COURSE OBJECTIVES LIST: CALCULUS
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- Whitney Osborne
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1 COURSE OBJECTIVES LIST: CALCULUS Clculus Honors n/or AP Clculus AB re offere. Both courses hve the sme prerequisites, n cover the sme mteril. Stuents enrolle in AP Clculus AB hve the following itionl requirements: ctul AP problems will be regulr prt of homework, quizzes, n tests stuents re require to tke the Avnce Plcement Test in My there is n extr clss meeting ech week to llow time to explore ies in greter epth thn the norml clss scheule llows PREREQUISITES: All skills from Algebr I, Geometry, Algebr II, n Preclculus re ssume. A prerequisites test is given uring the first week of clss to ssess knowlege of these prerequisite skills n to locte eficiencies. In prticulr, the Avnce Plcement Progrm Course Description for MAY 2002 MAY 2003 lists the following prerequisites: Before stuying clculus, ll stuents shoul complete four yers of seconry mthemtics esigne for college-boun stuents: courses in which they stuy lgebr, geometry, trigonometry, nlytic geometry, n elementry functions. These functions inclue those tht re liner, polynomil, rtionl, exponentil, logrithmic, trigonometric, inverse trigonometric, n piecewise efine. In prticulr, before stuying clculus, stuents must be fmilir with the properties of functions, the lgebr of functions, n the grphs of functions. Stuents must lso unerstn the lnguge of functions (omin n rnge, o n even, perioic, symmetry, zeros, intercepts, n so on) n know the vlues of the trigonometric functions of numbers such s 0, π 6, π 4, π 3, n π 2. The course objectives re elborte s follows. The orer in which the objectives re liste is not necessrily the orer in which they will be tught. The following informtion is tken from the Avnce Plcement Progrm Course Description for MAY 2002 MAY PHILOSOPHY: Clculus AB is primrily concerne with eveloping the stuents unerstning of the concepts of clculus n proviing experience with its methos n pplictions. The course emphsizes multirepresenttionl pproch to clculus, with concepts, results, n problems being expresse geometriclly, numericlly, nlyticlly, n verblly. The connections mong these representtions lso re importnt. Bro concepts n wiely pplicble methos re emphsize. The focus of the course is neither mnipultion nor memoriztion of n extensive txonomy of functions, curves, theorems, or problem types. Thus, lthough fcility with mnipultion n computtionl competence re importnt outcomes, they re not the core of the course. Course Objectives List: Clculus pge 1
2 Technology shoul be use regulrly to reinforce the reltionships mong the multiple representtions of functions, to confirm written work, to implement experimenttion, n to ssist in interpreting results. Through the use of the unifying themes of erivtives, integrls, limits, pproximtion, n pplictions n moeling, the course becomes cohesive whole rther thn collection of unrelte topics. These themes re evelope using ll the functions liste in the prerequisites. GOALS: GOAL1. GOAL2. GOAL3. GOAL4. GOAL5. GOAL6. GOAL7. GOAL8. GOAL9. Stuents shoul be ble to work with functions represente in vriety of wys: grphicl, numericl, nlyticl, or verbl. They shoul unerstn the connections mong these representtions. Stuents shoul unerstn the mening of the erivtive in terms of rte of chnge n locl liner pproximtion n shoul be ble to use erivtives to solve vriety of problems. Stuents shoul unerstn the mening of the efinite integrl both s limit of Riemnn sums n s the net ccumultion of rte of chnge n shoul be ble to use integrls to solve vriety of problems. Stuents shoul unerstn the reltionship between the erivtive n the efinite integrl s expresse in both prts of the Funmentl Theorem of Clculus. Stuents shoul be ble to communicte mthemtics both orlly n in wellwritten sentences n shoul be ble to explin solutions