AP CalculusAB Test 2003 with anwers

AP CalculusAB Test 2003 with anwers - @0@@@@@Hf)@@

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Unformatted text preview: @0@@@@@Hf)@@ ,@@@@@)<9@@00@0@@@@@@0G@000B ~@@@@~~ee@@@@@@@~~0~@ D@@@@@~0e@&@@@~@@v@0~0B @@@@@@@0~@0~@$@@0~0 @@@@@@0e@@~@@@@@@0& If !!!2i z i I~ 0 ~ The Cakuhw AB Exam CALCULUS AB !r~ I'I~~II ~::::::::::::::::::::::::: 'i)tii'>m\(0ltW'l""''''''r<nr;:. ... ''>'>c;<)(;J)@@@@@@@<B@0CD0@ . , )@@@@@@@<2:J0& I I ~<a@}@@,@@\B@@> ~~!~~~~~5~~~2~~~~5s~g~~~~~ 0000000000000000000000000 \lll(lg;;l'j::l~-:l:>l;:;~~~:;:~<:.:;O:il.:i~;i'I;;;iiI @6@@@)@@@@) 0 :;: A CALCULATOR CANNilT BE USED ON PART A OF SECTION I. A GRAPHING CALCULATOR FROM THE APPROVED liST IS REQUIRED FOR PART II OF SECTION I AND FOR PART A OF SECTION II OF THE E.{AMINATION. CALCULATOR MEMORIES NEED NOT BE CLEARED. COMPUTERS, NONGRAPHINO SCIENTIFIC ~ALC1JLATORS. CALCULATORS WlTH QWERTY KEYBOARDS, AND ELECTRONIC WRlTlNO PADS ARE NOT ALLOWED. CALCULATORS MAY NOT BE SHARED AND COMMUNICATION BETWEEN CALCULATORS IS PROHIBITED DumNO THE EXAMINATION. ATTEMPTS TO REMOVE TEST MATERIALS FROM rtrs ROOM BY ANY ME11I0D WILL RESULT IN THE INVAliDATION OF TEST SCORES. SECTION I ~~~~58t~gi~IQ~~~~~5~~~~3~SS 000000000000000000000000000 ..... ~ .., .. r'" ~ := :: ~ ~, ;! \!! ~ c :! ~ ~ .. I'l l:.l ;:; <:l l'3 ::. ~@@@~@@0~@&@@@@@<B@@0011 @@@@@@OG@0~@@@W@0@0@&1 @@@@@~e0e@@@0@0~~ IllHh HtH ~~ J===~~=~- =~==~ @,?@~'ff):@@(i:)~c@. @G@@)@l@ _!:!)@@@@@00@@(%l@@@@@(~8~(il<1)@@ -'=======.0-.= "" U Time- 1 hour and 45 minutes All questions are given equal weight Percent of 101u 1 grade- 50 Pan A: 55 minutes, 28 multiple-choice questions A calculator is NOT allowed. _ _@@@@@@@,@000@@@@0@Q:l00B @@@@@@0e0~@@@@~@01 @e@$@@~@@ @E)@@G)@@@@@ ~~~~~~~~~~ ~@~~@@e0@@~@@@80tD 1 @8@@@@@e@@ @@@@@@@@G6J@GCI)@@@@@@!E>@0@@0 Ie ~@@@@@86J000@@@<B@0~ ~@@@@@<:J?6)0~)@@@.@@.l<'@0(1)@:)&11I==~,-:"~~~~ ~@@@@@G0@G~@@@@0@0Q:l@ !CV@)@@l@@@0G)@@tI;l,:f:j@@Cl l <B(D@c<JQ:Ue 1@@@@@@@@00@@@@@@@@@0@@@ \ I @~~J?~~'~'{El@(j) @6@@@@@0 Part B: 50 minutes, 17 multiple-choice questions A graphing calculator is required. Pans aTe inA and 8 of Section package.(his exmninalion booklet; Paris A and B of Section II, which consist of longer problems. a separate, sealed I are in 1@@@@@<!)eG)0@@@@@0~@$@ ;@@@@@\!)<i)0e@@Qj@@@@@@(f)@0<D@GB @0@@@J@@)<B@@0 f9@($)@@iSI@86J@@~~(i,D~@'f!e@0~.&@@E:>@l~G;J~@e:~~-~ j@@@@@@@06)@0@@@@@@8@Q:l0& @G@@@@@(J@l@<S'> 1@@@@e00@@@@0<B@00@ @~@),~,-@~-(P.!pi~~ @e@@)@)@J@@@ General Instructions DO NOT OPEN THIS BOOKLET UNTIL YOU ARE INSTRUCTED TO DO SO. INDICATE YOUR ANSWERS TO QUESTIONS IN PART A ON PAGE 2 OF THE SEPARATE ANSWER SHEET. THE ANSWERS TO QUESTIONS IN PART B SHOULD BE INDICATED ON PAGE 3 OF THE ANSWER SHEET. No credit will be given for anything written in this examination booklet, but you may USethe booklet for notes or scrarchwork. After you have decided which of the suggested answers is best, COMPLETEL'l fill in the corresponding avalon the answer sheet. Give only one answer to each qnestion. 1f you change an answer, be sure that the previous mark is erased completely. Exam~~ f@@@@@~ee0~~@@~@@e~Q:l@~:::::i;::::::::f~dI~~~~~~~~~~ ~~~@<)@ ~e<!D@e1$'e~@l 1~$@)@~@)@eG>@Ql{j>~o/@@@@~@~_~~ 'i II Ii What is the arithmetic mean of the numbers 1, 3, and 6 ? (A) (B) Sample Answer CD9 !' 'I" ! I (C) (D) ~ .'II I "I' Iii lii'l l ~ @@@@)~G~~0~@@@@00~0~0@8@@@@@S@@0 :: ;~:: ::::::: :::;:::: :;;; : . :;;;:~::::~ ",'@@@(;@)@E.l{s.>G<1)@@()~''0@~_Gi~(.'Et@- :'@'e~~(~H~@-lB@)~6? < - @ @ @@ @$ ~~ ~~0 ~ @ ~ ~ ~~ @ <B@ ~~ ~ ~ ' ~ ~~~$~@ ~~~ ~ o i _ @@@<)@)~~_G6J@Q)@@~.@)0~@G)0 ,@~@~~_~~~~~,Sl ~ ~ ~~ i~ 1 I I, ~ ~ I~ " [I ~ ~:::::::::::::::::::::::::~::::::::::: ii~ @@l@@@@G0@0~@~(Y~@0~,~~ ~e@@@~~~@~,~ !~; @@~@$~~~e~@Q~~~@@~e~~~~~ 0@:w@@@@e0@G@(!)@@@@@0@0~_~~ @@~@)qg:@@e00G~Q)@@W?@:~o;E:J@~~0(EJ @~~~~@S~@@0 @,G~@@_~_~?@~ @e@e<@-~@@.l-@<Sl -2 ~-: i~~ @~~@@@@(0)00@@~~@<B@@tD@ @e@,@(~)(~@)(&,@,? ~ ~l _~~@@~~@@0~~~@@@@~~~~~~ @e~~~@~~~~~ :~~ 1@@)@@@.@d)~6J.@~(!)(2)~@,~@:V~(~)0@. ~e:@e@.@~~,~,':' f ~ g. 1@~~~@r~)@~~~@0~~~@@_~e~(~h~~", @e@@.~~@~,~,~,:? ~@~~~~e~@0~~@~~@~~@@~@~@ " . @0@@@@@@@@ " f a; ~@@@.@lf)@(~)~e~Q>~~()~_~~!){~@il'@@~;ID '~'aH3>~.~-!il@P?~5~'J~i ~e~~~~~~@~~ -S ~ i ~ ~~~ Ii I~ f :2~ flO) Many candidates wonder whether or not to guess the answers to questions about which they are not certain. In this section of the examination, as a correction for haphazard guessing, one-fourth of the number of questions you answer incorrectly will be subtracted from the number of questions you answer correctly. It is improbable, therefore, [hal mere guessing will Improve your score significantly; it may even lower your score, and it does take time. If, however, you are not sure of the best answer but have some knowledge of the question and are able to eliminate one or mare of the answer choices as wrong, your chance of answering correctly is improved, and it may be to your advantage to answer such a question, Use your time effectively, working as rapidly as you can without losing accuracy. Do not spend lao much lime on questions that lire 100 difficult. Go on to other questions and came back 10 the difficult ones later if you have time It is not expected chat everyone will be able to answer all the multiple-choice questions 1' :J1 16 17 Calculus AB f,~1 I Part A J l..ty= x +1 I (A) (3x (;B_~Cl';l;~,H:H (J 2 )2 ,tell ely = h dX (B) 2(.1'3 l + 1) (C) 2(3.1'2 + I) (D) 3x 2(x 3 + I) (E) 6x 2 (.1'3 + 1) ~al:'TI CALCULUS AB SECTION L Part A. Tirn02-55 minutes 2. J~e-4Xdx (A) ::: (B) _4e- 4 Number of questiaus-2H -e -4 -4 (C) e-4 - 1 1 (D) 4' - s.: 4 -4 (E) 4 - 4e- 4 A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION. Dh"ections: Solve each of the following problems, using the available space for scratchwork. After examining the form of the choices, decide which is the best of the choices given and fill in the corresponding ova! on the answer sheet. No credit will be given for anything written in the test book. Do not spend too much time on anyone problem. In tbis test: (I) Unless otherwise specified, the domain of a function f(x) is a real number. f is assumed to be the set of all real numbers x for which (2) The inverse of illrigollometric: function f may be indicated using the inverse function notation I-lor with (he prefix "arc" (e.g., sin- l r = arcsin x). Unauthorized copying or reuse 01 any part or this page Is illegaL GO ON TO THE NEXT PAGE. 19 I;: eil i I ',1 I :1, , I" Ii ! i,",I:~, i,}I! 18 Unauthorized I:opying or reuse of ~~~~~~~ Is 1Ilesal. . GO ON TO THE NEXT PAG Calculus AB ~ ~ 5. (A) J:. sin x tit = (B) _-J2 2 -J2 2 (e) _ -J2 - 1 2 (D) - 1: +1 (E) -J2 - 1 2 ~,,~tAJ For x li,,;nl~'~'r:d"H6 ...~;'1~ ~: 0, the horizontal line y ::;;: 2 i:-, an aSylllpl(Hc for Ibe graph ot the function} Which of the fcol!o\'!ing. statements must be true? (Ai (B) flO) = 2 f(x) '" 2 for all x" 0 6. }:,n.;, 4x 3 (A) 4 . x 3 - 2x 2 + 3x - 4 3x 2 + 2x - 1 (B) 1 (C) rm is undefined. ~ ID) lim fix) _'t---12 = (C) 4 1 (D) 0 (6) -1 (E) lim fix) \'--jOOq =2 4. If y = 3:<+2' then 2x + 3 dy dx = (B) (A) 12x + 13 (3x + 2)2 12x - 13 (3x + 2)2 5 (C) (3x + 2)2 -5 (D) (3x + 2)2 (E) 3" 2 Uns.uthori:zedcopying or nilUa4!l of any pari ot lhla page Is Illegal. GO ON TO THE NEXT PAGE. 21 20 .my !"Ill! "' tilt" pay~ Is IIk,ga) Unlulhori"z:ed COF-V1ng or rouse 01 GO ON TO THE NEXT PAl Calculus AB ~mmmt1I ~ 8. Jx2cos(x3)dx~ (A) - ~sin(x3) + C + C l.~ Calr.;"t: tu ~.:; f (8) 4sin(x3) 3 -'--~L~ar! ~~ (C) - ; sin(x (D) x sin(x 3 ) 3 3) +C 3 +C ) 3 4 -r (E) x Sin( x 3 4 +C --:::r---=r 0 /~. Graph off' 7. The graph of I ', the derivative of the function [. is shown above. Which of the following statements is true aboiu f .) (A) (B) 9. Tfl) ~ In(x + 4 + e-3' (A) ), then /,(0) is (C) (C) f is decreasing for -1 S ., $ l. f is increasing for -2 ::; x ::; o. f is increasing for 1 S; x $; 2. 2 -3 (B) 3 1 "4 1 (D) ~ 5 (E) nonexistent (D) (E) I f has" local minimum at x is not differentiable at .r ~ ~ O. ~ -1 and .r 1 ! I UnauthQrized ~opylng or reuse of an'll part of lhls pago I~ Illegal. GO ON TO THE NEXT PAGE. 23 2.2 ~Y_p'~~lis p~~lh'!J~ fU;"l-authoril:ed copying o,r reuse of GO ON TO THE NEXT PAG Calculus AB tmt""' A i Part 11. Using the substitution u 1 (A) ? _ = 2x + 1, (B) J: .J2x + 1 dx is equivalent to (C) I r "2 JI jU du l JIi' jU du -1/2 1 "2 f.' jU du 0 (D) J: jU du (E) r jU du . ~ '. .. {;alc-,.dug J-~]3 PclIl-.A.- J !( v} f:~ i.], and In, The function f has the property thal flr( r) are negative for all real values Y. Which of the following could be the graph of (A) f ? (B) y y 12. The rate of change of the volume, V, of water in a tank with respect to time, t. is directly proportional to the square root of the volume. Which of the following is a differential equation that describes this relationship? (A) V(t) = k.Ji (B) V(t) = k../V ____----.,,+ o ~. x __________ 0 I .r (C) (D) 'Z = lit dV = k.Ji k ../V (E) dV dt (C) (D) = k../V y o I x o J p x (E) y Unauthorized copvlng or reuse any port 01 this page hllmega!. or o ii It .. x GO ON TO THE NEXT PAGE. 25 \ l<l{'__ ~ldUlh.Orl-.lCd ca~'Jlng or rouue of p~r~~~:J~~ll<Jgal- GO ON TO THE NEXT PP Calculus AB 1L~1I 1 PartAJ 15. Let f be the function with derivative given by f'C<) = x 2 is ~. .r On which of the following intervals f decreasing? (-00, -1] only (-00,0) (A) lih~ '(;~11Gl:JAl1~' J-"!,.I> (B) GllW~"~~~:~- i\] (C) [-1,0) ou1y (D) (E) (o,Vi] [112,00) / 16. If the line tangent to the graph of the function then /,(1) is Graph of f 13. The graph of a function (A) a (B) b (A) -5 (B) 1 f at the point (1, 7) passes through the point (-2, -2), (E) undefined (C) 3 (D) 7 f is shown above. At which value of .r is f continuous, but not differentiable? (C) c (D) d (E) e 14. It y ~ \:- . -"). Sin 2x, then dx ~ ely (A) 2x cos 2x (B) 4xcos2x (C) 2x(sin 2x -l- cos 2x) (D) 2x(sin 2x - .r cos 2x) (E) 2x(sin 2x + x cos 2x) Unauthcrlzed copying or reuse ot any part of this page Iii Illegal. GO ON TO THE NEXT PAGE. 27 , '" u.nOlulhorized copying or reuse 2.6 ~~~~~!