### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# CALCULUS Math 2

UCI

GPA 3.73

### View Full Document

## 9

## 0

## Popular in Course

## Popular in Mathematics (M)

This 55 page Class Notes was uploaded by Adam Crona on Saturday September 12, 2015. The Class Notes belongs to Math 2 at University of California - Irvine taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/201866/math-2-university-of-california-irvine in Mathematics (M) at University of California - Irvine.

## Reviews for CALCULUS

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/12/15

Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 1 Notes Introduction In your previous calculus course vou learned about di erential calculus which is the study of the rate of change of one quantity with respect to another We briefly review the main ideas of differential calculus before introducing the closely related branch of mathematics known integral calculus which is the subject of this course Suppose that an object movs in a straight line during some interval in time How can we compute the velocity of the object at any particular point in time within this interval If we know that the object s velocity does not change over time then we can use the simple formula I change in position final position initial position 1 ve ocitv s quot change in time final time initial time Since the velocity is known to be constant during the given time interval we can conclude that this formula yields the velocity of the object at any point in time between the initial time and the final time What if the object is not necessarily moving at a constant velocity In this case equation 1 is only useful for computing the average velocity of the object between the initial time and the final time To compute the velocity at a particular instant to we must be more resourceful We can approximate this quantity which is called the instantaneous velocity by using equation 1 over some small interval in time that contains to such the interval 10 to Ii where Ii is some small positive number For concreteness we let st be the function that describes the relationship between the position of the object and time Specifically for each real number 1 between the initial time and the final time st is the position of the object at time t on the straight line along which it is moving Then the velocity of the object at time to can be approximated by sto Ii sto sto Ii sto to ll to ll 2 From equation 1 we can see that this quantity is actually the average velocity of the object on the interval 10 to Ii but since Ii is small it is reasonable to assume that the object s velocitv cannot much during such a short period of time and we can conclude that the instantaneous velocity at time to can be approximated by this average velocity The smaller the value of Ii the less the object s velocity can over the interval 10 to Ii Therefore it is reasonable to conclude that we choose smaller and smaller values of Ii the average velocity over the interval 10 to It becomes a better approximation of the instantaneous velocity of the object at time to We can therefore de ne the instantaneous velocity at time to to be the limit of the average velocity the width of the interval 10 to It which is It approaches zero Specifically if we denote this velocity by 1310 then 510 Ii 510 W0 1ng h 5 Informallv we are defining the instantaneous velocity at time to to be the average velocity over an infinitely smallquot interval in time containing the instant to In general if two quantities J and y are related by the equation 9 fJ where f is a given function then the instantaneous rate of change of y with respect to J when J Jo is given by the derivative of fJ at J J0 The derivative of fJ at J J0 which we denote by f J0 is defined by f I f 0 l f0 f at 3335 4 provided this limit exists The derivative of a function f J is itself a function denoted by f J that is defined at every point J0 at which the limit in equation 4 exists and whose value at J0 is the quantity f J0 defined in equation 4 This notion of the derivative a function allows us to compute derivatives of many functions using di erentiation rules which is much more ef cient than using the definition of the derivative directly Example 1 Suppose that a ball is thrown straight up with an initial velocity of 100 ft and that the initial height of the ball is 6 ft Then the height in feet of the ball 1 seconds after it has been thrown can be described by the function 51 where 51 1612 1001 6 12 0 39 L41 V Our goal is to compute the velocity of the ball 5 seconds after it has been thrown After 5 seconds the height of the ball is 55 2 106 ft Since the height of the ball has changed by 106 6 100 ft in 5 seconds it is tempting to use the simple formula for velocity in equation 1 to conclude that the velocity of the ball during this time is 1005 2 20 ft s but since the initial velocity of the ball is 100 fts we see that the velocity is clearly not constant so this formula cannot be used directly to compute the velocity at any particular point in time Instead we can approximate the velocity of the ball after 5 seconds by using the formula in equation 1 to compute the average velocity between t 5 and t 6 which is 56 55 30 106 6 1 Howe 39er we can obtain a more accurate approximation by computing the average velocity between t 5 and t which is 76 fts 6 A 1 68 fts l V 72 106 We can use this approach to obtain the exact velocity at t 5 using a little abstraction If we approximate this velocity by computing the average velocitv between t 5 and t 5 Ii where It is assumed to be some small positive number then we have 165 102 1005 ll 6 106 It 1625 10h 12 500 100 6 106 It 400 1601i 16h2 500 1001i 6 106 It 16i2 60h Ii 16l 60 8 Taking the limit Ii approaches zero we find that the velocity of the ball after 5 seconds is 60 ft In this case however the function st that describes the height of the ball is a function whose derivative can be t using Wquot rules In particular we can use the Power Rule the Sum Rule and the Constant Multiple Rule to obtain 131 s t 32t 100 9 Since velocity is the rate of change of position and st describes the position or height of the ball a function of time its derivative 131 s t describes the velocity of the ball a function of time That is 131 for t 2 0 repre ents the velocity of the ball 1 seconds after it has been thrown Substituting 1 5 into 131 yields 135 60 so we conclude before that the velocity of the ball after 5 seconds is 60 ft s E Based on this example we can summarize the advantage that the derivative provides us with regard to the use of quotient formulas such the formula in equation 1 The derivative being the limit of a quotient of di erences allows quotient formulas to be applied to more general problems For example the cost to produce a single unit of a product known marginal cost can be obtained by the simple quotient formula cost to produce all units 10 number of units produced marginal cost but this formula is valid only if the cost to produce each unit is the same This is rarely the case due to the distinction between xed costs such the cost of buying manufacturing equipment that are independent of the number of units and variable 60 such workersquot wages or costs to maintain equipment which increase the number of units increases Therefore in order to compute the marginal cost it is important to recognize that marginal cost is the rate of change of total production cost with respect to the number of units produced Therefore gi a function C 1 that describes the total production cost in terms of the number of units which is denoted by 1 the marginal cost after producing J units is given by the derivative C 1 This derivative can be interpreted the cost of producing one additional unit given that J units have already been produced It is natural to ask whether other types of formulas can be generalized to more dif cult problems using limits For example consider the formula for the area A of a rectangle of width 11 and height ll A wit 11 How can we use this formula to compute the area of a shape that is not a rectangle We can proceed by approximating the given shape by a number of rectangles Then we can approximate the area of the shape by using equation 11 to compute the area of each rectangle and then adding all of these areas together The accuracy of this approach depends on how well we have approximated the shape using rectangles If we use more smaller rectangles instead of fewer large ones then the approximate area obtained by adding the areas of these rectangles is likely to be more accurate Therefore we can define the area of the shape to be the limit of this approximate area the number of rectangles becomes infinite