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# DIFFERENTIAL CALCULUS MTH 251

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This 95 page Class Notes was uploaded by Mrs. Dedric Little on Monday October 19, 2015. The Class Notes belongs to MTH 251 at Oregon State University taught by A. Faridani in Fall. Since its upload, it has received 30 views. For similar materials see /class/224441/mth-251-oregon-state-university in Mathematics (M) at Oregon State University.

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Contents Syllabus for MTH 251 iii Introduction and Notes for Students iv MTH 251 Sample Symbolic Differentiation Test vii Lesson 0 7 Ch 1 Review of Functions 1 LIMITS Lesson 1 7 21 The Idea of Limits 4 Lesson 2 7 22 De nition of Limits 5 Lesson 3 7 323 Techniques for Computing Limits 6 Lesson 4 7 24 In nite Limits 8 Lesson 5 7 325 Limits at In nity and Horizontal Asymptotes 9 Lesson 6 7 26 Continuity 11 DERIVATIVES Lesson 7 7 31 Introducing the Derivative 12 Lesson 8 7 3321 Rules of Differentiation 14 Lesson 9 7 33 Product and Quotient Rules 16 Lesson 10 7 334 Derivatives of Trigonometric Functions 17 Lesson 11 7 35 Derivatives as Rates of Change 19 Lesson 12 7 5336 The Chain Rule 21 Lesson 13 7 37 Implicit Differentiation 23 Lesson 14 7 5338 Logarithmic and Exponential Functions 25 Lesson 15 7 39 Derivatives of Inverse Trigonometric Functions 27 APPLICATIONS OF DERIVATIVES Lesson 16 7 310 Related Rates 30 Lesson 17 7 41 Maxima and Minima 33 Lesson 18 7 42 What Derivatives Tell Us 34 Lesson 19 7 343 Graphing Functions 37 Lesson 20 7 44 Optimization Problems 38 Lesson 21 7 345 Linear Approximations and Differentials 40 Lesson 22 7 46 Mean Value Theorem 42 Lesson 23 7 347 L7Hopital7s Rule 44 Laboratory Manual 47 Instructions 47 Laboratory 1 7 Graphing 48 Laboratory ll 7 Limits 54 Laboratory 111 7 The lntermediate Value Theorem 60 Laboratory 1V 7 Velocity and Tangent Lines 63 Laboratory V 7 The Chain Rule 68 Laboratory Vl Derivatives in Action 72 Laboratory Vll Higher Derivatives7 Exponential Functions 77 Laboratory Vlll Curve Sketching 81 Laboratory lX Logarithmic Functions and Newton s Method 84 SYLLABUS FOR MTH 251 Lesson 0 Ch 1 Review of Functions Lesson 1 21 Idea of Limits Lesson 2 22 De nition of Limit Lesson 3 23 Computing Limits Lesson 4 24 In nite Limits Lesson 5 25 Limits at In nity Lesson 6 26 Continuity of Functions Lesson 7 31 De nition of Derivative Lesson 8 32 Basic Rules of Differentiation Lesson 9 33 Product and Quotient Rules Lesson 10 34 Derivatives of Trigonometric Functions Catch up Review Lesson 11 35 Rates of Change Lesson 12 36 Chain Rule Lesson 13 37 Implicit Differentiation Lesson 14 38 Logarithmic and Exponential Functions Lesson 15 39 Inverse rI39rigonometric Functions Lesson 16 310 Related Rates Lesson 17 41 Maxima and Minima Lesson 18 42 Locating Extreme Values Lesson 19 42743 Graphing Functions Catch up Review Lesson 20 44 Optimization Problems Lesson 21 45 Linear Approximations and Differentials Lesson 22 46 Mean Value Theorem Lesson 23l 47 L7Hopital7s Rule Catch up Review JOPTIONAL iii Introduction and Notes for Students Introduction This is a study guide for MTH 251 Differential Calculus It is intended to be used in conjunction with the rst edition of the text Calculus 7 Early Transcendentals by Briggs and Cochran The study guide is designed for a ten week term with 29 lectures and 9 laboratory sessions with two class hours reserved for exams four for review and catch up and 23 lessons devoted to new material from the text The lesson plan for the course by and large follows the organization of the material in the text Some of the recitation sessions may involve scheduled laboratory activities as de signed by your instructor and the remaining time is set aside for review and for questions and answers You should consult your instructor for a detailed syllabus for your section of the course Comments and suggestions on study habits In this course much new material will be presented at a rapid pace You will also be expected to understand and apply mathematical concepts and reasoning not merely perform calculations Therefore developing good study habits from the outset in order to keep up with the course is particularly important Listed below are a number of points to heed Time commitment It is essential to devote enough time on a daily basis to the course You should plan on spending at least two to three hours studying the ma terial and solving assignments for each hour of lecture If you have encountered some of this material before it is easy to fall into the habit of not dedicating enough time to the course at the beginning of the term Then when more chal lenging topics are presented later on you could nd yourself too far behind to catch up Algebra skills Applying the proper algebraic manipulations taught in precalculus courses is one of the most common hurdles for students taking a calculus class If your skills are rusty practice now and seek help before it will be too late This course is taught with the premise that you master the basics of algebra and trigonometry covered in your previous mathematics courses Attendance Your instructors will from time to time introduce a new viewpoint or amplify on the material set forth in the text or the study guide Beside the calculus text exams for the course will be based on the lectures recitations lab activities and assignments and you are accountable for all the materials It is then in your own interest to plan on attending all the class meetings and to get notes from another student for any that you might have missed Homework Exercises in the text are divided into several categories Review Ques tions test your conceptual understanding of the narrative7 while solving Basic Skills exercises will improve your computational dexterity Exercises under the Further Explorations and Applications heading are built on the Basic Skills prob lems and are more demanding Finally7 Additional Epereises will challenge your thinking and often involve mathematical proofs Each chapter in the text con cludes with Review Epereises which will help you to synthesize the contents of the entire chapter Most students can only excel in this course by solving a large number of problems and you should therefore aim to work through most of the exercises in the text As a starting point7 you will nd lists of basic Starter Problems and the more challenging Recommended Problems at the end of each lesson in this study guide Keep in mind that simply getting the answer at the back of the book should not be your only goal7 but that understanding the principles and methods needed to solve an exercise is the primary purpose of an assignment7 and also a requisite for solving similar problems in the exams and in real life situations Consult your instructor for further details on graded assignments7 which may also include online homework administered via the CourseCompass platform Laboratory Activities There are detailed group activities presented in the Labo ratory Manual of this study guide During the term you may be assigned to work on a subset of these as part of the weekly recitation sessions Your instructor will provide a detailed schedule for the activities Other Resources The Mathematics Learning Center MLC provides drop in help for all lower division mathematics courses The MLC is located on the ground oor of Kidder Hall in room 1087 and is normally open Monday through Thursday from 9 am to 5 pm and on Fridays from 9 amto 4 pm7 from the second week of the term through the Dead Week The MLC also provides evening tutoring in the Valley Library7 in general Sunday through Thursday from 7 pm till 10 pm Current hours can be found at the MLC homepage By purchasing the course text you will gain access to MyMathLab7 an online cal culus portal maintained by the publisher In addition to homework problems and tests with automated grading7 the site offers a number of useful tools7 including PowerPoint and video lectures7 review cards7 and tutorial exercises7 for organizing your studies and to facilitate learning the course material Acknowledgments This study guide relies on the previous Mathematic Depart rnent7s MTH 251 study guides7 with material written by Bob Burton7 TeVian Dray7 Christine Escher7 and Dennis Garity Valuable comments were also received from Dianne Hart and Stephen Scarborough To all of these people I extend my heartfelt thanks Vi MTH 251 Sample Symbolic Differentiation Test NAMF Student lD Show only your answers on this page Do your work on scratch paper and turn in your scrap paper with this page You must show your work to receive full credit NO CALCULATORS OR NOTES ARE ALLOWED Compute the following derivatives You do not need to simplify your answers d 1 gm 7 4x4 7 16 dxg E23 i cosx2 dx 9 d sinx3 7 7 U 03 d 5 43 7 i 14 MW tan m 5 d cosm d1 ii dx 3 d i 10 s1nx cos2 3 Lesson 0 Ch 1 Review of Functions The chapter contains a summary of some of the background material required for this course You should carefully review it on your own as it will not be thoroughly covered in class Section 11 Browse through the section to ensure that your understand the basic concepts and are able to solve the exercises in the text You should be familiar with the following terminology 0 Domain of a function 0 Range of a function 0 Independent variable 0 Dependent variable 0 Graph of a function 0 Composite function 0 Even function 0 Odd function An even function satis es ix x for all x in the domain of x while an odd function is characterized by ix i x For example any even powered monomial x x n n 1 2 3 yields an even function and any odd powered monomial x xzn n 012 an odd function Also as you can easily x if x Z 0 7 verify the absolute value function is even 7x ifx lt 0 Section 12 treats various types of elementary functions and their graphs For you the most