Calculus And Analytical Geometry Handouts MTH101 PDF

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This document is a set of handouts for a calculus and analytical geometry course (MTH 101) at the Virtual University of Pakistan. It covers various topics including coordinates, graphs, lines, absolute value, functions, limits, and derivatives. The handouts are organized into lessons; each lesson is assigned a number.

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Calculus And Analytical Geometry MTH 101 Virtual University of Pakistan Knowledge beyond the boundaries 45-Planning Production Levels: Linear Programming VU TABLE OF CONTENTS : Lesson 1 :Coordinates, Graphs, Lines...

Calculus And Analytical Geometry MTH 101 Virtual University of Pakistan Knowledge beyond the boundaries 45-Planning Production Levels: Linear Programming VU TABLE OF CONTENTS : Lesson 1 :Coordinates, Graphs, Lines 3 Lesson 2 :Absolute Value 15 Lesson 3 :Coordinate Planes and Graphs 24 Lesson 4 :Lines 34 Lesson 5 :Distance; Circles, Quadratic Equations 45 Lesson 6 :Functions and Limits 57 Lesson 7 :Operations on Functions 63 Lesson 8 :Graphing Functions 69 Lesson 9 :Limits (Intuitive Introduction) 76 Lesson 10:Limits (Computational Techniques) 84 Lesson 11: Limits (Rigorous Approach) 93 Lesson 12 :Continuity 97 Lesson 13 :Limits and Continuity of Trigonometric Functions 104 Lesson 14 :Tangent Lines, Rates of Change 110 Lesson 15 :The Derivative 115 Lesson 16 :Techniques of Differentiation 123 Lesson 17 :Derivatives of Trigonometric Function 128 Lesson 18 :The chain Rule 132 Lesson 19 :Implicit Differentiation 136 Lesson 20 :Derivative of Logarithmic and Exponential Functions 139 Lesson 21 :Applications of Differentiation 145 Lesson 22 :Relative Extrema 151 Lesson 23 :Maximum and Minimum Values of Functions 158 Lesson 24 :Newton’s Method, Rolle’s Theorem and Mean Value Theorem 164 Lesson 25 :Integrations 169 Lesson 26 :Integration by Substitution 174 Lesson 27 :Sigma Notation 179 Lesson 28 :Area as Limit 183 Lesson 29 :Definite Integral 191 Lesson 30 :First Fundamental Theorem of Calculus 200 Lesson 31 :Evaluating Definite Integral by Subsitution 206 Lesson 32 :Second Fundamental Theorem of Calculus 210 Lesson 33 :Application of Definite Integral 214 Lesson 34 :Volume by slicing; Disks and Washers 221 Lesson 35 :Volume by Cylindrical Shells 230 Lesson 36 :Length of Plane Curves 237 Lesson 37 :Area of Surface of Revolution 240 Lesson 38:Work and Definite Integral 245 Lesson 39 :Improper Integral 252 Lesson 40 :L’Hopital’s Rule 258 Lesson 41 :Sequence 265 Lesson 42 :Infinite Series 276 Lesson 43 :Additional Convergence tests 285 Lesson 44 :Alternating Series; Conditional Convergence 290 Lesson 45 :Taylor and Maclaurin Series 296 1-Coordinates, Graphs and Lines VU Lecture 1 Coordinates, Graphs and Lines What is Calculus?? Well, it is the study of the continuous rates of the change of quantities. It is the study of how various quantities change with respect to other quantities. For example, one would like to know how distance changes with respect to (from now onwards we will use the abbreviation w.r.t) time, or how time changes w.r.t speed, or how water flow changes w.r.t time etc. You want to know how this happens continuously. We will see what continuously means as well. In this lecture, we will talk about the following topics: -Real Numbers -Set Theory -Intervals -Inequalities -Order Properties of Real Numbers Let's start talking about Real Numbers. We will not talk about the COMPLEX or IMANGINARY numbers, although your text has something about them which you can read on your own. We will go through the history of REAL numbers and how they popped into the realm of human intellect. We will look at the various types of REALS - as we will now call them. So Let's START. The simplest numbers are the natural numbers Natural Numbers 1, 2, 3, 4, 5,… They are called the natural numbers because they are the first to have crossed paths with human intellect. Think about it: these are the numbers we count things with. So our ancestors used these numbers first to count, and they came to us naturally! Hence the name NATURAL!!! The natural numbers form a subset of a larger class of numbers called the integers. I have used the word SUBSET. From now onwards we will just think of SET as a COLLECTION OF THINGS. This could be a collection of oranges, apples, cars, or politicians. For example, if I have the SET of politicians then a SUBSET will be just a part of the COLLECTION. In mathematical notation we say A is subset of B if ∀x ∈ A ⇒ x ∈ B.Then we write A ⊆ B. Set The collection of well defined objects is called a set. For example © Copyright Virtual University of Pakistan 3 1-Coordinates, Graphs and Lines VU {George Bush, Toney Blair, Ronald Reagoan} Subset A portion of a set B is a subset of A iff every member of B is a member of A. e.g. one subset of above set is {George Bush, Tony Blair} The curly brackets are always used for denoting SETS. We will get into the basic notations and ideas of sets later. Going back to the Integers. These are …, -4, -3, -2, -1, 0, 1, 2, 3, 4,… So these are just the natural numbers, plus a 0, and the NEGATIVES of the natural numbers. The reason we didn’t have 0 in the natural numbers is that this number itself has an interesting story, from being labeled as the concept of the DEVIL in ancient Greece, to being easily accepted in the Indian philosophy, to being promoted in the use of commerce and science by the Arabs and the Europeans. But here, we accept it with an open heart into the SET of INTEGERS. What about these NEGATIVE Naturals??? Well, they are an artificial construction. They also have a history of their own. For a long time, they would creep up in the solutions of simple equations like x+2 = 0. The solution is x = - 2 So now we have the Integers plus the naturals giving us things we will call REAL numbers. But that's not all. There is more. The integers in turn are a subset of a still larger class of numbers called the rational numbers. With the exception that division by zero is ruled out, the rational numbers are formed by taking ratios of integers. Examples are 2/3, 7/5, 6/1, -5/2 Observe that every integer is also a rational number because an integer p can be written as a ratio. So every integer is also a rational. Why not divide by 0? Well here is why: If x is different from zero, this equation is contradictory; and if x is equal to zero, this equation is satisfied by any number y, so the ratio does not have a unique value a situation that is mathematically unsatisfactory. x/0 =y⇒ x= 0. y ⇒ x = 0 For these reasons such symbols are not assigned a value; they are said to be undefined. So we have some logical inconsistencies that we would like to avoid. I hope you see that!! Hence, no division by 0 allowed! Now we come to a very interesting story in the history of the development of Real numbers. The discovery of IRRATIONAL numbers. Pythagoras was an ancient Greek philosopher and mathematician. He studied the properties of numbers for its own sake, not necessarily for any applied problems. This was a major change in mathematical thinking as © Copyright Virtual University of Pakistan 4 1-Coordinates, Graphs and Lines VU math now took on a personality of its own. Now Pythagoras got carried away a little, and developed an almost religious thought based on math. He concluded that the size of a physical quantity must consist of a certain whole number of units plus some fraction m / n of an additional unit. Now rational numbers have a unique property that if you convert them to decimal notation, the numbers following the decimal either end quickly, or repeat in a pattern forever. Example: 1/2 =0.500000… =0.5 1/3= 0.33333… This fit in well with Pythagoras’ beliefs. All is well. But this idea was shattered in the fifth century B.C. by Hippasus of Metapontum who demonstrated the existence of irrational numbers, that is, numbers that cannot be expressed as the ratio of integers. Using geometric methods, he showed that the hypotenuse of the right triangle with base and opposite side equal to 1 cannot be expressed as the ratio of integers, thereby proving that 2 is an IRRATIONAL number. The hypotenuse of this right triangle can be expressed as the ratio of integers. Other examples of irrational numbers are Cos 190 ,1 + 2 The rational and irrational numbers together comprise a larger class of numbers, called REAL NUMBERS or sometimes the REAL NUMBER SYSTEM.So here is a pictorial summary of the hierarchy of REAL NUMBNERS. © Copyright Virtual University of Pakistan 5 1-Coordinates, Graphs and Lines VU Pictorial summary of the hierarchy of REAL NUMBNERS COORDINATE Line In the 1600’s, analytic geometry was “developed”. It gave a way of describing algebraic formulas by geometric curves and, conversely, geometric curves by algebraic formulas. So basically you could DRAW PICTURES OF THE EQUATIONS YOU WOULD COME ACROSS, AND WRITE DOWN EQUATIONS OF THE PICTURES YOU RAN INTO! The developer of this idea was the French mathematician, Descartes.The story goes that he wanted to find out as to what Made humans HUMANS?? Well, he is said to have seated himself in a 17th century furnace (it was not burning at the time!) and cut himself from the rest of the world. In this world of cold and darkness, he felt all his senses useless. But he could still think!!!! So