with each rectangle becoming infinitely small Example 2 Figure 1 shows how rectangles can be used to approximate a given shape so that the area of the shape can easily be approximated by computing the areas of all of the rectangles using equation 11 Using four rectangles we obtain an approximate area of 57 square units whereas with eight rectangles we obtain an approximate area of 5187 square units Later in this course we will learn how to compute the exact area of this shape which is 46 square units Note that the area obtained using eight rectangles is much more accurate than that obtained using only four rectangles E As another example suppose we rearrange the formula for velocity described in equation 1 to obtain the formula distance traveled velocity X elapsed time 12 As with equation 1 this formula is valid when the velocity is constant but what if it is not Again we need to be more resourceful We can approximate the distance traveled by dividing the interval of time into small subintervals and then using equation 12 to approximate the distance traveled during each of these subintervals Then we can add all of these distances together to approximate the distance traveled during the entire interval of time This approach gtlds a reasonable approximation to the distance traveled because previously mentioned the velocity cannot much during very short periods of time and therefore equation 12 yields an accurate approximation of the distance traveled during each subinterval As we divide the original time interval into more and more subintervals our approximation of the distance becomes more and more accurate Therefore 39e can define the exact distance traveled to be the limit of this approximate distance the number of subintervals becomes infinite provided that each subinterval becom infinitely small In both cas s we Wished to compute some quantity 2 using a product formula of the form 2 Jg In the first example 2 is area J is width and g is height in the second 2 is distance J is time and g is velocity However in both cases the formula 2 Jg is only valid if J and g are constant values that are independent of one another If this is not the case we can still use this formula on smaller instances of the same problem and add the results together The process of approximating the quantity z in this manner and taking the limit the number of smaller problems becomes infinite is called integration and the exact value of z is known a de nite integral The study of definite integrals and their computation is called integral calculus We now relate the definite integral to the derivative by the following statement that summarizes the usefulness of the definite integral in the same way that the usefulness of the derivative described earlier The de nite integral being the limit of a sum of products allows product formulas to be applied to more general problems In this course we will use the definite integral to expand the usefulness of a number of product formulas including not only the formulas for area and distance that we have discussed but also formulas for computing quantities such the volume of a solid object or the work that is required to move an object a given distance As we will see the concept of the definite integral closely parallels the concept of the derivative Both concepts are defined using a limit which is the essential ingredient that allows us to obtain an exact value for the quantity we wish to compute instead of a mere approximation Furthermore both derivatives and definite integrals can be viewed functions which leads to the development of rules that can be used to compute these exact values with much greater ef ciency than by computing the appropriate limit directly The significance of this one advantage cannot be overstated for it has served one of the most fundamental catalysts for scientific advancement over the past few centuries This benefit is a direct result of the f and of the who t and integral calculus in order to solve the problems of their day ef ciently and accurately possible We would do well to emulate their virtues in our efforts to solve the problems of today v x Jim Lambers Math 2A Winter Quarter 200304 Lecture 6 Examples These examples correspond to Section 23 and 25 in the text Example Compute a solution of the equation f cc 0 where f cc cc2 cc 1 Solution The graph of this function is shown in Figure 1 To nd a solution we can first look for two points a and I such that f a and f b have opposite signs For example f 1 1 and f 2 4 2 1 1 Since f is continuous everywhere we can apply the Intermediate Value Theorem and conclude that f cc 0 for some cc in the interval 1 2 We can apply the Intermediate Value Theorem repeatedly to narrow down our search for a solution using a procedure called bisection We first compute the midpoint of the interval 1 2 which is 122 32 Since f2 gt 0 and f32 94 32 1 14 lt 0 we can conclude that f cc 0 for some cc in the interval 32 2 Repeating this process on the interval 32 2 we obtain the midpoint m 3 22 2 74 We then evaluate f at this midpoint to obtain f74 4916 74 1 516 Since f32 lt 0 and f 74 gt 0 we can apply the Intermediate Value Theorem a third time to conclude that f cc 0 for some cc in the smaller interval 32 74 We can continue this process for as long as we wish until we are searching for a solution within an interval that is so small that we could choose any point in the interval and have an appml39imate solution that is close enough to the exact solution For example the interval 3 2 74 has a width of 14 If we halved this interval eight more times using bisection then the resulting interval would have a width of 1 1024 If we only wanted a solution that is correct to two decimal places then we could choose any point in this small interval and consider it our approximate solution In this example the solution that would eventually be obtained using bisection correct to four decimal places is approximately 16180 It should be noted that this particular equation cc2 cc 1 0 can easily be solved using the quadratic formula However there are countless equations of the form fcc 0 for which there is no simple formula for obtaining the solution For example there is no formula for the solution of f cc 0 where f is a polynomial of fourth degree or higher With such a function a procedure like bisection must be used to obtain an approximate solution B Example Compute lim cc cos1cc xao Solution For this function we cannot use limit laws to conclude 2 2 v milc DIEM meow30gt Bisection using the Intermediate Value Theorem 2 1 1 1 Figure 1 Graph of fx x2 quotc 1 Note that f15 is negative while f175 is positive Therefore it follows from the fact that f is continuous that fx 0 for some 30 between 15 and 175 because cos1cc does not have a limit as cc approaches 0 Like sin1cc it oscillates more and more rapidly between 1 and 1 as cc approaches 0 eventually oscillating in nitely rapidly Limit laws such as the law that the limit of a product is equal to the product of the limits only apply if all of the limits involved actually exist Therefore we rely on the Squeeze Theorem Because 1 g cos1cc g 1 for all cc 0 we can multiply through by cc2 and obtain cc2 g cc2 cos1cc g ccg since cc2 gt 0 for all cc 0 Now since cc2 0 and c2 0 we can apply the Squeeze Theorem and conclude that lim cc cos1cc 0 xao This has to be true because cc2 cos1cc is squeezed between cc2 and c2 two functions that bound cc2 cos1cc above and below near cc 0 and have the same limit as cc approaches 0 Those are the two conditions that must be satis ed in order to apply the Squeeze Theorem D Example Given that 3 3cc25cc 3 cc3 3cc23cc x gcc2 2cc2 cc 1 for cc near 1 compute cc3 3cc25cc 3 11m xal I 1 Solution For convenience let cc3 3cc25cc 3 2 30 1 hcc cc 2x2 cc3 3cc2 3cc Then we are given that flx S We S We We could compute limx l gcc directly but this is not easy We cannot use the Direct Substitution Property because g is not de ned at 1 We could try to manipulate g algebraically and arrive at the limit but given the above bounds on 950 we can instead apply the Squeeze Theorem since we can use the Direct Substitution Property to obtain lim m lim 03 3x2 3x 13 31 2 311 3 3 1 mal mal and linlMx linle 2 c2 1 212 1 22 1 mal xgt1 Since 950 is bounded by two functions that have the same limit as quotc approaches 1 we can apply the Squeeze Theorem to conclude that lim 1 ma The graphs of all three functions are shown in Figure 2 D Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 21 Notes These notes correspond to Section 83 in the text Trigonometric Substitution Recall the substitution rule fywy xgt dx m la 1 In applying this rule one makes the substitution u gJ which yields du g J1J Using this relation