Calculus Refresher, version c , Paul Garrett, [email protected] garrett/

Transcription

1 Calculus Refresher, version c , Paul Garrett, [email protected] garrett/

2 Contents () Introuction (2) Inequalities (3) Domain of functions (4) Lines (an other items in Analytic Geometry) (5) Elementary limits (6) Limits with cancellation (7) Limits at infinity (8) Limits of exponential functions at infinity (9) The iea of the erivative of a function (0) Derivatives of polynomials () More general power functions (2) Quotient rule (3) Prouct Rule (4) Chain rule (5) Tangent an Normal Lines (6) Critical points, monotone increase an ecrease (7) Minimization an Maximization (8) Local minima an maxima (First Derivative Test) (9) An algebra trick (20) Linear approximations: approximation by ifferentials (2) Implicit ifferentiation (22) Relate rates (23) Intermeiate Value Theorem, location of roots (24) Newton s metho (25) Derivatives of transcenental functions (26) L Hospital s rule (27) Exponential growth an ecay: a ifferential equation (28) The secon an higher erivatives (29) Inflection points, concavity upwar an ownwar (30) Another ifferential equation: projectile motion (3) Graphing rational functions, asymptotes (32) Basic integration formulas (33) The simplest substitutions (34) Substitutions (35) Area an efinite integrals (36) Lengths of Curves (37) Numerical integration 2

3 (38) Averages an Weighte Averages (39) Centers of Mass (Centrois) (40) Volumes by Cross Sections (4) Solis of Revolution (42) Surfaces of Revolution (43) Integration by parts (44) Partial Fractions (45) Trigonometric Integrals (46) Trigonometric Substitutions (47) Historical an theoretical comments: Mean Value Theorem (48) Taylor polynomials: formulas (49) Classic examples of Taylor polynomials (50) Computational tricks regaring Taylor polynomials (5) Prototypes: More serious questions about Taylor polynomials (52) Determining Tolerance/Error (53) How large an interval with given tolerance? (54) Achieving esire tolerance on esire interval (55) Integrating Taylor polynomials: first example (56) Integrating the error term: example 3

4 . Introuction The usual trouble that people have with calculus (not counting general math phobias) is with algebra, not to mention arithmetic an other more elementary things. Calculus itself just involves two new processes, ifferentiation an integration, an applications of these new things to solution of problems that woul have been impossible otherwise. Some things which were very important when calculators an computers in t exist are not so important now. Some things are just as important. Some things are more important. Some things are important but with a ifferent emphasis. At the same time, the essential ieas of much of calculus can be very well illustrate without using calculators at all! (Some not, too). Likewise, many essential ieas of calculus can be very well illustrate without getting embroile in awful algebra or arithmetic, not to mention trigonometry. At the same time, stuy of calculus makes clear how important it is to be able to o the necessary algebra an arithmetic, whether by calculator or by han. 2. Inequalities It is worth reviewing some elementary but important points: First, a person must remember that the only way for a prouct of numbers to be zero is that one or more of the iniviual numbers be zero. As silly as this may seem, it is inispensable. Next, there is the collection of slogans: positive times positive is positive negative times negative is positive negative times positive is negative positive times negative is negative Or, more cutely: the prouct of two numbers of the same sign is positive, while the prouct of two numbers of opposite signs is negative. Extening this just a little: for a prouct of real numbers to be positive, the number of negative ones must be even. If the number of negative ones is o then the prouct is negative. An, of course, if there are any zeros, then the prouct is zero. Solving inequalities: This can be very har in greatest generality, but there are some kins of problems that are very o-able. One important class contains problems like Solve: 5(x )(x + 4)(x 2)(x + 3) < 0 That is, we are asking where a polynomial is negative (or we coul ask where it s positive, too). One important point is that the polynomial is alreay factore: to solve this problem we nee to have the polynomial factore, an if it isn t alreay factore this can be a lot of aitional work. There are many ways to format the solution to such a problem, an we just choose one, which oes have the merit of being more efficient than many. 4

5 We put the roots of the polynomial P (x) = 5(x )(x + 4)(x 2)(x + 3) = 5 (x ) (x ( 4)) (x 2) (x ( 3)) in orer: in this case, the roots are, 4, 2, 3, which we put in orer (from left to right)... < 4 < 3 < < 2 <... The roots of the polynomial P break the numberline into the intervals (, 4), ( 4, 3), ( 3, ), (, 2), (2, + ) On each of these intervals the polynomial is either positive all the time, or negative all the time, since if it were positive at one point an negative at another then it woul have to be zero at some intermeiate point! For input x to the right (larger than) all the roots, all the factors x + 4, x + 3, x, x 2 are positive, an the number 5 in front also happens to be positive. Therefore, on the interval (2, + ) the polynomial P (x) is positive. Next, moving across the root 2 to the interval (, 2), we see that the factor x 2 changes sign from positive to negative, while all the other factors x, x + 3, an x + 4 o not change sign. (After all, if they woul have one so, then they woul have ha to be 0 at some intermeiate point, but they weren t, since we know where they are zero...). Of course the 5 in front stays the same sign. Therefore, since the function was positive on (2, + ) an just one factor change sign in crossing over the point 2, the function is negative on (, 2). Similarly, moving across the root to the interval ( 3, ), we see that the factor x changes sign from positive to negative, while all the other factors x 2, x + 3, an x + 4 o not change sign. (After all, if they woul have one so, then they woul have ha to be 0 at some intermeiate point). The 5 in front stays the same sign. Therefore, since the function was negative on (, 2) an just one factor change sign in crossing over the point, the function is positive on ( 3, ). Similarly, moving across the root 3 to the interval ( 4, 3), we see that the factor x + 3 = x ( 3) changes sign from positive to negative, while all the other factors x 2, x, an x + 4 o not change sign. (If they woul have one so, then they woul have ha to be 0 at some intermeiate point). The 5 in front stays the same sign. Therefore, since the function was positive on ( 3, ) an just one factor change sign in crossing over the point 3, the function is negative on ( 4, 3). Last, moving across the root 4 to the interval (, 4), we see that the factor x + 4 = x ( 4) changes sign from positive to negative, while all the other factors x 2, x, an x + 3 o not change sign. (If they woul have one so, then they woul have ha to be 0 at some intermeiate point). The 5 in front stays the same sign. Therefore, since the function was negative on ( 4, 3) an just one factor change sign in crossing over the point 4, the function is positive on (, 4). In summary, we have In particular, P (x) < 0 on the union of the intervals (, 2) an ( 4, 3). That s it. P (x) = 5(x )(x + 4)(x 2)(x + 3) > 0 on (2, + ) P (x) = 5(x )(x + 4)(x 2)(x + 3) < 0 on (, 2) P (x) = 5(x )(x + 4)(x 2)(x + 3) > 0 on ( 3, ) P (x) = 5(x )(x + 4)(x 2)(x + 3) < 0 on ( 4, 3) P (x) = 5(x )(x + 4)(x 2)(x + 3) > 0 on (, 4) (, 2) ( 4, 3) 5

6 As another example, let s see on which intervals P (x) = 3( + x 2 )(x 2 4)(x 2 2x + ) is positive an an on which it s negative. We have to factor it a bit more: recall that we have nice facts so that we get x 2 a 2 = (x a) (x + a) = (x a) (x ( a)) x 2 2ax + a 2 = (x a) (x a) P (x) = 3( + x 2 )(x 2)(x + 2)(x )(x ) It is important to note that the equation x 2 + = 0 has no real roots, since the square of any real number is non-negative. Thus, we can t factor any further than this over the real numbers. That is, the roots of P, in orer, are 2 << (twice!) < 2 These numbers break the real line up into the intervals (, 2), ( 2, ), (, 2), (2, + ) For x larger than all the roots (meaning x > 2) all the factors x + 2, x, x, x 2 are positive, while the factor of 3 in front is negative. Thus, on the interval (2, + ) P (x) is negative. Next, moving across the root 2 to the interval (, 2), we see that the factor x 2 changes sign from positive to negative, while all the other factors + x 2, (x ) 2, an x + 2 o not change sign. (After all, if they woul have one so, then they woul have be 0 at some intermeiate point, but they aren t). The 3 in front stays the same sign. Therefore, since the function was negative on (2, + ) an just one factor change sign in crossing over the point 2, the function is positive on (, 2). A new feature in this example is that the root occurs twice in the factorization, so that crossing over the root from the interval (, 2) to the interval ( 2, ) really means crossing over two roots. That is, two changes of sign means no changes of sign, in effect. An the other factors ( + x 2 ), x + 2, x 2 o not change sign, an the 3 oes not change sign, so since P (x) was positive on (, 2) it is still positive on ( 2, ). (The rest of this example is the same as the first example). Again, the point is that each time a root of the polynomial is crosse over, the polynomial changes sign. So if two are crosse at once (if there is a ouble root) then there is really no change in sign. If three roots are crosse at once, then the effect is to change sign. Generally, if an even number of roots are crosse-over, then there is no change in sign, while if an o number of roots are crosse-over then there is a change in sign. 6

7 #2. Fin the intervals on which f(x) = x(x )(x + ) is positive, an the intervals on which it is negative. #2.2 Fin the intervals on which f(x) = (3x 2)(x )(x + ) is positive, an the intervals on which it is negative. #2.3 Fin the intervals on which f(x) = (3x 2)(3 x)(x + ) is positive, an the intervals on which it is negative. 3. Domain of functions A function f is a proceure or process which converts input to output in some way. A traitional mathematics name for the input is argument, but this certainly is confusing when compare with orinary English usage. The collection of all legal reasonable or sensible inputs is calle the omain of the function. The collection of all possible outputs is the range. (Contrary to the impression some books might give, it can be very ifficult to figure out all possible outputs!) The question What s the omain of this function? is usually not what it appears to be. For one thing, if we are being formal, then a function hasn t even been escribe if its omain hasn t been escribe! What is really meant, usually, is something far less mysterious. The question usually really is What numbers can be use as inputs to this function without anything ba happening?. For our purposes, something ba happening just refers to one of trying to take the square root of a negative number trying to take a logarithm of a negative number trying to ivie by zero trying to fin arc-cosine or arc-sine of a number bigger than or less than Of course, iviing by zero is the worst of these, but as long as we insist that everything be real numbers (rather than complex numbers) we can t o the other things either. For example, what is the omain of the function f(x) = x 2? Well, what coul go wrong here? No ivision is inicate at all, so there is no risk of iviing by 0. But we are taking a square root, so we must insist that x 2 0 to avoi having complex numbers come up. That is, a preliminary escription of the omain of this function is that it is the set of real numbers x so that x 2 0. But we can be clearer than this: we know how to solve such inequalities. Often it s simplest to see what to exclue rather than inclue: here we want to exclue from the omain any numbers x so that x 2 < 0 from the omain. We recognize that we can factor x 2 = (x )(x + ) = (x ) (x ( )) 7

8 This is negative exactly on the interval (, ), so this is the interval we must prohibit in orer to have just the omain of the function. That is, the omain is the union of two intervals: (, ] [, + ) #3.4 Fin the omain of the function f(x) = x 2 x 2 + x 2 That is, fin the largest subset of the real line on which this formula can be evaluate meaningfully. #3.5 Fin the omain of the function f(x) = x 2 x2 + x 2 #3.6 Fin the omain of the function f(x) = x(x )(x + ) 4. Lines (an other items in Analytic Geometry) Let s review some basic analytic geometry: this is escription of geometric objects by numbers an by algebra. The first thing is that we have to pick a special point, the origin, from which we ll measure everything else. Then, implicitly, we nee to choose a unit of measure for istances, but this is inee usually only implicit, so we on t worry about it. The secon step is that points are escribe by orere pairs of numbers: the first of the two numbers tells how far to the right horizontally the point is from the origin (an negative means go left instea of right), an the secon of the two numbers tells how far up from the origin the point is (an negative means go own instea of up). The first number is the horizontal coorinate an the secon is the vertical coorinate. The ol-fashione names abscissa an orinate also are use sometimes. Often the horizontal coorinate is calle the x-coorinate, an often the vertical coorinate is calle the y- coorinate, but the letters x, y can be use for many other purposes as well, so on t rely on this labelling! The next iea is that an equation can escribe a curve. It is important to be a little careful with use of language here: for example, a correct assertion is The set of points (x, y) so that x 2 + y 2 = is a circle. It is not strictly correct to say that x 2 + y 2 = is a circle, mostly because an equation is not a circle, even though it may escribe a circle. An conceivably the x, y might be being use for something other than horizontal an vertical coorinates. Still, very often the language is shortene so that the phrase The set of points (x, y) so that is omitte. Just be careful. The simplest curves are lines. The main things to remember are: Slope of a line is rise over run, meaning vertical change ivie by horizontal change (moving from left to right in the usual coorinate system). The equation of a line passing through a point (x o, y o ) an having slope m can be written (in so-calle point-slope form) y = m(x x o ) + y o or y y o = m(x x o ) 8

9 The equation of the line passing through two points (x, y ), (x 2, y 2 ) can be written (in so-calle twopoint form) as y = y y 2 x x 2 (x x ) + y...unless x = x 2, in which case the two points are aligne vertically, an the line can t be written that way. Instea, the escription of a vertical line through a point with horizontal coorinate x is just x = x Of course, the two-point form can be erive from the point-slope form, since the slope m of a line through two points (x, y ), (x 2, y 2 ) is that possibly irritating expression which occurs above: m = y y 2 x x 2 An now is maybe a goo time to point out that there is nothing sacre about the horizontal coorinate being calle x an the vertical coorinate y. Very often these o happen to be the names, but it can be otherwise, so just pay attention. #4.7 Write the equation for the line passing through the two points (, 2) an (3, 8). #4.8 Write the equation for the line passing through the two points (, 2) an (3, 8). #4.9 Write the equation for the line passing through the point (, 2) with slope 3. #4.0 Write the equation for the line passing through the point (, 5) with slope. 5. Elementary limits The iea of limit is intene to be merely a slight extension of our intuition. The so-calle ε, δ-efinition was invente after people ha been oing calculus for hunres of years, in response to certain relatively pathological technical ifficulties. For quite a while, we will be entirely concerne with situations in which we can either irectly see the value of a limit by plugging the limit value in, or where we transform the expression into one where we can just plug in. So long as we are ealing with functions no more complicate than polynomials, most limits are easy to unerstan: for example, lim x 3 4x2 + 3x 7 = 4 (3) (3) 7 = 38 4x 2 + 3x 7 lim x 3 2 x 2 = 4 (3)2 + 3 (3) 7 2 (3) 2 = 38 7 The point is that we just substitute the 3 in an nothing ba happene. This is the way people evaluate easy limits for hunres of years, an shoul always be the first thing a person oes, just to see what happens. #5. Fin lim x 5 2x 2 3x

10 #5.2 Fin lim x 2 x+ x #5.3 Fin lim x x Limits with cancellation But sometimes things blow up when the limit number is substitute: x 2 9 lim x 3 x 3 = 0 0????? Ick. This is not goo. However, in this example, as in many examples, oing a bit of simplifying algebra first gets ri of the factors in the numerator an enominator which cause them to vanish: x 2 9 lim x 3 x 3 = lim (x 3)(x + 3) (x + 3) (3 + 3) = lim = = 6 x 3 x 3 x 3 Here at the very en we i just plug in, after all. The lesson here is that some of those arn algebra tricks ( ientities ) are helpful, after all. If you have a ba limit, always look for some cancellation of factors in the numerator an enominator. In fact, for hunres of years people only evaluate limits in this style! After all, human beings can t really execute infinite limiting processes, an so on. #6.4 Fin lim x 2 x 2 x 2 4 #6.5 Fin lim x 3 x 2 9 x 3 #6.6 Fin lim x 3 x 2 x 3 Next, let s consier 7. Limits at infinity 2x + 3 lim x 5 x The hazar here is that is not a number that we can o arithmetic with in the normal way. Don t even try it. So we can t really just plug in to the expression to see what we get. On the other han, what we really mean anyway is not that x becomes infinite in some mystical sense, but rather that it just gets larger an larger. In this context, the crucial observation is that, as x gets larger an larger, /x gets smaller an smaller (going to 0). Thus, just base on what we want this all to mean, lim x x = 0 an so on. lim x lim x x 2 = 0 x 3 = 0 0

11 This is the essential iea for evaluating simple kins of limits as x : rearrange the whole thing so that everything is expresse in terms of /x instea of x, an then realize that lim x is the same as lim x 0 So, in the example above, ivie numerator an enominator both by the largest power of x appearing anywhere: 2x + 3 lim x 5 x = lim x x 5 x = lim 2 + 3y y 0 5y = = 2 The point is that we calle /x by a new name, y, an rewrote the original limit as x as a limit as y 0. Since 0 is a genuine number that we can o arithmetic with, this brought us back to orinary everyay arithmetic. Of course, it was necessary to rewrite the thing we were taking the limit of in terms of /x (rename y ). Notice that this is an example of a situation where we use the letter y for something other than the name or value of the vertical coorinate. #7.7 Fin lim x x+ x #7.8 Fin lim x x 2 +3 x+. #7.9 Fin lim x x x 2 +x+. #7.20 Fin lim x x 2 5x 2 +x+. 8. Limits of exponential functions at infinity It is important to appreciate the behavior of exponential functions as the input to them becomes a large positive number, or a large negative number. This behavior is ifferent from the behavior of polynomials or rational functions, which behave similarly for large inputs regarless of whether the input is large positive or large negative. By contrast, for exponential functions, the behavior is raically ifferent for large positive or large negative. As a reminer an an explanation, let s remember that exponential notation starte out simply as an abbreviation: for positive integer n, 2 n = (n factors) 0 n = (n factors) ( ) n ( ) ( ) ( ) ( ) =... (n factors) From this iea it s not har to unerstan the funamental properties of exponents (they re not laws at all): a m+n = a a a... a (m + n factors) } {{ } m+n = (a a a... a) (a a a... a) = a m a n } {{ } } {{ } m n

