconsin the snow melts at a rate of.4 times the square of the number of inches of snow on the ground. How would we model this situation? Let A(t) be the number of inches of snow. Since the snow is melting at a rate of.4 times the square of A(t), our differential equation is given by A da dt.4a. Observe here that k.4 because the snow is melting which means A(t) is decreasing and so it must have a negative derivative. Ex 7. Assume the cost of ice cream changes inversely proportional to the temperature outside. Write down a differential equation to describe this situation. Suppose C(x) is the cost of ice cream where x is degrees Fahrenheit. Our differential equation is given by C (x) k C(x). Ex 8. Suppose in a particular group of, people, the change in number of people in a Ponzi is jointly proportional to the number of people in the scheme and the number of people not in the scheme. Use a differential equation to describe this situation. Let P (t) be the number of people in the Ponzi scheme at time t. Then, the number of people not in the scheme is given by, P (t). Joint proportionality means Examples. P (t) kp (t)(, P (t)). 3. Basic Examples. Find the particular solution to the differential equation y ky given y(), y () 4. Our first step for solving to y is to find k (which is the proportionality constant). We are told that y() and y () 4. Because our differential equation is y (t) ky(t), this is true for any t, including, 4 y () ky() k. Hence, we have 4 k k. Substituting y y. This is a separable differential equation and we can apply the same technique as in Example. Write y dy dt y

73 3. BASIC EXAMPLES 59 y dy dt dy dt y y dy dt ln y t + C There are two equally valid ways to proceed from here: we can first solve for y and then solve for C, or we can first solve for C and then solve for y. In this situation, we will first solve for C and then solve for y. The initial condition y() implies that ln y(t) t + C ln y() () + C if t ln( }{{} y() Next, we need only solve for y. ) C To undo ln, we apply e to both sides. So, we write ln y t + ln() }{{} C e ln y e t+ln() y e t+ln() So we have y e t. e t e ln() e t () e t. The absolute value here leaves us with a choice: either y e t or y e t. But, because y(), we conclude that y e t. 3. Suppose P (t) is the mass of a radioactive substance at time t. If P (t) 3 P (t), find the half-life of the substance. Definition. The half-life of a substance is the amount of time it takes for half of the substance to disappear.

74 6 LESSON 6: DIFFERENTIAL EQUATIONS: SOLUTIONS, GROWTH, AND DECAY This may appear to be an impossible problem because we are not told how much of the substance we initially have. For the time being, let A be the original amount of the substance. By definition of half-life, we want to find the t such that P (t) A. We are given P (t) 3 P (t), which is separable. So we write dp dt 3 P dp P dt 3 P dp 3 dt ( P dp 3 ) dt ln P 3 t + C e ln P e 3 t+c P e 3 t+c Because we never expect a substance to have a negative amount, we can assume that P is never negative and write P e 3 t+c. Here, we need to make the following observation: we assumed that at time t, we have an amount A. This means P () A. But, by our equation above, P () e 3 ()+C }{{} e e C e C. Therefore, A e C. Why is this useful? This lets us write e 3 t+c A e 3 t }{{} e C A A e 3 t A A e 3 t ln e 3 t ln ( )

75 3. BASIC EXAMPLES 6 ) 3 t ln ( ) t 3 ln ( Thus, the half-life of this substance is t 3 ln ( ). Here, we note that the half-life is independent of the initial amount of substance. Further, this t might appear to be a negative number (which wouldn t make sense) but note that ln x < whenever x <. Note (Useful Info). If P (t) kp (t), then half-life is given by ln ( ). k The general solution to P (t) kp (t) is P (t) Ae kt where P () A. 4. Find the general solution to dy dt We assume here that 5 y >. k(5 y). This is a separable differential equation and looks very similar to Example 3. However, we need to be very careful when we integrate. Write dy k(5 y) dt dy 5 y dt k dy k dt 5 y 5 y dy k dt Observe that dy is a u-substitution problem: let u 5 y, then 5 y du dy (note the negative). Hence 5 y dy du ln u ln 5 y. u

76 6 LESSON 6: DIFFERENTIAL EQUATIONS: SOLUTIONS, GROWTH, AND DECAY So ln 5 y kt + C ln 5 y kt + C We now want to solve for y. Write ln 5 y kt + C e ln 5 y e kt+c 5 y e kt+c. Since we assumed that 5 y > we may drop the absolute values and so our equation becomes Then, solving for y, we have 5 y e kt+c. 5 e kt+c + y 5 e kt+c y. If we let C e C, then our general solution is y 5 + Ce kt. Examples. 4. Additional Examples. Suppose a pot roast was 75 F when removed from an oven and set in a 7 F room. If after minutes the pot roast is 6 F, what is its temperature after an hour? Round your answer to the 4 th decimal place. This problem requires Newton s Cooling Formula: dt dt k(t S) where T (t) is the temperature function and S is the ambient (surrounding) temperature. We can choose to measure t in minutes or hours. Here, we measure in minutes. We are given T () 75, T () 6, S 7. Write We need to compute dt k(t 7) dt dt k dt T 7 T 7 dt T 7 dt. k dt.

77 Let u T 7, then du dt. So T 7 dt u du 4. ADDITIONAL EXAMPLES 63 ln u + C ln T 7 + C. Thus T 7 dt k dt ln T 7 kt + C We apply e to both sides to undo the natural log. We get ln T 7 e}{{} T 7 T 7 Ce kt. e kt+c e kt e C }{{} C Here we drop the absolute value because e x is always positive. Evaluating at t, we see that T () 7 }{{} e k() C C 75 7 C 5 C. Thus, which becomes T 7 5e kt T 5e kt + 7. We re not done yet since to get a particular solution and we still haven t found k. Evaluating at t, we have Putting this together, 6 T () 5e k() e k ek ( ) 6 ln k 7 ( ) 6 ln k. 7 T (t) 5e ln( 6 7)t + 7.

78 64 LESSON 6: DIFFERENTIAL EQUATIONS: SOLUTIONS, GROWTH, AND DECAY Now, to find the temperature after an hour, take t 6. Then T (6) 5e ln( 6 7)(6) Solve the initial value problem for y as a function of t when y t n with y() 8 where n is a constant and n. We are told that y is a function of t, which means that y dy dt. Thus, our differential equation becomes Next, we integrate: y t n dy dt tn dy t n dt. y dy t n dt. Here, we need to be careful. Our integration rules tell us that we must consider two cases: either n or n. We observe that if n, then we would have y t dt ln t. However, we are told that y() 8 but ln() is not defined. Therefore, n else our function y doesn t make sense. Thus, Since n, Now, since y() 8, we have We conclude that t n dt n + tn+ + C. y n + tn+ + C. 8 n + ()n+ +C C. }{{} y(t) n + tn Practice Quiz Questions. Suppose that y ky, y() 5, and y () 3. What is y as a function of t?

79 . Solve for y as a function of t given 5. PRACTICE QUIZ QUESTIONS 65 y 35 t y. 3. If P (t) is the mass of a radioactive substance at time t such that find the half-life of the substance. P (t) 4P (t), 4. Solve for y as a function of t given y ln t. 5. A bacterial culture grows at a rate proportional to its population. If the population is, at t hour and, at t hours, find the population of the culture as a function of time t. 6. Determine if dy dx 5ey x3 is a separable differential equation.

80

81 Lesson 7: Differential Equations: Separation of Variables (I) Examples.. Separable Differential Equations Examples. Find the general solution to dy dt + 5y. Since we are finding a general solution, we are not asked to solve for all the unknowns. Write dy dt + 5y dy dt 5y dy 5 dt y y dy ( 5) dt ln y 5t + C e ln y e 5t+C y e 5t+C. Here, we have a choice for y: either y e 5t+C or y e 5t+C. For this class, we re always going to want our functions to be non-negative, so we drop the absolute values and write Hence, y e 5t+C e 5t e C }{{} C y Ce 5t. Ce 5t.. A bacterial culture grows at a rate proportional to its population. Suppose the population is initially, and after hours the population has grown to 5,. Find the population of bacteria as a function of time. 67

82 68 LESSON 7: DIFFERENTIAL EQUATIONS: SEPARATION OF VARIABLES (I) This is asking us to find a particular solution. Let P (t) be the population of bacteria at time t hours. We are told that population grows at a rate proportional to its population, which means we have the following differential equation: P (t) kp (t) for k the proportionality constant which we will need to find. Further, we are told P (), and P () 5,. We will first find the general solution to P (t) kp (t) and then use P () and P () to find the particular solution. We write P (t) kp (t) dp dt kp dp k dt P P dp k dt ln P kt + C e ln P e kt+c P e kt+c P e kt+c e C e kt Ce kt where we drop the absolute values because we assume the population is never negative. Thus, P (t) Ce kt. Next, P (), implies Further, since P () 5,, C }{{} e k(), C,. 5,,e k().5 e k ln(.5) ln e k ln(.5) k ln(.5) k Therefore, P (t), e ln(.5) t.

83 . SEPARABLE DIFFERENTIAL EQUATIONS EXAMPLES 69 Note 3. We could have also remembered that the general solution to P (t) kp (t) is P (t) Ce kt where C P (). Then we could have skipped to the step that used P () to find k. 3. Find the particular solution to the equation da dt ( A) such that A(), A < for all t. The fact that A < for all t tells us that A >. So, da dt A da dt A A da The LHS is a u-sub problem. If u A, then du da. So A da du ln u ln A. u dt Thus, A da dt ln A t + C ln A t C We have assumed that A >, so we may drop the absolute values. Hence, ln A t C ln( A) t + C e ln( A) e t+c A e t+c e t+c A So our solution is of the form }{{} e C e t A. C A Ce t. We were told that A(), which means Ce C C.

84 7 LESSON 7: DIFFERENTIAL EQUATIONS: SEPARATION OF VARIABLES (I) Therefore, A e t. 4. Find the general solution to dy dx x6 (6 + y). Write dy dx x6 (6 + y) 6 + y dy x6 dx 6 + y dy x 6 dx ln 6 + y 7 x7 + C 6 + y e 7 x7 +C But as we did in Example, we may assume that 6 + y is non-negative so we drop the absolute value. This means we have Thus, 6 + y e 7 x7 +C Ce 7 x7. y Ce 7 x Suppose the volume of a balloon being inflated satisfies dv dt 5 V where t is time in seconds after the balloon begins to inflate. If the balloon pops when it reaches a volume of 4 cm 3, after how many seconds will the balloon pop? Round your answer to 3 decimal places. We are given dv dt 5 V (V ) /5 V /5. So, dv dt V /5 dv dt V /5 V /5 dv dt

85 . SEPARABLE DIFFERENTIAL EQUATIONS EXAMPLES 7 V /5 dv dt 5 3 V 3/5 t + C. Since V (), Therefore, we want to solve for t. But this is just 5 3 ()3/5 () + C C. 5 3 (4)3/5 t t 6 (4)3/ seconds. 6. A wet towel hung on a clothesline to dry outside loses moisture at a rate proportional to its moisture content. After hour, the towel has lost 5% of its original moisture content. After how long will the towel have lost 8% of its original moisture content? Round your answer to the nearest hundredth. Let M(t) be the percentage of the original moisture content the towel has after t hours. Then M (t) km(t) for k the proportionality constant. We know M() because we assume at time t that the towel has not lost any moisture content and M() 5 85 because after hour the towel is less 5% of its moisture. Our goal is to find the time t such that M(t) 8. is The general solution to a differential equation of the form M (t) km(t) So, we need to solve for k. Since M() 85, M(t) M()e kt. M(t) e kt which implies 85 e k() e k,.85 e k ln(.85) k. We want to find the t such that M(t) e ln(.85)t. Write. e ln(.85)t,

86 7 LESSON 7: DIFFERENTIAL EQUATIONS: SEPARATION OF VARIABLES (I) which, after applying ln to both sides, becomes ln(.) ln(.85)t t ln(.) ln(.85). Practice Quiz Questions. Find the general solution to dy + y. dx 9.9 hours.. Find the particular solution to dy dt that y > for all t. y such that y(). Assume 3. Find the general solution to dy dx x e x3. 4. Find the particular solution to dy dt + y sin t where y(π). 5. Find the particular solution to dy dx x + y such that y 6 when x.

87 Lesson 8: Differential Equations: Separation of Variables (II). Solutions to In-Class Examples Example. Find the general solution to x 3 y y + x e y. Solution: We first want to move y to one side and everything else to the other. We have x 3 y y + x e y x 3 y y x e y (x 3 )y x e y y x e y x 3 dy dx x e y x 3 Now, we want to separate the function. Multiply both sides by e y. Then we get e y dy dx x e y e y x 3 e y dy dx x e y+y x 3 e y dy dx x e x 3 e y dy dx x x 3 e y dy x x 3 dx. Our next step is integration, that is, e y dy x x 3 dx. The RHS is a u-substitution problem. Let u x 3, then du 3x dx du 3 x dx. So 73

88 74 LESSON 8: DIFFERENTIAL EQUATIONS: SEPARATION OF VARIABLES (II) x x 3 dx u 3 ( ) 3 du u du ln u + C 3 3 ln x3 + C. Thus, e y dy x x 3 dx e y 3 ln x3 + C. Next, we need to get y by itself so apply ln to both sides. We get ( ) ln e y ln 3 ln x3 + C ( ) y ln 3 ln x3 + C. Note 4. Because this is a general solution, we don t need to find C; we can leave the unknowns as unknowns. Moreover, ln(a + b) ln a + ln b. So the solution is NOT the same as ( ) ln 3 ln x3 + C. Example. Find y() if y is a function of x such that xy 6 y and y when x. Solution: This is a separable function, so our first step should be to move x to the RHS. Write xy 6 y y 6 y x y 6 dy dx x y 6 dy x dx. Now, we integrate:

89 . SOLUTIONS TO IN-CLASS EXAMPLES 75 y 6 dy x dx 7 y7 ln x + C. We are given that y when x, so to find C we can write: 7 y7 ln x + C 7 ()7 ln +C }{{} Hence, Thus C 7 7 y7 ln x + 7. The question asks us to find y(). So we need to get y by itself first. Write 7 y7 ln x + 7 y 7 4 ln x + y 7 4 ln x +. y() 7 4 ln() +. Example 3. Suppose during a chemical reaction, a substance is converted into a different substance at a rate inversely proportional to the amount of the original substance at any given time t. If, there were initially grams of the original substance and after an hour only half remained, how much of the original substance is there after hours? Solution: Let y(t) be the amount of the original substance in grams at time t hours. We are told that the rate of change is inversely proportional to the amount of the substance. That means our differential equation is given by dy dt k y where k is an unknown constant that we will need to find. Ultimately, the question asks us to find y() given the following information: dy dt k y y() y() 5.

90 76 LESSON 8: DIFFERENTIAL EQUATIONS: SEPARATION OF VARIABLES (II) We see our differential equation is separable, so we write Next, we integrate: dy dt k y y dy k dt. y dy k dt y kt + C. Because we are told y(), we have So our equation becomes Moreover, since y() 5, () k() + C C 5. y kt + 5. (5) k() k k 75 k. Solving for y, y 75 t + 5 y 75t + We are asked to find y(), so y 75t +. y() 75() which doesn t exist. What does this tell us? That all of the original substance is gone and the model has fallen apart. This sometimes happens where a model works only for certain input.

91 . SOLUTIONS TO IN-CLASS EXAMPLES 77 What does this mean answer-wise? It means that the answer is because all the substance is gone. Example 4. A 5-gallon tank initially contains 5 gallons of brine, a salt and water combination. Brine containing pounds of salt per gallon flows into the tank at a rate of 4 gallons per minute. Suppose the well-stirred mixture flows out of the tank at a rate of gallons per minute. Set up a differential equation for the amount of salt (in pounds) in the tank at time t (minutes). Solution: We are asked to set up the equation but not to solve it. For this type of problem, units are very important. Let A(t) be the pounds of salt in the tank at time t minutes. By how we have defined our function, the units associated to da dt are lbs/min. Hence da dt Rate of salt in lbs/min Rate of salt out lbs/min. Rate of salt in: Every minute, 4 gallons flow into the tank and each gallon contains pounds of salt. So ( ) ( ) lbs 4 gal Rate of salt in salt in per minute gal min }{{}}{{} salt in water in per gal per min 8 lbs/min Rate of salt out: The difficulty here is understanding what well-stirred means. Well-stirred means that each gallon in the tank has as much salt in it as any other gallon. Thus, to correctly interpret well-stirred, we need to take the total amount of salt in the tank and divide it by the total amount of liquid in the tank. There are initially 5 gallons of brine and each minute, gallons are added to the tank (to see that gallons are added to the tank each minute, consider 4 gal/min gal/min + gal/min). So, we see that ( ) ( ) A(t) lbs gal Rate of salt out salt out per minute 5 + (4 )t gal min }{{}}{{} salt out water out per gal per min A(t) 5 + t lbs/min. Therefore, our differential equation is in out da dt Rate of salt in Rate of salt out 8 A(t) 5 + t.

92 78 LESSON 8: DIFFERENTIAL EQUATIONS: SEPARATION OF VARIABLES (II) Note 5. This is not a separable equation. To actually solve for this A(t), we need a different technique which we will cover in Lesson 9. Fortunately, we are only asked to set up the differential equation. Example 5. A 7-gallon tank initially contains 4 gallons of brine containing 5 pounds of dissolved salt. Brine containing 6 pounds of salt per gallon flows into the tank at a rate of 3 gallons per minute, and the well-stirred mixture flows out of the tank at a rate of 3 gallons per minute. Find the amount of salt in the tank after minutes. Round your answer to 3 decimal places. Solution: Let A(t) be pounds of salt in the tank at time t minutes where we are given A() 5. Our goal is to find A(). As above, we have da dt Rate of salt in: Rate of salt in lbs/min Rate of salt out lbs/min. ( ) ( ) 6 lbs 3 gal 8 lbs/min salt in per minute. gal min }{{}}{{} salt in water in per gal per min Rate of salt out: ( ) ( ) A(t) lbs 3 gal 3A(t) 4 (3 3)t gal min 4 }{{}}{{} salt out water out per gal per min lbs/min salt out per minute. Hence, our differential equation becomes This is a separable equation. da dt 8 in 3A(t) 4 out 7 3A. 4 Note 6. The difference between this example and Example 4 is that the amount of water coming into the tank and the amount of water leaving the tank cancel each other out and this eliminates the t on the RHS. This transforms this problem into a separation of variables differential equation. We can write On your own verify that ( ) 4 da dt. 7 3A A Ce 3t/ (Use u-sub with u 7 3A, du 3 da.)