the original integrand fgJg J is then rewritten f ii In rewriting the integrand in this fashion one can write the substitution in such a way that J is expressed a function of 1 instead of the other way around J gquot 1 Then we have dJ dug fl u This is not alw the easiest approach to applying the substitution rule but can be very effective when the substitution to be employed is more complicated or if the function gquot is simpler to work with than g This approach to substitution is known inverse substitution but it should be recognized that it is nothing more than an alternative approach to u substitution This is precisely the case when evaluating an integral of the form f Va2 J3921739 2 Suppose that we use the substitution J a sin 9 Clearly it is preferable to write the substitution in this manner than expressing u a function of From the relation dJ acosBdB we obtain the new integral a a sin2 90 cos 9 d8 12 cos2 9 119 3 which can easily be evaluated using a half angle formula For more complicated integrands involving the expression Va J2 the same substitution can sometimes be used Example 1 Evaluate 2 J4 J2 dJ 4 l 0 Solution One approach is to use the substitution u 4 J2 which implies du 2J1J The new limits can be obtained by substituting the limits J 0 and J 2 into the relation u 4 J2 which yields u 4 and u 0 We then have 2 J4 J2d1 0 Alternatively we can use an inverse substitution J then have 1 0 v 7 ulZdu 2 4 1 4 7 ulZdu 2 0 1 32 gt E 32 0 12 3 7 x2 23quot 4 1 1 432 4123 5 2sinf9 which implies 11 Ecos d We 0310003 Val J2 4 2sin92 4 4sinz9 2 1 sinZ92Vcos28 2cos8 6 It follows that 2 J4 J2d7 0 WTZ 2 sin92 cos82 cos8d9 l 0 WTZ 8 sinBcosZBdB l 0 0 8 uz du u cos 9 la sianB i 1 1 8 uzdu i0 31 U5 8 3 0 Similar substitutions can be used for integrands containing expressions of the form a J2 or J392 12 In the first case the substitution J a tan 9 can be used in conjunction with the identity 1 tan2 9 sec2 9 In the second case the substitution J a sec 9 can be used in conjunction with the identity sec2 9 1 tan2 9 Example 2 Evaluate dJ Z 8 Solution We use the inverse substitution J 2 tan 9 which implies dJ 2 sec2 9 19 Furthermore J392 2tan92 4tai1294 2 tanz1 2Vsec29 2sec9 Therefore dJ since C represents an arbitrary constant C 9 2 sec2 9 19 2 sec 9 sec919 ln sec9 tan 9 C V2 4 ln J gC 2 1 ln 724JC 1 lnJ24Jln C lnJ24J ln2C lnJ24JC 10 Often these substitutions must be used in conjunction with other integration techniques We illustrate this with a simple example Suppose we wish to evaluate From the discussion above we should use the substitution J a tan 9 This yields 11gt a 1 dJ a sec39d9 12 However in order to evaluate the new integral we must resort to integration by parts which yields sec39d9 sec9tan9 sec9tanz9d9 13 Using the identity tan2 9 sec2 9 1 we obtain sec3 9 19 sec9tan9 sec3 919 sec919 14 Rearranging algebraically and using the integration rule for sec 9 we finally obtain a x2 11 12 sec3 9 19 9 9 2 2 0 sec9tan9 ln sec9 tan 9 C 2 J2a2ln x2a2xc 15 The key lesson to be learned from this example is that one should be prepared to use any integration technique at his or her disposal since for many integrands some combination of such techniques is necessary to evaluate the given integral Unlike the more straightforward process of differentiation anti differentiation often requires substantial ingenuity and open mindedness Example 3 Evaluate f t i w um Solution Before we can use a trigonometric substitution we need to write the expression 1 J J2 a sum or difference of two squares Completing the square we have 1 1 A 5 2 7 7 27 x x 1 x 2 1 4 Jfac12 4 1 and therefore 1 2 C3f m QQs m Since we now have a difference of two squares under the radical sign and since the first term is a constant we try to write the radical in the form a a sin2 9 for some choice of a and some substitution involving a sin 9 Matching 12 12 sinZ9 to 54 J 122 we find that we must have a 5 E and also J E sin 19 Using this inverse substitution we have 11 goos d 2 i sin 9 2 1 J J2IJ cos d9cos29d9 2 and rm L H1 4 2 39 It follows that A 2 Using the half angle formula 1 cos 29 2 loos29d9 lwalg lt9 gtc cos2 9 yields and therefore 7 7 1 J J2IJ 9 16sinE9C From our original substitution we obtain and using the doubleangle formula sin 29 2 2 sin 9 cos 9 along with the relations and 4sinz9 v1 sinz goos 20 23 24 26 yields sin 29 2 2 sin 9 cos 9 w M 2 lter 2 4 2 8 1 J i 1 J J2 43 Now we can assemble our final answer 1 J J2IJ asin l 2 1 J J2C Cl Example 4 Evaluate J2 J2 a dJ where a is a positive constant Solution We use the inverse substitution J a sec 9 which implies dJ a sec 9 tan 9 19 and J2 a a sec 92 a a2 sec2 9 a LS C2 9 1 W tan2 9 atan9 Therefore J2 J2 a dJ a sec 92atan 9a sec9tan 9 19 a4 sec3 9tan2 9d9 This integral worked out in the previous lecture Using that result we obtain 9 2 2 2 4 1 3 1 1 V J J a dJ a asec9tan 9 gsec9tan9 glnsec9 tan9 C Using the relations J a sec 9 30 31 Jim Lambers Math 2A Winter Quarter 200304 Lecture 4 Notes These notes correspond to Section 23 in the text Calculating Limits of Functions Now that we have precisely de ned what the limit of a function is we turn our attention to actually computing limits of certain functions While we can always resort to the de nition to compute the limit of a given function x as x approaches a given value a this is not the most practical course of action Instead it is best to use the de nition to establish some laws that show how limits of more complicated functions can be obtained in terms of limits of simpler functions Then after learning how to compute limits for some very simple functions using the de nition we can use the laws to handle more complicated functions We begin by learning how to compute limits of the simplest functions of all the constant function x c where c is any constant and the identity function x x For the constant function x c the limit of x as x approaches a must be equal to c Since x c for all x it is impossible for f to approach any other value In summary lim c c xaa Example Consider the constant function x 4 and suppose that we wish to compute its limit as x approaches 3 For any open interval 1 containing 4 we can easily nd an interval 2 containing 3 such that x is in 1 for x in 2 because no matter what interval 2 we choose x 4 on 2 and therefore x is in 1 It follows from the de nition of a limit that x must approach 4 D As for x x as x approaches a x must also approach a since x x That is lim x a xaa Proof Let 1 be any interval containing a According to the de nition limmnax a if there is some interval 2 containing a such that x is in 1 whenever x is in 2 We can simply choose 1 and 2 to be the same interval since they are both required to contain a D We now state some of the limit laws These laws can be proven using the de nition of a limit but we do not concern ourselves with that in this course These laws show how the problem of computing the limit of a function with a complicated formula can be reduced to the problem of computing limits of simpler functions We have mm gm giglt1 3319 hm W 7 9W 733311102 731m 17m hm cfzl egg mo 17 hmhltxgtgltzgti 3331 Hz gigglyx m7 m 7 limwagm 1 mg 7 hmwagx7 gg m 3quot 0 Example From the previous discussion7 we have limm2 lim33 z72 z72 Using the above limit laws7 we can compute the following limits limz3 limx lim32357 z72 z72 z72 limm23x limmm3 limx limz325107 z72 z72 z72 z72 limm23m73 limz23m7 hm310737 z72 z72 z72 hm 42 1235 hm 4z2 335 4 hm m2 3m 410 40 z72 z72 z72 and limmmg z 2 1 1 7 335 hmmmzm3 5 D Example In this example7 we use a combination of several of the limit laws m2 7 3m limwmg m2 7 3m lim z72 z 3 limmmg z 3 limmmg m2 7 3m 5 1 7 hm m2 7 hm 3m 5 17 72 1 7limmz731imz 5 172 172 1 1 i 3 lmx 131235 3lt2gtl 2 A direct consequence of the limit law for products is the limit law for exponentiation7 1 n 1 n we 7 1311 ml 7 where n is a positive integer Applying this law to the function x x7 we obtain lim x a xaa This result7 in combination with the preceding limit laws7 can be used to prove the following result concerning limits of polynomial functions or rational functions recall that every polynomial function is also a rational function Theorem Direct Substitution Property Rational Functions If f is a rational function and is de ned at x a then gym M In other words7 we can compute the limit of a rational function x as x approaches a by simply substituting x a into the formula for Example Let x x3 3x 1 Using the limit laws7 we have lim mg lim x3 3x 1 x46 x46 3 limx 3limxlim1 x46 x46 x46 33 33 1 34 Alternatively7 we can reach the same conclusion by computing 3 33 33 1 34 and applying the Direct Substitution Property D Example Let Hm To compute limmnA m we use the Direct Substitution Property to obtain 713 2 71 2 1 