12 an also = (a a a... a) } {{ } m a mn = (a a a... a) = } {{ } mn... (a a a... a) } {{ } m } {{ } n = (a m ) n at least for positive integers m, n. Even though we can only easily see that these properties are true when the exponents are positive integers, the extene notation is guarantee (by its meaning, not by law) to follow the same rules. Use of other numbers in the exponent is something that came later, an is also just an abbreviation, which happily was arrange to match the more intuitive simpler version. For example, a = a an (as consequences) a n = a n ( ) = (a n ) = a n (whether n is positive or not). Just to check one example of consistency with the properties above, notice that a = a = a ( ) ( ) = a = /a = a This is not suppose to be surprising, but rather reassuring that we won t reach false conclusions by such manipulations. Also, fractional exponents fit into this scheme. For example a /2 = a a /3 = [3]a a /4 = [4]a a /5 = [5]a This is consistent with earlier notation: the funamental property of the n th root of a number is that its n th power is the original number. We can check: a = a = (a /n ) n = a Again, this is not suppose to be a surprise, but rather a consistency check. Then for arbitrary rational exponents m/n we can maintain the same properties: first, the efinition is just a m/n = ( [n]a) m One hazar is that, if we want to have only real numbers (as oppose to complex numbers) come up, then we shoul not try to take square roots, 4 th roots, 6 th roots, or any even orer root of negative numbers. For general real exponents x we likewise shoul not try to unerstan a x except for a > 0 or we ll have to use complex numbers (which wouln t be so terrible). But the value of a x can only be efine as a limit: let r, r 2,... be a sequence of rational numbers approaching x, an efine a x = lim i a ri We woul have to check that this efinition oes not accientally epen upon the sequence approaching x (it oesn t), an that the same properties still work (they o). 2

13 The number e is not something that woul come up in really elementary mathematics, because its reason for existence is not really elementary. Anyway, it s approximately e = but if this ever really mattere you have a calculator at your sie, hopefully. With the efinitions in min it is easier to make sense of questions about limits of exponential functions. The two companion issues are to evaluate lim x + ax lim x ax Since we are allowing the exponent x to be real, we better eman that a be a positive real number (if we want to avoi complex numbers, anyway). Then lim x + ax = + if a > if a = 0 if 0 < a < 0 if a > lim x ax = if a = + if 0 < a < To remember which is which, it is sufficient to use 2 for a > an 2 for 0 < a <, an just let x run through positive integers as it goes to +. Likewise, it is sufficient to use 2 for a > an 2 for 0 < a <, an just let x run through negative integers as it goes to. 9. The iea of the erivative of a function First we can tell what the iea of a erivative is. But the issue of computing erivatives is another thing entirely: a person can unerstan the iea without being able to effectively compute, an vice-versa. Suppose that f is a function of interest for some reason. We can give f some sort of geometric life by thinking about the set of points (x, y) so that f(x) = y We woul say that this escribes a curve in the (x, y)-plane. (An sometimes we think of x as moving from left to right, imparting further intuitive or physical content to the story). For some particular number x o, let y o be the value f(x o ) obtaine as output by plugging x o into f as input. Then the point (x o, y o ) is a point on our curve. The tangent line to the curve at the point (x o, y o ) is a line passing through (x o, y o ) an flat against the curve. (As oppose to crossing it at some efinite angle). The iea of the erivative f (x o ) is that it is the slope of the tangent line at x o to the curve. But this isn t the way to compute these things Derivatives of polynomials There are just four simple facts which suffice to take the erivative of any polynomial, an actually of somewhat more general things. 3

14 First, there is the rule for taking the erivative of a power function which takes the nth power of its input. That is, these functions are functions of the form f(x) = x n. The formula is x xn = n x n That is, the exponent comes own to become a coefficient in front of the thing, an the exponent is ecrease by. The secon rule, which is really a special case of this power-function rule, is that erivatives of constants are zero: x c = 0 for any constant c. The thir thing, which reflects the innocuous role of constants in calculus, is that for any functions f of x x c f = c The fourth is that for any two functions f, g of x, the erivative of the sum is the sum of the erivatives: x f (f + g) = x x f + x g Putting these four things together, we can write general formulas like x (axm + bx n + cx p ) = a mx m + b nx n + c px p an so on, with more summans than just the three, if so esire. An in any case here are some examples with numbers instea of letters: x 5x3 = 5 3x 3 = 5x 2 x (3x7 + 5x 3 ) = 3 7x x 2 0 = 2x 6 + 5x 2 x (2 3x2 2x 3 ) = 0 3 2x 2 3x 2 = 6x 6x 2 x ( x4 + 2x 5 + ) = 4x x = 4x 3 + 0x 4 Even if you o catch on to this iea right away, it is wise to practice the technique so that not only can you o it in principle, but also in practice. #0.2 Fin x (3x7 + 5x 3 ) #0.22 Fin x (x2 + 5x 3 + 2) #0.23 Fin x ( x4 + 2x 5 + ) #0.24 Fin x ( 3x2 x 3 ) 4

15 . More general power functions It s important to remember some of the other possibilities for the exponential notation x n. For example x 2 = x x = x x 2 = x an so on. The goo news is that the rule given just above for taking the erivative of powers of x still is correct here, even for exponents which are negative or fractions or even real numbers: Thus, in particular, x xr = r x r x = x x x 2 = 2 x 2 x x = x x = x 2 = x 2 When combine with the sum rule an so on from above, we have the obvious possibilities: x (3x2 7 x + 5 x 2 = x (3x2 7x 2 + 5x 2 ) = 6x 7 2 x 2 0x 3 The possibility of expressing square roots, cube roots, inverses, etc., in terms of exponents is a very important iea in algebra, an can t be overlooke. #.25 Fin x (3x7 + 5 x ) #.26 Fin x ( 2 x x + 3) #.27 Fin x (7 5 x 3 + 5x 7 ) 2. Quotient rule The quotient rule is one of the more irritating an goofy things in elementary calculus, but it just couln t have been any other way. The general principle is x ( ) f = f g g f g g 2 The main hazar is remembering that the numerator is as it is, rather than accientally reversing the roles of f an g, an then being off by ±, which coul be fatal in real life. ( ) = x x 2 x (x 2) x (x 2) 0 (x 2) (x 2) 2 = (x 2) 2 = (x 2) 2 5

16 x ( ) x = (x ) (x 2) (x )(x 2) (x 2) (x ) x 2 (x 2) 2 = (x 2) 2 x = ( 5x 3 ) + x 2 x 7 (x 2) (x ) (x 2) 2 = (x 2) 2 = (5x3 + x) (2 x 7 ) (5x 3 + x) (2 x 7 ) (2 x 7 ) 2 = (5x2 + ) (2 x 7 ) (5x 3 + x) ( 7x 6 ) (2 x 7 ) 2 an there s harly any point in simplifying the last expression, unless someone gives you a goo reason. In general, it s not so easy to see how much may or may not be gaine in simplifying, an we won t make ourselves crazy over it. #2.28 Fin x ( x x 2 ) #2.29 Fin x ( x 2 ) #2.30 Fin x ( x x 2 5 ) #2.3 Fin x ( x3 2+ x ) 3. Prouct Rule Not only will the prouct rule be of use in general an later on, but it s alreay helpful in perhaps unexpecte ways in ealing with polynomials. Anyway, the general rule is x (fg) = f g + fg While this is certainly not as awful as the quotient rule just above, it is not as simple as the rule for sums, which was the goo-souning slogan that the erivative of the sum is the sum of the erivatives. It is not true that the erivative of the prouct is the prouct of the erivatives. Too ba. Still, it s not as ba as the quotient rule. One way that the prouct rule can be useful is in postponing or eliminating a lot of algebra. For example, to evaluate ( (x 3 + x 2 + x + )(x 4 + x 3 + 2x + ) ) x we coul multiply out an then take the erivative term-by-term as we i with several polynomials above. This woul be at least milly irritating because we have to o a bit of algebra. Rather, just apply the prouct rule without feeling compelle first to o any algebra: ( (x 3 + x 2 + x + )(x 4 + x 3 + 2x + ) ) x = (x 3 + x 2 + x + ) (x 4 + x 3 + 2x + ) + (x 3 + x 2 + x + )(x 4 + x 3 + 2x + ) = (3x 2 + 2x + )(x 4 + x 3 + 2x + ) + (x 3 + x 2 + x + )(4x 3 + 3x 2 + 2) 6

17 Now if we were somehow still oblige to multiply out, then we still have to o some algebra. But we can take the erivative without multiplying out, if we want to, by using the prouct rule. For that matter, once we see that there is a choice about oing algebra either before or after we take the erivative, it might be possible to make a choice which minimizes our computational labor. This coul matter. #3.32 Fin x (x3 )(x 6 + x 3 + )) #3.33 Fin x (x2 + x + )(x 4 x 2 + ). #3.34 Fin x (x3 + x 2 + x + )(x 4 + x 2 + )) #3.35 Fin x (x3 + x 2 + x + )(2x + x)) 4. Chain rule The chain rule is subtler than the previous rules, so if it seems trickier to you, then you re right. OK. But it is absolutely inispensable in general an later, an alreay is very helpful in ealing with polynomials. The general assertion may be a little har to fathom because it is of a ifferent nature than the previous ones. For one thing, now we will be talking about a composite function instea of just aing or multiplying functions in a more orinary way. So, for two functions f an g, x ((f(g(x))) = f (g(x)) g (x) There is also the stanar notation (f g)(x) = f(g(x)) for this composite function, but using this notation oesn t accomplish so very much. A problem in successful use of the chain rule is that often it requires a little thought to recognize that some formula is (or can be looke at as) a composite function. An the very nature of the chain rule picks on weaknesses in our unerstaning of the notation. For example, the function F (x) = ( + x 2 ) 00 is really obtaine by first using x as input to the function which squares an as to its input. Then the result of that is use as input to the function which takes the 00th power. It is necessary to think about it this way or we ll make a mistake. The erivative is evaluate as x ( + x2 ) 00 = 00( + x 2 ) 99 2x To see that this is a special case of the general formula, we nee to see what correspons to the f an g in the general formula. Specifically, let f(input) = (input) 00 g(input) = + (input) 2 The reason for writing input an not x for the moment is to avoi a certain kin of mistake. But we can compute that f (input) = 00(input) 99 g (input) = 2(input) 7

18 The hazar here is that the input to f is not x, but rather is g(x). So the general formula gives x ( + x2 ) 00 = f (g(x)) g (x) = 00g(x) 99 2x = 00( + x 2 ) 99 2x More examples: 3x + 2 = x x (3x + 2)/2 = 2 (3x + 2) /2 3 It is very important to recognize situations like for any constants a, b, n. An, of course, this inclues x (3x5 x + 4) = (3x 5 x + 4) 0 (5x 4 ) x (ax + b)n = n(ax + b) n a ax + b = x 2 (ax + b) /2 a x ax + b = (ax + a b) 2 a = (ax + b) 2 Of course, this iea can be combine with polynomials, quotients, an proucts to give enormous an excruciating things where we nee to use the chain rule, the quotient rule, the prouct rule, etc., an possibly several times each. But this is not har, merely teious, since the only things we really o come in small steps. For example: x by the quotient rule, which is then ( ) + x + 2 ( + 7x) 33 = ( + x + 2) ( + 7x) 33 ( + x + 2) (( + 7x) 33 ) (( + 7x) 33 ) 2 ( 2 (x + 2) /2 ) ( + 7x) 33 ( + x + 2) (( + 7x) 33 ) (( + 7x) 33 ) 2 because our observations just above (chain rule!) tell us that x + 2 = x 2 (x + 2) /2 (x + 2) = (x + 2) /2 2 Then we use the chain rule again to take the erivative of that big power of + 7x, so the whole thing becomes ( 2 (x + 2) /2 ) ( + 7x) 33 ( + x + 2) (33( + 7x) 32 7) (( + 7x) 33 ) 2 Although we coul simplify a bit here, let s not. The point about having to o several things in a row to take a erivative is pretty clear without oing algebra just now. #4.36 Fin x (( x2 ) 00 ) #4.37 Fin x x 3 8

19 #4.38 Fin x (x2 x 2 3) #4.39 Fin x ( x 2 + x + ) #4.40 Fin x ( 3 x 3 + x 2 + x + ) #4.4 Fin x ((x3 + x + ) 0 ) 5. Tangent an Normal Lines One funamental interpretation of the erivative of a function is that it is the slope of the tangent line to the graph of the function. (Still, it is important to realize that this is not the efinition of the thing, an that there are other possible an important interpretations as well). The precise statement of this funamental iea is as follows. Let f be a function. For each fixe value x o of the input to f, the value f (x o ) of the erivative f of f evaluate at x o is the slope of the tangent line to the graph of f at the particular point (x o, f(x o )) on the graph. Recall the point-slope form of a line with slope m through a point (x o, y o ): y y o = m(x x o ) In the present context, the slope is f (x o ) an the point is (x o, f(x o )), so the equation of the tangent line to the graph of f at (x o, f(x o )) is y f(x o ) = f (x o )(x x o ) The normal line to a curve at a particular point is the line through that point an perpenicular to the tangent. A person might remember from analytic geometry that the slope of any line perpenicular to a line with slope m is the negative reciprocal /m. Thus, just changing this aspect of the equation for the tangent line, we can say generally that the equation of the normal line to the graph of f at (x o, f(x o )) is y f(x o ) = f (x o ) (x x o) The main conceptual hazar is to mistakenly name the fixe point x, as well as naming the variable coorinate on the tangent line x. This causes a person to write own some equation which, whatever it may be, is not the equation of a line at all. Another popular boo-boo is to forget the subtraction f(x o ) on the left han sie. Don t o it. So, as the simplest example: let s write the equation for the tangent line to the curve y = x 2 at the point where x = 3. The erivative of the function is y = 2x, which has value 2 3 = 6 when x = 3. An the value of the function is 3 3 = 9 when x = 3. Thus, the tangent line at that point is The normal line at the point where x = 3 is y 9 = 6(x 3) y 9 = (x 3) 6 So the question of fining the tangent an normal lines at various points of the graph of a function is just a combination of the two processes: computing the erivative at the point in question, an invoking the point-slope form of the equation for a straight line. 9

20 #5.42 Write the equation for both the tangent line an normal line to the curve y = 3x 2 x + at the point where x =. #5.43 Write the equation for both the tangent line an normal line to the curve y = (x )/(x + ) at the point where x = Critical points, monotone increase an ecrease A function is calle increasing if it increases as the input x moves from left to right, an is calle ecreasing if it ecreases as x moves from left to right. Of course, a function can be increasing in some places an ecreasing in others: that s the complication. We can notice that a function is increasing if the slope of its tangent is positive, an ecreasing if the slope of its tangent is negative. Continuing with the iea that the slope of the tangent is the erivative: a function is increasing where its erivative is positive, an is ecreasing where its erivative is negative. This is a great principle, because we on t have to graph the function or otherwise list lots of values to figure out where it s increasing an ecreasing. If anything, it shoul be a big help in graphing to know in avance where the graph goes up an where it goes own. An the points where the tangent line is horizontal, that is, where the erivative is zero, are critical points. The points where the graph has a peak or a trough will certainly lie among the critical points, although there are other possibilities for critical points, as well. Further, for the kin of functions we ll eal with here, there is a fairly systematic way to get all this information: to fin the intervals of increase an ecrease of a function f: Compute the erivative f of f, an solve the equation f (x) = 0 for x to fin all the critical points, which we list in orer as x < x 2 <... < x n. (If there are points of iscontinuity or non-ifferentiability, these points shoul be ae to the list! But points of iscontinuity or non-ifferentiability are not calle critical points.) We nee some auxiliary points: To the left of the leftmost critical point x pick any convenient point t o, between each pair of consecutive critical points x i, x i+ choose any convenient point t i, an to the right of the rightmost critical point x n choose a convenient point t n. Evaluate the erivative f at all the auxiliary points t i. Conclusion: if f (t i+ ) > 0, then f is increasing on (x i, x i+ ), while if f (t i+ ) < 0, then f is ecreasing on that interval. Conclusion: on the outsie interval (, x o ), the function f is increasing if f (t o ) > 0 an is ecreasing if f (t o ) < 0. Similarly, on (x n, ), the function f is increasing if f (t n ) > 0 an is ecreasing if f (t n ) < 0. It is certainly true that there are many possible shortcuts to this proceure, especially for polynomials of low egree or other rather special functions. However, if you are able to quickly compute values of (erivatives of!) functions on your calculator, you may as well use this proceure as any other. Exactly which auxiliary points we choose oes not matter, as long as they fall in the correct intervals, since we just nee a single sample on each interval to fin out whether f is positive or negative there. Usually we pick integers or some other kin of number to make computation of the erivative there as easy as possible. 20

21 It s important to realize that even if a question oes not irectly ask for critical points, an maybe oes not ask about intervals either, still it is implicit that we have to fin the critical points an see whether the functions is increasing or ecreasing on the intervals between critical points. Examples: Fin the critical points an intervals on which f(x) = x 2 + 2x + 9 is increasing an ecreasing: Compute f (x) = 2x + 2. Solve 2x + 2 = 0 to fin only one critical point. To the left of let s use the auxiliary point t o = 2 an to the right use t = 0. Then f ( 2) = 2 < 0, so f is ecreasing on the interval (, ). An f (0) = 2 > 0, so f is increasing on the interval (, ). Fin the critical points an intervals on which f(x) = x 3 2x + 3 is increasing, ecreasing. Compute f (x) = 3x 2 2. Solve 3x 2 2 = 0: this simplifies to x 2 4 = 0, so the critical points are ±2. To the left of 2 choose auxiliary point t o = 3, between 2 an = 2 choose auxiliary point t = 0, an to the right of +2 choose t 2 = 3. Plugging in the auxiliary points to the erivative, we fin that f ( 3) = 27 2 > 0, so f is increasing on (, 2). Since f (0) = 2 < 0, f is ecreasing on ( 2, +2), an since f (3) = 27 2 > 0, f is increasing on (2, ). Notice too that we on t really nee to know the exact value of the erivative at the auxiliary points: all we care about is whether the erivative is positive or negative. The point is that sometimes some teious computation can be avoie by stopping as soon as it becomes clear whether the erivative is positive or negative. #6.44 Fin the critical points an intervals on which f(x) = x 2 + 2x + 9 is increasing, ecreasing. #6.45 Fin the critical points an intervals on which f(x) = 3x 2 6x + 7 is increasing, ecreasing. #6.46 Fin the critical points an intervals on which f(x) = x 3 2x + 3 is increasing, ecreasing. 7. Minimization an Maximization The funamental iea which makes calculus useful in unerstaning problems of maximizing an minimizing things is that at a peak of the graph of a function, or at the bottom of a trough, the tangent is horizontal. That is, the erivative f (x o ) is 0 at points x o at which f(x o ) is a maximum or a minimum. Well, a little sharpening of this is necessary: sometimes for either natural or artificial reasons the variable x is restricte to some interval [a, b]. In that case, we can say that the maximum an minimum values of f on the interval [a, b] occur among the list of critical points an enpoints of the interval. An, if there are points where f is not ifferentiable, or is iscontinuous, then these have to be ae in, too. But let s stick with the basic iea, an just ignore some of these complications. Let s escribe a systematic proceure to fin the minimum an maximum values of a function f on an interval [a, b]. Solve f (x) = 0 to fin the list of critical points of f. Exclue any critical points not insie the interval [a, b]. A to the list the enpoints a, b of the interval (an any points of iscontinuity or non-ifferentiability!) At each point on the list, evaluate the function f: the biggest number that occurs is the maximum, an the littlest number that occurs is the minimum. Fin the minima an maxima of the function f(x) = x 4 8x on the interval [, 3]. First, take the erivative an set it equal to zero to solve for critical points: this is 4x 3 6x = 0 2