93 Since we are given A() 5,. PRACTICE QUIZ QUESTIONS 79 Thus, So, 5 C } e 3()/4 {{} +4 C + 4 C 35. A 35e 3t/ A() 35e 3/ lbs.. Practice Quiz Questions. Find the general solution to 3x y y + 7xe y.. Find y() if y is a function of x such that xy 5 y and y when x. 3. In a particular chemical reaction, a substance is converted into a second substance at a rate proportional to the square of the amount of the first substance present at any time t. Write down a differential equation that describes this situation. 4. A 5-gallon tank initially contains 4 gallons of pure distilled water. Brine containing 4 pounds of salt per gallon flows into the tank at a rate of 5 gallons per minute, and the well-stirred mixture flows out of the tank at a rate of 5 gallons per minute. Find the amount of salt in the tank after minutes. (Round your answer to three decimal places.)

94

95 Lesson 9: First Order Linear Differential Equations (I). First Order Linear Differential Equations Today we introduce a new type of differential equation called a first order linear differential equation (FOLDE) and describe the method by which we solve them. FOLDE are of the form (6) dy dt + P (t)y Q(t) Observe the following about equation (6): (a) the + is very important. Any must be included in the P (t) (b) the y being multiplied by the P (t) is to the first power (c) P (t) and Q(t) do not include any y, they are functions only of t If you cannot write the differential equation exactly in the form of equation (6), then it is not a FOLDE. Ex. dy dt + 3y t t is a FOLDE with P (t) 3 t and Q(t) t y x y sin x is a FOLDE with P (x) x and Q(x) sin x Note 7 (FOLDE vs Separation of Variables). FOLDE is related to, but different, from separation of variables. Sometimes these methods overlap but most of the time only one method will apply. Take time to practice identifying when a particular method will apply. The key to solving this type of differential equation is an integrating factor, that is, the function (7) u(t) e P (t) dt. The solution of a FOLDE is found via (8) y u(t) Q(t)u(t) dt. y is the actual solution so, after we integrate the RHS, we must divide through by u(t). 8

96 8 LESSON 9: FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS (I) To apply the FOLDE method, take the following steps:. Find P, Q. Find the integrating factor 3. Set up equation (6) Examples.. Find the general solution to Step : Find P, Q P (x) x, Q(x) x Step : Find the integrating factor dy dx + y x x. u(x) e P (x) dx e x dx ln x e x ( ) x Note 8 ( ). We are cheating here and using the simplifying assumption that x >, which would mean that x x. Step 3: Set up equation (8) y u(x) y }{{} x u(x) Q(x)u(x) dx (x) }{{} Q(x) x dx (x) dx }{{} u(x) 3 x3 + C We divide both sides by x to get y 3 x + C x. Note 9. It is important that when you divide through by x, you also divide the C as well. C absorbs constants, not functions.

97 . FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS 83. Given t y + ty 5 and y(), find y(6). This is a FOLDE but it isn t in the right form (we want it to look like equation (6)). Write t y + ty 5 y + ty t 5 t y + y t 5 t Now, since our new equation looks like equation (6), we can proceed. Step : Find P, Q P (t) 5, Q(t) t t Step : Find the integrating factor u(t) e P (t) dt e t dt ln t e t ( ) t ( ) Again, we cheat and assume that t >. Step 3: Set up equation (8) y u(t) y }{{} t u(t) y t Q(t)u(t) dt ( ) 5 t }{{} Q(t) 5 t dt 5 ln t + C (t) dt }{{} u(t) 5 ln(t) + C since t > So we have which means y t 5 ln(t) + C y 5 ln(t) t + C t.

98 84 LESSON 9: FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS (I) We have found the general solution, but we were asked to find y(6) given y(). Hence, Since y(), Finally, 3. Find the general solution to 5 ln() + C }{{ } C. y y(6) 5 ln(t). t 5 ln(6) 6 dy dt + y t. Again, this is not in the same form as equation (6). So we rewrite: dy dt + y t dy dt y t. Now, we can apply our method. Step : Find P, Q P (t), Q(t) t Step : Find the integrating factor u(t) e P (t) dt e dt e t Step 3: Set up equation (8) y u(t) y }{{} e t u(t) ye t Q(t)u(t) dt ( t) }{{} Q(t) (e t ) dt }{{} u(t) te t dt }{{} integration by parts ( )

99 . FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS 85 We find ( ) using integration by parts. By LIATE, u t du dt dv e t dt v e t dt e t So, te t dt }{{} t ( e t ) }{{} u v te t e t dt ( e t ) ( dt) }{{}}{{} v du te t + e t + C. Therefore, we have ye t t ( ) te te t + e t + C which means y t + + Ce t. Observe that the only unknown constant here is C. The is not absorbed by an arbitrary constant because it is not arbitrary. 4. Find the general solution to (y ) sin x dx dy. This equation is both a FOLDE and separable. We find the solution using the FOLDE method. However, it is certainly not in the proper form for either method. We write (y ) sin x dx dy (y ) sin x dx dy (y ) sin x dy dx y sin x sin x dy dx sin x dy (sin x)y dx Now this is in the form as given in equation (6). Step : Find P, Q P (x) sin x, Q(x) sin x Step : Find the integrating factor u(x) e P (x) dx e sin x dx e cos x

100 86 LESSON 9: FIRST ORDER LINEAR DIFFERENTIAL EQUATIONS (I) Step 3: Set up equation (8) y u(x) y }{{} e cos x u(x) ye cos x Q(x)u(x) dx ( sin x) (e cos x ) dx }{{}}{{} Q(x) u(x) (sin x)e cos x dx } {{ } u-sub ( ) We find ( ). Let u cos x, then du sin x. So (sin x)e cos x dx e u du e u + C e cos x + C We write which means ye cos x e cos x + C y + C e cos x Again, the only unknown here is C and you may not replace the with an arbitrary constant because it is not arbitrary.. Practice Quiz Questions. Find the general solution to dy dx y x x4.. Find the general solution to y + (tan x)y cos x for < x < π. 3. Find y(4) if t y + ty 7 and y() Find the general solution to 5 dy dt + y t. 5. Find the general solution to (y ) sin x dx dy.

101 Lesson : First-Order Linear Differential Equations (II). Solutions to In-Class Examples Example. Suppose a silo contains 5 tons of grain and that a farmer is moving the grain to another silo. If the amount of grain in the second silo changes at a rate proportional to the amount of grain in the first silo, find a differential equation that represents this situation. Solution: Let y(t) be the amount of grain in the second silo. The amount of grain in the first silo is given by 5 y(t). Hence, our differential equation becomes dy dt k(5 y). Note 3. Write k (5 y) in Loncapa. Example. Find the integrating factor of (sin x)y (cot x)y cos x, < x < π 4. Solution: The < x < π is only meant to indicated where a valid solution exists 4 as a function. We do not explicitly use this information. cos x First, recall that cot x. So we are really looking at sin x ( ) cos x (sin x)y y cos x. sin x Second, this is a FOLDE but it isn t in quite the correct form. Divide both sides by sin x to get ( y cos(x) (sin(x))(sin(x)) ( cos(x) sin (x) y ) y cos(x) sin(x) ) y cos(x) sin(x) In this form, it is clear that ( ) cos(x) P (x) sin. (x) 87

102 88 LESSON : FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS (II) Because we are only asked to find the integrating factor, we need only determine P (x). The integrating factor u(x) is given by the formula So we will need to find This is a substitution problem. P (x) dx u(x) e P (x) dx. ( ) cos(x) sin dx. (x) Let t sin(x), then dt cos(x) dx, which means ( ) cos(x) sin dx (x) t du Therefore, our integrating factor is t sin(x) csc(x) u(x) e P (x) dx e csc x. Example 3. A store has a storage capacity for 5 printers. If the store currently has 5 printers in inventory and the management determines they sell the printers at a daily rate equal to % of the available capacity, when will the store sell out of printers? Solution: Let N(t) be the number of printers in the store s inventory at t days. The available capacity of printers is given by 5 N (which is the total capacity minus the amount of printers actually in the store). Moreover, our proportionality constant is k. (this is negative because we want to sell printers until there are none left which is to say the number of printers is decreasing). Our differential equation is given by dn.(5 N). dt This is a separation of variables problem. We need to get N all on one side and t all on the other. Write dn. dt 5 N 5 N dn. dt ln 5 N.t + C ln 5 N.t C e ln 5 N e.t+c

103 . SOLUTIONS TO IN-CLASS EXAMPLES 89 5 N Ce.t Now, since we are assuming that we will never have more printers than the store s capacity, 5 N. So 5 N Ce.t 5 Ce.t N. We are told that the store originally has 5 printers, which means N() 5. Thus 5 N() 5 Ce.() 5 C C 5. So, N(t) 5 5e.t. The question asks us to find when the store will sell out of printers, which is to say we need to find the t such that N(t). Write Therefore, our answer is 5 5e.t 5 5e.t e.t ln ln e.t ln.t ln t. t ln 6.93 days. Example 4. An 85-gallon tank initially contains 5 gallons of brine containing 5 pounds of dissolved salt. Brine containing 4 pounds of salt per gallon flows into the tank at a rate of 5 gallons per minute. The well-stirred mixture then flows out of the tank at a rate of gallons per minute. How much salt is in the tank when it is full? (Round your answer to the nearest hundredth.) Solution: Let A(t) be the pounds of salt in the tank at time t minutes. Then da dt ( 4 lbs Rate of salt in: gal ( Rate of salt out: Rate of salt in Rate of salt out. ) ( ) 5 gal lbs/min min A(t) lbs 5 + (5 )t gal Hence, our differential equation is ) ( ) gal A(t) min 5 + 3t lbs/min da dt A(t) 5 + 3t.

104 9 LESSON : FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS (II) To find the solution, we use the FOLDE method. However, our differential equation isn t quite in the correct form. So we write da dt A(t) 5 + 3t da dt + A 5 + 3t da ( ) dt + A t Now, since this is now in the right form we can go through the steps. Step : Find P, Q P (t), Q(t) 5+3t Step : Find the integrating factor Recall the integrating factor is given by u(t) e P (t) dt. We need to integrate P (t). Write P (t) dt 5 + 3t dt ln 5 + 3t 3 ln(5 + 3t) /3. Thus u(t) e ln(5+3t)/3 (5 + 3t) /3. Step 3: Set up solution equation A u(t) Q(t)u(t) dt. Plugging in what we know, we have A (5 + 3t) /3 }{{}}{{} (5 + 3t) /3 dt }{{} u(t) Q(t) u(t) Dividing both sides by u(t) (5 + 3t) /3, A 4(5 + 3t) + ( 3 5 C (5 + 3t) /3. ) ( ) (5 + 3t) 5/3 + C. 3 We need to find C. We are told that the initial amount of salt in the tank is 5 pounds, so 5 A() 4(5) + C 5 /3 C 5/3 (95). We leave this in the exact form and our equation becomes A(t) 4(5 + 3t) (5)/3 (95) (5 + 3t) /3.

105 . ADDITIONAL EXAMPLES 9 Now, we aren t quite done. We want to find A(t) for the t at which the tank is full. Since the amount of liquid in the tank is given by 5 + 3t, we write t 6 3t t. Thus, the tank is full when t. Finally, A() 4(5 + 6) (5)/3 (95), lbs. (5 + 6) /3 Examples.. Additional Examples. Suppose the height of a particular plant is given by the function h(t) where t is measured in days. If the plant grows at a rate of h th+t inches per day, how long will it take for the plant to grow to 3 feet? (Round your answer to the nearest hundredth.) Observe that the height of the plant is measured in inches, which means we are asked to find the t such that h(t) 36. Now, h th + t h th t is a FOLDE and so we apply our method. Step : Find P, Q P (t) t, Q(t) t Step : Find the integrating factor u(t) e P (t) dt e t dt e t / Step 3: Set up the solution equation h u(t) Q(t)u(t) dt he t / t e t / dt e t / + C h + Ce t / Since we may assume that h(), we see that + Ce + C C. Thus, h(t) + e t /.

106 9 LESSON : FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS (II) We need to solve h(t) 36 for t: 36 + e t / 37 e t / ln 37 t / ln 37 t ( ln 37) / t. We conclude it takes t ( ln 37) /.69 days.. A, ft 3 room initially has a radon level of 9 picocuries/ft 3. A ventilation system is installed that brings in 45 ft 3 of air per hour which contains picocuries/ft 3, while an equal quantity of well-mixed air leaves the room each hour. Set up and use a differential equation to determine how long it will take for the room to reach a safe-to-breathe level of picocuries/ft 3. (Round your answer to the nearest hundredth.) This is an extremely tricky problem and it takes a lot of patience for both the set up and solution. First, we should notice that, in principle, this should be very similar to the tank problem because it involves quantities entering, mixing, and leaving a space. Let r(t) be the amount of radon (in picocuries) per ft 3 per hour (note that this is different than the total amount of radon in the room per hour). So then our setup should look like r (t) Rate of Radon/ft 3 in Rate of Radon/ft 3 out. This is a little difficult to work with because none of our information is really given in picocuries/ft 3 per hour. So let s write it as r (t) Rate of Total Radon in Rate of Total Radon out Total ft 3 in Room Rate of Total Radon in Rate of Total Radon out., Rate of Total Radon in: There are picocuries of radon entering the room per ft 3 of air per hour, we write this as ( ) ( picocuries 45 ft 3 ) 4,5 picocuries ft 3 4,5 picocuries/hour. hour hour This is the total amount of radon entering the room per hour. Rate of Total Radon out: We are told that 45 ft 3 leaves the room per hour. This is represented as of well-mixed air

107 . ADDITIONAL EXAMPLES 93 ( ) ( total radon in room in picocuries 45 ft 3 ), ft 3. hour But observe that r(t) total radon in room in picocuries, ft 3. Hence, the rate out is just 45r(t) picocuries/hour. Putting this all together, we have r (t) Rate of Total Radon in Rate of Total Radon out, 4,5 45r(t), 4,5, 45r(t), 9 9r(t) Therefore, our differential equation is r (t) 9 9r(t). But we aren t done yet, because we are asked to find the t such that r(t). So we must solve this differential equation. This is a FOLDE, but it isn t quite in the correct form yet, so we write r (t) + 9r(t) 9. Now, we can use our steps to solve it. Step : Find P, Q P (t) 9, Q(t) 9 Step : Find the integrating factor u(t) e P (t) dt e 9 dt e 9t/ Step 3: Set up the Solution r(t) u(t) Q(t)u(t) dt

108 94 LESSON : FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS (II) ( 9 r(t) e} 9t/ {{} u(t) ) }{{} Q(t) e 9t/ }{{} u(t) dt r(t)e 9t/ e 9t/ + C r(t) + C e 9t/ We were told that r() 9, hence we see that C 9. So we conclude that r(t) + 9e 9t/. Finally, we want to find t such that r(t). We write + 9e 9t/ 9e 9t/ 9 e 9t/ ( ) ln(e 9t/ ) ln 9 ( ) t ln t 9 ln ( ) 5.7 hours. 3. Practice Quiz Questions. Find the general solution to t dy dt + y t where t >.. Find the integrating factor of (cos x)y + (tan x)y sec x where < x < π Suppose a bank account contains $, and that the account s owner is moving the money to another account. If the money is being moved from the first account to the second at a rate inversely proportional to the amount of money in the first account, write down a differential equation that describes the change in the amount of money in the second account. 4. A 5-gallon tank initially contains gallons of brine containing 85 pounds of dissolved salt. Brine containing 3 pounds of salt per gallon flows into the tank at a rate of 4 gallons per minute and the well-stirred mixture flows out of the tank at a rate of gallon per minute. How much salt is in the tank when it is full? (Round your answer to the nearest hundredth.)

109 Lesson : Area of a Region between Two Curves. Area between Two Curves Given a function, we know how to find the area under its curve. A natural next question is: if we have two functions, how do we determine how much area is trapped between them? Ex. We know how to find the area underneath f(x) x. This is just the integral ( x ) dx. But what if we wanted to find the area between f(x) x and g(x) x? Taking a moment to think about it, it should be clear that if we take the area under the curve f(x) x and subtract the area under the curve g(x) x, then we should get the purple area. We can represent this visually as follows: 95

110 96 LESSON : AREA OF A REGION BETWEEN TWO CURVES Algebraically, this translates to ( x ) dx x dx ( x x ) dx ( x ) dx x 3 x3 ( ) 3 ( ) Alternatively, if sketching the graph is difficult, then we can approach this without sketching a graph. To find the area between curves purely algebraically, we need two things: () where the functions intersect (if they do at all) and () which function is larger. () To determine where the functions intersect, set them equal to each other: x x x x x ±. Thus, our functions intersect at ±. Next, we need to determine which is function is larger on the interval,. () To determine which function is larger, we need only check a single point between and (think about why this is true). For convenience, we can check x. Plugging in x, the function x is clearly larger than x.

111 . AREA BETWEEN TWO CURVES 97 From this information, we can setup our integral: ( x ) dx x dx x x dx. larger function smaller function Check that this integral is indeed equal to 3 we did it). In general, the area between f(x) and g(x) is given by xr x L f(x) dx xr x L g(x) dx where f(x) > g(x) on the interval x L x x R. (which is the area we got the first way xr x L f(x) g(x) dx Note 3. We need to choose the correct larger function, else our area will be negative. Ex. We can also find the area between two functions of y. Consider the functions F (y) 8 y and G(y) y. Since these are functions of y rather than of x, these are harder to graph. So we should go about this algebraically. We apply the same two steps as above: () We set our functions (8 y and y ) equal to each other to see where the intersect. y ± y 8 y y 8 y 4 () We check which function is larger on the interval,. Taking the convenient point y, we see that 8 y is larger on this interval (note that since functions of y give x-values, the larger function is the one to the right). Our integral is then (8 y ) dy y dy (8 y y ) dy. Note 3. When setting up integrals, consider the following comments: (a) Functions of x have x-values for bounds and are integrated with respect to x (b) Functions of y have y-values for bounds and are integrated with respect to y Examples.. Find the area between the curves y x, y x + 6, x 9.

