Calculus I. Lecture Notes for MATH 114. Richard Taylor. Department of Mathematics and Statistics. c R. Taylor 2007

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1 Calculus I Lecture Notes for MATH 4 Richar Talor Department of Mathematics an Statistics c R. Talor 2007 last revision April 5, 2007

2 MATH 4 Calculus I Page 2 of 26 Contents Preinaries 6 2 Functions 6 2. Preinar Discussion Domain an Range Functions whose graphs ou shoul know Transformations of graphs The Derivative an Limits: Motivation 7 3. Instantaneous Velocit Slopes of Graphs Limits 8 4. Intuitive & Graphical Approach to Limits Algebraic Tricks for Computing Limits Limit Laws Continuit Limits Involving Infinit Derivatives 5. Definition of the Derivative Calculating Derivatives Using the Definition Derivative Functions from Graphs Power Rule Derivative Rules Derivatives of Important Functions Implicit Differentiation 4 6. Inverse Trigonometric Functions Logarithmic Differentiation Applications 6 7. Relate Rates Etreme Values Applie Optimization Linear Approimation Differentials Error Analsis Newton s Metho 22 0 Shapes of Graphs Increasing an Decreasing Functions

3 MATH 4 Calculus I Page 3 of Concavit Curve Sketching L Hôpital s Rule 24 2 Parametric Curves 25

4 MATH 4 Calculus I Page 4 of 26 About this Document These lecture notes are intene for m use onl, an the are ver much a work in progress. I make them available for free an without an warrant, epresse or implie. In fact ou probabl shouln t use them at all. This is not a tetbook. Please, bu the require tetbook for the course an use it. M aim is to prouce a conense set of notes to help me organize an improve upon m lecture material. Thus, ver few of m eamples inclue full solutions; these are to be worke out in class, along with the sunr other etails, iagrams, humorous anecotes, an everthing else that makes attening lectures worthwhile. Do not use these as a substitute for taking notes in class. You are free to cop an istribute this ocument, in whole or in part. I on t even min if ou teach from these notes because ou re too laz to make up our own. In either case, please creit the author.

5 MATH 4 Calculus I Page 5 of 26 Calculus I Lecture Notes for MATH 4 Course Sllabus Section numbers refer to the assigne tet: James Stewart, Single Variable Calculus: Concepts an Contets, 3r eition, Thomson, Functions omain an range.... sketching & manipulating graphs.... composition/ecomposition of functions Calculus motivation velocit an tangent line problems efinition of erivative erivative functions from graphs Limits graphical its one-sie its an eistence of its it laws its involving infinit continuit Intermeiate Value Theorem Derivatives erivative calculations from the efinition ifferentiabilit power rule sum/ifference rules; polnomials prouct an quotient rules trig, eponential an logarithmic functions... 3., 3.4, 3.7 chain rule implicit ifferentiation inverse trig functions Applications ma/min values an optimization...4.2, 4.6 relate rates linear approimation an ifferentials Newton s metho L Hôpital s rule curve sketching parametric curves...3.5

6 MATH 4 Calculus I Page 6 of 26 Lec. # Lec. #2 Preinaries - aminister Calculus Prepareness Test - mark test an begin iscussing - finish iscussing Calculus Prepareness Test 2 Functions 2. Preinar Discussion Lec. #3 What is a function? - rule that sas how to calculate output given input - a transformation of (i.e. action on) a number Functions are everwhere. In the sciences, almost all of what ou learn or research is about relationships between quantities (e.g. pressure an temperature in an ieal gas, carring capacit an population growth, genetic information an the form/function of an organism,...). These relationships can be epresse as functions. Often we on t know a formula for the relationship, but nevertheless it is the relationship (hence a function) that is the main object of stu. Was to specif functions - formula f() = - table of values (not ver etaile) - graph (not ver precise) Eample 2.. Epress the surface area of a sphere as a function of its circumference, C. Eample 2.2. A bo is to be esigne so it has a square base an an open top, with a volume of 00 cm 3. Epress the surface area as a function of the with of the base. 2.2 Domain an Range Definition 2.. For a given function f() the omain of f, written D(f), is the set of all numbers such that f() is efine. Definition 2.2. For a given function f() the range of f, written R(f), is the set of all numbers such that = f() for some number in the omain of f. Eample 2.3. Fin the omain range of: f() = 2 g() = h() = Functions whose graphs ou shoul know 2, 3, 4,, 2, sin, cos, tan, e, ln 2.4 Transformations of graphs Given graph of f(), how to graph f( a), f( + a), af(), f(a).