22 or, more simply, iviing by 4, it is x 3 4x = 0. Luckily, we can see how to factor this: it is x(x 2)(x + 2) So the critical points are 2, 0, +2. Since the interval oes not inclue 2, we rop it from our list. An we a to the list the enpoints, 3. So the list of numbers to consier as potential spots for minima an maxima are, 0, 2, 3. Plugging these numbers into the function, we get (in that orer) 2, 5,, 4. Therefore, the maximum is 4, which occurs at x = 3, an the minimum is, which occurs at x = 2. Notice that in the previous example the maximum i not occur at a critical point, but by coincience i occur at an enpoint. You have 200 feet of fencing with which you wish to enclose the largest possible rectangular garen. What is the largest garen you can have? Let x be the length of the garen, an y the with. Then the area is simply xy. Since the perimeter is 200, we know that 2x + 2y = 200, which we can solve to express y as a function of x: we fin that y = 00 x. Now we can rewrite the area as a function of x alone, which sets us up to execute our proceure: area = xy = x(00 x) The erivative of this function with respect to x is 00 2x. Setting this equal to 0 gives the equation 00 2x = 0 to solve for critical points: we fin just one, namely x = 50. Now what about enpoints? What is the interval? In this example we must look at physical consierations to figure out what interval x is restricte to. Certainly a with must be a positive number, so x > 0 an y > 0. Since y = 00 x, the inequality on y gives another inequality on x, namely that x < 00. So x is in [0, 00]. When we plug the values 0, 50, 00 into the function x(00 x), we get 0, 2500, 0, in that orer. Thus, the corresponing value of y is = 50, an the maximal possible area is = #7.47 Olivia has 200 feet of fencing with which she wishes to enclose the largest possible rectangular garen. What is the largest garen she can have? #7.48 Fin the minima an maxima of the function f(x) = 3x 4 4x on the interval [ 2, 3]. #7.49 The cost per hour of fuel to run a locomotive is v 2 /25 ollars, where v is spee, an other costs are $00 per hour regarless of spee. What is the spee that minimizes cost per mile? #7.50 The prouct of two numbers x, y is 6. We know x an y. What is the greatest possible sum of the two numbers? #7.5 Fin both the minimum an the maximum of the function f(x) = x 3 + 3x + on the interval [ 2, 2]. 8. Local minima an maxima (First Derivative Test) A function f has a local maximum or relative maximum at a point x o if the values f(x) of f for x near x o are all less than f(x o ). Thus, the graph of f near x o has a peak at x o. A function f has a local minimum or relative minimum at a point x o if the values f(x) of f for x near x o are all greater 22

23 than f(x o ). Thus, the graph of f near x o has a trough at x o. (To make the istinction clear, sometimes the plain maximum an minimum are calle absolute maximum an minimum.) Yes, in both these efinitions we are tolerating ambiguity about what near woul mean, although the peak/trough requirement on the graph coul be translate into a less ambiguous efinition. But in any case we ll be able to execute the proceure given below to fin local maxima an minima without worrying over a formal efinition. This proceure is just a variant of things we ve alreay one to analyze the intervals of increase an ecrease of a function, or to fin absolute maxima an minima. This proceure starts out the same way as oes the analysis of intervals of increase/ecrease, an also the proceure for fining ( absolute ) maxima an minima of functions. To fin the local maxima an minima of a function f on an interval [a, b]: Solve f (x) = 0 to fin critical points of f. Drop from the list any critical points that aren t in the interval [a, b]. A to the list the enpoints (an any points of iscontinuity or non-ifferentiability): we have an orere list of special points in the interval: a = x o < x <... < x n = b Between each pair x i < x i+ of points in the list, choose an auxiliary point t i+. Evaluate the erivative f at all the auxiliary points. For each critical point x i, we have the auxiliary points to each sie of it: t i < x i < t i+. There are four cases best remembere by rawing a picture!: if f (t i ) > 0 an f (t i+ ) < 0 (so f is increasing to the left of x i an ecreasing to the right of x i, then f has a local maximum at x o. if f (t i ) < 0 an f (t i+ ) > 0 (so f is ecreasing to the left of x i an increasing to the right of x i, then f has a local minimum at x o. if f (t i ) < 0 an f (t i+ ) < 0 (so f is ecreasing to the left of x i an also ecreasing to the right of x i, then f has neither a local maximum nor a local minimum at x o. if f (t i ) > 0 an f (t i+ ) > 0 (so f is increasing to the left of x i an also increasing to the right of x i, then f has neither a local maximum nor a local minimum at x o. The enpoints require separate treatment: There is the auxiliary point t o just to the right of the left enpoint a, an the auxiliary point t n just to the left of the right enpoint b: At the left enpoint a, if f (t o ) < 0 (so f is ecreasing to the right of a) then a is a local maximum. At the left enpoint a, if f (t o ) > (so f is increasing to the right of a) then a is a local minimum. At the right enpoint b, if f (t n ) < 0 (so f is ecreasing as b is approache from the left) then b is a local minimum. At the right enpoint b, if f (t n ) > (so f is increasing as b is approache from the left) then b is a local maximum. The possibly bewilering list of possibilities really shouln t be bewilering after you get use to them. We are alreay acquainte with evaluation of f at auxiliary points between critical points in orer to see whether the function is increasing or ecreasing, an now we re just applying that information to see whether the graph peaks, troughs, or oes neither aroun each critical point an enpoints. That is, the geometric meaning of the erivative s being positive or negative is easily translate into conclusions about local maxima or minima. Fin all the local (=relative) minima an maxima of the function f(x) = 2x 3 9x 2 + on the interval [ 2, 2]: To fin critical points, solve f (x) = 0: this is 6x 2 8x = 0 or x(x 3) = 0, so there are two critical points, 0 an 3. Since 3 is not in the interval we care about, we rop it from our list. Aing the enpoints to the list, we have 2 < 0 < 2 23

24 as our orere list of special points. Let s use auxiliary points,. At the erivative is f ( ) = 24 > 0, so the function is increasing there. At + the erivative is f () = 2 < 0, so the function is ecreasing. Thus, since it is increasing to the left an ecreasing to the right of 0, it must be that 0 is a local maximum. Since f is increasing to the right of the left enpoint 2, that left enpoint must give a local minimum. Since it is ecreasing to the left of the right enpoint +2, the right enpoint must be a local minimum. Notice that although the processes of fining absolute maxima an minima an local maxima an minima have a lot in common, they have essential ifferences. In particular, the only relations between them are that critical points an enpoints (an points of iscontinuity, etc.) play a big role in both, an that the absolute maximum is certainly a local maximum, an likewise the absolute minimum is certainly a local minimum. For example, just plugging critical points into the function oes not reliably inicate which points are local maxima an minima. An, on the other han, knowing which of the critical points are local maxima an minima generally is only a small step towar figuring out which are absolute: values still have to be plugge into the funciton! So on t confuse the two proceures! (By the way: while it s fairly easy to make up story-problems where the issue is to fin the maximum or minimum value of some function on some interval, it s harer to think of a simple application of local maxima or minima). #8.52 Fin all the local (=relative) minima an maxima of the function f(x) = (x + ) 3 3(x + ) on the interval [ 2, ]. #8.53 Fin the local (=relative) minima an maxima on the interval [ 3, 2] of the function f(x) = (x + ) 3 3(x + ). #8.54 Fin the local (relative) minima an maxima of the function f(x) = 2x + x 3 on the interval [ 3, 3]. #8.55 Fin the local (relative) minima an maxima of the function f(x) = 3x 4 8x 3 + 6x on the interval [ 3, 3]. 9. An algebra trick The algebra trick here goes back at least 350 years. This is worth looking at if only as an aitional review of algebra, but is actually of consierable value in a variety of han computations as well. The algebraic ientity we use here starts with a prouct of factors each of which may occur with a fractional or negative exponent. For example, with 3 such factors: f(x) = (x a) k (x b) l (x c) m The erivative can be compute by using the prouct rule twice: f (x) = = k(x a) k (x b) l (x c) m + (x a) k l(x b) l (x c) m + (x a) k (x b) l m(x c) m Now all three summans here have a common factor of (x a) k (x b) l (x c) m 24

25 which we can take out, using the istributive law in reverse: we have f (x) = = (x a) k (x b) l (x c) m [k(x b)(x c) + l(x a)(x c) + m(x a)(x b)] The minor miracle is that the big expression insie the square brackets is a mere quaratic polynomial in x. Then to etermine critical points we have to figure out the roots of the equation f (x) = 0: If k > 0 then x = a is a critical point, if k 0 it isn t. If l > 0 then x = b is a critical point, if l 0 it isn t. If m > 0 then x = c is a critical point, if m 0 it isn t. An, last but not least, the two roots of the quaratic equation are critical points. k(x b)(x c) + l(x a)(x c) + m(x a)(x b) = 0 There is also another issue here, about not wanting to take square roots (an so on) of negative numbers. We woul exclue from the omain of the function any values of x which woul make us try to take a square root of a negative number. But this might also force us to give up some critical points! Still, this is not the main point here, so we will o examples which avoi this aitional worry. A very simple numerical example: suppose we are to fin the critical points of the function f(x) = x 5/2 (x ) 4/3 Implicitly, we have to fin the critical points first. We compute the erivative by using the prouct rule, the power function rule, an a tiny bit of chain rule: f (x) = 5 2 x3/2 (x ) 4/3 + x 5/2 4 (x )/3 3 An now solve this for x? It s not at all a polynomial, an it is a little ugly. But our algebra trick transforms this issue into something as simple as solving a linear equation: first figure out the largest power of x that occurs in all the terms: it is x 3/2, since x 5/2 occurs in the first term an x 3/2 in the secon. The largest power of x that occurs in all the terms is (x ) /3, since (x ) 4/3 occurs in the first, an (x ) /3 in the secon. Taking these common factors out (using the istributive law backwar ), we rearrange to f (x) = 5 2 x3/2 (x ) 4/3 + x 5/2 4 (x )/3 3 ( 5 = x 3/2 (x ) /3 2 (x ) + 4 ) 3 x = x 3/2 (x ) /3 ( 5 2 x x) = x 3/2 (x ) /3 ( 23 6 x 5 2 ) Now to see when this is 0 is not so har: first, since the power of x appearing in front is positive, x = 0 make this expression 0. Secon, since the power of x + appearing in front is positive, if x = 0 then the whole expression is 0. Thir, an perhaps unexpectely, from the simplifie form of the complicate factor, if 23 6 x 5 2 = 0 then the whole expression is 0, as well. So, altogether, the critical points woul appear to be x = 0, 5 23, 25

26 Many people woul overlook the critical point 5 23, which is visible only after the algebra we i. #9.56 Fin the critical points an intervals of increase an ecrease of f(x) = x 0 (x ) 2. #9.57 Fin the critical points an intervals of increase an ecrease of f(x) = x 0 (x 2) (x + 2) 3. #9.58 Fin the critical points an intervals of increase an ecrease of f(x) = x 5/3 (x + ) 6/5. #9.59 Fin the critical points an intervals of increase an ecrease of f(x) = x /2 (x + ) 4/3 (x ) / Linear approximations: approximation by ifferentials The iea here in geometric terms is that in some vague sense a curve line can be approximate by a straight line tangent to it. Of course, this approximation is only goo at all near the point of tangency, an so on. So the only formula here is secretly the formula for the tangent line to the graph of a function. There is some hassle ue to the fact that there are so many ifferent choices of symbols to write it. We can write some formulas: Let f be a function, an fix a point x o. The iea is that for x near x o we have an approximate equality f(x) f(x o ) + f (x o )(x x o ) We o not attempt to clarify what either near or approximate mean in this context. What is really true here is that for a given value x, the quantity f(x o ) + f (x o )(x x o ) is exactly the y-coorinate of the line tangent to the graph at x o The aproximation statement has many paraphrases in varying choices of symbols, an a person nees to be able to recognize all of them. For example, one of the more traitional paraphrases, which introuces some slightly silly but oh-so-traitional notation, is the following one. We might also say that y is a function of x given by y = f(x). Let x = small change in x y = corresponing change in y = f(x + x) f(x) Then the assertion is that y f (x) x Sometimes some texts introuce the following questionable (but traitionally popular!) notation: y = f (x) x = approximation to change in y x = x an call the x an y ifferentials. An then this whole proceure is approximation by ifferentials. A not particularly enlightening paraphrase, using the previous notation, is y y Even though you may see people writing this, on t o it. More paraphrases, with varying symbols: f(x + x) f(x) + f (x) x 26

27 f(x + δ) f(x) + f (x)δ f(x + h) f(x) + f (x)h f(x + x) f(x) f (x) x y + y f(x) + f (x) x y f (x) x A little history: Until just 20 or 30 years ago, calculators were not wiely available, an especially not typically able to evaluate trigonometric, exponential, an logarithm functions. In that context, the kin of vague an unreliable approximation furnishe by ifferentials was certainly worthwhile in many situations. By contrast, now that pretty sophisticate calculators are wiely available, some things that once seeme sensible are no longer. For example, a very traitional type of question is to approximate 0 by ifferentials. A reasonable contemporary response woul be to simply punch in, 0, on your calculator an get the answer immeiately to 0 ecimal places. But this was possible only relatively recently. Example: For example let s approximate 7 by ifferentials. For this problem to make sense at all imagine that you have no calculator. We take f(x) = x = x /2. The iea here is that we can easily evaluate by han both f an f at the point x = 6 which is near 7. (Here f (x) = 2 x /2 ). Thus, here x = 7 6 = an 7 = f(7) f(6) + f (6) x = 6 + = Example: Similarly, if we wante to approximate 8 by ifferentials, we again take f(x) = x = x /2. Still we imagine that we are oing this by han, an then of course we can easily evaluate the function f an its erivative f at the point x = 6 which is near 8. Thus, here x = 8 6 = 2 an 8 = f(8) f(6) + f (6) x = = Why not use the goo point 25 as the nearby point to fin 8? Well, in broa terms, the further away your goo point is, the worse the approximation will be. Yes, it is true that we have little iea how goo or ba the approximation is anyway. It is somewhat more sensible to not use this iea for numerical work, but rather to say things like an x + x + 2 x x + h x + h 2 x This kin of assertion is more than any particular numerical example woul give, because it gives a relationship, telling how much the output changes for given change in input, epening what regime (=interval) the input is generally in. In this example, we can make the qualitative observation that as x increases the ifference x + x ecreases. Example: Another numerical example: Approximate sin 3 o by ifferentials. Again, the point is not to hit 3,, sin on your calculator (after switching to egrees), but rather to imagine that you have no calculator. An we are suppose to remember from pre-calculator ays the special angles an the values of 27

28 trig functions at them: sin 30 o = 2 an cos 30o = 3 2. So we use the function f(x) = sin x, an we imagine that we can evaluate f an f easily by han at 30 o. Then x = 3 o 30 o = o = o 2π raians 360 o = 2π 360 raians We have to rewrite things in raians since we really only can compute erivatives of trig functions in raians. Yes, this is a complication in our suppose computation by han. Anyway, we have sin 3 o = f(3 o ) = f(30 o ) + f (30 o ) x = sin 30 o + cos 30 o 2π 360 = Eviently we are to also imagine that we know or can easily fin 3 (by ifferentials?) as well as a value of π. Yes, this is a lot of trouble in comparison to just punching the buttons, an from a contemporary perspective may seem senseless. Example: Approximate ln(x + 2) by ifferentials, in terms of ln x an x: This non-numerical question is somewhat more sensible. Take f(x) = ln x, so that f (x) = x. Then an by the formulas above 2π 360 x = (x + 2) x = 2 ln(x + 2) = f(x + 2) f(x) + f (x) 2 = ln x + 2 x Example: Approximate ln (e + 2) in terms of ifferentials: Use f(x) = ln x again, so f (x) = x. We probably have to imagine that we can easily evaluate both ln x an x at x = e. (Do we know a numerical approximation to e?). Now x = (e + 2) e = 2 so we have ln(e + 2) = f(e + 2) f(e) + f (e) 2 = ln e + 2 e = + 2 e since ln e =. #20.60 Approximate 0 by ifferentials in terms of 00 = 0. #20.6 Approximate x + by ifferentials, in terms of x. #20.62 Granting that x ln x = x, approximate ln(x + ) by ifferentials, in terms of ln x an x. #20.63 Granting that x ex = e x, approximate e x+ in terms of e x. #20.64 Granting that x cos x = sin x, approximate cos(x + ) in terms of cos x an sin x. 2. Implicit ifferentiation There is nothing implicit about the ifferentiation we o here, it is quite explicit. The ifference from earlier situations is that we have a function efine implicitly. What this means is that, instea of a clear-cut (if complicate) formula for the value of the function in terms of the input value, we only have a relation between the two. This is best illustrate by examples. 28