112 98 LESSON : AREA OF A REGION BETWEEN TWO CURVES Here, we are given the bounds over which to integrate, so we need only determine which function is larger on that interval. Considering a graph of our functions: By the graph, we see that x + 6 is the larger function (because it is a function of x and is on top). Therefore, the area between the curves is given by 9 x ( + 6 x dx 3 ) (x + 6) 3/ ( 3 (9 + 6)3/ 3 (9)3/ 3 (5)3/ 3 (7) 3 (64) (34) 68 3 ( ) 9 (x) 3/ 3 ) 3 (6)3/ Remark 33. Observe that if we went about this algebraically, we would have written x x + 6 which has no solutions. What does this mean? It means the functions never intersect and thus one function is always larger than the other function.. Find the area between y sin x +, y cos x +, x π. The problem has given us the interval over which we should integrate, so we may focus only on x-values in the interval, π. For this problem, sketching the picture gives the most intuition about what is happening although we will ultimately need to use algebra to set up the integral. Consider the following graph:

113 . AREA BETWEEN TWO CURVES 99 We see that on one part of the interval, cos x + is larger than sin x +, but not on the rest of the interval. We need to determine where these functions intersect (so that we know the point at which the functions switch positions). Write cos x + sin x + cos x sin x Now, if we think about the unit circle (Appendix A.4), cos x sin x for x π when x π. If we had not graphed this function, this extra 4 point of intersection (on top of the interval they gave us) would imply that we need to split the integral into two different intervals and, on each interval, check which function is larger. Quickly checking which function is larger where (try the points x and x π ), our area is given by π/4 (cos x + ) (sin x + ) dx + π/4 (cos x sin x) dx + sin x + cos x π/4 ( sin π 4 + cos π ) 4 ( ) + 4 π π/4 π π/4 + ( cos x sin x) (sin x + ) (cos x + ) dx (sin x cos x) dx π π/4 (sin + cos ) Find the area bounded by ( ( cos π sin π) ) x y y, x + y. ( cos π 4 sin π ) 4

114 LESSON : AREA OF A REGION BETWEEN TWO CURVES We quickly rewrite the given information as x y y, x y. Here, we are not given the bounds over which we should integrate, which means we need to integrate between where the functions intersect. To find where they intersect, write y y y y y ± Thus, we are integrating from to. We need to see which function is larger than the other in this interval. Check y : x() and x(). Thus, we see x y is larger than x y y between y and y. So the area is given by ( y) (y y) dy y dy y 3 y3 3 ()3 ( ) 3 ( )

115 . ADDITIONAL EXAMPLES. Additional Examples. Find the equation of the horizontal line that divides the area of the region bounded by y x, y in half. Round your answer to decimal places. A horizontal line is an equation of the form y, y 5, etc. To tackle this problem, we need to draw graph of y x and y. In this case, it s pretty easy. Then, the question asks us to figure out for what line y a is the area of the region divided in half. Figure 3. Horizontal line y a dividing our figure in half.

116 LESSON : AREA OF A REGION BETWEEN TWO CURVES Since we are looking for a horizontal line, and as such an equation y a, we need to take an integral with respect to y, not x. Our first goal is to rewrite our equations in terms of y. If y x, then we see x + y x y x ± y. But we don t want to have to deal with both the + and the, so we observe that if y a divides the above figure in half, then the same line will divide in half as well since the region is symmetric. In particular, this means we can focus on the situation where x. Therefore, our goal is to solve for a. a We integrate both sides: a y dy y dy 3 ( y)3/ a y dy ( 3 y dy ) ( y) 3/ a 3 ( y)3/ 3 ( y)3/ a ( y) 3/ ( y)3/ ( a) 3/ ( ) 3/ ( )3/ ( )3/ ( a) 3/ ( a) 3/

117 3. PRACTICE QUIZ QUESTIONS 3 a /3 a / Suppose a person s current phone plan saves x dollars per month but it costs 3x dollars per month where x is the number of months since the person signed up for the plan. Find the accumulated savings that will occur before the person should change plans. Accumulated savings means we need to find the area between the curves y x and y 3x (The term accumulated should make us think integration). Graphing is useful but we can also approach this algebraically: () Setting the functions equal to each other, we have x 3x 5x x Because we are accumulating savings since the phone plan began, our bounds are then x and x. () We determine which function is larger on the interval,. Checking the point x, we see that x is a larger function. Our integral is then ( x) (3x) dx ( 5x) dx x 5x () 5() () 5() 5(4). 3. Practice Quiz Questions. Find the area bounded by y x and y x + 5 for x.. Find the area bounded by y x x and y x. 3. Find the area bounded by y 3 x and y x Find the area bounded by x 4y y and x + y 4.

118 4 LESSON : AREA OF A REGION BETWEEN TWO CURVES 5. Find the area bounded by y sin x, y cos x, x π 4, and x π.

119 Lesson : Volume of Solids of Revolution (I) Disk Method. Disk Method Solids of revolution are 3-dimensional shapes that come from graphs in the plane revolved about a line. Spheres, cones, and cylinders are all examples of solids of revolution. A particular interesting aspect of solids of revolution is that we can compute their volume. In fact, using the techniques learned today, we can derive the formula for volume of a generic sphere or cone (although this particular application is not explored in this class). Ex. Consider the function f(x) sin x on left on the interval from to π revolved about the x-axis on right: Figure 4. f(x) sin(x) for x π revolved about the x-axis Our solid of revolution on the right is a sort of lemon shape but not an object for which we have a volume formula. Our goal is to find a method of determining volume for these types of objects. The idea is this: slice the solid into very thin disks. The volume of any disk is given by πr w where r is the radius and w is the width of the disk. Because we are slicing our object very thinly, the width of each disk will be dx (which is something we always think of as very small). Once we find the volume of all these thin disks, we add them up using an integral. But, because we have has sliced the object so thinly, we know that this integral is actually the volume of the shape. The technical details really only involve determining the radius of each disk (which will have to be different depending on from where the disk has been sliced). For this example, the radius of each disk is sin x because that s the height of the function. 5

120 6 LESSON : VOLUME OF SOLIDS OF REVOLUTION (I) DISK METHOD This means, our volume is given by Figure 5. Thin disk sliced into our shape. Vol π π(sin x) dx. }{{} radius }{{} width This method of finding the volume is called the disk method and is given by the formula xr x L π(f(x)) dx where x L is the bound to the left, x R is the bound to the right, and f(x) is the function we are revolving about the x-axis. Examples.. Find the volume of the solid obtained by revolving the region enclosed by the curves y x, x, x, and y about the x-axis. If we sketch a picture, this becomes quite easy. The lower bound is x, the upper bound is x, and the function is f(x) x, so the volume

121 . DISK METHOD 7 Vol π π π 3 x3 ( x π 3 () }{{} radius x dx ) dx }{{} width. Find the volume of the solid generated by revolving the given region in the first quadrant about the y-axis: y x, x, and y 6. There is a big difference between this problem and Example. Here, we are revolving about the y-axis, not the x-axis. Be sure to read each question carefully so you know where you are revolving the function. We sketch a picture of the graph: Because we are revolving about the y-axis, the radius is not x. In fact, the radius will be an x-value that depends on the height y, that is to say, we think of x as a function of y. How do we determine this x? Well, y x x y. So our sketch should really look like

122 8 LESSON : VOLUME OF SOLIDS OF REVOLUTION (I) DISK METHOD Now, we write Vol π π π y 6 π( y ) dy π (6) 8π }{{} radius ( y) dy y dy }{{} width The disk method about the y-axis is given by y T y B π(g(y)) dy where y B is the bottom bound and y T is the top bound. Note 34. If we are revolving around the x-axis, the radius is y f(x). If we are revolving around the y-axis, the radius is x g(y). But just as before with integration: functions of x have bounds in x and functions of y have bounds in y. 3. Find the volume of the solid generated by revolving the region enclosed by the curves about the x-axis. y sec x, y, x, x π 4 This graph is tougher to sketch:

123 . DISK METHOD 9 The radius is sec x and we have bounds x, x π. We write 4 V π π/4 π/4 π(sec x) dx sec x dx π/4 π tan x π tan π 4 tan π π 4. Find the volume of the solid obtained by revolving the region bounded by y x x and the x-axis about the x-axis. We want to be able to sketch these types of functions quickly so here are some things to notice about y x x : we can rewrite it as x x x( x) which means it has roots at x, x. Further, the coefficient of x is negative, so the parabola will open down. With all this information, we can sketch Therefore, it becomes clear that the volume is given by the integral Vol π(x x ) dx.

124 LESSON : VOLUME OF SOLIDS OF REVOLUTION (I) DISK METHOD Write Vol π π π(x x ) dx (x x )(x x ) dx (x x 3 + x 4 ) dx

125 . DISK METHOD ( π 3 x3 4 x4 + 5 x5 ( π 3 + ) 5 ) π 3 5. Find the volume of the solid that results by revolving the region bounded by the curves y x, y, and x about the y-axis. We are told that y x, so squaring both sides we get y x x + y. Hence, our function is part of the circle of radius centered at the origin. Because we are given the curves y and x, we assume we are only talking about the region in the first quadrant. Therefore, our picture is going to look like But because we are rotating about the y-axis, we need to have our radius providing x-values. We know that x + y, so x y x y. So our sketch should really look like

126 LESSON : VOLUME OF SOLIDS OF REVOLUTION (I) DISK METHOD Finally, we write Vol π π ( y ) dy ( y ) dy π (y 3 ) y3 ( π ) 3 π 3. Practice Quiz Questions. Find the volume that results by revolving the region enclosed by the curves y x, x, x 6, and y about the x-axis.. Find the volume that results by revolving the region enclosed by the curve y x x and the x-axis about the x-axis. 3. Find the volume of the solid generated by revolving the region in the st quadrant enclosed by y x, x, and y 8 about the y-axis. 4. Find the volume of the solid that results by revolving the region enclosed by the curves y 36 x, y, and x about the y-axis.

127 Lesson 3: Volume of Solids of Revolution (II) Washer Method. Washer Method We continue discussing finding the volume of solids of revolution. We introduce the washer method. This method is used whenever there is a gap between what we are revolving and where we are revolving. Ex. Suppose we have two functions f(x) x and g(x) x and we want to revolve the area between them about the x-axis. We need to take the volume of the disks obtained from revolving f(x) x about the x-axis and subtract the volume of the disks obtained from revolving g(x) x about the x-axis: The volume of a disk with radius x is / / π( x ) dx 3

128 4 LESSON 3: VOLUME OF SOLIDS OF REVOLUTION (II) WASHER METHOD and the volume of a disk with radius x is / / π(x ) dx. Thus, the volume of this solid of revolution is given by / π( x ) dx / π(x ) dx / / / / Note 35. This is not the same as / / π( x x ) dx π ( x ) (x ) dx. because (a + b) a + b. To see this, observe that ( + 3) Writing (a + b) a + b is called the Freshman s Dream and it is incorrect. The washer method when about the x-axis is given by (9) Volume xr x L π (Outer Radius) (Inner Radius) dx. In Ex above, we see the outer radius was x and the inner radius was x. The outer radius is the further function and the inner radius is the closer function to where we are revolving. Examples.. Find the volume of the solid obtained by revolving the given region about the x-axis: y x, y x. We sketch a quick graph of this region:

129 . WASHER METHOD 5 Our bounds for this region will be x. This leaves us with determining the outer radius and inner radius. Consider The outer radius is x and the inner radius is x. Therefore, by equation (9), Volume π π ( π π ( x ) ( x) dx (x 4 x + ) ( x + x ) dx (x 4 3x + x) dx 5 x5 3 3 x3 + x ( ) π 5 + ) π 5. Find the volume of the solid generated by revolving the region inside the circle x + y 4 and to the right of the line x about the y-axis. The region should look like:

130 6 LESSON 3: VOLUME OF SOLIDS OF REVOLUTION (II) WASHER METHOD To derive this picture, observe that x +y 4 describes a circle of radius centered at the origin. Note 36. x + y r is a circle of radius r centered at the origin. Since we are revolving around the y-axis, our bounds must be in terms of y. So we need to find the y-values where x + y 4 intersects x. We are intersection x, so substituting this into x + y 4, () + y 4 y 3. This means our bounds will be 3 y 3. Next, we need to find our outer radius and inner radius. If we look at our sketch,

131 . WASHER METHOD 7 We need to figure out the outer radius as a function of y because it is clear that the inner radius is x. We have x + y 4 and we need to solve for x (because then we would have a function of y). Write x + y 4 x 4 y x ± 4 y. So we have a choice here, we take either x 4 y or x 4 y. Well, we want only x, so we take x 4 y as our outer radius. Finally, we can set up our volume function: 3 ( ) Volume π 4 y () dx. 3 Our final step is to solve this integral. 3 ( ) Volume π 4 y () dy π π (4 y ) dy (3 y ) dy π 3y 3 3 y3 3 π (3( 3) 3 ) ( 3) 3 (3( 3) 3 ) ( 3) 3

132 8 LESSON 3: VOLUME OF SOLIDS OF REVOLUTION (II) WASHER METHOD π ( 3) 3 π (3 3) π π Note 37. When we are rotating about the y-axis, the washer method is () Volume yt y B π (Outer Radius) (Inner Radius) dy. 3. Find the volume of the solid that results from revolving the region enclosed by y 4x, x, x, y about the y-axis. Round your answer to the nearest hundredth. Firstly, since we are revolving around the y-axis, we need to rewrite our function so that it gives us x-values. But this is easy, we write y 4x y 4 x. Now, we want to sketch a graph. Although this region is very simple, if you don t draw the picture correctly, you are going to miss an important point in the setup. We now need to find the outer radius and inner radius. Observe that depending on the y-values, the inner radius is given by different functions. This is difficult to detect algebraically. For problems about solids of revolution, it is best to sketch a picture (this will require taking time to practice sketching functions). Not drawing a quick graph can lead to missing key features for this type of question.

133 . WASHER METHOD 9 To address the issue of different inner radii, we break the region into two areas such that each area has only one function for its inner radius: For Area, we have Outer Radius : x Inner Radius : x y 4 Bounds : 4 y 8 Observe that these bounds are in terms of y because we are revolving about the y-axis. For Area,

134 LESSON 3: VOLUME OF SOLIDS OF REVOLUTION (II) WASHER METHOD we have Outer Radius : x Inner Radius : x Bounds : y 4 Thus, we get our formula for volume: 4 Volume π () () 8 ( y dy + π () dy 4 4) }{{}}{{} Area Area 4 8 ) π 3 dy + π (4 y dy ) π 3 dy + (4 y π 3y + (4y 48 ) 8 y3 π + 4(8) 48 (8)3 π π π dy (4(4) 48 (4)3 ) Examples.. Additional Examples

135 . ADDITIONAL EXAMPLES. The equation x 9 + y describes an ellipse. Find the volume of the solid 6 generated by this region being revolved around (a) the x-axis and (b) the y-axis. Note 38. An ellipse centered at the origin is always given by something of the form x a + y b. Further, this ellipse will go through the points ( a, ), ( a, ), (, b), (, b). In this case, this means our ellipse will pass through the points So, our picture should look like (3, ), ( 3, ), (, 4), (, 4). (a) We want to revolve about the x-axis, which means we should focus on this region: Here, we don t need to use the washer method because the inner radius is just. We do, however, need to determine the outer radius. Because we are revolving about the x-axis, we need a function of x. So,

136 LESSON 3: VOLUME OF SOLIDS OF REVOLUTION (II) WASHER METHOD we need to take x 9 + y 6 x 9 + y 6 and solve for y. Write y 6 x 9 y 6 6 y 9 x x where we take the positive y-values because we are above the x-axis. We observe that our bounds are 3 x 3. So, our volume is given by Volume π ( ) π x dx ( π 6 6 ) 9 x dx 6x x3 3 ( π 6(3) 6 ) 7 (3)3 π 48 6 ( ) π3 ( 3) 64π ( 6( 3) 6 ) 7 ( 3)3 (b) Because we are revolving around the y-axis, we need to focus on this region:

137 3. PRACTICE QUIZ QUESTIONS 3 Again, we can just use the disk method because we only have one radius. We are revolving around the y-axis, which means we need a function of y. So x 9 + y 6 x 9 y 6 x y x y where we take the positive x-values because we are to the right of the y-axis. by Our bounds here are 4 y 4, which means our volume is given Volume π π ( 9 96 y ) π (9 96 ) y dy 9y 9 6 π 9y y3 4 dy ( ) 4 y π (9(4) 36 ) (4)3 (9( 4) 36 ) ( 4)3 π (36 36 ) (64) ( ) (64) π π Remark 39. The values are for (a) and (b) not the equal even though they both concern the same ellipse. It s a very special situation when revolving a region around the x- and y-axis gives identical volume in both cases. Be careful to always know where you are asked to revolve the region. 3. Practice Quiz Questions. Consider the region bounded by y 3 and y x + 4. x

138 4 LESSON 3: VOLUME OF SOLIDS OF REVOLUTION (II) WASHER METHOD (a) Find the volume of the solid obtained by revolving the region about the x-axis. (b) Find the volume of the solid obtained by revolving the region about the y-axis.. Find the volume of the solid obtained by revolving the region enclosed by y 4 x and y 4 x about the x-axis. 3. Find the volume of the solid generated by revolving the region enclosed by y ln x, y, and x e about the y-axis. 4. Find the volume of the solid that results from revolving the region enclosed by y x, x, x, and y about the y-axis.

