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# DIFFERENTIAL CALCULUS MTH 251

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This 21 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 Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/224444/mth-251-oregon-state-university in Mathematics (M) at Oregon State University.

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Date Created: 10/19/15

Section 2 5 Continuity Continuity describes gradual change both in everyday life and in mathemate ics Many properties of continuity seem obvious but they are not always easy to prove For now we will rely on our geometric intuition accept the results and see how they are used as we continue our journey through calculus In this section all functions are de ned either on a single interval or a union of intervals An interval may be open closed or halfeopen See Appendix A in Stewart if you need a review of intervals A function f is continuous at 1 if mmqm In order for this limit statement to make sense it is understood that 1 1 must be in the domain of f 2 f must have a twosided limit at Z The related notions of continuous from the right resp left at Z are de ned by the oneesided limit relations 133 as f a and 3133 f as f a respectively Apparently a function f is continuous at 1 if and only if it is both con tinuous from the right and continuous from the left at 1 function is continuous on its domain if it is continuous at each point of its domain with oneesided continuity at endpoints of intervals in its domain Since limx nap p a for any polynomial p and 39 7 a for any rational function 7 and any 1 in its domain we see that All polynomials and all rational functions are continuous Likewise since W A 5 as at A Z for any 1 in the domain of W All neth root functions are continuous A function f is discontinuous at 1 if it is not continuous at Z The simplest type of discontinuity occurs when the oneesided limits exist but are not equal at 1 Then there is a jump or gap in the graph at Z and f is said to have a jump discontinuity at Z Algebraic Properties of Continuous thctions The algebraic limit laws and the de nition of continuity immediately give corresponding useful laws of continuity H The sum of two continuous functions is a continuous function N The difference of two continuous functions is a continuous function 03 A constant times a continuous function is the continuous function g The product of two continuous functions is a continuous function 01 The quotient of two continuous functions is a continuous function on its domain These algebraic laws of continuity apply to nite sums differences prod7 ucts and quotients of continuous functions They also are valid with continuity replaced by continuity at a point and with continuity replaced by one7sided continuity Composite Mnctions Complicated functions are frequently built up from simpler functions by substitution of one function into another For example substitution of the polynomial 1 7 12 into the square root function V yields the more complicated function 1 7 12 If f and g are functions we de ne a new function denoted by f o g and called the the composition of f and g by f 09 x f 9 1 for all at for which the right side makes sense that is for all at such that at is in the domain ofg and g is in the domain of f For example if f and 935 1 7 352 then fogavf9avf17zz What is the domain of f o g in this case In this example both f and g are continuous functions and it should be reasonably clear that their composition f o 9 also is continuous This illustrates a general fact The composition of continuous functions is continuous The Intermediate Value Theore The graph of a continuous function de ned on an interval I has no breaks or gaps A precise mathematical version of this fundamental property of continuity is the Intermediate Value Theorem IVT Let f be a continuous function on an in7 terval I and let a and b be any two points in I Let 7 be any number between f a and f b Then there is a number at between a and I such that f 7 lnformally stated a continuous function does not skip over any values A Substitution Limit Law lim 9 b and f is continuous at 117 xgta then lim f lim xgta xgta f b Section 23 Limit Laws This section is about effective evaluation of limits The ideas is to divide and conquer Once you know a few basic limits many other limits can be evaluated virtually by inspection from the basic limits algebraic limit laws and some other tricks of the trade coming later Algebraic Limit Laws The limit of a sum is the sum of the limits 4 The limit of a difference is the difference of the limits E0 The limit of a constant times a function is the constant times the limit of the function 9quot The limit of a product is the product of the limits He The limit of a quotient is the quotient of the limits provided the limit in the denominator is not zero 