29 For example, suppose that y is a function of x an y 5 xy + x 5 = an we are to fin some useful expression for y/x. Notice that it is not likely that we be able to solve this equation for y as a function of x (nor vice-versa, either), so our previous methos o not obviously o anything here! But both sies of that equality are functions of x, an are equal, so their erivatives are equal, surely. That is, 5y 4 y x y x y x + 5x4 = 0 Here the trick is that we can take the erivative without knowing exactly what y is as a function of x, but just using the rules for ifferentiation. Specifically, to take the erivative of the term y 5, we view this as a composite function, obtaine by applying the take-the-fifth-power function after applying the (not clearly known!) function y. Then use the chain rule! Likewise, to ifferentiate the term xy, we use the prouct rule x (x y) = x x y + x y x = y + x y x since, after all, x x = An the term x 5 is easy to ifferentiate, obtaining the 5x 4. The other sie of the equation, the function, is constant, so its erivative is 0. (The fact that this means that the left-han sie is also constant shoul not be mis-use: we nee to use the very non-trivial looking expression we have for that constant function, there on the left-han sie of that equation!). Now the amazing part is that this equation can be solve for y, if we tolerate a formula involving not only x, but also y: first, regroup terms epening on whether they have a y or not: Then move the non-y terms to the other sie y (5y 4 x) + ( y + 5x 4 ) = 0 y (5y 4 x) = y 5x 4 an ivie by the coefficient of the y : y = y 5x4 5y 4 x Yes, this is not as goo as if there were a formula for y not neeing the y. But, on the other han, the initial situation we ha i not present us with a formula for y in terms of x, so it was necessary to lower our expectations. Yes, if we are given a value of x an tol to fin the corresponing y, it woul be impossible without luck or some aitional information. For example, in the case we just looke at, if we were aske to fin y when x = an y =, it s easy: just plug these values into the formula for y in terms of both x an y: when x = an y =, the corresponing value of y is y = = 4/4 = 29

30 If, instea, we were aske to fin y an y when x =, not knowing in avance that y = fits into the equation when x =, we have to hope for some luck. First, we have to try to solve the original equation for y with x replace by its value : solve y 5 y + = By luck inee, there is some cancellation, an the equation becomes By further luck, we can factor this by han : it is y 5 y = 0 0 = y(y 4 ) = y(y 2 )(y 2 + ) = y(y )(y + )(y 2 + ) So there are actually three real numbers which work as y for x = : the values, 0, +. There is no clear way to see which is best. But in any case, any one of these three values coul be use as y in substituting into the formula y = y 5x4 5y 4 x we obtaine above. Yes, there are really three solutions, three functions, etc. Note that we coul have use the Intermeiate Value Theorem an/or Newton s Metho to numerically solve the equation, even without too much luck. In real life a person shoul be prepare to o such things. #2.65 Suppose that y is a function of x an Fin y/x at the point x =, y = 0. #2.66 Suppose that y is a function of x an Fin y/x at the point x =, y = 2. Fin 2 y x 2 y 5 xy + x 5 = y 3 xy 2 + x 2 y + x 5 = 7 at that point. 22. Relate rates In this section, most functions will be functions of a parameter t which we will think of as time. There is a convention coming from physics to write the erivative of any function y of t as ẏ = y/t, that is, with just a ot over the functions, rather than a prime. The issues here are variants an continuations of the previous section s iea about implicit ifferentiation. Traitionally, there are other (non-calculus!) issues introuce at this point, involving both story-problem stuff as well as requirement to be able to eal with similar triangles, the Pythagorean Theorem, an to recall formulas for volumes of cones an such. Continuing with the iea of escribing a function by a relation, we coul have two unknown functions x an y of t, relate by some formula such as x 2 + y 2 = 25 30

31 A typical question of this genre is What is ẏ when x = 4 an ẋ = 6? The funamental rule of thumb in this kin of situation is ifferentiate the relation with respect to t: so we ifferentiate the relation x 2 +y 2 = 25 with respect to t, even though we on t know any etails about those two function x an y of t: 2xẋ + 2yẏ = 0 using the chain rule. We can solve this for ẏ: ẏ = xẋ y So at any particular moment, if we knew the values of x, ẋ, y then we coul fin ẏ at that moment. Here it s easy to solve the original relation to fin y when x = 4: we get y = ±3. Substituting, we get ẏ = 4 6 ±3 = ±8 (The ± notation means that we take + chosen if we take y = 3 an if we take y = +3). #22.67 Suppose that x, y are both functions of t, an that x 2 + y 2 = 25. Express x t y y x t. When x = 3 an y = 4 an t = 6, what is t? in terms of x, y, an #22.68 A 2-foot tall og is walking away from a streetlight which is on a 0-foot pole. At a certain moment, the tip of the og s shaow is moving away from the streetlight at 5 feet per secon. How fast is the og walking at that moment? #22.69 A laer 3 feet long leans against a house, but is sliing own. How fast is the top of the laer moving at a moment when the base of the laer is 2 feet from the house an moving outwar at 0 feet per secon? 23. Intermeiate Value Theorem, location of roots The assertion of the Intermeiate Value Theorem is something which is probably intuitively obvious, an is also provably true: if a function f is continuous on an interval [a, b] an if f(a) < 0 an f(b) > 0 (or vice-versa), then there is some thir point c with a < c < b so that f(c) = 0. This result has many relatively theoretical uses, but for our purposes can be use to give a crue but simple way to locate the roots of functions. There is a lot of guessing, or trial-an-error, involve here, but that is fair. Again, in this situation, it is to our avantage if we are reasonably proficient in using a calculator to o simple tasks like evaluating polynomials! If this approach to estimating roots is taken to its logical conclusion, it is calle the metho of interval bisection, for a reason we ll see below. We will not pursue this metho very far, because there are better methos to use once we have invoke this just to get going. For example, we probably on t know a formula to solve the cubic equation x 3 x + = 0 But the function f(x) = x 3 x + is certainly continuous, so we can invoke the Intermeiate Value Theorem as much as we like. For example, f(2) = 7 > 0 an f( 2) = 5 < 0, so we know that there is a root in the interval [ 2, 2]. We like to cut own the size of the interval, so we look at what happens at the mipoint, bisecting the interval [ 2, 2]: we have f(0) = > 0. Therefore, since f( 2) = 5 < 0, we can conclue that there is a root in [ 2, 0]. Since both f(0) > 0 an f(2) > 0, we can t say anything at this point about whether or not there are roots in [0, 2]. Again bisecting the interval [ 2, 0] where we 3

32 know there is a root, we compute f( ) = > 0. Thus, since f( 2) < 0, we know that there is a root in [ 2, ] (an have no information about [, 0]). If we continue with this metho, we can obtain as goo an approximation as we want! But there are faster ways to get a really goo approximation, as we ll see. Unless a person has an amazing intuition for polynomials (or whatever), there is really no way to anticipate what guess is better than any other in getting starte. Invoke the Intermeiate Value Theorem to fin an interval of length or less in which there is a root of x 3 + x + 3 = 0: Let f(x) = x 3 + x + 3. Just, guessing, we compute f(0) = 3 > 0. Realizing that the x 3 term probably ominates f when x is large positive or large negative, an since we want to fin a point where f is negative, our next guess will be a large negative number: how about? Well, f( ) = > 0, so eviently is not negative enough. How about 2? Well, f( 2) = 7 < 0, so we have succeee. Further, the faile guess actually was worthwhile, since now we know that f( 2) < 0 an f( ) > 0. Then, invoking the Intermeiate Value Theorem, there is a root in the interval [ 2, ]. Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its egree. We can use the Intermeiate Value Theorem to get an iea where all of them are. Invoke the Intermeiate Value Theorem to fin three ifferent intervals of length or less in each of which there is a root of x 3 4x + = 0: first, just starting anywhere, f(0) = > 0. Next, f() = 2 < 0. So, since f(0) > 0 an f() < 0, there is at least one root in [0, ], by the Intermeiate Value Theorem. Next, f(2) = > 0. So, with some luck here, since f() < 0 an f(2) > 0, by the Intermeiate Value Theorem there is a root in [, 2]. Now if we somehow imagine that there is a negative root as well, then we try : f( ) = 4 > 0. So we know nothing about roots in [, 0]. But continue: f( 2) = > 0, an still no new conclusion. Continue: f( 3) = 4 < 0. Aha! So since f( 3) < 0 an f(2) > 0, by the Intermeiate Value Theorem there is a thir root in the interval [ 3, 2]. Notice how even the ba guesses were not entirely waste. 24. Newton s metho This is a metho which, once you get starte, quickly gives a very goo approximation to a root of polynomial (an other) equations. The iea is that, if x o is not a root of a polynomial equation, but is pretty close to a root, then sliing own the tangent line at x o to the graph of f gives a goo approximation to the actual root. The point is that this process can be repeate as much as necessary to give as goo an approximation as you want. Let s erive the relevant formula: if our blin guess for a root of f is x o, then the tangent line is y f(x o ) = f (x o )(x x o ) Sliing own the tangent line to hit the x-axis means to fin the intersection of this line with the x-axis: this is where y = 0. Thus, we solve for x in 0 f(x o ) = f (x o )(x x o ) to fin x = x o f(x o) f (x o ) Well, let s call this first serious guess x. Then, repeating this process, the secon serious guess woul be x 2 = x f(x ) f (x ) 32

33 an generally, if we have the nth guess x n then the n + th guess x n+ is x n+ = x n f(x n) f (x n ) OK, that s the formula for improving our guesses. How o we ecie when to quit? Well, it epens upon to how many ecimal places we want our approximation to be goo. Basically, if we want, for example, 3 ecimal places accuracy, then as soon as x n an x n+ agree to three ecimal places, we can presume that those are the true ecimals of the true root of the equation. This will be illustrate in the examples below. It is important to realize that there is some uncertainty in Newton s metho, both because it alone cannot assure that we have a root, an also because the iea just escribe for approximating roots to a given accuracy is not foolproof. But to worry about what coul go wrong here is counter-prouctive. Approximate a root of x 3 x + = 0 using the intermeiate value theorem to get starte, an then Newton s metho: First let s see what happens if we are a little foolish here, in terms of the blin guess we start with. If we ignore the avice about using the intermeiate value theorem to guarantee a root in some known interval, we ll waste time. Let s see: The general formula becomes x n+ = x n f(x n) f (x n ) x n+ = x n x3 x + 3x 2 If we take x = as our blin guess, then plugging into the formula gives x 2 = 0.5 x 3 = 3 x 4 = This is iscouraging, since the numbers are jumping aroun somewhat. But if we are stubborn an can compute quickly with a calculator (not by han!), we see what happens: x 5 = x 6 = x 7 = x 8 = x 9 = x 0 = x = x 2 = x 3 = x 4 = x 5 = x 6 = x 7 = x 8 = x 9 = x 20 = x 2 =

34 Well, after quite a few iterations of sliing own the tangent, the last two numbers we got, x 20 an x 2, agree to 5 ecimal places. This woul make us think that the true root is approximate to five ecimal places by The stupi aspect of this little scenario was that our initial blin guess was too far from an actual root, so that there was some wacky jumping aroun of the numbers before things settle own. If we ha been computing by han this woul have been hopeless. Let s try this example again using the Intermeiate Value Theorem to pin own a root with some egree of accuracy: First, f() = > 0. Then f(0) = + > 0 also, so we might oubt that there is a root in [0, ]. Continue: f( ) = > 0 again, so we might oubt that there is a root in [, 0], either. Continue: at last f( 2) = 5 < 0, so since f( ) > 0 by the Intermeiate Value Theorem we o inee know that there is a root between 2 an. Now to start using Newton s Metho, we woul reasonably guess x o =.5 since this is the mipoint of the interval on which we know there is a root. Then computing by Newton s metho gives: x = x 2 = x 3 = x 4 = so right away we have what appears to be 5 ecimal places accuracy, in 4 steps rather than 2. Getting off to a goo start is important. Approximate all three roots of x 3 3x+ = 0 using the intermeiate value theorem to get starte, an then Newton s metho. Here you have to take a little care in choice of beginning guess for Newton s metho: In this case, since we are tol that there are three roots, then we shoul certainly be wary about where we start: presumably we have to start in ifferent places in orer to successfully use Newton s metho to fin the ifferent roots. So, starting thinking in terms of the intermeiate value theorem: letting f(x) = x 3 3x +, we have f(2) = 3 > 0. Next, f() = < 0, so we by the Intermeiate Value Theorem we know there is a root in [, 2]. Let s try to approximate it pretty well before looking for the other roots: The general formula for Newton s metho becomes x n+ = x n x3 3x + 3x 2 3 Our initial blin guess might reasonably be the mipoint of the interval in which we know there is a root: take x o =.5 Then we can compute x = x 2 = x 3 = x 4 = x 5 = x 6 = So it appears that we have quickly approximate a root in that interval! To what looks like 9 ecimal places! Continuing with this example: f(0) = > 0, so since f() < 0 we know by the intermeiate value theorem that there is a root in [0, ], since f() = < 0. So as our blin gues let s use the mipoint of this 34

35 interval to start Newton s Metho: that is, now take x o = 0.5: x = x 2 = x 3 = x 4 = x 5 = x 6 = x 7 = x 8 = so we have a root eviently approximate to 3 ecimal places after just 2 applications of Newton s metho. After 8 applications, we have apparently 5 correct ecimal places. #24.70 Approximate a root of x 3 x + = 0 using the intermeiate value theorem to get starte, an then Newton s metho. #24.7 Approximate a root of 3x 4 6x 3 + 8x 2 + = 0 using the intermeiate value theorem to get starte, an then Newton s metho. You might have to be sure to get sufficiently close to a root to start so that things on t blow up. #24.72 Approximate all three roots of x 3 3x+ = 0 using the intermeiate value theorem to get starte, an then Newton s metho. Here you have to take a little care in choice of beginning guess for Newton s metho. #24.73 Approximate the unique positive root of cos x = x. #24.74 Approximate a root of e x = 2x. #24.75 Approximate a root of sin x = ln x. Watch out. 25. Derivatives of transcenental functions The new material here is just a list of formulas for taking erivatives of exponential, logarithm, trigonometric, an inverse trigonometric functions. Then any function mae by composing these with polynomials or with each other can be ifferentiate by using the chain rule, prouct rule, etc. (These new formulas are not easy to erive, but we on t have to worry about that). The first two are the essentials for exponential an logarithms: The next three are essential for trig functions: x x The next three are essential for inverse trig functions x ex = e x x ln x = x sin x = cos x cos x = sin x x tan x = sec2 x x arcsin x = x 2 x arctan x = +x 2 x arcsec x = x x 2 35

36 The previous formulas are the inispensable ones in practice, an are the only ones that I personally remember (if I m lucky). Other formulas one might like to have seen are (with a > 0 in the first two): x ax = ln a a x x log a x = ln a x x x sec x = tan x sec x csc x = cot x csc x x cot x = csc2 x x arccos x = x 2 x arccot x = +x 2 x arccsc x = x x 2 (There are always some ifficulties in figuring out which of the infinitely-many possibilities to take for the values of the inverse trig functions, an this is especially ba with arccsc, for example. But we won t have time to worry about such things). To be able to use the above formulas it is not necessary to know very many other properties of these functions. For example, it is not necessary to be able to graph these functions to take their erivatives! #25.76 Fin x (ecos x ) #25.77 Fin x (arctan (2 ex )) #25.78 Fin x ( ln (x )) #25.79 Fin x (e2 cos x+5 ) #25.80 Fin x (arctan ( + sin 2x)) #25.8 Fin x cos(ex x 2 ) #25.82 Fin x [3] ln 2x #25.83 Fin e x x e x + #25.84 Fin x ( ln ( x )) 26. L Hospital s rule L Hospital s rule is the efinitive way to simplify evaluation of limits. It oes not irectly evaluate limits, but only simplifies evaluation if use appropriately. In effect, this rule is the ultimate version of cancellation tricks, applicable in situations where a more own-to-earth genuine algebraic cancellation may be hien or invisible. Suppose we want to evaluate f(x) lim x a g(x) where the limit a coul also be + or in aition to orinary numbers. Suppose that either lim f(x) = 0 an lim g(x) = 0 x a x a 36

37 or lim f(x) = ± an lim g(x) = ± x a x a (The ± s on t have to be the same sign). Then we cannot just plug in to evaluate the limit, an these are traitionally calle ineterminate forms. The unexpecte trick that works often is that (amazingly) we are entitle to take the erivative of both numerator an enominator: f(x) lim x a g(x) = lim f (x) x a g (x) No, this is not the quotient rule. No, it is not so clear why this woul help, either, but we ll see in examples. Example: Fin lim x 0 (sin x)/x: both numerator an enominator have limit 0, so we are entitle to apply L Hospital s rule: sin x cos x lim = lim x 0 x x 0 In the new expression, neither numerator nor enominator is 0 at x = 0, an we can just plug in to see that the limit is. Example: Fin lim x 0 x/(e 2x ): both numerator an enominator go to 0, so we are entitle to use L Hospital s rule: x lim x 0 e 2x = lim x 0 2e 2x In the new expression, the numerator an enominator are both non-zero when x = 0, so we just plug in 0 to get x lim x 0 e 2x = lim x 0 2e 2x = 2e 0 = 2 Example: Fin lim x 0 + x ln x: The 0 + means that we approach 0 from the positive sie, since otherwise we won t have a real-value logarithm. This problem illustrates the possibility as well as necessity of rearranging a limit to make it be a ratio of things, in orer to legitimately apply L Hospital s rule. Here, we rearrange to ln x lim x ln x = lim x 0 + x 0 /x In the new expressions the top goes to an the bottom goes to + as x goes to 0 (from the right). Thus, we are entitle to apply L Hospital s rule, obtaining lim x 0 + x ln x = lim ln x x 0 /x = lim x 0 /x /x 2 Now it is very necessary to rearrange the expression insie the last limit: we have lim x 0 /x /x 2 = lim x 0 x The new expression is very easy to evaluate: the limit is 0. It is often necessary to apply L Hospital s rule repeately: Let s fin lim x + x 2 /e x : both numerator an enominator go to as x +, so we are entitle to apply L Hospital s rule, to turn this into 2x lim x + e x But still both numerator an enominator go to, so apply L Hospital s rule again: the limit is lim x + 2 e x = 0 37