he concluded that his ability to think is what made him human, and then he uttered the famous line : “ I THINK, THEREFORE I AM”.In analytic geometry , the key step is to establish a correspondence between real numbers and points on a line. We do this by arbitrarily designating one of the two directions along the line as the positive direction and the other as © Copyright Virtual University of Pakistan 6 1-Coordinates, Graphs and Lines VU the negative direction. So we draw a line, and call the RIGHT HAND SIDE as POSITIVE DIRECTION, and the LEFT HAND SIDE as NEGATIVE DIRECTION. We could have done it the other way around too. But, since what we just did is a cultural phenomenon where right is + and left is -, we do it this way. Moreover, this has now become a standard in doing math, so anything else will be awkward to deal with. The positive direction is usually marked with an arrowhead so we do that too. Then we choose an arbitrary point and take that as our point of reference. We call this the ORIGIN, and mark it with the number 0. So we have made our first correspondence between a real number and a point on the Line. Now we choose a unit of measurement, say 1 cm. It can be anything really. We use this unit of measurement to mark of the rest of the numbers on the line. Now this line, the origin, the positive direction, and the unit of measurement define what is called a coordinate line or sometimes a real line. With each real number we can now associate a point on the line as fo1lows: Associate the origin with the number 0. Associate with each positive number r the point that is a distance of r units (this is the unit we chose, say 1 cm) in the positive direction from the origin. Associate with each negative number ‘- r ’ the point that is a distance of r units in the negative direction from the origin. The real number corresponding to a point on the line is called the coordinate of the point. Example 1: In Figure we have marked the locations of the points with coordinates -4, -3, -1.75, -0.5, π , 2 and 4. The locations of π and 2 which are approximate, were obtained from their decimal approximations, π = 3.14 and 2 = 1.41 © Copyright Virtual University of Pakistan 7 1-Coordinates, Graphs and Lines VU It is evident from the way in which real numbers and points on a coordinate line are related that each real number corresponds to a single point and each point corresponds to a single real number. To describe this fact we say that the real numbers and the points on a coordinate line are in one-to-one correspondence. Order Properties In mathematics, there is an idea of ORDER of a SET. We won’t go into the general concept, since that involves SET THEORY and other high level stuff. But we will define the ORDER of the real number set as follows: For any two real numbers a and b, if b-a is positive, then we say that b > a or that a < b. Here I will assume that we are all comfortable working with the symbol “” which is read as “greater than.” I am assuming this because this stuff was covered in algebra before Calculus. So with this in mind we can write the above statement as If b - a is positive, then we say that b > a or that a < b.A statement involving < or > are called an INEQUALITY.Note that the inequality a < b can also be expressed as b > a. So ORDER of the real number set in a sense defines the SIZE of a real number relative to another real number in the set. The SIZE of a real number a makes sense only when it is compared with another real b. So the ORDER tells you how to “ORDER” the numbers in the SET and also on the COORDINATE LINE! A little more about inequalities. The inequality a ≤ b is defined to mean that either a < b or a = b. So there are two conditions here. For example, the inequality 2 ≤ 6 would be read as 2 is less than or it is equal to 6. We know that it’s less than 6, so the inequality is true. SO IF ONE OF THE CONDITIONS IS TRUE, THEN THE INEQUALITY WILL BE TRUE, We can say a similar thing about. The expression a < b < c is defined to mean that a < b and b < c. It is also read as “b is between a and c”. As one moves along the coordinate line in the positive direction, the real numbers increase in size. In other words, the real numbers are ordered in an ascending manner on the number line, just as they are in the SET of REAL NUMBERS. So that on a horizontal coordinate line the inequality a < b implies that a is to the left of b, and the inequality a < b < c implies that a is to the left of b and b is to the left of c. The symbol a < b < c means a < b and b < c. I will leave it to the reader to deduce the meanings of such symbols as ≤ and ≥. Here is an example of INEQUALITIES. a≤b −1 > −3 Some incorrect inequalities are: 2 ≥ 4, π ≤ 0, 5 < −3 REMARK: To distinguish verbally between numbers that satisfy a ≥ 0 and those that satisfy a > 0, we shall call a nonnegative if a ≥ 0 and positive if a > 0. Thus, a nonnegative number is either positive or zero. The following properties of inequalities are frequently used in calculus. We omit the proofs, but will look at some examples that will make the point. THEOREM 1.1.1 a ) If a < b and b < c, then a < c b) If a < b and a + c < b + c, then a − c < b − c c) If a < b and ac < bc, when c is positive and ac > bc when c is negative. d ) If a < b and c < d , then a + c < b + d e) If a and b are both positive or both negative 1 1 and a < b then > a b REMARK These five properties remain true if < and > are replaced by ≤ and ≥ INTERVALS We saw a bit about sets earlier. Now we shall assume in this text that you are familiar with the concept of a set and fully understand the meaning of the following symbols. However, we will give a short explanation of each. Given two sets A and B a ∈ A : a is an element of the set A, 2 ∈ {1, 2, 3, 4} © Copyright Virtual University of Pakistan 9 1-Coordinates, Graphs and Lines VU a ∉ A : a is NOT an element of the set A 5 ∉{1, 2, 3, 4} ∅ represents the Empty set, or the set that contains nothing. A ∪ B represents the SET of all the elements of the Set A and the Set B taken together. Example: A = {1,2,3,4}, B = {1,2,3,4,5,6,7}, then, A ∪ B = {1,2,3,4,5,6,7} A ∩ B represents the SET of all those elements that are in Set A AND in Set B. Example: A = {1,2,3,4}, B = {1,2,3,4,5,6,7}, then A ∩ B = {1,2,3,4} A = B means the A is exactly the same set as B Example: A = {1,2,3,4} and B = {1,2,3,4}, then A = C and A ⊂ B means that the Set A is contained in the Set B. Recall the example we did of the Set of all politicians! {George Bush, Tony Blair} ⊂ {George Bush, Toney Blair, Ronald Reagoan} One way to specify the idea of a set is to list its members between braces. Thus, the set of all positive integers less than 5 can be written as { 1, 2, 3, 4} and the set of all positive even integers can be written as { 2, 4, 6, …} where the dots are used to indicate that only some of the members are explicitly and the rest can be obtained by continuing the pattern. So here the pattern is that the set consists of the even numbers, and the next element must be 8, then 10, and then so on. When it is inconvenient or impossible to list the members of a set, as would be if the set is infinite, then one can use the set- builder notation. This is written as { x : _________ } which is read as “the set of all x such that ______” , In place of the line, one would state a property that specifies the set, Thus, { x : x is real number and 2 < x < 3} is read, "the set of all x such that x is a real number and 2 < x < 3," Now we know by now that © Copyright Virtual University of Pakistan 10 1-Coordinates, Graphs and Lines VU 2 < x < 3 means that all the x between 2 and 3. This specifies the “description of the elements of the set” This notation describes the set, without actually writing down all its elements. When it is clear that the members of a set are real numbers, we will omit the reference to this fact. So we will write the above set as Intervals. We have had a short introduction of Sets. Now we look particular kind of sets that play a crucial role in Calculus and higher math. These sets are sets of real numbers called intervals. What is an interval? { x: 2< x< 3} Well, geometrically, an interval is a line segment on the co-ordinate line. S if a and b are real numbers such that a < b, then an interval will be just the line segment joining a and b. But if things were only this simple! Intervals are of various types. For example, the question might be raised whether a and b are part of the interval? Or if a is, but b is not?? Or maybe both are? Well, this is where we have to be technical and define the following: The closed interval from a to b is denoted by [a, b] and is defined as [a, b] = { x : a ≤ x ≤ b} Geometrically this is the line segment So this includes the numbers a and b, a and b a are called the END- POINTS of the interval. The open interval from a to b is denoted by and is defined by (a, b) = { x: a < x < b} This excludes the numbers a and b. The square brackets indicate that the end points are included in the interval and the parentheses indicate that they are not. Here are various sorts of intervals that one finds in mathematics. In this picture, the geometric pictures use solid dots to denote endpoints that are included in the interval and open dots to denote endpoints that are not. © Copyright Virtual University of Pakistan 11 1-Coordinates, Graphs and Lines VU As