to problems. Stuents shoul be ble to moel written escription of physicl sitution with function, ifferentil eqution, or n integrl. Stuents shoul be ble to use technology to help solve problems, experiment, interpret results, n verify conclusions. Stuents shoul be ble to etermine the resonbleness of solutions, incluing sign, size, reltive ccurcy, n units of mesurement. Stuents shoul evelop n pprecition of clculus s coherent boy of knowlege n s humn ccomplishment. I. FUNCTIONS, GRAPHS, n LIMITS Anlysis of Grphs. FGRL1. With the i of technology, grphs of functions re often esy to prouce. The emphsis is on the interply between the geometric n nlytic informtion n on the use of clculus both to preict n to explin the observe locl n globl behvior of function. Course Objectives List: Clculus pge 2
3 Limits of functions (incluing one-sie limits). FGRL2. FGRL3. FGRL4. An intuitive unerstning of the limiting process. Clculting limits using lgebr. Estimting limits from grphs or tbles of t. Asymptotic n unboune behvior. FGRL5. FGRL6. FGRL7. Unerstning symptotes in terms of grphicl behvior. Describing symptotic behvior in terms of limits involving infinity. Compring reltive mgnitues of functions n their rtes of chnge. (For exmple, contrsting exponentil growth, polynomil growth, n logrithmic growth.) Continuity s property of functions. FGRL8. FGRL9. An intuitive unerstning of continuity. (Close vlues of the omin le to close vlues of the rnge.) Unerstning continuity in terms of limits. FGRL10. Geometric unerstning of grphs of continuous functions (Intermeite Vlue Theorem n Extreme Vlue Theorem). II. DERIVATIVES Concept of the erivtive. DER1. DER2. DER3. DER4. Derivtive presente geometriclly, numericlly, n nlyticlly. Derivtive interprete s n instntneous rte of chnge. Derivtive efine s the limit of the ifference quotient. Reltionship between ifferentibility n continuity. Derivtive t point. DER5. Slope of curve t point. Exmples re emphsize, incluing points t which there re verticl tngents n points t which there re no tngents. DER6. Tngent line to curve t point n locl liner pproximtion. DER7. DER8. Instntneous rte of chnge s the limit of verge rte of chnge. Approximte rte of chnge from grphs n tbles of vlues. Derivtive s function. DER9. Corresponing chrcteristics of grphs of f n f. DER10. DER11. DER12. Reltionship between the incresing n ecresing behvior of f n the sign of f. The Men Vlue Theorem n its geometric consequences. Equtions involving erivtives. Verbl escriptions re trnslte into equtions involving erivtives n vice vers. Course Objectives List: Clculus pge 3
4 Secon erivtives. DER13. Corresponing chrcteristics of the grphs of f, f, n f. DER14. Reltionship between the concvity of f n the sign of f. DER15. Points of inflection s plces where concvity chnges. Applictions of erivtives. DER16. DER17. DER18. DER19. DER20. Anlysis of curves, incluing the notions of monotonicity n concvity. Optimiztion, both bsolute (globl) n reltive (locl) extrem. Moeling rtes of chnge, incluing relte rtes problems. Use of implicit ifferentition to fin the erivtive of n inverse function. Interprettion of the erivtive s rte of chnge in vrie pplie contexts, incluing velocity, spee, n ccelertion. Computtion of erivtives. DER21. DER22. DER23. Knowlege of erivtives of bsic functions, incluing power, exponentil, logrithmic, trigonometric, n inverse trigonometric functions. Bsic rules for the erivtive of sums, proucts, n quotients of functions. Chin rule n implicit ifferentition. III. Integrls Interprettions n properties of efinite integrls. INT1. INT2. INT3. Computtion of Riemnn sums using left, right, n mipoint evlution points. Definite integrl s limit of Riemnn sums over equl subivisions. Definite integrl of the rte of chnge of quntity over n intervl interprete s the chnge of the quntity over the intervl: b f (x) x = f(b) f() INT4. Bsic properties of efinite integrls. (Exmples inclue itivity n linerity.) Course Objectives List: Clculus pge 4