38 since now the numerator is fixe while the enominator goes to +. Example: Now let s illustrate more ways that things can be rewritten as ratios, thereby possibly making L Hospital s rule applicable. Let s evaluate lim x 0 xx It is less obvious now, but we can t just plug in 0 for x: on one han, we are taught to think that x 0 =, but also that 0 x = 0; but then surely 0 0 can t be both at once. An this exponential expression is not a ratio. The trick here is to take the logarithm: ln( lim xx ) = lim ) x 0 + x 0 + The reason that we are entitle to interchange the logarithm an the limit is that logarithm is a continuous function (on its omain). Now we use the fact that ln(a b ) = b ln a, so the log of the limit is lim x ln x x 0 + Aha! The question has been turne into one we alreay i! But ignoring that, an repeating ourselves, we first rewrite this as a ratio ln x lim x ln x = lim x 0 + x 0 + /x an then apply L Hospital s rule to obtain /x lim x 0 + /x 2 = lim x = 0 x 0 + But we have to remember that we ve compute the log of the limit, not the limit. Therefore, the actual limit is lim x 0 + xx = e log of the limit = e 0 = This trick of taking a logarithm is important to remember. Example: Here is another issue of rearranging to fit into accessible form: Fin lim x2 + x + x 2 + x + This is not a ratio, but certainly is ineterminate, since it is the ifference of two expressions both of which go to +. To make it into a ratio, we take out the largest reasonable power of x: lim x2 + x + x 2 + = lim x ( x + x + = lim x + + x + x + 2 x 2 /x + x + x 2 + x 2 ) The last expression here fits the requirements of the L Hospital rule, since both numerator an enominator go to 0. Thus, by invoking L Hospital s rule, it becomes = lim x + 2 x 2 2 x 3 2 x3 + x + x 2 + x 2 /x 2 38

39 This is a large but actually tractable expression: multiply top an bottom by x 2, so that it becomes 2 = lim + x x + x + + x + x + 2 x 2 At this point, we can replace every x by 0, fining that the limit is equal to = It is important to recognize that in aitional to the actual application of L Hospital s rule, it may be necessary to experiment a little to get things to settle out the way you want. Trial-an-error is not only ok, it is necessary. #26.85 Fin lim x 0 (sin x)/x #26.86 Fin lim x 0 (sin 5x)/x #26.87 Fin lim x 0 (sin (x 2 ))/x 2 #26.88 Fin lim x 0 x/(e 2x ) #26.89 Fin lim x 0 x ln x #26.90 Fin #26.9 Fin #26.92 Fin #26.93 Fin lim x 0 (ex ) ln x + lim x lim x + lim x + ln x x ln x x ln x x 2 #26.94 Fin lim x 0 (sin x) x 27. Exponential growth an ecay: a ifferential equation This little section is a tiny introuction to a very important subject an bunch of ieas: solving ifferential equations. We ll just look at the simplest possible example of this. The general iea is that, instea of solving equations to fin unknown numbers, we might solve equations to fin unknown functions. There are many possibilities for what this might mean, but one is that we have an unknown function y of x an are given that y an its erivative y (with respect to x) satisfy a relation y = ky 39

40 where k is some constant. Such a relation between an unknown function an its erivative (or erivatives) is what is calle a ifferential equation. Many basic physical principles can be written in such terms, using time t as the inepenent variable. Having been taking erivatives of exponential functions, a person might remember that the function f(t) = e kt has exactly this property: t ekt = k e kt For that matter, any constant multiple of this function has the same property: t (c ekt ) = k c e kt An it turns out that these really are all the possible solutions to this ifferential equation. There is a certain buzz-phrase which is suppose to alert a person to the occurrence of this little story: if a function f has exponential growth or exponential ecay then that is taken to mean that f can be written in the form f(t) = c e kt If the constant k is positive it has exponential growth an if k is negative then it has exponential ecay. Since we ve escribe all the solutions to this equation, what questions remain to ask about this kin of thing? Well, the usual scenario is that some story problem will give you information in a way that requires you to take some trouble in orer to etermine the constants c, k. An, in case you were wonering where you get to take a erivative here, the answer is that you on t really: all the calculus work was one at the point where we grante ourselves that all solutions to that ifferential equation are given in the form f(t) = ce kt. First to look at some general ieas about etermining the constants before getting embroile in story problems: One simple observation is that c = f(0) that is, that the constant c is the value of the function at time t = 0. This is true simply because from properties of the exponential function. f(0) = ce k 0 = ce 0 = c = c More generally, suppose we know the values of the function at two ifferent times: y = ce kt y 2 = ce kt2 Even though we certainly o have two equations an two unknowns, these equations involve the unknown constants in a manner we may not be use to. But it s still not so har to solve for c, k: iviing the first equation by the secon an using properties of the exponential function, the c on the right sie cancels, an we get y y 2 = e k(t t2) Taking a logarithm (base e, of course) we get ln y ln y 2 = k(t t 2 ) Diviing by t t 2, this is k = ln y ln y 2 t t 2 40

41 Substituting back in orer to fin c, we first have Taking the logarithm, we have Rearranging, this is Therefore, in summary, the two equations y = ce ln y ln y 2 t t 2 t ln y = ln c + ln y ln y 2 t t 2 t ln c = ln y ln y ln y 2 t t 2 t = t ln y 2 t 2 ln y t t 2 y = ce kt y 2 = ce kt2 allow us to solve for c, k, giving k = ln y ln y 2 t t 2 c = e t ln y 2 t 2 ln y t t 2 A person might manage to remember such formulas, or it might be wiser to remember the way of eriving them. Example: A her of llamas has 000 llamas in it, an the population is growing exponentially. At time t = 4 it has 2000 llamas. Write a formula for the number of llamas at arbitrary time t. Here there is no irect mention of ifferential equations, but use of the buzz-phrase growing exponentially must be taken as inicator that we are talking about the situation f(t) = ce kt where here f(t) is the number of llamas at time t an c, k are constants to be etermine from the information given in the problem. An the use of language shoul probably be taken to mean that at time t = 0 there are 000 llamas, an at time t = 4 there are Then, either repeating the metho above or plugging into the formula erive by the metho, we fin c = value of f at t = 0 = 000 k = ln f(t ) ln f(t 2 ) ln 000 ln 2000 = t t Therefore, = ln = ln 2 4 = (ln 2)/4 f(t) = 000 e ln 2 4 t = t/4 This is the esire formula for the number of llamas at arbitrary time t. Example: A colony of bacteria is growing exponentially. At time t = 0 it has 0 bacteria in it, an at time t = 4 it has At what time will it have 00, 000 bacteria? Even though it is not explicitly emane, we nee to fin the general formula for the number f(t) of bacteria at time t, set this expression equal to 00, 000, an solve for t. Again, we can take a little shortcut here since we know that c = f(0) an we are given that f(0) = 0. (This is easier than using the bulkier more general formula for fining c). An use the formula for k: k = ln f(t ) ln f(t 2 ) ln 0 ln 2, 000 = = t t ln 0 2,000 4 = ln 200 4

42 Therefore, we have ln 200 f(t) = 0 e 4 t = t/4 as the general formula. Now we try to solve ln , 000 = 0 e 4 t for t: ivie both sies by the 0 an take logarithms, to get ln 0, 000 = ln t Thus, t = 4 ln 0, 000 ln #27.95 A her of llamas is growing exponentially. At time t = 0 it has 000 llamas in it, an at time t = 4 it has 2000 llamas. Write a formula for the number of llamas at arbitrary time t. #27.96 A her of elephants is growing exponentially. At time t = 2 it has 000 elephants in it, an at time t = 4 it has 2000 elephants. Write a formula for the number of elephants at arbitrary time t. #27.97 A colony of bacteria is growing exponentially. At time t = 0 it has 0 bacteria in it, an at time t = 4 it has At what time will it have 00, 000 bacteria? #27.98 A colony of bacteria is growing exponentially. At time t = 2 it has 0 bacteria in it, an at time t = 4 it has At what time will it have 00, 000 bacteria? 28. The secon an higher erivatives The secon erivative of a function is simply the erivative of the erivative. The thir erivative of a function is the erivative of the secon erivative. An so on. The secon erivative of a function y = f(x) is written as The thir erivative is y = f (x) = y = f (x) = 2 x 2 f = 2 f x 2 = 2 y x 2 3 x 3 f = 3 f x 3 = 3 y x 3 An, generally, we can put on a prime for each erivative taken. Or write n x n f = n f x n = n y x n for the nth erivative. There is yet another notation for high orer erivatives where the number of primes woul become unwiely: n f x n = f (n) (x) as well. The geometric interpretation of the higher erivatives is subtler than that of the first erivative, an we won t o much in this irection, except for the next little section. 42

43 #28.99 Fin f (x) for f(x) = x 3 5x +. #28.00 Fin f (x) for f(x) = x 5 5x 2 + x. #28.0 Fin f (x) for f(x) = x 2 x +. #28.02 Fin f (x) for f(x) = x. 29. Inflection points, concavity upwar an ownwar A point of inflection of the graph of a function f is a point where the secon erivative f is 0. We have to wait a minute to clarify the geometric meaning of this. A piece of the graph of f is concave upwar if the curve bens upwar. For example, the popular parabola y = x 2 is concave upwar in its entirety. A piece of the graph of f is concave ownwar if the curve bens ownwar. For example, a flippe version y = x 2 of the popular parabola is concave ownwar in its entirety. The relation of points of inflection to intervals where the curve is concave up or own is exactly the same as the relation of critical points to intervals where the function is increasing or ecreasing. That is, the points of inflection mark the bounaries of the two ifferent sort of behavior Further, only one sample value of f nee be taken between each pair of consecutive inflection points in orer to see whether the curve bens up or own along that interval. Expressing this as a systematic proceure: to fin the intervals along which f is concave upwar an concave ownwar: Compute the secon erivative f of f, an solve the equation f (x) = 0 for x to fin all the inflection points, which we list in orer as x < x 2 <... < x n. (Any points of iscontinuity, etc., shoul be ae to the list!) We nee some auxiliary points: To the left of the leftmost inflection point x pick any convenient point t o, between each pair of consecutive inflection points x i, x i+ choose any convenient point t i, an to the right of the rightmost inflection point x n choose a convenient point t n. Evaluate the secon erivative f at all the auxiliary points t i. Conclusion: if f (t i+ ) > 0, then f is concave upwar on (x i, x i+ ), while if f (t i+ ) < 0, then f is concave ownwar on that interval. Conclusion: on the outsie interval (, x o ), the function f is concave upwar if f (t o ) > 0 an is concave ownwar if f (t o ) < 0. Similarly, on (x n, ), the function f is concave upwar if f (t n ) > 0 an is concave ownwar if f (t n ) < 0. Fin the inflection points an intervals of concavity up an own of f(x) = 3x 2 9x + 6 First, the secon erivative is just f (x) = 6. Since this is never zero, there are not points of inflection. An the value of f is always 6, so is always > 0, so the curve is entirely concave upwar. 43

44 Fin the inflection points an intervals of concavity up an own of f(x) = 2x 3 2x 2 + 4x 27 First, the secon erivative is f (x) = 2x 24. Thus, solving 2x 24 = 0, there is just the one inflection point, 2. Choose auxiliary points t o = 0 to the left of the inflection point an t = 3 to the right of the inflection point. Then f (0) = 24 < 0, so on (, 2) the curve is concave ownwar. An f (2) = 2 > 0, so on (2, ) the curve is concave upwar. Fin the inflection points an intervals of concavity up an own of f(x) = x 4 24x 2 + the secon erivative is f (x) = 2x Solving the equation 2x 2 48 = 0, we fin inflection points ±2. Choosing auxiliary points 3, 0, 3 place between an to the left an right of the inflection points, we evaluate the secon erivative: First, f ( 3) = > 0, so the curve is concave upwar on (, 2). Secon, f (0) = 48 < 0, so the curve is concave ownwar on ( 2, 2). Thir, f (3) = > 0, so the curve is concave upwar on (2, ). #29.03 Fin the inflection points an intervals of concavity up an own of f(x) = 3x 2 9x + 6. #29.04 Fin the inflection points an intervals of concavity up an own of f(x) = 2x 3 2x 2 + 4x 27. #29.05 Fin the inflection points an intervals of concavity up an own of f(x) = x 4 2x Another ifferential equation: projectile motion Here we encounter the funamental iea that if s = s(t) is position, then ṡ is velocity, an ṡ is acceleration. This iea occurs in all basic physical science an engineering. In particular, for a projectile near the earth s surface travelling straight up an own, ignoring air resistance, acte upon by no other forces but gravity, we have Thus, letting s(t) be position at time t, we have acceleration ue to gravity = 32 feet/sec 2 s(t) = 32 We take this (approximate) physical fact as our starting point. From s = 32 we integrate (or anti-ifferentiate) once to uno one of the erivatives, getting back to velocity: v(t) = ṡ = ṡ(t) = 32t + v o where we are calling the constant of integration v o. (No matter which constant v o we might take, the erivative of 32t + v o with respect to t is 32.) Specifically, when t = 0, we have v(o) = v o Thus, the constant of integration v o is initial velocity. An we have this formula for the velocity at any time in terms of initial velocity. 44

45 We integrate once more to uno the last erivative, getting back to the position function itself: s = s(t) = 6t 2 + v o t + s o where we are calling the constant of integration s o. Specifically, when t = 0, we have s(0) = s o so s o is initial position. Thus, we have a formula for position at any time in terms of initial position an initial velocity. Of course, in many problems the ata we are given is not just the initial position an initial velocity, but something else, so we have to etermine these constants inirectly. #30.06 You rop a rock own a eep well, an it takes 0 secons to hit the bottom. How eep is it? #30.07 You rop a rock own a well, an the rock is going 32 feet per secon when it hits bottom. How eep is the well? #30.08 If I throw a ball straight up an it takes 2 secons for it to go up an come own, how high i it go? 3. Graphing rational functions, asymptotes This section shows another kin of function whose graphs we can unerstan effectively by our methos. There is one new item here, the iea of asymptote of the graph of a function. A vertical asymptote of the graph of a function f most commonly occurs when f is efine as a ratio f(x) = g(x)/h(x) of functions g, h continuous at a point x o, but with the enominator going to zero at that point while the numerator oesn t. That is, h(x o ) = 0 but g(x o ) 0. Then we say that f blows up at x o, an that the line x = x o is a vertical asymptote of the graph of f. An as we take x closer an closer to x o, the graph of f zooms off (either up or own or both) closely to the line x = x o. A very simple example of this is f(x) = /(x ), whose enominator is 0 at x =, so causing a blow-up at that point, so that x = is a vertical asymptote. An as x approaches from the right, the values of the function zoom up to +. When x approaches from the left, the values zoom own to. A horizontal asymptote of the graph of a function f occurs if either limit or lim f(x) x + lim f(x) x exists. If R = lim x + f(x), then y = R is a horizontal asymptote of the function, an if L = lim x f(x) exists then y = L is a horizontal asymptote. As x goes off to + the graph of the function gets closer an closer to the horizontal line y = R if that limit exists. As x goes of to the graph of the function gets closer an closer to the horizontal line y = L if that limit exists. So in rough terms asymptotes of a function are straight lines which the graph of the function approaches at infinity. In the case of vertical asymptotes, it is the y-coorinate that goes off to infinity, an in the case of horizontal asymptotes it is the x-coorinate which goes off to infinity. 45

46 Fin asymptotes, critical points, intervals of increase an ecrease, inflection points, an intervals of concavity up an own of f(x) = x+3 2x 6 : First, let s fin the asymptotes. The enominator is 0 for x = 3 (an this is not cancelle by the numerator) so the line x = 3 is a vertical asymptote. An as x goes to ±, the function values go to /2, so the line y = /2 is a horizontal asymptote. The erivative is f (x) = (2x 6) (x + 3) 2 (2x 6) 2 = 2 (2x 6) 2 Since a ratio of polynomials can be zero only if the numerator is zero, this f (x) can never be zero, so there are no critical points. There is, however, the iscontinuity at x = 3 which we must take into account. Choose auxiliary points 0 an 4 to the left an right of the iscontinuity. Plugging in to the erivative, we have f (0) = 2/( 6) 2 < 0, so the function is ecreasing on the interval (, 3). To the right, f (4) = 2/(8 6) 2 < 0, so the function is also ecreasing on (3, + ). The secon erivative is f (x) = 48/(2x 6) 3. This is never zero, so there are no inflection points. There is the iscontinuity at x = 3, however. Again choosing auxiliary points 0, 4 to the left an right of the iscontinuity, we see f (0) = 48/( 6) 3 < 0 so the curve is concave ownwar on the interval (, 3). An f (4) = 48/(8 6) 3 > 0, so the curve is concave upwar on (3, + ). Plugging in just two or so values into the function then is enough to enable a person to make a fairly goo qualitative sketch of the graph of the function. #3.09 Fin all asymptotes of f(x) = x x+2. #3.0 Fin all asymptotes of f(x) = x+2 x. #3. Fin all asymptotes of f(x) = x2 x 2 4. #3.2 Fin all asymptotes of f(x) = x2 x Basic integration formulas The funamental use of integration is as a continuous version of summing. But, paraoxically, often integrals are compute by viewing integration as essentially an inverse operation to ifferentiation. (That fact is the so-calle Funamental Theorem of Calculus.) The notation, which we re stuck with for historical reasons, is as peculiar as the notation for erivatives: the integral of a function f(x) with respect to x is written as f(x) x The remark that integration is (almost) an inverse to the operation of ifferentiation means that if f(x) = g(x) x then g(x) x = f(x) + C The extra C, calle the constant of integration, is really necessary, since after all ifferentiation kills off constants, which is why integration an ifferentiation are not exactly inverse operations of each other. 46

47 Since integration is almost the inverse operation of ifferentiation, recollection of formulas an processes for ifferentiation alreay tells the most important formulas for integration: x n x = n+ xn+ + C unless n = e x x = e x + C x x = ln x + C sin x x = cos x + C cos x x = sin x + C sec 2 x x = tan x + C +x x = arctan x + C 2 An since the erivative of a sum is the sum of the erivatives, the integral of a sum is the sum of the integrals: f(x) + g(x) x = f(x) x + g(x) x An, likewise, constants go through the integral sign: c f(x) x = c f(x) x For example, it is easy to integrate polynomials, even incluing terms like x an more general power functions. The only thing to watch out for is terms x = x, since these integrate to ln x instea of a power of x. So 4x 5 3x + 7 x + 3 x Notice that we nee to inclue just one constant of integration. Other basic formulas obtaine by reversing ifferentiation formulas: a x a x = x loga x x = ln a + C x 2 4x6 x = 6 3x2 7x3/2 + x + 3 ln x + C 2 3/2 x + C x = arcsin x + C ln a x x = arcsec x + C x 2 Sums of constant multiples of all these functions are easy to integrate: for example, 5 2 x 23 x x 2 + 5x2 x = 5 2x 5x3 23 arcsec x + ln C #32.3 4x 3 3 cos x + 7 x + 2 x =? #32.4 3x 2 + e 2x + cos x x =? #32.5 sec 2 x x =? # x 2 x =? #32.7 6x 7 x + 3 x x =? # sin x 2 x x =? 2 47