-~~~~~~ j GO ON TO THE NEXT PM Calculus AB t'f'fli!lJWlB , Part A I 19. A curve has slope 2x + 3 at each point (x, y) on the curve. Which of the following is an equation for this curve if it passes through the point 0, 2) ? (A) y = 5x - 3 ~ - a. 0, . , 'Ca.A(;"L~ln"s _ i (B) y = x' + 1 + 3x + 3x - 2 ; , '. Pall: A.J (C) Y = x' (D) y = x' (E) y = 2x' + 3x - 3 2.\t' ~ . The grJph of II') .\ .c -I J 7. LeI ./ he the function given by (1\) /(1) :::; f is ClJfll.:;\\'e dO\\'L1 when x 'c IB) .Y'> -2 (Dl .\ -, -] (E), -: 0 X f(x) = { +2 4x - 7 if x 0; 3 if x> 3 r-;l~GEH~R-4l 20. Let f be the function given above. Which of the following statements are true about f ~B~II~E-~-ED 1. lim f(x) exists. ,-.3 f II. f is continuous at x = 3. 1lI. is differentiable at x 18. The derivative s' of a function g is continuous and has exactly two zeros. Selected values of g' are given in the table above. If the domain of g is the set of all real numbers, then g is decreasing on which of the followi intervals? = 3. (A) None (B) I only (C) IT only (D) I and IT only (A) -2 0; x 0; 2 only (B) -10; r 0; I only Ie) x "' -2 (D) x", 2 only (E) x S; -'2 or x :2: 2 (E) I, II, and ill UnauthorLzed copying Of reuse 01 Qny part ollhls PQgo Iii Illegal. GO ON TO THE NEXT PAGE. 29 28 rt:Jt;<111.lhOriZed. copying or reuse al l~~'~~~~~~ lJilgU i:;;lIIe9~ GO ON TO THE NEXT F Calculus AB .. y ~. PartA I 6 Calc~ld,('LS ~~f'(x) I,~. '~_l',,~t~] ' o (A) 0 (B) 3 (C) 6 I '\, 22. The graph of j', the derivative of f. is the line shown in the figure above. If frO) ~ ~ 5, then fell ~ ("(X) (D) 8 (E) 11 21. The second derivative of the function f is given by fN(x) ee x(x - alex - b)'. The graph of fN is shown above. For what values of x does the graph of f have a point of Inflection? (El b, j, and k (D) 0, a, and b (e) b andj only (B) 0 and 111 only (A) 0 and a only 23. d ax l'Jro x2 . sin(13)dl ) ~ (A) -cos(x 6 ) (B) sin(x 3 ) (C) sin(,6) (D) 2xsin(x 3 ) (E) 2x sin( x 6 ) UnauthQrized copyIng or reuse any part of thl~ page Is Illegal. or GO ON TO THE NEXT PAGE. 31 Unauthorized copying or reuse 30 <lriV Pdct of this page I,. illegal. ;t1 ~ GO ON TO THE NEXT P Calculus AB "~11 ~ 26. What is the slope of the line tangent to the curve 3y' - 2x' = 6 - 2xy at the point (3, 2)? (A) 0 (B) 9 4 (C) 9 7 (D) "7 6 (E) 5 3 Cakuj:us Af: ~-~~}~] 2-1 Let l be the function defined by to the grJph of (A) \1 ::: f~:y) ~ .:.\-.t l - :1 \ ~- 3. Which of the following is an equntion of the line tangent f at the point where v - I? 7 r -:, 27. LetJbethefunctiondefinedby J(x) = x 3 + x.1f g(x) = J-I(x) and g(2) = I, what is the value of g'(2)? (A) (B) .\' = 'lx (e) v (D) v +7 13 1 (B) "4 1 (C) "4 7 (D) 4 (E) 13 = 7x -l- 11 = -5x .- 1 = -5x 5 (E) v 1 25. A particle moves along the x-axis so that at time t '2: 0 its position is given by x(t) ::: 2(3 - 21t At what time t is the particle at rest? + 72t - 53. (A) t = 1 only (B) t = 3 only (e) t = 7 "2 only 7 (Dlt=3andt=2 (E) t = 3 and t = 4 aNy UflaulhDrlzl!ld copying or reuse of Pllrt of (hI.. page I.. Illegal. GO ON TO THE NEXT PAGE. 33 3~ ru;:;-aUlhori:z.ed c:op~illg or reuse 01 t~~~~ this ~a.ge Is megal. GO ON TO THE NEXT P.~ Calculus AB I:mim1iU B .... i Part I CALCULUSAB SECTION I, Part B Time-50 minutes Number of questions-17 . " : ' Part ~ i:"J :: : 0 ' " (~<5jCUJ'8J,:j A GRAPHING CALCULATOR IS REQUIRED FOR SOME QUESTIONS ON TillS PART OF THE EXAMINATION. ~~?) Let g be a twice-differentiable function with g'(x) > 0 and g"(x) > 0 for all real numbers g(4) ~ 1'2 and g(5) = 18. Of the following, which is a possible value for g~6)? t A) l, such thi-II J5 lB) l8 (C) 21 (D) 2-\ tEl 27 Directions: Solve each of the following problems, using the available space for scratch work. After examining the form of the choices, decide which is the best of the choices given and fill in the corresponding avalon the answer sheet. No credit will be given for anything written in the test book, Do not spend too much time on anyone problem. BE SURE YOU ARE USING PAGE 3 OF THE ANSWER SHEET TO RECORD YOUR ANSWERS TO QUESTIONS NUMBERED 76-92. YOU MAY NOT RETURN TO PAGE 2 OF THE ANSWER SHEET. In this test: (1) The exact numerical value of the correct answer does not always appear among the choices given. When this happens, select from among the choices the number that best approximates the exact numerical value. (2) Unless otherwise specified, the domain of a function I(x) is a real number. (3) The inverse of a trigonometric function prefix "arc" (e.g., sin -I x ~ arcsin x), I is assumed to be the set of all real numbers x for which I may be indicated using the inverse function notation I-lor with the Unaulhorized copying or reuse of any part of (his pago 1,9 lIIoga!. GO ON TO THE NEXT PAGE. 35 END OF PART A OF SECTION I UnmlthoJr!zad copying Of reuse of :J-t ~~~t~~~page~_ Calculus AB 1biillI1IJ1 ~ 78. The radius of a circle is increasing at a constant rate of 0.2 meters per second. What is the rate of increase in the area of the circle at the instant when the circumference of the circle is 20n meters? (A) 004n m'lsec Pa;r~Bl ~ ",." "'" . . e8lif~Td;i]E 1'::,.,. (B) O.4n m'/sec (C) 4n m'lsec (D) 20" ill 21sec (E) lOOn m 'Isec 76, A particle moves along the r-axis so that at any time t ~ 0, its velocity is given by 1'(r) = 3 -t- 4.1 cos(0.90 What is the acceleration of the particle at time I .::: 4 '? (A) -2.016 (13) -0677 (C) 1.63.1 (D) 1814 (R) 2.978 79. For wbich of the following does lim I(x) exist? , ..... 1. y j y U. y ill. y 1 4 Graph of'j' 0" :/ 4 Graph ofj' J I). 0" >-G x (A) I only (B) Uonly 4 Graph ofj' (C) III only 77. The regions A, B, and C in the figure above are bounded by the graph of the function area of each region is 2, what is the value of (A) -2 I and the .r-axis. If the (D) I and U onIy J3 (J (x) -3 + I)dx ? (E) 12 (E) I and III onI y (13) -I (C) 4 (D) 7 Unauthorized copying or Ieu"e of flny Pllrt of (his page Is Illegal. GO ON TO THE NEXT PAGE. 37 Unauthorized (;opying or rouse. 01 36 .. ~ pall .. 11!lJIl paW8 I_ 11l~9UI. GO ON TO THE NEXT P. Calculus AB ~ ~ 82. The rate of change ofthe altitude of a hot-air balloon is given by ,-(t) = t 3 - 4(2 + 6 for 0 .:::; t ~ 8. Which of the following expressions gives the change in altitude of the balloon during the time the altitude is decreasing? (A) J, 1.572 r3..514 r(t)dt ~ --'-L~,,'~=:J 00. The fuucriou CaJcHl'us l~] (E) J: r(J)dt J (2.667 (e) o 3.5 14 r(l) dt f is continuous for -2 S; x ~ 1 and differentiable for -'2 < .c < 1. If f l-''2) = -S and f(l) = -L (D) which of the following statements could be false? (A) There exists c, where --2 < c < l, such that f(c) ~ 1 Jo 1.572 r'(I) dt o. o. (E) (2.667 r'(t)dt (B) There exists r, where -2 < c' < I, such that ric) ric) ~ (e) There exists c, where -2 < (' < 1, ",eIllIlat fCc) ~ l. (D) There exists c, where -2 < c < J, such that ~ 3 (E) There exists c, where -2 < c <; 1, such that {(c) ~ fix) for all .r on the closed interval -2 ,; .r ,; 1. 81. Let f be the function with derivative given by (E) Two rex) ~ sin(x 2 + I). How many relative extrema does f (D) Four have on the interval 2 < x < 4 ? (A) One (C) Three (El Five UnauthQrlzed copylnlil or reUSIlI of any part of lhlll page 1&Illegal. GO ON TO THE NEXT PAGE. 39 rUllilulharizod copying er reuse of 38 ~!, ..rl ;;of1J~ p<lyo I", I"~eal .. GO ON TO THE NEXT PJ Calculus AB L'MUtmiil - I Part B I 85. If a trapezoidal sum overapproximares f>(X) dx, and a right Riemann sum underapproxirnates f:f(x) dx, which of the following could be the graph of Y ~ f(x)? ~---c:--\ u~L_~CH:t 13 Ga!c'~1Jn; 1-~lY Y Y (A)bm 4 4 3 (B)k 3 2 2 1 x I x o s ~ The velocity, in fr/sec, of d 1 2 3 4 o 1 2 3 4 panicle moving along the .v-axis is given by [he tunerion l = V(I I= ;!/ -t- tel. What is dk average velocity or the particle from time (A) ~O.086 0 to time I :::: 3? (C) Y ft/sec (Fl) ::'6447 ftlsec (e) 3::'.809 ft/sec (D) 40.671 ftlsec (E) 79.342 ftlscc (E) ID.. lk:Y 4 3 Y (D)k 4 3 2 I x o I 2 3 4 2 1 x I 234 84. A pizza, heated to a temperature of 350 degrees Fahrenheit e'F), is taken out of all oven and placed in a 75F room at time t :::: 0 minutes. The temperature of [he pizza is changing at a rate of -1I0e-0.'1/ degrees Fahrenheit per minute. To the nearest degree, what is the temperature of the pizza at time (A) 112F t :::: 5 minutes? (B) I J90F (e) 147F (Il) 238F (E) 335'F Unauthorized copvlng or reuse 01 Qny part 01 thl!ll pe.gu I" Illegal. GO ON TO THE NEXT PAGE. 41 411 UnaulhorJzcd copying or reuse 01 n~~.f~~~"" 1'''9''' I". i~:.g ..l. GO ON TO THE NEXT p_~ Calculus AB IffifiiiM .11 ~ 88. On the closed interval [2, 4], which of the following could be the graph of a function 1 4 _ 2 /(1) di = 1 ? J' f with the property that ~ ... ; G .. , :'1;"' ,; - - __ ~~_~rt :t~ J (;aknlus !\J (A) y 4 (B) Y 4 3 3 2 86 The base of it solid is the region in the first quadrant bounded by rhev-uxis, the graph of .v = tan -I .r. the horizontal line y =:; 3, and the vertical line .v = L For this solid, each cross section perpendicular [0 the .v-uxis i~ a square. What is the volume of the solid? 1 t I 0 ( \ 1 2 3 4 ~x (D) 2 I t I 0 1 2 Ad 3 4 tA) 2.561 (B) 6.612 (C) 8.046 (D) 8755 (E) 20773 (C) y 4 3 2 Y 4 J 2-1 / I 0 1 ! 1. V I .. X 0 87. The function f has first derivative given by rex) = 2 3 4 x point of the graph of (A) 1.008 f 1+x -IX 3' + x What is the x-coordinate of the inflection (E) ? (e) 0 Y 4 (B) 0.473 (D) -0.278 (E) The graph of f has no inflection point 2 o I 2 4 1 r Uhaut!JorizEld copylhg or reuse 01 any part of thl, page 1&Illegal. GO ON TO THE NEXT PAGE. 43 42 ~1:'.I.arl of till:> lla9tll:.