5 Applictions of integrls. Approprite integrls re use in vriety of pplictions to moel physicl, biologicl, or economic situtions. Although only smpling of pplictions cn be inclue in ny specific course, stuents shoul be ble to pt their knowlege n techniques to solve other similr ppliction problems. Whtever pplictions re chosen, the emphsis is on: INT5. using the integrl of rte of chnge to give ccumulte chnge; or INT6. using the metho of setting up n pproximting Riemnn sum n representing its limit s efinite integrl. To provie common fountion, specific pplictions shoul inclue fining: INT7. INT8. INT9. INT10. the re of region the volume of soli with known cross sections the verge vlue of function the istnce trvele by prticle long line Funmentl Theorem of Clculus. INT11. INT12. Use of the Funmentl Theorem to evlute efinite integrls. Use of the Funmentl Theorem to represent prticulr ntierivtive, n the nlyticl n grphicl nlysis of functions so efine. Techniques of ntiifferentition. INT13. INT14. Antierivtives following irectly from erivtives of bsic functions. Antierivtives by substitution of vribles (incluing chnge of limits for efinite integrls). Applictions of ntiifferentition. INT15. INT16. Fining specific ntierivtives using initil conitions, incluing pplictions to motions long line. Solving seprble ifferentil equtions n using them in moeling. In prticulr, stuying the eqution y = ky n exponentil growth. Numericl pproximtions to efinite integrls. INT17. Use of Riemnn n trpezoil sums to pproximte efinite integrls of functions represente lgebriclly, geometriclly, n by tbles of vlues. The following objective will be inclue in the topic outline for Clculus AB for the cemic yer for the 2004 AP Exmintions: INT18. Geometric interprettion of ifferentil equtions vi slope fiels n the reltionship between slope fiels n solution curves for ifferentil equtions. Course Objectives List: Clculus pge 5
6 MUST-KNOW MATERIAL FOR CALCULUS MISCELLANEOUS: intervl nottion: (, b), [, b], (, b], (, ), etc. Rewrite ricls s frctionl exponents: 3 x = x 1/3, x3 = x 3/2 etc. An impliction If A then B is equivlent to its contrpositive If (not B) then (not A) To go from grph of y = f(x) to grph of x = f(y) : tke the (fmilir) grph of y = f(x), rotte 90 egrees clockwise, then flip bout the horizontl xis. TEST POINT METHOD: for solving f(x) > 0 : there re only two types of plces where function cn chnge from positive to negtive (or vice vers): where it equls zero, or t brek. Locte ll such plces, n check the resulting subintervls. GEOMETRY: Circle with rius r : AREA = πr 2, CIRCUMFERENCE = 2πr, DIAMETER = 2r Sphere with rius r : VOLUME = 4 3 πr3, SURFACE AREA = 4πr 2 Are of tringle: bse b n height h, AREA = 1 2 bh sies n b with inclue ngle θ : AREA = 1 2b sin θ Tringles: ngles sum to 180 ; longest sie is opposite biggest ngle, etc. Consier n rbitrry tringle with ngles A, B, C n opposite sies, b, c : sin A Lw of Sines: = sin B = sin C b c Lw of Cosines: 2 = b 2 + c 2 2bc cos A Similr tringles: hve the sme ngles; scling fctor in going from one to the other Right tringles: hve 90 ngle; longest sie is clle the hypotenuse; the Pythgoren Theorem: 2 + b 2 = c 2 Trpezoi with bses b 1 n b 2 n height h : AREA = (b 1+b 2 ) 2 h (verge the bses n multiply by the height) cyliner (2 prllel congruent plne figures of re A, perpeniculr istnce between plnes is h): VOLUME = Ah right circulr cyliner: VOLUME = πr 2 h cone ( plne figure of re A, point, ll lines connecting; h is perpeniculr istnce from point to plne): VOLUME = 1 3 Ah right circulr cone: VOLUME = 1 3 πr2 h Must-Know Mteril - 1