important ones to be able to work with are c Polynomials o Rational functions 0 Algebraic functions 0 Exponential functions 0 Logarithmic functions 0 Trigonometric functions A linear function x mx b where m b are constants is a polynomial function of degree 1 The square root function x and the cubic root function x x13 are familiar examples of algebraic functions The slope function gx of a function x speci es the slope ofthe graph ofy x at x You will discover in this class that for example the slope function of the parabola y x2 is gx 2x while the slope function of the sine function y sinx is gx cos x You can construct new functions by shifting and scaling the graph of a given function x in the x and y directions As an example graph by hand the x shift sinx2 the y shift sin x2 the x scaling sin2x and the y scaling 2 sin x of the sine function in the same grid Make sure you understand how the graphs of these are related to the graph of their progenitor Section 13 Brush up on inverse functions and the horizontal line test for checking whether a function is one to one As an exercise determine the largest intervals on which the function f 2 7 2x is one to one and nd its inverse function on each of the intervals you have identi ed Review the de nition of the exponential function and the logarithmic function as its inverse You will be expected to understand their basic properties and to be able to apply the eccponential and logarithmic rules as spelled out in the margin of the text Section 14 Start by reviewing the de nition of the radian measure and for prac tise express the angle measurements 15 30 45 60 90 180 in radians You should be familiar with the graphs of the basic trigonometric functions sin 9 cos 9 sec 9 csc 9 tan 9 cot t9 and to be able to nd their precise values for the angles 90 7T6 7r4 7T2 7139 without a calculator The various trigonometric identities often prove handy in simplifying complicated trigonometric expressions The inverse trigonometric functions will be covered in Lesson 15 Starter Problem List Section 11 6 13 22 31 36 41 47 Section 12 3 7 8 13 16 21 30 Section 13 4 8 10 12 14 17 19 25 29 35 49 Section 14 23 7 16 17 18 25 31 Recommended Problem List Section 11 8 14 18 25 29 33 39 44 45 52 64 Section 12 10 2024 28 32 37 47 Section 13 11 15 21 22 24 28 30 36 40 44 54 Section 14 4 8 20 21 26 28 32 33 66 Lesson 1 21 The Idea of Limits The limit of a function which describes the behavior of the function near a xed point is a key concept in calculus Roughly the limit is the number that the values of a function approach as the input gets close but is not equal to the xed point In this section the importance of limits is underscored by the way of examples in volving average and instantaneous velocities and the secant and tangent lines You will revisit these type of examples after learning about the derivative of a function Starter Problem List 2 4 5 7 13 Recommended Problem List 6 9 12 16 19 20 23 24 Lesson 2 22 De nition of Limits lntuitively the limit of a function f at m 1 equals L grill 1 06 L if the values of f are arbitrarily close to L when x is suf ciently close but not equal to 1 Hence the limit of a function may exist at z 1 even when the function is not de ned at z a You can nd the precise 6 6 de nition of a limit in Section 27 of the text However the description given in Section 23 will by and large suf ce for the purposes of this course although you might want to study gures 2577259 to gain a deeper understanding of the limit concept When computing the limit of a function at z a one considers values of f on both sides of 1 Thus the limit of a function is often referred to as a two sided limit A useful variation to the basic limit concept is that of me sided limits lim f lim fx zgta maa where in the rst instance one only considers values of z larger than a and in the second one smaller than a The limit liming f exists if and only ifthe one sided limits limmnaJr f limmnai f exist and are equal Hence a one sided limit may exist at a point even when the two sided limit does not Can you construct an example You can often estimate the limit limmna x by zooming in on the graph of f near z 1 However graphing utilities can also lead you awry due to rounding errors as you will discover in Laboratory assignment I Starter Problem List 26711 15 17 31 35 Recommended Problem List 13 14 16 21 25 27 29 37 Lesson 3 23 Techniques for Computing Limits In this lesson you will learn a number of useful rules for computing limits for the most common types of functions you will encounter in this course You should study the examples in the text carefully as these will help you understand how to apply the limit laws The basic rules for computing limits are as follows Suppose that c is a constant7 n a positive integer7 and that the limits limmna x and limmnagw exist Then 1 lim 0 c maa 2 lim z a mgta 3 Elana ciggfm igfxgz f31 5 hm imam 3331 f96i1331996 maa 4 03 lim f lim L provided that lim 91 31 0 maa llm maa wall 5 hm om 33 mo w a 8 lim 1fx nlim f provided that f Z 0 for z near a when n is even EH04 EH04 As an example7 if lin f 73 and lingx 27 it then follows from limit laws 37 mm zgt 47 and 7 that 3312 we 7 39I1312fw2 engage 3 All the above rules also hold for one sided limits with the obvious modi cations The limits laws can be used to nd the limits of polynomial and rational functions lfp and q are polynomials7 then lim p 10a7 lim Z provided that qa 3A 0 wall The computation of a limit liming fx of the quotient of two functions Where 9a 0 often calls for algebraic manipulations Two basic techniques cancel ing common factors and multiplying by the algebraic conjugate are illustrated in Example 6 in the text Yet another technique for nding limits is afforded by the Squeeze Theorem lf f S 92 g h for all z near 1 except possibly at a and if limmna x liming Ms L then i gw i L You should try to decipher the content of the Squeeze Theorem in terms of the graphs of x 9z and Starter Problem List 4 7 10 11 15 17 23 25 33 37 49 Recommended Problem List 22 31 34 38 41 44 45 47 51 55 60 64 Lesson 4 24 In nite Limits A function f possesses an in nite limit at x a when its values grow larger and larger without bound as X approaches a In this case one writes 13 f 00 As an example limmn01z2 00 In analogy ifthe values of f are negative and grow larger and larger in magnitude as x approaches a then the limit of f at z a is negative in nity or 13 f foo One de nes in nite limits for the one sided limits limmnai f in a like fashion For in nite limits one has the following variant of the Squeeze Theorem Suppose that fx 2 g for all z near but not equal to a and limmnagw 00 then also liming f 00 As an exercise formulate an analogous statement involving a negative in nity limit If at least one of the one sided limits at m a is either 00 or foo the line 1 a is called a vertical asymptote of lf f is a rational function in reduced form that is p q share no common factors then the vertical asymptotes are situated at the zeros of the denominator The trigonometric functions tan 9 cot 9 sec 9 csc possess an in nite number of vertical asymptotes Can you locate all of them Starter Problem List 4 5 6 8 13 17 27 29 39 Recommended Problem List 10 15 18 20 23 25 32 34 36 37 43 Lesson 5 25 Limits at In nity and Horizontal Asymptotes The limit 113 f 95gt of a function f as m approaches 00 describes the values of fx as x becomes larger and larger without bound For example7 since 1M x 0 when x is large7 one can conclude that 1 ng s1n 0 On the other hand7 lim 2 7 s 007 00 since the term 2 grows much faster than x Finally7 lim sinx 1200 does not exist Can you see why The limit limmnnoo f of fx at negative in nity is similarly found by letting z become negative and larger and larger in magnitude If either one of limmnioO f L exists and is nite then the line y L is a horizontal asymptote for Thus7 as we saw above7 the line y 0 is a horizontal asymptote for f sin1z Suppose that f is a rational function with p am m amn mil alx 107 bn n bnil nil l 231 l bov where am7 on 7E 07 then 1996 m 7 07 if m lt 717 m ioo 2 1300 gogm amb 1f m n 3 lirin ioo7 if m gt 717 depending on whether m 7 n is even or odd ma 0 Hence the rational function f has a horizontal asymptote precisely when the degree of qx is equal or larger than the degree of One can use the exponent rules for the natural exponential function to conclude that lim em oo lim em 0 mace maioo Consequently the logarithm function as the inverse of the exponential function must satisfy limln7oo lim lnzoo 1A0 1 as you can also verify by graphing the functions Starter Problem List 4 9 11 15 21 31 35 38 Recommended Problem List 12 13 18 20 23 25 27 33 36 40 44 53 Lesson 6 26 Continuity A function f is continuous at x a if liin u fa That is fx is continuous atzaif 1 u is de ned on some interval containing z a 2 the limit limmna z exists and 3 the limit is equal to the value fa lntuitively speaking if u is continuous at z a then the graph of u contains no holes or gaps at z a A function f is continuous on an interval if it is continuous at every point con tained in that interval All algebraic functions and the basic transcendental functions sin n cos x em lnx are continuous in their respective domains of de nition The Limit Laws imply that the sum the difference the product and the quotient of two continuous functions are again continuous in their domains of de nition More over the composition of two continuous functions is continuous as is the inverse function of a continuous function when it exists The Limit Laws also yield many other continuous functions Can you see why the function 1 1 06 W2 smlt1 2 is continuous for all values of x The Intermediate Value Theorem can be used to locate solutions to equations For example you can apply the 1VT to conclude that the seemingly complicated equation x775x5z210 must