~~~~J Unautbcstaed t:;opylng or reu.-~ GO ON TO THE NEXT PJ Calculus AB ~artB I 91. A particle moves along the x-axis so that at any time I > 0, its acceleration is given by a(l) ~ In(1 + 2'). If the velocity of the particle is 2 at time (A) 0.462 (B) 1.609 I ~ I, then the velocity of the particle at time (D) 2.886 I ~ 2 is (C) 2.555 (E) 3346 P':;;;:tB _.TI ~ _ . . - . .- , .... Cal.culm; hie; 89. Let j be a differentiable function with j(2) ~ 3 and j '(2) ~ -5, and let g be the function defined by g(x) :::: .r lCd Which of the following is an equation of the line tangent to the graph of g at the point where \' :::: 2 ? (A) )' ~ 92. Let g be the function given by g(x) 3x ~ J: sin(t 2 )dl for -1"; .c ,,; 3. On which of the following intervals is g decreasing? (H) y - 3 ~ -5(x - 2) (A) -I"; x"; 0 (B) 0 (e) v - 6 ~ -5(x - 2) (D) y - 6 ~ -7(x - 2) s .c ,,; 1.772 (C) 1.253,,; .c ,,; 2.171 (D) 1.772 s .c (E) 2.802 (E) y - 6 ~ -10(, - 2) s x s 2.507 s3 END OF SECTION I 90. For all .r in the closed interval [2, 5], the function j has a positive first derivative and a negative second derivative. Which of the following could be a table ofvalnes for j? (A) x 2 j(x) 7 (B) x j(x) (C) x 2 j(x) (D) x 2 3 j(x) (E) 3 9 12 r2- 3 4 5 \--1_ II 14 16 3 4 4 5 16 5 16 12 9 7 4 5 16 14 II 7 '+ 3 4 j(x) 16 13 5 10 7 AFTER TIME HAS BEEN CALLED, TURN TO THE NEXT PAGE AND ANSWER QUESTIONS 93-96. Unau.lhorlzed copying or reuse 01 any part 01 this page Is illegaL 45 44 Unilulhcrlz.ad copying or UlU9C of on" purl of thl" pagEl 13 1I1e1J1l1. GO ON TO THE NEXT PAG Calculus AB mtr:!Wl'l!lJ CALCULUSAB SECTION II Time - I hour and 30 minutes Percent of total grade - 50 Part A: 45 minutes, 3 problems Part B: 45 minutes, 3 problems I~~~_J 93 Which graphing calculator did you use during the examination? 'C;a](;,l';I.iH~; l-,.. PART A (A graphing calculator is required for some problems or parts of problems.) During the timed portion for Part A, you may work only on the problems in Pan A. The problems for Part A are priuted in the green insert only. When you are told to begin, open your booklet. carefully tear out the green insert, aud write your solution to each part of each problem in the space provided for that part in the pink test booklet. On Part A, you are permitted to use your calculator to solve an equation, find the derivative of a function at a point, or calculate the value of a definite integral. However, you must clearly indicate the setup of your problem, namely the equation, function, Of integral you are using. If you use other built-in features or programs, you must show the mathematical steps necessary to produce your results. PART B (No calculator is allowed for these problems.) The problems for Part B are printed in the blue insert only. When you are told to begin, open the blue insert, and write your solution to each part of each problem in the space provided for that part in the pink test booklet During the timed portion for Part B, you may keep the green insert and continue to work on the problems in Part A without the use of any calculator. (A) Casio 6300, Casio 7300, Casio 7~()0, Casio nllll, 1'1,73, 1'1,80, or 1'1,81 (31 Casio 9700, Casio 9800, Sharp 9200, Sharp 9300, 1'1,82, or 1'1,85 (C) Casio 9850, Casio FX 1,0, Sharp 9600, Sharp 9900, 11,83/1'1,83 Plus, or 1'1,86 (1)) Casio 9970, Casio Algebra FX HI' 380, HI' 390, HP 400, HP 48 series, HP 49 series, or 1'1,89 (E) Some other graphing calculator z.o. 94. During your Calculus AB course, which of the following best describes your calculator use'? (A) (E) (C) (D) (E) I used my own graphing calculator. 1 used a graphing calculator furnished by my school, both in class and at home. 1 used a graphing calculator furnished by my school only in class. 1 used a graphing calculator furnished by my school mostly in class, but occasionally at home. I did not use a graphing calculator. 95. During your Calculus AB course, which of the following describes approximately how often a graphing calculator was used by you or your teacher in classroom teaming activities? (A) (B) (C) (D) (E) Almost every class About three-quatters of the classes About one-half of the classes About one-quarter of the classes Seldom or never GENERAL INSTRUCTIONS FOR SECTION II PARTAAND PARTB For each part of Section II, you may wish to look over the problems before starting to work On them, since it is not expected that everyone will be able (0 complete all parts of all problems. All problems are given equal weight, but the paris of a particular problem are not necessarily given equal weight. YOU SHOULDWRITEALLWORKFOR EACHPARTOF EACHPROBLEMINl1IE SPACEPROVIDED FORTHAT PART IN THE PINK TEST BOOKLET. Be sure to write clearly and legibly. If you make an error, you may save time by crossing it out rather than tryiug to erase it. Erased or crossed-out work will not be graded. Show all your work. Clearly label any functions, graphs. tables, or other objects that you use. You will be graded on the correctness and completeness of your methods as well as your answers. Answers without supporting work may not receive credit. Justifications require that you give mathematical (noncalculator) reasons. Your work must be expressed in standard mathematical notation rather than calculator syntax. x 2 dx may not be written as fnInt(X 2, X. 1.5). 1 Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If you use decimal approximations in calculations, you will be graded On accuracy. Unless otherwise specified, your fiual answers should be accurate to three places after the decimal point. Unless otherwise specified, the domain of a function f is assumed to be the Set of all real numbers x for which f(x) is a real number. For example, r J 5 96 During your Calculus AB course, which of the following describes the portion of testing time you were allowed to use a graphing calculator? (A) (B) (C) (D) (E) All or almost all of the time About three-quarters of the time About one,half of the time About one-quarter of the time Seldom or never 47 ~lhorl2l:dcap~lng or reuse of 46 ~~~~f'<.\g .._~~~~~ Calculus AB 2. A particle moves along the .r-axis so that its velocity at time t is given by ~artA I vet) = -(t + 1) sin( ~ ) t = At time t = 0, the particle is at position x = 1. (a) Find the acceleration of the particle at time t why not? = 2. Is the speed of the particle increasing at 2 ? Why or ' .. ~ . " U I" I Calcuiu,,~ p,.1 !~~~J CAIlJLlJ~; All (b) Find all times t in the open interval 0 < t < 3 when the particle changes direction. Justify your answer. (c) Find the total distance traveled by the particle from time t =0 until time t = 3. SlECTION II, Part A Time-..J.5 minutes Number of (lroolo""s-3 (d) During the time interval 0 ~ t ~ 3, what is the greatest distance between the particle and the origin? Show the work that leads to your answer. A gru phing calculator is required for some problems or parts of problems. y ~I =:t= .JX ~x 1. Let R be the shaded region bounded by the graphs of y = and y = e -3.< and the vertical line x = I, as shown in the figure above, (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the horizontal line y = 1. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a rectangle whose height is 5 times the length of its base in region R. Find the volume of this solid. Unauthorind copying 01 reuse or any part 01 this page Is Illegal. GO ON TO THE NEXT PAGE. 49 48 ~t~~! U",,~~~l~~~_ unautncnzed cop.ylng or reuse 0'1 GO ON TO nlE NEXT PI!,! Calculus AB ~ill CALCULUSAB SECTION 11,ParI B Time-45 minutes Number of problems-a-S No calcnlotor is allowed for these problems. ~iD Ru) 1(;c,J'(:"(i!,~l:iL }\} y I [60 _ 7() l :~~t / g 1"< / ~.-----v-- / ~;~~=::~ I) 2" [Nf) (g~IIJ()llS per rniuutc) (-3, 1) 21J 30 40 55 65 70 JIJ 30 ,/ ~ 40 50 70 90 lflJ"t -3 'c-.. _I, I -1 ........- oc:::::::: ! .. X 2 4 'c 20 -- '" ;;; 10 ) ~(--+-!Ijl" Time -2' Graph of (4, -2) 10 20 30 40 50 60 70 RO 90 I' 4. Let f be a function defined on the closed interval -3 $ x $ 4 with f(O) ~ 3. The graph of of 1, consists of one line segment and a semicircle, as shown above. (a) On what intervals, if any, is f increasing? Justify your answer. (b) Find the x-coordinate of each point of inflection of the graph of your answer. (c) Find an equation for the line tangent to the graph of !" the derivative 3. The rate of fuel consumption, in gallons per minute, recorded during an airplane flight is given by a twice differentiable and strictly increasing function R of tune t. The graph of R and a table of selected values of R(t), for the time interval 0 ::; t S 90 minutes, arc shown above. (a) Use data from the table to find an approximation for R'(45). Show the computations that lead to your answer. Indicate units of measure. (b) The rate of fuel consumption is increasing fastest at time t '-- 15 minutes. What is the value of R"(45) ? Explain your reasoning. (c) Approximate the value of 0 R(t) lit using a kfi Riemann sum with tile five subintervals indicated by the 90 f on the open interval -3 < x < 4. Justify f at the point (0,3). (d) Find f(-3) andf(4}. Show the work that leads to your answers. J. data in the table. Is this numerical approximation less than the value of (d) For 0 < b 5" 90 minutes, explain the meaning of Explain the meaning of So 90 R(t) lIt? Explain your reasoning. i J:R(t) dt J: R(t)dl ill terms of fuel consumption for the plane in terms offue! consumption for the plane. Indicate units of measure in both answers Unituthorll:lld copying ,,( reuse or any part bf this page Is Illegal. GO ON TO THE NEXT PAGE. 51 END OF PART A OF SECTION II I~!:.~~~~~~j.;~~i:~~1:ol~1~:~4 I"'~i ---:-_I ~5iu:.