7 TRIGONOMETRY: RADIAN MEASURE: the rin mesure of n ngle is the length of the rc on the unit circle: positive is counterclockwise. Right tringle efinitions: SOHCAHTOA Unit circle efinitions; ly off ngle x sin x is the y-vlue of the point cos x is the x-vlue of the point tn x = sin x cos x cot x = 1 tn x = cos x sin x sec x = 1 cos x csc x = 1 sin x sin 1 x = rcsin x is the ngle between π 2 n π 2 whose sine is x. cos 1 x = rccos x is the ngle between 0 n π whose cosine is x. tn 1 x = rctn x is the ngle between π 2 n π 2 Note: sin 2 x mens ( sin x ) 2 etc. whose tngent is x. Double-ngle formuls: sin 2x = 2 sin x cos x ; cos 2x = cos 2 x sin 2 x the Pythgoren Ientity: sin 2 x + cos 2 x = 1 Two specil tringles: n FUNCTIONS: f(x) is the output from the function f when the input is x. Functions hve the property tht ech input hs exctly one corresponing output. ZERO of function: n input whose output is zero DOMAIN of function: set of llowble inputs RANGE of function: its output set GRAPH of function: the picture of its (input,output) pirs GRAPHS of BASIC MODELS: constnt (y = k) y = x 2 n higher powers y = x 3 n higher powers y = 1 x y = x y = x Must-Know Mteril - 2
8 y = e x y = ln x y = sin x y = cos x y = tn x y = sec x y = [[x]], the gretest integer function, [[x]] is the gretest integer less thn or equl to x COMPOSITIONS OF FUNCTIONS: f(g(x)) mens g cts first, f cts lst EVEN function: f( x) = f(x) ; when inputs re opposites, outputs re the sme ODD function: f( x) = f(x) ; when inputs re opposites, outputs re opposites ONE-TO-ONE FUNCTION: Ech output hs exctly one corresponing input; grph psses both horizontl n verticl line test; the inputs n outputs cn be tie together with strings INVERSE FUNCTIONS: If f is 1-1, then its inverse f 1 unoes wht f i: f(f 1 (x)) = x n f 1 (f(x)) = x the omins n rnges of f n f 1 re switche the grphs of f n f 1 re reflections bout the line y = x if (, b) is on the grph of f, then (b, ) is on the grph of f 1 TRANSFORMATIONS of functions: strt with y = f(x) working with y is intuitive: y = f(x) + 3 moves up 3 y = 3f(x) multiplies ll y-vlues by 3 (verticl stretch) y = f(x) multiplies the y-vlues by 1 ; reflects bout the x-xis working with x is counter-intuitive: y = f(x 3) ; replce every x with x 3 ; moves to the RIGHT 3 y = f(3x) ; replce every x with 3x ; (, b) ( 3, b) ; horizontl compression y = f( x) ; replce every x with x ; reflects bout the y-xis LINES: liner functions: y = mx + b or x + by + c = 0 ; chnges in y equl chnges in x give rise to equl slope: m = y 2 y 1 x 2 x 1 ; if m = 3, then the y-vlues re chnging 3 times s fst s the x-vlues point-slope form: y y 1 = m(x x 1 ) prllel lines hve the sme slope; perpeniculr lines hve slopes tht re opposite reciprocls horizontl lines: y = c ; hve zero slope verticl lines: x = c ; hve no slope Must-Know Mteril - 3
9 QUADRATIC FUNCTIONS: f(x) = x 2 + bx + c ; grph s prbols; > 0 hols wter, < 0 shes wter vertex: t x = b 2 POLYNOMIALS: Let P (x) = x + 2 x n x n. egree of P : highest exponent As x ±, polynomil looks like its highest power term. The following re equivlent: c is zero of P x c is fctor of P (x) P (c) = 0 the point (c, 0) is on the grph of P the grph of P crosses the x-xis t c x c goes into P (x) evenly (reminer = 0) EXPONENTIAL FUNCTIONS: Allowble bses: b > 0, b 1 y = b t Incresing when b > 1 ; ecresing when 0 < b < 1 Common form: A(t) = A 0 e kt ; A 0 is the mount t time 0 Every exponentil function cn be written with ANY llowble bse, so use whtever bse is convenient. For equl chnges in x, y gets MULTIPLIED by constnt (tht epens both on the bse of the exponentil function, n the chnge in x) Doubling time: for n incresing exponentil function, it lwys tkes the sme mount of time for quntity to ouble hlf-life: for ecresing exponentil function, it lwys tkes the sme mount of time for quntity to be cut in hlf LOGARITHMS: y = log b x Allowble bses: b > 0, b 1 Incresing when b > 1 ; ecresing when 0 < b < 1 Lws work for ll llowble bses: ln x = log e x is the nturl logrithm ln xy = ln x + ln y (the log of prouct is the sum of the logs) ln x y = ln x ln y (the log of quotient is the ifference of the logs) ln x y = y ln x (you cn bring powers own) chnge-of-bse formul: log b x = log x log b Must-Know Mteril - 4