9quot These algebraic limit laws apply to nite sums differences products and quo tients of functions each of which has a limit It is easiest to remember these laws from the foregoing verbal statements but when you apply them you usu ally express them in mathematical notation If f and g have limits as x approaches a and c is a constant express the ve algebraic limit laws in limxna notation Check your work by looking at the box on page 120 in Stewart Basic Limit Laws 1 and c are real numbers BL 1 limxna c 0 BL 2 limxna ac a BL 3 limxna xquot aquot for n any positive integer BL 4 limxna if 76whenever 5 makes sense BL 5 If 1111114 f x exits then whenever the right member makes sense BL 3 comes from repeated use of the algebraic product law for limits and BL 2 Check this out for yourself for n 2 BL 4 is a more subtle limit It should seem plausible but the proof takes some care In BL 4 if n is even then the law only makes sense if a gt 07 while if n is odd then the law makes sense for any value of 1 Repeated use of BL 3 and the algebraic limit laws yield 1 limxna 7 aquot for any a 0 limxnap p a for any polynomial p 39 7 a for any rational function 7 and any a in the domain of 7 The Squeeze Law If h S f S g for all so near a but different from a and if limh limg L then there exists liir1fxL The squeeze law is used to establish the existence and value of several tricky limits such as sm ac lim 1 1 0 which is the key limit related to the differential calculus of trigonometric functions Stay tuned Mth 251 Exam This test requires a scantron Bent Petersen 2 5 1f 2 0 0 4 exam tex Dec 2 2004 Time 110 minutes Instructions gt o This test is multiplechoice You must mark your answer on the provided scantron Be If you do not read the in fore you begin ll in all the required information on the scantron StruCtionSgt then how Will 0 Fill in the appropriate bubbles for your information and for your answers on the scant you know what to do ron very carefully Read them now 0 You may use one 85 X 11 inch note sheetprepared in advance You may write on both sides ofyour note sheet Be sure to enter all Note sheets may not be shared Ifyou do not bring a note sheet you will have to do required information on without any help notes the scantron 0 You may not use any books notebooks additional note sheets nor note cards Section number 060 0 You are expected to have a simple scienti c calculator available for use on this test Calculators and other equipment may not be share 0 You may use a simple graphics calculator but not a laptop computer nor any device capable of extensive symbolic manipulation other than your own brain There are XXproblems Problem 1 Let f l 5013 Compute the TaylorMaclaurin polynomial Taylor polynomial centered at the origin or Maclaurin polynomial of degree 2 for f A lw7w2 B lw3w29 C 1w37w29 D 1w372w29 None ofthe above l eWrite letter corresponding to your answer here and mark it on the scantron Problem I Problem 2 Let 9a l 5043 The TaylorMaclaurin polynomial Taylor polynomial centered at the origin or Maclaurin polynomial of degree 2 for 9a is l 2 2 l 7 g cc 5 cc Use this Taylor polynomial to approximate g0l Choose the number closest to your result from the list below A 0968022 B 0968729 C 0968803 D 0968888 0968999 B eWrite letter corresponding to your answer here and mark it on the scantron Problem 2 Problem 3 Find the solution 3 of the ordinary differential equation 613 2 72 i 2 dw 3 w which satis es 3 1 when cc 1 Then nd the value of 3 which corresponds to ac 2 A 7 B 7 C D g None ofthe above l eWrite letter corresponding to your answer here and mark it on the scantron Problem 3 Problem 4 A rectangular box with open top and square base is to be made from 48 square inches of material Find the maximum volume that can be obtained A 31in3 B 32in3 C 29 in3 D 26in3 None of the above l lt7Write letter corresponding to your answer here and mark it on the scantron Problem 4 Problem 5 If f sinw tan then the derivative f is given by A sinw tan2w B sinw 1 sec2w C cos sec2w D sinw 1 secw None of the above l lt7Write letter corresponding to your answer here and mark it on the scantron Problem 5 Problem 6 The tangent line at 112 to the graph of A y54w734 B y45w7310 C y 3w 7 52 D 3 32w 7 1 None ofthe above l lt7Write letter corresponding to your answer here and mark it on the scantron Problem 6 Problem 7 Find the maximum value of w 7 1y 7 3 ifwy 12 and O S x S 4 A 12 B 9 C 6 D 3 None of the above l lt7Write letter corresponding to your answer here and mark it on the scantron Problem 7 Problem 8 Compute the limit w 7 sinw 11m 3 I70 0 A 0 B 13 C 1 12 D unde ned None of the above l lt7Write letter corresponding to your answer here and mark it on the scantron Problem 8 Problem 9 For a certain function f on the interval 737 3 the TaylorMaclaurin polynomial of degree 3 is given by 1 P3w 37w2 1303 and we have lflt4gt I S 12 X 10 