~1 __ ~_~'i~~~_~_~_ l radius r and height h is V (a) Show dli = h in 1 5. A coffeepot has the shape of a cylinder with radius 5 inches, as shown in the figure above. Let h. he the depth of the coffee in the pot, measured in inches, where h is a function of time t, measured in seconds. The volume V of coffee in the pot is changing at the.rate of -51!.Jh cubic inches per second. (The volume V of a cylinder with nr 2 h.) -Jh that di = - 5 . = 17 at time f (b) Given that Ii = 0, solve the differential equation <i.- = " z i t ) .J!; for h as a function of I. (e) At what time I is the coffeepot empty? 6. Let f be the function defined by fix) = {h:+l 5 - .x for 0 ,; .r ,; 3 for 3 < x ,; 5. (a) Is f continuous at x =3? Explain why or why nor. (b) Find the average value of fix) on the closed interval 0 ,; x S 5. (c) Suppose the function g is defined by g(x) = {kE+I mx + 2 for 0 s x ,; 3 for 3 < x ,; 5, where k and m are constants. If g is differentiable at .r = 3, what are the values of k and m ? ENDOF EXAM 52 Unaujhcrlaed copying or rc~--;;fl ~~~_Qt Ihl9 ll.. go l~~~j ""G'. '.",", ! Vel'C'c::li.'ir. ,,",< '."'I",j-'(U Galf;"nh 1I,s ./-iB Item 1 l-:-)flJ'~ ,------ CHlr:>tJi"uE; rPEdct A Correct "" __ E ~[---o-'--'--l R('; 7') -13 ~ c 8 8 75 I u.~ +3 Jb 76 67 I-I -!9 86 1-1 15 76 75 E 0 ,,~ u r <:I"::' 78 83 -11 69 16 18 84 73 39 16 77 ~~o T! 8-1 91 133 19 20 21 -----5~5 2~ :25 5" G ~I) /4 6-[ 82 -n 7~ 113 _131-1 i 0:; fi5 LJT!3,3 -1 C A 86685:2..,l010 83 5334 ---' II 18 _._-- 41 52 38 fiG 32 13 GO 19 28 45--1 73 .-:] 5 ._ _"_-:_ : 4 ::'1 10 14 18 19 20 21 22 23 A 0 --0-- --A--- --0- --E- 75 69 - 78 79 ~ 80 55 61 c15 53 52 Jl 37 41 ~ 23 29 ~ ----~-5- 5'1 52 J5 87 8--1 76 JJ 13--9-'---3g 3,] 22 [3 ~~"-- 24-17--j:2- --:29~1 63 70 'J6 51 2-1 24. 66 - - - - D D B -1-1 79 -13 53 to 47 l3 ':::0 '3 36 1] 2-1 ~l:; _~_ J I ~ ~- , 25 23 14 ~ E B 18 88 24 75 14 64 8 .:19 ..J 28 2-1 C I 25 ~-B----9-O I E I 95 9-1 0 63 -86, J8 65 28 ~~O -~~ _ 19 ~~ -='K S?-~-l'-------s3-' I l1! ~ -16 72 -18--9-8~-.J 18 'I :=:'26==_ 0 - ~~-79 27 ~o ~9 67 J2 23 ~ __l ~lj 23 11 G _1=~-3i 2,3 E D 71 62 46 36 -:-.i'==:'3i:::~5 22 15 1-1 33 .2. -17 J -~- g"-- ~ Calculus AB Part B Item No. 76 Calculus BC Part B Total Percent Correct 79 Correct Answer C Percent Correct by Grade 5 4 3 2 1 97 94 87 --73-3-8 - 55 9,1 Qt:: Item Correct Answer E Percent Correct by Grade 5 Percent Correct 75 ~ - 77- No. 77 78 'Jr'l C C r-, 26 81 69 38 13 63 52 25 8 45 37 18 [0 33 23 11 23 65 55 31 78 -I-SO 81 82 83 84 85 86 88 A ~9 r; ------ce r, 63 A A A E e-=-+--=--+--'~~~-'-----'.:::::-~ ~~ ~~ '33 ~~ 15 ~~ 50 t 8584 - 83 A 79 6G ~ - 88 : C E E 0 A A 65]4 77 57 ,51 17 9 -43--30 ..J2 24 6 19 26 46 -- c A --~ ~~:~~::~:~ 82 62 80 26 70 12 17 9 ~~ 46 ""87' 89 90 91 92 C C 92 ~ --~ '-'" 89 90 92 _,,_ 0 , 84 --- - - - - - - - - - - - - - - - ---- --88 -!9 32 18 53 92 54 56 30 eo -13 37 12 7 ~9 59 -~2-4- 0 B B 88 88 95 88 98 79 76 96 86 88 60 77 84 74 80 61 81 4 78 3 74 87 71 94 6J 57 87 71 64 75 5.5 87 52 42 2 61 1 44 '""' ~ Ie! 63 73 41 42 68 44 64 ~1 76 47 60 32 25 56 26 49 57 61 36 78 41 23 59 3..J 45 27 'j3 54 41 80 66 20 58 28 17 89 6~ 5-! 20 ~2 ., - I ! 37 23 29 24 69 "!4 41 9 22 7 16 65 -15 50 12 32 20 17 C ~ 54 30 35 18 18 13 13 ~_.~-- -r: 52 - Jr . ;',;,,:",1.il,,",,".';:.;.- i"-' ~C; ,.: -,j' , ,11:, I1lclil,-c:' on the lVJnlU.pif:J (~:h{).b:;,s ;::;.ectil-'::1 J(;'(lll Sl\'c' ) ou. ':Jl il:kllL, {',l!clJillJ t, III j\Cl Ill! rhc ~()Il.\ t.,lku! ,c,.-\L', or on th.it I,) c ,li<-'gn! \, l' Til 8 1 ' cv.uu tOI"11!";!'."lilx. :'illl l11..lV \\',lnt 1<' :lll,l]1 .c the rc..:s1dts to lind overall "lrcn~nh~. .md 'I\":clhkS:::'c> In IhciJ undc'rS(,lllding, (,L\P c,ll(lt!ll,S. T'IF: i",l!l,)\,jng the i l' dv nur.il L,bL' (:'1" ,J[ ,1: cn.J i,J lI . I )i.lgllustie I ;lll,-k. d!<H~J\nsli,: \\,\)r!,s!lccl.) v..-ill hell' 1,l)LI till 1:li~. You nrc Y\llil [1cr~lljlll'd ')llldl:ll l" 111 \~,lLb [n .r.Lluion, under e.ich item. .msw-rcd L'\IITc'nly i., SbIJ\\'!1,'1.:"1 Ilh~ l'~rll-'l1l studcnt., cnn III AJ) SLlltlUIl> \1'!1(J thc.. III l,hlij\)((Jpy .in.l di-rribuu Ihc'lJl (" formane- (111 inc.ividu.tl items. Thi, inform.uiou \\ 1111,(" that cue qllc::,llUll m.ty ,ll-Ipcar in ::,c\\~r,11 I~jr rornplc t ion. hcloful m dcciJing hovv srudc.u, s!luuld pl.mrhcir study jlLl"rl .1 ,,'ll";l."j:, mar]. L"clteg\lry, ::,tllll~l1iS should rune. Plc,lse nol c' hH' each correct answer. ! Adlllogell1er the luLl1 number oj' correct ~ln:;\"'LTS Vi)! diftercn. (lle~Llrjt':-', ,L, Ljllc::.li(lll:> \~;H] cover different topics. t"h.:h categorv. Diagnostic Guide for the 2003 AP Calculus AB Exam Differential Calculus (Average correct = 17.2) ---_._--- -1 I 1 .\ 0 Question -# - - _ - - - - . -- ... 9 \0 12 1J 14 15 16 17 18 ---- 20 Correct/Incorrect --------- - - - -- - fio - -56 .. --_ .._. - Percent or SI udertts Answering Correctly 93 u SJ < S" ----- 6~ ~-1 75 76 75 77 "12 52 -15 --Question # 21 Correct! Incorrect Percent of Students Answering Correctly :'4 :'5 261:'7 ----- 28 70 7S 79 80 --- 81 87 89 90 92 39 66 6; 58 --- 18 31 79 65 55 31 5-1 ~7 3~~ 86 Integral Calculus (Average correct = 6.0) .----~---- Question # Correct/Incorrect --_._- Percent of Students Answering Correctly :' r----~--- 1-- . 5 __. 8 11 19 22 23 77 8:' 83 8.\ 85 --e- -- 1-- e- - - - - ~ ---I - ~5 58 65 37 38 ~1 29 23 39 50 26 ,16 37 1J 93 i~ f) :ldff'tU);"3 ,l.ic; (c(1ntintl(?,I.:}j Part Calculator (f!'Iveclg,e l.'ilrreci = i 55) -,---,-- ---,------,--- <2l1cstiOl,~e- Correct/Incorrect - -_.'-_.,._----~--- Perren. of Students Answering Correct! -----.~--_._- ~] "I" ~56r56 76 77 17 18 ISl , ;u" 82 - 23 .2-! ~- 25 - 26 -- 4] 52 JR ;~141129 83 84 59 50 26 66 67 58 Part Be-With Graphing Calculator (Average correct = 7.7) ~I~I 88 89 ~---r-- :llle=tiOi # Correct/ Incorrect 78 79 80 81 _- 1 ... 85 86 87 90 .~- 91 ~- 92 1 t _._-- - 79 23 - I -I--- -- 'J6 57 '17 Percent )f Students Answer! 19 Correctly 65 55 ._- 51 54 46 39 59 2'J 53 Calculator-Active Items (Average correct = 4.1) ~-~---_. Question # Correct/Incorrect 76 81 82 83 84 86 87 91 92 -- Percent of Students Answering Correctly Number Correct 79 54 39 50 26 37 47 24 53 Differential Calculus -~~--~~--- Integral Calculus 15 6.0 (40.0%) Pari A-No Calculator 28 15.5 (55.4%) PartE-With Calculator-Active Graphing Calculator Items 17 9 4.1 (45.6%) Number of Questions --~~--~~- 30 17.2 (57.3%) Average Number Correct My Number Correct ~- 7.7 (45.3%) 94 Scoring Guideline for AB/BC Ouestion 1 Let R be the shaded region bounded by the graphs of y = the vertical line x (<:1.) (b) E' and y = e-: i ' and = 1, as shown in the figure above Find the area. of R Find the volume of the solid generated when R is revolved about the horizontal line y = 1. The region R is the base of a solid. For this solid, each cross section perpendicular to the .e-axis is a rectangle whose height 1S . ,.t:'::..B/BC l}\L[<[3:Dtion 1 (c) 5 times the length of its Commentary This question presented a region bounded by two graphs and J vertical line. Studeurs needed to use integration to find aii erea and two volumes. .Agraphing calculator was required to find the point of lnrersection of the two graphs; PartIu) r~,quin:~l the use ,of q detl1~jtc inl:egr~l to find the' area of the, region. Part (b) required another application Q,f i~~:~g.l'~;f~)n. ro find:~le volt1~e,~ene~ated by revol~ins. this region about a horizontal line. The resulting solid had cross se1'iqWsin the shape ofwashers.Part (e) again,required integration, this time to find the volume of a solid whose base base in region R. Find t.he volume of this solid. Point of intersection I: Correct limits in an integral in ,-3, = -JX at (T, 'l 5) = (0.238734, 0488604) (a), (b), or (e) L: integrand .: ;' ~. 5,81 f?~.Calculu,~ Be. Students llid not need to compute antideriva 'pUtelh defli;i~int~grals iri p~~t.(a) an,\, tb);thecalc"lator maybe nsed tocalculate the value ofthe defi ~tn'ou~~furthef.;jXrlantti6\1\lnce the,etu~lofihedet1niteinte~ro1 Is shp'(?n,Antidirferel;tiation was not the con'ktint<;grandin part (c), "0 a calculator had to be l~ed to evaluate that definlteiutegral. MB1jystll dr:tf~g~~.ll\'~ i!ltegl:a~~ill're,~~i~nfor.th.'~olul1legenerat~d by revclving'rhe gicenregion aboutthe x-axis rather than . .' t~e.~~tiz,oritalliqe ?'",l,snidentf .need.a';"l.\ch prccticeas possiple wltliserting up Integrals forregions revolved ar~~~d horizontal Hne,oth~rtlla~ the x:a~is,This ';" the first time Oil the free-response seclioh that a solid was given !i~6re ,;a.3,90. t~.f C:~kq\Ui~Band Ul_~~(' whose9i~'~s~edi?ns w~r;e~~ectangles., _' , , (a) Area = Jr (-JX - e- 3000 )d.T ') . .., . { 1 : answer = 0.442 or 0.443 (b) Volume = 7f :~~'QSe~~stos~secti6ns w~~,e' re~t~,rigles,,~ c?JTl~?n ,T~o~ was . 