Calculus Class Notes for the Combine Calculus an Physics Course Semester I Kelly Black December 14, 2001 Support provie by the National Science Founation - NSF-DUE-9752485 1
Section 0 2 Contents 1 Average Rate of Change........................... 3 2 Sequences................................... 8 3 The Derivaitve................................ 13 4 Derivatives of Polynomials.......................... 17 5 Anti-Derivatives................................ 22 6 Shifte Functions an Their Derivatives.................. 27 7 Slope an Concavity............................. 31 8 Prouct an Quotient Rules......................... 37 9 Composition of Functions an the Chain Rule............... 42 10 Exponential Functions............................ 46 11 Exponential Functions............................ 52 12 Inverse Functions............................... 58 13 Taylor Polynomials.............................. 68 14 The Area Function.............................. 77 15 The Integral.................................. 82 16 The Funamental Theorem of Calculus................... 88 17 Center of Mass................................ 95 18 Work Integrals................................ 102 19 Unconstraine Optimization......................... 110 20 Constraine Optimization.......................... 114 21 The Impulse-Momentum Theorem...................... 120 22 The Unit Circle................................ 124 23 Derivatives of Trigonometric Functions................... 131 24 Working with Vectors............................. 135 25 Inverse Trigonometric Functions an Their Derivatives.......... 140
Section 1 3 1 Average Rate of Change The efinition of a function is given an explore on this ay. The efinition of average velocity is also given, an the iea of the erivative is briefly explore. Almost all of the examples that are explore in the first thir of this semester are functions given in terms of time, t, not x. This is important since almost everything in the physics class is in terms of time. In the first mini-lecture the efinition of a function is given as well as the efinition for the average velocity. In the secon mini-lecture the iea of ecreasing the change in the omain is briefly explore leaing up to the iea of the slope of the tangent line at a point on a function. Finally, in the thir mini-lecture an overview of the average velocity is given. This is a relatively light ay, an the activities are relatively straight-forwar. As the course progresses the activities will get much more ifficult, but in these first couple of ays the activities are not too ifficult so that stuents who are not familiar with this approach are not simply thrown to the wolves. 1.1 Mini-Lecture I The efinition of function is given as well as the efinition for the average velocity. Note the the efinition for the omain an the range is given in the thir mini-lecture. Definition: The wor function is a generic term. A function is a metho or a rule. A function efines the way to provie one particular item given some initial item. Most of the functions that we will eal with in this course will provie a number given another number. For example, the following metho efines a function: Given a number between 0 an 1, multiply the number by 3 an a one to the result. Mathematicians are big into compression an prefer simpler representations. The efiniton above coul also be written as f(t) = 3t + 1, for 0 < t < 1. For example, given the number 1 2, the function returns 5 2. Note
Section 1 4 that we gave this function a name, f(t). This is so that we can refer to it later. Definition: The average velocity is change in istance change in time. Example: It takes 30 mintes to go 5 miles. average velocity? 5 miles = 10miles hours hour 1 2 (Note the units.) What is the If x(t) is the position (in metres) at a given time, t (in hours), then we can also take a graphical view of the average rate of change. The average velocity is the slope of the line through two points. En Class Notes
Section 1 5 1.2 Mini-Lecture II The average velocity is examine again, an the basic iea of the erivative is given. The average velocity is also calle the average rate of change, x t. (The is shorthan for change. ) We can also think of this as the slope of a secant line. (A secant line is a line that goes through two specific points on a curve.) The question that is being aske is what happens to x t as t gets closer an closer to zero. Note that both x an t both get closer an closer to zero, but their ratio oes something else. Note: We have alreay seen two ifferent ways to express a function. The first is with a formula, an the secon is using a graph. There are many other ways. One way that we will
Section 1 6 make use of is a table. t (sec.) x (m.) 0.0 0.0 0.5 2.0 1.0 3.0 1.5 2.0 Given a specific time, 0, 0.5, 1.0, or 1.5, the istance can be foun from the cooresponing value in the right column. Note that when you graph this function you o not connect the ots. The only information that we have is for specific points. If you connect the ots you are implying information that is not given. En Class Notes 1.3 Mini-Lecture III The final mini-lecture provies an overview of function an average rate of change as well as some more formal efinitions. Given the average velocity we can get the total change in isance. Example: The average spee for a 3 hour trip was 45 miles per hour. Total istance = 45mph 3hr. = 135 miles. We nee some efinitions an will refer to some things throughout the year. We assume that x(t) is the position given a time, t:
Section 2 7 Domain = Vali times that we can put into the function. Range = Vali istances that can be returne by the function. Total Change = Difference in a function over a given interval. Average rate = Change in function s range Change in the omain. Tangent Line = The line that just touches a curve at one point. (Locally) En Class Notes
Section 2 8 2 Sequences The efinition of the erivative will be the limit efinition. In orer to motivate limits we will first examine sequences. This ay will begin by efining an working with sequences of numbers an will en with the formal efinition of the limit. The first mini-lecture will focus on sequences an will provie a small amount of motivation as to why we are looking at sequences. The secon mini-lecture begins to examine the limit, an the final mini-lecture relates this work back to the erivative. 2.1 Mini-Lecture I This is a relatively straight-forwar introuction to sequences. We will be looking at the average rate of change. If we are fining the slopes of these lines we are really generating a sequence of numbers:..333,.400,.444,.472,.485,... The question that we will ask toay is what oes it mean to examine a sequence of these slopes? How o we analyze the sequence? A sequence of numbes is an orere list of num- Definition: bers. A more formal efinition can be foun in the book that
Section 2 9 Example: n n 2 1 n=2 = 2 3, 3 8, 4 15, 5 24, 6 35,... These numbers can be graphe: Note that they can be easily graphe in matlab: >> n = 2:1:10; >> f = n./(n.^2-1); >> plot(n,f, * ) En Class Notes 2.2 Mini-Lecture II Examples with the efinition of the limit are given here.