48 The simplest kin of chain rule application 33. The simplest substitutions x f(ax + b) = a f (x) (for constants a, b) can easily be run backwars to obtain the corresponing integral formulas: some an illustrative important examples are cos(ax + b) x = a sin(ax + b) + C e ax+b x = a eax+b + C ax + b x = a (ax+b)3/2 3/2 + C ax+b x = a ln(ax + b) + C Putting numbers in instea of letters, we have examples like cos(3x + 2) x = 3 sin(3x + 2) + C e 4x+3 x = 4 e4x+3 + C 5x + x = 5 ( 5x+)3/2 3/2 + C 7x 2 x = 7 ln(7x 2) + C Since this kin of substitution is pretty unramatic, an a person shoul be able to o such things by reflex rather than having to think about it very much. #33.9 e 3x+2 x =? #33.20 cos(2 5x) x =? #33.2 3x 7 x =? #33.22 sec 2 (2x + ) x =? #33.23 (5x 7 + e 6 2x x ) x =? #33.24 cos(7 x) x =? 34. Substitutions The chain rule can also be run backwar, an is calle change of variables or substitution or sometimes u-substitution. Some examples of what happens are straightforwar, but others are less obvious. It is at this point that the capacity to recognize erivatives from past experience becomes very helpful. Example Since (by the chain rule) then we can anticipate that x esin x = cos x e sin x cos x e sin x x = e sin x + C 48

49 Example Since (by the chain rule) then we can anticipate that x5 + 3x = x 2 (x5 + 3x) /2 (5x 4 + 3) 2 (5x4 + 3)(x 5 + 3x) /2 x = x 5 + 3x + C Very often it happens that things are off by a constant. This shoul not eter a person from recognizing the possibilities. For example: since, by the chain rule, then 5 + ex = x 2 (5 + ex ) /2 e x e x (5 + e x ) /2 x = 2 2 ex (5 + e x ) /2 x = e x + C Notice how for bookkeeping purposes we put the 2 into the integral (to make the constants right there) an put a compensating 2 outsie. Example: Since (by the chain rule) x sin7 (3x + ) = 7 sin 6 (3x + ) cos(3x + ) 3 then we have = 2 cos(3x + ) sin 6 (3x + ) x 7 3 cos(3x + ) sin 6 (3x + ) x = 2 sin7 (3x + ) + C #34.25 cos x sin x x =? # x e x2 x =? # x 5 e x6 x =? #34.28 cos x sin x x =? #34.29 cos xe sin x x =? # x e x x =? #34.3 cos x sin 5 x x =? #34.32 sec 2 x tan 7 x x =? #34.33 (3 cos x + x) e 6 sin x+x2 x =? #34.34 e x e x + x =? 49

50 35. Area an efinite integrals The actual efinition of integral is as a limit of sums, which might easily be viewe as having to o with area. One of the original issues integrals were intene to aress was computation of area. First we nee more notation. Suppose that we have a function f whose integral is another function F : f(x) x = F (x) + C Let a, b be two numbers. Then the efinite integral of f with limits a, b is b a f(x) x = F (b) F (a) The left-han sie of this equality is just notation for the efinite integral. The use of the wor limit here has little to o with our earlier use of the wor, an means something more like bounary, just like it oes in more orinary English. A similar notation is to write [g(x)] b a = g(b) g(a) for any function g. So we coul also write b a f(x) x = [F (x)] b a For example, As another example, 5 0 x 2 x = [ x3 3 ]5 0 = = x + x = [ 3x2 2 + x]3 2 = ( ) ( ) = All the other integrals we ha one previously woul be calle inefinite integrals since they in t have limits a, b. So a efinite integral is just the ifference of two values of the function given by an inefinite integral. That is, there is almost nothing new here except the iea of evaluating the function that we get by integrating. But now we can o something new: compute areas: For example, if a function f is positive on an interval [a, b], then b a f(x) x = area between graph an x-axis, between x = a an x = b It is important that the function be positive, or the result is false. For example, since y = x 2 is certainly always positive (or at least non-negative, which is really enough), the area uner the curve (an, implicitly, above the x-axis) between x = 0 an x = is just 0 x 2 x = [ x3 3 ] 0 = =

51 More generally, the area below y = f(x), above y = g(x), an between x = a an x = b is area... = b a f(x) g(x) x right limit = left limit (upper curve - lower curve) x It is important that f(x) g(x) throughout the interval [a, b]. For example, the area below y = e x an above y = x, an between x = 0 an x = 2 is 2 0 since it really is true that e x x on the interval [0, 2]. e x x x = [e x x2 2 ]2 0 = (e 2 2) (e 0 0) = e 2 + As a person might be wonering, in general it may be not so easy to tell whether the graph of one curve is above or below another. The proceure to examine the situation is as follows: given two functions f, g, to fin the intervals where f(x) g(x) an vice-versa: Fin where the graphs cross by solving f(x) = g(x) for x to fin the x-coorinates of the points of intersection. Between any two solutions x, x 2 of f(x) = g(x) (an also to the left an right of the left-most an rightmost solutions!), plug in one auxiliary point of your choosing to see which function is larger. Of course, this proceure works for a similar reason that the first erivative test for local minima an maxima worke: we implicitly assume that the f an g are continuous, so if the graph of one is above the graph of the other, then the situation can t reverse itself without the graphs actually crossing. As an example, an as an example of a certain elicacy of woring, consier the problem to fin the area between y = x an y = x 2 with 0 x 2. To fin where y = x an y = x 2 cross, solve x = x 2 : we fin solutions x = 0,. In the present problem we on t care what is happening to the left of 0. Plugging in the value /2 as auxiliary point between 0 an, we get 2 ( 2 )2, so we see that in [0, ] the curve y = x is the higher. To the right of we plug in the auxiliary point 2, obtaining 2 2 2, so the curve y = x 2 is higher there. Therefore, the area between the two curves has to be broken into two parts: area = since we must always be integrating in the form 0 (x x 2 ) x + 2 right higher - lower x left (x 2 x) x In some cases the sie bounaries are reunant or only implie. For example, the question might be to fin the area between the curves y = 2 x an y = x 2. What is implie here is that these two curves themselves enclose one or more finite pieces of area, without the nee of any sie bounaries of the form x = a. First, we nee to see where the two curves intersect, by solving 2 x = x 2 : the solutions are x = 2,. So we infer that we are suppose to fin the area from x = 2 to x =, an that the two curves close up aroun this chunk of area without any nee of assistance from vertical lines x = a. We nee to fin which curve is higher: plugging in the point 0 between 2 an, we see that y = 2 x is higher. Thus, the esire integral is area... = 2 (2 x) x 2 x 5

52 #35.35 Fin the area between the curves y = x 2 an y = 2x + 3. #35.36 Fin the area of the region boune vertically by y = x 2 an y = x+2 an boune horizontally by x = an x = 3. #35.37 Fin the area between the curves y = x 2 an y = 8 + 6x x 2. #35.38 Fin the area between the curves y = x an y = x Lengths of Curves The basic point here is a formula obtaine by using the ieas of calculus: the length of the graph of y = f(x) from x = a to x = b is b ( ) 2 y arc length = + x x Or, if the curve is parametrize in the form a x = f(t) y = g(t) with the parameter t going from a to b, then arc length = b a (x ) 2 + t ( ) 2 y t t This formula comes from approximating the curve by straight lines connecting successive points on the curve, using the Pythagorean Theorem to compute the lengths of these segments in terms of the change in x an the change in y. In one way of writing, which also provies a goo heuristic for remembering the formula, if a small change in x is x an a small change in y is y, then the length of the hypotenuse of the right triangle with base x an altitue y is (by the Pythagorean theorem) hypotenuse = x 2 + y 2 = + ( ) 2 y x x Unfortunately, by the nature of this formula, most of the integrals which come up are ifficult or impossible to o. But if one of these really mattere, we coul still estimate it by numerical integration. #36.39 Fin the length of the curve y = x 2 from x = 0 to x =. #36.40 Fin the length of the curve y = 4 (e2x + e 2x ) from x = 0 to x =. #36.4 Set up (but o not evaluate) the integral to fin the length of the piece of the parabola y = x 2 from x = 3 to x = 4. 52

53 37. Numerical integration As we start to see that integration by formulas is a much more ifficult thing than ifferentiation, an sometimes is impossible to o in elementary terms, it becomes reasonable to ask for numerical approximations to efinite integrals. Since a efinite integral is just a number, this is possible. By constrast, inefinite integrals, being functions rather than just numbers, are not easily escribe by numerical approximations. There are several relate approaches, all of which use the iea that a efinite integral is relate to area. Thus, each of these approaches is really essentially a way of approximating area uner a curve. Of course, this isn t exactly right, because integrals are not exactly areas, but thinking of area is a reasonable heuristic. Of course, an approximation is not very valuable unless there is an estimate for the error, in other wors, an iea of the tolerance. Each of the approaches starts the same way: To approximate b f(x) x, break the interval [a, b] into a smaller subintervals [x 0, x ], [x, x 2 ],..., [x n 2, x n ], [x n, x n ] each of the same length an where x 0 = a an x n = b. Trapezoial rule: This rule says that x = b a n b a f(x) x x 2 [f(x 0) + 2f(x ) + 2f(x 2 ) f(x n 2 ) + 2f(x n ) + f(x n )] Yes, all the values have a factor of 2 except the first an the last. (This metho approximates the area uner the curve by trapezois inscribe uner the curve in each subinterval). Mipoint rule: Let x i = 2 (x i x i ) be the mipoint of the subinterval [x i, x i ]. Then the mipoint rule says that b a f(x) x x[f(x ) f(x n )] (This metho approximates the area uner the curve by rectangles whose height is the mipoint of each subinterval). Simpson s rule: This rule says that b a f(x) x x 3 [f(x 0) + 4f(x ) + 2f(x 2 ) + 4f(x 3 ) f(x n 2 ) + 4f(x n ) + f(x n )] Yes, the first an last coefficients are, while the inner coefficients alternate 4 an 2. An n has to be an even integer for this to make sense. (This metho approximates the curve by pieces of parabolas). In general, the smaller the x is, the better these approximations are. We can be more precise: the error estimates for the trapezoial an mipoint rules epen upon the secon erivative: suppose that f (x) M for some constant M, for all a x b. Then error in trapezoial rule 53 M(b a)3 2n 2

54 M(b a)3 error in mipoint rule 24n 2 The error estimate for Simpson s rule epens on the fourth erivative: suppose that f (4) (x) N for some constant N, for all a x b. Then error in Simpson s rule N(b a)5 80n 4 From these formulas estimating the error, it looks like the mipoint rule is always better than the trapezoial rule. An for high accuracy, using a large number n of subintervals, it looks like Simpson s rule is the best. 38. Averages an Weighte Averages The usual notion of average of a list of n numbers x,..., x n is average of x, x 2,..., x n = x + x x n n A continuous analogue of this can be obtaine as an integral, using a notation which matches better: let f be a function on an interval [a, b]. Then average value of f on the interval [a, b] = b a f(x) x b a For example the average value of the function y = x 2 over the interval [2, 3] is average value of f on the interval [a, b] = 3 2 x2 x 3 2 = [x3 /3] = (3 2) = 9/3 A weighte average is an average in which some of the items to be average are more important or less important than some of the others. The weights are (non-negative) numbers which measure the relative importance. For example, the weighte average of a list of numbers x,..., x n with corresponing weights w,..., w n is w x + w 2 x w n x n w + w w n Note that if the weights are all just, then the weighte average is just a plain average. The continuous analogue of a weighte average can be obtaine as an integral, using a notation which matches better: let f be a function on an interval [a, b], with weight w(x), a non-negative function on [a, b]. Then weighte average value of f on the interval [a, b] with weight w = b a w(x) f(x) x b w(x) x a Notice that in the special case that the weight is just all the time, then the weighte average is just a plain average. For example the average value of the function y = x 2 over the interval [2, 3] with weight w(x) = x is average value of f on the interval [a, b] with weight x 54

55 = 3 2 x x2 x 3 2 x x = [x4 /4] 3 2 [x 2 /2] 3 = 2 4 ( ) 2 ( ) 39. Centers of Mass (Centrois) For many (but certainly not all!) purposes in physics an mechanics, it is necessary or useful to be able to consier a physical object as being a mass concentrate at a single point, its geometric center, also calle its centroi. The centroi is essentially the average of all the points in the object. For simplicity, we will just consier the two-imensional version of this, looking only at regions in the plane. The simplest case is that of a rectangle: it is pretty clear that the centroi is the center of the rectangle. That is, if the corners are (0, 0), (u, 0), (0, v) an (u, v), then the centroi is ( u 2, v 2 ) The formulas below are obtaine by integrating up this simple iea: For the center of mass (centroi) of the plane region escribe by f(x) y g(x) an a x b, we have x-coorinate of the centroi = average x-coorinate An also b a x[g(x) f(x)] x = b [g(x) f(x)] x a right right left x[upper lower] x left x[upper lower] x = = right left [upper lower] x area of the region y-coorinate of the centroi = average y-coorinate right left = right left = b a b a 2 [upper2 lower 2 ] x [upper lower] x 2 [g(x)2 f(x) 2 ] x [g(x) f(x)] x right left = area of the region 2 [upper2 lower 2 ] x Heuristic: For the x-coorinate: there is an amount (g(x) f(x))x of the region at istance x from the y- axis. This is integrate, an then average iviing by the total, that is, iviing by the area of the entire region. For the y-coorinate: in each vertical ban of with x there is amount x y of the region at istance y from the x-axis. This is integrate up an then average by iviing by the total area. For example, let s fin the centroi of the region boune by x = 0, x =, y = x 2, an y = 0. x-coorinate of the centroi = 0 x[x2 0] x 0 [x2 0] x = [x4 /4] 0 [x 3 /3] = /4 0 0 /3 0 =

56 An y-coorinate of the centroi = = 2 [x5 /5] 0 [x 3 /3] = (/5 0) = 3 / [(x2 ) 2 0] x 0 [x2 0] x #39.42 Fin the center of mass (centroi) of the region 0 x an 0 y x 2. #39.43 Fin the center of mass (centroi) of the region efine by 0 x, 0 y an x + y. #39.44 Fin the center of mass (centroi) of a homogeneous plate in the shape of an equilateral triangle. 40. Volumes by Cross Sections Next to computing areas of regions in the plane, the easiest concept of application of the ieas of calculus is to computing volumes of solis where somehow we know a formula for the areas of slices, that is, areas of cross sections. Of course, in any particular example, the actual issue of getting the formula for the cross section, an figuring out the appropriate limits of integration, can be ifficult. The iea is to just a them up : right limit volume = left limit (area of cross section at x) x where in whatever manner we escribe the soli it extens from x =left limit to x =right limit. We must suppose that we have some reasonable formula for the area of the cross section. For example, let s fin the volume of a soli ball of raius. (In effect, we ll be eriving the formula for this). We can suppose that the ball is centere at the origin. Since the raius is, the range of x coorinates is from to +, so x will be integrate from to +. At a particular value of x, what oes the cross section look like? A isk, whose raius we ll have to etermine. To etermine this raius, look at how the soli ball intersects the x, y-plane: it intesects in the isk x 2 + y 2. For a particular value of x, the values of y are between ± x 2. This line segment, having x fixe an y in this range, is the intersection of the cross section isk with the x, y-plane, an in fact is a iameter of that cross section isk. Therefore, the raius of the cross section isk at x is x 2. Use the formula that the area of a isk of raius r is πr 2 : the area of the cross section is Then integrate this from to + to get the volume: = + cross section at x = π( x 2 ) 2 = π( x 2 ) volume = right left area of cross-section x π( x 2 ) x = π[x x3 3 ]+ = π[( ( )3 ) ( )] = =

57 #40.45 Fin the volume of a circular cone of raius 0 an height 2 (not by a formula, but by cross sections). #40.46 Fin the volume of a cone whose base is a square of sie 5 an whose height is 6, by crosssections. #40.47 A hole 3 units in raius is rille out along a iameter of a soli sphere of raius 5 units. What is the volume of the remaining soli? #40.48 A soli whose base is a isc of raius 3 has vertical cross sections which are squares. What is the volume? 4. Solis of Revolution Another way of computing volumes of some special types of soli figures applies to solis obtaine by rotating plane regions about some axis. If we rotate the plane region escribe by f(x) y g(x) an a x b aroun the x-axis, the volume of the resulting soli is volume = b a π(g(x) 2 f(x) 2 ) x right limit = left limit π(upper curve 2 lower curve 2 ) x It is necessary to suppose that f(x) 0 for this to be right. This formula comes from viewing the whole thing as slice up into slices of thickness x, so that each slice is a isk of raius g(x) with a smaller isk of raius f(x) remove from it. Then we use the formula area of isk = π raius 2 an a them all up. The hypothesis that f(x) 0 is necessary to avoi ifferent pieces of the soli overlap each other by accient, thus counting the same chunk of volume twice. If we rotate the plane region escribe by f(x) y g(x) an a x b aroun the y-axis (instea of the x-axis), the volume of the resulting soli is volume = b a 2πx(g(x) f(x)) x right = 2πx( upper - lower) x left This secon formula comes from viewing the whole thing as slice up into thin cylinrical shells of thickness x encircling the y-axis, of raius x an of height g(x) f(x). The volume of each one is (area of cyliner of height g(x) f(x) an raius x) x = 2πx(g(x) f(x)) x an a them all up in the integral. As an example, let s consier the region 0 x an x 2 y x. Note that for 0 x it really is the case that x 2 y x, so y = x is the upper curve of the two, an y = x 2 is the lower curve of the two. Invoking the formula above, the volume of the soli obtaine by rotating this plane region aroun the x-axis is right volume = π(upper 2 lower 2 ) x left 57