have a solution in the interval 01 You will nd other applications of the lVT in the exercises below and in Laboratory Activity lll Starter Problem List 45911 1317 253051 Recommended Problem List 15 24 28 33 35 41 45 49 54 55 61 74 77 11 Lesson 7 31 Introducing the Derivative The di erenee quotient of a function fx on the interval 1 a h is given by f a h i f a h 7 which can be interpreted as the slope of the secant line through the points a fa and a hfa h or alternately as the average rate of change of f on the interval a a h A function f is di erentiable at the point z a if the limit of the difference quotient fa lim hgt0 f a h 7 f a It exists and is nite One frequently also writes W hm we 7 u man 7 1 Check that the two de nitions are equivalent The value of the limit f a is called the derivative of fx at z a If f is differentiable at every point on an interval I then the differentiation process determines a new function f the derivative of n on I One often employs the Leibniz notation to indicate the derivative f m 1ff is differentiable at z a then it is also continuous at z 1 Thus if f is not continuous at z a then it can not be differentiable at that point either But keep in mind that there are many continuous functions that are not differentiable as for example is the case with f at z 0 The computation of the derivative using the de nition is illustrated in Examples 3 4 and 5 in the text and you should carefully go through the steps so that you will be able to carry out similar calculations on your own Basic applications of the derivative include the slope of the tangent line to the graph of a function and the instantaneous rate of change of a function Thus given the slope in f a of the tangent line at z 1 its equation can be written as y ma 1 1n the same vein if z 5t is the position of a particle moving along the x axis d5 its velocity is given by Vt Note that the time t is now the independent variable so the derivative must be computed with respect to t Starter Problem List 6 13 18 26 36 39 41 46 Recommended Problem List 11 15 19 22 27 34 43 50 53 65 71 Lesson 8 32 Rules of Differentiation The following rules for the derivative can be derived directly from the de nition d 1 Constant function rule die 0 a d 71 2 Power rule is nz n123 dx 3 Constant multiple rule cf a 4 Sum rule f 91 d As you will see in section 387 the power rule div rxr l in fact7 holds true for any real exponent r One often needs to combine the above rules to compute the derivative of a given function For example7 by applying both the constant multiple and sum rules we see that immemmvnwwn Notice that in order to compute a derivative at a particular point using any of the four rules above7 you only need to know the value of the derivatives of the constituent functions at that point So if we know that f 1 71 g 1 37 then g uoewmm wnrw ver Euler s number e 2718 is de ned by the property that h e 71 l 7 1 ugh h As a consequence7 d 7e dx that is7 the natural exponential function is its own derivative m 7 16 The higher order derivatives d2 d3 fe f x We f a of a function fx are obtained by repeatedly differentiating For example d2 7 d d 7 d 6 5 7 7 7 77 42 dzz dxdz dmx z i d4 i i As a exerc1se compute 74x3 What do you notice Can you generalize your observation into a rule for higher order derivatives of a polynomial function Starter Problem List 7 9 13 17 19 27 29 31 35 39 42 50 54 Recommended Problem List 10 14 18 22 25 32 38 41 45 48 56 66 Lesson 9 33 Product and Quotient Rules In this section you will learn rules for computing the derivatives of the product and quotient of two functions in terms of the values of the functions and their derivatives 1 Product rule fWMW f f9 dx f 9x 7 IVE96 39 d 2Quot1ent rule gay lmportantly7 these rules allow one to compute the derivative of the product and quo tient of two functions at a point from the values of the functions and their derivatives at that point only As an example7 suppose that the following information is given f1 3 f 1 1 94 2 9 1 3 Then by the quotient rule7 1 WC f 191f19 1 7 dx g 9 1 In more complicated derivative computations you will need to be able to combine these and other rules of differentiation see Example 7 in the text for a typical application The quotient rule can also be used to extend the power rule of differentiation of Lesson 8 to negative values of the exponent The horizontal scaling f a fkz changes the instantaneous rate of change of the d function by the factor k Thus cf1m so7 in particular7 m d dx The function 6 appears in models for population growth and radioactive decay 6 k6 for any real number k Starter Problem List 67 87 137 187 237 277 317 377 417 47 547 64 Recommended Problem List 97 117 217 267 297 357 397 44 507 557 607 627 667 77 81 Lesson 10 34 Derivatives of Trigonometric Func tions The following limits7 which are important on their own right7 are used to compute the derivatives of the sine and cosine functions s 7 1 lim SIM 1 lim L 95 zgt0 z weo m 0 Both expressions are in indeterminate form7 so nding the required limits by a geo metric argument7 as is done in the text7 takes some work It is important to bear in mind that in these limit statements the z variable must be expressed in radians You should commit to memory the derivative formulas of the basic trigonometric functions cataloged in the table below d d 1 ismxc0sx 2 icosxism dx dx d 2 72 d 2 x 72 3 itanzsec xcosm 4 7cotz7csc 7smx dx dx d sinx d cosm 5 isecxsecztanz 2 6 icscmicscxcoti 2 dx cos x dx sm z Note the antisymmetry in the formulas for the derivatives of the sine and cosine functions The remaining derivatives 3 6 can be obtained from these two with the help of the quotient rule7 as you should verify on your own Typically the derivatives of tanx7 cot x secs and cscx are given in terms of the same7 but keep in mind that when manipulating or simplifying a complicated trigonometric expression7 it is often helpful to convert it rst to a form involving sine and cosine only Caveat In all the derivative formulas for trigonometric functions z must be expressed in radians If you use other units degrees7 grads7 7 you will need to modify the formulas accordingly Thus7 for example7 if z is given in degrees7 then d sin 2 7T 7 7 cos 2 dz 7 180 as you should con rm on your own Starter Problem List 6 7 9 15 19 23 27 32 33 37 45 51 54 Recommended Problem List 10 13 17 22 30 35 40 47 50 56 58 68 69 Lesson 11 35 Derivatives as Rates of Change The key concepts of this lesson are the average rate of change and the instantaneous rate of change of a quantity that varies in time t Suppose that an object moves along the x axis so that its position is given by x ft7 where t denotes the time For example7 x cost describes an instance of harmonic motion Then the displacement of the object during a time period from t a to t a At is Ax fa At 7 fa7 so the average velocity of the object is given by Ax fa At 7 fa 111 ave 7 7 l V At At where7 as you notice7 the right hand side is simply the difference quotient of the function ft on the interval a a 1 At As can be easily checked7 the average velocity in harmonic motion x cost over the interval 027T is zero even though the object moves back and forth between 1 and 71 Thus the average velocity does not often accurately characterize motion A more precise description is obtained by computing the average velocity on increasingly smaller intervals7 that is7 by letting At 7 0 in 111 But the limit once again results in the derivative and so the instantaneous velocity of the particle is simply V ft The instantaneous rate of change of velocity is the acceleration7 so at Vt Warning Dont be confused by the texts use of the same symbol a for a xed value of the x coordinate and the acceleration of an object The nal example of the section treats a business application Suppose that the cost of producing x items in a manufacturing plant can be computed from the cost function Then the average cost of producing x items is C39xx7 while the marginal cost measures the expense of manufacturing one additional item The increase in production from x to x Ax items incurs the extra average expense of Cx Ax 7 Cx Ax 19 per item The marginal cost is obtained by taking the limit as Ax a 0 and so is given by the derivative C z of the cost function Of course in real life situation z and Ax take on integral values so the marginal cost will only give an approximation to the actual increase in the cost of producing one more item Cz1 702 Cs Starter Problem List 6 7 9 12 17 21 26 38 Recommended Problem List 10 15 19 24 28 31 37 41 48 51 Lesson 12 36 The Chain Rule The chain rule allows you to compute the derivative of the composition of two func tions fgx in terms of the derivatives f and g Chain Rule f g9 You will see many variants of this basic formula such as d 4 dune dg dtfhtifhtht7 dfdm and dtidm but they all express the same fact as the boxed formula7 only in different notation The chain rule is one ofthe most useful tools for nding the derivatives of complicated functions as it lets you break down the computation into the differentiation of simpler functions Perhaps the following argument will help you to understand and remember the chain rule Recall that if z changes by a small amount Ax then the change in the values of the function g is approximately gltz M 7 gm 9mm Thus7 in the composition fgz7 the argument or the input of y changes by Ay g xA where we have written y So the value of fgz changes by fy A107 y w f yAy w f 99 9 m But this must be equal to Ax so by comparing the two expressions for x the change of fgz7 you will recover the chain rule f gzg z x Example If y y with the derivative f y nyn l then the chain rule reduces to ber namelyz For n 71 this7 combined with the product rule7 yields an alternate useful form d hx d 7 7 am ame 1 Image 1 e hltzgtgltzgt we 21 of the quotient rule D Example A