Using the de nition of a limit7 we can compute the limit of the root of a function We have lim f 5 zaa where n is a positive integer and a is assumed to be positive ifn is even This rule can be generalized to obtain the limit of the root of a general function lim quotfm rolim m maa maa where n is a positive integer and limmna x gt 0 if n is even This rule can be used to extend the Direct Substitution Property to a larger class of functions Theorem Direct Substitution Property Algebraic Functions If f is an algebraic function and is de ned at m a then hm W u maa 3 2 f 2139 To compute limmnA m we can use the law for roots to obtain Example Let 111M 95gt minim Alternatively we can use the Direct Substitution Property to obtain 2 43 2 T 2 l 71 4 7 39 minim V 71 1 2 2 In some cases7 it may be dif cult to compute the limit of a given function x using the limit laws7 but we can learn something about the limit by comparing 1 to simpler function g This leads to the following result Theorem If gx for x near a except possibly at x a itself and limmma x and limmaagx both exist then lim x lim zaa xaa A direct consequence of this result is as follows Theorem Squeeze Theorem If gx hx for x near a except possibly at x a itself and lim x lim hx L xaa xaa then i133 990 L This theorem is called the Squeeze Theorem because the function gx is squeezed between x and hx and since both of these functions approach the same value L as x approaches a 9 must also approach L Example Consider the function x x2 tan x For err4 x 7r4 we have 71 tanx 1 Multiplying through this inequality by x2 we nd that 7x2 x x2 Since limmao x2 limmao x2 0 it follows that limmao 7x2 0 as well so we can apply the Squeeze Theorem and conclude that lim x2 tanx O an This application of the Squeeze Theorem is illustrated in Figure 1 D Because we have established all of these laws of limits we are able to avoid resorting to the de nition of a limit to compute limits of a wide variety of functions The process of establishing these laws is illustrative of the fact that in mathematics two general types of problems need to be solved 0 Speci c problems typically those obtained by modeling real world problems for which math ematics was developed in the rst place and 0 General abstract problems typically for the purpose of developing more powerful problem solving techniques It is recommended that one be comfortable with solving both types of problems in order to be able to apply the power of mathematics effectively and ef ciently Complement to Section 156 Theorem 1 Chain Rule for Functions and Curves Let f be a di erentiable function of two three variables 0t a di erentiable parametric curve in the plane in space and ht fat the composite function Then d E Vf0t 39 WW 1 De nition 1 Alternative de nition of directional derivative Let fr7 y be a function of two variables7 r0 107y0gt a point in its domain7 and u abgt a unit vectori Moreover let 0t r0 tu be a parametric curvei Then the derivative d Duflt 07y0 Dufr0 fatlto 2 is called the directional derivative of f in the direction of u Theorem 2 Directional derivative in terms of the gradient vector If f is a di erentiable function of z and y then f has directional derivative in the direction of any unit vector u a7 12gt and DufzyVfui 3 Proof Use the de nition of directional derivative given above and the chain rule Then d Dufr E a hlwo WWW 39 0t 4 where r zygt and the last equality holds because of the chain rule Moreover since 0t r tu then at ui This concludes the proof NiBi if f is not differentiable you cannot use the chain rule D The theorem above is important because it allows to write the directional derivative in a simple way However the Theorem does not hold if the function f is not differentiable This is illustrated by the following example Example Let ySIQ y2 f00 0 a Compute fghfy7 1250707 fy00i b Show that for any 9 the directional derivative d 100 cos 9 r s1n t9lro exists c Show that the directional derivatives are not all given by dotting the direction vector with the gradient vectori Why does this not contradict the chain rule Solution 6 f 2 E I2 y22 2 4 y y 3 7 7 2 7 fy 12 y2 I2 y22 Moreover fz0 0 for I f 07 f0y y for y f 0 Therefore 7 f0h707f00 7 0707 fwlt070gtem h 75337 0 and f00 h f0 0 h 0 7 7 7 7 fy0701L13 h il ih 71 Jim Lambers Math 2A Winter Quarter 200304 Lecture 2 Examples These examples correspond to Sections 13 and 21 in the text Example Suppose that Wile E Coyote has once again been duped by the Road Runner into heading over a cliff As usual he is well past the edge of the cliff before realizing what he has done at which time he falls straight down What is his velocity after two seconds of free fall Solution At time t where t is measured in seconds his position or altitude in feet is given by the function 1 2 p 75975 7 where p 0 corresponds to the level of the cliff and g is the constant of gravity 32 ftsZ Since his altitude is decreasing p will be negative Substituting the value of 9 we have p 716252 This function is shown in Figure 1 Note that the graph of this function can be obtained from the graph of p t2 by rst multiplying t2 by 71 to re ect its graph across the p axis and then multiplying it by 16 to stretch the graph vertically by a factor of 16 His velocity is the instantaneous rate of change at t 2 Since we don t have a mathematical de nition for the velocity at a particular instant yet we instead work with what we do know how to compute the average velocity over an interval in time lntuitively our approach to this problem is as follows we will gure out how to compute the average velocity over an interval in time of the form 2 2 h that is the interval 2 g t g 2 h and nd out what happens to the average velocity as the length of the time interval h approaches zero Why does this approach make sense The reason is that the average velocity over an interval in time is equal to the instantaneous velocity at every point in the interval if the object is traveling at constant speed The smaller the interval the less the object s speed should vary so for a very small interval the average velocity should be a very good approximation to the instantaneous velocity If the interval is in nitely small77 ie h 0 then we should have the exact value of the instantaneous velocity We now recall what is the average velocity on an interval in time it is the total distance traveled during the interval divided by the length of the interval the elapsed time In general the average velocity of an object that has traveled from position yl at time 251 to position yg at time 252 is distance traveled 7 yz 241 elapsed time 252 7 t1 39 Altitude of Wile E Coyote p 1612 100 cL150 200 250 300 0 Figure 1 Graph Ofp 7162 In this case the average velocity over the interval 2 2 h is given by 7162 h2 71622 2 h 7 2 39 Ideally we would like to simply set It 0 to get the instantaneous velocity immediately but we cannot do that in this case because we would be dividing by zero Instead we will try to simplify this expression and see if that helps We have 7162 h2 7 71622 7164 4h h2164 2h72 h 764 7 64h 716112 64 h 764h 716h2 h 764 7 16h Therefore the average velocity of free fall over the interval from t 2 to t 2 h for any length of time h is 764 7 16h fts2 Setting h O we nd that the instantaneous velocity at time t 2 is 764 fts2 If we were to substitute various values for h we would nd that the average velocity over the interval 2 2 h converges to the instantaneous velocity at time t 2 For this reason we de ne the instantaneous velocity at a given instant to be the limit of the average velocity over an interval containing that instant as the length of the interval in this example It approaches zero The following table illustrates this convergence of the average velocity to the instantaneous velocity h Avg vel 2 796 1 780 05 772 01 7656 001 76416 We now interpret average velocity and instantaneous velocity geometrically The average ve locity over the interval 22 l h is the change in p between these two times divided by the change in time This is equal to the slope of the line passing through the points 2764 and 2 h 7162 h2 This line is called a secant line for the curve p 716t2 A few such secant lines are shown in Figure 2 Note that as h approaches zero these secant lines converge to a line that is tangent to the curve at the point 2 764 that is it touches the curve but does not cross it The instantaneous velocity is equal to the slope of this tangent line Since we know the slope of the tangent line and we know a point on the line we can obtain an equation for the line We use the point slope form P P0mtt0 Secant lines converging to tangent line 100 120 1407 160 180 200 i 220 i 240 e 260 15 2 25 Figure 2 Various secant lines in