58 = 0 π((x) 2 (x 2 ) 2 ) x = π[x 3 /3 x 5 /5] 0 = π(/3 /5) On the other han, if we rotate this aroun the y-axis instea, then = 0 right volume = 2πx(upper lower) x left 2πx(x x 2 ) x = π 0 2x 3 3 2x4 4 x = [2x3 3 2x4 4 ] 0 = = 6 #4.49 Fin the volume of the soli obtaine by rotating the region 0 x, 0 y x aroun the y-axis. #4.50 Fin the volume of the soli obtaine by rotating the region 0 x, 0 y x aroun the x-axis. #4.5 Set up the integral which expresses the volume of the oughnut obtaine by rotating the region (x 2) 2 + y 2 aroun the y-axis. 42. Surfaces of Revolution Here is another formula obtaine by using the ieas of calculus: the area of the surface obtaine by rotating the curve y = f(x) with a x b aroun the x-axis is area = b a 2πf(x) + ( ) 2 y x x This formula comes from extening the ieas of the previous section the length of a little piece of the curve is x2 + y 2 This gets rotate aroun the perimeter of a circle of raius y = f(x), so approximately give a ban of with x 2 + y 2 an length 2πf(x), which has area 2πf(x) x 2 + y 2 = 2πf(x) Integrating this (as if it were a sum!) gives the formula. + ( ) 2 y x x As with the formula for arc length, it is very easy to obtain integrals which are ifficult or impossible to evaluate except numerically. Similarly, we might rotate the curve y = f(x) aroun the y-axis instea. The same general ieas apply to compute the area of the resulting surface. The with of each little ban is still x 2 + y 2, but now the length is 2πx instea. So the ban has area with length = 2πx x 2 + y 2 58

59 Therefore, in this case the surface area is obtaine by integrating this, yieling the formula area = b a 2πx + ( ) 2 y x x #42.52 Fin the area of the surface obtaine by rotating the curve y = 4 (e2x + e 2x ) with 0 x aroun the x-axis. #42.53 Just set up the integral for the surface obtaine by rotating the curve y = 4 (e2x + e 2x ) with 0 x aroun the y-axis. #42.54 Set up the integral for the area of the surface obtaine by rotating the curve y = x 2 with 0 x aroun the x-axis. #42.55 Set up the integral for the area of the surface obtaine by rotating the curve y = x 2 with 0 x aroun the y-axis. 43. Integration by parts Strangely, the subtlest stanar metho is just the prouct rule run backwars. This is calle integration by parts. (This might seem strange because often people fin the chain rule for ifferentiation harer to get a grip on than the prouct rule). One way of writing the integration by parts rule is f(x) g (x) x = f(x)g(x) f (x) g(x) x Sometimes this is written another way: if we use the notation that for a function u of x, then for two functions u, v of x the rule is u = u x x u v = uv v u Yes, it is har to see how this might be helpful, but it is. The first theme we ll see in examples is where we coul o the integral except that there is a power of x in the way : The simplest example is x e x x = x (e x ) = x e x e x x = x e x e x + C Here we have taken u = x an v = e x. It is important to be able to see the e x as being the erivative of itself. A similar example is x cos x x = x (sin x) = x sin x sin x x = x sin x + cos x + C 59

60 Here we have taken u = x an v = sin x. It is important to be able to see the cos x as being the erivative of sin x. Yet another example, illustrating also the iea of repeating the integration by parts: x 2 e x x = x 2 (e x ) = x 2 e x e x (x 2 ) = x 2 e x 2 x e x x = x 2 e x 2x e x + 2 e x x = x 2 e x 2x e x + 2e x + C Here we integrate by parts twice. After the first integration by parts, the integral we come up with is xe x x, which we ha ealt with in the first example. Or sometimes the theme is that it is easier to integrate the erivative of something than to integrate the thing: ln x x = ln x (x) = x ln x x (ln x) We took u = ln x an v = x. = x ln x x x x = x ln x x = x ln x x + C Again in this example it is easier to integrate the erivative than the thing itself: since we shoul recognize the arctan x x = = xarctan x arctan x (x) = xarctan x x + x 2 x = xarctan x 2 = xarctan x 2 ln( + x2 ) + C 2x + x 2 as being the erivative (via the chain rule) of ln( + x 2 ). x (arctan x) 2x + x 2 x #43.56 ln x x =? #43.57 xe x x =? #43.58 (ln x) 2 x =? #43.59 xe 2x x =? #43.60 arctan 3x x =? #43.6 x 3 ln x x =? #43.62 ln 3x x =? #43.63 x ln x x =? 60

61 44. Partial Fractions Now we return to a more special but still important technique of oing inefinite integrals. This epens on a goo trick from algebra to transform complicate rational functions into simpler ones. Rather than try to formally escribe the general fact, we ll o the two simplest families of examples. Consier the integral x(x ) x As it stans, we o not recognize this as the erivative of anything. However, we have Therefore, x x (x ) = = x x(x ) x(x ) x(x ) x = x x = ln(x ) ln x + C x That is, by separating the fraction /x(x ) into the partial fractions /x an /(x ) we were able to o the integrals immeiately by using the logarithm. How to see such ientities? Well, let s look at a situation cx + (x a)(x b) = A x a + B x b where a, b are given numbers (not equal) an we are to fin A, B which make this true. If we can fin the A, B then we can integrate (cx + )/(x a)(x b) simply by using logarithms: cx + (x a)(x b) x = To fin the A, B, multiply through by (x a)(x b) to get A x a + B x = A ln(x a) + B ln(x b) + C x b cx + = A(x b) + B(x a) When x = a the x a factor is 0, so this equation becomes c a + = A(a b) Likewise, when x = b the x b factor is 0, so we also have That is, c b + = B(b a) A = c a + a b B = c b + b a So, yes, we can fin the constants to break the fraction (cx + )/(x a)(x b) own into simpler partial fractions. Further, if the numerator is of bigger egree than, then before executing the previous algebra trick we must firstivie the numerator by the enominator to get a remainer of smaller egree. A simple example is x 3 + 4x 2 x + =? x(x ) 6

62 We must recall how to ivie polynomials by polynomials an get a remainer of lower egree than the ivisor. Here we woul ivie the x 3 + 4x 2 x + by x(x ) = x 2 x to get a remainer of egree less than 2 (the egree of x 2 x). We woul obtain x 3 + 4x 2 x + x(x ) = x x + x(x ) since the quotient is x + 5 an the remainer is 4x +. Thus, in this situation x 3 + 4x 2 x + x = x(x ) Now we are reay to continue with the first algebra trick. In this case, the first trick is applie to We want constants A, B so that As above, multiply through by x(x ) to get an plug in the two values 0, to get That is, A = an B = 5. Putting this together, we have 4x + x(x ) 4x + x(x ) = A x + B x 4x + = A(x ) + Bx x x + x(x ) x = A 4 + = B x 3 + 4x 2 x + x(x ) = x x + 5 x Thus, x 3 + 4x 2 x + x = x(x ) x x + 5 x x = x x ln x + 5 ln(x ) + C In a slightly ifferent irection: we can o any integral of the form ax + b + x 2 x because we know two ifferent sorts of integrals with that same enominator: x = arctan x + C + x2 2x + x 2 x = ln( + x2 ) + C where in the secon one we use a substitution. Thus, we have to break the given integral into two parts to o it: ax + b + x 2 x = a 2x 2 + x 2 x + b + x 2 x 62

63 = a 2 ln( + x2 ) + barctan x + C An, as in the first example, if we are given a numerator of egree 2 or larger, then we ivie first, to get a remainer of lower egree. For example, in the case of x 4 + 2x 3 + x 2 + 3x + + x 2 x we ivie the numerator by the enominator, to allow us to write x 4 + 2x 3 + x 2 + 3x + + x 2 = x 2 + 2x + x + + x 2 since the quotient is x 2 + 2x an the remainer is x +. Then x 4 + 2x 3 + x 2 + 3x + + x 2 x = x 2 + 2x + x + + x 2 = x3 3 + x2 + 2 ln( + x2 ) + arctan x + C These two examples are just the simplest, but illustrate the iea of using algebra to simplify rational functions. #44.64 x(x ) x =? # x +x 2 x =? # x 3 +4 x(x+) x =? # x+x 2 +x 2 x =? # x 3 +4 x 2 x =? # x +x 2 x =? #44.70 x 3 + (x )(x 2) x =? #44.7 x 3 + x 2 + x =? 45. Trigonometric Integrals Here we ll just have a sample of how to use trig ientities to o some more complicate integrals involving trigonometric functions. This is just the tip of the iceberg. We on t o more for at least two reasons: first, harly anyone remembers all these tricks anyway, an, secon, in real life you can look these things up in tables of integrals. Perhaps even more important, in real life there are more sophisticate viewpoints which even make the whole issue a little silly, somewhat like evaluating 26 by ifferentials without your calculator seems silly. The only ientities we ll nee in our examples are cos 2 x + sin 2 x = Pythagorean ientity 63

64 sin x = cos 2x 2 + cos 2x cos x = 2 half-angle formula half-angle formula The first example is sin 3 x x If we ignore all trig ientities, there is no easy way to o this integral. But if we use the Pythagorean ientity to rewrite it, then things improve: sin 3 x x = ( cos 2 x) sin x x = ( cos 2 x)( sin x) x In the latter expression, we can view the sin x as the erivative of cos x, so with the substitution u = cos x this integral is ( u 2 ) u = u + u3 3 + C = cos x + cos3 x + C 3 This iea can be applie, more generally, to integrals sin m x cos n x x where at least one of m, n is o. For example, if n is o, then use cos n x = cos n x cos x = ( sin 2 x) n 2 cos x to write the whole thing as sin m x cos n x x = sin m x ( sin 2 x) n 2 cos x x The point is that we have obtaine something of the form (polynomial in sin x) cos x x Letting u = sin x, we have cos x x = u, an the integral becomes (polynomial in u) u which we can o. But this Pythagorean ientity trick oes not help us on the relatively simple-looking integral sin 2 x x since there is no o exponent anywhere. In effect, we ivie the exponent by two, thereby getting an o exponent, by using the half-angle formula: cos 2x sin 2 x x = x = x sin 2x C

65 A bigger version of this application of the half-angle formula is sin 6 cos 2x x x = ( ) 3 x = cos 2x cos2 2x 8 cos3 2x x Of the four terms in the integran in the last expression, we can o the first two irectly: 8 x = x 8 + C 3 3 cos 2x x = sin 2x + C 8 6 But the last two terms require further work: using a half-angle formula again, we have 3 3 3x 8 cos2 2x x = ( + cos 4x) x = sin 4x + C 64 An the cos 3 2x nees the Pythagorean ientity trick: 8 cos3 2x x = 8 Putting it all together, we have ( sin 2 2x) cos 2x x = 8 [sin 2x sin3 2x ] + C 3 sin 6 x x = x x sin 2x sin 4x + 8 [sin 2x sin3 2x ] + C 3 This last example is typical of the kin of repeate application of all the tricks necessary in orer to treat all the possibilities. In a slightly ifferent vein, there is the horrible sec x x There is no ecent way to o this at all from a first-year calculus viewpoint. A sort of rationalize-inhinsight way of explaining the answer is: sec x x = sec x(sec x + tan x) sec x + tan x All we i was multiply an ivie by sec x + tan x. Of course, we on t preten to answer the question of how a person woul get the iea to o this. But then (another miracle?) we notice that the numerator is the erivative of the enominator, so sec x x = ln(sec x + tan x) + C x There is something istasteful about this rationalization, but at this level of technique we re stuck with it. Maybe this is enough of a sample. There are several other tricks that one woul have to know in orer to claim to be an expert at this, but it s not really sensible to want to be expert at these games, because there are smarter alternatives. #45.72 cos 2 x x =? #45.73 cos x sin 2 x x =? #45.74 cos 3 x x =? 65

66 #45.75 sin 2 5x x =? #45.76 sec(3x + 7) x #45.77 sin 2 (2x + ) x =? #45.78 sin 3 ( x) x =? 46. Trigonometric Substitutions This section continues evelopment of relatively special tricks to o special kins of integrals. Even though the application of such things is limite, it s nice to be aware of the possibilities, at least a little bit. The key iea here is to use trig functions to be able to take the square root in certain integrals. There are just three prototypes for the kin of thing we can eal with: x 2 + x 2 x2 Examples will illustrate the point. In rough terms, the iea is that in an integral where the worst part is x 2, replacing x by sin u (an, corresponingly, x by cos u u), we will be able to take the square root, an then obtain an integral in the variable u which is one of the trigonometric integrals which in principle we now know how to o. The point is that then x2 = sin 2 x = cos 2 x = cos x We have taken the square root. For example, in x2 x we replace x by sin u an x by cos u u to obtain cos2 x2 x = sin 2 u cos u u = u cos u u = = cos u cos u u = cos 2 u u Now we have an integral we know how to integrate: using the half-angle formula, this is + cos 2u cos 2 u u = u = u sin 2u + + C An there still remains the issue of substituting back to obtain an expression in terms of x rather than u. Since x = sin u, it s just the efinition of inverse function that u = arcsin x To express sin 2u in terms of x is more aggravating. We use another half-angle formula sin 2u = 2 sin u cos u Then 4 sin 2u = 4 2 sin u cos u = 4 x x 2 66

67 where of course we use the Pythagorean ientity to give us Whew. cos u = sin 2 u = x 2 The next type of integral we can improve is one containing an expression + x 2 In this case, we use another Pythagorean ientity + tan 2 u = sec 2 u (which we can get from the usual one cos 2 u + sin 2 u = by iviing by cos 2 u). So we let x = tan u x = sec 2 u u (mustn t forget the x an u business!). For example, in we use an turn the integral into = x = tan u sec2 u tan u sec2 u u = by rewriting everything in terms of cos u an sin u. + x 2 x x x = sec 2 u u + x 2 + tan 2 u x = sec 2 u u = x tan u sec u tan u sec2 u u = sin u cos 2 u u For integrals containing x 2, use x = sec u in orer to invoke the Pythagorean ientity sec 2 u = tan 2 u so as to be able to take the square root. Let s not execute any examples of this, since nothing new really happens. Rather,, let s examine some purely algebraic variants of these trigonometric substitutions, where we can get some mileage out of completing the square. For example, consier 2x x2 x The quaratic polynomial insie the square-root is not one of the three simple types we ve looke at. But, by completing the square, we ll be able to rewrite it in essentially such forms: 2x x 2 = (2x + x 2 ) = ( + + 2x + x 2 ) = ( + ( + x) 2 ) = ( + x) 2 Note that always when completing the square we take out the coefficient in front of x 2 in orer to see what s going on, an then put it back at the en. So, in this case, we let sin u = + x 67 cos u u = x

68 In another example, we might have 8x 4x2 x Completing the square again, we have 8x 4x 2 = 4( 2 + x 2 ) = 4( + 2x + x 2 ) = 4( + (x ) 2 ) Rather than put the whole 4 back, we only keep track of the ±, an take a +4 outsie the square root entirely: 8x 4( 4x2 x = + (x )2 ) x Then we re back to a familiar situation. = 2 ( + (x )2 ) x = 2 (x )2 ) x #46.79 Tell what trig substitution to use for x 8 x 2 x #46.80 Tell what trig substitution to use for x 2 x #46.8 Tell what trig substitution to use for x 2 x #46.82 Tell what trig substitution to use for 9 + 4x 2 x #46.83 Tell what trig substitution to use for x 9 x 2 + x #46.84 Tell what trig substitution to use for x 8 x 2 x 47. Historical an theoretical comments: Mean Value Theorem For several reasons, the traitional way that Taylor polynomials are taught gives the impression that the ieas are inextricably linke with issues about infinite series. This is not so, but every calculus book I know takes that approach. The reasons for this systematic mistake are complicate. Anyway, we will not make that mistake here, although we may talk about infinite series later. Instea of following the traition, we will immeiately talk about Taylor polynomials, without first tiring ourselves over infinite series, an without fooling anyone into thinking that Taylor polynomials have the infinite series stuff as prerequisite! The theoretical unerpinning for these facts about Taylor polynomials is The Mean Value Theorem, which itself epens upon some fairly subtle properties of the real numbers. It asserts that, for a function f ifferentiable on an interval [a, b], there is a point c in the interior (a, b) of this interval so that f (c) = f(b) f(a) b a Note that the latter expression is the formula for the slope of the chor or secant line connecting the two points (a, f(a)) an (b, f(b)) on the graph of f. An the f (c) can be interprete as the slope of the tangent line to the curve at the point (c, f(c)). In many traitional scenarios a person is expecte to commit the statement of the Mean Value Theorem to memory. An be able to respon to issues like Fin a point c in the interval [0, ] satisfying the conclusion of the Mean Value Theorem for the function f(x) = x 2. This is pointless an we won t o it. 68

69 48. Taylor polynomials: formulas Before attempting to illustrate what these funny formulas can be use for, we just write them out. First, some reminers: The notation f (k) means the kth erivative of f. The notation k! means k-factorial, which by efinition is k! = (k ) k Taylor s Formula with Remainer Term first somewhat verbal version: Let f be a reasonable function, an fix a positive integer n. Then we have f(input) = f(basepoint) + f (basepoint) (input basepoint)! + f (basepoint) 2!... + f (n) (basepoint) n! for some c between basepoint an input. (input basepoint) 2 + f (basepoint) (input basepoint ) 3 3! (input basepoint) n + f (n+) (c) (input basepoint)n+ (n + )! That is, the value of the function f for some input presumably near the basepoint is expressible in terms of the values of f an its erivatives evaluate at the basepoint, with the only mystery being the precise nature of that c between input an basepoint. Taylor s Formula with Remainer Term secon somewhat verbal version: Let f be a reasonable function, an fix a positive integer n. f(basepoint + increment) = f(basepoint) + f (basepoint) (increment)! + f (basepoint) 2! (increment) 2 + f (basepoint) (increment ) 3 3!... + f (n) (basepoint) (increment) n + f (n+) (c) n! (n + )! (increment)n+ for some c between basepoint an basepoint + increment. This version is really the same as the previous, but with a ifferent emphasis: here we still have a basepoint, but are thinking in terms of moving a little bit away from it, by the amount increment. An to get a more compact formula, we can be more symbolic: let s repeat these two versions: Taylor s Formula with Remainer Term: Let f be a reasonable function, fix an input value x o, an fix a positive integer n. Then for input x we have f(x) = f(x o ) + f (x o )! for some c between x o an x. (x x o ) + f (x o ) 2! (x x o ) 2 + f (x o ) (x x o ) !... + f (n) (x o ) (x x o ) n + f (n+) (c) n! (n + )! (x x o) n+ Note that in every version, in the very last term where all the inices are n +, the input into f (n+) is not the basepoint x o but is, instea, that mysterious c about which we truly know nothing but that it lies 69

70 between x o an x. The part of this formula without the error term is the egree-n Taylor polynomial for f at x o, an that last term is the error term or remainer term. The Taylor series is sai to be expane at or expane about or centere at or simply at the basepoint x o. There are many other possible forms for the error/remainer term. The one here was chosen partly because it resembles the other terms in the main part of the expansion. Linear Taylor s Polynomial with Remainer Term: Let f be a reasonable function, fix an input value x o. For any (reasonable) input value x we have for some c between x o an x. f(x) = f(x o ) + f (x o )! (x x o ) + f (c) (x x o ) 2 2! The previous formula is of course a very special case of the first, more general, formula. The reason to inclue the linear case is that without the error term it is the ol approximation by ifferentials formula, which ha the funamental flaw of having no way to estimate the error. Now we have the error estimate. The general iea here is to approximate fancy functions by polynomials, especially if we restrict ourselves to a fairly small interval aroun some given point. (That approximation by ifferentials circus was a very crue version of this iea). It is at this point that it becomes relatively easy to beat a calculator, in the sense that the methos here can be use to give whatever precision is esire. So at the very least this methoology is not as silly an obsolete as some earlier traitional examples. But even so, there is more to this than getting numbers out: it ought to be of some intrinsic interest that pretty arbitrary functions can be approximate as well as esire by polynomials, which are so reaily computable (by han or by machine)! One element uner our control is choice of how high egree polynomial to use. Typically, the higher the egree (meaning more terms), the better the approximation will be. (There is nothing comparable to this in the approximation by ifferentials ). Of course, for all this to really be worth anything either in theory or in practice, we o nee a tangible error estimate, so that we can be sure that we are within whatever tolerance/error is require. (There is nothing comparable to this in the approximation by ifferentials, either). An at this point it is not at all clear what exactly can be one with such formulas. For one thing, there are choices. #48.85 Write the first three terms of the Taylor series at 0 of f(x) = /( + x). #48.86 Write the first three terms of the Taylor series at 2 of f(x) = /( x). #48.87 Write the first three terms of the Taylor series at 0 of f(x) = e cos x. 49. Classic examples of Taylor polynomials Some of the most famous (an important) examples are the expansions of x, ex, cos x, sin x, an log( + x) at 0: right from the formula, although simplifying a little, we get x = + x + x2 + x 3 + x 4 + x 5 + x e x = + x! + x2 2! + x3 3! + x4 4!