derivative computation often requires a repeated application of the chain rule For example sincosx2 coscos2i cosx2 coscosx2i sinz2xz coscos27 sin22x 72x coscosx2 sin2 where at rst we used the chain rule with y siny g cosx2 and in computing the derivative of cosx2 with y cosy g 2 Example 5 in the text highlights the same type of computation D Example In light of the chain rule the rate of change of the composition gx at z a is the product of the rate of change of y at y 9a and ofg at z a As an illustration suppose we know that 90 1 g 0 3 f 1 71 Then d f996 lm0 f 909 0 f 13 3 Thus the rate of change of gx at z 0 is 73 D Starter Problem List 2 6 9 12 20 30 36 38 49 59 Recommended Problem List 14 16 22 26 31 34 39 45 50 55 58 62 67 75 Lesson 13 37 Implicit Differentiation Suppose your task is to compute the slope ofthe tangent line to the ellipse 4z2y2 1 at the point 1 You can of course solve the equation for y to nd that 47 7 y ix1 7 4 which can then be differentiated to obtain the slope But as an alternate approach note that the equation for the ellipse de nes up to a i sign y as a function of x Write y f for the function so that 4x2 fx2 1 But this equation simply states that the function 4x2 fx2 on the left hand side must be identically constant 1 Hence its derivative must vanish so by the chain rule for powers we obtain 8x 2ff 96 07 H96 if Thus the slope at i will be We 7 One typically forgoes writing f for y and directly differentiates both sides of the de ning equation 42 y2 1 while keeping in mind that y is a function of x This gives 4m y It is now a simple matter to nd the slope by plugging in the z and y coordinates of the base point 8x2yy 0 gt y 7 This process of differentiating y without knowing its explicit expression in terms of z is called implicit di erentiation As we saw it simpli ed the task of nding the slope of the tangent line to the ellipse but implicit differentiation really becomes an indispensable tool in problems in which one can not analytically solve y in terms of x It can be shown that the equation y em2y 1 de nes y as a function of x but can you nd an explicit expression for y in terms of 7 How about computing dydz in terms of z and y Caveat Common mistakes in carrying out implicit differentiation are to forget to apply the chain rule when differentiating terms involving y which is to be considered a function of m and to forget to differentiate constant terms 23 Two lines 1 2 intersects perpendicularly or at 90 angle if their slopes satisfy 5152 71 provided that 51 7E 000 You can use this property to nd the normal We to a plane curve at a point since the slope of the tangent line is dydm the slope of the normal line will be 71dydx The normal line is vertical when dydx 0 Example Example 5 in the text asks for the slope of the tangent line to the curve 2 11 y at 4 4 As in the text differentiate the equation implicitly to get 2 723 dy dy 1 7 7 3 a y lt dx dx Since the goal is to nd the slope at a given point there is no need to solve the equation to derive a general formula for dydx 7 one can instead directly substitute the coordinates of the base point to see that dy idy 3 723 51171 7 344 1d761did 75 from which dydz 15 Hence in particular the tangent line to the curve at 44 is given by y 25 165 and the normal line by y 75x 24 You can exploit the same shortcut to facilitate solving several of the exercises in this section B Starter Problem List 189 15 23 27 35 41 61 65 Recommended Problem List 10 17 21 29 31 39 45 50 52 64 67 69 Lesson 14 38 Logarithmic and Exponential Func tions You recall that the natural exponential function is one to one so it has an inverse which is called the natural logarithmic function lna The domain of lnx is 000 and the range is foo 00 as you should verify by graphing lnx using the graph of em as the starting point Thus by de nition 6 x for z gt 0 and ln 6 x for all x The power rules for the exponential function have their counterparts the logarithm rules which are summarized below 1 emey em ltgt lnzy lnz lny 2 662 6w ltgt lnxy yln x These formulas also yield the expression for the derivative of the natural logarithm function Differentiate the equation x eh with respect to z and don7t forget to apply the chain rule to see that d d 1 elm lnz mi lnx where in the second step we used the relation z oh This can be easily solved As we will see later the above computation is just a special case of a general algorithm for determining the derivative of an inverse function The logaiithm function in base b b gt 0 is de ned by the relations ylogbz ltgt xby wherezgt0 You should derive the identities lnx bm 6 lo z 7 7 El lnb 7 25 which by differentiation give d m m d 1 lnbb logbz L z ln b 7 for the derivatives of the exponential and logarithm function in base b Logarithmic dz erentz39ation is founded on the observation that f 96 f96 Thus one can nd the derivative of a function f by starting with differentiating ln f and then multiplying the result by For certain type of functions this will simplify computing the derivative Can you think of any examples d d Eln z ltgt fzfzlnf d Example As an application of logarithmic differentiation we compute d7sin s We have that d d d sin xv sin lnsin mm sinxV Q ln sin 2 COS gt sin zw ln sinx xcos xsin 2V4 sinzwlnsin z sinx Starter Problem List 6 7 9 13 17 26 31 38 44 59 Recommended Problem List 11 15 19 21 23 32 35 39 47 51 55 68 71 76 85 Lesson 15 39 Derivatives of Inverse Trigono metric Functions Suppose that x is invertible with the inverse function f 1z Then by the very de nition fx and f 1x satisfy ff 1z x We can differentiate this identity to see that d 71 i 71 139 f f 96gtdxf I Can you see why we again needed to use the chain rule Solving for the derivative of the inverse function we obtain In other words the derivative of f 1p is the reciprocal of f composed with f AWN Example Suppose that fx is invertible and that you are given the following information f137 f327 f 167 f 34 d d Find d7f71 m3 Can you compute dif 1zlm1 based on the above data p p d In order to nd dif 1xlm3 we apply the basic identity for the derivative of the x inverse function For this we will need f 13 which since f1 3 equals 1 Thus 1 1 1 d 1 i i if of WV rm m 639 d In order to nd dif 1plm1 we would need need to know the value of f 11 that p is an x value so that fp 1 but this is not given so the answer to the latter question is negative D The sine function is 27T periodic so it fails the horizontal line test and can not be invertible However its principal part the restriction of sinx to the interval 771717172 is one to one and possesses an inverse function which is denoted by 1 sin z or often also by arcsin Thus i 7T if i 1 ys1np iagzgg ltgt xs1n y ilgpgl 27 As an exercise graph sin lx starting from the graph of sin x carefully identifying the domain and range of sin 1 z in your plot You should note that the inverse function of sinx can be de ned on any interval for example on 737T2 7r2 7r237r2 where it is one to one The choice of the particular interval is usually dictated by the problem at hand Caveat 1 On account of the de nitions sinsin 1x x for all 71 g x g 1 but 7T 7T sin 1sin s only when 75 lt m g 5 Can you derive an expression for sin 1sin m that holds for any x Caveat 2 Make sure that you can distinguish between the notation used for the inverse sine and for the reciprocal of the sine function 1 sin 1 z arcsinm sin x 1 7 sinx In order to nd the derivative of sin 1 we apply the general procedure for differ entiating inverse functions This gives d 1 1 ism s dx cossin 1z We still need to simplify the composition cossin lx which will be based on the basic trigonometric identity cosz l sinz l 1 If you substitute 9 sin lx into this and use the identity sinsin 1 s you see that cos2sin 1 m x2 1 When solving for the cosine term keep in mind that the values of 9 sin 1z fall on the interval WQ TQ where cost is positive so the above equation yields cossin 1x V1 7 2 In conclusion 1 v 1 7 x2 39 The inverse functions of the other basic trigonometric function are similarly de ned by a restriction to a suitable interval The derivatives of the inverse functions are then computed using the general process as explained above see the table below sin 1 28 H sinx 2 003 03 tan 4 COt 5 secx 6 030 Domain and Range of Principal Part igvaL 7171 OJ 711 7 7 7 700700 077139 70000 075U 77T 700771UH7OO 757 0 U 0757 7007 71 U 1700 Starter Problem List 37 57 7 137 207 317 357 407 447 52 Recommended Problem List 97 157 267 297 347 377 417 47 537 577 66 Lesson 16 310z Related Rates Suppose that two quantities changing with time t are related by an equation Then if we know the rate of change of one of the quantities we can nd the rate of change of the other quantity by differentiating the relating equation with respect to t As a speci c example consider the z and y coordinates of an object moving along the ellipse 4x2 112 1 Then z and y are functions of time so by taking the t derivative of the equation of the ellipse and again applying the chain rule we see that dx dy i 87 2 fie dtydt dy 774xdz dt 7 y dt39 Thus if we know the position of the object and the rate of change of the x coordinate we can readily compute the rate of change of the y coordinate This simple example underscores the general fact that related rate problems typically reduce to differentiating an equation involving two or more quantities changing in time The differentiation step entails applying the chain rule very much like in problems requiring implicit differentiation Many of the exercises in the text involve basic geometry and call for a bit of creativity but steps for solving related rates problems outlined in the section will help you to get on the right track A typical example involving related rates is presented below Example A circuit consists of three resistors with resistance R1 R2 and R3 set in parallel and