10 y = e x n y = ln x re inverse functions; use this ie to solve exponentil n logrithmic equtions A log is n exponent! log 3 5 is the POWER tht 3 must be rise to, to get 5 EXPONENTIAL FUNCTIONS grow fster thn POWER FUNCTIONS grow fster thn LOGARITHMIC FUNCTIONS ABSOLUTE { VALUE: x if x 0 x = x if x < 0 For c > 0, x < c c < x < c x > c x > c or x < c x = c x = ±c 0 < x c < δ x (c δ, c) (c, c + δ) (puncture neighborhoo bout c) LIMITS: Consier the limit sttement: lim f(x) = l x c low-level unerstning: when x is close to c, f(x) is close to l higher level: we cn mke the vlues of f(x) s close to l s we like, by tking x to be sufficiently close to c, but not equl to c Precisely: ɛ > 0 δ > 0 s.t. if 0 < x c < δ then f(x) l < ɛ When we evlute limit s x c, we never let x equl c x c + mens x pproches c from the right-hn sie x c mens x pproches c from the left-hn sie LIMIT LAWS: Work nicely! Proviing the iniviul limits exist, the limit of sum is the sum of the limits (sme for ifference, proucts, quotients, etc.) If you hve continuous function (see below) then evluting limit is s esy s DIRECT SUBSTITUTION. BE CAREFUL! If you re working with iscontinuous function (e.g., gretest integer function, some piecewise-efine functions), then irect substitution MAY NOT WORK. Try l Hospitl s rule, renming, grphing, etc. sin x An importnt limit: lim x 0 x = 1 Must-Know Mteril - 5
11 CONTINUITY: Low-level unerstning: no breks in the grph higher level: when inputs re close, outputs re close The following re equivlent: f is continuous t c lim x c f(x) = f(c) (when f is continuous t c, then evluting the limit is s esy s irect substitution) lim h 0 f(c + h) = f(c) As x c, f(x) f(c) INTERMEDIATE VALUE THEOREM: Suppose f in continuous on [, b], n N is number between f() n f(b). Then there exists number c between n b for which f(c) = N. (If grph hs no breks, n you trvel long the grph from one point to nother, you must pss through ALL the y-vlues in between; i.e., ll the intermeite vlues.) EXTREME VALUE (MAX/MIN) THEOREM: Let f be continuous on close intervl [, b]. Then f ttins both n bsolute mximum vlue f(c) n bsolute minimum vlue f() for some c n in [, b]. (This theorem GUARANTEES the existence of highest n lowest point on grph uner pproprite conitions.) MEAN VALUE THEOREM: Let f be ifferentible on [, b]. Then there exists number c between n b for which f f(b) f() (c) =. b (This theorem gurntees plce where the instntneous rte of chnge is the sme s the verge rte of chnge uner pproprite conitions.) Must-Know Mteril - 6
12 DERIVATIVES: The following re equivlent: f (c) = m f(c + h) f(c) lim = m h 0 h The slope of the tngent line to the grph of f t the point (c, f(c)) is m f is ifferentible t c (n the vlue of the erivtive is m) f(x) f(c) lim = m x c x c The instntneous rte of chnge of f t (c, f(c)) is m When x = c, the function vlues re chnging m times s fst s the inputs Suppose f (2) = 5. Roughly: when x chnges by 1 (from 2 to 3), we expect y to go up by ABOUT 5. Or, when x chnges by 1 (from 2 to 1), we expect y to go own by ABOUT 5. The UNITS of f (c) re the units of f(x) (the outputs from f) ivie by the units of x (the inputs to f) If function is ifferentible, then its grph is SMOOTH: it hs non-verticl tngent lines everywhere. A function is NON-DIFFERENTIABLE t: verticl tngent lines; kinks; iscontinuities Theorem: If f is ifferentible t x, then f is continuous t x Contrpositive: If f is not continuous t x, then f is not ifferentible t x Leibnitz nottion versus prime nottion: y = y x, y = 2 y x etc. 2 Liner Approximtion (lineriztion): function is best pproximte t point by its tngent line: t (c, f(c)) we hve: L(x) = f(c) + f (c)(x c) f(x) If f (x) > 0 then f is incresing If f (x) < 0 then f is ecresing If the SLOPES re INCREASING (f incresing; f > 0) then f is concve up If the SLOPES re DECREASING (f ecresing; f < 0) then f is concve own Remember: function cn increse in bsiclly three ifferent wys: linerly, concve up; concve own Inflection point: where the concvity chnges (from concve up to own, or own to up): cnites re where f (c) = 0 or f (c) oes not exist. LOCAL MAX/MIN: A locl mx/min for function cn only occur t three types of plces (clle the criticl points ): where f (c) = 0 where f (c) oes not exist t ENDPOINTS of omin of f Must-Know Mteril - 7