71 cos x = x2 2! + x4 4! x6 6! + x8 8!... sin x = x! x3 3! + x5 5! x7 7! +... log( + x) = x x2 2 + x3 3 x4 4 + x5 5 x where here the ots mean to continue to whatever term you want, then stop, an stick on the appropriate remainer term. It is entirely reasonable if you can t really see that these are what you get, but in any case you shoul o the computations to verify that these are right. It s not so har. Note that the expansion for cosine has no o powers of x (meaning that the coefficients are zero), while the expansion for sine has no even powers of x (meaning that the coefficients are zero). At this point it is worth repeating that we are not talking about infinite sums (series) at all here, although we o allow arbitrarily large finite sums. Rather than worry over an infinite sum that we can never truly evaluate, we use the error or remainer term instea. Thus, while in other contexts the ots woul mean infinite sum, that s not our concern here. The first of these formulas you might recognize as being a geometric series, or at least a part of one. The other three patterns might be new to you. A person woul want to be learn to recognize these on sight, as if by reflex! 50. Computational tricks regaring Taylor polynomials The obvious question to ask about Taylor polynomials is What are the first so-many terms in the Taylor polynomial of some function expane at some point?. The most straightforwar way to eal with this is just to o what is inicate by the formula: take however high orer erivatives you nee an plug in. However, very often this is not at all the most efficient. Especially in a situation where we are intereste in a composite function of the form f(x n ) or f(polynomial in x) with a familiar function f, there are alternatives. For example, looking at f(x) = e x3, if we start taking erivatives to expan this at 0, there will be a big mess pretty fast. On the other han, we might start with the familiar expansion for e x e x = + x + x2 2! + x3 3! + ec 4! x4 with some c between 0 an x, where our choice to cut it off after that many terms was simply a whim. But then replacing x by x 3 gives e x3 = + x 3 + x6 2! + x9 3! + ec 4! x2 with some c between 0 an x 3. Yes, we nee to keep track of c in relation to the new x. So we get a polynomial plus that funny term with the c in it, for the remainer. Yes, this gives us a ifferent-looking error term, but that s fine. So we obtain, with relative ease, the expansion of egree eleven of this function, which woul have been horrible to obtain by repeate ifferentiation an irect application of the general formula. Why eleven?: well, the error term has the x 2 in it, which means that the polynomial itself stoppe with a x term. Why in t we see that term? Well, eviently the coefficients of x, an of x 0 (not to mention x, x 2, x 4, x 5, x 7, x 8!) are zero. 7

72 As another example, let s get the egree-eight expansion of cos x 2 at 0. Of course, it makes sense to use cos x = x2 2! + x4 4! + sin c 5! with c between 0 an x, where we note that sin x is the fifth erivative of cos x. Replacing x by x 2, this becomes cos x 2 = x4 2! + x8 4! + sin c x 0 5! where now we say that c is between 0 an x 2. x 5 #50.88 Use a shortcut to compute the Taylor expansion at 0 of cos(x 5 ). #50.89 Use a shortcut to compute the Taylor expansion at 0 of e (x2 +x). #50.90 Use a shortcut to compute the Taylor expansion at 0 of log( x ). 5. Prototypes: More serious questions about Taylor polynomials Beyon just writing out Taylor expansions, we coul actually use them to approximate things in a more serious way. There are roughly three ifferent sorts of serious questions that one can ask in this context. They all use similar wors, so a careful reaing of such questions is necessary to be sure of answering the question aske. (The wor tolerance is a synonym for error estimate, meaning that we know that the error is no worse than such-an-such) Given a Taylor polynomial approximation to a function, expane at some given point, an given a require tolerance, on how large an interval aroun the given point oes the Taylor polynomial achieve that tolerance? Given a Taylor polynomial approximation to a function, expane at some given point, an given an interval aroun that given point, within what tolerance oes the Taylor polynomial approximate the function on that interval? Given a function, given a fixe point, given an interval aroun that fixe point, an given a require tolerance, fin how many terms must be use in the Taylor expansion to approximate the function to within the require tolerance on the given interval. As a special case of the last question, we can consier the question of approximating f(x) to within a given tolerance/error in terms of f(x o ), f (x o ), f (x o ) an higher erivatives of f evaluate at a given point x o. In real life this last question is not really so important as the thir of the questions liste above, since evaluation at just one point can often be achieve more simply by some other means. Having a polynomial approximation that works all along an interval is a much more substantive thing than evaluation at a single point. It must be note that there are also other ways to approach the issue of best approximation by a polynomial on an interval. An beyon worry over approximating the values of the function, we might also want the values of one or more of the erivatives to be close, as well. The theory of splines is one approach to approximation which is very important in practical applications. 72

73 52. Determining Tolerance/Error This section treats a simple example of the secon kin of question mentione above: Given a Taylor polynomial approximation to a function, expane at some given point, an given an interval aroun that given point, within what tolerance oes the Taylor polynomial approximate the function on that interval? Let s look at the approximation x2 2 + x4 4! to f(x) = cos x on the interval [ 2, 2 ]. We might ask Within what tolerance oes this polynomial approximate cos x on that interval? To answer this, we first recall that the error term we have after those first (oh-so-familiar) terms of the expansion of cosine is sin c x 5 5! For x in the inicate interval, we want to know the worst-case scenario for the size of this thing. A sloppy but goo an simple estimate on sin c is that sin c, regarless of what c is. This is a very happy kin of estimate because it s not so ba an because it oesn t epen at all upon x. An the biggest that x 5 can be is ( 2 ) Then the error is estimate as This is not so ba at all! sin c x 5 5! 2 5 5! We coul have been a little clever here, taking avantage of the fact that a lot of the terms in the Taylor expansion of cosine at 0 are alreay zero. In particular, we coul choose to view the original polynomial x2 2 + x4 4! as incluing the fifth-egree term of the Taylor expansion as well, which simply happens to be zero, so is invisible. Thus, instea of using the remainer term with the 5 in it, we are actually entitle to use the remainer term with a 6. This typically will give a better outcome. That is, instea of the remainer we ha must above, we woul have an error term cos c x 6 6! Again, in the worst-case scenario cos c. An still x 2, so we have the error estimate cos c x 6 6! 2 6 6! This is less than a tenth as much as in the first version. But what happene here? Are there two ifferent answers to the question of how well that polynomial approximates the cosine function on that interval? Of course not. Rather, there were two approaches taken by us to estimate how well it approximates cosine. In fact, we still o not know the exact error! The point is that the secon estimate (being a little wiser) is closer to the truth than the first. The first estimate is true, but is a weaker assertion than we are able to make if we try a little harer. This alreay illustrates the point that in real life there is often no single right or best estimate of an error, in the sense that the estimates that we can obtain by practical proceures may not be perfect, but represent a trae-off between time, effort, cost, an other priorities. #52.9 How well (meaning within what tolerance ) oes x 2 /2 + x 4 /24 x 6 /720 approximate cos x on the interval [ 0., 0.]? 73

74 #52.92 How well (meaning within what tolerance ) oes x 2 /2 + x 4 /24 x 6 /720 approximate cos x on the interval [, ]? #52.93 How well (meaning within what tolerance ) oes x 2 /2 + x 4 /24 x 6 /720 approximate cos x on the interval [ π 2, π 2 ]? 53. How large an interval with given tolerance? This section treats a simple example of the first kin of question mentione above: Given a Taylor polynomial approximation to a function, expane at some given point, an given a require tolerance, on how large an interval aroun the given point oes the Taylor polynomial achieve that tolerance? The specific example we ll get to here is For what range of x 25 oes (x 25) approximate x to within.00? Again, with the egree-one Taylor polynomial an corresponing remainer term, for reasonable functions f we have f(x) = f(x o ) + f (x o )(x x o ) + f (c) (x x o ) 2 2! for some c between x o an x. The remainer term is remainer term = f (c) (x x o ) 2 2! The notation 2! means 2-factorial, which is just 2, but which we write to be forwar compatible with other things later. Again: no, we o not know what c is, except that it is between x o an x. But this is entirely reasonable, since if we really knew it exactly then we be able to evaluate f(x) exactly an we are eviently presuming that this isn t possible (or we wouln t be oing all this!). That is, we have limite information about what c is, which we coul view as the limitation on how precisely we can know the value f(x). To give an example of how to use this limite information, consier f(x) = x (yet again!). Taking x o = 25, we have x = f(x) = f(xo ) + f (x o )(x x o ) + f (c) (x x o ) 2 = 2! = 25 + (x 25) ! 4 (c) (x 3/2 25)2 = = (x 25) (x 25)2 8 c3/2 where all we know about c is that it is between 25 an x. What can we expect to get from this? Well, we have to make a choice or two to get starte: let s suppose that x 25 (rather than smaller). Then we can write 25 c x From this, because the three-halves-power function is increasing, we have 25 3/2 c 3/2 x 3/2 Taking inverses (with positive numbers) reverses the inequalities: we have 25 3/2 c 3/2 x 3/2 So, in the worst-case scenario, the value of c 3/2 is at most 25 3/2 = /25. 74

75 An we can rearrange the equation: x [5 + 0 (x 25)] = (x 25)2 8 c3/2 Taking absolute values in orer to talk about error, this is Now let s use our estimate c 3/2 /25 to write x [5 + 0 (x 25)] = 8 c (x 3/2 25)2 x [5 + 0 (x 25)] 8 25 (x 25)2 OK, having one this simplification, now we can answer questions like For what range of x 25 oes (x 25) approximate x to within.00? We cannot hope to tell exactly, but only to give a range of values of x for which we can be sure base upon our estimate. So the question becomes: solve the inequality 8 25 (x 25)2.00 (with x 25). Multiplying out by the enominator of 8 25 gives (by coincience?) so the solution is 25 x 26. x 25 2 So we can conclue that x is approximate to within.00 for all x in the range 25 x 26. This is a worthwhile kin of thing to be able to fin out. #53.94 For what range of values of x is x x3 6 within 0.0 of sin x? #53.95 Only consier x. For what range of values of x insie this interval is the polynomial + x + x 2 /2 within.0 of e x? #53.96 On how large an interval aroun 0 is x within 0.0 of /( + x)? #53.97 On how large an interval aroun 00 is 0 + x within 0.0 of x? 54. Achieving esire tolerance on esire interval This thir question is usually the most ifficult, since it requires both estimates an ajustment of number of terms in the Taylor expansion: Given a function, given a fixe point, given an interval aroun that fixe point, an given a require tolerance, fin how many terms must be use in the Taylor expansion to approximate the function to within the require tolerance on the given interval. For example, let s get a Taylor polynomial approximation to e x which is within 0.00 on the interval [ 2, + 2 ]. We use e x = + x + x2 2! + x3 3! xn n! + e c (n + )! xn+ 75

76 for some c between 0 an x, an where we o not yet know what we want n to be. It is very convenient here that the nth erivative of e x is still just e x! We are wanting to choose n large-enough to guarantee that e c (n + )! xn for all x in that interval (without knowing anything too etaile about what the corresponing c s are!). The error term is estimate as follows, by thinking about the worst-case scenario for the sizes of the parts of that term: we know that the exponential function is increasing along the whole real line, so in any event c lies in [ 2, + 2 ] an e c e /2 2 (where we ve not been too fussy about being accurate about how big the square root of e is!). An for x in that interval we know that x n+ ( 2 )n+ So we are wanting to choose n large-enough to guarantee that Since e c (n + )! ( 2 )n e c (n + )! ( 2 2 )n+ (n + )! ( 2 )n+ we can be confient of the esire inequality if we can be sure that That is, we want to solve for n in the inequality 2 (n + )! ( 2 )n (n + )! ( 2 )n There is no genuine formulaic way to solve for n to accomplish this. Rather, we just evaluate the lefthan sie of the esire inequality for larger an larger values of n until (hopefully!) we get something smaller than So, trying n = 3, the expression is which is more like 0.0 than So just try n = 4: which is better than we nee. 2 (3 + )! ( 2 )3+ = (4 + )! ( 2 )4+ = The conclusion is that we neee to take the Taylor polynomial of egree n = 4 to achieve the esire tolerance along the whole interval inicate. Thus, the polynomial + x + x2 2 + x3 3 + x4 4 approximates e x to within for x in the interval [ 2, 2 ]. 76

77 Yes, such questions can easily become very ifficult. An, as a reminer, there is no real or genuine claim that this kin of approach to polynomial approximation is the best. #54.98 Determine how many terms are neee in orer to have the corresponing Taylor polynomial approximate e x to within 0.00 on the interval [, +]. #54.99 Determine how many terms are neee in orer to have the corresponing Taylor polynomial approximate cos x to within 0.00 on the interval [, +]. # Determine how many terms are neee in orer to have the corresponing Taylor polynomial approximate cos x to within 0.00 on the interval [ π 2, π 2 ]. #54.20 Determine how many terms are neee in orer to have the corresponing Taylor polynomial approximate cos x to within 0.00 on the interval [ 0., +0.]. # Approximate e /2 = e to within.0 by using a Taylor polynomial with remainer term, expane at 0. (Do NOT a up the finite sum you get!) # Approximate 0 = (0) /2 to within 0 5 using a Taylor polynomial with remainer term. (Do NOT a up the finite sum you get! One point here is that most han calculators o not easily give 5 ecimal places. Hah!) 55. Integrating Taylor polynomials: first example Thinking simultaneously about the ifficulty (or impossibility) of irect symbolic integration of complicate expressions, by contrast to the ease of integration of polynomials, we might hope to get some mileage out of integrating Taylor polynomials. As a promising example: on one han, it s not too har to compute that T On the other han, if we write out then we coul obtain T 0 0 x x x = [ log( x)]t 0 = log( T ) x = + x + x2 + x 3 + x ( + x + x 2 + x 3 + x ) x = [x + x2 2 + x ]T 0 = = T + T T T Putting these two together (an changing the variable back to x ) gives log( x) = x + x2 2 + x3 3 + x (For the moment let s not worry about what happens to the error term for the Taylor polynomial). 77

78 This little computation has several useful interpretations. First, we obtaine a Taylor polynomial for log( T ) from that of a geometric series, without going to the trouble of recomputing erivatives. Secon, from a ifferent perspective, we have an expression for the integral T 0 x x x without necessarily mentioning the logarithm: that is, with some suitable interpretation of the trailing ots, T x x x = T + T T T Integrating the error term: example Being a little more careful, let s keep track of the error term in the example we ve been oing: we have x = + x + x x n + (n + ) xn+ ( c) n+ for some c between 0 an x, an also epening upon x an n. One way to avoi having the ( c) n+ blow up on us, is to keep x itself in the range [0, ) so that c is in the range [0, x) which is insie [0, ), keeping c away from. To o this we might eman that 0 T <. For simplicity, an to illustrate the point, let s just take 0 T 2. Then in the worst-case scenario ( c) n+ ( 2 )n+ = 2n+ Thus, integrating the error term, we have T 0 n + ( c) n+ xn+ x T n + 2n+ x n+ x = 2n+ x n+ x n + 0 = 2 n+ n + [ xn+2 n + 2 ]T 0 = 2n+ T n+2 (n + )(n + 2) Since we have cleverly require 0 T 2, we actually have That is, we have T 0 n + ( c) n+ xn+ x 2n+ T n+2 (n + )(n + 2) log( T ) [T + T 2 2n+ ( 2 )n+2 (n + )(n + 2) = 2(n + )(n + 2) for all T in the interval [0, 2 ]. Actually, we ha obtaine log( T ) [T + T 2 an the latter expression shrinks rapily as T approaches T n n ] 2(n + )(n + 2) T n n ] 2 n+ T n+2 2(n + )(n + 2) 78