7 MATH 4 Calculus I Page 7 of 26 Lec. #4 3 The Derivative an Limits: Motivation 3. Instantaneous Velocit Basic Problem: Suppose an object is moving along the -ais, an that its position at an time t is given b a function = f(t). How o we fin a function that gives the object s velocit at time t? Eample 3.. Suppose f(t) = t 2. How to fin the velocit at time t = 3? Make table of values for = f(t) for t near 3. Consier average velocit over some time interval starting at t = 3. The velocit we get epens on the averaging interval t; smaller t gives more accurate velocit. But what is the eact, instantaneous velocit at t = 3? Clearl it shouln t rel on an particular value of t. Iea: as t gets smaller, the average v we calculate gets closer an closer to some value (which we suspect is 6)... this value gives the instantaneous velocit. We can justif the velocit 6 more rigorousl Notation: t 0 ( ) means the iting value, as t gets closer an closer to 0, of the epression ( ). Then at t = 2 the velocit is In general, the velocit at time t is given b 3.2 Slopes of Graphs (3 + t) 2 (3) 2 v = t 0 t t + ( t) 2 9 = t 0 t 6 t + ( t) 2 = t 0 t = (6 + t) t 0 = 6. v(t) = t 0 f(t + t) f(t). t Basic Problem: Given a function f(), how o we fin the slope of the graph (i.e. the slope of the tangent line to the graph) at a particular point? Eample 3.2. Suppose f() = 2. How to fin the slope at the point where = 3? Start with something close to the tangent line: the secant line joining (, f()) to a nearb point ( + h, f( + h)): m = rise f( + h) f() = run h Of course the slope we get epens on h. In our case, with f() = 2 an = 2:

8 MATH 4 Calculus I Page 8 of 26 f(2 + h) f(2) m = h = (2 + h)2 2 2 h = 4 + 4h + h2 4 h = 4 + h So with h = 0. we get a slope m = 4.. With h = 0.0 we get m = 4.0. As h gets smaller, the secant line gets closer to the tangent line. In fact, the slope of the tangent line is the iting value that the slope of the secant line approaches as h goes to 0: m = h 0 (4 + h) = 4 Lec. #5 4 Limits This course is reall about the erivative, but to calculate erivatives we nee to be able to unerstan an calculate its, like the h 0 that appears in the efinition. Notation: f() = L means, roughl, as approaches the number a the value of f() approaches the number L. 4. Intuitive & Graphical Approach to Limits Eample = 9 (interpret in terms of graph) - graphical eample for a function with a hole - graphical eample for a function with a point iscontinuit - graphical eample for a function with a jump iscontinuit: the it oesn t eist Notation: f() means the it as a from the left. Similarl, f() means the it + as a from the right. Theorem 4.. The it f() eists an equals L if an onl if f() = f() = L. + Eample 4.2. Evaluate 5 f(), if it eists, for the function { 2 if < 5 f() = 25 if > 5. (interpret in terms of the graph) 4.2 Algebraic Tricks for Computing Limits Common tricks involving factoring, reciprocals an roots: Eample 4.3. Evaluate (a) (b)

9 MATH 4 Calculus I Page 9 of 26 Eample 4.4. Evaluate (a) (b) Eample 4.5. Evaluate (a) (b) +2 3 Lec. #6 4.3 Limit Laws Guessing values of its isn t goo enough; we nee some rules. Suppose that f() an g() both eist; then:. c = c 2. n = a n 3. [f() ± g()] = f() ± g() 4. [cf()] = c f() 5. [f()g()] = f() 6. g() = 7. [f()] n = f() [ ] f() [ ] g() g() provie [ ] n f() 8. n f() = n f() g() 0. Eample 4.6. In a previous eample we ha to evaluate = = ( 4)( + 2) 4 4 = 4 ( + 2). Of course we suspect the answer is = 4, but we nee to justif this using the it laws: ( + 2) = + 2 (rule 3) = (rules & 8) = (rule 2) = Continuit - intuitive graphical eamples (where f(a) ne, or the it ne, or both) Definition 4.. A function f() is continuous at = a if f() = f(a). - justif our conclusions in previous eamples { if < Eample 4.7. Let f() = 2. Is f() continuous at =? if