red for the which they converge as h approaches 0 curve p 716252 and the tangent line in black to where m is the slope and 250190 is a point on the line In this example m 764 to 2 and p0 764 Therefore the equation for the tangent line is 19 64 76425 7 2 which can be rearranged to obtain an equation in slope intercept form 19 mt b or in this case p 7642f 64 Note You are likely accustomed to seeing the equation of a line written using the letters z and y for the independent variable and dependent variable respectively In fact often these letters will be used for those purposes in this course However it is important to keep in mind that one should not get too attached to speci c letters being used for speci c purposes because it s the intemretatz39on of the letter or other symbol that is important Just as you are concerned with the meaning of words as you are reading a book in order to comprehend what is being written you should keep in mind the meaning of mathematical symbols in order to comprehend their usage D Example My car is traveling along a dark country road at night at 45 mph All of a sudden a deer darts out into the road and l slam on my brakes and manage to stop in time and avoid hitting the deer The following table lists my velocity over time in mph where t 0 corresponds to the instant at which I rst apply the brakes 25 seconds 1 mph 0 45 1 24 2 9 3 0 At what rate am I decelerating after 2 seconds in mhr2 Solution Just as velocity is the rate of change of distance with respect to time acceleration is the rate of change of velocity with respect to time In this case the acceleration will be negative since I am slowing down so we will actually nd out how quickly I am decelerating First we need a function that models the given data There are many ways to construct such a function that are beyond the scope of this course so we will simply work with one such function u 3w e 4 71 where 1 denotes velocity in mph and t denotes time in seconds It should be noted that the graph of this function can easily be obtained from the graph of v t2 by applying the following transformations 1 First the graph of v t2 is shifted by 4 units to the right which yields the curve 1 t 7 42 2 Second the graph is shifted one unit down which yields the curve 1 t 7 42 7 1 3 Finally the graph is stretched vertically by a factor of 3 which yields the curve 1 3t 7 7 1 This sequence of transformations is illustrated in Figure 3 As before we will compute the average deceleration over the interval in time 2 2 h and nd out what happens as the elapsed time h seconds approaches zero This average deceleration is given by change in velocity 32 h 7 42 7 1 7 32 7 42 7 1 change in time 2 h 7 2 which can be simpli ed as follows 32 h 7 42 7 1 7 32 7 42 71 7 3h 7 22 71 7 3722 71 2 h 7 2 7 h 7 3h274h47173471 f 30 7 4h 3 7 33 h 37127 12h979 h 371271211 h 31712 Setting h O we obtain the instantaneous deceleration at time t 2 which is 712 miles per hour per second since if is measured in seconds To obtain the deceleration in mihr2 we must multiply this result by the number of seconds in an hour 3600 Our nal answer is therefore 743200 mihr2 This seems like an incredibly large number under the circumstances but keep in mind that this is equivalent to only 7176 ftsZ D Example Let be a function that describes the temperature in degrees Celsius in a rod that is 4 m long where x denotes the distance in meters between any given point on the rod and its left endpoint which corresponds to z 0 Speci cally sin As we travel along the rod from left to right what is the instantaneous rate of change of the temperature with respect to distance at the point that is 1 m from the left endpoint of the rod Also what is the equation of the tangent line to at z 1 vt vt 42 50 50 40 40 30 30 gt gt 20 20 10 10 0 0 4 2 0 2 4 4 2 0 2 4 t t vt421 v3t421 50 50 40 40 30 30 gt gt 20 20 10 10 0 0 4 2 0 2 4 4 2 0 2 4 t t Figure 3 Transformation of v 252 upper left by shifting to the right 4 units upper right7 shifting down one unit lower left7 and stretching vertically by a factor of 3 lower right Solution The point that is 1 m from the left endpoint of the rod corresponds to z 1 The function that models the temperature in the rod7 sinn39z47 is chosen so as to satisfy the condition that the temperature is held xed at 0 C The graph of this function can be obtained from the graph of sinz by stretching it horizontally by a factor of 47139 This stretching is accomplished by multiplying the independent variable7 x by the reciprocal of the factor by which we want to stretch7 ie7 7r4 The graph of is shown in Figure 4 In order to compute the instantaneous rate of change of temperature with respect to distance at z 17 we proceed as in the previous examples and compute the average rate of change over the interval 17 1 h where h is allowed to vary We obtain change in temperature sin7r1 h4 7 sin7r4 change in distance 1 h 7 1 To simplify this7 we rst use the identity sinz y sin z cos y cos x sin y which yields sin7r4 cos7rh4 cos7r4 sin7rh4 7 sin7r4 h sin7r4cos7rh4 7 1 cos7r4 sin7rh4 h We then use the half angle formula i 2 1 7 cos 2x sin z 2 to obtain 2 sin7r4 sin27rh8 cos7r4 sin7rh4 h Rearranging yields h 8 h 4 72sinltw4gtsinltwh8gth sown 8m gt We multiply and divide the rst term by 39Ir87 and multiply and divide the second term by 7r4 to obtain sin7rh8 7r sin7rh4 7r 72 s1n7r4s1n7rh8 ll8 g cos7r4T44 We now use the fact that as 9 approaches 07 the expression sin 9 0 8 Temperature in a rod 15 r r ux degrees Celsius 0 0 05 1 15 2 25 3 35 4 x meters Figure 4 Graph of sin7rz4 The red circle marks the point z 17 u The red line is tangent to the curve at that point Jim Lambers Math 2A Winter Quarter 200304 Lecture 18 Notes These notes correspond to Section 42 in the text The Mean Value Theorem In upcoming lectures we will be learning how the derivatives of a function can be used to understand the behavior of the function on its domain While the derivative describes the behavior of a function at a point we often need to determine a function s behavior on an interval as we have seen from our consideration of the problem of nding the maximum or minimum value on an interval Our study of how the local behavior described by the derivative in uences the global behavior of the function on an interval relies largely on a single theoretical result called the Mean Value Theorem In this lecture we will present the statement and proof of this theorem as well as some of its most basic interpretations and applications In later lectures we will discover even more applications of the theorem Before we can state and prove the Mean Value Theorem we need the following simpler result Theorem Rolle s Theorem If f is continuous on a closed interval hub and is di erentiable on the open interval a b and if fa fb then f c 0 for some number c in a b Proof There are two cases to consider a If is equal to a constant on cub then f c 0 for all a in a b o If f is not constant on m b then we first apply the Extreme Value Theorem to conclude that f has an absolute maximum and an absolute minimum on a b Since f is not constant on mb there must exist some number a in ab such that f gt f a or f lt f a This means that fa cannot be both the maximum value and minimum value on m b and since u f b it follows that either the absolute maximum or the absolute minimum must occur in the open interval a b Since an absolute maximum or an absolute minimum occurring in an open interval must also be a local maximum or local minimum we can apply Fermat s Theorem to conclude that there exists at least one critical number 0 in a b By the de nition of critical number either f 0 or f does not exist However f is assumed to be differentiable on a b so f must exist We conclude that f 0 El Example In Figure 1 the function c2 255 1 satisfies f0 f 2 and has a horizontal tangent at a 1 In other words f 1 0 as suggested by Rolle s Theorem D fx x22x1 f 10 fX 1 f01 f21 Y Figure 1 The function c2 251 1 satis es f0 f2 1 so by Rolle s Theorem it must satisfy f c 0 for some c in 0 2 In fact f 1 0 We can now state and prove the Mean Value Theorem itself as the proof will require Rolle s Theorem Theorem Mean Value Theorem If f is continuous on a closed interval mb and is di erentiable on the open interval a b then m l b a f c for some number 0 in ab Proof De ne the function by ge a i9 39e al Then because f is continuous on a b and differentiable on a b so is g Furthermore mo m 3 9m o n and f b f a mm m igiw mfw uwrdmn o Since ga gb Rolle s Theorem applies to g and we can conclude that there is some number c in a b such that g 0 However since waMa r e afm g i we must have b m f 5 0 m m o f c b a El Remark The expression M b a is the slope of the secant line passing