shown in the table, an interval can extend indefinitely in either the positive direction, the negative direction, or both. The symbols −∞ (read "negative infinity") and +∞ (read , 'positive infinity' ') do not represent numbers: the +∞ indicates that the interval extends indefinitely in the positive direction, and the −∞ indicates that it extends indefinitely in the negative direction. An interval that goes on forever in either the positive or the negative directions, or both, on the coordinate line or in the set of real numbers is called an INFINITE interval. Such intervals have the symbol for infinity at either end points or both, as is shown in the table An interval that has finite real numbers as end points are called finite intervals. A finite interval that includes one endpoint but not the other is called half-open (or sometimes half-closed). [a, + ∞), (a, − ∞), (−∞, b], (−∞, b) Infinite intervals of the form [a, + ∞) and (−∞, b] are considered to be closed because they contain their endpoint. Those of the form (a, − ∞) and (−∞, b) has no endpoints; it is regarded to be both open and closed. As one of my Topology Instructors used to say: “A set is not a DOOR! It can be OPEN, it can be CLOSED, and it can be OPEN and CLOSED!! Let's remember this fact for good!” Let's look at the picture again for a few moments and digest the © Copyright Virtual University of Pakistan 12 1-Coordinates, Graphs and Lines VU information. PAUSE 10 seconds. SOLVING INEQUALITIES We have talked about Inequalities before. Let's talk some more. First Let's look at an inequality involving and unknown quantity, namely x. Here is one: x < 5,x = 1, is a solution of this inequality as 1 makes it true, but x = 7 is not. So the set of all solutions of an inequality is called its solution set. The solution set of x < 5 will be It is a fact, though we wont prove this that if one does not multiply both sides of an inequality by zero or an expression involving an unknown, then the operations in Theorem 1.1.1 will not change the solution set of the inequality. The process of finding the solution set of an inequality is called solving the Inequality. Let's do some fun stuff, like some concrete example to make things a bit more focused Example 4. Solve 3 + 7x ≤ 2x − 9 Solution. We shall use the operations of Theorem 1.1.1 to isolate x on one side of the inequality 7 x ≤ 2 x − 12 Subtracting 3 from both sides 5 x ≤ − 12 Subtracting 2 x from both sides x ≤ −12 / 5 Dividing both sides by 5 Because we have not multiplied by any expressions involving the unknown x, the last inequality has the same solution set as the first. Thus, the solution set is the interval shown in Figure 1.1.6. Example Solve 7 ≤ 2 − 5x < 9 Solution ; The given inequality is actually a combination of the two inequalities 7 ≤ 2 − 5 x and 2 − 5 x < 9 We could solve the two inequalities separately, then determine the value of x that satisfy both by taking the intersection of the solution sets , however, it is possible to work with the combined inequality in this problem © Copyright Virtual University of Pakistan 13 1-Coordinates, Graphs and Lines VU 5 ≤ −5 x < 7 Subtracting 2 from both sides −1 ≥ x < −7 / 5 Dividing by − 5 inequality symbols reversed −7 / 5 < x ≤ −1 Re writing with smaller number first (−7 / 5, −1] Thus the solution set is interval shown the figure Example Similarly, you can find © Copyright Virtual University of Pakistan 14 2-Absolute Values VU Lecture # 2 Absolute Value In this lecture we shall discuss the notation of Absolute Value. This concept plays an important role in algebraic computations involving radicals and in determining the distance between points on a coordinate line. Definition The absolute vale or magnitude of a real number a is denoted by |a| and is defined by a if a ≥ 0, that is, a is non − negative |a|=  −a if a < 0, that is, a is positive. Technically, 0 is considered neither positive, nor negative in Mathematics. It is called a non-negative number. Hence whenever we want to talk about a real number a such that a ≥ 0, we call a non-negative, and positive if a > 0. Example 4 4 4 5 =5 , − = −( − ) = , 0 =0 7 7 7 since 5>0 since -4/7 < 0 since 0 ≥ 0 Note that the effect of taking the absolute value of a number is to strip away the minus sign if the number is negative and to leave the number unchanged if it is non-negative. Thus, |a| is a non-negative number for all values of a and − a ≤ a ≤ a , if ‘a’ is itself is negative, then ‘–a’ is positive and ‘+a’ is negative. a + b ≥ 0 or a + b < 0 a+b= a+b a+b ≤ a + b a+b =−( a + b) © Copyright Virtual University of Pakistan 15 2-Absolute Values VU Caution: Symbols such as +a and –a are deceptive, since it is tempting to conclude that +a is positive and –a is negative. However this need not be so, since a itself can represent either a positive or negative number. In fact , if a itself is negative, then –a is positive and +a is negative. Example: Solve x−3 = 4 Solution: Depending on whether x-3 is positive or negative , the equation |x-3| = 4 can be written as x-3 = 4 or x-3 = -4 Solving these two equations give x=7 and x=-1 Example Solve 3x − 2 = 5 x + 4 Because two numbers with the same absolute value are either equal or differ only in sign, the given equation will be satisfied if either 3x − 2 = 5 x + 4 3 x − 5 x =4 + 2 −2 x = 6 x = −3 Or 3 x − 2 =−(5 x + 4) 3 x − 2 =−5 x − 4 3 x + 5 x =−4 + 2 1 x= − 4 © Copyright Virtual University of Pakistan 16 2-Absolute Values VU Relationship between Square Roots and Absolute Values : Recall that a number whose square is a is called a square root of a. In algebra it is learned that every positive real number a has two real square roots, one positive and one negative. The positive square root is denoted by a For example, the number 9 has two square roots, -3 and 3. Since 3 is the positive square root, we have 9 = 3. In addition, we define 0 = 0. It is common error to write a 2 = a. Although this equality is correct when a is nonnegative, it is false for negative a. For example, if a=-4, then a 2 = (−4) 2 = 16 =4 ≠ a The positive square root of the square of a number is equal to that number. A result that is correct for all a is given in the following theorem. Theorem: For any real number a, a2 = a Proof : Since a2 = (+a)2 = (-a)2, the number +a and –a are square roots of a2. If a ≥ 0 , then +a is nonnegative square root of a2, and if a < 0 , then -a is nonnegative square root of a2. Since a2 denotes the nonnegative square root of a2, we have a2 = +a if a ≥ 0 if a 2 = −a if a < 0 That is, a2 = a Properties of Absolute Value Theorem If a and b are real numbers, then (a) |-a| = |a|, a number and its negative have the same absolute value. (b) |ab| = |a| |b|, the absolute value of a product is the product of absolute values. (c) |a/b| = |a|/|b|, the absolute value of the ratio is the ratio of the absolute values © Copyright Virtual University of Pakistan 17 2-Absolute Values VU Proof (a) : | −a=| ( − a )= 2 a= 2 |a| Proof (b) : = ab = (ab) 2 = a 2b 2 a= 2 b2 a b This result can be extended to three or more factors. More precisely, for any n real numbers, a1,a2,a3,……an, it follows that |a 1 a 2 …..a n | = |a 1 | |a 2 | …….|a n | In special case where a 1 , a 2 ,…….,a n have the same value, a, it follows from above equation that |an|=|a|n Example (a) |-4|=|4|=4 (b) |(2)(-3)|=|-6|=6=|2||-3|=(2)(3)=6 (c) |5/4|=5/4=|5|/|4|=5/4 Geometric Interpretation Of Absolute Value The notation of absolute value arises naturally in distance problems, since distance is always nonnegative. On a coordinate line, let A and B be points with coordinates a and b, the distance d between A and B is b − a if a < b  = d a − b if a > b 0 if a = b  © Copyright Virtual University of Pakistan 18 2-Absolute Values VU As shown in figure b-a is positive, so b-a=|b-a| ; in the second case b-a is negative, so a-b = -(b-a) = |b-a|. Thus, in all cases we have the following result: Theorem (Distance Formula) If A and B are points on a coordinate line with coordinates a and b, respectively, then the distance d between A and B is d = |b-a| This formula provides a useful geometric interpretation Of some common mathematical expressions given in table here Table EXPRESSION GEOMETRIC INTERPRETATION ON A COORDINATE LINE |x-a| The distance between x and a |x+a| The distance between x and -a |x| The distance between x and origin Inequalities of the form |x-a|k arise often, so we have summarized the key facts about them here in following table © Copyright Virtual University of Pakistan 19 2-Absolute Values VU Example Solve |x-3| < 4 Solution: This inequality can be written as -4 < x-3 < 4 adding 3 throughout we get -1 < x < 7 This can be written in interval notation as (-1,7) © Copyright Virtual University of Pakistan 20 2-Absolute Values VU Example Solve x + 4 ≥ 2 Solution: The given inequality can be written as  x + 4 ≤ −2  or x + 4 ≥ 2  or simply  x ≤ −6  or  x ≥ −2  Which can be written in set notation as ( −∞, −6]U [ −2, +∞ ) The Triangle Inequality © Copyright Virtual University of Pakistan 21 2-Absolute Values VU It is not generally true that a + b = a + b For example , if a =2 and b =−3 , then a + b =−1 so that a + b =−1 =1 whereas a = b = 2 + −3 = 2 + 3 = 5 so a + b ≠ a + b It is true, however, that the absolute value of a sum is always less than or equal to the sum of the absolute values. This is the content of the following very important