10 MATH 4 Calculus I Page 0 of 26 if < 0 Eample 4.8. Let f() = 2 if > 0. For what value(s) of C is f() continuous at = 0? C if = 0. The following functions are continuous everwhere on their omains: 2, 3,..., sin, cos, tan, /,, e, ln,... in fact ever elementar function ou know. Theorem 4.2. Suppose f() an g() are both continuous at = a. Then the following are also continuous at = a:. f() ± g() 2. f()g() 3. f() g() (provie g(a) 0) 4. (f g)() (provie f() is continuous at = g(a)) Eample 4.9. At what value(s) of are the following not continuous? (a) f() = sin (b) g() = + 2 (c) h() = tan + 2 () F() = sin Lec. #7 4.5 Limits Involving Infinit Up to this point, we woul have sai that 0 oes not eist. However, it s more precise to sa 2 that 0 2 = +, in that the function / 2 gets arbitraril large (an positive) as approaches 0 from either the left or the right. Similarl, we have = + an =. The following eample introuces a notation that is helpful in evaluating its involving infinit. ( + 2)( 5) Eample 4.0. Evaluate 3 + ( + )(3 ). Strictl speaking our its laws o not appl, because the enominator has a it of zero. However, it is help to write: ( + 2)( 5) 3 + ( + )(3 ) = (5)( 2) (4)(0 ) = +. Here the smbol 0 is use to represent the negative quantit 3, which approaches 0 from the left as 3 +. Overall the function evaluates to a positive quantit (because of the 2 in the numerator) that becomes arbitraril large as 3 +, hence the iting value if +. 5 Eample 4.. Evaluate: (a) (b) 0 5(2 + ) (3 ) Another kin of it involving infinit is the following: (c)

11 MATH 4 Calculus I Page of 26 which means the iting value of the function / as becomes arbitraril large (an positive). Thus = 0. The following eample illustrates a useful trick in evaluating this kin of it. + 2 Eample 4.2. Evaluate We can rewrite the it as follows, b iviing numerator an enominator b the highest power of : Now appl it laws: = = ( ( + 2) ( 2 + 8) ) + 2 ( ) = 2 2 = = = Eample 4.3. Evaluate: (a) (b) (c) + 5 Derivatives Lec. #8 5. Definition of the Derivative In both instantaneous velocit an slopes of tangent lines, we ene up with an epression like: f( + h) f() h 0 h (In the velocit problem, the s were t s, an h was a t, but in form the epressions are the same.) This epression arises in so man ifferent applications that is has a name, an a shorthan notation: Definition 5.. The erivative of a function f() is given b an is efine for all for which this it eists. f () = h 0 f( + h) f() h Note that f () is itself a function of. For a given number, the number f () can be interprete as a formula for the slope of the tangent line to the graph = f() at. It is important to memorize this efinition an to unerstan where it comes from. Notation: The following alternative notations for the erivative are useful in ifferent situations: The smbol f () = f = f() means the erivative (with respect to ) of the epression that follows.

12 MATH 4 Calculus I Page 2 of Calculating Derivatives Using the Definition Eample 5.. Use the efinition of the erivative to fin f () for: (a) f() = 2 (b) f() = 3 (c) f() = () f() = Eample 5.2. For each function in the previous eample, fin the slope of = f() at = Derivative Functions from Graphs Sketch some graphs of = f(), an from these sketch graphs of = f (). = f() = f() (a) (b) = f() (c) = f() = f() () = f() (e) (f) Lec. #9 5.4 Power Rule Calculating erivatives using the efinition can be ver ifficult. Consier fining f () for the complicate function f() = cos 2 (ln) + sec ; eventuall we ll see how, but it s much too har to o it irectl from the efinition. Instea, we ll use some general properties of the erivative that allow us avoi the it calculation. We alrea use the efinition to prove the 2 = 2. Similar calculations show that 3 = 3 2, 4 = 4 3, an in fact we coul prove the general rule that n = n n So, for eample, we now sa (without going to the trouble of using the efinition of the erivative) that 0 = Derivative Rules From now on we will calculate erivatives smbolicall, b recognizing patterns an using the following smbolic rules, rather than use the efinition irectl.. C = 0