through the points a f a and bf The Mean Value Theorem therefore states that under the given assumptions the slope of this secant line is equal to the slope of the tangent line of f at the point e f 0 where c is in a b This is illustrated in Figure 2 yx24x6 l secant line yx2 tangent line yx14 2 l l l l l l l l l Figure 2 Illustration of the Mean Value Theorem For the function f 52 41 6 the slope of the secant line passing through the points 1 3 and 4 6 has slope 1 as does the tangent line passing through the point 52 In other words 1 f 52 The Mean Value Theorem has the following practical interpretation the average rate of change of y f with respect to a on an interval nub is equal to the instantaneous rate of change 3 with respect to a at some point in a b The following example illustrates how this interpretation can be applied Example Suppose that you are driving along the freeway and you see a Highway Patrol car on the side of the road looking for speeders You slow down in time and the of cer s radar gun determines that you are traveling at a reasonable speed After passing you speed up again and after traveling three miles within the next two minutes you encounter another of cer and slow down only to speed up after passing once again The second of cer does not pursue you and you think you are safe but after a few minutes you are pursued by the of cer stopped and given a ticket for driving in excess of 90 mph How could it be determined that you were speeding if neither of cer s radar gun indicated that you were The of cers could collaborate and determine that you traveled three miles in two minutes which is equivalent to 90 mph Therefore your average speed during those two minutes was 90 mph By the Mean Value Theorem your instantaneous speed must have been 90 mph at some point during that time To see this mathematically let f t be your position along the road at time t where t 0 corresponds to the time that you passed the first of cer 5 is measured in hours and fft indicates position in miles Then since two minutes is equal to 130th of an hour we have f130 f0 3 130 0 130 and by the Mean Value Theorem we must have f 90 for some c between 0 and 1 30 In this situation your wise choice is to fight the ticket since calculus may not necessarily be admissible in court I 90 We know that if a function f is equal to a constant then f 0 One of the most basic consequences of the Mean Value Theorem is the converse of this statement if a function s derivative is equal to zero on some domain then the function must be constant on that domain We now state this precisely and show how the Mean Value Theorem can be used to prove it Theorem If f c 0 on an open interval ab then f is constant on a b Proof Let c and d be any two numbers in a b with c d Then f is differentiable on c d and continuous on 0 d By the Mean Value Theorem there is a number it in c d such that fc d m However since u is in c d u is in a b and so f u 0 Since c d it follows that f f Because this is true for any numbers 0 and d in a b f must be constant on a b D Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 9 Notes These notes correspond to Section 63 in the text Volume by Shells In the previous lecture we learned how we could compute the volumes of a solid S that lay between the planes J a and J l by integrating its cross sectional AJ area over the interval aw Unfortunately this technique of computing volume quotby slices where each slice is a cylinder of ctional area is relatively to determine If this is not the case then we can instead try to determine whether the solid can be approxi mated by several concentric cylindrical shells For example suppose that l gt a 2 0 and we have a solid that can be obtained by rotating the region bounded by y fJ where fJ 2 0 J a J l and y 0 Then the volume of resulting solid can be approximated by first dividing the interval 0 b into n subintervals of equal width AJ 2 I an with the ith subinterval having endpoints 721 and Ji where 73 lAJ for l 0 n For each subinterval we denote the midpoint by Ji1 Ji2 Then we compute the volumes of n shells of thickness AJ 2 I an height f J inner radius J54 and outer radius 75 for l 1 n The volume of the ith shell is 1 27rJff J AJ and therefore the volume V of the entire solid is approximated by R R v a Z V Z2mffmmx 1 i1 i1 Letting n gt so this approximation converges to the exact volume with the summation converging to the definite integral v b 2wfltxgtdx 2 More generally if the solid can be viewed a collection of concentric cylindrical shells of radius 391 J39 and height f J for a 3 J g b the volume of the solid is given by J 12 27W Jf7d7 3 If the solid is formed by rotating a region around the t a then 39I J However a different 39I J must be used if the solid is obtained by rotating a region around a different line For example if the line around which the region is rotated is J c where 0 must be outside the interval 0 b then J c if c g a while c J if c 2 l Example 1 In the previous lecture we used the disc method to compute the volume of the solid obtained by revolving the region bounded by y 2 9 0 J 0 and J 1 around the yaxis Now we will compute the volume of the same solid using the shell method The average radius of the shell at J is equal to J and the height of this shell is equal to the height of the region at J which is 2 It follows that the volume of the solid is l 1 1 u 2W 2 6 27rJ2 dJ27r 2J J32dJ27r JZ Jf 2rr1 T 4 l 0 i0 3 39 C Example 2 Let fJ J2 Consider the region bounded by the curve 9 fJ the horizontal line 9 0 and the vertical lines J 1 and J 2 Compute the volume of the solid obtained by revolving this region around the taxis Solution We use the method of cylindrical shells The given region is shown in Figure 1 while the solid obtained by revolving the region around the bands is shown in Figure 2 This solid can be approximated by a collection of concentric cylindrical shells The approximation proceeds follows first we divide the interval 12 into n subintervals of equal width AJ 2 171 These intervals have endpoints J0J1 J1J2 Jn1Jn where J 1 lAJ for l 012 n Then we approximate the region below 9 by rectangles of width AJ and height where is any point in the interval Ji1Ji for i 1 Then by revolving each rectangle around the tax s we obtain n cylindrical shells that ap proximate the solid just the rectangles approximate the region This process of revolving each rectangle around the t axis to obtain a shell is illustrated in Figure 3 For each i 1 n the ith shell has thickness AJ height f J inner radius Ji1 and outer radius Ji The volume Vi of this shell is given by 14 27 fanny By adding the volume of all of these shells we obtain a Riemann sum that yields an approximation to the volume of the solid As the number of subintervals n becomes infinite this approximation converges to the exact volume The volume V of the solid is therefore given by R lim E V naooA 71 R 14 JH g 320 Z1 27 MW 1 A m V v n AJ 6 2H 27 a 2 mnm yX 5 x 4 7 3 gt 27 1 7 G 1 l l l l l l 1 05 0 05 1 15 2 25 3 x Figure 1 Region bounded by y x2 9 0 J 1 and J 2 shaded i l 2 Ermf 1 11 Evaluating the resulting definite integral yields V 2 27F7J217 i 1 Solid of revolution C Example 3 Let fJ J2 Consider the region from the previous example that is bounded by the curve 1 f J the horizontal line 9 0 and the vertical lines J 1 and J 2 Compute the volume of the solid obtained by revolving this region around the vertical line J 2 Solution In this case we can approximate the solid by cylindrical shells before but the center and radii of the shells is different Because the center of the solid is the line J 2 the inner radii of the ith shell corresponding to the subinterval Ji1Ji is J54 2 since that is the distance between the inner boundary of the shell and the center Similarly the outer radius of the ith shell is J 2 Proceeding in the previous example we can determine that the volume V of the solid is given by R v 7331010219 7 Shell method Figure 3 Cylindrical shell obtained by revolving rectangle around the bands The rectangle is one of several that is used to approximate the region bounded by y 1 y 0 J 1 J 2 whose outline is shown R mm 2 15 2 f 733101072127 f fUilAJ n A1 6 7331010 2 127 17 2 2 fUHAJ 1 2 27rJ 2 f7 11 i 1 Evaluating the resulting definite integral yields 92 V 27rJ 2J2 11 i 1 2 277 J3212d1 ll C 15 14 2 i T 4 1017 9 6 Example 4 Let fJ J2 Consider the region from the previous example that is bounded by the curve 1 f J the horizontal line 9 0 and the vertical lines J 1 and J 2 Compute the volume of the solid obtained by revolving this region around the horizontal line 1 Solution We will compute the volume of this solid using both the washer method that is volume by slices and the shell method