5 for each at in the interval 733 Ifwe use P32 l to approximate f2 then we make an error no larger than choose the best estimate A 500gtlt10 7 B 800x10 5 C 120gtlt10 5 D 192x104 E 198x10 2 l eWrite letter corresponding to your answer here and mark it on the scantron Problem 9 Problem 10 If a person 6 feet tall is walking at a rate of 4 ftsec away from a lamp which is 30 feet off the ground at what rate does the person s shadow lengthen A lftsec B 2ftsec C 3 ftsec D 4 ftsec None of the above l eWrite letter corresponding to your answer here and mark it on the scantron Problem 10 Problem 11 Let f be a differential function on 311 satisfying f3 2 and fll 6 Then there must be a point c such that 3 S c S 11 and A f cgt1 3 f cgt12 C fC 14 D fc 18 None ofthe above l eWrite letter corresponding to your answer here and mark it on the scantron Problem 11 Problem 12 If g w S 004 for O S at S 3 then g2225 7 92100 S a where pick the best value A a 0002 B a 0003 C a 0004 D a 0005 None ofthe above l eWrite letter corresponding to your answer here and mark it on the scantron Problem 12 Problem 13 If x 35 7 10902 4 then f has a critical point which is a local minimum Use the second derivative test to locate this critical point A 0 B 2V C l D The test fails None of the above l eWrite letter corresponding to your answer here and mark it on the scantron Problem 13 Problem 14 The equation 56434 avg3 2 d determines y as a function of x such that y 1 when w 1 Compute dig when w l and y l w A 757 B 737 C 717 D 712 None of the above l eWrite letter corresponding to your answer here and mark it on the scantron Problem 14 Use the rest of this page and the backs of all the pages for scratch work Rates of Change in the Natural and Social Sciences Sec 33 The evolution of any process physical biological economic whatever in volves studying rates of change of key components of that process These may be average rates or change or instantaneous rates of change AKA derivatives Several examples follow The basic set up is always the same Let y f be a function and x and x h be two values in its domain Then The change in ac Ax h The corresponding change in the function f is f 90 h 7 f which is often denoted by Ay The change inyAyfxh 7fac The average rate of change of f with respect to so over the interval from so to 90 h is g fh f Ax h 39 The instantaneous rate of change of f with respect to so can be expressed as dy 1 Ag 7 1m 7 dac A140 A90 Example 1 Motion Let s t the position of an object at time t on the s axis Then ds velocity 1 a is the rate of change of position with respect to time d1 acceleration a E is the rate of change of veloc1ty With respect to time ds speed M a Newton s second law of motion one of the bedrock principles of physics states mEF where F is the force that acts on a mass m To begin a serious study of physics requires calculus Galileo s Law H a body moves vertically near the surface of the Earth and the positive s axis is directed upward then the position 8 t of the body at any time t is given by s t 1 2 eigt U0t so where g the gravitational accleleration near the Earth 110 the initial velocity of the object at t 0 so the initial position of the object at t 0 Example 2 Atmospheric Pressure The pressure force per unit area exerted by the atmosphere on object in it varies with the height H p p h is the density of the atmosphere h meters above the surface of the earth then dp dh measure the rate of change of pressure per vertical foot at height h A reasonable model of the atmosphere leads to ELI dh poeiah where p0 101 gtlt105Nm2 and a 0116 km l So at an altitude of h 05 km dp Nm2 7 7110567 dh m This means that at the elevation of 12 kilometer atmospheric pressure drops the minus sign by approximately 11 Newtons per square meter for each meter of elevation gain Example 3 Economics 1n a typical month of operation XYZ Steel has total costs of production C dollars when 90 tons of steel are produced per month and its revenue from sales is R dollars 1ts monthly pro t is P R00 7 C Then dC dac the rate at which cost changes per ton produced at production level 90 For example if dCdx 200 per ton when the monthly production is x 17 000 tons then it would cost about 200 to produce one additional ton of steel ln economics C 7 is called mammal cost 90 R 7 is called mammal revenue ac dP 7 is called mammal pro t dac What is the economic meaning of marginal cost marginal revenue and marginal pro t How Derivatives Affect the Shape of a Graph Sec 43 In this section you learn how to determine certain key properties a function may have by nding its critical points andor by examining the sign of its rst or second derivative Further information is obtained from the critical points of the derivative of the function which are not the same as the critical points of the function Throughout this section I stands for an interval open closed half