13 MATH 4 Calculus I Page 3 of n = n n [power rule] ( ) Cf() = Cf () [ ] f() ± g() = f () ± g () [ ] f()g() = f ()g() + f()g () [prouct rule] f() g() = f ()g() f()g () [g()] 2 f( g() ) = f ( g() ) g () [quotient rule] [chain rule] The following eamples involve applications of the first four rules. Eample 5.3. Differentiate smbolicall: (a) ( ) (b) 53 (c) ( 2 ) () ( + 2)2 Eample 5.4. Fin the slope of the tangent line to = + 2 at = 3. The following eamples involve the prouct an quotient rules. Eample 5.5. Differentiate: (a) (3 + 5 )(2 + ) (b) 2 ( 8) 2 (c) 4 () The following eamples involve the chain rule. Eample 5.6. Differentiate: (a) (3 + 2) 6 (b) (52 ) 0 Eample 5.7. Fin the slope of = 5 at = 4. (c) 2 () Derivatives of Important Functions You will nee to know the erivatives of the following important elementar functions (all of the functions that are important enough to be available on a scientific calculator) sin = cos 4. ln = cos = sin 5. ln = tan = sec2 The following eamples combine these erivative formulas with the various erivative rules. Eample 5.8. Differentiate the following: (a) 2 sin (b) e + 2 (c) sin(2 ) () e3 (e) e

14 MATH 4 Calculus I Page 4 of 26 Lec. #0 6 Implicit Differentiation Eample 6.. Fin the erivatives of: (a) f() = 2 (b) f() = (sin) 2 (c) f() = (ln) 2 () f() = ( + 3 ) 2 Notice the pattern? Suppose () is a given ifferentiable function. Write a formula (in terms of ()) for the erivative of f() = (()) 2. Eample 6.2. Suppose () is a given ifferentiable function. Write formulas (in terms of ()) for the erivatives of: (a) f() = (()) 3 (b) f() = 2 () (c) f() = 3 + () () f() = ()sin(()) The technique above is useful for fining erivatives of implicitl efine functions. Eample 6.3. Suppose the function () satisfies = 0. (The graph of () is a semicircle.) Fin the slope of the graph of () at the point (, 3). We know the function () satisfies the equation 2 + (()) 2 = 0. We coul solve this for (), then ifferentiate. But there is an easier wa. Differentiate both sies of the equation (with respect to ): 2 + 2() () = 0 or simpl = 0. We can solve this for (): So, at the point (, 3), the slope is 2 = 2 = =. = 3. In the previous eample, the function () is implicitl efine b the relationship = 0. That is, the equation coul, in principle, be solve for (). Until the etails are worke out an the equation is actuall solve, the precise efinition of () remains implicit. If we solve the equation we get () = 0 2, which is an eplicit representation of (). The technique of implicit ifferentiation use above makes it possible to ifferentiate implicitl efine functions, without actuall solving to get an eplicit formula for (). Eample 6.4. Fin a formula for if the function () is efine implicitl b the equation Eample 6.5. Fin a formula for : =. (a) = (b) = + e (c) cos + cos = () sin Eample 6.6. Fin the slope of the curve at the point (3, ). 2( ) 2 = 25( 2 2 ) ( ) =

15 MATH 4 Calculus I Page 5 of 26 Lec. # 6. Inverse Trigonometric Functions So far we have formulas for the erivatives of ever important function on the scientific calculator ecept the inverse trigonometric functions arcsin, arccos an arctan. Recall that arcsin returns the angle (between π/2 an π/2) whose sine is. That is, = arcsin = sin. The other inverse trig functions are efine similarl. Notation: The functions arcsin, arccos an arctan are sometimes written sin, cos an tan respectivel. But be careful: while the notation sin 2 is use to represent (sin) 2, in the special case of sin we o not mean (sin ). Here, the refers to the inverse of the sine function, not its reciprocal. That is, sin sin We can fin the erivatives of inverse trig functions using implicit ifferentiation. Eample 6.7. To erive a formula for arcsin, let () be efine b = arcsin. Or equivalentl, Appl implicit ifferentiation: sin =. (cos ) = = = cos = cos(arcsin ). This formula can be simplifie using the following trick. Recall that = arcsin is the angle whose sine is. This relationship between an is illustrate in the figure below. To evaluate cos we can use Pthagoras to solve for the ajacent sie, 2. Then So we have cos = 2 = 2. = cos = 2. Use the same metho to fin the erivatives of the other inverse trig functions, an memorize them: arcsin = 2 arccos = 2 arctan = + 2 Lec. #2 6.2 Logarithmic Differentiation We have no erivative rules that tell us how to ifferentiate something like ( sin )