Section 2 10 Given any value of ɛ > 0, if there is an N such that every number in the sequence falls within the interval awe say that the sequence converges. Example: 1 1 2 n n=0 = 0,, 1 2, 3 4, 7 8, 15 16,... This sequence converges to 1. Even though none of the numbers actually is 1, the numbers in the sequence get arbitrarily close to 1. Since none of the numbers in the sequence are 1 we have to show that they get close. We ask if the numbers are close to one 1 1 n 1 2 We can simplify the left han sie ( 1 ( 1 2 ) n ) 1 = 1 ( 1 2 < ɛ. ) n 1 = ( 1 2 ) n = ( 1 ) n 2. Can I get this number to be smaller that ɛ? If so, is it smaller than ɛ for all numbers bigger than some value? Given ɛ can I fin a value of N such that this value is less than ɛ for all values of n > N? 1 n < ɛ 2 ln 1 n < ln(ɛ) 2
Section 2 11 So for any n > ln(ɛ) ln( 1 2) n ln 1 2 n > we have that < ln(ɛ) 1 1 n 1 2 En Class Notes ln(ɛ) ln ( 1 2 ). < ɛ. 2.3 Mini-Lecture III Limits are examine again, only this time it is in the context of the erivative. Suppose that we have a function, f(t) = t 2, an we want to examine what happens to the average rate of change as the change in time gets closer an closer to zero. The average rate of change of f(t) is f(t + t) f(t). t Does this look like 2t as t gets close to zero? Let t = ( 1 ) n 2. From the activity we have that ( ( ) t + 1 n ) 2 t 2 ( 2) 1 n 2t < ɛ, for every n > ln(ɛ). This means that as n gets large, the ln( 1 2) average rate of change gets closer to 2t. Note that the both the numerator an the enominator of the average rate of change goes to zero, but the ratio oes not.
Section 2 12 Given a parabola, what is the slope of the tangent line at any time, t? As t gets close to zero, the slope of the secant lines approach 2t. For example at t = 5 the slope of the tangent line is 10. En Class Notes
Section 3 13 3 The Derivaitve The limit efinition of the erivative is motivate but is not given on this ay. The average rate of change is examine at the beginning of the ay using some numerical examples. During the secon mini-lecture the numerical values are relate to the slope of the tangent line. The principle aim of this approach is to emonstrate that the ieas behin the limit are about the process of fining a limit. During the thir mini-lecture a efinition of the erivative is given an some examples are also given. 3.1 Mini-Lecture I The average rate of change is examine once again. The iea that this is the slope of a secant line is emphasize. The average velocity is the change in the istance ivie by the change in the time. We have the following observations: Tells us how istance changes as time changes. For us it has been iscrete in nature. (This isn t really the case in general.) We woul like to quantify what happens uring one point in time.
Section 3 14 The average rate of change is the change in the range ivie by the change in the omain. This is the same as the slope of the secant line. To get at the rate of change at a point in time, we can look at what happens as the change in the omain gets closer an closer to zero. At this point I like to o the pre-class work on an overhea emphasizing the ifference between total change an average rate of change. En Class Notes 3.2 Mini-Lecture II The average rate of change for f(t) = t 2 is erive, an the iea of the tangent line is introuce. The average rate of change for f(t) = t 2 is (t + h) 2 t 2 h = t2 + 2th + h 2 t 2 h 2th + h2 = h = 2t + h. What happens as h gets closer an closer to zero?
Section 3 15 Graphically, we are fining secant lines. As h gets closer an closer to zero the secant lines get closer an closer to the tangent line. The tangent line at a point, (f(t 0 ), t 0 ), is the straight line that just touches the curve at the point but oes not cross through the curve. The secant lines above get closer an closer to the tangent line, an the average rates of changes get closer an closer to the slope of the tangent line. En Class Notes 3.3 Mini-Lecture III A efinition of the erivative is given an several examples are given. If f(t) = t 2 then f(t + h) f(t) h = 2t + h. The slope of the tangent line at any point is 2t. Some stuents get confuse between the ifference between the tangent line an the erivative. This is the reason that I try to emphasize the ifference between the tangent line an the slopes of the secant lines.
Section 4 16 If g(t) = t 3 then g(t + h) g(t) h = 3t 2 + 3th + h 2, The slope of the tangent line at any point is 3t 2. In general, the slope of the tangent line to the function t n is nt n 1. Definition: The erivative of a function at t is the slope of the tangent line at t. Notation: The erivative of a function, x(t) is enote ẋ(t), x (t), x t (t). it is important to note that this is just notation. For example, the notation x t is not x ivie by t. This is just a way to inicate that we are fining the slope of the erivative. Also, the notation t (g(t)) inicates that you are being aske to fin the erivative of the function g(t). Examples: What is ( t t 4 )? What is t What is ( t t 10 )? What is t En Class Notes ( t 7 )? ( t 15 )?
Section 4 17 4 Derivatives of Polynomials The erivative of a polynomial is foun on this ay. During the first mini-lecture the general notion of the erivative is reinforce. During the secon mini-lecture the iea that the erivative of the sum of two functions is the sum of the erivatives of the two functions is examine. Finally, in the final mini-lecture the general form of the erivative of a polynomial is given. 4.1 Mini-Lecture I The graphical nature of the erivative is re-examine. An example of a function that is not ifferentiable is also given. The average rate of change of f(t) from t to t + h is f(t + h) f(t). h The average rate of change is the change in the range ivie by the change in the omain. This is the same as the slope of the secant line. Moreover, as h gets close to zero, the secant line approaches the tangent line.
Section 4 18 Does a tangent line always exist? Does a unique tangent line exist at t = 0? In the next activity we will look at what happens when you try to fin the erivative of the sum of two functions, f(t) + g(t). We will start out by looking at a specific example. Recall that the averarge rate of change for a function is f(t + h) f(t). h Quick Note: The factorial notation is efine to be n! = n(n 1)(n 2)(n 3) 3 2 1. For example 5! = 5 4 3 2 1 = 120. En Class Notes 4.2 Mini-Lecture II The iea that the erivative of the sum of two functions is the sum of the erivative of both functions. What is the erivative of t 2 + t?
Section 4 19 If I have f(t) + g(t) what is t (f(t) + g(t))? t (f(t) + g(t)) = f (t) + g (t). Note that this implies that t (f 1(t) + f 2 (t) + f 3 (t) + f n (t)) = f 1(t) + f 2(t) + f 3(t) + f n(t). Also note that when looking at the average rate of change for a function multiplie by a constant, cf(t), we get that cf(t + h) cf(t) h = c f(t + h) f(t). h This implies that the erivative of cf(t) is cf (t). Example: t ( 4t 2 ) = 8t. Example: t ( t + 4t 5 ) = 1 + 20t 4. Example: t ( 1 + 8t 9 ) = 72t 8. En Class Notes 4.3 Mini-Lecture III Derivatives of non-trivial polynomials are examine. In the activity we looke at the polynomial ˆQ(t) = a 0 + a 1 t + a 2 t 2 + a 3 t 3 + + a n t n. Note that ˆQ(0) = a 0.