open that may be either nite or in nite in length is the interval obtained from I by removing any endpoints of I that belong to I For example if I a b then I 11 Determining Intervals Where a Function Increases or Decreases De nition A function f de ned on an interval I is increasing on I if f c lt f d for allclt dinI A function f de ned on an interval I is decreasing on I if f c gt f d for allclt dinI Test for Increasing or Decreasing 1 If f is continuous on I and f gt 0 on I then f is increasing on I 2 If f is continuous on I and f lt 0 on I then f is decreasing on I The foregoing de nition immediately yield the following convenient test to determine local extreme values of a function Test for Local Extreme Values 1 If f is continuous and changes from increasing to decreasing as 90 increases through c then f has a local maximum at c 2 If f is continuous and changes from decreasing to increasing as 90 increases through c then f has a local minimum at c The derivative test now gives a companion test for extreme values First Derivative Test Let c be a critical point of f 1 If f changes from positive to negative as 90 increases through c then f has a local maximum at c 3 If f changes from negative to positive as 90 increases through c then f has a local miniman at c The next two theorems further reveal the importance and usefulness of critical points Theorem CP 1 Let be continuous on an interval I Then f either increases or decrease on each interval between successive critical points of f or between endpoints and adjacent critical points of How can you determine easily in the context of CP 1 whether f increases or decreases on each interval When f is a complicated formula it can be di cult to determine directly from the formula when f gt 0 and when f lt 0 Here again the critical points come to the rescue Theorem CP 2 Let be continuous on an interval I Let 046 be an interval between successive critical points of f or between an endpoint and an adjacent critical point of Then f gt 0 on 046 iff is increasing on 046 and f lt 0 on 046 iff is decreasing on 046 In many practical applications the f in question has only a nite number of critical points If you can nd the critical points Theorems CP 1 and CP 2 enable you to determine directly from the critical points the intervals on which the function increases or decrease and the intervals on which its derivative is positive or negative Determining Intervals of Concavity De nition A continuous function f de ned on an interval I is concave up on I if f is increasing on I A continuous function f de ned on an interval I is concave down on I if f is decreasing on Second Derivative Test for Concavity 1 If f is continuous on I and f gt 0 on i then f is concave up on 2 If f is continuous on I and f lt 0 on i then f is concave down on I The geometric interpretations of concave up and concave down suggest the following useful result The Second Derivative Test Let f be continuous near 0 a If f c 0 and f c gt 07 then f c is a local minimum of b If f c 0 and f c lt 07 then f c is a local maximum of De nition We say that f has an in ection point at c or that the graph of f has an in ection point at cf if f changes its sense of concavity as 90 passes through c In ection Point Test ln ection points of f may only occur at critical points of f that is if c is an in ection point of f then either f c does not exist or f c 0 Section 21 The Tangent and Velocity Problems Well let s start the journey This section gives a rst glimpse of Derivative 2 Rate of Change 2 Slope We will meet rate of change rst in the context of velocity and second in nding the slope of curve The slope of a curve at a point is the same as the slope of the tangent line drawn to the curve at the point In calculus rate of change means instantaneous rate of change The ad jective instantaneous is sometimes used to emphasize the distinction between rate of change in calculus and the more elementary and familiar notion of average rate of change For example if you travel a distance of 21 miles in 20 minutes your average speed which is a rate of change for the trip is 21 mi 13 hr 63 mihr This is useful information but it does not tell the whole story Did your speedometer register exactly 63 mi hr during the entire trip Or did your instantaneous speed vary during the trip We use average rates of change to get at instantaneous rates of change For example instantaneous velocity is the limiting value the limit of average velocities over shorter and shorter time intervals The slope of the tangent line to a curve at a point is the limit of slopes of shorter and shorter sec ant lines connecting the point to nearby points on the curve Limits are at the very heart of calculus It takes time to fully appreciate and understand limits but it is well worth your effort to do so Finally a key goal of this lesson is to begin to understand