'ta~~ ~?r students \~O their-w?r~ on the free-response section of the exam, especially the setup for a definite integral , whose value is computed wi.tlplcalculator; ' It~i'ghf~f5.))edm.a~/ms\..,~f~;:r'i~h SUPPol'~'ine work received no c(~dit on this qu~stion).evenif correct. It is lmpor no to assUln~ ~h~~ each ret1a.ntr-~~ar cross section had a constant (1Jl( l' e" ') - 32 ") (1- -JX)- d 3 : ~ 2 : integrand = OA53 7f or 1.423 or 1.424 < -1 > reversal sho\V << 1 > error with constant -1> omits 1 in one radius < - 2 > other errors 1 : answer Student Response 1 (Excellent: 9 points) The student presented all required work and answers ;1$ directed. in part (c), the student clearly indicated the dimensions of a tvprca! rectangular cross section and the resulting area. Student Response 3 (Fair: 5 points) The student determined the correct region to use for the inLegration for 1 point. The integral setup in part (:01) is correct. The studenr computed the definite: integral ead function separately La three decimal places. This risked tlH possibility at premature rounding leading to a final answe that was not accurate to the third decimal place, but in rhi question that did not happen and the student earned 2 points for part {a]. The integrand ill part (b ) is incorrect because it does not consider rotation around the horizonr or (e) Length = Height = Student Response 2 (Good: 7 points) The student determined the correct region to use for the integration for 1 point and also earned both points in part (a). The integrand in part (b) is reversed, however, losing 1 point, and no answer was computed for a total of only lout 3 points. Both points could have been earned if the student had gone on to correctly compute the value of the integral and then reported the volume as the absolute value. The student earned all 3 points for the setup and answer ill part (c), -JX - e- 3., 5(-JX -- e- h I ( : integrand ) < - 1 > incorrect but 3: has .J'X1 : answer e- h Volume = or line y:::: 1. The student was not eligible for the answer pail and hence earned no points ill part (oj. The setup is cone in p,Ht (c), but this time the premature rounding of the calculation of the definite integral did result in a final answer that was not accurate to the third decimal place, so the student did not earn the answer point. J 5 -JX - e-") d = 1.554 T 3' , as a factor 99 98 [Work for question T(b5] \I pl<;r<- =-1TL p,1._('-)h 1T Si;r",rD't" ;~~ t):i.I~-'J~~, 1H, 'If:f;~]P{) /J :;Jcl~S j~ I .J.?"I \-e-3' ) 2_ ( 1- (,< )2J d t :f(J"i( ~-'tlJ.dent fi>2Sr,\:m.SE~ '_~ {E.i{( 9H.snt: ~1 poiTr:::.5:~) y ~ I), 4C;3lT .r [WOrk for question 1(0) I = L\\"'1i' for question 1(aJj - - - - ~ - LD '312 se.1.. Q.::. q) ~ \ ({1 - e-")d;Z . 1..'I (1)< 2 e-:''':) S ~ 0.4-.<:\3 ~ 5 1 ( -IX - e-3 ;") d j{ .H"I I. ss 4 101 100 i;S ~ J0 \) ~ g' ~ 5. ('[l :;; ~ ~ :oJ:: 1...--0 0 - "';-J \..,..J P f a '< (1) U: '1:' ;,; ffi (Ii ~'\j o 5 c 0 ~ I (1J !'" .... 1 'c; 8. g. !!2.. .x ;X I 'J-' '------" >b ~f 0 II s-, ) .c: -c: '-.? - 0 In 0 < w o I~ \ (h, .;. 7' 1\ ~c::J ~ ~ ~ .g " ;. 0 e n Q' * ~ * Q' ~ ~ I (1) J H 1\ '" <. 1\ -:::::\ Ii ~ ~ .0 R 5 ~ " ::.:> \I ?~~ < II '? II ... ~ \ i; L..---J - ~ '"\ \" OJ r- 0 ;1 I (1), ..) :;: ... c.>> (J :::t .---... t <:> ------. tJ1 I ~ t J 1/ ~ "-../ ,.., >< If'! ~ -+ ffl ----./ ~ ~ ;J II ~ J' ;>< (1), ~ .:: I' vl :::j t'J >7 t" ;L-J ""(I> I <; ==! 1>< X Scoring Guideline for AB Question 2 A particle moves along the 1.;,,&.,\:i8 5D that its velocity at. time 'v(t) ~ -(' I is given by + l j sin 2, [tll At time t = 0, the particle is at position x = l. (a) (!.r:l.,,"J>::lt~{J{j Find the acceleration of the particle at time f. = 2. Is the speed of the particle increasing at t = 27 Why or why not? (b) Find all times t in the open interval 0 < t < ~1 when the particle changes direction. Justify your answer. Commentary ThLi question presented the vslociry function for a.particle moving along the x-axis and asked questions about accelera tion, change of direction, distance traveled, and greatest distance from the origin-s-questions that involved interprets- , lion of velocity and knowledge of its' relationshipto the position ofthe particle. Both differentiationat a point and integration over a' sp.c~~~ed interval w~n:'. llse~ in solvb~g the question, Part (a). req~ire~ students todifferentiatethe velocity function at aspec,ific tIi~~ In oroer to find~h~I'C~~~~~~l,tio_n. ~tudents needed tOlmd,ers~an~~~e~~ff~~en~~ between velocity and spe~d. qneway to d~~_ide Wh~~,~~r'or,:nPtfbe speed was increasing ~3S, to rec?&ni.:Z~ _iryat"pd~iti~e aCLele~'ati9n and n~g~4Ye.~eloc~~ iI11P,~Y de~r~_~J~~rg sp,~e~... :f~!~.(b) required ,~tudents t~. use ~e ~b,<tnge in the .SigL~ of the velocity functiontQ,qe,~~mfne.Wh~l:etheJ'a~Hcl~~Ii~~~~qjr~cti(Jr\.P'lft(c) askedstudei)!;;.IOsetup.aIldcoml'ute ( definite integral that gave the distance .traveled over-a time jiif~r;aLThis could.be done ,itper hy lfsi~&'t he turn lrlg point from part (b) Or by integratIDg the speed Part (d) brol{ghte;erythingtogether~"sltingor thefur(hest\~ist~nce f from the origin Becausethere'was no closed formula Ji>rthepositicin,students neededtnlaiqwh?w to use the definite. integral of the velocity to determine posltipn'Ttieanswer to part (b) )Jto,;idedone candidate for a time at which!},e ..Studentsalso' particle c~llld be furthest from the origi'JiJty/as !;ecessaryo>6nd the position .of the particle ~tthistilTle (c) Find the total distance traveled by the particle from time t = 0 until time t (d) origin? Show the work that leads to your answer. = 3. During' the time interval 0::; t ::; ;3, what is the greatest distance between the particle and the (a) a(2) = v'(2) = 1.587 Or 1.588 2; u(2) = -3 sin(2) < 0 Speed is decreasing since .(2) I i i 0(') 1: speed decreasing with reason > 0 and u(2) < o. j : (b) u(t) = 0 when t I' "2 = IT t = 2: .J2if only = .J2if or 2.506 or 2.507 1: justification neededtocompare this distance tothed/stance from thenrigin attime~ I;" o.and 1",,3;, ' .. y . . <, . : ' The mean score was 3,00, ThiswaS,"'difficult question for students as only 0.4 percent earned all ~p6ints. Students .. did okay on those parts of the question for which nUpleriS") '1I1SWe" could be obtained from the calculator.Most diffi culties occurred with the justllk,tiall; po~t\?m of the 'lue~tion, Port(e) .could be correctly done even if the student computed an illcotre~tturniii~ point ,,; TIQ turning pointin.part (b),Th~!!1ostwl~llIlori~r'ror in part (c)j"~S noUnte' gr"tiug the absolute value of the velocity fuiiction. In Pill'.t(d), thenull)e~icalan1'"e;s""ere easy to ,,,,In'ttl,buL the most common error was not using the -initial condition an0/ or not reporting distance as il positive Since v(l) < [) for 0 < I < .J2if and u(l) > 0 for .J2if- < t < 3 the particle changes directions at I = .J2if. I (c) Dis Lance number. ' = fa 3 [v(t)ldl = 4.333 or 4.334 l : limits 3 : J 1: integrand Student Response 1 (Excellent: 9 points) The student repoi Led the correct answers in parts (a) and (b) and provided apuropri.ue justifications. The calculation done in part (c) is the easiest: way to compute total distance usiog the calculator to do the numerical mtegrarion. The student constructed the position function in part (d) and then duplicated some of the work from part (b} but tied everything together nicely to determine the correct answer. Student Response 3 (Fair: 5 points) In part (a), the student found the symbolic derivative of tl velocity function and correctly calculated the value. The student failed to provide a reason, however, fat" why the speed is decreasing and sa earned only 1 out olZ pointsIr part (b), the student set I'(t):::: 0, committed to the mot in the interval (0,3), drew a labeled sign chart, and correctly interpreted the sign chart in words to justify why the particle changes direction. This earned both points in part (b). In part (c), the student did not use the turning point in finding distance and earned only 1 point for the limits on the definite integral. In part (d), the student tour the displacement, earning the first point, but did nut consider the initial position of the particle. (d) Jo .J2ii v(l) dt: = = -3.265 x( -J27i) x(O) + 1 0 .J2i 2: I'," 1; answer (distance particls tr ev els while velocity is negative) v(l)dl = -2.265 1: answer Since the total distance from I = [) to t = 3 is 4.334, the particle is still to the left of the origin at t = 3, Hence the greatest distance from the Student Response 2 (Good: 7 points) The student earned lout of 1.points in part (a) for com puting the correct acceleration value, but the explanation for why the speed is decreasing had the signs of the accel eration and velocity reversed and therefore did not earn the second point. The solution in part (b) earned 2 out of 2 points. The notation v(t) is close enough to be considered a label for the sign chart. This was reinforced by the written interpretation of the sign chart. In part (c), the studen t correctly handled the turning point to compute the total distance and earned all 3 points. In part (d), however, the student computed the correct displacement but failed to make use of the initial pcsinon and earned only 1 out of 2 points. 106 origin is 2.265. 107 ~ ~ , Vj -::-- c; ~ a., cc, ,-...,. ::r-:, , .J ~ r ~ , ~ o r- ~ ~ -r- I Q C rr- -' , ~ ~ ,---..,. J.A-+", ~ <. ~ S- " e..-' ';. [ ) ~ < Q,:' VI j' '-" i ~'I r-: -+'---" IF 1;0r ~ 10 ~ g, '-i ru ~fJ ;;1 '" ~ (;. ~ ;;.' ~, t't' i~! ':!.-l i~1 I~i ~ c;"';; s~ - > II hj i gl >~ "'''1 es ea t"';";~; g:, ~ :~ ....! It\ " ;r-- 0 -- .r , ..... ~ r -c -r- " ----L ' - ' ........ + 0 -... ,.. '" -e- r-; . --5" j c --J ... ~ :~ fo' 'Q e-, VI ml ~, ~ . " a V> -3 ( P- ..., "p "1: ~ -e- 0.' "- z- r; 1..1 '-- I I (" \ \ r>: ?' IV II c r-; hj ,~, "'(" r>; <:: ~ ':t-' ch ~? ~; to' (~; <: " <:: r-, T '---' '.-''<, "' (:"; f"" W I~. ~ '--' ~ ;i? (Q ,r..... !: '---' (ft ....... ~ -+, .... r ~ ;""" ~, :~ (e) 1- ;0 l/\ ., +. tT- po ~ oJ, co ~ it ,-;--.. ~~r"'"""' <:> ~." I"" o -..J e .... ~ 1 ( ~ ...., Q \I \ ~ \ "" ~'-~, 0 /' C )' ~ ->- ->- '" 0 -.J :: p ~ s -.. t. c.. r r ---.. I r- )' ~ ~ (l ('h l.J' ;:: " VI ~ o r-- V) ------- "" k Q L @ ~ ~ 'I ~,~ --1l -c Q ;::' " .... ::. l'\ t: 1 S< :e; " Iv ~ "" " ~ ; ~ r CI ~ ~ * ~ s ~ ---.J , r-. .... T 2 + 0 ,I hi ...., r-; -rV1 "'- '" .... r 1 ~ V> -, Iv " T ... r:-: I r>; .---, U' ;;,. p IC 1 I " c I>~ ...~ r;; f \ ..j e ,.. , r-, !\.. !~ [.. " c~ C.> r-. r I I r; .... 't- "!: '-" P .: r- > '-' U\ ) ;, t r " \I "---. -, )..) '" -\-. $I 7 V I IV) I '" ;- ~ -. >, ~ V. ,.-.... Ii' V) r if'--.. \ -0/ ; o '-- II f' ( ~ > ~ ~ ,,\... ~ "---' 0... . "-----' -:p " :1:: <.., LV r-.. i ~-1- . -1::. "-'" .-,/..J "-...