13 So, to fin mx/min, locte ll cnites, n check them. Creful: criticl point is not necessrily mx or min! FIRST DERIVATIVE TEST: Check signs of first erivtive to the left/right of cnite (where the function is continuous) to ecie if it is mx or min. Why is continuity neee? See the sketch below the test woul tell us tht there s locl mx t c! SECOND DERIVATIVE TEST: If concve up t cnite; it s min. If concve own t cnite, it s mx. AVERAGE RATE OF CHANGE: The verge rte of chnge of f on the intervl f(b) f() [, b] is ; this is the slope of the line between (, f()) n (b, f(b)) b DIFFERENTIATION FORMULAS: Be ble to GENERALIZE ll these formuls: replce x by f(x), n multiply by f (x) ) x xn = nx n 1 n ( ) n 1 generlize: x( f(x) = n f(x) f (x) x cf(x) = c xf(x) (you cn slie constnts out) the erivtive of sum/ifference is the sum/ifference of the erivtives x ex = e x (the y-vlue of the point tells you how fst the function is chnging t tht point) PRODUCT RULE: x f(x)g(x) = f(x)g (x) + g(x)f (x) (the erivtive of prouct is NOT!! NOT!! NOT!! the prouct of the erivtives) f(x) QUOTIENT RULE: x g(x) = g(x)f (x) f(x)g (x) (g(x)) 2 NOT!! NOT!! NOT!! the quotient of the erivtives) (the erivtive of quotient is x MUST BE MEASURED IN RADIANS FOR THESE FORMULAS TO BE CORRECT: x x sin x = cos x cos x = sin x x tn x = sec2 x x cot x = csc2 x x x sec x = sec x tn x csc x = csc x cot x Chin Rule: x x = x ln x f(g(x)) = f (g(x) g (x) ; how to ifferentite composite functions x rcsin x = x sin 1 x = 1 1 x 2 Must-Know Mteril - 8
14 x rccos x = 1 x cos 1 x = 1 x 2 x rctn x = x tn 1 x = x 2 x ln x = 1 x x log x = 1 x ln erivtive of n inverse function: (f 1 ) (x) = 1 f (f 1 (x)) So if (, b) is on the grph of f with slope of tngent line m, then (b, ) is on the grph of f 1 with slope of tngent line 1 m! IMPLICIT DIFFERENTIATION: Whenever you see y, tret it s function of x n ifferentite ccoringly. For exmple: For exmple: x y2 = 2y y x y xxy = x x + y LOGARITHMIC DIFFERENTIATION: Use this to ifferentite complicte proucts or quotients; lso to ifferentite vrible stuff rise to vrible power. First tke logs, then ifferentite! PARTICLE MOVING ON A NUMBER LINE: Let s(t) enote the position t time t. Then, s (t) = v(t) is the velocity; positive is moving to the right; negtive to the left. s (t) = v (t) = (t) is the ccelertion. Speeing up mens moving to the right fster n fster (v(t) > 0 n (t) > 0) or moving to the left fster n fster (v(t) < 0 n (t) < 0). Thus, the prticle spees up when velocity n ccelertion hve the sme sign (both positive, or both negtive). Note: spee = v(t) Suppose you re given the velocity of prticle trveling long number line, v(t). Then, totl istnce trvele from t 1 to t 2 is given by t 2 t 1 v(t) t ; i.e., integrte the spee. However the totl DISPLACEMENT is t 2 t 1 v(t) t = s(t 2 ) s(t 1 ). Notice tht if you strt t 0, move to the right 5 n then to the left 5, your totl isplcement is 0 but the totl istnce trvele is 10. RELATED RATE PROBLEMS: Ask: Wht is chnging with time? Rtes re erivtives! Write own SOMETHING THAT IS TRUE tht involves wht you re intereste in. (Look for: similr figures, right tringles, etc.) Remember: if x is chnging with time, then the erivtive of x 2 is 2x x t. Must-Know Mteril - 9