that the problem of velocity and the problem of tangents are really the same This is the rst step to understanding that Derivative 2 Rate of Change 2 Slope Geometrically an average rate of change is the slope of a secant line and an instantaneous rate of change is the slope of a tangent line What is the derivative anyway Stay tuned Check for homework on the course web site Further Techniques of Differentiation Sec 36 37 38 In class we will cover the following topics once over lightly The ideas are relatively easy and or extensions of what you already know What is needed to master the following topics is for you to practice solving problems with them We will use them whenever they are helpful in Chapter 4 H Implicit Differentiation Sec 36 E0 The Inverse Function Rule It is not in text but it should be 9quot Higher Order Derivatives Sec 37 gt Motion Along a Straight Line Sec 37 and already covered some time ago 9quot Differentiation of Terms Involving Logarithms Sec 38 mostly covered al ready 9 Logarithmic Differentiation Sec 38 Implicit Differentiation The idea here is very simple An equation in the variable x and y or other letters often de nes y implicitly as a differentiable function of 90 When this happens you can differentiate the equation with respect to x and calculate the derivative dydac even though you do not know the explicit formula that gives y as a function of x The only hard part of the method is remembering that y y is a differentiable function of 90 so that when you differentiate the equation with respect to x the chair rule is often needed to calculate derivatives that involve y The Inverse Function Rule If y f is differentiable at x and has inverse function x f 1y then x f 1 y is differentiable at y and div 7 1 dy T dac provided dydx a 0 For example since y e is differentiable with dydac 6quot its inverse function x lny is differentiable and dac 1 1 1 d7 WE 1m 1 dy 24 which recovers a result we learned some time ago expressed with y as the variable instead of with Applying the same rule to the inverse trigonometric function yields the follow ing two results you should know d 1 arcsin x W d t 90 1 7 arc an dac 1 902 For which 90 is each formula valid Higher Order Derivatives Most differentiable function you meet in calculus can be differentiated more than once Naturally enough the derivative of the derivative of a function is called its second derivative If y f is a twice differentiable function then its rst two derivatives are denoted by y f y f or by dy dgy E W The superscripts do NOT mean raised to the second power This is just the way Leibniz denoted a second derivative You can guess how third and higher derivatives are expressed in Leibniz notation 1n the prime notation a switch takes place at n 4 because the writing of primes becomes cumbersome y f 90 y f 90 y f 90 94 f 90 etc where the superscripts mean differentiate n times for n 2 4 Then are now exponents For motion along the s axis with our standing notation we now have position 2 s S 13 ds 1 39t 2 l t i ve oc1 y U 5 dt d1 dgs 1t 2 lt tii acce era 1011 a U 5 dt dt2 Logarithmic Differentiation The idea here is very simple It is sometimes simpler to differentiate a func tion especially one that involves powers and or roots by rst differentiating the logarithm of that function and taking advantage of the properties of logarithms before differentiating Example 1 Diferentiate 9012 x 25 1 902 Solution Express this function by 9012 90 25 1 902 Then 9012 x 25 my 111W 1 Elnac5lnx2 iln 1ac2 why d 1 5 2 lny 7 g 90 2 7 1 202 via which rules But and here is the key to logarithmic di erentmtz39on d y 71 7 h dx ny y wy andhence 1 5 220 y 7777 y 290 x2 1ac2 x12x25 11L 5 2x 1ac2 2x ac2 1ac2 Depending on the situation it may be advantageous to nd a common denomi nator and simplfy Check these steps 7x3 7 6x2 1190 2 y i1225 lt17376211x2 7 1 x24 1x2 2 xx21x2 2 1x22 Differentiation Rules 11 Sec 34 The six trigonometric functions are differentiable To show this it suf ces to establish that the sine and cosine are differentiable Then the quotient rule implies that the other four trigonometric functions are differentiable Then judicious use of a few trigonometric identities yields the following table isinxcossc icosac isinac if 2 dz 2 itansec ac icotsc icsc ac asecscsecxtanx Hcscsc icscxcotsc Notice that the entries in the second column can be obtained from those in the rst column by replacing each function in the rst column by its cofunction and inserting a minus sign in the right member of the equality This observation may help make it easier for you to remember these six basic results You must know all six just like two plus two equals four All the results in the foregoing paragraph stem from two fundamental limit results sinh l5 h and licosh 3337 Either of these limits implies the other

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