-- " '\--. ...... ., \) til J ",,' \ ~ , /I ... ----- ----. , '--- Q... \ tJ, \ .., I I. 0 Iv -+ Gf:: (f\ 4 I 0 6' 7 I:! V. vI .' ...J ~t _c>~ ... J i ~ 1-~' r > ""t '.I o >-> <D [ Work fOfques/iou ~ J o 7 _(HI),;,(t)o\~ = - 21~' 8tr~di;nt Response :, (F-c.d:r~ 5 poirscs) IW"r1"[,)I_'1UCOli011-2~;il Y'(i)'22+" II -_._-_ ... _ - - - - - - - \1 ( ~ ) -~ ( -10 {- \ ) l; " (J!-') \ -(( +11(0<; \I'(t)~ - 1 -I t'l) ,( (ACt) = -eU\)CO~(2 i: - ')1" \.,~) -'[2 \ (e)/2(ut.=.ft2)) -1- '),V)(1- 2 \ . ~i;--I :; n'{f?\ --(u I )(0'7(1'L) -t - '1' \~ -2-) v(z) :: -rz I) &,:" (, .~ \ 1 [0., ( 1' ) 2., J '11" 'l 1 \ :: _ ( 1 .+ 0\ ( -) L t\ 1') 2. - ~I" ~ . (,"') Y(z): -J~'''Z VI' 0<. . () (2)::- -2,129 I '. '::1fl'C<' ,L, O'1e(v"""~''''j IWork for question 2(d) I yC +\ ('.2 j-; - 6 (Ole:?) - 'ji"U) ~ (1)-: 2.Qn - . q07 I 'i\'t> 0\(2');' ." +- t=2 ~ \j - L{ t ') '" ~ ~ C}2) 1.'0 ~ 1 lhc: Ct[(e(lr~-\-;." ;~ X-::o Y-='2.";Of 0 I ! ",,[v-eo. ..,r-"l '0;' +\'e ~reJ IJ Jec"-Si~) t~) L-:'.2,5 I + { ') ['vV()rk!or quest;',,; ~~blj \j[+)~ _(-\--\'\) 1;'" cf) 'I. =.0 y. =: 2. jOt 1 1 t lor ,I; (te 2, 5,,::1 o - (-I-t-I)'7;" C~~-=- -.J, '- Z 6? i--, cn",P\ ,I ..,-re o: r ~.l.,L -' c/') ianet' *e..' ov,,;'-' -tAf' P<>.f+;c Ie. '((~)/~o _ 1.,0 '3 '7 + -II, r(ll, (\.c)" o..-t vl"rn ( ") - (H I) s'lh ( if )JI-:: )1.. ';0 '} , I. 0 {; ~ ~E't' ;;;'" 2,]o1,.rc orJ ;(I';1 Ii "I: of '1(1-). vCr) r-e.,,,,-l-iljje "'0 -/0 0. C\'-'9\'')'" 1 f~" --\v<r'" ~o",,+iV( ~ X'" 2.')0 7 113 112 Scoring Guideline for AB Question 3 The rate of fuel consumption, in gallons per minute, recorded during an airplane flight twice-differentiable and strictly increasing function R of time t. The graph of R. and a table of selected values of flU), for the time interval 0 minutes, are shown above. ~,U3 (~\U.&BaOi!'; R(f) 113 g-iven by a s: t S g ;~l!: ~. 50 -10 2\J gO o JO 50 I zo 30 40 ~ ~ ~ 3 1:: 30 (a) Use data hom the table to find an approximation for R'(45). Show the computations that lead to '" 55 10 70 65 70 YO 0102030-\03060708090 Time Commentary This question presented the rate of fuel consumption, R(t) ~ of an airplane. The rate was given in both graphical and tabular form, Students were asked questions that addressed the values and meanings of the derivative, the second derivative, and. the integra] of the functiondescribing the rate of consumption. Units ofmeasure were important in your answer. Indicate units of measure. (b) The rate of fuel consumption is increasing fastest at time t = 45 minutes. What is the value of RI/(45)? Explain your reasoning. (c) Approximate the value of parts (")anll (d). In parr(a),stndents were required to Imow how' to use the values in the table to find an approxima tion to thederivative Of Rat time I~. 45, Students.should have used those values closest to I ~ 45. Port (b) required ~tud~n~s. t~ recog~.ize that. when the ,tat,e.of jqC;fease of Ri"s m~111i~ed) the second derivative of ~,will p+ze~o. :l~,:,r (c).~t:ldelltsneedefto cal,ul~te a I~ft~iemdnl1!:\ni~ppro,pl11atiOl1tQ a definiteintegral ".nd I? kl1ow~h~t fO.:"li il1,te~~il1WA0~~on,this '~'~)Wl.!!11<leJ;,stiJ11~!~1ihe }~ey?1u~0(t~"intg]'a( Part(d)a~kedfoq~e meaningofthe. , definiie'iniegral of R;Irin~i\rtedth'etis~al qnestlonalioutthe ~ver~gevaliie ofa fnnctiop bypresentingtliedefinite inte~~al diyi~d qttheIe!i~tllllfthe tiJ11~ifJtery.l an~a1khlgthesllldent forit~ me:minll in terms of thefllcl COn~Ulll\ par (d) 1 R.(t)dt using a left Riemann sum with the five subintervals indicated by the data in the table. Ttl this numerical approximation less than the value of 1 R(t) dt ? o Explain your reasoning. For 0 < :S 90 minutes, explain the meaning of 1 in terms of fuel consumption for the o 0 90 90 b b R(t)dt plane. Explain the meaning of ~lb R(t)dt b 0 in terms of fuel consumption for the plane. Indicate "Tbe itie~ll'icOrewa1~.'12:iil'~j,;j'tirig pQorperf("1nm~eby }~e'A1l~~~ents,F6\lf onhe'nine poil1ts9n this ques-. l tionl'1vo1ved expla'1ation~. One qitbe pdmaiy goals of the changes tt><ItiAp.cakulVs hag undergone inthe past ~evero ye~rs is t9woV~aW"y froUltel.tlng rote. !na~ipll!aiio[1andwwa'dqueStillnstharprobe understanding of the funda mel\talco~c~p:~,Thisque~ti~n was an example of one that required it~<i~nts tom"co~.nectibn; between calculus a}'dthe )eal:'\vorld, and between toplcs.withtn calculus itself.Students /leedaddi\icinaIpractice working with .1\1I1C'; rionst!lilt ~r" givell numerical PI' tabular-form, and in nsil1gappropriate'l1fthematIcol terminblogy",hcll. explaining the meaning of Inathematicalexpressions; They should also see more ,examples of calculating Riemann . sums with subintervals of different lerigths. . .' ti~l1of:lleairplani'1'7fel'en~e!t~}the~peqftci~lellitir;'~l",a~"niUlP?r\a~tPart?ft~eeXf]anatiq~,.,. l 'f;' >.;; . (a) units of measure in both answers. R'( 5) '" R(50) - R(40) = 55 - 40 4 50-40 = 1.5 gal/min" 10 1 : a difference quotient using 2; numbers from table and interval that contains 45 I : 1.5 gal/min' onlY.ln (b) R"(45) = 0 since R'(t) has a maximum at I: R"(45) = 0 2 { 1 : reason t = 45. Student Response 1 (Excellent: 9 points) The student used the appropriate difference quotient in part (a) with the correct units. Part (b) provides a nice explanation forwhy RI'(45)~O. The correct Riemann sum is given in part (c). The student's written explanation and drawing of the inscribed rectangles under the graph earned the reason point. The meaning of the integrals in part (d) is correct when the phrase "until time: b" is interpreted as ,1 reference 1'0 the time interval (0, b). Student Response 3 (Fair: 5 points) The student computed an appropriate difference quo tiel to earn the first point in part (a) but lost the second poi! because the units are incorrect. Part (b) is blank. A right Riemann sum is computed in part (c), thus losing the first point, but the student did earn the second point for asserting that the right RIemann sum is i:H1 overestimate and providing an oppropriaic reason, Part (d) earned ,iiI 3 points. (c) J o R(t)dt '" (30)(20) + (10)(30) + (10)(40) +(20)(55) + (20)(65) = 3700 (90 : value of left Riemann Sum 2: { : "less" with reason Yes, this approximation is less because the graph of R is increasing on the interval. (el) fob R(t) dt ! Jo R(t)dt (b b is the total amount of fuel in : meanings gallons consumed for the first b minutes. is the average value of the rate of Student Response 2 (Good: 7 points) Parts (a) and (b) each earned 2 out of2 points. The correct Riemann sum is given in part (c), and the student's drawing supported the reason provided for why the' numerical approximation is less than the value of the definite integral. The meaning of [he second integral in part (d) is incorrect, as are the units for that integral, so the student earned only 1 ouLof:> points. 3: fo R(t)dt I I : meaning of b fa R(t)dt I: meaning of b b fuel consumption in gallons/min during the first b minutes. < - 1 > if no reference to time b 1 : units in both answers 115 114 IWork for question 3S0J S E!an~l;,L~ ~:~"~'i!lf:l,a"lcJ.~', nespi[1<~j.3ei::: tDt ~O I> ~Lt) J.-\.::' z: 30 (IO) t- tlO (30) -t !fa ('-{o) +- CD l$5)+ "Z()(b~ /:JB (H':h~Jsij,!j1!J:2i 600 -r 300 + '-100 + /fAOO +- 11")00 S'uident Hespons'3 O~;:neenen.t: H pvlnts) z: "31- 00 R(i) '" 70- .[ 60 - J/ i H '5 ~SS I 'S'M.P-lllLr 20 30 40 55 65 70 be.u-'-LSL E ~ 50 & B"' s OJ '10 ~ 20 ~ '" 10 0 :Ji / - / " --'" (minutes) (gallons per minnie) 0 30 ~~ ~ ~ rtLl,.o...."iUs lMJla.... e, +L e>-...c1-\).. r;J o...tf.p. WA.hr ~ -t~ lr lXf'" . 40 50 70 I j I I f1I>.t 90 10 20 30 40 50 60 70 80 90 IWork for question 3(d) I Tune 01 [\ '1 k for question 3(aJ .. _-~-- 56 Rl-l)JJ. 0 is -tl... "-""'--Ou"'I",.!;c...o\l\..$'.A..~ -_ J ';;: o~ IJ...'A..l:\ l fc>-C <> \ ;."'-. ,oc....li.. c!A.S R' (l(S) ss - 'II) -SO-~O \--\,-"J45 ,(0 !1,S ~,,-ll..o"'-S fU ."N..",-u..\-L< +I.u.. \~ \s ~ \lA. +;"",---,,-- b. ~ ~: Klt)~~ \>-'A.+ .- \ o..Ue..r Cl....l10~ rlL~ ~ u:J"'~\J...If'-e 1,.; 0 '" hM...L lo. 6a..~lo\.o..s fU t\.Jl'-c-k _ [?l0rl; fiJI' question 3~IJ (Z'I ('-jS) :; 0 S\",-UL Q. IZ (t) Q~ ~S ~ 'J'.. e..~{'> _ _ C__'S.l v~ \ o."",\.<s\- c>-\ t" (t.) is leo l\S , \<-.\U -\-Ir...c.. \u, 1 117 \V\..J}..\<', ""'-LLVv'- a..\-- t= 4S . 0- ~(l'jQ~lJiL (' Ct) Q.~ 'fo,'-.f0<,' ~\'~LL'.AA. +ko... ~01Ja.\\\JQ.. ..) I.A-.nj S o. 116 [Work for question 3(c) I -I- 30(fl,lO)) ~ 10 (12. (30) } 112.('"10)) + 20((2(SV)) -+ ;20(f?(70 ) ) 3c(20) + 10 (iil)"I 10('-10)7 20(5S)-+ 20(6S)" [3700\ S,CfH:-ient H.SSlJOli.1S 2: (GOO.(.1: '7 points) less -maY\, fheY'e -fhCli t Me ~-fill area s of' -fhe C\ cUY\Je. R(t) c; ClI" >loi I"f\cluc/Qd by etol'\1~ \.eft QI'ewcClhv1 ~VW1, 70~" "a E ~ o 60~ (minutes) (gallons per nunute R(r) , ) 50 40" 30 0 .J 20 30 40 55 65 70 [Work for questinn 3(d) " 8 g u, ~ 30 ) ) 40 \,) 50 70" lu 90 ' 'O20~ s 10 o '--. 10 20 30 4() 50 60 70 80 90 TIme I c.oV\sul\-led -Fro"," [WO~~ oletf.)d-f.. ,'.s-fhe lX~\JVii af-tU\ 1.5 8,,1 P''' "',",vi~ pv ""~ b I b t=O -10 i=b i~jClIIDY1S, -+="0"'-1 -e(so) - R(40) s Lj o 5'5"-QO -::: jbQ(-L)d-t,-s -L~o lo-t~b f",e IvI CM'.tl'Clje (\l'IIou~t of -fUel CDI'I';Ufl-uJ stJ-40 jcd[of1s. IWork for question 3(b) I R"(Y5";=O H' r+ I'::, ihWClS1V1j y\\CtXlflIUM I -fr<'Ste<.;,1 (Ii -toYS, R'(L15) MClbvti) R"(LJSJ ~ 0. I'::, o<t '" 119 118 [WortCfOrCtuestion 3( c) <l.0 I oj Stll-fl,ent H.eSpGHS-a '3 (F"E<i:r: 6 points) R(+-)d1- ':8 6o)(~o) -+L.jCj5"6 'J..\\,,-\~ ..... :, GO)( L.JD) +- ~O)(55-) rQO)(H')+ @o)(q~ j o t R(t) C-'~-""G;~'-"--~'-' '1.0 Ref!J::- ~ R(I) 'f '" o r~~r,,::::::?'1 (JllinUleS~ (gallons per mi~lIle) n;5 "r?'....i ....,\-." " 7~ 0 ~r~e.:r~( th.. ", t~ ~ O.~~ ...\ ., 30 [.L ~V r;F ~:i ~ 10 o 30 40 20 30 40 "...-".. . e ~(t.".~ f~ R(I.) cl. Go oeG<o. "'\~ fec.l-<,.," J",kr e!(ktl.o.eo. ... 