139 Lesson 4: Volume of Solids of Revolution (III). Revolving about Horizontal and Vertical Lines We continue to expand our understanding of solids of revolution. The key takeaway from today s lesson is that finding the volume of a solid of revolution is all about determining the radius. Specifically, we will be revolving around horizontal and vertical lines that are not the x- and y-axes. Here, we need to think very geometrically about how we determine the radii of our disks. Although this may appear to be a daunting task (revolving about lines that are not the axes), the idea is essentially the same: the radius is the difference between the function we are revolving and where we are revolving. The biggest difficulty is matching x-values with x-values and y-values with y-values. Examples.. Consider the region bounded by the curves y, y, x 5, x. x (a) Find the volume of the solid generated by revolving the region about the line y 5. We sketch a graph to get a geometric understanding of what is going on. We are not revolving around the x-axis, but instead the line y 5 although this is still a horizontal line. We need to be careful with how we choose our outer radius and inner radius because now it is relative to y 5 and not the x-axis. 5

140 6 LESSON 4: VOLUME OF SOLIDS OF REVOLUTION (III) Our outer radius is the difference between y 5 and y and the inner radius is the difference between y 5 and y (think about x how you would draw the disks and determine their radii). Then we use exactly the same formula as we have been using for the washer method: Volume 5 π (5 ) outer ( 5 x inner ) dx. The only difference between this and the previous methods is determining the radii, every other aspect remains the same. So, Volume π π 5 5 ( π 5 5 x dx x ln x + x ( π ln + 5 x + ) dx x ) π ln + ln 5 π (ln ln 5) + π ln ( ln 5 + ) 5 (b) Find the volume of the solid generated by revolving the region about the line x 5. We go about this with very much the same spirit as in part (a). However, there is a major difference because of where we are revolving. Observe that this is a vertical line. Consider the graph

141 . REVOLVING ABOUT HORIZONTAL AND VERTICAL LINES 7 Additionally, we observe that, depending on the y-value, we have different outer radii. So we need to break the graph into Area and Area based on the two different outer radii functions. We do want to observe that for either Area or Area, the inner radius is because there s no gap between where we are revolving and what we are revolving. So we may use the disk method for both areas. Since we are revolving around the y-axis, we need to solve for x (see note (4)). Given y, we get x. For Area, we find x y Radius : x y 5 Bounds : y

142 8 LESSON 4: VOLUME OF SOLIDS OF REVOLUTION (III) We emphasize that we think of x as a function of y and the bounds are in terms of y. This follows whenever we revolve about a vertical line. For Area, we get Radius : x 5 Bounds : y Remark 4. These radii are the differences between the functions and the line we are revolving about. When in doubt, take the function you are revolving and where you are revolving and subtract them. The squaring will take care of any issues with the sign. Therefore, our volume is given by ( ) Volume π( 5) dy + π y 5 dy }{{}}{{} Area Area ( ) π(5) dy + π y 5 dy π 5 dy + π 5y + ( y y + 5 ( ln y + 5y y ) dy )

143 . REVOLVING ABOUT HORIZONTAL AND VERTICAL LINES 9 ( ) ( π (5 ) ln + 5() ) ln + 5() π 5 5 ln π ln π( ln ). Let S be the region bounded above by x 4 y 6, below by y, on the left by x, and on the right by x. (a) Find the volume of the solid generated by revolving S around the line y. Because we are revolving about the line y, we are revolving about a horizontal line, which means our radius function must be a function of x. Given x 4 y 6, solving for y, we get y 6 x 4. Now, the radius is given by 6 (which is the difference between x4 the function and the line) and the bounds are x. Note that we don t need to use the washer method here because there is no space between the region and the line we are revolving about. So our volume is given by ( ) 6 Volume π x dx 4 ( 56 π x 3 ) 8 x + dx 4 π 56 7x x + x 3 ( π 56 7() + 3 ) 7 3() + 3 ( 56 7() + 3 ) 7 3() + 3

144 3 LESSON 4: VOLUME OF SOLIDS OF REVOLUTION (III) 587π (b) Find the volume of the solid generated by revolving S around the line x. Here, we are revolving around a vertical line which means that our radius must be a function of y. We solve for x: x 4 y 6 x 4 6 y x 4 6 y x y /4 Thus, our graph looks like This does require the washer method because there is a gap between the region and the line we are revolving about. This means Outer Radius : x Inner Radius : x y /4 Bounds : y 6 Therefore, our volume is given by 6 ( Volume π ( ) ) dy y /4 6 π () ( ) dy y /4

145 . REVOLVING ABOUT HORIZONTAL AND VERTICAL LINES 3 π π π 6 6 ( y 8 / ( 3 4 y + 8 ) / y /4 3y 8y / y3/4 y /4 6 π 3(6) 8(6) / + 3 ) dy dy 3 (6)3/4 π 48 8(4) () π 3 ( 3() 8() / ()3/4 3. Find the volume of the solid generated by the region enclosed by y x, y, x, x 5 revolved about the line x 5. As usual, we sketch a graph of this region. ) This is certainly a problem involving the washer method since there is a gap between the region and where we are revolving. Moreover, the inner radius differs depending on what y-value we choose. Thus, we need to divide this region into two areas:

146 3 LESSON 4: VOLUME OF SOLIDS OF REVOLUTION (III) We also want to note that because we are revolving about a vertical line, we need to have radius functions as functions of y. So, we solve for x: Our picture is then y x x y. For Area, we see that Outer Radius : x 5 Inner Radius : x 5 ( y) Bounds : 5 y and for Area,

147 . ADDITIONAL EXAMPLES 33 Outer Radius : x 5 Inner Radius : x 5 5 Bounds : y 5 We put this together to compute the volume: 5 Volume π (5 ) (5 5) dy + π (5 ) (5 ( y)) dy 5 }{{}}{{} 5 Area π (5) () dy + 5 π (5 ) dy + 5 π (5) dy + π 5y 5 π 5(5) π (5) (5 + y) dy Area ( 5 (5 + y + y ) ) dy ( y y ) dy + y y 3 y3 5 ( () 5() 3 ()3 π π 3 3 Note 4. Revolving about a horizontal line function of x and bounds in x vertical line function of y and bounds in y ((5) 5(5) 3 (5)3 )). Additional Examples. A propane tank is in the shape that is generated by revolving the region enclosed by the right half of the graph of x + 4y 6 and the y-axis about the y-axis. If x and y are measured in meters, find the depth of the propane in the tank when it is filled to one-quarter of the tank s volume. Round you answer to 3 decimal places. Our first step is to put the equation we are given into a more useful form. Divide through by 6 on both sides to get x 6 + 4y 6 x 6 + y 4.

148 34 LESSON 4: VOLUME OF SOLIDS OF REVOLUTION (III) By note (38), we see the ellipse passes through (5, ), ( 5, ), (, 7), (, 7). Thus, our picture looks like The blue area is the region enclosed by the ellipse and the y-axis. Now, what are we asked to find? We are asked to find the horizontal line y a such that revolving the region enclosed by x 6 + y, the y-axis, 4 and the line y a has only a fourth of the volume as the original tank. Let s first determine the total volume of the tank. Since we are rotating around the y-axis, our function needs to be in the form x something. So x 6 + y 4 x + 4y 6 x 6 4y x 6 4y. We take the positive root because we are considering only x. Hence, the total volume of the tank is given by π ( 6 4y ) dy.

149 . ADDITIONAL EXAMPLES 35 Integrating: ( ) π 6 4y dy π(6 4y ) dy π (6y 43 ) y3 π 6() 4 3 ()3 π π π 3. (6( ) 43 ( )3 ) But we aren t done here. We don t want the total volume, we want onequarter of the volume. This is where the line y a comes in. We need to find the a that solves the following equation: a ( ) ( π 6 4y dy π 6 4y ) dy. 4 }{{} one-quarter of the volume By above, we know that 4 Hence, we solve for a. So, ( ) π 6 4y dy 4 ( ) 8π 3π a ( 3 π 6 4y ) dy 3π a ( 3 π 6 4y ) dy π (6y 43 ) a y3 π 6a 4 3 a3 3 3π 3. (6( ) 43 ( )3 ) π 4 3 a3 + 6a

150 36 LESSON 4: VOLUME OF SOLIDS OF REVOLUTION (III) π 4 3 a3 + 6a a3 + 6a a3 + 6a This is a cubic equation which you can t factor and so you can t solve it by hand. So we find a root via a graphing calculator or Wolfram Alpha. Here, we have three possible solutions for a: a a a We use the approximate forms here because each exact form here has an imaginary number in it and we don t address those in this class. We need to think about which of these answers makes sense for how we have setup our problem: a and a are outside our bounds for y (because we are only looking at y ), so we throw those out. Thus, we are left with a But how does it make sense for the depth from the bottom of the tank to be a negative number? It doesn t, because a is not the depth from the bottom of the tank. a is just a line. The depth from the bottom of the tank is the difference between y a and y. So really, our answer is the difference between a and the bottom of the tank which is at y. Hence, ( ) Practice Quiz Problems. Consider the region enclosed by the curves y 3 x and y x + 4. (a) Find the volume obtained by revolving the region about the line y 4. (b) Find the volume obtained by revolving the region about the line x 3.. Consider the region enclosed by the curves y x, y, x, and x. (a) Find the volume obtained by revolving the region about the line x. (b) Find the volume obtained by revolving the region about the line y.

151 3. PRACTICE QUIZ PROBLEMS Find the volume obtained by revolving the region enclosed by y x, y, and x 4 about the line x 4.

152

153 Lesson 5: Improper Integrals. Motivation After practicing integration through a variety of applications, we return to some more theoretical aspects of the topic. Improper integrals address some of the deficiencies with typical definite integration. Before when addressing definite integrals, we limited ourselves to intervals of finite length and a function defined everywhere on that interval (i.e., the function made sense for every point in that interval). Improper integrals allow us to address () where a function might exist for very large numbers or () if the function doesn t exist at a point in the interval. Ex. Suppose we are modeling how certain particles decay over time and we know that the energy given off by the particles at any time a is modeled by a e t dt. We might ask: how much energy is given off from now until the end of time? Well, that means a is increasing toward which is exactly what limits were created to handle: lim a a e t dt : e t dt. This is exactly an improper integral, the improper-ness coming from the fact that, ) is an interval of infinite length. The symbol : means is defined to be. Comments on : is not a number, it is an upper bound. All this means is that is larger than every real number. +( ) is undefined this means there is no consistent way to define what + ( ) should be. If you are taking the limit and you get to the point where you have + ( ) or +, then you have to go back and redo the limit. + 7, 3, 7 Takeaway: is essentially different than finite things and so it should not be treated like finite things 39

154 4 LESSON 5: IMPROPER INTEGRALS Improper integrals always involve limits.. Review of Basic Limits Let f(x) be a function and a some real number. Recall that lim x a +f(x) is a right hand limit where we only consider x > a lim x a f(x) is a left hand limit where we only consider x < a We say a limit exists if lim x a f(x) lim x a +f(x). We call this shared value lim x a f(x). If f(x) is continuous at a (which, for this class, will more or less mean a is in the domain of f), then Examples.. Compute lim x x lim x xn lim x n x e x e x lim x n x lim x a limf(x) f(a). x a lim e x x lim ln x x lim ln x x + (cf(x) + g(x)) c(limf(x)) + lim g(x) x a x a x dx 3. Improper Integrals We make sense of improper integrals via limits. We make the following definition: t dx : lim x t x dx. So, to compute this integral, we first evaluate t x dx Suppose we let + ( ) :. What issues might this raise? Well, 7 + but So this doesn t make any sense ( ) + ( ) 7 + () 7

155 and then take the limit as t. Write 3. IMPROPER INTEGRALS 4 t x dx ln x t ln t ln }{{} ln t Here, we drop the because we are assuming that t >. Next, we take the limit We say this integral diverges. t dx lim x t x dx lim t ln t Definition 4. An integral diverges if the limit is ± or DNE. Note 43. Take time to review the techniques of finding limits of functions.. Find x dx By definition, So, we compute t t x lim t x dx. dx and then take its limit as t. Write x t x dx t x ( t ) t + Next, we take the limit as t : lim ( t ) + () +. t

156 4 LESSON 5: IMPROPER INTEGRALS Therefore, we say the integral converges and write x dx. Definition 44. An integral converges if its limit is a real number. 3. Determine if 3x e x3 dx converges or diverges. If it converges, find its value. We know t 3x e x3 dx : lim 3x e x3 dx. t }{{} u-sub So Let u x 3, then du 3x. Further, u() () 3 and u(t) (t) 3 t 3. t 3x e x3 dx t 3 e u du Therefore, 3x e x3 dx lim t t e u t3 e t3 e e t3 e t3 ( ) 3x e x3 dx lim t e t3 So the interval converges and its value is. 4. Determine if x dx converges or diverges. If it converges, find its value. This is different from Examples -3 because we are not dealing with an interval of infinite length. How is this an improper integral? Well, the function doesn t exist at x (because then we would be dividing x

157 3. IMPROPER INTEGRALS 43 by ). For a definite integral to make sense the function must exist on the entire interval. So what do we do here? We take a limit as x. Write Recall that lim s + dx : lim x s + s x dx. means we are looking at numbers that are getting very close to but which are all bigger than. Why do we care if our numbers are bigger than? Our interval is,, which means we don t care about anything that is less than. As before, we first evaluate We have s x s dx ln x dx and then take the limit. x ln ln s ln() ln s s ln(s ) Why do we drop the? Because we are only interested in s >, which means that s >. Now, we take the limit: dx lim x s + x dx ( ) lim ln(s ) s + ( ) ( ) follows because as s +, s. But as x, ln(x). Therefore, we conclude the integral diverges. 5. Determine if π sec (x) dx converges or diverges. If it converges, find its value. Again, we are looking at an interval of finite length. Why is this an ( π ) improper integral? Recall that sec(x) cos(x) and that cos. Thus, sec (x) does not exist on all of the interval, π. We address this

158 44 LESSON 5: IMPROPER INTEGRALS by breaking the interval into two halves:, π exists on both of these intervals sans the point π. and π, π because sec (x) π/ We have π sec (x) dx π/ sec (x) dx + π π/ sec (x) dx where both integrals on the right are improper integrals. Hence, sec (x) dx + π π/ sec (x) dx : lim s π s The antiderivative of sec (x) is tan(x). So lim s π s sec (x) dx + lim t π + s lim tan(x) s π π t π + lim tan(x) t π + t sec (x) dx + lim t π + sec (x) dx lim (tan(s) tan() ) + lim s π }{{} (tan(π) tan(t)) t π + }{{} lim s π tan(s) lim tan(t). t π + π t sec (x) dx. To determine what tan(x) is doing at x π, we consider its graph π Interpreting the graph, we conclude that lim s π Putting this together, sec (x) dx lim s π tan(s) and lim tan(t). t π + tan(s) lim tan(t) ( ) +. t π + We conclude that the integral diverges.

159 6. Determine if 3. IMPROPER INTEGRALS 45 x(ln x) dx converges or diverges. If it converges, find its value. We write Let u ln x, then du dx with x So t t dx lim x(ln x) t x(ln x) dx. }{{ } u-sub u() ln() and u(t) ln t. ln t x(ln x) dx ln u du ln t u Taking the limit, we see that ( lim t ln ) ln t ln ln t ln ln t. ( ) ln ln ln. 7. Compute We conclude the integral converges and Write x e x dx. x e x dx x(ln x) dx ln. xe x dx lim t t This is an integration by parts integral: by LIATE xe x dx. u x du dx dv e x dx v e x

160 46 LESSON 5: IMPROPER INTEGRALS Hence, t t xe x dx xe x t t t xe x e x ( e x ) dx t ( xe x e x ) te t e t ( (e ) e ) te t e t + Taking the limit, lim t ( te t e t + ) lim ( te e ) + t t t + +. We conclude x e x dx. 4. Practice Quiz Questions. Determine whether its value. e t dt converges or diverges. If it converges, find. Determine whether find its value. e dx converges or diverges. x(ln x) 5 If it converges, 3. Determine whether value. 4. Determine whether its value. te t dt converges or diverges. If it converges, find its dx converges or diverges. If it converges, find x 5. Determine whether its value. 4 4t + 7 dt converges or diverges. If it converges, find 6. Determine whether find its value. π/ sec (θ) dθ converges or diverges. If it converges,

161 Lesson 6: Geometric Series and Convergence (I). Introduction to Series Series are just sums of things, like numbers or functions. Ex. 3 n + 3 n + } 3 {{ } n n }{{} n }{{} n3 This is something in series notation. The (called sigma ) means to add a bunch of things together. The n is called the index and is used to put an ordering on the sum (so that we can keep what were adding together organized). The bottom number under tells us when to start counting and the top number above tells us when to stop counting. We can also use series to talk about the sum of an infinite number of things.. Ex. n + n : lim t n t n +, n a n : lim t a n t n Definition 45. We say the n th partial sum is the sum of the first n terms. Ex 3. Consider n + n The 3 rd partial sum is }{{ + } 3 terms 6 and the 5 th partial sum is }{{ 4 + } 5 terms We say a series converges if the partial sums limit to a finite number. We say the series diverges otherwise. Series Facts: 47

162 48 LESSON 6: GEOMETRIC SERIES AND CONVERGENCE (I) () c a n ca n n n Ex: 3 3 n n n n () a n n Ex: a n m, nm n a n nm.6 n n + n.6 n n a n+m n, n3 4(3 n ) 5 n n 4(3 (n+3) ) 5 n+3. Geometric Series A geometric series is a series that looks like 3 7 and n 8. n n Geometric series are special because we can actually compute what the infinite sum is (which is actually very difficult for any other type of series). In fact, we even have a formula for the geometric series: () cr n n n c r. A geometric series converges if r < and diverges otherwise. Remark 46. Simply applying the formula doesn t mean the series converges. For example, we may apply equation () to the series ( ) n but this series diverges because r. Note 47. Make particular note of where the series starts and what power we are raising r to. To use equation (), the geometric series must look exactly like equation ()). Examples.. If continues as a pattern, write it in series 5 notation. n We think about the pattern in the numbers. Write ( ) 5