Section 4 20 What happens to the erivatives? ˆQ (t) = a 1 + 2a 2 t + 3a 3 t 2 + 4a 4 t 3 + 5a 5 t 4 + + na n t n 1. So ˆQ (0) = a 1. ˆQ (t) = 2a 2 + 3 2a 3 t + 4 3a 4 t 2 + 5 4a 5 t 3 + + n (n 1)a n t n 2 ˆQ (0) = 2a 2. ˆQ (t) = 3 2a 3 + 4 3 2a 4 t + 5 4 3a 5 t 2 + + n (n 1) (n 2) a n t n 3, ˆQ (0) = 3 2a 2. ˆQ IV (t) = 4 3 2a 4 + 5 4 3 2a 5 t + + n (n 1) (n 2) (n 3) a n t n 4, ˆQ IV (0) = 4 3 2a 4. If I know that ˆQ(0) = 1, ˆQ (0) = 4, ˆQ (0) = 3, ˆQ (0) = 12, ˆQ IV (0) = 6, can I fin a formula for Q(t)? If there is time I like to ask the stuents to graph out an play with the function 1 + t + t2 2 + t3 3! + t4 4! + + t9 9!.
Section 4 21 It is interesting to note that for t close to zero the erivative of this function looks a lot like the original function. En Class Notes
Section 5 22 5 Anti-Derivatives The anti-erivative is introuce on this ay. It is important to introuce the antierivative as soon as possible because the physics class is making use of the kinematic equations. The first mini-lecture introuces the iea of going backwars given that you know the erivative of a function but not the function itself. This iea is aresse again in the secon mini-lecture an the efinition of the anti-erivative is given. An example of the erivation of the kinematics equations is given uring the secon mini-lecture. Finally, an introuction to ifferential equations is given in the thir mini-lecture. it is unusual to iscuss ifferential equations so soon in a calculus class, but this is one of the continuing themes that brings the physics an the calculus class together. We will emphasize the role that Newton s secon law has on eveloping mathematical moels for physical phenomenon. Newton s secon law is use to evelop the ifferential equations that escribe how things move. 5.1 Mini-Lecture I A brief introuction to the anti-erivative is given. The basic rule is not given, rather this is just a brief way to introuce the iea that it is possible to go backwars. The iea is re-examine in the secon mini-lecture so this is just a brief overview of the basic iea. What are the following erivatives? ( t 6 + t 3 t ) =? t ( t 1 2 t 2 ) + 1 =? t ( t 2 ) 3 4t + 7 =? t ( ) t 4 3 t + 323 =? t Note that the aition of a constant oes not change the slope of the graph, it only shifts the graph up or own. Suppose that I know that (f(t)) = 1. t
Section 5 23 What is f(t)? Notice that (t + c) = 1, t where c is any constant. Suppose that I know that (f(t)) = t + 1, t what is f(t)? Notice that 1 t 2 t2 + t + c = t + 1, where c is any constant. This iea is explore in the activity, an we will iscuss how to fin f(t) given its erivative in the secon mini-lecture. En Class Notes 5.2 Mini-Lecture II The anti-erivative is efine an further explore at this time. An example is given in which the kinematic equations for constant acceleration are erive. If you know that t (f(t)) is a given function the process of funing f(t) itself is calle fining the anti-erivative. For example, if then t (f(t)) = t3, f(t) = 4 t 4 + c
Section 5 24 where c is a constant. If then t (f(t)) = t6, f(t) = 7 t 7 + c, where c is a constant. In general if then where c is a constant. t (f(t)) = tn, f(t) = 1 n + 1 tn+1 + c, Example: Assume that an object has a constant ownwar acceleration. Can I fin its velocity an position if its initial velocity is v 0 an its initial position is y 0? From Netwon s secon law we get that ma = mg, a = g.
Section 5 25 Fining the anti-erivative we have that v(t) = gt + c 1, but v(0) = v 0 = g(0) + c 1 c 1 = v 0, v(t) = gt + v 0. Fining the anti-erivative again we have that y(t) = 1 2 gt2 + v 0 t + c 2 but y(0) = y 0 = 1 2 g(0)2 + v 0 (0) + c 2 c 2 = y 0, y(t) = 1 2 gt2 + v 0 t + y 0. These formulas are the kinematic equations for constant acceleration. En Class Notes 5.3 Mini-Lecture III An overview of Newton s secon law is given. The connection between the secon law an ifferential equations is iscusse as well. Newton s Secon Law: The mass times the acceleration of an object is equal to the sum of all of the forces (Note that the acceleration an the forces are all vectors.) This is expresse symbolically as m a = i F i. The thing to notice is that the erivative of the velocity is the acceleration. Example: A car of mass m kg has a force of t 2 t N acting on it. Fin the position of the car if its initial velocity is 2 metres per secon, an its initial position is 0 m. ma x = t 2 t, t v(t) = 1 ( t 2 t ) m
Section 6 26 v(t) = 1 m 1 3 t3 1 2 t2 + c 1 v(0) = 1 m (0 0) + c 1 2 = c 1 v(t) = 1 m 1 3 t3 1 2 t2 The position is foun in the same manner t x(t) = x(t) = 1 m 1 m 1 3 t3 1 2 t2 + 2 1 12 t4 1 6 t3 + 2. + 2t + c 2 x(0) = 1 m (0 0) + 0 + c 2 0 = c 2 x(t) = 1 m 1 12 t4 1 6 t3 + 2t When you have a relation that inclues the erivative of a function the relationship is calle a ifferential equation. The ifferential equation that we starte out with above was ma x = t 2 t, t v(t) = 1 ( t 2 t ). m Newton s secon law relate the erivative of the velocity with the forces acting on the system. Newton s secon law is a ifferential equation. En Class Notes
Section 6 27 6 Shifte Functions an Their Derivatives The erivative of functions that are shifte to the left or right in their omain is explore. The main goal for this ay is to fin the erivatives an anti-erivatives of P (t a) where P (t) is a polynomial. The first mini-lecture is a very simply overview of what happens to a function when there is a shift in the omain. This is mainly a graphical interpretation. The secon mini-lecture covers the erivative of a polynomial that has been shifte while the thir mini-lecture focuses on the anti-erivative of a shifte polynomial. 6.1 Mini-Lecture I A graphical overview of what it means to shift a function is given. This is a very short an very straightforwar iscussion. The stuents are expecte to graph the tangent lines to the shifte graphs in the activity an euce that t f(t a) = f (t a). Given a function, f(t), how o you fin f(t a)? The graph of f(t a) is just like the graph of f(t) only it is shifte to the right a units. En Class Notes 6.2 Mini-Lecture II This is another unusually short mini-lecture. The basic iea is given an several examples are state. Given f(t) the graph of f(t a) is just shifte a units to the right.