theorem, known as the triangle inequality. This TRIANGLE INEQUALITY is the essence of the famous HISENBURG UNCERTAINITY PRINCIPLE IN QUANTUM PHYSICS, so make sure you understand it fully. THEOREM 1.2.5 (Triangle Inequality) If a and b are any real numbers, then a+b ≤ a + b PROOF Remember the following inequalities we saw earlier. − a ≤ a ≤ a and − b ≤ b ≤ b Let's add these two together. We get −a ≤a≤ a + − b ≤b≤ b = ( − a ) + (− b ) ≤ a + b ≤ a + b (B) Since a and b are real numbers, adding them will also result in a real number. Well, there are two types of real numbers. What are they?? Remember!!!!! They are either > = 0, or they are < 0! Ok!! SO we have a + b ≥ 0 or a + b < 0 In the first of these cases where a+b ≥ 0 certainly a+b= a+b by definition of absolute value. so the right-hand inequality in (B) gives a+b ≤ a + b In the second case © Copyright Virtual University of Pakistan 22 2-Absolute Values VU a+b =−( a + b) But this is the same as a + b =− a + b So the left-hand inequality in (B) can be written as −( a + b ) ≤ − a + b Multiplying both sides of this inequality by - 1 give a+b ≤ a + b © Copyright Virtual University of Pakistan 23 3- Coordinate Planes and Graphs VU Lecture # 3 In this lecture we will discuss Graphs in the coordinate plane. Intercepts. Symmetry Plane. We begin with the Coordinate plane. Just as points on a line can be placed in one-to-one correspondence with the real numbers, so points in the PLANE can be placed in one-to-one correspondence with pairs of real numbers. What is a plane? A PLANE is just the intersection of two COORDINATE lines at 90 degrees. It is technically called the COORDINATE PLANE, but we will call it the plane also whenever it is convenient. Each line is a line with numbers on it, so to define a point in the PLANE, we just read of the corresponding points on each line. For example I pick a point in the plane By an ordered pair of real numbers we mean two real numbers in an assigned order. Every point P in a coordinate plane can be associated with a unique ordered pair of real numbers by drawing two lines through P, one perpendicular to the x-axis and the other to the y-axis. © Copyright Virtual University of Pakistan 24 3- Coordinate Planes and Graphs VU For example if we take (a,b)=(4,3), then on coordinate plane To plot a point P(a, b) means to locate the point with coordinates (a, b) in a coordinate plane. For example, In the figure below we have plotted the points P(2,5), Q(-4,3), R(-5,-2), and S(4,-3).Now this idea will enable us to visualise algebraic equations as geometric curves and, conversely, to represent geometric curves by algebraic equations. Labelling the axes with letters x and y is a common convention, but any letters may be used. If the letters x and y are used to label the coordinate axes, then the resulting plane is also called an xy-plane. In applications it is common to use letters other than x and y to label coordinate axes. Figure below shows a uv-plane and a ts- plane. The first letter in the name of the plane refers to the horizontal axis and the second to the vertical axis. © Copyright Virtual University of Pakistan 25 3- Coordinate Planes and Graphs VU Here is another terminology. The COORDINATE PLANE and the ordered pairs we just discussed is together known as the RECTANGULAR COORDINATE SYSTEM. In a rectangular coordinate system the coordinate axes divide the plane into four regions called quadrants. These are numbered counter clockwise with Roman numerals as shown in the Figure below. Consider the equations 5 xy = 2 x2 + 2 y 2 = 7 = y x3 − 7 We define a solution of such an equation to be an ordered pair of real numbers(a,b) so that the equation is satisfactory when we substitute x=a and y=b. Example 1 The pair (3,2) is a solution of 6 x − 4 y = 10 © Copyright Virtual University of Pakistan 26 3- Coordinate Planes and Graphs VU since this equation is satisfied when we substitute x = 3 and y = 2. That is 6(3) − 4(2) = 10 which is true!! However, the pair (2,0) is not a solution, since 6(2) − 4(0) = 18 ≠ 10 We make the following definition in order to start seeing algebraic objects geometrically. Definition. The GRAPH of an equation in two variables x and y is the set of all points in the xy-plane whose coordinates are members of the solution set of the equation. Example 2 Sketch the graph of y = x 2 When we plot these on the xy-plane and connect them, we get this picture of the graph © Copyright Virtual University of Pakistan 27 3- Coordinate Planes and Graphs VU IMPORTANT REMARK. It should be kept in mind that the curve in above is only an approximation to the graph of y = x 2. When a graph is obtained by plotting points, whether by hand, calculator, or computer, there is no guarantee that the resulting curve has the correct shape. For example, the curve in the Figure here pass through the points tabulated in above table. © Copyright Virtual University of Pakistan 28 3- Coordinate Planes and Graphs VU INTERCEPTS Points where a graph intersects the coordinate axes are of special interest in many problems. As illustrated before, intersections of a graph with the x-axis have the form (a, 0) and intersections with the y-axis have the form (0, b). The number a is called an x-intercept of the graph and the number b a y-intercept. Example: Find all intercepts of Solution is the required x-intercept is the required y-intercept Similarly you can solve part (b), the part (c) is solved here In the following figure, the points (x,y),(-x,y),(x,-y) and (-x,-y) form the corners of a rectangle. © Copyright Virtual University of Pakistan 29 3- Coordinate Planes and Graphs VU SYMMETRY Symmetry is at the heart of many mathematical arguments concerning the structure of the universe, and certainly symmetry plays an important role in applied mathematics and engineering fields. Here is what it is. As illustrated in Figure the points (x, y ) , ( − x, y ), ( x, − y ) and ( − x, − y ) form the corners of a rectangle. © Copyright Virtual University of Pakistan 30 3- Coordinate Planes and Graphs VU For obvious reasons, the points (x,y) and (x,-y) are said to be symmetric about the x-axis and the points ( x, y) and ( -x, y) are symmetric about the y-axis and the points (x, y) and ( -x, -y) symmetric about the origin. SYMMETRY AS A TOOL FOR GRAPHING By taking advantage of symmetries when they exist, the work required to obtain a graph can be reduced considerably. Example 9 Sketch the graph of the equation 1 4 =y x − x2 8 1 Solution. The graph is symmetric about the y-axis since substituting − x for x yields y= (− x) 4 − (− x) 2 8 which simplifies to the original equation. As a consequence of this symmetry, we need only calculate points on the graph that lies in the right half of the xy-plane ( x >= 0). The corresponding points in the left half of the xy-plane ( x a f(x), I will assume that f(x) will have a limit that matches from both sides and so the LIMIT EXISTS for f (x). So I won’t distinguish between left and right hand limits. We begin with a table of LIMITS of two basic functions The functions are f ( x) = k g ( x) = x Here is the table of the limits and the same information from the graph f ( x) = k g ( x) = x © Copyright Virtual University of Pakistan 84 10-Limits and Computational Techniques VU Here we have a theorem that will help with computing limits. Won’t prove this theorem, but some of the parts of this theorem are proved in Appendix C of your text book. THEOREM 2.5.1 Let Lim stand for one of the limits lim, lim , lim , lim , lim x→a x→a− x→a+ x →+∞ x →−∞ if L1 = lim f ( x) and L2 = lim g ( x) both exists, then a ) lim[ f ( x) + g ( x)] = lim f ( x) + lim g ( x) = L1 + L2 b) lim[ f ( x) − g ( x)] = lim f ( x) − lim g ( x) = L1 − L2 c) lim[ f ( x). g ( x)] = lim f ( x). lim g ( x) = L1.L2 f ( x) lim f ( x) d ) lim[ ]= g ( x) lim g ( x) L = 1 ( L2 ≠ 0) L2 For the Last theorem, say things like “Limit of the SUM is the SUM of the LIMITS etc. Parts a) and c) of the theorem apply to as many functions as you want Part a) gives lim[ f1 ( x) + f 2 ( x) +... + f ( xn )] = lim f1 ( x) + lim f 2 ( x) +...lim f ( xn ) Part c) gives © Copyright Virtual University of Pakistan 85 10-Limits and Computational Techniques VU lim[ f1 ( x) ⋅ f 2 ( x) ⋅... ⋅ f n ( x)] = lim f1 ( x) ⋅ lim f 2 ( x) ⋅... ⋅ lim f n ( x) Also if f1= f 2=...