a 10 volt power source The value of R1 is increasing at the rate of 04 95 and that of R2 is decreasing at the rates of 07 95 while R3 stays constant Find the rates of change of the total resistance and the current in the circuit when R1209 R2 Q and R35 Solution First draw a picture to ensure you comprehend the physical set up and understand the question You may recall from your physics courses I I that the total resistance R of three resis 10 V r Rl RZ th tors set in parallel is determined by the 4 4 1 equation l 1 i 1 1 1 162 E E l g i gv and that the total current across the three resistors is given by I VR These two equations express the basic relationships between the variables in the problem 30 We will start with nding the rate of change of the current as this is somewhat easier computation of the two to carry out Since voltage V is assumed to be constant the equation for current after differenti ating with respect to time t and applying the chain rule yields d V dB 163 i 777 t dt R2 dt Hence we can determine the rate of change of the current I once we know the value of R ZdRdt But this quantity can be derived by differentiating equation 162 for the total resistance We have 16 4 1 dB 7 1 de 1 ng 1 ng 39 det Pgdt Rgdt Rgdt39 We will omit units in the subsequent computations After substituting the given values de dRZ dRS 165 R20 R35 R5 704 7707 70 gt123ddd into 164 we nd that 1dRi 04 70707 3 N43X1074 R2 dt 7 202 352 7 7000 N 39 39 Consequently by equation 163 d 3 7 if z 743 10 3A dt 700 X 5 Next by the equation 162 for the total resistance 1 39 140 E m i R E whence by 164 dB 7 w 55 10 39 dt gtlt s D Example Example 4 in the text deals with the rate of change of the angle t between the horizontal and an observer7s line of sight to a rising balloon This 31 problem can be disposed of in short order by applying the ideas you learned in Lesson 13 From the picture in the text 7 W tanlt6lttgtgt 7 200 which can be directly differentiated to yield d0 d0 1 dy 2 i 2 i if 166 sec tdtt 1tan tdtt 200 dt d By the statement of the problem 4 so y30 120 and tan 30 35 Now 166 can be solved for the rate of Change of 9 to give d1 4 1 E80 7 x 00147 rads 2001352 68 Starter Problem List 2 58 12 1419 27 32 Recommended Problem List 11 13 17 23 29 31 35 37 Lesson 17 41 Maxima and Minima Many applications of calculus to concrete problems call for nding the maximum or the minimum value of a pertinent function which for example could be the revenue or cost function the acceleration of an object air drag the size of a population and so on A value c is a local mauimum of a function u if there is some interval cihc h h gt 0 so that c is the largest value of u in this interval so c Z u for any u between c 7 h and c h but there can be u values outside the interval with u gt c One similarly de nes a local miriimum A function assumes an absolute mauimum value at u c if c Z u for any u in the domain of the function Again the absolute miriimum of u is de ned similarly Note that a function needs not possess any local or absolute maxima or minima as you can see for example by graphing the function u u3 or u tanu However a continuous function de ned on a riite arid closed interval always attains an absolute maximum and minimum on that interval which could also be realized at the endpoints of the interval This fact is known as the eutreme value theorem and as is easily veri ed by the way of examples continuity is a necessarily condition for the theorem to hold lf c yields a local maximum for a differentiable function u then the slope of the graph of u must be non negative immediately to the left of c and non positive immediately to the right of c Thus at u c the derivative of u must vanish f c 0 An analogous statement holds for local minima Hence in general if c is a local mauimum or miriimum of u theri either f c does riot euist or f c 0 Points contained in the domain of u that satisfy either one of these two conditions are collectively called critical points of u and they can often be effectively used to locate local maxima and minima and with the help of these also the absolute maximum and minimum of a given function Starter Problem List 5 11 15 17 23 31 33 43 48 66 Recommended Problem List 19 22 25 29 34 42 49 53 61 65 69 Lesson 18 42 What Derivatives Tell Us The rst and second derivatives are intimately related to the behavior of a function and consequently they provide further tools for analyzing extremum value problems discussed in the previous lesson Roughly speaking the derivative f measures the steepness of the graph or its direction and the second derivative gauges the rate at which this direction is changing Thus intuitively speaking f z yields a measurement for the curvature of the graph A function f is strictly increasing on an interval I if for all x2 gt 1 on I ag gt fz1 and strictly decreasing on I if for all 2 gt 1 on I zz lt f1 One can detect intervals on which f is increasing or decreasing by the sign of the derivative f gt 0 for all x on I gt f is increasing on I f lt 0 for all x on I gt f is decreasing on I lntuitively f x gt 0 implies that the slope ofthe tangent line is pointing upward so the graph of the function is tending higher when x moves to the right that is f is increasing Analogously f lt 0 implies that the graph is tending lower so the function must be decreasing But note that the converse of the above derivative test does not hold a function increasing on an interval I can have f 0 at some in fact even at in nitely many x on I Next assume that c is a critical point of n and that f gt 0 on some interval immediately to the left of c and f lt 0 on some interval immediately to the right of c Then as x approaches c from the left x increases and as x moves past c it decreases It follows that a must have a local maximum at c see the diagram below Critical f gt 0 Point f lt 0 l f increasing Mig im fx decreasing Similarly the conditions f lt 0 immediately to the left of c and f gt 0 immediately to the right of c imply that fc must be a local minimum for fx Critical I f lt 0 Point f gt 0 l f decreasing L cal fx increasing Minimum Finally if the derivative f does not change sign at z c fc will not be a local maximum or minimum The above arguments can also be ap plied to the function f provided of course that its derivative df dz f z exists Thus if f x gt 0 on an interval I the derivative f is increas x ing In this situation the tangent lines drawn along the graph of f turn coun terclockwise with increasing z ithe func tion n is said to be concave up on I If in turn f z lt 0 on an interval I then the tangent lines turn clockwise along the graph ofthe function and f is said to be concave down on I Graph ofa mmon concave up A point c in the domain of f is called an in ection point for f if concavity changes at c In particular you can locate in ection points by rst nding all points c at which f c does not exist or at which f c 0 If f z changes sign at c then c will be an in ection point But bear in mind that a point c in the domain of f can be an in ection point also when f c does not exist These type of arguments also yields the second derivative test for local eatrema Suppose that f z exists on an interval I and z c is an critical point on I with f c 0 If f c gt 0 then fc is a local minimum for fx while if f c lt 0 then fc is a local maccimum for The fact that positive second derivative is associated with a local minimum might seem counter intuitive at rst but the second derivative test is easy to remember in its correct form by considering the parabola f x2 for which f 0 gt 0 at the local in fact absolute minimum at z 0 35 Finally the second derivative test is inconclusive if f c f c 0 and in this case you should investigate the sign of the derivative f on each side of 0 Starter Problem List 8 11 15 17 24 31 34 39 43 47 59 64 67 Recommended Problem List 13 22 26 29 33 37 40 45 53 55 60 61 71 Lesson 19 43 Graphing Functions As we saw in the previous lesson the rst and second derivatives are closely associ ated with the properties of a function It then comes as no surprise that they also yield a wealth of information that is valuable in analyzing the graph of the function Caveat When plotting a function do not rely on your graphing utility to do the work for you but perform the steps outlined below on your own and by hand Only when you are done with the problem should you double check your results against the graph produced by software In a typical graphing problem you should follow the steps outlined below i 9 P03 9quot 7 Identify the interval on which the function f is to be graphed This can be a subset of the domain of Check for special properties of ls the function even es or odd ex ifx or more generally symmetric with respect to the line z c f2c 7 m or antisymmetric with respect to it f20 7 m 7fx7 ls it periodic fz L or antiperiodic fz L 7fx7 How would you graph the function in each instance Find the intercepts of the graph with the z and y axes Compute the derivative f and locate all critical points ldentify intervals on which f is increasing or decreasing and classify all local and absolute extrema Compute the second derivative f x and locate all in ection points ldentify intervals on which f is concave up or down You may also double check your classi cation of local extrema in step 3 by the second derivative test Compute the limits limmnioO f and locate vertical and horizontal asymptotes if any Graph the function with the help of all the information you have gathered in the above steps And keep in mind that there is no substitute for doing it yourself Practice makes perfect Starter Problem List 2 68 9 15 18 23 29 38 42 51 Recommended Problem List 13 17 25 31 33 35 41 43 47 56 61 Lesson 20 44 Optimization Problems Finding the best possible solution in practical problems is one of the principal ap plications of calculus