Section 6 28 Given f (t) how can we fin t (f(t))? It is just f evaluate at t a. Examples: Fin the following erivatives: ( t ) 1. t 3. ( (t + 4) 75 ). 2. t 3. t ( (t 14) 2 ). En Class Notes 6.3 Mini-Lecture III This is another straight-forwar mini-lecture. The antierivative for general polynomials is given. Note that ( 3 + (t 1) 2 + (t 1) 15 + 6t 4) = 2(t 1) + 5(t 1) 4 + 24t 3. t Example from the activity: v(t) = (t 1) 2 + t, x(0) = 3 At t = 0, x(0) = 3. x(t) = 1 3 (t 1)3 + 1 2 t2 + c. x(0) = 1 3 ( 1)3 + 0 + c, = 1 3 + c.
Section 6 29 To satisfy the bounary conitions 3 = 1 3 + c, c = 3 + 1 3, = 10 3. x(t) = 1 3 (t 1)3 + 1 2 t2 + 10 3. Example from the activity: v(t) = t + 4, x(0) = 1. At t = 0, x(0) = 1. x(t) = (t + 4)3/2 3/2 + c. x(0) = 43/2 3/2 + c, = 8 3/2 + c, = 16 3 + c. To satisfy the bounary conitions 1 = 16 3 + c, c = 1 16 3, = 13 3.
Section 6 30 x(t) = (t + 4)3/2 3/2 13 3. In General: If t (f(t)) = c 0 + c 1 (t a) + c 2 (t a) 2 + c 3 (t a) 3 +... + c n (t a) n, where c n an a are all constants, then f(t) = k + c 0 (t a) + c 1 (t a) 2 (t a) n+1 c n, n + 1 where k is a constant. 2 + c 2 (t a) 3 3 + c 3 (t a) 3 4 +... + Example: What is the antierivative of (t 1) 4 + (t 7) 18 6 t 3 2 4 3? En Class Notes
Section 7 31 7 Slope an Concavity The erivative is examine on this ay as well as concavity. The iscussions in class focus on the meaning that can be attribute to the values of the erivatives. In particular the shape of a curve given knowlege about the erivatives of the function is one of the important aspect of this ay s activities. During the first mini-lecture the iea that a positive erivative means that the function increases is iscusse as well as how the concavity affects the shape of the graph of a function. The secon mini-lecture focuses on how erivative information can be use to etermine the shape of the original function. The thir mini-lecture a broa overview of the ay s activities is given. 7.1 Mini-Lecture I The relationship between the erivative an whether or not a function is increasing or ecreasing is given. These ramificiations are examine in terms of the concavity. What is happening to a function if the erivative is positive? What is happening to a function if the erivative is negative? Note that if f (t) goes from positive to negative the function reaches a local max. If f (t) goes from negative to positive the function reaches a local min.
Section 7 32 If f (t) is positive then the function is increasing. If f (t) is ecreasing then the function is ecreasing. If f (t) is positive then f (t) is increasing so... if the erivative is positive then the function is increasing at a faster rate, if the erivative is negative then the function is ecreasing at a slower rate. These things imply that the graph of the function is concave up. If f (t) is negative then the function is concave own.
Section 7 33 En Class Notes 7.2 Mini-Lecture II The iea that the erivaitve gives you a great eal about the original function is explore uring this mini-lecture. This is mainly a graphical view of what information the erivaitve gives you. Given the erivative you know a lot about the original function. If the erivative is postive the function is increasing. If the erivative is negative the function is ecreasing. If the erivative is increasing the function is concave up. If the erivative is ecreasing the function is concave own. If the erivative changes sign you have a local max or local min.
Section 7 34 Note that if there is a cusp in the original function the erivative is iscontinuous. En Class Notes 7.3 Mini-Lecture III An overview of the ay s activities is given. The main goal is to simply reinforce some of the ieas of what the erivative is an the ifference between the erivative an concavity. There is a ifference between concavity an the erivative.
Section 7 35 If there is time o the following graphical problem. Plot the velocity an then iscuss how to get the acceleration an the position. Provie hints such as o the acceleration first since it can help with the shape of the position.