= f n then lim[ f ( x)]n = [lim f ( x)]n From this last result we can say that = lim( = x n ) [lim xn ] an x→a x→a This is a useful result and we can use it later. Another useful result follows from part c) of the theorem. Let f(x) = k in part c), where k is a constant (number). lim[kg ( x)] = lim(k ) ⋅ lim g ( x) = k ⋅ lim g ( x) So a constant factor can be moved through a limit sign LIMITS OF POLYNOMIAL Polynomials are functions of the form f ( x)= bn x n + bn −1 x n −1 +.... + b1 x + b0 Where the a’s are all real numbers Let's find the Limits of polynomials and x approaches a numbers a Example lim( x 2 − 4 x + 3) x →5 = lim x 2 − lim 4 x + lim 3 x →5 x →5 x →5 = lim x − 4 lim x + lim 3 2 x →5 x →5 x →5 = (5) − 4(5) + 3= 8 2 Theorem 2.5.2 Proof lim p (= x) lim(c0 + c1 x +... + cn x n ) x→a x→a = lim c0 + lim c1 x +... + lim cn x n x→a x→a x→a = lim c0 + c1 lim x +... + cn lim x n x→a x→a x→a = c0 + c1a +... + cn a = p (a ) n © Copyright Virtual University of Pakistan 86 10-Limits and Computational Techniques VU 1 Limits Involving x 1 Let's look at the graph of f ( x) = x Then by looking at the graph AND by looking at the TABLE of values we get the following Results 1 lim+ = +∞ x →0 x 1 lim− = −∞ x →0 x 1 lim = 0 x →+∞ x 1 lim = 0 x →−∞ x © Copyright Virtual University of Pakistan 87 10-Limits and Computational Techniques VU For every real number a, the function 1 1 g ( x) = is a translation of f ( x) =. x−a x So we an say the following about this function LIMITS OF POLYNOMIALS AS X GOES TO +INF AND –INF From the graphs given here we can say the following about polynomials of the form lim x n = +∞ n = 1, 2,3,... x →+∞ +∞ n = 2, 4, 6,... lim x n =  x →−∞ −∞ n = 1,3,5,... © Copyright Virtual University of Pakistan 88 10-Limits and Computational Techniques VU EXAMPLE lim 2 x 5 = +∞ x →+∞ lim − 7 x 6 = −∞ x →+∞ n 1  1  = lim =lim  0 x →+∞ x n  x →+∞ x n  n 1  1  = lim n = lim n  0 x →−∞ x  x →−∞ x  HERE are the graphs of the functions y = f ( x) =1/ x n ( n is positive integer) Limit as x goes to +inf or –inf of a polynomial is like the Limits of the highest power of x lim (c0 + c1 x +... + cn x n ) =lim cn x n x →+∞ x →+∞ Motivation © Copyright Virtual University of Pakistan 89 10-Limits and Computational Techniques VU c0 c (c0 + c1 x +... + cn x= n + n1−1 +... + cn ) ) xn ( n x x n Factor out x , and then from what we just saw about 1 the limit of n , everything goes to 0 as x → +∞ x or x → −∞ except cn Limits of Rational Functions as x goes to a A rational function is a function defined by the ratio of two polynomials Example 5 x3 + 4 Find lim x→2 x − 3 Sol: lim 5 x3 + 4 = x→2 lim x − 3 x→2 5(2)3 + 4 = = −44 2−3 We used d) of theorem 2.5.1 to evaluate this limit. We would not be able to use it if the denominator turned out to be 0 as that is not allowed in Mathematics. If both top and bottom approach 0 as x approaches a, then the top and bottom will have a common factor of x – a. In this case the factors can be cancelled and the limit works out. Example x2 − 4 lim x→2 x − 2 ( x + 2)( x − 2) = lim x→2 ( x − 2) = lim( x + 2)= 4 x→2 Note that x is not equal to two after Simplification for the two functions to be the same. Nonetheless, we calculated the limit as if we were substituting x = 2 using rule for polynomials That’s ok since REALLY LIMIT means you are getting close to 2, but not equaling it!! What happens if in a rational functions, the bottom limit is 0, but top is not?? It’s like the limit as x goes to 0 of f(x) = 1/x. © Copyright Virtual University of Pakistan 90 10-Limits and Computational Techniques VU The limit may be + inf The limits may be –inf +inf from one side and –inf from another Example 2− x Find lim+ x→4 ( x − 4)( x + 2) The top is –2 as x goes to 4 from right side. The bottom goes to 0, so the limit will be inf of some type. To get the sign on inf, Let's analyze the sign of the bottom for various values of real numbers Break the number line into 4 intervals as in The important numbers are the ones that make the top and bottom zero. As x approaches 4 from the right, the ratio stays negative and the result is –inf. You can say something about what happens from the left. Check yourselves by looking at the pic. 2− x So lim+ = −∞ x→4 ( x − 4)( x + 2) LIMITS of Rational Functions as x goes to +inf and -inf Algebraic manipulations simplify finding limits in rational functions involving +inf and –inf. Example 4x2 − x lim x →−∞ 2 x 3 − 5 Divide the top and the bottom by the highest power of x 4x2 x 4 1 4 1 3 − 3 − 2 lim ( − 2 ) x x x x x →−∞ x x = lim = lim x →−∞ 2 x 3 5 x →−∞ 5 5 − 3 2− 3 lim (2 − 3 ) 3 x x →−∞ x x x 1 1 4 lim − lim 2 x →−∞ x x →−∞ x = 1 2 − 5 lim 3 x →−∞ x © Copyright Virtual University of Pakistan 91 10-Limits and Computational Techniques VU 4(0) − 0 = = 0 2 − 5(0) Quick Rule for finding Limits of Rational Functions as x goes to +inf or –inf c0 + c1 x +... + cn x n cn x n lim = lim x →+∞ d + d x +... + d x n x →+∞ d x n 0 1 n n c0 + c1 x +... + cn x n cn x n lim = lim x →−∞ d + d x +... + d x n x →−∞ d x n 0 1 n n Not true if x goes to a finite number a. Example 4x2 − x 4x2 2 lim = lim = = 0 lim x →−∞ 2 x − 5 3 x →−∞ 2 x 3 x →−∞ x Same answer as the one we got earlier from algebraic manipulations © Copyright Virtual University of Pakistan 92 11-Limits: A Rigorous Approach VU Lecture # 11 Limits: A Rigorous Approach In this section we will talk about -Formal Definition of Limit - Left-hand and Right-hand Limits So far we have been talking about limits informally. We haven't given FORMAL mathematical definitions of limit yet. We will give a formal definition of a limit. It will include the idea of left hand and right hand limits. We intuitively said that lim f ( x) = L x→a means that as x approaches a, f (x) approaches L. The concept of “approaches” is intuitive. The concept of “approaches” is intuitive so far, and does not use any of the concepts and theory of Real numbers we have been using so far. So let’s formalize LIMIT Note that when we talked about “f (x) approaches L” as “x approaches a” from left and right, we are saying that we want f( x) to get as close to L as we want provided we can get x as close to a as we want as well, but maybe not equal to a since f (a) maybe undefined and f (a) may not equal L. So naturally we see the idea of INTERVALS involved here. I will rephrase the statement above in intervals as For any number ε > 0 if we can find an open interval ( x0 , x1 ) on the x − axis containing a po int a such that L − ε < f ( x) < L + ε for each x in ( x0 , x1 ) except possibly x = a. Then we say lim f ( x) = L x→a So, f ( x) is in the interval ( L − ε , L + ε ) Now you may ask, what is this ε all about?? Well, it is the number that signifies the idea of “f (x) being as close to L as we want to be” could be a very small positive number, and that why it Let's us get as close to f (x) as we want. Imagine it to be something like the number at the bottom is called a GOOGOLPLEX!! 1 100 1010 © Copyright Virtual University of Pakistan 93 11-Limits: A Rigorous Approach VU So ε > 0 but very close to it, and for ANY such ε we can find an interval on the x-axis that can confine a. Let's pin down some details. Notice that When we said L − ε < f ( x) < L + ε holds for every x in the interval (x0, x1) ( except possibly at x= a), it is the same as saying that the same inequality hold of all x in the interval set ( x0 , a ) ∪ ( a, x1 ). But then the inequality L − ε < f ( x) < L + ε holds in any subset of this interval, namely ( x0 , a ) ∪ ( a, x1 ) ε is any positive real number smaller than a – x 0 and x 1 - a. Look at figure below L − ε < f ( x) < L + ε can be written as | f ( x) − L | < ε and ( a − δ , a ) ∪ ( a, a + δ ) as 0 < | x − a | < δ. Are same sets by picking numbers close to a and a-delta. Have them look at definition again and talk. Let's use this definition to justify some GUESSES we made about limits in the previous lecture. Example Find lim(3 x − 5) = 1 x→2 Given any positive number ε we can find an δ such that (3 x − 5) − 1 < ε if x satisfies 0 < x − 2 < δ In this example we have f ( x) =− 3x 5 L= 1 a= 2 © Copyright Virtual University of Pakistan 94 11-Limits: A Rigorous Approach VU So our task is to find out the ε for which will work for any δ we can say the following (3 x − 5) − 1 < ε if 0< x−2 0 and differentiable at x. d 1 du [log= b (u )] ⋅ dx u ln(b) dx Using the chain rule here d 1 du [ln(u )]= ⋅ dx u dx © Copyright Virtual University of Pakistan 140 20-Derivative of Logarithmic and Exponential Functions VU Example d Find ln( x 2 + 1)  dx d 1 d ln( x 2 + 1) =  ⋅ [ x 2 + 1] dx ( x + 1) dx 2 1 2x = ⋅= 2x ( x + 1) 2 ( x + 1) 2 Example d   x 2 sin( x)   ln   dx   1 + x   d  d  1  =  1 + x  ln[ x 2 sin( x)] − ln =  2 ln( x) + ln(sin x) − ln(1 + x)  dx dx  2  2 cos( x) 1 = + − x sin( x) 2(1 + x) Logarithmic Differentiation Using log function properties, we can simplify differentiation of messy functions. Example x 2. 3 7 x − 14 y= (1 + x 2 ) 4 Apply ln to both sides and simplify the right side using ln properties  x 2. 3 7 x − 14  1 ln y= ln  = 2 ln x + ln(7 x − 14) − 4 ln(1 + x ) 2  (1 + x ) 2 4  3 Differentiating both sides w. r. t. x gives 7 1 dy 2 8x = + 3 − y dx x 7 x − 14 1 + x 2  7  dy  2 8x  ⇒ = + 3 − ⋅ y dx  x 7 x − 14 1 + x 2     7  dy  2 3 8 x  x 2. 