ldentifying these optimal solutions is based on the minmax problems of Lesson 177 but7 besides calculus7 some experience with general problem solving will be called for The following example illustrates the general procedure Example A farmer has 1200 m of fencing and wants to fence off a rectangular pasture that borders a straight river He needs no fence along the river Find the maximum area the pasture i D 03 P U The problem asks for the maximum area of a rectangular eld7 so the rele W vant quantities in this problem are the RiVer W width w along the river7 the length l7 W and the area A of the pasture l Pasture l The quantity to be maximized is the area which is given in terms of the other W quantities by A wl This is a func tion of two variables and we have not learned yet how to nd the extreme values of such functions However7 the problem imposes a constraint between w and l The total length of the three sides of the pasture is 12007 so w 2l 1200 Recall that we are assuming that the river runs along the width of the pasture We will also omit units from now on The constraints between w and l can be solved7 say7 for w to yield w 1200 7 2l This expression7 when substituted into the expression for A7 gives A 120072ll7 a function of one variable Both l and w are be positive7 so l must be constrained to the interval 07 600 We have reduced our optimization problem to nding the absolute maximum of the function A 1200 7 2ll 1200l 7 2l2 on the interval 07 600 This function has only one critical point at l 300 as you can verify by solving for the zeros of the derivative dAdl At the endpoints A0 07 A600 07 so A300 187 000 yields the maximum area in square meters that the farmer can fence off To check the reasonableness of our answer7 we compare it to the area of a square pasture with three sides made out of a total of 17 200 m of fencing We nd that 38 the area of the square pasture7 167 000 m27 is less than the answer we obtained for the maximum area7 lending support to our solution of the problem B One can solve a typical optimization problem by completing steps that are similar to the ones carried out in the above example Procedure for Solving Optimization Problems 1 Read the problem carefully to identify all the quantities involved Give them names or designate each one by a symbol Try drawing a sketch to visualize the problem Make sure you understand what the question is 2 Identify the quantity to be optimized and nd a mathematical expression for it in terms of the quantities uncovered in Step 1 This is the objective function in the terminology of the text 3 Write down all the relationships you can come up with between the quantities found in step 1 4 Use the conditions discovered in step 3 to eliminate all but one variable in the function to be optimized Find the relevant interval for the remaining variable By requiring that the length7 area7 and volume must be positive7 etc 5 Use calculus cf Lesson 17 to nd the absolute minimum or maximum value of the function to be optimized7 and interpret your result in terms of the original quantities in the problem Finally7 double check that your answer is reasonable by verifying that it is physically sensible and satis es the constraints imposed by the problem and7 for example7 by making a comparison with some easily computable special cases of the problem Starter Problem List 4 6 7 10b 12 19 26 27 39a 48 55a Recommended Problem List 8 15 21 28 31 37 41 47 50 56a7c 59 Lesson 21 45 Linear Approximations and Dif ferentials If a function f is differentiable at z a then the limit fa hm 7 17m 7 1 gives the approximation or f96 fa f a a7 for z near a The expression M96 fa f a9 a on the right hand side is called the linear appmm39matz39on to f at z a As you recognize the graph of the linear approximation is simply the tangent line to the graph of f at z 1 Thus geometrically using linear approximation amounts to replacing the graph of the function by the graph of the tangent line As a rule of thumb the linear approximation gives a good estimate for the values of f when the second derivative f is small near 1 Example We use linear approximations to estimate sin 005 Note that the angle measurement is in radians Now f sins We need a point near z 005 for which we know the precise value for sin x The obvious choice is a 0 Thus the linear approximation gives L005 sin0 cos 0005 7 0 005 As you can check on your calculator sin 005 004998 to 5 signi cant gures so the linear approximation yields a surprisingly good estimate for sin 005 But this was more or less expected as f z 7sin x which is small near the base point a 0 Writing Ax z 7 a for the change in z and Ag fx 7 fa for the change in the values of x we can rewrite the expression for linear approximation as Ay f aA 40 By replacing Ax and Ag by their dz erentz39als dz dy which in traditional calculus represent in nitesimal changes77 in z and y the above formula becomes dy fadx The custom of denoting in nitesimal variations by differentials might seem odd at rst but the notion can be made fully rigorous in the setting of modern differential geometry Starter Problem List 2 6 7 11 13 23 25 32 43 Recommended Problem List 10 17 19 22 27 34 36 40 44 45 Lesson 22 46 Mean Value Theorem This section treats an important theorem that can be used to prove rigorously many of the claims justi ed on intuitive grounds in the previous lessons Roughly the theorem asserts that given an interval 1 b contained in the domain of a differentiable function fx there is some 0 between a and b at which the tangent line is parallel to the line connecting the points a fa and b fb on the graph of the function The Mean Value Theorem MVT Let fx be continuous on the closed interval ab and differentiable on the open interval ab Then there is some 0 a lt c lt b so that b m f 21 Rolle s Theorem is a special case of the MVT obtained by assuming fa fb 0 Note that for a given f and an interval 1 b nding the value of c can be dif cult if not impossible consider for example f sinx 7 2 a 0 b 1 but in spite of this the MVT yields a surprising amount of useful information about the behavior of differentiable functions For example you can now see once and for all without having to appeal to intuitive geometric arguments that if f gt 0 on an interval I then fx is strictly increasing on I Simply choose any a b in I b gt 1 Then by the MVT fb fafCbia for somealtcltb But as both f c gt 0 and b 7 a gt 0 by assumption fb fa f cb i a gt 0 Consequently fb gt fa But the conclusion holds for any a lt b on I so f must be increasing on I The MVT also implies that if two functions have the same derivatives f g z on some interval I then there is a constant C so that gx f C identically on I Thus the graph of g can be obtained from the graph of fx by a shift in the y direction As you recall cf the table on page 29 the derivatives of sin 1 z and cos 1 x are opposite What do you conclude about the two functions in light of the MVT Can you derive similar identities between tan l z and cot l m and sec 1 z and csc l x 42 Starter Problem List 3 6 7 12 16 21 27 29 31 36 Recommended Problem List 9 11 14 17 19 24 28 30 35 Lesson 23 47 L Hopital s Rule Suppose that your task is to compute the limit ewil lim zao Sinm The substitution z 0 yields 00 so the limit is in so called indeterminate form and you wont be able to use the quotient rule for limits to solve the problem As a workaround divide both the numerator and denominator by z and insert sin0 0 in the denominator after which the original limit problem becomes ewil m lim 1H0 s1n 7 s1n0 m Now you can recognize the difference quotients of em and sin x at m 0 in the resulting expression so taking the limit as m a 0 amounts to computing the derivatives of these functions at z 0 But both derivatives equal 1 so now the quotient rule for limits can be used to compute 1 5 7 w 1m7 i emil mgt0 z Che lm0 lim f 1 we0 s1nx s1nz7s1n0 d i 11m isms lFO mgt0 z dx The same process can be applied to any limit lim f in the indeterminate form 1 00 that is fa 9a 0 If the derivmafives f a g a exist and g a 31 0 th e we 1 gm 39 This is the basic form of l7Hopital7s rule which is one of the most useful techniques for nding limits of the quotient of two functions in indeterminate form Caveat When putting l7Hospital7s rule into practice you need to make sure that the limit is in indeterminate form and that you rst differentiate fx and g separately and only then compute the limit of the quotient of the derivatives L7Hopital7s rule also applies in the same form to limits lim in the inde mall terminate forms ioooo that is the limits of f and g are both either 00 or 44 700 In each of the cases 00 and ioooo the limit point a can as well be ioo L7Hopital7s rule also holds for one sided limits with the obvious modi cations If an application of l7Hopital7s rule results in indeterminate form the method can be applied repeatedly until the value of the limit is found provided of course that the required derivatives exist 1 7 Example Compute the limit lim w 1A0 i The limit is in indeterminate form 00 so we resort to l7Hopital7s rule By computing derivatives we get 1 7 cos x sin z lim 7 m 7 mgt0 z mgt0 2x The resulting limit is still in indeterminate form 00 so to proceed we apply l7Hopital7s rule again This gives 1 licoszilA cosxil will 952 33 2 7 2 where in the last step we were able to compute the limit by substitution D L7Hopital7s rule can also be used to compare the growth rates of functions at ioo For example 11mm hm W hm nan 96gt A A 7 mace mace mace 100 where in each step save the last the limit is in indeterminate form Thus when x approaches in nity z becomes much larger than ln x for any integer n that is z dominates ln There are many variants of the basic version of l7Hopital7s rule designed to compute the limit in the case of the indeterminate forms 0 ioo oo 7 oo 1i 00 000 The