15 OPTIMIZATION PROBLEMS: Fin the CANDIDATES for locl mx/min: enpoints, plces where the erivtive is zero, plces where the erivtive oesn t exist. Use the First Derivtive Test or Secon Derivtive Test to check whether they re mx or min. If you wnt n ABSOLUTE mx/min, fin ALL the locl mx/min, n choose the highest/lowest from these. Remember, you CAN T USE YOUR CALCULATOR to locte mx/mins; this is NOT n llowble opertion! ANTIDERIVATIVES: A function F (x) is n ANTIDERIVATIVE of f(x) if n only if F (x) = f(x) ; i.e., F is function whose erivtive is f. Antierivtives uno erivtives. An ntierivtive hs specifie erivtive, n this erivtive etermines the shpe, but not the verticl trnsltion. So, if you hve ONE ntierivtive, then you hve n infinite number they ll iffer by constnt. The symbol f(x) x enotes ll the ntierivtives of f(x). x n x = 1 n+1 xn+1 + C for n 1 1 x = ln x + C sin x x = cos x + C cos x x = sin x + C e kx x = 1 k ekx + C 1 1+x 2 x = tn 1 x + C 1 1 x 2 x = sin 1 x + C Any CONTINUOUS function f hs n ntierivtive: the function x f(t) t is n ntierivtive of f(x). Tht is, the function tht fins AREA uner the grph of f is n ANTIDERIVATIVE of f! DEFINITE INTEGRALS: the efinite integrl of f from to b is enote by b f(x) x n is efine s follows: Divie [, b] into n equl subintervls, ech of length x = b n. Choose x i from the ith subintervl. Then, b f(x) x = lim n n f(x i ) x You cn pproximte efinite integrls with rectngles (left-hn; right-hn; mipoint), with trpezois, even with prbols. The efinite integrl gives informtion bout the (signe) re trppe between the grph of f n the x-xis: re bove is trete s positive; re below is negtive. i=1 Must-Know Mteril - 10
16 Cution: if b f(x) x = 0, this only mens tht there is the sme mount of re ABOVE the x-xis s BELOW on the intervl [, b]. When you hve efinite integrl problem tht cn be solve with simple geometry formuls (tringles, trpezois, circles) then USE GEOMETRY to fin the efinite integrl it s much more efficient! EVALUATION THEOREM: If F is ny ntierivtive of f, then b f(x) x = F (b) F (). TOTAL CHANGE THEOREM: When you integrte rte of chnge, you get totl chnge: Rewrite this s b f (x) x = f(b) f() f(b) = f() + b f (x) x n think of it like this: If you wnt to know the vlue of f t b, first fin the vlue of f t someplce you know (), n then see how much f hs CHANGED BY in going from to b ( b f (x) x). SUBSTITUTION METHOD for ntiifferentition/integrtion: Choose u to be something whose erivtive is in the integrn, perhps off by constnt. Often, u is something in prentheses, the rgument of function, something in n exponent, etc. Be sure to chnge the limits if you hve efinite integrl. APPLICATIONS OF INTEGRATION: AREAS BETWEEN CURVES: Fin the intersection points, write the re of typicl slice n sum ppropritely. verticl slices: AREA = b( ) f(x) g(x) x horizontl slices: AREA = ( ) c f(y) g(y) y (Will nee to solve for x in terms of y) VOLUMES OF REVOLUTION: DISK METHOD: revolve y = f(x) bout the x-xis on [, b] ; volume of the resulting soli is b π( f(x) ) 2 x. SHELL METHOD: revolve y = f(x) on [, b] bout the y-xis ; volume of the resulting soli is b 2πxf(x) x. 1 AVERAGE VALUE OF A FUNCTION: the verge vlue of f(x) on [, b] is b f(x) x ; you re summing up the outputs from to b (the integrl), n then iviing by how mny you hve (the length of the intervl). If you smush the re into rectngulr shpe, the verge vlue gives the height of the rectngle. Cution: Don t mix up verge rte of chnge n verge vlue! CONNECTION BETWEEN verge vlue n the verge rte of chnge: b 1 f f(b) f() (x) x = : verging the vlues of f (x) on [, b] gives the verge b b rte of chnge of f on [, b] b Must-Know Mteril - 11
17 SEPARABLE DIFFERENTIAL EQUATIONS: Get ll the y s on one sie, n ll the x s on the other sie. Integrte. Don t forget the constnt of integrtion. Use given conition to solve for this constnt. SLOPE FIELDS: Slope fiels help us to visulize the solutions to first-orer ifferentil equtions. Get formul for the erivtive in terms of x n y; fin the slope t mny ifferent points. The resulting fiel of slopes helps us to see the shpes of the solution curves. Must-Know Mteril - 12
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