3 7 x − 14 =+ − ⋅ dx  x 7 x − 14 1 + x 2  (1 + x 2 ) 4   Derivatives of Irrational powers of x Remember that we proved the power rule for positive integers, then for all integers, then for rational numbers? © Copyright Virtual University of Pakistan 141 20-Derivative of Logarithmic and Exponential Functions VU Now we prove it for all REAL numbers by proving it for the remaining type of real numbers, namely the irrationals. Power Rule d r  x = r ⋅ x r −1 dx We will use ln to prove that the power rule holds for any real number r Let y = x r , r any real number Proceed by differentiating this function using ln = ln y ln= x r r ln x d d [ln y ] = [r ln x] dx dx 1 dy r = y dx x dy r r = ⋅ y = ⋅ x r = r ⋅ x r −1 dx x x Derivatives of Exponential functions f ( x) = b x = Now we want to find the derivative of the function y f= ( x) b x To do so, we will use logarithmic differentiation on y = b x = ln y ln= b x x ln b d d [ln y ] = [ x ln b] dx dx 1 dy = ln b y dx dy =⋅ y ln b = b x ⋅ ln b dx d x So b= b x ⋅ ln b dx In general d u du [b ] =bu ⋅ ln b ⋅ dx dx If b = e d x e =e x ⋅ ln e =e x dx In general d u du du [e ] =eu ⋅ ln e ⋅ =eu ⋅ dx dx dx Inverse Functions We have talked about functions. © Copyright Virtual University of Pakistan 142 20-Derivative of Logarithmic and Exponential Functions VU One way of thinking about functions was to think of a function as a process that does something to an input and then throws out an output. So this way, when we put in a real number x, it comes out as some new number y, after some ACTION has taken place. Now that question is, that is there another ACTION, that undoes the first action? If f is a function that performs a certain action on x, is there a function g that undoes what f does? Sometimes there is and sometime not. When there is a function g undoes what f does, then we say that g is an inverse function of f. Here is how to determine if two given function are inverses of each other or not. Definition 7.4.1 Example 1 =f ( x) 2= x and g ( x) x 2 are inverse functions. This is obvious, but we can use def 7.4.1 1 f (= = g ( x)) 2( x) x 2 1 g(= f ( x)) = (2 x) x 2 Inverses don’t always exists for a given function. There are conditions for it. Most important one is that a function must be ONE to ONE to have an inverse. Do you recall what One to One is?? Its when a function does not have two values for the same x value in the domain. Review this. The inverse function to a function f is usually denoted as f −1 read as “f inverse” and =y f −1 ( x) ⇒ =x f ( y) Derivatives of Inverse Functions Theorem 7.4.7 © Copyright Virtual University of Pakistan 143 20-Derivative of Logarithmic and Exponential Functions VU The formula in this theorem can be simplified if you write =y f −1 ( x) ⇒ =x f ( y) dy dx Then =( f −1 ) '( x) ⇒ = f '( y ) = f '( f −1 ( x)) dx dy Plugging these values into the theorem we get a simpler formula dy 1 = dx dx dy © Copyright Virtual University of Pakistan 144 21 – Applications of Differentiation VU Lecture # 21 Applications of Differentiation Related Rates Increasing Functions Decreasing Functions Concavity of functions Related Rates Related Rates are real life problems. These involve finding the rate at which one quantity changes w.r.t another quantity. For example, we may be interested in finding out how fast the polar ice caps are melting w.r.t the changes in temperature. We may want to know how fast a satellite is changing altitude w.r.t changes in time, or w.r.t changes in gravity. To solve problems involving related rates, we use the idea of derivatives, which measure the rate of change. Example Assume that oil spilled from a ruptured tanker spread in a circular pattern whose radius increases at a constant rate of 2 ft/sec. How fast is the area of the spill increasing when the radius of the spill is 60ft? Let t = number of seconds elapsed from the time of the spill Let r = radius of the spill in feet after t seconds Let A = area of the spill in square feet after t seconds dA dr We want to find given that = 2 ft / sec dt r =60 dt The spill is circular in shape so A = π r 2 dA dr dA = 2π r ⇒ = 2π (60)(2) = 240π ft 2 / sec dt dt dt r =60 The following steps are helpful in solving related rate problems 1. Draw a figure and label the quantities that change 2. Identify the rates of change that are known and those that are to be found 3. Find an equation that relates the quantity whose rate of change is to be found to those quantities whose rates of change are known. 4. Differentiate the equation w.r.t the variable that quantities are changing in respect to. Usually Time. 5. Evaluate the derivative at appropriate points. Example © Copyright Virtual University of Pakistan 145 21 – Applications of Differentiation VU A five foot ladder is leaning against a wall. It slips in such a way that its base is moving away from the wall at a rate of 2 ft/sec at the instant when the base is 4ft from the wall. How fast is the top of the ladder moving down the wall at that instant? Let t = number of seconds after the ladder starts to slip Let x = distance in feet from the base of the ladder Let y = distance in feet from the top of the ladder to the floor. dx = rate of change of the base of the ladder (horizontal movement) dt dy = rate of change of the top of the ladder (vertical movement) dt dy dx We want Given That = 2 ft / sec dt x =4 dt x=4 How do we relate the x and the y? Look at the picture again I see Pythagoras's theorem in here! x2 + y 2 =25 Differentiating wrt t and using chain rule gives dx dy 2x + 2y = 0 dt dt dy x dx = − dt y dt dy 4 8 = − (2) = − ft / sec dt x=4 3 3 When x = 4, use the above equation to find corresponding y. Increasing and decreasing functions We saw earlier in the lectures that we can get an idea of the graph of a function by plotting a few values. But remember that we also said that this graph was an approximation as a few points may not give all the info. Now we will see that we can use derivatives to get accurate info about the behavior of the graph in an interval when we move from left to right. Increasing function on an interval means that as we move from left to right in the x-direction, the y- values increase in magnitude. Decreasing function on an interval means that as we move from left to right in the x-direction, the y- values decrease in magnitude. An interval means that as we move from left to right in the x-direction, the y-values decrease in magnitude. The graph of the function in this figure shows that the function is increasing on the intervals and (−∞, 0) and [2, 4] decreasing on the interval [0,2]. © Copyright Virtual University of Pakistan 146 21 – Applications of Differentiation VU Let’s make this idea concrete. Definition 4.2.1 As shown in the figures below Let's take a few points on the 3 graphs in above figures and make tangent lines on these points. This gives Note that incase where the graph was increasing, we get tangent line with positive slopes, decreasing we get negative slope, and constant gives 0 slope. © Copyright Virtual University of Pakistan 147 21 – Applications of Differentiation VU Let's formalize this idea Theorem 4.2.1 Example Find the intervals on which the function is increasing and those on which its decreasing. Since f is continuous on (2,+∞), the function is actually increasing on the interval [2,+∞) Similarly it is decreasing on the interval (-∞,2] The derivative is 0 at the point x=2 Since this is the only point not the interval we think of the point x=2 as the point where the transition occurs from decreasing to increasing in f. © Copyright Virtual University of Pakistan 148 21 – Applications of Differentiation VU © Copyright Virtual University of Pakistan 149 21 – Applications of Differentiation VU © Copyright Virtual University of Pakistan 150 22-Relative Extrema VU Lecture # 22 Relative Extrema Relative Maxima Relative Minima Critical Points First Derivative test Second Derivative test Graphs of Polynomials Graphs of Rational functions Relative Maxima Most of the graphs we have seen have ups and downs, much like Hills and valleys on earth. The Ups or the Hills are called relative Maxima. The downs or the Valleys are called relative Minima The reason we use the word relative is that just like a given Hill in a mountain range need not necessarily be the Highest point in the range. Similarly a given maxima in a graph need not be the maximum possible value in the graph. Same goes for the relative minima. In general, we may say that a given Hill is the highest one in some area. Look at relative maxima in a given interval. Again, same is true for valleys and relative minima. So when we talk about relative maxima and relative minima, we talk about them in the context of some interval. Definition 4.3.1 Definition 4.3.2 A function is said to have a relative minimum at x 0 if f(x 0 ) ≤ f(x) for all x in some open interval containing x 0. Definition 4.3.3 © Copyright Virtual University of Pakistan 151 22-Relative Extrema VU Example Here is a graph of a function f. This has a relative maximum in the interval (a, b) because from the graph its obvious that f ( x0 ) ≥ f ( x) Critical Points It so happens that relative extrema can be viewed as transition points that separate the regions where a graph of a function is increasing from those where a graph is decreasing. Here is a figure. This shows that relative extrema of a function occur at points where f has a horizontal tangent, or where the function is not differentiable. Horizontal tangent means derivative = 0. Non-differentiable means corners. Theorem 4.3.4 Definition 4.3.5 © Copyright Virtual University of Pakistan 152 22-Relative Extrema VU So theorem 4.3.4 can be read as now with this new terminology As “The relative extrema of a function, if any, occur at critical points.” Example a) x0 here is a critical and stationary point as tangent line has slope 0 f) x0 here is a critical point and it has minimum value at that point but the tangent line is not defined at that point. g) x0 is a critical point but not stationary as derivative does not exist Here are the figures of the situations. First Derivative test and Second Derivative test Note that in (g) of the last figure, x o was a critical point, but there was not relative extrema there! This can happen. So how do we know at which critical point a relative extrema occurs or not? Here is a theorem for that. In short: “ The relative extrema, if any, on an open interval where a function f is continuous and not constant occurs at those critical points where f / changes sign” © Copyright Virtual University of Pakistan 