163 . GEOMETRIC SERIES 49 7 ( ) ( ) ( ) 7 + ( )3 + 3( )4 + 4( )5 + 5( ) n( ) n+ 7. n 3 n. Compute n ( ) n This is not in the correct form since our formula is given by c. So we need to mess with the index: r n ( ) n n n ( ) n+ ( ) ( ) n cr n n 3. Compute n 4e n ( ) ( ) ( ) + 3 This is a geometric series, although it might not seem like it. We rewrite to make this fact more apparent. Consider 4e n 4(e ) n. n n Now, because e < (since e > ), this geometric series converges. e By the geometric series formula,

164 5 LESSON 6: GEOMETRIC SERIES AND CONVERGENCE (I) 4. Compute n 3 n+ 4 n 4(e ) n n 4 4e e e. This is also a geometric series, but not in the correct form to directly apply equation (). We write 5. Compute n n 3 n+ 4 n n n n 3 3 n 4 n 9 3n 4 n ( ) n ( ( 3) + 6 ) n 4 n+ by the equation () We tackle these sorts of problems by splitting up the summation. Write ( 3) ( ) n n 3 n n n ( n+ n n ( ) 4 5 ) ( ) 4 n ( ) ( ) n

165 Thus, ( ( 3) + 6 ) n 4 n+ n 6. Compute 3( ) n n 5 n 3. PRACTICE QUIZ PROBLEMS 5 ( 3) + 6 n 4 3 n n Again, this is not in the correct form: n n n 3( ) n (5 ) 3( ) n+ n (5 ) n+ 3 ( 5 ) n+ n 3 ( 5 ) ( 5 ) n n 3 ( ) ( ) Using summation notation, rewrite the number 6.3. This problem comes down to interpreting what decimals mean: 6.3 has 6 ones, 3 tenths, hundredths, etc. We might even write this as tenth + hundredth n+3 n n n3, thousandth +, + 3. Practice Quiz Problems. Assuming the following pattern continues indefinitely, rewrite it using summation notation:

166 5 LESSON 6: GEOMETRIC SERIES AND CONVERGENCE (I) Note: There are several equivalent ways to write this, so don t worry about writing it exactly as you think I want it written. Just write it a correct way and I can translate to different forms as necessary.. Find the 4 th partial sum of. Round your answer to the nearest hun- n dredth. 3. Find the 3 rd partial sum of hundredth. 4. Compute n ( 3) n. n 4. Round your answer to the nearest 3n + n3 5. Compute n 3 n+ 5 n. 6. Rewrite 4.6 is series notation. 7. Does n 4 n+ converge? If so, find its value. Give a reason for your answer. 3n+

167 Lesson 7: Geometric Series and Convergence (II). Solutions to In-Class Examples Example. A ball has the property that each time it falls from a height h onto the ground, it will rebound to a height of rh for some < r <. Find the total distance traveled by the ball if r 3 and it is dropped from a height of 9 feet. Solution: We draw a picture to get a feel for what is going on. Notice that other than when we originally drop the ball, at each step the distance traveled by the ball is doubled because we must include the height the ball rebounds to and the distance the ball travels as it falls to the ground. Observe ) 3 (9) ( 3 (9) 3 (3) ( 3 (9) ) 3 }{{} 3 ( ) 9 3 ( 3 ) ( ) (9) 3 ( ). 3 From this we can determine a pattern: the distance the ball travels is described by ( ) n 9 + (9) n 53 n ( ) n. 3

168 54 LESSON 7: GEOMETRIC SERIES AND CONVERGENCE (II) This is clearly a geometric series so we use the geometric series formula to compute this sum. But our series starts at n (not n ), so we can t apply our formula just yet. Instead, write Hence, n ( 3 ) n n ( ) n+ 3 n n ( 3 ( ) ( n ) ( ) n 3 3 n ) n ( 3 3 n ( ) ( ) ( ) n ( ) ) n 3 ( ) n. 3 Example. Suppose that in a country, 75% of all income the people receive is spent and 5% is saved. What is the total amount of spending generated in the long run by a $ billion tax rebate which is given to the country s citizens to stimulate the economy if saving habits do not change? Include the government rebate as part of the total spending. Solution: The question is asking us to determine what is spent from now to the end of time (assuming the pattern holds). Since we are including the government rebate as part of the spending, we see at time n, $ billion is spent. But, according to what they tell us, the citizens then spend 75% of the $ billion. So at time t, $(.75) billion is spent. At time n, the citizens spend $(.75)(.75) $ (.75) billion and we continue on in this way. We assume the pattern holds indefinitely. Our goal is to find the total amount spent (measured in billions), which is the sum of all that is spent over time n,,,... This is described by the sum }{{} n + (.75) + (.75) + }{{}}{{} n n n (.75) n n (.75) n. Because n and.75 <, we can apply our formula for the geometric series to determine that the total amount spent (in billions) is

169 . SOLUTIONS TO IN-CLASS EXAMPLES 55 ( ) ( ) (.75) n (4) 4 billion n Example 3. How much should you invest today at an annual interest rate of 4% compounded continuously so that in 3 years from today, you can make annual withdrawals of $ in perpetuity? Round your answer to the nearest cent. Solution: The question is asking: what do we need to invest today so that every year, we have $ in the bank. The formula for continuously compounded annual interest is given by A P e rt where r is the interest rate, t is time in years, A is the amount we have in the bank after t years, and P is the investment we make today. Let P 3 be the amount we invest today so that in 3 years, we have $. Then, at the interest rate we are given, P 3 e.4(3) P 3 e.4(3). Let P 4 be the amount we invest today so that in 4 years, we have $. Write P 4 e.4(4) P 4 e.4(4). Similarly, for any year n > 3 we can let P n be the amount we invest today so that after n years, we have $. Then P n e.4(n) P n e.4(n). Where does this leave us? Well, the sum of all these P n gives the total amount we need to invest today so that we will always have $ in the bank each year beginning 3 years from now. So Total P 3 + P 4 + P 5 + e.4(n). To determine how much we need to invest today, we need to find the value of e.4(n). We will need to use the formula for the geometric series but our n3 series is not in the correct form. So e.4(n) n3 n3 ( e.4) n n3 ( e.4) n+3 n ( e.4) 3 ( ) e.4 n n ( e ).4(3) e.4 n. n

170 56 LESSON 7: GEOMETRIC SERIES AND CONVERGENCE (II) Now that this is in the correct form and e.4 <, we can apply the geometric formula. Our total is ( e ) ( ).4(3) e.4 n e.4(3) $45, e.4 n Example 4. 5 people are sent to a colony on Mars and each subsequent year 5 more people are added to the population of the colony. The annual death proportion is 5%. Find the eventual population of the Mars colony after many years have passed, just before a new group of 5 people arrive. Solution: Let P k be the population of the colony on Mars at the start of year k. Then P 5 because 5 people were sent to Mars initially. Moreover, Further, P }{{} 5 + (P.5P ). }{{} people sent population already to Mars on Mars P }{{} 5 + (P.5P ) }{{} people sent population already to Mars on Mars and we continue on in this pattern. But we want a nicer way to write this. Try and P 5 + (P.5P ) P (5) P 5+(P.5P ) 5+.95P 5+.95(5+.95(5)) 5+.95(5)+(.95) (5). So our pattern is given by 5(.95) n. n This is in the correct form to apply the geometric series formula. So we can write n 5(.95) n ,. We aren t quite done though. We were asked to find the population just before a new group of 5 people arrive. So we need to subtract 5. Thus, our answer is 9,5.. Additional Examples. In a right triangle, a series of perpendicular line segments are drawn starting with the altitude using the vertex of the right angle in the right triangle then subsequently continuing to draw altitudes from the right angles in the new right triangles created which always include the vertex from the smallest angle in the original right triangle. The series of altitudes are drawn so they move closer and closer to the smallest angle in the original right triangle.

171 . ADDITIONAL EXAMPLES 57 Find the sum of all these perpendicular line segments if one of the angles of the triangle is 6 and the side of the triangle adjacent to this angle is. This is an extremely difficult problem and one of the first issues we need to address is vocabulary. What is the altitude of a right triangle? What do they mean by the vertex of the right angle in the right triangle? An altitude is a line segment that goes from the vertex to the hypotenuse of a right triangle. We even have a very nice formula for the length of an altitude of a right triangle: Also note that the altitude creates two more right triangles, the green one and the blue one. Now, let s focus on the question. We are given the following triangle: (The smaller angle is going to be 3 because the angles of a triangle add up to 8 and we re told one of the angles is 6.) The question tells us to draw a series of altitudes. How do we do this? First, draw the initial altitude.

172 58 LESSON 7: GEOMETRIC SERIES AND CONVERGENCE (II) Then draw an altitude in the right triangle that contains the smaller angle of the original right triangle. Draw your next altitude in the right triangle that contains the smaller angle of the original right triangle. Keep going in this pattern. Our question asks us to find the sum of the lengths of all these colored lines. We approach this line by line. Red line: By our formula for the length of the altitude, we have

173 . ADDITIONAL EXAMPLES 59 since in this case a. We need to determine b and c. This involves using trig. We know that tan 6 Opp Adj b b. Now, 6 corresponds to π 3 radians, so b tan 6 tan π Note 48. Make sure that your calculator is set to degrees if you plug 6 into tan x. If it is set to radians, you will get tan 6.3 which is not If we know that b 3, then by the Pythagorean Theorem, c + ( 3) Therefore, our triangle is and so the length of the red line is 3.

174 6 LESSON 7: GEOMETRIC SERIES AND CONVERGENCE (II) Orange line: Our next goal is to find the length of the orange line. This is a tougher situation than above because our triangle is no longer the entire triangle but rather just In particular, the length of the side (c ) shown is not known. So we will need to find this. Hence, By what we know about right triangles, we know that cos(3 ) adj hyp c 3. c 3 cos(3 ). At this point it is tempting to simplify, but the key to determining the pattern of this problem is to not simplify this. Leave this as it is. By our formula for the length of the altitude, the length of the orange line is 3 c 3 ( 3 cos(3 ) ) cos(3 ).

175 . ADDITIONAL EXAMPLES 6 3 (Recall that the came from the length of the red line.) Therefore the sum we have so far is red line + orange line cos(3 ). Yellow line: Our picture now looks like this: Observe that we are in a similar situation as we were in with the orange line, except this time we don t know b : So we need to determine b. Hence, Again, we know that cos(3 ) adj hyp b c. b c cos(3 ) 3 cos(3 ) }{{} c cos(3 ) 3 (cos(3 )). Therefore, by our formula for the length of an altitude, the length of the yellow line is ) ( 3 ) ( 3 cos(3 ) b cos(3 ) (cos(3 )) ) 3 c 3 cos(3 ) (cos(3 )) ( 3 where 3 cos(3 ) is the length of the orange line.

176 6 LESSON 7: GEOMETRIC SERIES AND CONVERGENCE (II) Our sum so far is red line + orange line + yellow line cos(3 ) + (cos(3 )). Now, a clear pattern has emerged. The sum we are trying to find is: 3 (cos(3 )) n. n Because cos(3 3 ) < and our series starts at n, we can apply the formula we have for the geometric series. Therefore, our answer is ( ) 3 3 (cos(3 )) n n 3 3 ( ) ( ) ( 3 + ) ( + 3) ( 3)( + 3) The general formula for this is as follows: let θ be the angle given (in degrees) and x be the length of the side adjacent to this angle, then sum x sin θ cos(9 θ). Because this might appear on an exam (only Exam 3 or the Final), I would recommend memorizing this formula to get easy points for an extremely difficult problem. 3. Practice Quiz Problems. Compute n ( ( 3) 7 ). Round your answer to the nearest hundredth. n 5 n

177 3. PRACTICE QUIZ PROBLEMS 63. Assuming the following pattern continues indefinitely, determine whether the following converges and, if it does, find its sum: Compute n e n. Round your answer to 4 decimal places. n 4. Assuming the following pattern continues indefinitely, determine whether the following converges and, if it does, find its sum: Round your answer to the nearest hundredth. 5. Suppose a particular laboratory has rats to be observed and each year another rats are added for observation. If the annual death proportion for the rats is %, find the eventual population of rats in the lab after many years have passed, just before a new group of rats is added.

178

179 Lesson 8: Functions of Several Variables Intro. Functions of Several Variables Most functions actually depend on several inputs, we call these functions of several variables. In our case, we think of functions with inputs, x and y. For example, how the weather feels to humans depends on both the temperature humidity which are two variables that doe not depend on each other. Geometrically, by adding another input, we are adding another dimension to our graphs which we label z. We write z f(x, y), which means that z depends on x and y. Note 49 (Caution). Before, we would write y f(x) which means that y is related to x. Now, we write z f(x, y), which means z is related to x and y BUT this does not mean x and y are related to each other. x and y are just the inputs and will act independently. Examples.. If f(x, y) x ln(y), find f ) (, e3. A good question here is to ask: which input is x and which is y? Fortunately, we will always write our multivariable functions as z f(x, y) so ( z f x 65, e 3 y )

180 66 LESSON 8: FUNCTIONS OF SEVERAL VARIABLES INTRO Thus, f ) (, e3 ln ( e 3 ) ln e Find the domain of f(x, y) x 3y +. x Just as with single variable functions, functions of several variables can have issues of domain which is why we ask these sorts of questions. Fortunately, nothing really changes from the variable case. Finding the Domain: We have to check the following three things to make sure our function is defined. () No dividing by zero Ex: doesn t exist when x + y x + y () Even roots have non-negative input Ex: + x + y has issues whenever x + y < because then the input is negative. But 3 + x + y has no issues whatsoever. (3) ln has positive input Ex: ln(x + y) doesn t exist when x + y Sometimes these 3 things can overlap, for example, f(x, y) ln(x + y) requires you to check () and (3). To not divide by zero, we can t have ln(x + y) x + y. To make sure ln(x + y) exists, we must have x + y >. This means our domain is See Appendix E for more details. {(x, y) : x + y > and x + y }. x 3y + Returning to f(x, y), we check () and (). That is, x we must have x and x 3y + x 3y. Our domain is then {(x, y) : x, x 3y }. 3. Find the domain of f(x, y) x ln(y ) 3.

181 . LEVEL CURVES 67 We need to check (), (), and (3). () If ln(y ) 3, then our function does not exist. This means that ln(y ) 3 y e 3 y e 3 +. So we must exclude y e 3 +. () We have an even root, so we need x x. (3) For ln(y ) to exist, we must have y > y >. Putting this all together, our domain is {(x, y) : y e 3 +, x, y > }.. Level Curves Now, we talk about how we work to understand functions of several variables. These functions are much more difficult to graph so we need to understand them using different techniques. One technique is called a level curve. The idea is to choose a point on the z-axis, then take a slice of the function to see what it s doing at that point. Ex. Let f(x, y) ln(x + y ) and suppose we want to see what is going on with this function. Without access to some graphing instrument, this will be very tricky to draw. Level curves help us analyze the function without needing to graph it in detail. Observe that if we choose z C for some arbitrary, but constant, C, then C ln(x + y ) }{{} f(x,y) e C x + y. This tells us each level curve value z C is simply a circle centered at the origin of radius e C. Hence, if we choose we get the following picture: C, ln 4, ln 9, ln 6, How do we use this to reconstruct the actual graph of f(x, y)? Think of level curves as what happens if we take a 3-dimensional image and smash it flat on the floor (so

182 68 LESSON 8: FUNCTIONS OF SEVERAL VARIABLES INTRO the level curves are a bird s eye view of the graph). To get the graph of f(x, y), we use C as labels for the height of the function: Observe that all the cross-sections of this graph are circles of increasing radius, which we see for our different z-values. Note 5. Recall that (x h) + (y k) r is a circle of radius r centered at (h, k). Level curves can also come in the following shapes: x + y z Figure 6. Lines x y z y x z Figure 7. Parabolas

183 . LEVEL CURVES 69 y x z x y z Figure 8. Hyperbolas This list isn t exhaustive as level curves can appear as any function. Examples. 4. Consider the function f(x, y) x y. (a) Find the level curves of f(x, y) for z, z. Because we have been given specific z-values, we don t need to consider any other values. So, really, we are looking at the functions x y and x y. We want to graph these functions. y rational function x y rational function x We observe that the symmetry is about the y-axis. (b) What are the vertical and horizontal asymptotes for these functions? Horizontal Asymptote: y Vertical Asymptote: x 5. Consider the function f(x, y) e x + y. (a) Find the level curves of f(x, y) for z,.

184 7 LESSON 8: FUNCTIONS OF SEVERAL VARIABLES INTRO Again, we need only focus on z,. This means we are looking at the functions e x + y and e x + y. y e x exponential function y e x + exponential function (b) Find the horizontal asymptotes of these functions. We have y and y. (c) Find the y-intercepts of these functions. The y-intercept is just where x. So y e y 3 and y e + y. 3. Practice Quiz Questions. Compute the function value of f(x, y) x + y 3x y at the point (, ).. Compute the function value of f(x, y) x ln(y) x at the point (3, e7 ). 3. Find the domain of the function f(x, y) ex y. Write your answer in set 3 exy builder notation. x + y 4. Find the domain of the function f(x, y). Write your answer ln(x + y ) in set builder notation. x 5. Find the domain of the function f(x, y). Write your answer in ln(y) set builder notation. 6. Let f(x, y) ln(9(x 3) + 9y ). The level curve of f(x, y) at C ln 36 is a circle centered at of radius.