Using the washer method we see that for each J in the interval 1 2 the corresponding washer has inner radius 1 and outer radius J2 1 It follows that the volume V of the solid is given by V 2 7FJ39212 7r12d7 r1 J42J21 11J 2 7 A 2 272 dJ 2J3 5 3 7T 2 7 3 3 3114 7 i 3 3 163 7 10 1a A sample washer is illustrated in Figure 4 Washer method Figure 4 The region bounded by y J y 0 J 1 and J 2 is to be revolved around the line 9 1 The washer corresponding to J 14 is shown Using the shell method we integrate with respect to 9 because we are revolving the region around a horizontal line For each 9 in the interval 1 4 we have a cylindrical shell centered at the line 1 1 with thickness 19 average radius 9 1 and height 2 since that is the horizontal distance between the line J 2 and the curve 1 2 or equivalent Furthermore for each 9 in the interval 01 we have a cylindrical shell centered at the line 1 1 with thickness 19 average radius 9 1 and height 1 It follows that the volume V is given by 1 3 V 27ry11dy 27ry12 dy in l 1 ol 4 27r y1dy 29 y3Z ylZF2dy r0 r1 4 2 y y 2 l y 2 71 1 T 2 J J 52 32 1 52 32 A M 1 H V 1 A 9 N w 1 H lt V w gt 3 64 16 2 7 247777 57 Tl2i li 3 64 16 2 2 2 7 217777 7 7 Tl2 355l 2 E 73 30 30 30 50 163 2 7 TM 1637f 15 A sample shell is illustrated in Figure C le H V A H O l w 0 w l lw mi 00 V l A H I Hm I Chalk m V 1 Jim Lambers Math 2A Winter Quarter 200304 Lecture 6 Notes These notes correspond to Section 26 in the text Tangents Velocities and Other Rates of Change In Lecture 2 we learned that the instantaneous rate of change of a function x at a point z a was equal to the slope of a line that passed through the point a fa and was tangent to the curve y In this lecture we will use the notation of limits to more precisely de ne a tangent line and the corresponding instantaneous rate of change Velocities Consider a function ft where the independent variable t denotes time and ft denotes the position of an object at time t From previous discussion we learned that we could compute the instantaneous velocity of the object at a particular time to by computing its average velocity over an interval t0 to h for some number h using the simple formula U ft0hft0 ft0h ft0 t to 7 to h where v is the average velocity from time to to time to h t is the elapsed time and d is the distance traveled during this interval of time As we chose smaller and smaller values of h the interval t0 to 71 shrunk to the single point to and the average velocity over the interval converged to a particular value which we then de ned to be the instantaneous velocity at time to We now use the notation of limits to more precisely de ne this concept De nition Let the function denote the position of an object at time t Then the instanta neous velocity of the object at time to denoted by vt0 is to h 750 h 7 t l vlt 0gt provided that this limit exists Example Suppose that the position of an object at time t where t is in seconds and position is measured in feet is described by the function ft t2 We wish to compute its velocity at time t 2 which we denote by 122 Using the de nition of instantaneous velocity along with the limit laws introduced in Lecture 4 we obtain 7 i f2hf2 2 53 h 2 2 MW haO h i 44hh274 lim haO h 7 1i 4hh2 T hlgl h lim4h haO lim 4 lim h haO haO 4 We conclude that the velocity at time t 2 is 4 fts D Tangent s In Lecture 2 we also discussed a geometric interpretation of the concepts of average velocity and instantaneous velocity Recall that the average velocity over the interval to7 to h is equal to the slope of the secant line that passes through the points t07 ft0 and to h7 ft0 As h approaches 07 the corresponding secant line converges to a line that is tangent to the curve y ft at the point to7 ft0 that is7 it touches the graph of f at this point but does not cross the graph The slope of this tangent line is equal to the instantaneous velocity at time to We can now give a precise de nition of a tangent line using the notation of limits In the following de nition7 we use the letter z for the independent variable instead of t7 and the letter a to denote the point of tangency instead of the symbol to This notation is used to emphasize the fact that the given function 1 need not be a function of time the independent variable can have any interpretation De nition The tangent line to the came y x at the point a7 fa is the line with slope mmpwme o haO h that passes through the point a7 fa provided that the above limit in exists If we let x a h7 then x approaches a as h approaches 0 We use this notation to give an alternative de nition that is consistent with the de nition given in the text De nition The tangent line to the came y x at the point a7 fa is the line with slope mhm i maa zia that passes through the point a7 fa provided that the above limit in exists Example We again consider the function ft t2 and compute the equation of the tangent line at the point 27 4 To compute the slope7 denoted by m7 we need to evaluate m hm at 7 H2 taZ t7 2 Using the limit laws7 we obtain m hm at e flt2gt taZ t7 2 l t2 7 22 111 25 2 7 1 t2 7 4 T 13 t7 2 hm t 2t 7 2 taZ t7 2 limt 2 taZ limt lim 2 taZ taZ 2 2 4 Therefore7 the tangent line is the line with slope 4 that passes through the point 27 4 It follows that the equation of the tangent line7 in point slope form7 is y744t72 or7 in slope intercept form7 y 4t 7 4 The tangent line is illustrated in Figure 1 D Other Rates of Change We have seen that the slope of the tangent line of a function y ft at a point to is equal to the instantaneous velocity at time to of the object whose position at time t is given by ft However7 since velocity at a particular time is de ned to be the instantaneous rate of change of position with respect to time7 and since the de nition of the tangent line is independent of any interpretation of the independent variable or dependent varialole7 it is natural to equate the slope of the tangent line of y x at a7 fa with the instantaneous rate of change of y with respect to x at z a7 regardless of the interpretation of z and y This leads to the following de nitions Tangent line to yt2 at 24 i Figure 1 Tangent line of y t2 at the point 27 4 The point of tangency is indicated by the red circle De nition Let y The increment of m as m changes from a to b denoted by Am is given by Am b 7 a Similarly the increment of y as m changes from a to b denoted by Ay is given by Av f0 7 N0 The average rate of change of y with respect to z over the interval 11 is given by the difference quotient Av 7 f0 7 1 E 7 b 7 a The instantaneous rate of change of y with respect to x at z a is given by Av N 7 1 lim 7 lim Am7gt0 Am b7nz b 7 a Example Suppose that one is studying the temperature within a rod that is 4 m long The temperature T can be described by a function Mm where x denotes the position on the rod at which the temperature is measured In this example7 we assume that the temperature is held xed at 0 C at both ends of the rod7 and that the temperature the point z is given by 90 7 2V 17 7 M 4 Note that for this choice of v v0 v4 07 as intended We wish to compute the instantaneous rate of change at the point that is 1 m from the right endpoint the rod ie7 at z 3 From the de nition of the instantaneous rate of change of v with respect to x at z 37 we obtain A390 E 13137 1 w 22 lt1 3 32 m 7 3 2 2 32 2 hm14 14 73 m 7 3 7 1A 7 1722 37422 7 113 m 7 3 7 7m7223722 7 Z m 7 3 7 7z274m41 7 Z m 7 3 Jim Lambers Math 2A Winter Quarter 200304 Lecture 5 Notes These notes correspond to Section 25 in the text Continuity In the previous lecture we learned that in many cases the limit of a function fz as x approached a could be obtained by simply computing fa lntuitively this indicates that f has to have a graph that is one continuous curve because any break or jump in the graph at z a is caused by f approaching one value as x approaches a only to actually assume a different value at a This leads to the following precise de nition of what it means for a function to be continuous at a given point De nition Continuity We say that a function f is continuous at a if hm W Ha maa Example Let fz m21 Because f is a polynomial it satis es the Direct Substitution Property that was introduced in the previous lecture In other words for any a at which f is de ned lim m 11mz21 limz2 lim1a21fa zaa zaa zaa wall and therefore f is continuous at a Since a is arbitrary and f is de ned everywhere it follows that f is continuous everywhere D Example Let fz 1z Being a rational function it also satis es the Direct Substitution Property so it is continuous everywhere that it is de ned However f is not de ned at z O and therefore f is not continuous at O D The above discussion also suggests how to de ne what it means for