153 22-Relative Extrema VU Example Locate the relative extrema of 5 2 f (= x) 3 x 3 − 15 x 3 2 1 1 − − 5 f '( x) = 5 x 3 − 10 x 3 = 5 x 3 ( x − 2) = 1 ( x − 2) 3 x Note that there are two critical points, namely x = 0 and x = 2 Because at x = 2, the derivative f / = 0, and at x = 0, the derivative does not exists. Now we need to know where there is a relative extrema by checking for the changing sign of f / at the 2 critical points. We use theorem 4.3.6 and draw a number line test We see that there is relative maximum at 0 We see that there is relative minimum at 2 There is another test for finding extrema easier than the first derivative test. Theorem 4.3.7 EXAMPLE Locate the relative extrema of f ( x= ) x4 − 2x2 f / ( x) = 4 x3 − 4 x = 4 x( x − 1)( x + 1) f / /= ( x) 12 x 2 − 4 Setting f / ( x) = 0 gives stationary points x = 0 and x = ±1 Also, f / / (0) =−4 < 0 f / / (1)= 8 > 0 f / / (−1) = 8 > 0 There is a relative maximum at x = 0, and relative minima at x = 1 and x = -1 Graphs of Polynomials © Copyright Virtual University of Pakistan 154 22-Relative Extrema VU In applied sciences and engineering, it is required many times to understand the behavior of a function. Graphs are a good way to understand function behavior. But many times it is hard to graph the function. So it is often necessary to understand the behavior in terms of maxima and minima and concavity etc. We will look at how the stuff from the last lecture helps us in graphing polynomial and rational functions In applied sciences and engineering, it is required many times to understand the behavior of a function. Graphs are a good way to understand function behavior. But many times it is hard to graph the function. So it is often necessary to understand the behavior in terms of maxima and minima and concavity etc. We will look at how the stuff from the last lecture helps us in graphing polynomial and rational functions Example Sketch the graph of P ( x) = y =x 3 − 3 x + 2 dy = 3 x 2 − 3= 3( x − 1)( x + 1) −2 0 dx −1 4 d2y = 6x 0 2 dx 2 1 0 You find the stationary points, inflection points. 2 4 © Copyright Virtual University of Pakistan 155 22-Relative Extrema VU Above figure Shows the intervals of increase/decrease and of concavity Y intercept at (0,2) Inflection point is at x = 0 Graph of Rational Functions Rational function, remember, is a function defined by the ratio of two polynomials P( x) R( x) = Q( x) Its obvious that if Q(x) = 0, then R(x) has discontinuity at those values of x where Q(x) = 0 Consider the following graph of x f ( x) = x−2 © Copyright Virtual University of Pakistan 156 22-Relative Extrema VU Vertical asymptotes occur where the denominator is 0 Example Find horizontal and vertical asymptotes of x2 + 2x f ( x) = x2 −1 © Copyright Virtual University of Pakistan 157 23-Maximum&Minimum Values of Functions VU Lecture # 23 Maximum and Minimum Values of Functions Absolute Extrema Finding Absolute Extrema for a continuous function Summary of extreme behaviors of functions over Applied maximum and minimum problems Problems involving finite closed intervals Problems involving intervals that are non-finite and closed Absolute Extrema Previously we talked about relative maxima, relative minima of functions. These were like the highest mountain and the deepest valley in a given vicinity or neighborhood. Now we will talk about absolute maximum and minimum values of functions. These are like the highest peak in a mountain range, and the deepest valley. Absolute maximum means that the value is the maximum one over the entire domain of the function. Absolute minimum means that the value is the minimum one over the entire domain of the function. If we think of the Earth’s surface as defining some function, then its absolute maximum will be Mt. Everest, and absolute minimum will be the Marianna Trench in the Pacific Ocean near Hawaii. Example Consider the following picture of the graph of f (x) = 2x+1 on the interval [0,3). The minimum value is 1 at x = 0. But is there a maximum? © Copyright Virtual University of Pakistan 158 23-Maximum&Minimum Values of Functions VU No. Because the function is defined on the interval [0,3) which excludes the point x = 3. So note that you can get very close to 7 as the maximum value as you get very close to x = 3, but this is in a limiting process, and you can always get more closer to 7, and yet never EQUAL 7! So 7 looks like a max, but its NOT! The question of interest given a function f(x) is: Does f (x) has a maximum (minimum) value? If f (x) has a maximum (minimum) , what is it? If f (x) has a maximum value, where does it occur? Won’t prove this as its difficult. Just use it. This theorem doesn’t tell us what the max and the min are, just the conditions on a function which will make it have a max or min Example Note that in the previous example we saw a function f (x) = 2x+1 which was defined over the interval [0, 3). This one is a continuous function on that interval, but had no maximum because the interval was not closed! Example Here is another function graphed over the interval [1,9]. Although the interval is closed, the function not continuous on this interval as we can see from the graph, and so has no maximum or minimum values on that interval. © Copyright Virtual University of Pakistan 159 23-Maximum&Minimum Values of Functions VU Step 1: Find the critical points of f in (a,b) Step 2: Evaluate f at all the critical points and the endpoint a and b Step 3: The largest of the values in Step 2 is the maximum value of f on [a,b] and the smallest is the minimum. Example Find the maximum and the minimum values of f ( x) =2 x 3 − 15 x 2 + 36 x on the interval [1, 5] Since f is a polynomial, it’s continuous and differentiable on the interval (1, 5) © Copyright Virtual University of Pakistan 160 23-Maximum&Minimum Values of Functions VU f / ( x) = 6 x 2 − 30 x + 36 = 0 f / ( x) = ( x − 3)( x − 2) = 0 So f / is 0 at x = 2 and x = 3. Max or min will occur at these two points or at the end points. Evaluate the function at the critical points and the endpoints and we see that max is 55 at x = 5 and min is 23 at x = 1 We want to know the max and min over (.−∞, +∞) Here is how to find them for continuous functions Summary of extreme behaviors of functions over (a,b) Find the max and min values if any, of the function f ( x) =x 4 + 2 x 3 − 1 on the interval (−∞, +∞) This is a continuous function on the given interval and lim ( x 4 + 2 x3 − 1) = +∞ and lim ( x 4 + 2 x3 − 1) = +∞ x →+∞ x →−∞ © Copyright Virtual University of Pakistan 161 23-Maximum&Minimum Values of Functions VU So f has a minimum but no maximum on (-inf, +inf). By Theorem 4.6.5, the min must occur at a critical point. So f '( x) = 4 x3 + 6 x 2 = 2 x 2 (2 x + 3) = 0 This gives x = 0, and x = -3/2 as the critical points. Evaluating gives min = -43/16 at x = -3/2 Applied maximum and minimum problems We will use what we have learnt so far to do some applied problems in OPTIMIZATION. Optimization is the way efficiency is got in business, machines and even in nature in terms of animals competing for resources. Problems involving continuous functions and Finite closed intervals These are problems where the function is defined over a closed interval. These problems always have a solution because it is guaranteed by the extreme values theorem. Example Find the dimensions of a rectangle with perimeter 100 ft whose area is as large as possible. Let x = length of the rectangle in feet y = width of the rectangle in feet A = area of the rectangle Then A = xy We want to maximize the area A = x(50 − x) = 50 x − x 2 Perimeter = 100ft = 2x + 2y or y = 5 – x Use this values of y in the equation A = xy to get A function of x Because x represents a length, it cannot be negative and it cannot be a value that exceeds the perimeter of 100 ft. So we have the following constraints on x 0 ≤ x ≤ 50 So the question is of finding the max of = A 50 x − x 2 on the interval [0,50]. By what we have seen so far, that max must occur at the end points of this interval or at a critical point Critical points: dA = 50 − 2 x = 0 ⇒ x = 25 dx So now we substitute x = 0, x = 25, and x = 50 into the function A to get the max 625 at the point x = 25. Note that y = 25 also for x = 25. So the rectangle with perimeter 100 with the greatest area is a square with sides 25ft. © Copyright Virtual University of Pakistan 162 23-Maximum&Minimum Values of Functions VU Example An open box is to be made from a 16 inch by 30 inch piece of cardboard by cutting out squares of equal size from the 4 corners and bending up the sides. What size should the squares be to obtain a box with largest possible volume? If we cut out the squares from the 4 corners of the cardboard, the resulting BOX will have dimensions (16 – 2x) by (30 – 2x) s So want to find max of the function on [0,8]. dV 10 = 480 − 184 x + 12 x 2 = 0 ⇒ x= and x = 12 dx 3 Because x = 12 is out of [0, 8], ignore it. Check V at the end points and at x = 10/3. We see then that V = 19600/27 is max when x = 10/3. © Copyright Virtual University of Pakistan 163 24-Newton’s Method, Rolle’s & Mean Value Theorem VU Lecture # 24 Newton’s Method, Rolle’s Theorem, and the Mean Value Theorem Newton’s method for approximating solutions to f(x)=0 Some difficulties with Newton’s method Rolle’s theorem Mean Value Theorem Newton’s method for approximating solutions to f (x) =0 b We have