16 MATH 4 Calculus I Page 6 of 26 because the function sin appears in the eponent (tr it an ou ll see). You can use implicit ifferentation to get aroun this problem. Let Take logarithms of both sies to get () = sin. ln = ln( sin ) then use algebraic rules for logarithms to write this as Now appl implicit ifferentiation: ln = sin ln(). = cos ln() + (sin) = = Substituting sin back in for we get [ = sin cos ln() + (sin) ]. [ cos ln() + (sin) ]. Eample 6.8. Differentiate: (a) ( ) (b) (2 ) (c) (ln) () (cos ) The previous eample (b) can be generalize to the following: (a ) = (a )ln(a) 7 Applications Lec. #3 7. Relate Rates Eample 7.. Consier a circle whose raius is 0 cm an growing at 2 cm/s. At what rate is the area of the circle increasing? If we represent the raius b an (unknown) function r(t) (where t is the time in secons) then the area as a function of time can be written A(t) = π ( r(t) ) 2. Use implicit ifferentiation to ifferentiate both sies with respect to t: A = π 2rr. Recall that A an r represent the instantaneous rates of change of the area an the raius, respectivel. Using the given ata r = 0 cm an r = 2 cm/s we get: A = π 2(0 cm)(2 cm/s) = 40π cm 2 /s 26 cm 2 /s The previous eample is a relate rates problem in that A(t) an r(t) are functions of time whose rates of change are relate through the equation A = πr 2. Implicit ifferentation is use to fin the relationship relating the rates of change A an r.

17 MATH 4 Calculus I Page 7 of 26 Eample 7.2. We can moel the human heart as a sphere of raius 5 cm. As the heart contracts its raius ecreases at about cm/0. s= 20 cm/s. At what rate is bloo epelle from the heart as it contracts? For a sphere heart the volume is V = 4 3 πr3 where V (t) an r(t) are both (unknown) functions of time. Differentiating with respect to t we get Using r = 5 cm an r = 0 cm/s we get V = 4 3 π 3r2 r = 4πr 2 r. V = 4π(5) 2 ( 0) 34 cm 3 /s 3. L/s. Eample 7.3. Susan is.5 m tall. As she walks awa from a 3 m-tall lamppost at 2 m/s, what is the spee of the tip of her shaow? [similar triangles] Eample 7.4. The clock on the clocktower has a minute han 80 cm long an an hour han 40 cm long. At 3:00, at what range is the istance between the ens of the hans changing? [cosine law] Eample 7.5. San is being ae to a conical pile at a rate of 5 m 3 /min. At all times the pile retains its shape so that the cone s height is alwas eactl twice its raius. At what rate is the height of the pile increasing when the height is 2 m? Lec. #4 7.2 Etreme Values Eample 7.6. A farmer has 00 m of fencing material an wants to buil a rectangular enclosure for his llamas. What shoul the imensions of the enclosure be, so that it has the maimum possible area? Let the enclosure have with an height. The farmer nees to maimize the area A =. Because he has a ite amount of fencing, he can t just make both an as large as he wants. In fact, if he chooses a value for the with,, then the height is automaticall etermine b the remaining fencing: = 00 = 2 = 00 2 = = = 50. We can use this to einate an write the area as a function of the length of just one of the sies: The graph of this function is shown below. A = A() = (50 ) = A() 50

18 MATH 4 Calculus I Page 8 of 26 Notice: the maimum of A() occurs where the graph has a horizontal tangent line, i.e. where A () = 0. Using this observation, we can fin the value of that gives the maimum of A(): A () = 0 = 50 2 = 0 = = 25. So, to maimize the area of the enclosure, the imensions shoul be = 25 an = = 25 (i.e. the enclosure shoul be a square). Some terminolog: Consier the graph of = f() below. = f() a b c f has an absolute or global maimum at = c f has an absolute or global minimum at = b f has a local maimum at = a an at = c f has a local minimum at = b an at = Definition 7.. A function f on in interval [a, b] has an absolute or global......maimum at = c if f(c) f() for all [a, b]....minimum at = c if f(c) f() for all [a, b]. There are onl two was for a function f() to have a local etremum (i.e. a local ma or min). These are illustrate below. = f() = f() This motivates the following efinition an theorem. c c Definition 7.2. A number c is a critical number for a function f if either f (c) = 0 or f (c) oes not eist. Theorem 7.. If f() has a local maimum or minimum at = c then c is a critical number for f. In applications we usuall want to fin an absolute etremum, not just the local etrema. If f is a continuous function on a close interval [a, b] then there are just two was for an absolute etremum to occur: either the etremum occurs at a local etremum insie the interval, or it occurs at one of the enpoints of the interval. These two possibilities are illustrate below.