Section 7 36 En Class Notes
Section 8 37 8 Prouct an Quotient Rules The prouct an quotient rules are introuce an examine on this ay. By the en of the ay the stuents shoul have some experience using both rules an shoul have worke out several examples. The examples inclue graphical an analytic views of the two methos of ifferentiation. During the first mini-lecture the prouct rule is given. The introuction relies on the stuents to finish the pre-class activity. The secon mini-lecture provies an example of the prouct rule, an the quotient rule is erive from the prouct rule. The focus of the final mini-lecture is the quotient rule as well as what happens when the enominator approaches zero. 8.1 Mini-Lecture I The prouct rule is introuce. This mini-lecture relies heavily on the pre-class activity. Suppose we multiply two functions h(t) = f(t)g(t). From the pre-class work we have that h(t 2 ) h(t 1 ) = g(t 2 ) f(t 2) f(t 1 ) + f(t 1 ) g(t 2) g(t 1 ). t 2 t 1 t 2 t 1 t 2 t 1 As t 2 gets closer an closer to t 1 what happens? If f(t) an g(t) are continuous an ifferentiable then t (f(t)g(t)) = f (t)g(t) + f(t)g (t). This is calle the Prouct Rule. Examples: 1. t ((t + 37)(t + 47))
Section 8 38 ( 2. t (4t + 7t 2 )(t + 300) 17) ( t ( + 3 t 2 + 1 )) 3. t 4. t ( (18t) t + 5 ) En Class Notes 8.2 Mini-Lecture II Examples from the prouct rule are given, an the quotient rule is erive. Note that the case when the enominator goes to zero is not examine here. That case will be examine in the thir mini-lecture. Prouct Rule: The har part is recognizing when to use it. For example: ( 1. t (t 2 3t + 1)(t 2 + 5t + 1) ) ( (t + 4)(t 2 + 3t + 2)(t + 1) 500) 2. t Note that the prouct rule implies that we can o the last example out in one step, t (f(t)g(t)h(t)) = f (t)g(t)h(t) + f(t)g (t)h(t) + f(t)g(t)h (t). Example: t h(t) = f(t) t g(t) Yikes! We can t o this. We can rearrange things so that we can use the prouct rule, though, h(t)g(t) = f(t),
Section 8 39 t (h(t)g(t)) = t f(t), h (t)g(t) + h(t)g (t) = f (t), h (t)g(t) = f (t) h(t)g (t), h (t)g(t) = f (t) f(t) g(t) g (t), h (t) = f (t) f(t) g(t) g(t) g (t) h (t) = f (t)g(t) f(t)g (t). g 2 (t), If f(t) an g(t) are continuous an ifferentiable then t f(t) = f (t)g(t) f(t)g (t). g(t) g 2 (t) This is calle the Quotient Rule. Examples: 1. t 2. t ( 3t 2 ) +5t 4+t. ( ) t+1 t 1. En Class Notes
Section 8 40 8.3 Mini-Lecture III Several examples are given that require the quotient rule. The har part about the quotient rule is to know how to recognize it: 1. t 2. t ( t 2 ) 3t+2 t 3 +6. ( ) 3 4 t t. 3. t ( ) 1 1 t = t 1 t t = t ( t 1 t ). Note: What happens when the enominator gets close to zero? ( ) t t t 1 = (t 1) t(1) = 1. (t 1) 2 (t 1) 2 As t gets close to 1, what happens? Moral of the story: be careful! enominator gets close to zero. Things break own when the
Section 9 41 Note: The anti-erivative of (t 1) 2 1 is t 1 this be consistant with the example above? 1 t 1 + c = 1 t 1 + ct c t 1 = ct c + 1. t 1 + c. How can If you let c = 1 you get the result from the example above. Moral of the story: (1) be careful an (2) the constant matters! En Class Notes
Section 9 42 9 Composition of Functions an the Chain Rule The main focus of this ay is the composition of functions. The first activity focuses on compositions, an the secon activity provies an example that emonstrates how composition of functions can be use to simplify a ifficult problem. The first mini-lecture provies a efinition of composition an is relatively straightforwar. The secon mini-lecture focuses on how to recognize which functions are being compose given a composition, an in the final mini-lecture the chain rule is state. 9.1 Mini-Lecture I The efinition for composition of functions is given. Several examples are given. The composition of two functions, f(t) an g(t), is efine to be g f(t) = g(f(t)). Example: g(t) = t + 3, f(t) = t 3 + 4, g f(t) = g(f(t)) = t 3 + 4 + 3 = t 3 + 7. Example: g(t) = t 2 + t + 1, f(t) = t, g f(t) = g(f(t)) = ( t ) 2 + t + 1 = t + t + 1. Note that f g(t) = t 2 + t 2 + 1 = ±t + t 2 + 1. A couple of things to note about this. First, f(g(t)) is not equal to g(f(t)). Also, sometimes their can be a bit of ambiguity about
Section 9 43 what the result is. In the example above, I can put in a + or a - in front of the t an the answer is correct. Given that t 2 = 4 is t = 2 or is t = 2? En Class Notes 9.2 Mini-Lecture II Compositions are examine once again. At this time the iea of ientifying the functions that are being compose is examine. Given that h(t) = t 2 + 3t + 5, can we fin an f(t) an a g(t) so that h(t) = f(g(t))? It is important to know which operation is one last. This will ientify the f(t) an everything else is g(t). You can think about this in terms of your calculator. If you were to program a calculator what woul you have to at the very en? In this example the last thing that is one is to take the square root of some number. This inicates that f(t) = t an g(t) = t 2 + 3t + 5.
Section 9 44 Example: h(t) = ( t + 4t + t 2 ) 3, Outsie Function = t 3, Insie Function = t + 4t + t 2. Example: h(t) = ( 15t 2 4t + t 1 ) 4, Outsie Function = t 4, Insie Function = 15t 2 4t + t 1. Again, you have to be very careful here because the orer matters. En Class Notes 9.3 Mini-Lecture III The chain rule is introuce an several examples are given. Be careful about units! If f(t) has a omain that is measure in secons an a range that is measure in metres, an g(t) has a omain that is measure in meteres an a range that is measure in metres per secon, then the composition g(f(t)) makes sense, but it cannot be one the other way. Note: Given that h(t) = f(g(t)) how can we fin the erivative? h(t+h) h(t) h = f(g(t+h)) f(g(t)) h = f(g(t+h)) f(g(t)) g(t+h) g(t) g(t+h) g(t) h.
Section 9 45 If f(t) an g(t) are continuous an ifferentiable then t (f(g(t))) = f (g(t))g (t). This is calle the Chain Rule. Examples: ( 1. t t2 + 1 ) 2. t 3. t 4. t 5. t 6. t ( (t 2 + 4t + 1 ) 6 ) ( (4t + 1) 18 ) ( t t2 1 ) ( (1 t) 30π ) ( 2t (1 4t) 10 )1 3 En Class Notes
Section 10 46 10 Exponential Functions On this ay exponential functions are introuce an examine. The basic properties are not iscusse in great etail. It is assume that some stuents know these an others have forgotten them. Stuents shoul be able to work on these on their own. The first mini-lecture is very brief. It is a simple introuction to exponentials. The secon mini-lecture examines ifferential equations an slope fiels. The graphs from this exercise shoul look a lot like the plots one in the first activity. The secon activity an the secon mini-lecture can take more time than usual. For this reason the first mini-lecture is kept very short, an it is a goo thing to cut off all of the stuents on this first activity. The last activity is not critical. If most of the stuents can o most of the work on the thir problem the activity can be stoppe. The final mini-lecture just focuses on the solutions approximate from the slope fiels. The similarity between the resulting approximations an exponential functions is given at the very en. 10.1 Mini-Lecture I This is an extremely brief mini-lecture because the secon activity an the secon mini-lecture are longer than usual. This mini-lecture is literally 2 minutes long. This is a very brief introuction to the exponential function. The main thing is that most of the stuents at least start on the thir problem in the first activity. In polynomial function we take a number an raise that number to a fixe egree. In an exponential function we take a fixe number an raise it to a variable power. For example f(t) = 2 t. In this function f(1) = 2, f(2) = 4, f(3) = 9. Note that f(3.5) = 2 3.5 = 2 3 2 1/2. Since we can fin integer powers an fin roots, it is no problem to just fin the exponential for any number. En Class Notes 10.2 Mini-Lecture II Differential equations are introuce an the general iea about slope fiels is introuce. Newton s Law is really a ifferential equation m a = i F.