a function to not be continuous at a given point lntuitively we want our de nition to include the notion that a function is de ned near the point at which it fails to be continuous that is the failure to be continuous at a point is an isolated occurrence This leads to the following de nition De nition Discontinuity We say that a function f is discontinuous at a if f is de ned in an open interval containing a except possibly at a itself and f is not continuous at a Altematiuely we say that f has a discontinuity at a Example Discontinuities can occur for a variety of reasons as illustrate in Figure 1 In Figure 1a x x2 7 7 1 is approaching the value 2 as x approaches 1 but 1 is not de ned at 1 so it cannot be continuous there In Figure 1b the function HM 1il is approaching 2 as x approaches 1 but 1 3 so again it is not continuous there In both cases the discontinuity can be removed by simply de ning x to be equal to 2 at x 1 In Figure 1c the Heaviside function 0 tlt0 Ht1tgt0 approaches 0 as t approaches 0 from the left while it approaches 1 as t approaches 0 from the right In other words the one sided limits lim Ht 0 and lim Ht 1 taO taOt do not agree This results in a jump in the graph of H Finally in Figure 1d the function x 1x2 fails to be continuous at x 0 because 1 is not de ned at 0 However this discontinuity cannot be removed by simply de ning f at 0 because 1 becomes in nite as x approaches 0 that is f has a vertical asymptote at O D Based on these examples the following de nitions are used to categorize various types of disconti nuities De nition Types of Discontinuities Suppose that f has a discontinuity at a 1 The discontinuity is removable if limmHaJr and limmaai both exist and are equal but are not equal to a or f is not de ned at a 2 The discontinuity is called a jump discontinuity if limmaa x and limmaai x both exist but are not equal 3 The discontinuity is in nite if f has a uertical asymptote at a In the previous example the Heaviside function Ht approaches H0 as t approaches 0 from the right but not from the left This is an example of one sided continuity which we now de ne precisely in terms of one sided limits De nition One sided Continuity We say that a function f is continuous from the right at a 239f hm rm na xaa Removable discontinuity Removable discontinuity 4 4 3 3 2 2 gt gt 1 1 0 0 1 1 2 0 2 4 2 0 2 4 x x Jump discontinuity Infinite discontinuity 2 5 15 4 1 3 gt 05 gt 2 0 1 05 0 1 1 2 1 0 1 2 3 2 1 0 1 2 3 t x Figure 1 Types of discontinuities In plot a7 upper left7 fz Henxi 1 has a discontinuity at z 1 that can be removed by de ning f1 2 In plot b7 upper right7 the function 1 de ned by fz z i 1 for z 7 17 f1 3 has a discontinuity at z 1 that can be removed by rede ning f1 to be equal to 2 In plot c7 lower left7 the Heaviside function has a jump discontinuity in its graph at t O In plot d7 lower right7 the function fz 1z2 is discontinuous at z 0 because of its vertical asymptote Similarly we say that f is continuous from the left at a if 133 ax M So far we have discussed continuity or lack thereof at a single point In describing where a function is continuous the concept of continuity over an interval is useful so we de ne this concept now De nition Continuity on an Interual We say that a function f is continuous on the interval 11 if f is continuous at euery point in 11 Similarly we say that f is continuous on 1 1 1 if f is continuous on 11 and continuous from the right at 1 2 11 if f is continuous on 11 and continuous from the left at 1 3 11 if f is continuous on 11 continuous from the right at 1 and continuous from the left at 1 Example The function fz z is continuous on the entire real number line which is the interval 700 D Example The Heaviside function is continuous on the interval foo 0 and on the interval 0 00 since it is is continuous at every point except 0 and it is continuous from the right at O D Because continuity is de ned using limits the limit laws introduced in the last lecture can be used to establish corresponding continuity laws These laws make it very easy to determine where complicated functions are continuous Just as the limit laws allow one to compute a limit of a complicated function in terms of limits of simpler functions these continuity laws allow one to decompose such a function into simpler functions and analyze their continuity instead Theorem Continuity Laws If f and g are continuous at 1 and c is any constant then the functions f 9 f 9 cf and fg are continuous at 1 Furthermore if g1 31 0 then fg is also continuous at 1 Example Because fz z and 9a 3 both satisfy the Direct Substitution Property and are de ned everywhere we can conclude that they are continuous everywhere It follows that the functions z 3 z 7 3 3 7 z and 3x are also continuous everywhere Since 9 is nonzero we can also conclude that fg mB is continuous everywhere Finally gf 3z is continuous at every point except at z 0 since 3z is not de ned at z O D In the previous lecture we stated that all algebraic functions a class of functions that includes all polynomials and rational functions satisfy the Direct Substitution Property This leads the following statement regarding the continuity of such functions Theorem Continuity of Algebraic Functions If f is an algebraic function then f is continuous on its entire domain Example Consider the function This is an algebraic function whose domain is all z except for negative z due to the square root and z 1 due to the denominator In other words its domain is the union of the intervals 0 71 and 1 00 On these intervals 1 is continuous D Although it is not so simple to prove the trigonometric functions also satisfy the Direct Sub stitution Property so we can comment on their continuity as well Theorem Continuity of Trigonometric Functions Each of the basic trigonometric functions sinz cos x tan z cot z sec z and csc z are continuous on their respectiue domains Example The functions sinz and cosz are continuous everywhere On the other hand secz is continuous at all points except for z 7r2 1 1m where k is any integer This is because cos x 0 at these points and therefore sec z 1cos x has vertical asymptotes at these same points Because sec z is de ned everywhere else it is also continuous everywhere else as can be seen from the fact that cosz is continuous everywhere and the law that states that the quotient of two continuous functions in this case 1 and cos x is continuous wherever the function in the denominator is nonzero D Previously we discussed how functions could be composed with one another to obtain new functions Speci cally given two functions x and 9a the composition of f and 9 written as f 09 is the function whose value at any point z in the domain ofg is given by f ogz We now ask when is such a composition continuous We would like to be able to answer this question using our knowledge of the continuity of f and g In order for f o g to be continuous at a we must have gm from mm Suppose that g is continuous at a and let I 9a Furthermore suppose that f is continuous at I Then as y approaches 1 y approaches fb lntuitively it would seem that we could replace y with gz because 9 is continuous at a and conclude that as x approaches a y gz approaches 1 9a and therefore y fgz approaches fb fga This is in fact the case as we formally state now Theorem Continuity of Compositions Ifg is continuous at a and f is continuous at 9a then f o g is continuous at a Example The function hz xm 1 is continuous for all z in the interval 7100 This is because the function gz z 1 is continuous everywhere and the function x is Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 22 Notes These notes correspond to Section 84 in the text Integration of Rational Functions In this lecture we consider the integration of rational functions A function f 1 is said to be rational if it can be written P J f 1 7 1 QM where PJ and 90 are polynomials If the degree of P is less than that of Q then f is said to be a proper rational function otherwise it is improper We shall only consider proper rational functions because if is improper then we can divide P by Q using long division of polynomials to obtain the representation RV W so 7 2 QM where S and are both polynomials and the rational function QJ is proper Therefore we can integrate by integrating each term of S using the power rule and then handling gJ separately Example 1 Evaluate 2133127J4 11 3 21 1 Solution Dividing the numerator and denominator we have 2x33x27x4 JZ2712727J4 JZJ211614 JZJ32111 4 and therefore 2x33xz7x4 2 1 3 7 n1 J J 211 1 which yields by the substitution u 21 1 23 2 quot 4 1 ww lyz3 u 271 211 3 2 J 1 1 7 3 7 7d 3 2 12 u u x2 f 1 t Jx 1nmuo 6

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.