185 Lesson 9: Partial Derivatives (I). Partial Derivatives We address how to take a derivative of a function of several variables. Although we won t get into the details, the idea is that we take a derivative with respect to a direction. What we mean is this: if we have a function of several variables, we choose variable and take the derivatives thinking of all the other variables as constants. But this type of derivative doesn t give the entire picture of what the function is doing so we call these partial derivatives. Ex. Let f(x, y) x + y and suppose we want to find its partial derivatives. First, we need to choose a variable, say x. Second, we think of the other variables (in this case just y) as constant with respect to x. This means we think of x and y as acting totally independently so x changing doesn t affect y. We use a special notation denote this concept: x. So the partial derivative with respect to (wrt) x is: f(x, y) (x + y) x x x (x) + x (y) +. }{{} y does not change wrt to x Notice that x (x) d (x). This is because with respect to x, dx derivative as we ve always done it. x is exactly the We use the same line of thinking when we take y and hold x fixed. Here, x is a constant with respect to y. Again, we have our own notation: y. The partial derivative with respect to y is: y f(x, y) (x + y) y y (x) + (y) +. y }{{} x does not change wrt y Again, we see that y (y) d (y) because with respect to y, dy derivative as before. y is the same y Remark 5. This is not a d, and we will call it del. is used exclusively for partial derivatives. 7

186 7 LESSON 9: PARTIAL DERIVATIVES (I) Note 5. We will use a variety of notation for partial derivatives but they will mean the same thing. For example, if our function is z f(x, y), we can write and f x f x (x, y) f f(x, y) x x z x f y f y (x, y) f f(x, y) y y z y. For f(x, y) x + y, our partial derivatives are f x and f y. Now is the time to review all of the differentiation rules you have forgotten (see Appendix B). Examples.. Find f x, f y if f(x, y) e x + ln y. Thus, f x (x, y) x (ex + ln y ) x (ex ) + x (ln y ) xe x }{{} f y (x, y) y (ex + ln y ) y (ex ) + y (ln y ) y y }{{} y f x xe x and f y y. Find f x, f y if f(x, y) y cos x. For this problem, we want to remember that whenever c was a constant d dx (cx3 ) c d dx (x3 ). We have the same property for partial derivatives: x (cx3 ) c x (x3 ). Even more, any function of y is constant with respect to x. So x ((sin y)x3 ) sin(y) x (x3 ).

187 With this in mind, we compute f x (x, y) (y cos x) x y f y (x, y) y. PARTIAL DERIVATIVES 73 }{{} constant wrt x (y cos x) cos x }{{} constant wrt y (cos x) y sin x x y (y) }{{} cos x Thus, f x y sin x and f y cos x. 3. Find f x (, ) and f y (, ) if f(x, y) 3x y y. The difference between this example and the examples above is that here we need to differentiate and then evaluate the derivative at the point (, ). Differentiating with respect to x, f x (x, y) ( ) 3x y x y y }{{} constant wrt x 3 (3x y) (3 ) x y y. Hence, f x (, ) 3 3. Now, to differentiation with respect to y, we will need to use the quotient rule (we could also use the product rule after a small rewrite). So f y (x, y) ( ) 3x y y y ( y) y (3x y) (3x y) ( y) y ( y) ( y)( ) (3x y)( ) ( y) y y + 3x ( y) 3x ( y). Thus, f y (, ) 3() ( ).

188 74 LESSON 9: PARTIAL DERIVATIVES (I) Finally, f x (, ) 3 and f y (, ). 4. Find f x, f y if f(x, y) e xy. Recall that if we were considering a function of a single variable, say e 3x+x, its derivative with respect to x is ( ) d d dx e3x+x dx (3x + x ) e 3x+x (3 + x)e 3x+x by the chain rule. The chain rule still apply for partial derivatives. f x x (exy ) x (x y)e xy y }{{} constant wrt x x (x )e xy y(x)e xy xye x y f y y (exy ) y (x y)e xy }{{} x y (y)exy x e x y constant wrt y So, f x xye xy and f y x e xy. 5. Find f x (, ) if f(x, y) ln(ln(y)x). Recall that d dx ln(g(x)) g (x) g(x). This rule still follows for partial derivatives: f x (x, y) x (ln(ln(y)x)) Thus, x (ln(y)x) ln(y)x f x (, ). ln(y) x (x) ln(y)x ln(y) ln(y)x x 6. Find f x, f y if f(x, y) xy sin(xy). We need to use the product rule and chain rule. f x (x, y) (xy sin(xy)) x xy x (sin(xy)) + (xy) sin(xy) x xy( y cos(xy)) + y sin(xy) }{{} x (xy)

189 . PRACTICE QUIZ QUESTIONS 75 Thus xy cos(xy) + y sin(xy) f y (x, y) xy y (sin(xy)) + (xy) sin(xy) y xy( x }{{} y (xy) cos(xy)) + x sin(xy) x y cos(xy) + x sin(xy) f x xy cos(xy) + y sin(xy) and f y x y cos(xy) + x sin(xy).. Practice Quiz Questions. Find f x and f y if f(x, y) x y.. Find f x and f y if f(x, y) x ln(xy). 3. Find f x and f y if f(x, y) e x +xy. 4. Find f x and f y if f(x, y) xy x + y. 5. Compute f x (, 5) if f(x, y) x + y y.

190

191 Lesson : Partial Derivatives (II). Second Order Partial Derivatives Just as with functions of a single variable, it makes sense to take higher derivatives of functions of several variables. We can take derivatives with respect to the same variable twice, which we would denote f xx f ( x) and f yy f ( y). But we can also take the derivative with respect to one variable and then with respect to another. For example, we might take the derivative with respect to x and then with respect to y. We denote this by (f x ) y f xy. Similarly, if we differentiate with respect to y and then with respect to x, we would write (f y ) x f yx. Fact 53 (Clairaut s Theorem). f xy f yx So it turns out the distinction doesn t actually matter so much. However, there are situations where it is easier to differentiate with respect to one variable first and then other second. Ex. If f(x, y) y sin(x) cos(x) and we want to find f xy, it is actually easier to differentiate with respect to y and then with respect to x. We call f xy and f yx the mixed partials. Examples.. Find the second order derivatives of f(x, y) x 3 y + xy 6. When we are asked to find the second order derivatives, this means we need to find f xx, f yy, f xy, i.e., all the second order derivatives. We start by finding the first order derivatives: f x (x, y) x (x3 y + xy 6 ) 3x y + y 6 f y (x, y) y (x3 y + xy 6 ) x 3 y + 6xy 5 77

192 78 LESSON : PARTIAL DERIVATIVES (II) Then we note that f xx (f x ) x, f xy (f x ) y, f yy (f y ) y. So, starting from f x, f y, we find f xx (x, y) x (3x y + y }{{} 6 ) 6xy f x f xy (x, y) y (3x y + y 6 }{{} f x ) 6x y + 6y 5 f yy (x, y) y (x3 y + 6xy 5 }{{} f y ) x 3 + 3xy 4 Therefore, our second order derivatives are f xx 6xy, f yy x 3 + 3xy 4, f xy 6x y + 6y 5. Find f uv if f(u, v) e 7u+v. Nothing is different here except how they have named the variables. Further, we need only find one second order derivative. Differentiating with respect to u, we get f u (u, v) u (e7u+v ) (7u + v) e 7u+v 7e 7u+v. u Then differentiating with respect to v, we get f uv (u, v) v (7e7u+v ) 7 (7u + v) e 7u+v 7e 7u+v. v }{{} 3. Find the second order derivatives of f(x, y) x ln(3xy). Again, when asked to find the second order derivatives, we are asked to find f xx, f xy, f yy. To start, f x (x, y) (x ln(3xy)) x x x (ln(3xy)) + ln(3xy) x x (3xy) + ln(3xy) 3xy x (x) }{{}

193 . SECOND ORDER PARTIAL DERIVATIVES 79 Next, ( ) 3y x + ln(3xy) 3xy + ln(3xy) f y (x, y) (x ln(3xy)) y x y (ln(3xy)) x y (3xy) 3xy ( ) 3x x x 3xy y f xx (x, y) ( + ln(3xy) ) x }{{} f x f xy (x, y) ( + ln(3xy) ) y }{{} f x f yy (x, y) ( ) x x y y }{{} f y Thus, y x (3xy) 3xy y (3xy) 3xy 3y 3xy x 3x 3xy y ( ) x ( y ) x y y f xx x, f yy x y, f xy y 4. If f xx (, ) 4 where f(x, y) ye ax, find a. Here, a is a constant that we need to find. But, because it is a constant, (a). x We write f x (x, y) x (yeax ) y x (ax ) e ax axye ax. }{{} ax

194 8 LESSON : PARTIAL DERIVATIVES (II) Differentiating again, f xx (x, y) x (axyeax ) axy x (eax ) axy(ax)e ax + aye ax 4a x ye ax + aye ax We are told that f xx (, ) 4, hence Therefore, a. + e ax x (axy) 4 4a () ()e a() +a() e }{{}} a() {{} 4a. 5. Find the second order derivatives of f(x, y) ye sin x. Write f x (x, y) x (yesin x ) y (sin x) e sin x x y(cos x)e sin x f y (x, y) y (yesin x ) e sin x Next, Thus, f xx (x, y) x (y(cos x)esin ) x }{{} f x y cos x x (esin x ) f xy (x, y) y y(cos x)e sin x y(sin x)e sin x (y(cos x)esin x } {{ } f x f yy (x, y) y (esin x }{{} f y ) + e sin x (y cos x) x ) (cos x)e sin x f xx y(cos x)e sin x y(sin x)e sin x, f xy (cos x)e sin x, f yy

195 . PRACTICE QUIZ QUESTIONS 8 Note 54. This homework is not too tough conceptually but it is still difficult because you need to be very careful with your algebra. Keeping track of all the little components is tricky and takes a lot of practice. When you are working on your homework, keep your work neat and detailed. If you are sure you have added each detail at each step, then checking for errors is significantly easier. It is important that you learn how to proofread your own work.. Practice Quiz Questions. Find all first and second partial derivatives of f(x, y) x 3 y + y 3.. If f xx (, ) 4 where f(x, y) 8ye ax, find a. 3. Compute f uv for f(u, v) uv + e u +v. 4. If f(u, v) u ln(uv) 7v, find f uu. 5. Find all first and second partial derivatives of f(x, y) x ln(y). 6. Compute all second order partial derivatives of f(x, y) ln(x + y).

196

197 Lesson : Differentials of Multivariable Functions 3. Quick Review of Differentials Ex. Consider the function f(x) x. We know that f(9) 9 3, but what is f(9.) 9.? Obviously, if you have a calculator this is easy. But there are some functions that even calculators have trouble handling, which is where differentials come in. Let x 9 and x + x 9., that is, x.. x is the actual change in the input x. Our goal is to approximate how this change in the input affects the output function, that is, f(9.) f(x + x). For this, we use calculus. Write y f(x + x) f(x) f(9.) f(9) y is the actual change the function f(x), which is our goal. In an ideal world, we could compute this directly for any given x. But, in general, this is difficult to compute even with a calculator so we settle for an approximation of y instead. () Observe that the equation y x f(x + x) f(x) x looks a lot like a derivative. In fact, the only difference between equation () and an actual derivative is that we need to take the limit as x. Because limits deal with things getting really close together, if our x is small we can make an approximation of y using this derivative. We can write this like x More helpfully, we have y x f(x + x) f(x) x (3) y f (x) x. f (x) dy dx. This just means that we can approximate the change in the function by taking the change in the input and multiplying it by the derivative of the function. Let s apply this to the example above. Since f(x) x, we have Hence, by equation (3), f (x) x y f (9) x 83 9 (.). (3) 6.

198 84 LESSON : DIFFERENTIALS OF MULTIVARIABLE FUNCTIONS So, if , we can add 9 to both sides to get 9. }{{} Using a calculator, we find So our approximation is pretty good. Note 55. We call dx and dy differentials. By the nature of derivatives (in particular, because we would assume that x ), the smaller x is, the better the approximation of y. Think of as the actual change and d as the infinitesimal change. This is why we use dx in an integral and not x because x is too big. 4. Differentials of Multivariable Functions We can apply much of this thinking to functions of more than variable as well. This time, however, we consider how changes in x and y affect z f(x, y). Our notation will be essentially the same and our goal will be to approximate The total differential is given by z f(x + x, y + y) f(x, y). z z z dx + x y dy f x(x, y)dx + f y (x, y)dy. We can use this formula to approximate z (remember, z is the actual change in z). As with before, we think of x dx and y dy especially for x and y small. Hence, (4) z z z x + x y y. We call this equation the incremental approximation formula for functions of two variables. Ex 3. Suppose we have z f(x, y) x + y. Then if x 3, y 4, f(3, 4) (3) + (4) What if we wanted to find f(3., 3.8)? Then, take x and y Next, note that f x (x, y) ( ) x + y x f y (x, y) ( ) x + y y x x + y x x + y y x + y y x + y.

199 Therefore, by equation (4) above, z 5. SOLUTION TO IN-CLASS EXAMPLES 85 x x + y x + y x + y y 3 (3) + (4) (.) + 4 (3) + (4) (.) 3 5 (.) (.) So we can write z f(x + x, y + y) f(x, y) f(3., 3.8) f(3, 4) and adding f(3, 4) to both sides, we get f(3., 3.8) f(3, 4) + z 5 + (.) 4.9. Plugging it into a calculator, (3.) + (3.8) So our approximation wasn t too far off. Note 56. Try to choose your x and y as simply as possible. This isn t exact to begin with (it s inherently an approximation) so make choices that make your life easier. In the previous Ex, we took x and y to be the whole numbers because that s easier than taking x 3. and y 3.8. Once you choose your x and y, your z is always given by z f(x + x, y + y) f(x, y). 5. Solution to In-Class Examples Example. Use increments to estimate the change in z at (, ) if z x 3x+y and z 9y given x. and y.. y Solution: We use our total differential formula. z z z (, ) x + (, ) y x y 3() + ( )(.) + 9( )(.) (.) 9(.). 9(.) 8(.).6.

200 86 LESSON : DIFFERENTIALS OF MULTIVARIABLE FUNCTIONS Example. Suppose that when a babysitter feeds a child x donuts and y pieces of cake, the child needs to run x y + 7 laps in the backyard to be able to go to bed before the parents get home. If one evening the babysitter gives the child 3 donuts and pieces of cake and the next time babysitting, 3.5 donuts and.5 pieces of cake, estimate the difference in the number of laps the child will need to run. Solution: Take x 3, y. Then x and y.5.5. Next, we need to find the derivatives with respect to x and y. Write Thus, z z x xy x y + 7 xy x y + 7 z y x x y + 7. xy x y + 7 x + x x y + 7 y (3)() (3) () + 7 (.5) + (3) (3) () + 7 (.5) 6 9 (.5) (.5) Example 3. A company produces boxes with square bases. Suppose they initially create a box that is cm tall and 4 cm wide but they want to increase the box s height by.5 cm. Estimate how they must change the width so that the box stays the same volume. Solution: Because we are told these boxes have a square base, the formula for volume is V hw where h is the height and w is the width. We are told h, w 4, h.5 and V (because we want the volume of the box to stay the same). Now, we know that So, applying our formula we have V h w and V w wh. V V V h + h w w V (w ) h + (wh) w (4) (.5) + ()(4) w w.

201 5. SOLUTION TO IN-CLASS EXAMPLES 87 So we need to solve for w given w. We conclude that w. This tells us that the width decreases by cm. Example 4. Suppose the function S W F + F W describes the number of fern spores (in millions) released into the air where F is the number of ferns in an area and W is the speed of the wind in miles per hour. Suppose F 56 and W with maximum errors of ferns and 3 miles per hour. Find the approximate relative percentage error in calculating S. Round your answer to the nearest percent. Solution: Here, we think of the relative errors as our. Let F ± and W ±3. We are essentially trying to figure out how changing the inputs (in the sense of correcting the error) changes the number of spores released. We know that By our formula, S F W + F W and S W W F + F. S (W + F W ) F + (W F + F ) W + (56)()(±) + ()(56) + 56 (±3) ±( + (56))() ± ((56) + 56 )(3) ±44 ±, 768. Now, we need to consider the 4 different possibilities that we get from the ± signs. Write 44 +, 768 5, 8 44, 768, , 768, 38 44, 768 5, 8. To find the maximum error, we re looking for is the largest number in absolute value. So we say S 5, 8. Finally, to determine the relative error, we take Thus, our answer is 4%. S S 5, 8 () (56) + (56) () 5, 8 36, This tells us that our formula is not very good as a model because small changes in the input (i.e., the errors) lead to large changes in the output.

202 88 LESSON : DIFFERENTIALS OF MULTIVARIABLE FUNCTIONS 6. Practice Quiz Questions. Use increments to estimate the change at (, ) if z z x 3 and x y 4y+ where the change in x is.5 and the change in y is... A soft drink can is a cylinder that is 5 cm tall and has a radius of 6 cm. Suppose the height of the can is increased by cm, estimate the change in the radius needed so that the volume of the can remains the same. Recall that the volume of a cylinder is given by V (h, r) πhr where h is the height of the cylinder and r is the radius. 3. Suppose a company produces P (x, y) x /3 y /3 thousand units where x is the number of employees and y is the amount of money invested in thousands of dollars. Approximate the change in the number of units produced if the number of employees is increased from 75 to and the amount of money invested is decreased from $5, to $,. Round your answer to the 3 rd decimal place.