Section 10 47 What makes this a ifferential equation is that the acceleration is really the erivative of the velocity. If you have an equation that has the erivative of a function in it, the equation is calle a ifferential equation. The basic iea is to fin an unkown function that satisfies the equation. There are times when the acceleration epens on the velocity. For example, if there is air resistance, the force ue to the air resistance is proportional to the velocity square. So if the velocity is ouble the force ue to air friction goes up by a factor of four. The question is how o we fin the function to the ifferential equation if we o not know the original equation? There are two ifferent things that we o: 1. Qualitative Aspect. We can try to get a feel for the behavior of the solution base on the equation. 2. Quantitative Aspect. Use the insight gaine in the first step an our knowlege of calculus to fin the analytic formula for the function satisfying the ifferential equation. Toay we will concentrate on the first aspect. Example: Suppose that we are given that a(t) = v 2 (t).
Section 10 48 Given the velocity at any time we can fin the acceleration. So if v(t 0 ) is equal to 2, then the acceleration is -4. Note that this means that the slope of the velocity is -4 if v(t 0 ) is 2. A tangent line to the curve at this time is y 2 = 4(t t 0 ). Locally, we know the shape of the function, v(t), near this point. Why not raw a potential tangent line at every point in the plane? If you can o that, you have the tangent line at any point which will tell you the shape of the curve at any point. The way
Section 10 49 that you o this is to start at some initial point an then move to the right in the irection inicate by the slope fiel. I like to go through an o a couple of examples on an overhea with the next graph. En Class Notes 10.3 Mini-Lecture III Some of the acitivities are examine. The slope fiel gives you the iea of what shape the solutions might have. If you have an iea of what the functions look like, then you can get an iea of what kin of function to look for in the solution of the original ifferential equation. Example: A ball is roppe, an the force of friction is proportional to one-thir of its velocity.
Section 10 50 From Newton s Secon Law we get ma = mg F f, a = g 1 3m v. If the acceleration is zero what is the velocity? 0 = g 1 3m v, v = 3mg. If the velocity approachs 3mg then the acceleration gets close to zero. This is the steay state for the ball. Sketch ifferent solutions on the following graph on an overhea:
Section 11 51 Note that the solutions increase or ecrease to the steay state. Is this consistant with the equation? a = g 1 3m v. If v > 3mg then the acceleration is negative so the velocity ecreases. Likewise, if v < 3mg then the acceleration is positive so the velocity increases. All solutions get closer to the steay state. En Class Notes
Section 11 52 11 Exponential Functions This ay represents a continuation of the previous ay s activities. The focus on this ay is on exponential functions. Special attention is pai to the range an omain of these functions as well as their inverses. The first mini-lecture focuses on exponential functions an provies several examples. The secon mini-lecture goes back to the ifferential equation that is motivating the use of exponential functions, an the last mini-lecture focuses on erivatives of exponential functions. 11.1 Mini-Lecture I Exponential functions are given an explore by looking at several examples. The main focus is on the ifference between exponential ecay an exponential growth. In terms of ecay, we look at functions that inclue an exponential ecay term an ask about the long-term behavior of the function. This is one because of stuent s confusion when face with functions that have only one ecay term in the later activities. Exponential Functions: f(t) = 2 t Notice that f(1) = 2, f(2) = 4, f(3) = 8, f(4) = 16. The function grows really fast in time. We call this exponential growth. f(t) = 3 t
Section 11 53 Notice that f(1) = 1 3, f(2) = 1 9, f(3) = 1 27, f(4) = 1 81. The function ecays really fast in time. We call this exponential ecay. Properties of exponential functions: 2 4 2 3 = 2 3+4 = 2 7. 2 t 3 t = (2 3) t = 6 t. 5 t+7 = 5 t 5 7. 7 t+c = 7 t 7 c. 6 t 2 t = ( 6 2 ) t = 3 t. Note that with exponential ecay, anything that looks like c t gets closer to zero as t gets large. For example, what happens to the function f(t) = 5 7 t in time? The function gets closer an closer to 5.
Section 11 54 En Class Notes 11.2 Mini-Lecture II The ifferential equation that is use to motivate exponential functions is re-examine. This is a very brief overview that inclues a reminer about how erivatives are foun. The actual erivatives of exponential functions are given in the last mini-lecture. This is just an overview to motivate how we fin the erivatives. Newton s Secon Law: m a = i m t v(t) = i F, or F i. Newton s Secon Law gives us a ifferential equation that provies a mathematical moel of some physical phenomena. The goal is to fin a function, v(t), that satisfies the equation. Example: The force ue to friction on a car rolling across a flat surface is one half its velocity, an this is the only force acting on the object, ma = 1 2 v, a = 1 2m v,
Section 11 55 1 v(t) = t 2m v(t). The goal here is to fin a function, v(t), whose erivative is irectly proportional to the function itself! Can we fin such a function? It turns out that exponential functions hol the key to this question. Before starting the activity we nee a couple of notes for the next activity. The average rate of change for a function from a to b is efine to be f(b) f(a). b a To get the erivative of the function we look at f(t + h) f(t) h an let h get closer an closer to zero. En Class Notes 11.3 Mini-Lecture III The erivatives of exponential functions are examine. Let f(t) = 2 t. Then f(t+h) f(t) h = 2t+h 2 t h = 2t 2 h 2 t h = 2 t [ 2 h ] 1 h. Note that the fraction on the right oes not epen on t in any way! The erivative looks like the original function multiplie by some constant. In general, b t+h b t h = bt b h b t h = b t [ b h ] 1 h.