6..,,~ l-lt I;" e. f=f(+) '0 20 50 70 90 55 65 70 ~k for question 3(d>] " ol , t I I I I I I. 10 20 30 40 50 60 70 80 90 TlillC ~k forqueslion 3ED ~\"\5 of f (b "/) Ra),\!- 1-~ is hJ ,j-k "to k.\ <,yV[''1. Y\d bl ,f....e: I is c:: ..,~ ..... d R'65)~ ~ pl5t-1Zll./0) 56-i-fO 0 ,..: b. co t1 ! .... r'>'-(. iJ ..l. 5R(t)c,\r. b a ,J-k MI;"""h:..o. AV<j y:..~ o l- J.. . (.\ J. T~ '1'\ ~,,\lM ~ ?ct' t) 1(45}~ ~'Lj5 -: - .;'5" -'-ICl S-O-<-t 0 \ tOM +.. ,.....e..~ \-=:6 I-t> b '"3 "3 'jc,\\'Vl ) ?d t"/,,~ ...\k [Work for question 3<b2J 121 120 Scoring Guideline for AB/BC Question 4 Let 1 be a function defined on the closed interval -3 Sz :; 4 with 1(0) = 3. The graph of [', the derivative of 1, consists of one line segment and a semicircle, as shown above. (a) On what intervals, if any, is (b) !~B/BC (}I"~~~8tl101J Ij (-3, I) 1 increasing? Justify your answer Find the z-coordinate of each point of inflection of the graph of on the open interval -3 1 -3 < x < 4. Justify your answer. (c) Find an equation for the line tangent to the graph of point f at the (0,3). 1( -3) and f ( 4). Show the work that leads to your answers -2 Graph off' (--I, -2) Commentary This question presented students with the grilph of the derivative off It explored students' understanding of the relationship between the behavior of the derivative and of the functinnf 111 part (n), students needed to know how to (d) Find iJ.s,e~lJF,:~~~n, qf ~e de:~iv'ativ~ to detennine:~.herL the function is increasing. Part (h) could be answered by knowing hoi (a) The function f is increasing on [--3, -2J since : interval to~e~~~~~ineth,e:con"~'a~ityoff,~)1 observing where the kno\viIlgJh"relationship- between the values of r and graph of f' is increasing or decreasing. Part (c) required I' > 0 (b) x for -3 'S x < -2. 2: : reason the slope of the tangent line off Part (d) required an applira t\or(,g:ft 1 Pundamental Theorem of Calculus, recognizing that feb) - f(a,) is the integral of w 1" over the dosed i"t,~ryal[a. b},and thennslng this and the fact that frO) ~ 3 to calculate the values off"t x ~:"3 and x ~ 4. The valu seti\!Fi~qle) and-subtracting this area from the value offat x ~ O. = 0 and z = 2 ,#t.~i;.~R 4js ,""qstea~jly"~P1p\ltcdDy finding the area bet\~een the graph of 1" and the x-uxis (a rectangle minus a If changes from decreasing to increasing at x = a and from increasing to decreasing at 2: 1: x = 0 and 7: = 2 only 1 : justification ,}1~i'rtf$~r9~.;~~t~~ ..g~e~timiWbs ~d~sf"Ilp~~tI1,l~t;':co~sid~ring the emphasis placed on analyzing the: behavior ~f a fl~nc:tlO g\\<~l1J}1#'~r~rhofjts >leri:hl\\,eiJl]pre'li~u.APl;:alC\,]ns Exams, Most students attempted the problemin part fa}.Jn udents \vereql"arlyc()nrfortable~ith oints of inflection where rex) ~ 0 such as occurred atX ~ 2, but a p . , (c) Tlierlifil!l ~corewas2,68iort1]et:B student~ and 4.1~ for the BC students. The number of students whoearned no x=2 /,(0) = -2 1 : equation ~',.,,:~c~;,' t , '~~~er:.of st~14eDts:were'confu~ed ahouthow to handle the inflection point at x -;::: 0 where the second -4e;;.iYativbAoe&not;;exi~. ~~?st's,tude~tse~wned th~ point for the tangent line 'in part (c). ~he'AB students, however, Tangent line is y = ~2x + 3. the area connection, many could caicllla~e the areas recognize the need to n.sethe Pundamentnl Theor~m of Calculus. 't,$\11aU'perentageofstuden ts attempted to solve part (d) analytically, Few were successful. C?Sl~~tlt~tJ.f~htl'e~:.in~,de-~ignerfqi~' nrfailed to ~.rt (d)' to beextremely difficult. Of those Who recognized (d) f(O) - f(-3) = L',!'(t)dt 1:(~-2) 2 = ~(l)(l) - ~(2)(2) =-~ 2 2 (difference of areas of triangle-s) Student Response 1 (Excellent: 9 points) The sruden t earned all Sl points. The half-open interval in pan ta) is accepted for the first point. In part (b), the student summarized the behavior of II using a sign chan and provided an interpretation of this chart that justified the answer for the .r-coordinate of each point of inflection Part (d) illustrates a clear application of the fundamental Theorem of Calculus to compute: each function value. cornpuuug the signed areas corresponding to the definite uucgrals hut did demonsrrare a correct understanding ol the use of the Pundament.il Theorem of Calculus to rela. those areas and the values of the function. The student earned '2 out LJI4 points in part (d). f(-3) = flO) + :2 = 3 :2 9 1 : answer for f(-3) using FTC 4: f(4) - flO) = = Jo' f'(t)dt 1: (8 -~(2)2Jf) - area of semicircle) Student Response 3 (Fair: 5 points) The student earned both the:interval and reason point ir part (a). The student lost both points in part (b) because only one inflection point was identified and the justificai "vas not correct. The correct tangent line equation is give. in part (c). This student attempted an analytic solution t part (d) by starting with the correct piecewise formula fc fl. The student earned the first point in (d) by finding s antiderivativc of the line segment from x;:: -3 to x = O. The second point was earned when the student deter mit the constant of integration and computed the correct function value at x;:: -3. The student lost the last two points in part (d) since the antiderivative for the semicirct part of the function was never found. -(8 -~(2)2TI) = -8 +27f + 27f = -5 (area of rectangle Student Response 2 (Good: 7 points) The student earned both the interval and reason point in part (a). The student earned the first point in part (b) by identifying .r > 0 and x;:: '2 as the .r-coordinates of the points of inflection; the second point was earned by stating that F(x) changes between increasing and decreasing at these points. A more specific argument is not needed to earn the justification point as long as the student IS reason ing from f'(.\") since its graph is provided. The correct tangent line equation is given in part (c). The student lost the first and third points in part (d) for incorrectly f(4) = f(O) - 8 + 27f : answer for f(4) using FTC 123 122 ~ " "'" "" r-, '; ? s: , 0 --I-, ....,., Jj' o X iii 15" 0 '1) ? r , "" s: ~ '- .<; ,;, <;~ I" s: , , ~ r , 1 1, . ~ r +" -+, () ~ 11:'" I' .s 11 + :r :5 ~. ~ +, ~ -' ~ I~ '" r I~ ,~ '~t rJ : r- > ~ .J 1 t, , > -I' .r x I' , L " -' - r -e ;F , ,~ 1. r, -,f' f . ) > r- J '{ D ~ i- i ~ s: 3'"' r ,..., v! I \ ~ '1'\'1;' ,'h (. \ l~ t J.' r: 7' s, r -.& D -s: !'-' }-J r '-> ..., " r: r r: , '- , " 'l. ~. ",-r, -t-, :;: s- " } J ~ li' ~. .... " l;: Yo ~ f ." e, ..). , ~ -'"\') r-- -\-" ("- I >i g-' J .c II W t vJ II -:: -+, Gv O~~ vJ + <--,~ r=c b e ~j <s-J -: ",- 11 I ~ ~ --< II ,., G \./ ~ " ~ D II vJ r-' ~ rX J )( ,..., <>':) I f'.I ["1 -n v: ~ ,, . tJ r; 11 " <c II rv g ",1- -" ::l. ~ I' '" \1 W / e- Ii ('-'lob VJ W ,0 VJ I V i r-' }( NI W I r;::' {;: J J: ( U 'f" + r;- -\' vJ oC .,. ~ rJ '4 =l 1'-'\ 1"1 r;' :{ rv '-' /1 c -"'<" .J r .J V" " ~ ':0< ;:::: --2 I:;:: ~IS: ~ I'" 1 ---' 2 ~ _'I";: ~ g I~ i~ _ I L:::.J cJ; \ I\' w I o -, C) "' I V1 C N \ ~ tv co ~ ;. { -i-" + l!' -r ~' ,D ~ -h :r p ,.:: " 'g. " e -. -l-" II" ~ I) 4J' ,0' ~ I~ ~ ~ [ " ~ 'll ~ tI', (0 -., (b ~6' 7 '" '1 P e ;f-' ~, ,.... C/ I[ is' " .: ~ s t/j m '" III I> ~, ~~ A--I- ~- c; It> .s-, : P :> ~ j7 ~t t-.) .w ~ I ;!;. m. ~ ~ a- ~ ~ (> ~ ('> .-1 ., 0 ... -'" ~ o -t--. r ". % -;7 U' I ~v F' E. ;7 '/ tt 0 5' L.. '" -7 .!... ~ o 'g. OJ 0 , ,!., .!... a \-..J '< + -:. +-'-0 ~ 0' ~ I!> , I" ~ v.' ," .>' "'" '" r :) t; ~" -\' . ,... t- ,J It> ,... " ~ """ Q u X ~ .sc'" ( ': rJ . <" :> ~ rp L 0 ll' ~~1 b ~T >: t. :,. () R' ... o iT \l' (\ . fI' ;> 0 ~ ~ ~ o -0 > -y o ~ -r o -h ~ I ,I ii: ~ ~ ~ ~ ii: ,D ,D ~ " " (\ " g. " .: c " II ~ .c, " is" L9"~ II ,---..--, I ~ c5 II N , , 7 .s, V ., , !Y r-' I ,..J rJ LN ........... .... ? (V '> ,:I" 0' co eo ~ Scoring Guideline for AB/BC Question 5 A coffeepot has the shape of a cylinder with radius 5 inches, a" shown in the figure above. Let h be the depth of the coffee in the pot I measured in inches, where It. is a Iuucuon of time l, measured in seconds. The volume V of coffee in the pot is changing at the rate of -5lr.Jh cubic inches per second. (The volume V of a cyliurler with radius l~BjHC': (lfh~i;:,': T 5 In awl height h is V ::...: 7rT".!.h. ) ;,on (a) Show that dt = -5' = dh .jJi (b) Given that h = 17 at time t 0, solve the differential equa.tion ~ dt = __ -Jh !j for Commentary 'fJlf~)jl~l~iti?n'presented a r~l'at~d rates se~Hng\ as well as a separable differential equation. The, VOl~lil1C of coffe~ l~ a c!lin~ri~al pot was changi~g ai a rate. that wasgiven as a function of the height. Part (a) asked students tofind the tat, of~B~nge of height as a function of the lleight. This required translating the information on the rate of change of the VO,~,~)l1e? into ll)ath~matical,l1ntationand showing how to use the chain rule to find the deslredresl\lt.T~le, s~,lUtion \,,~,,'giv~n so that al) students c-ould S~1ftpart(b)on an equal footing; Part (11) askedstudents to wive theseparable , lialeq!,ationthat was the soluti~n topart(a), Part (c) used the answer to part (bJam! r~q~iIedIe;ogilftiol~~h cepotwas empty preciselywhen the height was zero. .... ......., . .' .';.ii' .. ' "!"'h;~;fu",ir\ ~\,ore'w"s >.26 for the A~.stll<lent,alld 5.88 fOIthe Be students' Ab9ut:6percentofth~ A~stWleQis pomtS'lierhi!ps nMt,;opad f(iI a question that lnvolved buth a related r~te& question and s~lvhlg asepac Hid equation, {\bWt p perc~ntoLthe BC studentsearned all 9 points. Most students correctly transhit (c) h 85 a function of t. At what time t is the coffeepot empty? (a) V = 251fh = '251f rl!>. = -51f.Jh dt -57r.Jh & 3: dV = -51f.Jh dV dt dh : computes ----;It dV Tt= 2S;- =-5 .Jh : shows result i . : ..' lystu4ents.corre~Uy tre~tedli~s a ,,~(i~b\ebutlTIadclntstak~s wlili,th~.~iffereI1til\ti~~,The ~COI~"~ . , part. (b) t~prese!lite\jwhat has becomethetlS~al. approachto grading ,,;parable ~iff~r~;!t\~iequ~tio~ : ' .. :;,< :.