seen in algebra that the solution for the equation ax + b =0 is x = −. a Similarly we have algebraic formulas for polynomial equation up to degree 5. But there is no algebraic solution to the equation of the kind x − cos( x) = 0 For this equation and many like it, we settle for approximate solutions. How do we approximate the solutions? There are many methods, and one of them is Newton’s method. Here is how Newton's method works to find approximate solutions to equations. What does it mean for an equation to have a solution? or that f (x) = 0 ? It means that we are looking for those x values, for which the corresponding y value or f (x) is 0. This means that the solutions are those points where the graph of the function crosses the x axis. Suppose that x = r is the solution we are looking for. Let's approximate it by an initial guess called. We draw a line tangent to the graph of the given function at the point. If the tangent line is not parallel to the x-axis, then it will x2 eventually intersect the x-axis at some point x1 which will generally be closer to r than We will repeat the process, with a tangent line at x2 that meets x axis at x3, and so on… This is Newton’s method. We need a formula for this method. Note that the tangent line at x1 has the equation y − f ( x= 1) f '( x1 )( x2 − x1 ) If f '( x1 ) ≠ 0 , then the line meets the x axis at ( x2 , 0) Plug this coordinate into above equation, we get − f ( x= 1) f '( x1 )( x2 − x1 ) f ( x1 ) ⇒ x2 = x1 − f '( x1 ) Repeating this process for a third point ( x3 , 0) gives − f ( x= 2) f '( x2 )( x3 − x2 ) f ( x2 ) ⇒ x3 = x2 − f '( x2 ) © Copyright Virtual University of Pakistan 164 24-Newton’s Method, Rolle’s & Mean Value Theorem VU In general, then we have f ( xn ) xn += 1 xn − f '( xn ) There is limiting process involved here for finding the solutions. We get as close as we like to the solution. Example The equation x = cos( x) has a solution between 0 and 1. Approximate it using Newton’s method. Rewrite as x = cos( x) so our function is f ( x)= x − cos( x). The derivative is f / ( x) = 1 + sin( x) So we have xn − cos( xn ) xn += xn − 1 + sin( xn ) 1 As our approximation formula Here is a graph of the situation. From the graph it looks like the solution is closer to 1 than 0. So we will use x 1 = 1.So we get the following approximations. x − cos( x1 ) 1 − cos(1) x2 = x1 − 1 = 1− = 0.7503 1 + sin( x1 ) 1 + sin(1) x − cos( x2 ) 0.7503 − cos(0.7503) x3 = x2 − 2 = 0.7503 − = 0.7391 1 + sin( x2 ) 1 + sin(0.7503) You may continue if you will, but we will say that the solution is approximately x ≈ 0.7391 Some difficulties with Newton’s method Newton’s method does not always work. If for some values of n f / ( xn ) then the formula for Newton’s method involves division by 0 and we are out of business. © Copyright Virtual University of Pakistan 165 24-Newton’s Method, Rolle’s & Mean Value Theorem VU Such a case will occur if the tangent line for some approximation has slope 0 or is parallel to the x axis. Sometimes the approximations don’t converge to a solution. 1 Consider the equation x 3 = 0 The only solution is x = 0. Let's approximate it by Newton’s Method with initial approx x 1 = 1. We get the following Formula 1 xn +1 = x − ( xn ) 3 = −2 xn n 2 1 ( xn ) 3 − 3 Plug in x 1 =1 and then the following approx to see that the values do not converge Rolle’s theorem Rolle’s Theorem says essentially that for a certain kind of function, if it crosses the x-axis at two point, then there is one point between those two points where the derivative of f is 0. Example The function f ( x) = sin( x) is continuous and differentiable everywhere, hence continuous on [0 , 2π] and differentiable on (0, 2π). Also, f (0) = sin (0) = 0 and f(2π)=sin (2π) = 0 So the function satisfies the hypotheses of Rolle’s Theorem. So there exists a point c in the interval (0 , 2π) such that = = f / (c) cos( c) 0 f (b) − f (a ) y= − f (a) ( x − a) b−a ⇒ f (b) − f (a ) =y ( x − a) + f (a) b−a Here is a more tangible way to think of Roll’s theorem © Copyright Virtual University of Pakistan 166 24-Newton’s Method, Rolle’s & Mean Value Theorem VU I leave from Lahore to Islamabad. When i start driving from Lahore, my velocity is 0, and when i reach Islamabad, my velocity is 0 as well. Velocity is a continuous function on the interval [0 , 376]. Also, velocity is differentiable on (0 , 376) as its derivative acceleration is defined at each point on the velocity curve. Hence during my drive from Lahore to ISB, there is some point on the motorway where the acceleration of the car was 0. One could argue: What if you keep accelerating on the motorway! Mean Value Theorem This says basically that under the right conditions, a function will have the same slope for the tangent line at a point as that of a certain secant line. Proof of MVT From this figure we have the following: Slope of Secant line joining A and B:  f (b) − f (a )  v( x)= f ( x) −  ( x − a) + f (a)   b−a  Since f(x) is continuous [a , b] and differentiable on (a , b), so is v(x) by its formula involving f(x). Also note that =v(a ) 0= and v(b) 0 So v(x) satisfies the assumptions of Roll’s theorem on the interval [a , b]. So there is a point c in (a , b) such that v / ( c)=0. But note that © Copyright Virtual University of Pakistan 167 24-Newton’s Method, Rolle’s & Mean Value Theorem VU  f (b) − f (a )  v= / ( x) f / ( x) −    b−a  f (b) − f (a )  ⇒ v / (c ) = f / (c ) −    b−a So by this last formula, at the point where v / (c) = 0 , we have f (b) − f (a ) f / (c ) = b−a © Copyright Virtual University of Pakistan 168 25-Integration VU Lecture # 25 Integrations In this lecture we will look at the beginnings of the other major Calculus problem. The Area Problem Anti-derivatives (Integration) Integration formulas Indefinite Integral Properties of Indefinite Integral The Area Problem Given a continuous and non negative function on an interval [a , b], find the area between the graph of f and the interval [a , b] on the x-axis. Instead of trying to solve a particular case like the one in the picture we just saw, we will generalize to solve this problem where the right end point will be any number x greater than or equal to b instead of just b. We will denote the area we are trying to find as A(x) because this will be a function of x now as it depends on how far away x is from a. It was the idea of Newton and Leibniz that to find the unknown area A(x), first find it derivative A/ ( x) and use this derivative to determine what A (x) is ! Interesting approach. So we want to find out first. A( x + h) − A( x) A '( x) = lim h →0 h Let’s assume for now that h > 0 The top of the derivative quotient is the difference of the two areas A(x) and A(x + h) Let c be the midpoint of between x and x +h. Then the difference of areas can be approximated by the area of the rectangle with base length h and height f ( c). © Copyright Virtual University of Pakistan 169 25-Integration VU So we have A( x + h) − A( x) f (c) ⋅ h ≈ = f (c ) h h Note that the error in the approximation from this rectangle in A will approach 0 as h goes to 0. Then we have A( x + h) − A( x) = = lim f (c) A/ ( x) lim h →0 h h →0 As h goes to 0, c approaches x. Also, f is assumed to be a continuous function, so we have that f( c) goes to f (x) as c goes to x. Thus lim f (c) =f ( x) ⇒ A/ ( x) =f ( x) h →0 So: The derivative of the area function A (x) is the function whose graph forms the upper boundary of the region under which the area is to be found Example Find the area of the region under the graph of= y f= ( x) x 2 over the interval [0 , 1 ] Look at the situation over the interval [0 , x]. Then we have from the discussion that A '( x) = x 2 To find A(x) we look for a function whose derivative is x 2 This is called an antidifferentiation problem as we are trying to find A(x) by undoing a differentiation. A guess is the function 1 A( x) = x3 3 This is a formula for the areas function. So on the interval [0 , 1], we have x = 1 and our result is A(1) = 1/3 units Anti-derivatives Definition 5.2.1 A function F is called antiderivative of a function f on a given interval if F/(x) = f (x) for all x in the interval. © Copyright Virtual University of Pakistan 170 25-Integration VU Example 1 3 1 3 1 3 The functions x, x −π, x + C are all anti-derivatives of 3 3 3 =f ( x) x 2 on the interval (−∞, +∞) As the derivative of each is f ( x) = x 2 If F (x) is any anti-derivative of f (x), then so is F (x) +C where C is a constant. Here is a theorem Indefinite Integral The process of finding anti-derivatives is called anti-differentiation or Integrations. d If there is some function F such that [ F ( x)] = f ( x) dx Then function of the form F (x) +C are anti-derivatives of f (x). We denote this by ∫ f ( x= )dx F ( x) + C The symbol ∫ is called the integral sign and f (x) is called the integrand. It is read as the “Indefinite integral of f (x) equals F (x)” ∫ f ( x= )dx F ( x) + C The right side of the above equation is not a specific function but a whole set of possible functions. That’s why we call it the Indefinite integral. C is called the constant of integrations. Example As we saw earlier, the anti-derivatives of f ( x) = x 2 are functions of the form So we can write 1 3 F= ( x) x +C 3 The dx serves to identify the independent variable in the function involved in the integration.

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