19 MATH 4 Calculus I Page 9 of 26 = f() = f() a c b a b Together with the previous theorem, this observation suggests the following recipe for fining local etrema. Close Interval Metho: To fin the absolute minimum an maimum of a continuous function f() on an interval [a, b]:. Fin the critical numbers for f in the interval [a, b]. 2. Evaluate f at the critical numbers an at the enpoints. 3. Select the largest an the smallest values from step 2; these are the absolute maimum an minumum values of f on [a, b]. Eample 7.7. Fin the absolute maimum an minimum values of f() = on the interval [, 5]. First we fin the critical numbers for f: so f () = f () = 0 = = 0 = 3( ) = 0 = 3( )( 3) = 0 = = or = 3. Therefore the critical numbers for f are an 3. Evaluating f at these numbers we get f() = 6 f(3) = 2 an evaluating f at the enpoints of the interval [, 5] gives f( ) = 4 f(5) = 22. Selecting the largest of these, we fin the the absolute maimum of f() is 22 an occurs at = 5; the absolute minimum is 2 an occurs at = 3. Lec. #5 7.3 Applie Optimization Eample 7.8. A farmer has 400 m of fencing to buil a rectangular enclosure with a partition parallel to one of the sies. What shoul the imensions be so the maimum possible area is enclose? Eample 7.9. A farmer want to buil a rectangular enclosure with an area of 00 m 2. What shoul the imensions be so the minimum length of fencing is use? Eample 7.0. A bo with a square base an an open top must have a volume of 000 cm 3. What shoul the imensions be so that the minimum amount of material is use in the construction of the bo. Eample 7.. If 2400 cm 2 of carboar is available to make a close bo with a square base, what shoul the imensions be so the volume is as large as possible? Eample 7.2. Fin the point on the graph of = 2 that is closest to the point (, 0).

20 MATH 4 Calculus I Page 20 of 26 Lec. #6 8 Linear Approimation Suppose we want to estimate the value of 4. without a calculator. Since we know 4 = 2, we suspect the answer shoul be slightl greater than 2. The problem is to estimate how much greater. Note that this problem is equivalent to estimating the height of the graph of = at = 4.. Let the line tangent to the graph at = 4 be given b = L(). Because the graphs of = an = L() are ver close (at least for values of near 4, we shoul be able to estimate 4. b simpl evaluating L(4.). For this reason, the function L() is calle the linear approimation of f() base at = 4. 2 = L() = 4 To fin the function L() = m + b we nee the slope m which comes from evaluating the erivative at = 4: = 2 /2 = m = 2 (4) /2 = 4. So we know L() = 4 + b. To fin b we force this line to go through the point (4, 2): 2 = b = b =. So an we can use this to estimate L() = L(4.) = 4 (4.) + = (compare this with the eact value 4. = ) In general, the linear approimation to a function f(), base at = 0, is L() = f( 0 ) + f ( 0 )( 0 ) Keep in min: this is just the equation of the tangent line (in point-slope form) to the graph of = f() at 0. The main iea behin the linear approimation is that L() is close to f(), provie is close to 0. Eample 8.. Use a linear approimation to estimate What we want is a linear approimation of the function f() = 3 base at = 27 (since we know that 3 27 = 3). We have so that f () = 3 2/3 = f (27) = 3 (27) 2/3 = 27 L() = f(27) + f (27)( 27) = ( 27). Now we can use this to get our approimation. Since 28 is close to 27 we have 3 28 L(28) = (28 27) Eample 8.2. Use a linear approimation to estimate (a) (b) ln(.2) (c) sin(0.05)