203 Lesson : Chain Rule for Functions of Several Variables. Chain Rule for Multivariable Functions Sometimes functions are written as function of more than one variable but can actually come down to a single variable. This means that x and y are related through a third variable. In particular, this situation occurs when both of our variables are related to time. For example, to know how the weather feels to humans, we need the actual temperature and the humidity. But both of these variables depend on the time of day we are considering, even if otherwise temperature and humidity are unrelated. Ex. Suppose x(t) t 4 + describes the number of socks a store sells over time and y(t) 3t + 6 describes the price of the socks over time. Let z(t) xy be the revenue the store earns from the sale of the socks. How does the revenue change with respect to time? One way to do this is to write z entirely in terms of t and then differentiate, but the more complicated x, y, and z become, the more difficult this task is. Instead, we use the chain rule for multivariable functions: dz (5) dt z ( ) dx + z ( ) dy x dt y dt Note 57. Observe that this is dz dt of a single variable. So we need to use dz dt. In this Ex, we have z x y, z y x, z and not. As a function of t, z is a function t dx dt 4t3, dy dt 6t. Applying equation (5), we see that the change in the revenue over time is given by dz dt (y)(4t3 ) + (x)(6t). We leave it like this (this is how you should input your answer for the homework). Examples.. Find dz dt given z x y, x cos t, y 3t 3. 89

204 9 LESSON : CHAIN RULE FOR FUNCTIONS OF SEVERAL VARIABLES Differentiating, z x xy, z y x y, dx dt sin t, dy dt 9t. Applying equation (5), dz dt (xy ) ( sin t) + (x y) (9t ) xy }{{}}{{}}{{}}{{} sin(t) + 8x yt. z x dx dt z y dy dt Again, leave it like this for the homework.. Given z x + y, x ln t, and y dz, find t dt evaluated at t. We start this problem in the same manner as above: we differentiate. Write z x x x + y, z y y x + y, And as before, we apply equation (5), ( ) dz dt x + x + y }{{ t }}{{} z dx x dt y x + y } {{ } z y dx dt t, ( t ). }{{} dy dt dy dt t. Now, we are asked to evaluate at t, which means wherever we see t, we put instead. But what do we do about x and y? We return to our original expressions for x and y, and take t. So, Therefore, dz x (t ) dt x ln t and y t, x() ln ln() and y(). ( ) + x + y t ( ) + () + () (). ( y ) x + y t ( ) () + () ()

205 . CHAIN RULE FOR MULTIVARIABLE FUNCTIONS 9 3. The width of a box with a square base is increasing at a rate of in/min and a height decreasing at a rate of in/min. What is the rate of change of the surface area when the width is 8 inches and the height is 3 inches? Let w be the width and h be the height of the box. Because the box has a square base, the surface area is given by SA w + 4wh. Our goal is to find dsa when w 8 and h 3. So we need to use dt equation (5) and then evaluate at (w, h) (8, 3). Now, we are told that Further, SA w dw dt and dh dt. 4w + 4h and SA h 4w. By equation (5), we have dsa dt SA w ( ) dw + SA dt h Evaluating at (w, h) (8, 3), we get ( ) dh 4w + 4h() + 4w( ). dt dsa (8, 3) 4(8) + 4(3)() + 4(8)( ) 336 in/min. dt 4. P V nrt is the ideal gas law where P is pressure in Pascals (Pa), V is volume in liters (L), and T is temperature in Kelvin (K) of n moles of gas. R is the ideal gas constant. Suppose P is decreasing at a rate of Pa/min and the temperature is increasing at a rate of K/min. How is the volume changing? We are tasked with finding dv. We were given the formula dt P V nrt V nrt P. Now, R is a constant and P, T are variables. Written in this form, V is a function of P and T. So we may write V P P ( ) nrt nrt P P ( ) ( nrt ) nrt P P P and V T T ( ) nrt nr P P T (T ) nr P.

206 9 LESSON : CHAIN RULE FOR FUNCTIONS OF SEVERAL VARIABLES By equation (5), dv dt V P dp dt + V T ( nrt P nrt P dt dt ) ( ) + + nr P. ( nr P ) () We leave our answer in the above form because they have not given us enough information to be more specific. Note 58. This lesson may seem very similar to Lesson and rightly so as the only difference is that the incremental approximation formula is an approximation of the chain rule for multivariable functions. When do we know which formula applies? Think of incremental change as change over some period of time and the type of change discussed in this lesson as instantaneous (that is, a derivative). If the questions asks about a rate of change or something implying instantaneous change, then the chain rule for multivariable functions applies. However, if the question asks for an estimate or approximation or talks about change not in terms of a derivative function, then the incremental formula applies.. Practice Quiz Questions. Use the chain rule to compute dz dt if z x3 y 4, x ln(t), and y 3t +. Leave your function in terms of x, y, t. Do not simplify as a function of t.. Use the chain rule to compute dz dt if z ln(x + y ), x t 3 +, and y t + t +. Leave your function in terms of x, y, t. Do not simplify as a function of t. 3. Evaluate dz ( dt at t if z +3xy+y π ) +x ex, x cos t, and y ln t. 4. The surface area of a cylinder is given by SA(h, r) πr + πrh where h is the height of the cylinder and r is the radius. Suppose the height of the cylinder is increasing at a rate of 3 inches per minute and the radius of the cylinder is decreasing at a rate of 7 inches per minute. What is the rate of change of the surface area when the height is inches and the radius is inches? 5. P V nrt is the ideal gas law where P is pressure in Pascals (Pa), V is volume in liters (L), and T is temperature in Kelvin (K) of n moles of gas. R is the ideal gas constant. Suppose V is increasing at a rate of L/min and the temperature is decreasing at a rate of K/min. How is the pressure changing?

207 Lesson 3: Extrema of Functions of Two Variables (I). Extrema of Multivariable Functions Just like with functions of a single variable, we want to find the minima (plural of minimum) and maxima (plural of maximum) of functions of several variables. We call the minima and maxima the extrema (plural of extremum). Definition 59. A local (or relative) minimum/maximum point is the point (x, y) that makes the function the smallest/largest in some area. A global (or absolute) minimum/maximum point is the point (x, y) that make the function the smallest/largest on the whole graph. Note 6. The extrema, i.e. the minima and maxima, of a function are the function values, e.g., if (, 3) is a minima point of the function f(x, y), then the minima of f is the value f(, 3). However, because minima and maxima (function values) only occur at minima and maxima points (and vice versa) we often conflate these two topics which can lead to confusion. Whether the question is asking for a function value or an ordered pair often depends on context. Ex. Figure 9. Functions of a Single Variable A difference between functions of a single variable and functions of several variables is that functions of several variables have what are called saddle points. Figure. Functions of a Several Variables 93

208 94 LESSON 3: EXTREMA OF FUNCTIONS OF TWO VARIABLES (I) Definition 6. The points (x, y ) such that f x (x, y ) f y (x, y ) are called critical points. Definition 6. We call the function D(x, y) f xx (x, y)f yy (x, y) (f xy (x, y)) is called the discriminant. Second Derivative Test: Suppose (x, y ) is a critical point of f. If () D(x, y ) > and f xx (x, y ) <, (x, y ) is a local maximum point () D(x, y ) > and f xx (x, y ) >, (x, y ) is a local minimum point (3) D(x, y ) <, then (x, y ) is a saddle point (4) D(x, y ), the test is inconclusive (i.e., this test doesn t give you any information) Examples.. Find and classify the critical points of We apply the following steps. Step : Find critical points f(x, y) x3 3 + y3 3 y x. Critical points are points (x, y ) that make both f x and f y equal to. Write f x (x, y) x (x )(x + ) f y (x, y) y (y )(y + ) Hence, if f x then (x )(x + ) x ± and if f y then (y )(y + ) y ±. So, setting both f x and f y equal to zero, our critical points are (x, y ) (, ), (, ), (, ), (, ). Step : Find second derivatives Write f xx x, f yy y, and f xy. Step 3: Find discriminant Our formula for the discriminant is D f xx f yy (f xy ).

209 . EXTREMA OF MULTIVARIABLE FUNCTIONS 95 So, D(x, y) (x) (y) ( }{{}}{{}}{{} ) 4xy. f xx f yy f xy Step 4: Apply test We go through each critical point and apply the test. Critical Point D(x, y ) f xx (x, y ) Classification (, ) 4()() 4 > () > local min (, ) 4()( ) 4 < saddle point (, ) 4( )() 4 < saddle point (, ) 4( )( ) 4 > ( ) < local max. Find and classify the critical points of g(u, v) u v uv v. Again, we go through our steps. Step : Find critical points We have g u uv v v(u ) g v u u v Recall that our critical points are the (u, v ) that make both g u and g v equal to. We write g u v(u ). Here, we have a choice: either v or u. Moreover, we have g v u u v v u u. At this point, it is not clear what our points (u, v ) should be. This is where we break it into cases: Case. v If v, then v u u becomes u u u(u ). So u,. This means that two of our critical points are ( u, v ) and ( u, ). v (Because we were given g(u, v), the order is going to be (u, v).) Case. u

210 96 LESSON 3: EXTREMA OF FUNCTIONS OF TWO VARIABLES (I) If u, then v u u becomes which implies v ( ) 4 4, v 8. Hence, our last critical point is ( ),. 8 Putting this all together, our critical points are (, ), (, ), and Step : Find second derivatives We have ( ),. 8 g uu v, g vv, and g uv u. Step 3: Find discriminant The formula for the discriminant is given by which becomes D g uu g vv (g uv ) D(u, v) (v) ( ) (u }{{}}{{}}{{ } ) 4v (u ). g uu g vv g uv Step 4: Apply test Write ( ) D, 8 ( ) g uu, 8 ( 4 ) ( 8 ( ) 8 4 ( ) ) D(, ) 4() (() ) D(, ) 4() (() ) 4 5

211 . EXTREMA OF MULTIVARIABLE FUNCTIONS 97 Critical Point D(u, v ) g xx (u, v ) Classification ( ), 8 > < 4 local max (, ) < saddle point (, ) 5 < saddle point 3. Find the local minima and maxima of f(x, y) x + y x + y. Observe that this question is different than the previous examples. Before we were asked to classify the critical points, but now we are asked to find the actual function values at the critical points. Our process, fortunately, doesn t change too much. We still need to find and classify the critical points but then we need to plug them back into f(x, y) to determine the function value. Thus, and Step : Find critical points We have f x x and f y y +. f x x x f y y + y. This means our critical point is (, ). Step : Find second derivatives Step 3: Find discriminant f xx, f yy, f xy D(x, y) f xx f yy (f xy ) ()() () 4 Step 4: Apply test Since D(, ) 4 > and f xx (, ) >, we have a local min at (, ) and no local max. Step 5: Determine function values By the previous steps, we know that f(, ) () + ( ) () + ( ) + is a local minimum of f(x, y) (recall that a min or max is function value). 4. Find the local minima and maxima of g(x, y) x3 3 + xy y.

212 98 LESSON 3: EXTREMA OF FUNCTIONS OF TWO VARIABLES (I) We go through our steps and then plug our points to find the function values. Step : Find critical points g x x + y and g y x y So g x x + y x y and g y x y x y. Since we have that y x, we can substitute this into y x which means (x) x 4x x x or x 4. Then we break this down into cases. Case. x If x, then y (). Hence, one critical point is (, ). Case. x 4 If x 4, then y (4) 8. Thus, another critical point is (4, 8). Putting this together, our critical points are (, ) and (4, 8). Step : Find second derivatives g xx x, g yy, and g xy Step 3: Find discriminant D(x, y) g xx g yy (g xy ) ( x)( ) () x 4 Step 4: Apply test We write D(, ) () 4 4 D(4, 8) (4) g xx (4, 8) (4) 8 Hence, Critical Point D(x, y ) g xx (x, y ) Classification (, ) 4 < saddle point (4, 8) 4 > 8 < local max

213 . PRACTICE QUIZ QUESTIONS 99 Step 5: We only have one local maximum which is g(4, 8) (4) (4)(8) (8) Practice Quiz Questions. Find all the local minima and maxima points of f(x, y) +x y x y.. Find all the local minima and maxima points of f(x, y) x + y + 4x 9y. 3. Find and classify the critical points of f(x, y) 8x xy + 6y 35x + 9y. 4. Suppose g(x, y) is a function such that g x (x, y) 3x + y and g y (x, y) x y3. Determine all critical points and show how many points are minima, 7 maxima, and saddle points. 5. Find and classify the critical points of g(u, v) 6u v + 48uv v.

214

215 Lesson 4: Extrema of Functions of Two Variables (II). Solutions to In-Class Examples Example. We are tasked with constructing a rectangular box with a volume of 64 cubic feet. The material for the top costs 8 dollars per square foot, the material for the sides costs dollars per square foot, and the material for the bottom costs 4 dollars per square foot. To the nearest cent, what is the minimum cost for such a box? (Round your answer to decimal places.) Solution: Let w be the width, h be the height, and l be the length of this box. The volume of the box is given by V whl. The goal here is to minimize the cost function, not the volume. In fact, we are requiring that the volume be exactly 64 cubic feet. We ll call this the constraint. By the picture above, you should see the area of the top is lw, the area of the bottom is lw, and the total area of the sides is wh + lh (this is because we are not assuming the box has a square base so w and l may be different). Thus, by the information we are given above, our cost function is C(w, h, l) 8 (lw) }{{} area of top + (wh + lh) }{{} area of sides +4 (lw) lw + wh + lh. }{{} area of bottom Unfortunately, this is a function of 3 variables and our tools only work for functions of variables. We use the constraint (lwh 64) to rewrite the cost function as a function of variables. Write l 64. Then, substituting, we get wh ( ) ( ) C(w, h) w + wh + h wh wh

216 LESSON 4: EXTREMA OF FUNCTIONS OF TWO VARIABLES (II) 768 h + wh + 8 w. Since this is now a function of variables, we can find the critical points. Differentiating, C w h 8 w and C h 768 h + w. Recall that the critical points are the points (w, h) that make both C w and C h equal to zero. So and C w h 8 w h 8 w w h 8 64 C h w 768 h w 768 h h w Now, we observe that (64). This means that 5 h w (64) 3 5 w h. Since we are assuming the volume is 64 cubic inches, we must have w and h. So, we divide both sides by hw and our equation becomes h 3 5 w. Then, because w h 64, w ( 3 5 w ) 64 w 3 5(64) Thus, w 3 3 3

217 and so. SOLUTIONS TO IN-CLASS EXAMPLES 3 h 3 5 w Our critical point is then ( 3 3 (w, h) 3, 3 ) Therefore, our cost is minimized at ( 3 3 C 3, 3 ) ( ) ( ) Note 63. Technically, you should check that this is actually a minimum by going through the Second Derivative Test. But since this is the only critical point and it is much easier to maximize a cost rather than minimize it, we can assume that this is indeed a minimum. For word problems, if there is a single critical point, then this is probably the point we need to find. Example. The post office will accept packages whose combined length and girth is at most 5 inches (girth is the total perimeter around the package perpendicular to the length and the length is the largest of the 3 dimensions). What is the largest volume that can be sent in a rectangular box? (Round your answer to the nearest integer.) Solution: Our goal is to maximize the volume V lwh where l is the largest of the 3 dimensions. The girth is then g w + h. We are told that the combined length and girth can be at most 5 inches, to maximize the volume we will want to maximize the girth as well (think about why this is true). Set l + g 5. Now, our volume function is a function of more than variables so we need to rewrite it in terms of variables. We know that l + g 5 and that g w + h, so Thus, our volume function becomes l 5 g 5 (w + h) 5 w h. }{{} g V lwh (5 } w {{ h } )wh 5wh wh w h. l This is the function we want to maximize and so we find its critical points. Find critical points: V w 5h h 4wh which means V w 5h h 4wh h(5 h 4w). So either h or 5 h 4w. Because we can t let one of our dimensions be, we throw out the case where h. From 5 h 4w, we get 5 h 4w 5 h w

218 4 LESSON 4: EXTREMA OF FUNCTIONS OF TWO VARIABLES (II) V h 5w 4wh w means h 5 w V h 5w 4wh w w(5 4h w), so either w or 5 4h w. Again, we can t let one of our dimensions be so we throw out w. Hence, we have 5 4h w which means Thus, 5 4h w 5 h w w 5 h 5 (5 w) }{{} h w 5 + 4w 3w 5 w 5 3. ( ) 5 h 5 w 5 3 }{{} w ( 5 3, 5 3 Our critical point is then (w, h) 5 (5) 3 ) Finally, we need to plug this into our function for volume. Since l 5 w h, ( ) ( ) 5 5 l Thus, our maximum volume is ( ) ( ) ( ) V }{{}}{{}}{{} l w h 3, Example 3. A biologist must make a medium to grow a type of bacteria. The percentage of salt in the medium is given by S.x y z, where x, y, and z are amounts in liters of 3 different nutrients mixed together to create the medium. The ideal salt percentage for this type of bacteria is 48%. The costs of x, y, and z nutrient solutions are respectively, 6, 3, and 8 dollars per liter. Determine the minimum cost that can be achieved. (Round your answer to the nearest decimal places.) Solution: We want to minimize the cost function, given by C(x, y, z) 6x + 3y + 8z

219 . SOLUTIONS TO IN-CLASS EXAMPLES 5 such that S(x, y, z).48. We want to reduce our cost function to a function of variables, which we do using S. Write.48.x y z z 48 x y. Substituting this into our cost function, we get ( ) 48 C(x, y) 6x + 3y + 8 x y }{{ } z Our next step is to find the critical points. 6x + 3y x y. Find critical points: C x 6 (384) x 3 y x 3 y, so C x implies C y x y 3, so C y implies 768 x 3 y x 3 y 8 x 3 y 768 x y x y 3 56 x y 3 Now, since (8) 56, we see by our work above that (x 3 y ) x y 3, }{{}}{{} 8 56 Hence, either x or y or x y. Because we want S(x, y, z).48, we can t have x or y. So we have x y. Since 56 x y 3 56 x (x) 3 8x 5 3 x 5 x Then y () 4 implies our critical point is then (, 4). The function value at (, 4) is

220 6 LESSON 4: EXTREMA OF FUNCTIONS OF TWO VARIABLES (II) C(, 4) 6() + 3(4) () (4) Example 4. A manufacturer is planning to sell a new product at the price of 3 dollars per unit and estimates that if x thousand dollars is spent on development and y thousand dollars is spent on promotion, consumers will buy approximately 7y y x units of the product. If manufacturing costs for the product are x + 9 dollars per unit, how much should the manufacturer spend on development and how much on promotion to generate the largest possible profit? Round your answer to the nearest cent. Solution: Profit is the difference of revenue and cost. Here, the revenue is ( 7y.3 y x ) because it is the number of units sold times their price and x + 9 ( 7y we are measuring our dollars in thousands. The cost is. y x ) + x + y x + 9 because the cost is the cost of each unit times the number sold but we also need to consider what is spent on development and promotion. Thus, the function we want to maximize is ( 7y Profit P (x, y).3 y x ) ( 7y. x + 9 y x ) + x + y x + 9 ( 7y.3 y x ) ( 7y. x + 9 y x ) x y x + 9 ( 7y.9 y x ) x y x y y x x + 9 x y To maximize P (x, y), we need to find its critical points. Find the critical points: Differentiating with respect to x, we get and, with respect to y, P x P y (x + 9)(7) (7x)() (x + 9) 7x x (x + 9) 43 (x + 9) (y + 4)(4.3) (4.3y)() (y + 4)

221 . PRACTICE QUIZ QUESTIONS 7 and 4.3y y (y + 4) 97. (y + 4) Setting P x and P y equal to, we see that P x 43 (x + 9) 43 (x + 9) (x + 9) 43 x ± 43 9 P y 97. (y + 4) 97. (y + 4) (y + 4) 97. y ± Since it doesn t make sense for x or y to be negative, we conclude our critical point is (x, y) ( 43 9, 97. 4). This tells us that the developer should spend dollars on development and dollars on promotion.. Practice Quiz Questions. We are tasked with constructing a rectangular box with a volume of 3 cubic feet. The material for the top costs 9 dollars per square foot, the material for the four sides costs 4 dollars per square foot, and the material for the bottom costs 6 dollars per square foot. To the nearest cent, what is the minimum cost for such a box?. Find and classify the critical points of f(x, y) y 6x y 6x 4y.