instances involving powers typically require an application of the natural logarithm function in their solution Example Compute the limit lim sin mm mgt04r We write h sin sf and start by computing lim ln hx lim sin s ln sin x mgt04r zgt0 The resulting limit is in indeterminate form 0 foo so in order to apply l7Hopital7s rule we rewrite it as 1213 ln h lnsinx lim 7 mgt0 1 7 sinx which in the form foo00 We compute derivatives to see that COS 1 1h 13w lt95 i sin l1m mgt04r 7 COS lim isinm 0 mgt0 sin2 Thus the limit of ln h is 0 so by continuity of the natural exponential function limJr h lim elm elimaHOJr 1M 60 1 0 1A0 In conclusion we have shown that lim sin sh 7 1 mgtOJr Starter Problem List 7 11 13 15 27 31 39 49 75 Recommended Problem List 17 19 23 30 34 35 38 40 55 65 71 76 83 Mathematics Department Laboratory Manual for Mth 251 Differential Calculus Instructions 1 Overview of laboratory activities You work in small groups on each labora tory activity according to your recitation instructor7s directions You typically spend most of the recitation period on the scheduled lab and the remaining time may be used as a question and answer session about homework assignments7 for tests7 or for other class activities You may be required you to prepare a report observing the format speci ed for each lab Your instructor and recitation instructor will provide additional information about the expectations and about the grading policy for the laboratory activities 2 Procedure Discuss the problems included in the laboratory activity and work on them with your group You should print a paper copy of the lab in advance for recording your work done during the recitation hour Complete the assignment and7 if required7 polish up your report outside the class to turn it in to your recitation instructor by the designated deadline 3 Graphs Several of the laboratory activities require you to plot the graph of a function or functions on a given grid Before graphing7 carefully choose and label the variables and their ranges on your plot Unlabeled graphs may not be graded You may need to experiment on your graphing utility to nd ranges that correctly display the requisite information 47 Laboratory I Graphing Background and Goals This laboratory activity is designed to acquaint you with the graphing features of your calculator or computer and to reinforce the lectures on limits Speci c goals for this activity are 21 Use calculators or computers to graph functions in different windows b Use calculators or computers to identify limits from a graph and from numerical data c Compute limits analytically to determine the correct answer This activity is based on the material of Lesson 0 Problem I This problem is designed to investigate the limit linjiL f as x ap 1rgt proaches 1 of the function x 7 1 fa T 273x239 a Graph the function f for m in the interval 0515 Choose an appropriate y range for your graph Remember to label all your graphs b Graph the function f for z in the interval 09711 Choose an appropriate y range for your graph c Based on these two graphs7 estimate the limit lirnL zgt 1 Fill in the table of values for Give your answers correct to six decimal places e Estimate the limit lim f based on the values in the table zgt f Compute the limit by factoring the denominator and simplifying the expression for f when x 7 1 What is the limit Problem II This problem is designed to estimate the slope of the tangent line to the graph of f when x 1 Remember to label your graphs 3 Graph x for x in 0515 Choose an appropriate y range for your graph b Graph x for z in 0911 Choose an appropriate y range for your graph c Graph f for z in 099101 Choose an appropriate y range for your graph d On each of the above graphs7 try to draw a tangent line to the graph at the point 11 Estimate the slope of the tangent lines from your sketches e Fill in the chart with the values of the slope of the secant line through the points 11 and 1 h7 f1 h for small values of h Recall that the slope f1h f1 h is given by dh Give your answers correct to six decimal places f Based on the chart7 what does the slope of the tangent line appear to be g Why did you obtain a wrong answer with a calculator The correct slope is Problem III Describe in a few sentences what you have learned from this labora tory activity Laboratory II Limits Background and Goals This laboratory activity is designed give you more expe rience in working with limits The goals are to be able to use graphing calculators or computers to estimate limits from a graph and from numerical data This activity is based on the material in Lessons 175 Problem I 21 Why is the equation 2 i 6 L 95 r 3 z 7 2 incorrect b Explain7 however7 why the equation 2 i 6 1m lim z 3 1A2 2 w is correct Problem II Suppose that f satis es 710 x 10 for all z in its domain I 01 3 Explain Why lim f 0 zgt0 by applying the properties of limits from the text Justify the value of the above limit graphically using the grid below b Let fz 5 x Use your graphing calculator or computer to nd a small interval 07 6 With the property that for all values of z in this interval7 is Within 0001 of 0 Draw the graph of the for values of z in this interval on the grid 2 Problem III Consider the function f 1 000m for z 2 0 Try to nd the limit le f of f as x a 007 as follows 21 Chart several values of f for large values of z to estimate the limit What do you think the limit is X b Graph the function f on your calculator or computer and sketch the graph on the grid for m in the interval 07100 c Graph the function fz on your calculator or computer and sketch the graph on the grid for z in the interval 07 1000 1 Graph the function f on your calculator or computer and sketch the graph on the grid for x in the interval 07 57000 e Estimate the limit How can you be sure that you would not arrive at a different answer by increasing the graphing interval Laboratory III The Intermediate Value Theorem Background and Goals This laboratory activity is designed to acquaint you with one of the basic theorems in calculus the Intermediate Value Theorem and to help you gain appreciation for its applications in both theoretical and concrete problems The activity is based on the material in Lesson 6 The Intermediate Value Theorem states that given a function f continuous on a closed interval I ab and a number L between the values fa and fb then there is at least one c on I so that fc L In short a continuous function f on a closed interval I ab realizes every value between fa and fb on I or geometrically there will be no gaps in the graph of fx on I no matter how close you zoom in Problem I Let p 67327x1 be a 6th degree polynomial and let I 1 3 a Compute 131 and p3 Is there a point c on I so that 100 0 that is is there a root of the the polynomial p on I b Graph the polynomial p on I Does the graph agree with you conclusion in part a W if0 x 1 Problem II Consider the function f 7 x275x4 1f1 lt 3 a Compute f0 and f3 and note that these have opposite signs Next solve the equation f 0 on the interval I 03 How many solutions did you nd Does your result contradict the IVT Explain b Graph f on I and verify that the plot agrees with your conclusion in part b The IVT is often used to nd the rough location of a solution to an equation that can not be solved analytically This is then used to choose the seed or the starting value for a numerical algorithm such as Newton7s method treated in Laboratory activity IX for nding an approximate solution to high accuracy Note that the IVT under the assumptions of the theorem only guarantees the existence of a solution on a given interval but it does not provide a method for pinpointing its location Problem III Consider the equation sin zx6x 2 In order to apply IVT7 write f sinx x6 x Then nding the solutions of the above equation becomes tantamount to nding the x values for which f 2 Compute f0 and f1 Can you use the IVT to conclude that the equation has at least one root on the interval 01 Explain Plot the function Does the graph support your conclusion in part a Use your graphing utility together with the IVT to identify other intervals on which the equation has solutions How many solutions do you think there are A straightforward algorithm for approximating solutions to complicated equations is furnished by the bisection method7 which is illustrated in the following problem Problem IV In this problem we try to nd an approximate solution to the equation cosz a H v 3 7 sinx 07 accurate within 002 Write f cosz3 7 sinx so that solving the original equation becomes equivalent to nd the zeros of Why does the IVT apply to f on any nite interval Verify that the equation must have a solution on the interval 01 by computing f0 and f1 Recall that z is expressed in radians Next bisect the interval 07 1 into the intervals 07 05 and 0571 and compute f05 Then use IVT to show that our equation must have a solution on 0571 Next bisect this interval into the intervals 057075 and 075717 compute f075 Note that the bisection point 075 must be a solution to our equation or7 otherwise7 the IVT will guarantee the existence of a solution on exactly one of the subintervals Explain This process leads to a general algorithm for approximating solutions to equa tions Give a detailed step by step description of it Finally apply the algorithm to nd a solution to our original equation accurate within 002 Problem V The following problems involve typical applications of the IVT a Let f and gm be continuous functions on I 11 Suppose that fa lt 61 9a and fb gt gb Show that the graphs of fx and g rnust intersect at least once on the interval I Show that the polynomial p 325 7144z4 7163 456x2 2x 7105 must have at least ve zeros Hint compute 1007 101p717 1027 p72 EX actly how many zeros does p then have in light of the fundamental theorem of algebra A cyclist start at 8 am from Corvallis