21 MATH 4 Calculus I Page 2 of Differentials Consier again the problem of estimating 4.. An alternative approach, again base on the function f() =, is to write the erivative as f = 2 /2 an consier the epression f as the slope, i.e. rise over run, between two points on the graph, a ver small istance apart. That is, we can think of f as the infinitesimal rise corresponing to an infinitesimal run of. The quantities f an must be thought of as being infinitesimall small, an the are calle infinitesimals or ifferentials. Strictl speaking the equation above is onl true in the it 0. However, for finite but sufficientl small, it is still a goo approimation: f 2 /2 = f 2 /2. Thus, when = 4, if we increase b an amount = 0. then we epect the value of f() to increase b about f = 2 (4) /2 (0.) = so we estimate that = Eample 8.3. Use ifferentials to estimate (a) (b) ln(.2) (c) sin(0.05) Lec. #7 8.2 Error Analsis Linear approimations an ifferentials are use often in error analsis of calculations base on eperimental ata. Eample 8.4. The with of a cube measure to be 3.0 cm with an uncertaint of 0. cm. Calculate the volume an the associate uncertaint. The volume is clearl V = 3 = (3.0) 3 = 297 cm 3. To estimate the uncertaint, consier: how much oes the volume change if the true with is 3. cm (i.e. if our measurement of 3.0 cm is wrong b the full uncertaint amount)? We coul calculate this eactl, but an approimation will be goo enough (an simpler) so we ll use ifferentials: So we estimate that V = 32 = V = 3 2 = 3(3.0) 2 (0.) 5 cm 3. V = 297 ± 5 cm 3 Eample 8.5. You measure the raius of a sphere to be 0 cm wtih an uncertaint of 2 cm. What is the volume, an what is the associate uncertaint. Eample 8.6. You measure height of a tree b staning 40 m from its base an measuring the angle of elevation of our line of sight to the top of the tree to be 30 with an uncertaint of 2. What is the height of the tree an the associate uncertaint?

22 MATH 4 Calculus I Page 22 of 26 9 Newton s Metho Often when mathematics is applie to a problem in science, the solution of the problem is foun b solving an equation. But what to o if ou on t know how to solve the equation ou nee to solve? In fact, man equations simpl can t be solve eactl. In situations like this, an approimate solution (accurate to, sa, 6 ecimal places) is all ou reall nee. Newton s Metho is one wa of getting ver accurate approimate solutions of equations. Suppose we wante to solve the equation B rewriting this as cos =. } {{ cos } = 0 f() we can interpret the solution as the -intercept the graph of = f(). From the graph of f = f() we can rea off a number close to the -intercept; call this 0. We can then fin another point,, that is even closer to the -intercept b oing the following: construct the tangent line to the graph at 0 fin the point where the tangent line crosses the -ais; call this. = f() 0 If ou go through the algebra, ou will obtain the following formula for in terms of 0 : = 0 f( 0) f ( 0 ). B repeating this process, constructing a tangent line at an locating its -intercept at 2, we obtain a point even closer to the -intercept of the graph of = f(): 2 = f( ) f ( ). In fact, we can repeat this process an number of times, with increasing accurac, alwas iterating the formula n+ = n f( n) f for n = 0,, 2,... ( n ) Eample 9.. To solve the equation f() = cos = 0, we can start with an initial guess 0 = (which is close to the intersection of the graphs of = an = cos ). Newton s metho gives the formula n+ = n f( n) f ( n ) = n n cos( n ) + sin( n )

23 MATH 4 Calculus I Page 23 of 26 so = cos() + sin() cos(0.75) 2 = sin(0.75) cos(0.739) 3 = sin(0.739) If we continue iterating Newton s metho, the first 4 igits of n will never change, so the solution is = accurate to 4 ecimal places. Eample 9.2. Fin solutions of each of the following equations, accurate to 3 ecimal places: (a) 2 2 = 0 (b) sin = 0 Shapes of Graphs Lec. #8 0. Increasing an Decreasing Functions Since the erivative gives the slope of = f(), we can istinguish intervals where f() is increasing from intervals where it is ecreasing b eamining the sign of f (): Theorem 0.. Let f be a ifferentiable function. If f () > 0 on an interval then f is increasing on that interval. If f () < 0 on an interval then f is ecreasing on that interval. Eample 0.. Fin the intervals on which f() is increasing, an the intervals where it is ecreasing: (a) f() = (b) f() = e At a critical point where f changes from increasing to ecreasing, f() must attain a maimum. Similarl, if f changes from ecreasing to increasing, f() has a minimum. This gives us the first erivative test: Theorem 0.2 (First Derivative Test). Suppose c is a critical number for a continuous function f. If f () changes from positive to negative at c then f has a local maimum at c. If f () changes from negative to positive at c then f has a local minimum at c. If f () oes not change sign at c then f has neither a maimum nor a minimum at c. 0.2 Concavit Given a function f() an its erivative, f (), we can ifferentiate f () to obtain the secon erivative of f(). Notation: Differentiating again gives the thir erivative: (f ) = f () = 2 f 2 (f ) = f () = 3 f 3