222

223 Lesson 5: Lagrange Multipliers Constrained Min/Max (I). LaGrange Multipliers Lagrange multipliers is another method of finding minima and maxima of functions of more than one variable. This method applies when we are finding exterma that is subject to some constraint. The Method of Lagrange Multipliers: Suppose we want to minimize or maximize a function f(x, y) subject to the constraint g(x, y) C. Introduce a dummy variable, λ, and solve the system of equations for (x, y). f x (x, y) λg x (x, y) f y (x, y) λg y (x, y) g(x, y) C Remark 64. In this setup, our method only works for functions of variables. If a problem is presented with more than variables or if we are asked to classify critical points, then we need to use the method from the previous lessons. However, if the problem contains key words like subject to or has only variables and a constraint, then this method applies Ex. Maximize the area of a rectangular garden subject to the constraint that its perimeter is ft. Let x be the length and y the width of the garden. Then the function we are maximizing is f(x, y) xy. But this is subject to the constraint that x + y. }{{} perimeter By our method, we set up our system of equations: y λ () }{{}}{{} f x g x }{{} x λ () }{{} f y x + y }{{} g(x,y) g y }{{} C 9 λ λ

224 LESSON 5: LAGRANGE MULTIPLIERS CONSTRAINED MIN/MAX (I) We solve for x and y (which means we eliminate λ). Since x λ and y λ, we see that given x + y, Thus λ 8 5 ( }{{} λ ) + ( }{{} λ ) 4λ + 4λ 8λ. x y which means ( ) ( ) 5 5 x 5 and y 5. }{{}}{{} λ λ We conclude that the area is maximized when x 5 and y 5 and the maximum area of the garden is 5(5) 65 ft. Question: How do we know this is a maximum and not a minimum? Because the minimum area of the garden is ft. For these problems, you always need to consider whether your answer makes sense in context. Note 65. Lagrange multipliers will never tell you if there is a saddle point because that involved classifying critical points which are different than solutions to the system of equations for Lagrange multipliers. Examples.. Minimize f(x, y) (x + ) + (y ) subject to g(x, y) x + y 5. Taking derivatives, we see that f x (x + ), f y (y ), g x x, g y y. Setting up our equations (x + ) λ(x) λx (y ) λ(y) λy x + y 5 The method of Lagrange multipliers calls for a little creativity and the key is staying flexible when solving them. We focus on the first two equations. We have Thus, (x + ) λx x + λx (y ) λy y λy x + λx x λx + x( λ) + x( λ) y λy y λy y( λ) y( λ)

225 . LAGRANGE MULTIPLIERS From this we gather that x, y else these equations can t be true. This means that we may divide by x and y without losing solutions to the system. Hence, x λ y, but by cross-multiplication this becomes y x y x. By our constraint, g(x, y) x + y, we see 5 x + y x + ( x) x + 4x 5x 5 x We conclude x ±5, which implies y (±5). Thus, our extrema points are (5, ) and ( 5, ). Evaluating f(x, y) at these points, f(5, ) (5 + ) + ( ) 6 + ( ) Max f( 5, ) ( 5 + ) + ( ) ( 4) + (8) Min Therefore, the minimum is 8 because it is the smaller of the two values.. Find the minimum value of x e y subject to y + x 6. Whatever function we are finding the extrema for is our f(x, y) and the constraint is g(x, y). Thus, we have f(x, y) x e y and g(x, y) y + x 6. Next, we find derivatives: f x xe y, f y x ye y, g x, g y 4y. We set up our equations to get xe y λ x ye y 4λy y + x 6

226 LESSON 5: LAGRANGE MULTIPLIERS CONSTRAINED MIN/MAX (I) By the first equation, we see that xe y λ xe y λ. Substituting this into the second equation, we get yx e y 4 (xe y ) y. }{{} λ We may divide both sides by e y because this is never. Thus, this becomes x y xy. Subtracting xy from both sides we get x y xy (x x)y. Hence, either y or x x x(x ) x or x. We check all three of these cases: Case. y Case. x If y, then our constraint implies that Thus, one solution is (3, ). + x 6 x 3. If x, then by our constraint: y + 6 y 3 y ± 3. So two of our solutions are (, 3) and (, 3). Case 3. x If x, then y + () 6 y y y ±. This adds another two solutions: (, ) and (, ). Putting this all together, our solutions are (3, ), (, 3), (, 3), (, ), (, ). Finally, we check the function values: f(3, ) (3) e () 9() 9 f(, 3) () e ( 3) min f(, 3) () e ( 3) min f(, ) () e () 4e max

227 . LAGRANGE MULTIPLIERS 3 f(, ) () e ( ) 4e max Therefore, the function s minimum value is. 3. Find the extrema of f(x, y) e xy subject to 9x + 4y 7. This is a slightly different problem than what we have encountered so far. Here, our constraint is an inequality rather than an equality. Fortunately, this is not as daunting as this may appear. We break this problem into two parts: () we find the critical points of f(x, y) which are contained in the region described by g(x, y) 9x + 4y < 7 and () we apply the Lagrange multiplier method to f(x, y) subject to g(x, y) 9x + 4y 7. () The derivatives of f(x, y) are f x ye xy and f y xe xy. Setting these equal to, we see that x, y because e xy is never. Since the point (, ) satisfies g(x, y) 9x + 4y < 7, we include this in our list of solutions. () Now, we assume that g(x, y) 9x + 4y 7 and apply the Lagrange multiplier method. The derivatives of g(x, y) are g x 8x and g y 8y. Setting up our system of equations, Next, we solve for (x, y). ye xy 8λx xe xy 8λy 9x + 4y 7 We observe that if any of x, y, or λ are, then x and y. But this case has already been covered, so we assume that x, y, λ. This means we can divide by x and y to get λ by itself. Write Thus, y 8x e xy λ and x 8y e xy λ. y 8x e xy x 8y e xy. Because e xy is never zero, we may divide through on both sides to get y 8x x 8y. Cross-multiplying: 8y 8x y 8 8 x 9 4 x.

228 4 LESSON 5: LAGRANGE MULTIPLIERS CONSTRAINED MIN/MAX (I) Now, we return to our constraint and substitute for y, 7 9x + 4y ( ) 9 9x x }{{} y 9x + 9x 8x. Solving for x, we find x ±. Since y 9 4 x, we get y 9 y ±3. Therefore, we get the following 4 solutions: (, 3), (, 3), (, 3), (, 3). We need to check the function values at each of our solutions from both () and (): f(, ) e ()() f(, 3) e ()(3) e 6 min f(, 3) e ()( 3) e 6 max f(, 3) e ( )(3) e 6 max f(, 3) e ( )( 3) e 6 min Thus, the minimum function value is e 6 value is e 6. and the maximum function 4. Find the minimum value of f(x, y) y x 4x subject to y 8 x. Our f(x, y) y x 4x but we need to determine our g(x, y). We are told our constraint is y 8 x and, adding x to both sides, we have x + y 8. Hence g(x, y) x + y 8. Next, we differentiate: f x x 4, f y y, g x, g y. Setting up our equations, x 4 λ y λ x + y 8 We know immediately that y λ, so, substituting into the first equation, we get x 4 (y) 4y. }{{} λ Dividing both sides by 4, we get

229 . LAGRANGE MULTIPLIERS 5 y x. According to our constraint, 8 x + y x + ( ) x x x 3 x. }{{} y We solve 8 3 x for x: 8 3 x 9 3 x 8 3x 6 x Since x 6 and y x, we have y (6) 3 4. Thus, our solution is (6, 4). Plugging this into the function, f(6, 4) ( 4) (6) 4(6) Note 66. We should check that this is actually a minimum as opposed to a maximum. To do this, we check any other point that satisfies x+y 8, say (, 8). If 44 is a minimum, then we must have 44 < f(, 8) 8 () 4() 64. So we can rest easy knowing that this really is the minimum of the function subject to the given constraint. 5. Find the maximum value of f(x, y) 3 x3/ y subject to x y. Again, we need to rearrange our constraint to determine our g(x, y). Adding y to both sides of x y, we get x + y which means Next, we differentiate: g(x, y) x + y. f x x / y, f y 3 x3/, g x, g y. Now, we set up our equations: x / y λ 3 x3/ λ x + y

230 6 LESSON 5: LAGRANGE MULTIPLIERS CONSTRAINED MIN/MAX (I) Since λ x / y and λ 3 x3/, we can write x / y 3 x3/. Multiplying through by x /, we get xy 3 x. Subtracting xy from both sides and regrouping, this becomes ( ) 3 x xy x 3 x y. Hence, either x or 3 x y x y. We check both cases. 3 Case. x If x, our constraint implies that + y y. Hence, one solution is (, ). Case. 3 x y If x y, then 3 x + y x + 3 x 5 3 x 5 3 x x 6. Thus, since y (6) 4, one solution is (6, 4). 3 Putting this together, we have two solutions: (, ) and (6, 4). We check the function values at these points: f(, ) 3 ()3/ () min f(6, 4) 3 (6)3/ (4) 6 6 max Therefore, the maximum value is Practice Quiz Questions. Find the minimum value of f(x, y) x + y subject to the constraint 8y 3 x.. Find the maximum value of f(x, y) x 3/ y subject to x + y. 3. Find the maximum value of f(x, y) x y subject to x + y 9.

231 . PRACTICE QUIZ QUESTIONS 7 4. At what points does the minimum value of f(x, y) x e y subject to the constraint y + x occur? 5. Let f(x, y) e xy. Find the minimum and maximum values of f(x, y) subject to the constraint x + y 8.

232

233 Lesson 6: LaGrange Multipliers Constrained Min/Max (II). Solutions to In-Class Examples Example. There is an ant on a circular heated plate which has a radius of meters. Let x and y be the meters from the center of the plate measured horizontally and vertically respectively. Suppose the temperature of the plate is given by f(x, y) x y + 5 F and that the ant is walking along the edge of the plate. What is the warmest spot the ant can find? Solution: We want to maximize the function f(x, y) x y + 5 subject to the constraint x + y because the ant is walking around the edge of a plate with a radius of meters. Hence, g(x, y) x + y. Differentiating, So we set up our equations: f x x, f y y, g x x, g y y. x λx y λy x + y This is the system we need to solve. From the first equation, we get x λx x λx λx x x(λ ) Thus, the first equation implies that either x or λ. Similarly, the second equation implies y λy y λy y + λy y( + λ) which means either y or λ. 9

234 LESSON 6: LAGRANGE MULTIPLIERS CONSTRAINED MIN/MAX (II) Now, λ cannot be and at the same time, so we have a choice. If λ, then y. By our constraint, this implies x ±. So (±, ) are points we ought to check. If λ, then x. Thus, our constraint implies y ±. So the other points we need to check are (, ±). So, f(, ) (±) Max f(, ) (±) Min. This means that our answer is f(±, ) 5 F. Example. A rectangular box with a square base is to be constructed from material that costs $5/ft for the bottom, $4/ft for the top, and $/ft for the sides. Find the box of the greatest volume that can be constructed for $6. Round your answer to 4 decimal places. Solution: Since we are assuming the box has a square base, we see its volume is given by V w h where w is the width and h is the height. Our cost function is then given by C(w, h) 5w }{{} cost of bottom + 4w }{{} cost of top + (4wh) 9w + 4wh. }{{} cost of sides Moreover, we are told that our cost will be $6. Therefore, in this context our volume acts as our f and the cost function acts as our g. This is to say we are maximizing the volume subject to the constraint C(w, h) 6. Differentiating, we get V w wh, V h w, C w 8w + 4h, C h 4w. Thus, the system we need to solve is wh λ(8w + 4h) w λ(4w) 4λw 9w + 4wh 6. We note that because we need our box to have some volume, we must have w and h. Thus given w 4λw, we are able to divide through by w because we know it is nonzero. Therefore, w 4λ λ w. Going to the first equation, we 4 substitute: wh λ(8w + 4h) w (8w + 4h) 4 9 w + wh.

235 . SOLUTIONS TO IN-CLASS EXAMPLES Therefore, by subtracting wh from both sides we get wh 9 w. Again, w so h 9 w. Since we have a relationship between w and h that doesn t involve λ, we plug this back into our constraint to get ( ) 9 6 9w + 4w w }{{} h This means h 9 ( ) }{{} (, 9 V ( ) w 9w + 8w 7w w w Thus, the maximal volume is, 9 ) 9. Therefore, the point we need to check is ( ( ) 9 ) ft 3. Example 3. A rectangular building with a square front is to be constructed of materials that cost $ per ft for the flat roof, $ per ft for the sides and back, and $5 per ft for the glass front. We will ignore the bottom of the building. If the volume of the building is, ft 3, what dimensions will minimize the cost of materials? Solution: Observe that we are asked to find the dimensions which minimize the cost. We will use the same method but as an answer we need to state dimensions instead of a minimal cost. Because we assume the building has a square front, we know that two of the dimensions are the same. So we can write V wh where w is the width and h is the height. Then, our cost function is given by C(w, h) } wh {{} + (h + wh) + }{{}}{{} 5h. top sides and back front Simplifying, this becomes C(w, h) 35h + 5wh. Because this is subject to the constraint V wh,, we see that the cost acts as our f and the volume acts as our g, which is to say we are minimizing the cost subject to the constraint that the volume is, ft 3. Differentiating, C w 5h, C h 7h + 5w, V w h, V h wh.

236 LESSON 6: LAGRANGE MULTIPLIERS CONSTRAINED MIN/MAX (II) Now, we the system we need to solve is 5h λh 7h + 5w λwh wh, We note that because our volume is nonzero, w and h. Thus, 5h λh 5 λh. Moreover, λ 5. The second equation then implies h 7h + 5w λwh ( ) 5 7h + 5w wh w h Subtracting 5w from both sides, we get 7h 5w w 7 h. Returning to our 5 constraint,, wh 7 5 h(h ) 7 5 h3 5, 7 5, h 3 h 3. 7 So w 7 3 5,. 5 7 Thus, the dimensions that minimize the cost are w 7 5 5, 3 7 and h 3 5, 7. Example 4. On a certain island, at any given time, there are R hundred rats and S hundred snakes. Their populations are related by the equation (R 6) + (S 6) 8. What is the maximum combined number of rats and snakes that could ever be on the island at the same time? (Round your answer to the nearest integer). Solution: Let f(r, S) R + S (which is the total number of rats and snakes in hundreds) and g(r, S) (R 6) + (S 6) (which is our constraint function). Differentiating, f R, f S, g R (R 6), g S 4(S 6). Then the system we need to solve is λ(r 6) 4λ(S 6) (R 6) + (S 6) 8. The first two equations mean we can write λ(r 6) 4λ(S 6).

237 . PRACTICE QUIZ QUESTIONS 3 This implies that λ and so dividing through by λ we get R 6 (S 6). Therefore, returning to our constraint, we find 8 (R 6) + (S 6) ((S 6)) + (S 6) 4(S 6) + (S 6) 4(S 6). Thus, (S 6) 8 4 S 6 ± 9 (. Further, R 6 ± 9 ). 4 4 Finally, we get ( ) 9 R and S and ( ) 9 R and S Taking the larger R and larger S, the maximum combined number of rats and snakes will be ( ) ( ) f , }{{}}{{} R S Since we are measuring snakes in hundreds, this means that we have 4 snakes.. Practice Quiz Questions. Suppose Arjun has exactly 4 hours to study for an exam and, without preparation he will get points out of total exam points. Suppose he estimates that his exam score will improve by x(39 x) points if he reads lecture notes for x hours and y(48 y) if he solves review problems for y hours. However, due to fatigue, he will lose (x + y) points. What is the maximum score Arjun can obtain if he uses the entire 4 hours to study? Round your answer to the nearest hundredth.. Suppose an artist sells sketches of her cat and dog online and that she notices she can make a profit of p(x, y) x 3/ y / dollars/day if she offers x sketches of her cat and y sketches of her dog at the beginning of a day. If she is only able to create 3 total sketches per day, what is the maximum profit she can make per day? Round your answer to the nearest cent.

238

239 Lesson 7: Double Integrals, Volume, Applications (I). Integrating Functions of Several Variables Today we address how we integrate functions of several variables. The motivation is initially very geometric in that we have our function of two variables and we want to determine how much volume there is under the curve (just as we thought of integrals as the area under a curve). Figure. The volume under the function over the region R (blue square) is contained in the black and red box. Note 67. The regions over which we integrate will always be -dimensional. If R is the region we are considering, then the volume under f(x, y) over the region R is denoted (6) f(x, y) da. R The da simply stands for the area of the region R. We write it this when when it isn t immediately clear if we want to integrate with respect to x or y first. Ex. Consider the function f(x, y) x + y and the region R bounded by x and y, then the volume under f(x, y) over R is denoted f(x, y) da (x + y) dx dy. R 5