-;,:; .,>, :Tre:first zpojurs were 'for m~chariical.oper<ltions of separating the variables and 'alltJP.1~erentiati~~ the' iW~ .~I~~'s'The la~(3)loinis were-for !111ding'theparticular solution satisfying the given initial condition. 1nclllSi?nQfthe ~'?~Sl~~~.ofi:rit~gt,<;!tion w,\scri~ical because without it the student could not proceed from general solution to the l1artietlla'i'sQlution. . J'1fo~!DatlOnmlhestem 01 the quesuon mto ~ nwthematlcal~xpreSS1Ql1 fordt\Vben COtUPtltrI\gilt '. (b) 'I'.' : r. "';' . , ........ :','.:". (iV.. . .'ilV rl!>. = _ .Jh elt : separates variables : a.ntideriv atives 5 a ~dh = -.!:.dt .J7i J 5: : constant of integration : uses initial condition It = 17 when t = 0 Student Response 1 (Excellent: 9 points) The 51udent earned all 9 points with solutions that are easily followed. Student Response 3 (Fair: 5 points) The student earned all 3 points in part (a}. This student . I . d . dh. substituter the desire result fur di mto a correctly dV (amputeeif orrnul ' ([{ to vert fy t )tat -J[ = a tor dV -)l[ 2.J!' = _'!:'t + C 5 2m : solves for h = 0+ C Note: max 2/5 [1-1-0-0-0J if no constant 2 Student Response 2 (Good: 7 points) The student earned the first (trauslation ) point in part ta) --[f; ] h=(-fot+m) of integration b y identitving . I .,. dV -)Tf - HI dV. 1 as --:It. Ihe sru d ent did not compute I ----;Ii correctly and thus could not demonstrate the correct result for h !i- . All points are earned in parts j at claimed in the stem of the quest iou.Tu part (0), rhe stud separated variables and antiditferentiated correctly for 2 points but never introduced a constant of integration. This student was not eligible for the last 3 points of part (b). The work in part lc) did not earn the last poiru Note: 0/5 if no separation of variables 2 (e) (-fot+m) =0 1 : answer (b) and (e). t=lOm 1311 1']0 IWork for ques~ dH __-rH 6.1" t5,:::-:rfnl~)lB f)t.rR,d~3'.i!t H6~'Srp'(r~I.13e'f.: t(jl"t ~ ,n.'~2-~n:1{'i:,J 5 )~ ~)- ~ er J \-I-'h dl-\ '2 ,'\/~ H ~ ::: H~:~,ILdB~~li" rt2S'fHJlnSe (E::JH':8H,'cI1J: '9 poi:nts) :: -k-r +-C +- c \ -s T +- C 'L-ftl 5 in :::-\.0 t=21i7 l1H = -~ T -I- '2 {i1 -rH=-~I-t--rn I [W~,rk for question 5(~)l ________'!J o:: (-I~ T-t- {\,r) hin 1 -\ldV - ,_ 5i\-rrr Cb' (\I' (e= 5 v:: 11'.1. \-\ \f:: (2..51\ \-\ q~ [Work for question 5(c) J YI ::c 0 =t rDT+ in)1.. D:: \ ('\'\' Q" ~\\ 111 ciT :: '2. S rr cDl z: -..!...T +-:rrl 10 dT crr~ 5 -S1T~ d~:: ~ ~fJJ d\ /2511 .s '2511 cll-l ciT ib l :; -f\7 (T:: IO-rn} 133 -- - ~ I~ I ~ l~ II LJ) II J~ " }J ~ 11 I I V\ iJ1 "! e<-1 + C ,I N =l I < " .. ':j ~ I~ ~l 8'1 ~I 1:0 -:J r' I' r>: ::.r r-r-; 9=/C;- s:-\~ ~10- " U) -3 'tJ'} "! ~1~ 0 to ,-. ==t :::;" :'~ ~I~ ~I; 11 ~rs- ~F 1.-10 -r < \1 I I " lJ\ If\ "'l I i /\ '' ~ o..}Q-. + -, II ---I ~ ~ II t..I /> ~ ~ II P ::r_ i' II +-t " \1 I 1II1+t LP} 1 3] Vl,' r+ ot (1 o ~~ ~ '" L , ~~ t ~ + o lJ\ V> w en ~ co '" !2 '0 :!l;:~ ~ -<. " " "=1 :oj -;:.. r N ?-Is> "<: ", '-" 2~ ==l :<:: "' c, \1 "- ;':j J if "" ~ " ..z' I ",~, ft '0;' . '.\ tt, "::::j ~ N ~ i' VI '- ~ I, r---,. "" ":::\ If) '---- I, I \J\ , ~ ( '1 ~ ". ~ ........ 1 -t-<' Q' .0 <...-, ", ~ g. e V> ~ s: '" "=I :2 Qr~~ ~h :.. r?l. l \ U -flo.. 11 \ "'h -... c---> I ..:;,.: "'1 - r r ;':t/~ e, "'I~ " ?- II " ..J .. <-----> I V1(':;: " \J\b~ <---. :,- 7 :: I! r i t Scoring Guideline for AB Question 6 Let f he the function defined by . (a) Is.f cont.iuuous i,!; ((x) ~ ,,(t+1 [ 5-: fOI (J ~:I:.:S 3 for3<xS;.5. x ~ 3'{ Explain why or why not 011 (b) (c) Find the average value of .f(x) the closed interval 0 :S x :S .s Suppose: the function 9 is defined by Commentary l'h is question presented two pien:wise-ddlntd functions, It explored students' understanding of continl1itYl Jifft..'yentlabilHy, and iWel'<I~C value of a function. Part (<1) required students to.know that afunction is nJ.h 1noti ' ilaud ~)nly if the value of lite function agrees with both its limit from the left' and its 1hnit from the righ~'. Pad (l where k and 'In g('1:) = [ k~ rn:l' forOS;x:S;3 for 3 +2 <x S; ,5, arc consteu.ts If 9 is differentiable at x = co ---(~l) ;~, what arc the values of k and m? \Y<l~; J II average value problem. Students needed to know how to integrate a piecewise-defined function, Part (~) a piecewise-defined function with rum free parameters that needed to be:determined using the knowledge that tj fllnetlon W<I) differentiable <lnd hence also continuous at x::::: 3. These conditionsproduced two [inear equati9n~ thc:IWOUilkJH)WES, : _ " , , __ ",', ;, ,.' f is continuous at X = :{ because 1 : answers "yes') and equates the .r~r lim ((x) = ;J:---.3+ ((x) = 2, lim Therefore, ll~!'(x) = 2 ~ ((3), 2: values of the left- lind right-hand limits 1 : explanation involving limite 'Illls was the first time" question of this tyl,e invoIVingcont~uit1(an~dilfel,li(i,,~Hity ~Pt'e~r~d9!ttli~~,~1 response section in ah{\\\t 20 years. Nevertheless, m~ny'~iudtl)t~'did atternptthi"s 'questiOl~'as repodslQdiP:ted't~ tivclv few s,tudenn left the question blank (onl'Y abtYl~t ~,per,~nt ~;t;~~ived,_~}iC?:r,.~in~_icatin~ no:,J1~~th,~'mil,~~~~I_:~V9.~ the pagc r.The mean score, however; was only 2,68.It wasveiydiffiCl1lttbe.,~~1l9points; ~ly().7~e~f$Q;Of\K n dents achievedthis, Students cunid earn..the first point iJ;l .part (;)', I)y'g~:~.n'g d~e 'correctan-'iWer,a~1~ eqit-~t~'I~g~~~.': values of the left, and rlght-hand limits. However, the student needed to provide a more colnp\ete'eJ\Plan~tion,'; involving the usc of limits to caru the second point. Students did not have t~ provide jUstiticdti~.nsiI~p~~t ~C!}~~ "',~-,',, -: _; ,:,', ''-'',-;, ,,:'>',>""'.)'\ (b) fa 5 ((T)dT = Jo ((x) d," + 2, 51, = ~\,X + 1),21 3 3 11 L f(x)dl' ..5 \: k Ju ((x)dx + k r' J" f(x)dx 3 ", ( 5x 0 1 '-,C ')Ib (where k = 0) 2.'1 4: function g is differentiable on an open interval containing "~:::: Q ,and if hoth Iirn g'(x): and 1im~.l((i) ,~xis~; thi . X-)O' X~ = (~ - ~) + (~ , ~) = ~ 3 3 2 2 3 Average value: 1 -5 ; antiderivntive of JX""+T : ant iderivati ve of .5 - :t : cvaluo.tion and answer . " -. the two limits are equal and the common Yall1~ 'is g'(a). ~Th.is is a. consequence ()it~1e1vleallValue T~~?r~'ll1,:'~~e,; art.cle "Things I Have Learned at the Afl Reading" hy Dan Kennedy! College l\![~th~,matiC's JOtlrn;l~ Nov~0~~r,1999 page 3:!6 1 for related ideaa.] . ", , I 5 ((x)dx U =J 4 7 (c:) Since 9 is continuous at Student Response 1 (Excellent: 9 points) The student enrued both points ill part (a) bv ventving all the conditious in the definition or continuity. The calcula tion. i.i part (h\ is wd\ organized. file sl)\l.1til-il1 in pal'l (I is .1' = 3} 2k = 3711 + 2 :1: [ 1 ' 2k = 3m + 2 Student Response 3 (Fair: 5 points) The student earned both floinls ill part (a) for a com explanation involving limits. The lack (lCthe notouor n kft.- and right -nand limit in the tint two lines iJ nc penalized. The student did not spEt th~ inlegr;1tiol1 il part (b), thus losiug thr. tlrsl pocnL, and was not digil for th,: answcr poinl. Huwever, the .':itt/dent earned b.:: antidifferenliation points evel1 tbough the evaluation incorrect tor the tint integral. Ir, part (c), the studcnt earned the fIrst point For finding the linc:ar cqllatioll 1 Oil contimlity but n~vcr considel'eu. \he differentiabili ..::ondiuoll and had the incorrect values for k and m, 9 I (:r) = I'~ 2'>1 X + 1 m for 0 < x < 3 [or 3< x < 5 1:.10. ~ m '1 1 : values for k and m x---+~-- lim 9'(1') = Ie: 4 and lim g'(x) x___<,g+ =m 9 particuLlrlYdelJib'l Since these two lImits exist and differentiable at k X =: is St"dent Hesponse 2 (Good: 7 points) The silldl'nt c;lrned the firsl point in ~)<'lrt (11) fur declaring th-..\t {is (Ol1tinlll)lJ~ c{;lll indkatiug that the valul.:'s of the k{l- anel right-hand limits are both equal to 2. However, rhe sludellt lost the seconu poinl (()r r~liling to complete the ex:pt1llrltior, by' connecting the limit valae vvitI! the value of the fUlldion <It X 3, the two limjts are equaL Thus 4" = m m 8rn = ,1m + 2i =~ 5 and k =~ 5 = 3, The student earned 3 p(j~nls out of 4 in p<:rt (b) for correctly splitling the inkgrati ')l1 and fll1ding the antiderivatiw'$, but lhc' final an~WCl wa:) nol correcl delE' lo an Milhmdi( enOl' in the ev~lllat10n. The n.vo lille;J[' equations ,,<'ereclHreetly derived alld solved in part (c> 10 eJrn all 3 poinls 139 ,t, 13~, II yt \~G +(V ;1-;" t II II II 0........ J II L.-5= a: ~ ,... ? -! .., VOl , 0 ",I'" ~ V'ltJ .c:> -I 'f ~H~ II t'1~ -lN r> Q .... ".t! ;J -r ~ ;--"1 L.-I-> ----' '2' .... VT " -+ c ... \J{'" g.... \'7' .? '; c '-" r ~ S "" ; s ... V' ~ ~ ~ -M ... ....., V1 . ,. I "'It" II + I'I~ 0 fEJ~ ~-I~ -,.: -+v1 .' ,..... vJ ~ .J.i c: If '" t.-I: ~ :;, ~. -" P gc., ~ "'>::" ~;- ..... '1 t :; "' 3' ..bi r (i ::> ~ -+1 o ,p ~...;' ~ ~);- ~ t Cc.s> Q " --' It ) t- .- ,......~ " .."Iro> '-"..c: ~ '1\ ,::.., rr ,~- ",t- ~... ~ J"v ,... ?l oJ 1 >( -'" dJI .:{ .; :~ !f ....., {'I, i t s-: -=\j;: ill "'\ -\ Vl ul}J c!'r ~ r: ttJ- vJL I U1 I ~\~~~ Jo~~ 7 Vb \ll~ I CI1 et ~I + J \-I15l riA " c.Jt d:> 1""" II ~ II c.P :5 ~ ,-.... toll- -<- ~ r\~ t -t ,/J) I It - X' :;; .2 x i ~ ~ 3 i .--./ \i "\1 t' f }l 0\tJ II >( 11 \1 ~ x "'i (\ t ? - :t. 1"' r-, ~ { 0 tI1\d-' o\l~ ~ .... c#tJ v l' 'I ..,] ",I" r; .r\tr ~ \~ t 'j\ ~-t3 V ~ t x r -t 5 II V II \J rr ~ 1 ~ 1/1 r'. ~ t ~ ",Ie; rJ 1\ ~ ./' '\ I" 1. ? _. ~ J .tl" .r\t 4- \' ~InJ \.:I ~ ~ ~ -, ~ f W .0 '.";', )<. II uJ '3 II A II I ~ ~. I ...
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This note was uploaded on 02/03/2012 for the course MATH 101 taught by Professor Lee during the Spring '11 term at Université Saint-Esprit de Kaslik.

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