24 MATH 4 Calculus I Page 24 of 26 In general, the n th erivative is written f (n) () = n f n The secon erivative is closel relate to the shape of the graph of = f(). If f () > 0 then we known f () is an increasing function. A graph whose slope is increasing is sai to b concave up. On the other han, if f () < 0 then f () is ecreasing, an f() is concave own. Eample 0.2. Fin the intervals of concavit for the following functions: (a) f() = (b) f() = e A number c such that f (c) = 0 an f () changes sign at c is calle an inflection point of f. 0.3 Curve Sketching Arme with the intervals of increase, ecrease an concavit for a given function, we have all the essentiall information about the shape of its graph, an we can use this information to make an accurate sketch. Eample 0.3. Make tables of increase, ecrease an concavit for the following funtions, an use this information to make a sketch of the graph of = f(), labelling all local maima an minima an inflection points. (a) f() = (b) f() = e (c) f() = + 2 Lec. #9 L Hôpital s Rule We have looke at man it calculations similar to This kin of it is sai to be ineterminate form, since we can t simpl use the it laws to arrive at the answer (because the enominator is approaching 0). In general, its of ineterminate form are those that can be written f() g() where either f() = g() = 0 (an we sa the it has the form 0 0 ), or f() = g() = ± (an we sa the it has the form ). The following simple rule can be useful for calculating such its. L Hôpital s Rule: If then f() g() is of the form 0 0 or an both f an g are ifferentiable at a, f() g() = f () g (). 2 4 Eample has the form 0 0 so L Hôpital s Rule can be use: = 2 (2 4) ( 2) = 2 2 = 4.

25 MATH 4 Calculus I Page 25 of 26 Eample.2. Use L Hôpital s Rule to evaluate: (a) e (b) 2 e (c) 000 e () ln 2 (e) 0 ln L Hôpital s Rule is often use to fin horizontal asmptotes of graphs: Definition.. The line = b is calle a horizontal asmptote of the curve = f() if either f() = b or f() = b. For eample, we foun above the e = 0 so the graph of = e has the line = 0 as a horizontal asmptote. Eample.3. Fin intervals of increase, ecrease an convavit, as well as all horizontal an vertical asmptotes for = e, an use this to sketch the graph. Lec. #20 2 Parametric Curves We have been ealing with curves efine b the graph of some function = f(). But often in the sciences curves are escribe not b the graphs of functions, b parametricall. The following is a familiar eample from phsics. Eample 2.. A ball is thrown horizontall from the top of the builing, at spee 5 m/s. At time t its - an -coorinates (in meters), relative to the top of the builing, are = 5t, = 4.9t 2. The two equations above efine the path of the ball as a parametric curve. That is, for each value of t we can calculate an plot the position of the ball; b oing this for all values of t 0 we obtain the curve representing the trajector of the ball. In general, a parametric curve in two imensions is escribe b equations = f(t) = g(t) giving the - an -coorinates of a point on the curve at time t. The variable t is a parameter, an it label is arbitrar (it might as well be s or something else; this oesn t change the curve). This escription of a curve arises often in phsics. Problem: How to fin the slope of a parametric curve? Answer: For a curve efine b parametric equations = f(t), = g(t), the slope at the point ( (t), (t) ) is given b = /t /t = g (t) f (t) (if /t 0.) Eample 2.2. For the previous eample, fin the slope of the ball s trajector at time t = 2. Eample 2.3. The path of a ball thrown upwar at an angle to the groun is given b the parametric equations (t) = 5t (t) = 2 4.9t 2. (a) Fin an epression for /. (b) At what time oes the ball reach its maimum height?

26 MATH 4 Calculus I Page 26 of 26 Eample 2.4. For the parametric curve (t) = 2 sin2t (t) = 2 sint (a) fin the slope at the point where t = 5. (b) fin all points where the curve has a horizontal tangent.