Section 11 56 The erivative of an exponential is the original exponential multiplie by some constant. Let s see what bh 1 h is for ifferent values of b: 2 h h 1 3 h h h 1 2.7 h h h 1 h 1 1 1 2 1 1.7 1/2.83 1/2 1.46 1/2 1.29 1/4.76 1/4 1.26 1/4 1.13 1/8.724 1/8 1.18 1/8 1.06 1/16.708 1/32.701 1/64.700 1/16 1.14 1/32 1.12 1/64 1.10 1/16 1.03 1/32 1.01 1/64 1.00 There is a number, approximately 2.718 - oh heck let s just call it e, such that t et = e t. Examples: t t e2t ( e t ) t ( te t ) t ( t 2 + 7e πt) (e t2) t
Section 11 57 Note to the lecturer: we o not give the general formula for t bt just yet. It will be examine once the natural log is foun. En Class Notes
Section 12 58 12 Inverse Functions The efinition of a function is revisite as well as the range an omain of a function. The main thrust of the ay s activities is to relate these ieas to the inverse of a function. The examples focus on the inverses of exponential functions. Note that the activities on this ay are shorter than usual an there is more lecture time than normal. The first mini-lecture focuses on the efinition of a function an its inverse. secon mini-lecture focuses on inverse functions with examples of exponentials, an the last mini-lecture focuses on erivatives of inverse functions. 12.1 Mini-Lecture I The efinition of a function an its inverse are given. The range an omain of a function are formally efine. Suppose that f(t) is a function. Then for every number, t, that is acceptable to the function there is another number, y, for which y = f(t). The Example: f(t) is the square of t. (Given a number, t, the function returns t 2 ). Example: f(t) is the number, b, that satisfies b 2 = t.
Section 12 59 For any given positive number, t, there are two numbers that satisfy b 2 = t. This is not a function. We will have to be careful. So far we have playe things fast an loose an shoul have been more precise. We will have to clean up the mess now or else we will be in big trouble. Definition: The set of all possible values of t that can be use in f(t) is calle the omain of f(t). Definition: The set of all possible values that can be returne by f(t) is calle the range of f(t). Example: f(t) = t 2 + 1
Section 12 60 The omain is the set of all real numbers. The range is all numbers equal to or bigger than one. Example: f(t) = 3 t
Section 12 61 The omain is the set of all real numbers. The range is all numbers strictly bigger than zero. Example: f(t) = t
Section 12 62 The omain is the set of all non-negative numbers. The range is all numbers equal to or bigger than zero. Definition: A function, f 1 (t), is the inverse of f(t) if the following are true: f(f 1 (t)) = t, f 1 (f(t)) = t. Example: The inverse of t is t 2. (We will see later that we have to be more careful in terms of the range an omain of these two functions, though.) En Class Notes 12.2 Mini-Lecture II The inverse is explore further. The logarithm is efine uring this mini-lecture. Is the function f(t) = t 2 + 5t + 2 invertible?
Section 12 63 For almost any value of f, there are two ifferent values of t that return the same number. The inverse is not a function! Note: At the bottom of the parabola the erivative of the function is zero: f (t) = 2t + 5 t = 5 2. The omain of the function is all of the real numbers. The range of the function is all numbers greater than or equal to f ( 5 2). If I limit the omain to all t 5 2 then on this new omain I can fin an inverse since I am only looking at the right half of the parabola: y = t 2 + 5t + 2 t = 5+ 25 4(2 y) 2 = f 1 (y). The last equation is the inverse of f when the omain is restricte.
Section 12 64 Example: The exponential function y = e t. It is invertible, but how o we fin the inverse of this monster? Definition: Given that y = e t the function to fin t given a value for y is calle the natural logarithm. We will enote this function as ln(y). From this efinition we have that Note that f(t) = e t ln ( e t) = t, e ln(t) = t. Domain is all real numbers. Range is all positive numbers. f 1 (t) = ln(t) Domain is all positive numbers. Range is all real numbers. Properities of the natural log. ln(a b) =? = y, then e y = e ln(a b) = a b = e ln(a) e ln(b) = e ln(a)+ln(b), ln(a b) = ln(a) + ln(b).
Section 12 65 ln ( a b) =? = y, then e y = e ln (a b ) = a b = ( e ln(a)) b ln ( a b) = b ln(a). = e b ln(a), These two results imply that ln ( ) a b = ln(a) ln(b). Note: t ef(t) =? Do not forget the chain rule! Finally, in the activity it is important to ask if e f(t) = t, then what is f(t)? En Class Notes 12.3 Mini-Lecture III The erivative of the natural log is explore an several examples are given. We have that if e ln(t) = t,
Section 12 66 then t eln(t) = 1, or e ln(t) ln(t) = 1, t t ln(t) = 1, t t ln(t) = 1 t. In general we have that a t t at t at = ( e ln(a)) t = e t ln(a) ln(a) = a t ln(a). = e t ln(a) Examples: t 4t =? t (18.7)t =? t (.25)t =? Why shoul the erivative be negative for that last one? The general formula for logs can also be foun: y(t) = log a t a y(t) = t,
Section 13 67 (a y ln(a)) y(t) = 1, t t y(t) = 1 a y(t) ln(a), = 1 t ln(a). Examples: t ( t3 t ) =? t ( ( t + 1 log10 t 2 )) =? ( 14 t log t 37.1 (t) ) =? One last note: The anti-erivative of e t is e t + c. The anti-erivative of 1 t is ln(t) + c. En Class Notes
Section 13 68 13 Taylor Polynomials The focus for this ay is on the erivation of Taylor polynomials. The ay begins by graphing polynomials whose erivatives match the logarithm at t = 1, an by the en of the ay the general form for the Taylor series is erive. The first mini-lecture only covers the erivatives of polynomials. This is one in such a way as to motivate the use of polynomials in a form that is useful in the later parts of this ay s class. The secon mini-lecture is a review of the first activity, an the general form for the Taylor polynomial is given. In the final mini-lecture the iea of linearizations are examine with an eye towars the use of Newton s metho. 13.1 Mini-Lecture I The erivatives of a particular polynomial are examine. This is use to motivate the general form for the Taylor Series. Define a polynomial to be P (t) = 5 + 6(t 1) + 7 1 2 (t 1)2 + 8 1 (t 1)3 3! +9 1 4! (t 1)4 + 10 1 5! (t 1)5. We will make note of some of the properties of this polynomials. First what is the value of the polynomial at t = 1? P (1) = 5. What about the first erivative? P (t) = 6 + 7(t 1) + 8 1 2! (t 1)2 + 9 1 3! (t 1)3 + 10 1 4! (t 1)4, P (1) = 6. Okay, what about the secon erivative? P (t) = 7 + 8 1 1! (t 1) + 9 1 2! (t 1)2 + 10 1 3! (t 1)3, P (1) = 7.