and rides along highway 20 to Newport7 arriving there at 1 pm On the next day he returns along the same way7 starting from Newport at 8 am and arriving in Corvallis at 1 pm ls there a location on highway 20 that the cyclist passed at exactly the same time on both days Laboratory IV Velocity and Tangent Lines Background and Goals This laboratory activity is designed to give you more experience in working With the velocity of an object and with the tangent line as an approximation to a function The activity is based on the material in Lesson 7 Problem I A person gets in a car and drives 100 feet to the house next door The graph below represents the position of the person during the ride Recall that the instantaneous velocity is given by the slope of the tangent line to the graph of position as function of time a How far did the person travel between t 45 and t 60 What is the location of the car at those times b Estimate the velocity of the car at t 2 c At which of the times t 05 15 25 40 55 does the car have the highest velocity Explain 1 At which of the timest 05 15 25 40 50 is the acceleration ie the rate of change of velocity the highest Explain Problem II Consider the function y f 3x2 7 4x 3 The graph of this function near the point 12 is sketched below Check Whether your calculator or computer produces the same graph a On the graph above7 sketch the line that seems to you as the best approximation to the function through the point 17 2 b The exact slope of the tangent line is f 1 6 1 7 4 2 Estimate the slope of the line that you sketched in part a Compare the exact slope of the tangent line with the slope of your line c Any reasonable approximating line to y x near the point P 17 2 must pass through point P Below7 you will try to gather evidence for the fact that the best approximating line of this form is the tangent line Write down formulas for the four lines of slope 107 157 27 and 25 through P Line 1 of slope 1 through P Line 2 of slope 15 through P Line 3 of slope 2 through P Line 4 of slope 25 through P If you use these lines as an approximation to x7 the error is the absolute value of the difference between f and the corresponding y value of a point on the line y Lx Errorlfz 7 Write down formulas for the errors using each of the four lines Error for line 1 yll Error for line 2 121 Error for line 3 ysl Error for line 4 141 Compute the errors for the 4 indicated values of z for the lines 1 2 3 4 by lling in the tables below You should perform all calculations accurate to at least 6 decimal places y value for 1 y value for 2 y value for 3 y value for 4 error for 1 error for 2 error for 3 error for 4 f Which of the four lines seems to result in the smallest errors Laboratory V The Chain Rule Background and Goals This laboratory activity provides hands on practice on the chain rule covered in Lesson 12 Problem I Let 21 Write down an expression for the composition b Compute the derivative of fgx with respect to x using the expression in part a c Next compute the derivative of fgx with respect to x by using the chain rule Did you get the same answer as in part b 1 Compute the equation of the tangent line to fgz at the point 17 1 e Graph both fgz and the tangent line for z Values between 7T and 7139 on the grid below Dont forget to label your graph Problem II Let r 2 7 x 3 Compute the derivative of f by the Chain rule b Graph both x and f on the grids below for s Values between 72 and 2 Choose an appropriate y range for your graphs c Compare the graphs of f and f to see if your computation of the derivative seems reasonable How do you detect intervals on which f is increasing or decreasing d Are there any points at which f is not differentiable 71 Laboratory VI Derivatives in Action Background and Goals This laboratory activity7 which is based on the material in Lessons 8710 and 117157 is designed to give hands on practice on applying the various differentiation rules Problem I The values of functions f and g and their derivatives at z 07 i17i2 are collected in the table below Compute the following derivatives based on the given data a Let h gx 9 cos 2x Find h 0 b Let 29x3x2 fx2 Find j 72 c Let 7rgx2 Find 1471 1 Let lx f3fx 7 Find l 1 e Assume that 9x is also invertible Let 9 1x Find m 3 f Let arctan47rfz2 Find n 2 g Suppose p satis es px6 4p3 fgem 8 Find p 0 h Let q 2 7 6zg and suppose that q 3 10 Find g 3 Problem II The graphs of functions f and g are depicted below Using the information from the plot7 estimate the values of the given derivatives 4 35 0 02 04 06 08 1 12 14 16 18 2 X d agg awmwpi mgmmemw 74 c wm ms 1 gm m x m d f95 am H d QWW z0v4 Problem III The gas mileage of a car going at speed 1 in miles per hour is given by Suppose M55 25 and M 55 703 1 Let UV denote the amount of gas the car uses to travel one mile HOW are Mv and Uv related Compute U55 and U 55 2 Let G1 stand for the amount of gasoline the car consumes when it travels at constant speed 1 for one minute How are Mv and G1 related Compute G55 and G 55 3 What is the practical meaning of the derivatives M 557 U 557 and G 557 Laboratory VII Higher Derivatives Exponential Functions Background and Goals This laboratory activity explores the properties of the rst and second derivatives of a function7 and of exponential functions and their derivatives The activity is mainly based on the material in Lessons 14 and 18 Problem I Pictured below are the graphs of a function f and the rst two derivatives f and f z for values of z between 73 and 3 The graph of g is represented by the red curve7 and the graphs of Mr and by the blue and green curves7 respectively By analyzing where the tangent lines to the various graphs have positive and negative slope7 you should be able to determine which of the three function represents x f or f Circle the true statement on the next page For each statement that you deem false7 provide a reason7 based on the graphs below7 for your conclusion gx kx kw a 996 f967 390 996 N96 C M96 Wt d 7196 WC 6 M96 Wt f M96 WC M76 f W M96 f 96 mm m mm m 996 f 96 mm m Problem II 21 Use a calculator or computer to evaluate the quantity 4h 7 1 h for the values 017 0017 00017 000017 and 700001 of h Perform computations correct to ve decimal places b Use your answers from part a to estimate the limit 1 M71 if h to two decimal places c What does the limit in part b represent geometrically Problem III 21 Use a calculator or computer to estimate the limits 27h i 1 28h i 1 lim 7 and lim 7 hgt0 h we h to two decimal places Use a procedure similar to that employed in problem b What do these limits represent geometrically Laboratory VIII Curve Sketching Background This laboratory activity covers material about concavity in ection points and their applications to curve sketching covered in Lessons 17719 Problem I Consider the function 10zx 7 l4 95gtm39 a Find the z and y intercepts and all the asymptotes of this function b On the grid below sketch the graph by hand using asymptotes and intercepts but not derivatives c Use your sketch as a guide to producing a graph now with the graphing cal culator on the grid below that displays all major features of the curve7 ie7 asymptotes7 intercepts7 maxima7 minima and in ection points 1 Use your graph to estimate the maximum and minimum values of Problem II Consider the polynomial Pz x4 of z where c is a constant a For what values of c does the polynomial Pz have two in ection points One in ection point None b Illustrate what you discovered in part a by rst graphing Pz with c 17 c 07 c 71 and c 72 all in the same Viewing window on your graphing utility Then sketch these graphs on the grid below c Describe how the graph changes as 0 decreases Laboratory IX Logarithmic Functions and New ton s Method Background This laboratory activity covers material about derivatives of log arithmic functions and investigates an algorithm7 so called Newton7s method7 for approximating solutions to equations that can not be solved explicitly Problem I In this problem we analyze linear approximations to the natural loga rithm function lnx Recall that 7 lnx 7 dx x 21 Find the linear approximation to fx ln at the point 17 0 That is7 nd the equation of the tangent line to the graph of fx at that point b Graph both f and the linear approximation on the interval 6 g x g 16 Choose an appropriate y range for your graph c For what values of z is the linear approximation accurate to within 010 Give an answer accurate to two decimal places To determine this7 graph the functions ln7 lnz 0107 and your linear approximation in the same Viewing window Then zoom in to the places where the linear approximation meets the graph of lnz 010 Problem II Consider the equation lnz71 7 64 0 Write f lnx71 767m so that the roots of the equation correspond to the zeros of the function a Can you solve the equation analytically for x Explain b On the grid below7 graph the function f near the point where the graph crosses the x axis Choose an appropriate xirange and yirange for the graph and label the graph Your total xirange should be 1 to 3 units long c Use your graph from part a to estimate to the nearest tenth where the graph crosses the x axis Write 0 for your estimate and compute f0 d In order to improve your estimate for the x intercept of the graph7 nd the equation for the tangent line to the graph of f at 0 n and solve for its z intercept7 carrying out the computations on your calculator correct to at least 8 decimals Write 1 for the intercept of the tangent line this is the rst iteration in Newton7s method Then compute fz1 What do you notice e Next nd the tangent line to the graph of fx at x1fx1 and solve for its x intercept x2 Then compute f2 What do you notice f Continue this iterative process to compute 3 4 5 and 6 Then ll out the table below and analyze the results 2 n 0 1 2 3 4 5 6

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