Section 13 69 The thir erivative? P (t) = +8 + 9 1 1! (t 1)1 + 10 1 2! (t 1)2, P (1) = 8. The fourth erivative? P (t) = 9 + 10 1 1! (t 1)1, P (1) = 8. Finally, the fifth erivative? P V (t) = 10, P V (1) = 10. Wow! Notice how those erivatives match up with the coeficients above? En Class Notes 13.2 Mini-Lecture II An overview of the first activity is given, an the general form for Taylor polynomials is given. I like to put up the following plots:
Section 13 70
Section 13 71
Section 13 72
Section 13 73 Definition: The Taylor polynomial of egree N for a function f(t) aroun the fixe point t 0 is P N (t) = f(t 0 ) + f (t 0 )(t t 0 ) + f (t 0 ) 1 2 (t t 0) 2 +f (t 0 ) 1 3! (t t 0) 3 + f IV (t 0 ) 1 4! (t t 0) 4 +f V (t 0 ) 1 5! (t t 0) 5 + + f (N) (t 0 ) 1 N! (t t 0) N This is a polynomial. The function an its erivatives are evaluate at some constant, t 0, so the coefficients are all constants. If t 0 = 0 then the polynomial is also calle the Maclaren polynomial.
Section 13 74 Example: equation Fin the Taylor polynomial for the ifferential t y = 2y(t), y(1) = 1. y (t) = 2y(t) y (0) = 2, y (t) = 2y (t) = 4y(t) y (0) = 4, y (t) = 4y (t) = 8y(t) y (0) = 8, y (4) (t) = 8y (t). = 16y(t) y (4) (0) = 16, y (n) (t) = 2 N 1 y (t) = 2 N y(t) y (n) (0) = 1, P 6 (t) = 1 + t + 1 2 t2 + 1 3! t3 + 1 4! t4 + 1 5! t5 + 1 6! t6 Finally, note that if N = 1 you get a line. This line is calle the linearization of the function at a point. En Class Notes 13.3 Mini-Lecture III The secon activity is iscusse with an emphasis on linearizations an an introuction to Newton s metho. I like to have everyboy running Matlab or other software package at this time an step through the calculations as I o them on the boar. The linearization of of f(t) at a point t 0 is the first Taylor polynomial at the given point, P 1 (t) = f(t 0 ) + f (t 0 )(t t 0 ). This is the point-slope form for a line!
Section 13 75 We can use this to fin the points at which a function crosses the t-axis. For example, suppose I want to fin a value of t that satisfies t 3/2 7t 2 = 0? I cannot solve this analytically, so I have to play a game: t 3/2 = 7t + 2, t 3 = (7t + 2) 2, let g(t) = t 3 (7t + 2) 2. I can fin the linearization by fining the erivative of g g (t) = 3t 14(7t + 2) 3. The linearization about a point can now be foun. We will try to fin the point, t, which satisfies the original equation. First graph g(t) an make an intial estimate. Here we will let t = 48 an procee: >> t = 48; >> g = t.^3 - (7*t+2).^2 g = -3652 >> gp = 3*t.^2-14*(7*t+2) gp =
Section 13 76 2180 >> t = t - g/gp t = 49.6752 Cycle through this until t gets close to the root which shoul only take two more iterations. This process is calle the Newton-Raphson Metho. En Class Notes
Section 14 77 14 The Area Function The main iea for this iea is the area function. The integral is motivate in terms of calculating change in istance given the velocity. There is not an explicit connection mae to the anti-erivative, though. That will happen in the following class. The first mini-lecture is very short an introuces the way to fin the change in istance given the average velocity. This iea is expane upon in the secon minilecture in which the Riemann sum is introuce. Finally, the integral is efine in the last mini-lecture. 14.1 Mini-Lecture I The total change in istance is foun by using the average velocity. The basic question that is examine is how o we fin the change in istance given the velocity? The average velocity is efine to be Average Velocity = Change in istance Change in time. Given the average velocity we can fin the change in istance by multiplying by the change in time. Example: The times an velocities for a trip are given below: t (sec.) v (m/sec.) 0 1 1 3 2 4 3 5 We o not have the average velocities here, but we can approximate them by pretening that the velocities are constant over each time interval.
Section 14 78 Estimate of the change in istance: Low Estimate = 1 1 + 3 1 + 4 1 = 8m. High Estimate = 3 1 + 4 1 + 5 1 = 12m. En Class Notes 14.2 Mini-Lecture II The connection between the change in istance an the area uner the velocity graph is mae explicit. The basic iea behin Riemann sums is also iscusse. Notice! If the velocity is a piecewise constant function, then the change in istance is equal to the area uner the graph. Is this true in general? We will ask a ifferent question. Given a velocity plot, how can we approximate the change in istance? We will break up the omain of the function, the time, into small pieces an try to make an approximation over each small piece. To o this we will assume that the velocity is about constant over each little interval.
Section 14 79 Example: If the velocity of a car is v(t) = te t from t = 1 to t = 3, approximate the istance travele using 15 intervals. The with of each time interval is 3 1 15 = 2 15 : t 0 = 1 t 1 = 1 + 2 15 t 2 = 1 + 4 15 t 3 = 1 + 6 15 t n = 1 + n 2 15. The change in istance can be foun using two ifferent approximations. The first assumes that the velocity over each interval is the same as the left sie of each panel, x 14 n=0 (t n e t n ) 2 15 36.4m. (This is calle the left han Riemann sum.) The other approximation assumes that the velocity over each interval is the same as the right sie of each panel, x 15 n=1 (t n e t n ) 2 15 44.1m.