AP Calculus AB 2013 Scoring Guidelines


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1 AP Calculus AB 1 Scoring Guidelines The College Board The College Board is a missiondriven noforprofi organizaion ha connecs sudens o college success and opporuniy. Founded in 19, he College Board was creaed o expand access o higher educaion. Today, he membership associaion is made up of over, of he world s leading educaional insiuions and is dedicaed o promoing excellence and equiy in educaion. Each year, he College Board helps more han seven million sudens prepare for a successful ransiion o college hrough programs and services in college readiness and college success including he SAT and he Advanced Placemen Program. The organizaion also serves he educaion communiy hrough research and advocacy on behalf of sudens, educaors, and schools. The College Board is commied o he principles of excellence and equiy, and ha commimen is embodied in all of is programs, services, aciviies, and concerns. 1 The College Board. College Board, Advanced Placemen Program, AP, SAT and he acorn logo are regisered rademarks of he College Board. All oher producs and services may be rademarks of heir respecive owners. AP Cenral is he official online home for he AP Program: apcenral.collegeboard.org.
2 1 SCORING GUIDELINES Quesion 1 On a cerain workday, he rae, in ons per hour, a which unprocessed gravel arrives a a gravel processing plan is modeled by G ( ) = cos, 1 where is measured in hours and. A he beginning of he workday ( =, ) he plan has 5 ons of unprocessed gravel. During he hours of operaion,, he plan processes gravel a a consan rae of 1 ons per hour. (a) Find G ( 5. ) Using correc unis, inerpre your answer in he conex of he problem. (b) Find he oal amoun of unprocessed gravel ha arrives a he plan during he hours of operaion on his workday. (c) Is he amoun of unprocessed gravel a he plan increasing or decreasing a ime = 5 hours? Show he work ha leads o your answer. (d) Wha is he maximum amoun of unprocessed gravel a he plan during he hours of operaion on his workday? Jusify your answer. (a) G ( 5) = 4.5 (or 4. 57) The rae a which gravel is arriving is decreasing by 4.5 (or 4.57) ons per hour per hour a ime = 5 hours. 1 : G ( 5) : 1 : inerpreaion wih unis ons : { 1 : inegral (b) G( ) d = : answer (c) G ( 5) = < 1 A ime = 5, he rae a which unprocessed gravel is arriving is less han he rae a which i is being processed. Therefore, he amoun of unprocessed gravel a he plan is decreasing a ime = 5. (d) The amoun of unprocessed gravel a ime is given by A ( ) = 5 + ( Gs ( ) 1 ) ds. A ( ) = G ( ) 1 = = : compares G( 5 ) o 1 : 1 : conclusion : 1 : considers A ( ) = 1 : answer 1 : jusificaion A ( ) The maximum amoun of unprocessed gravel a he plan during his workday is 5.7 ons. 1 The College Board.
3 1 SCORING GUIDELINES Quesion A paricle moves along a sraigh line. For 5, he velociy of he paricle is given by ( ) ( ) 5 v = + +, and he posiion of he paricle is given by s ( ). I is known ha s ( ) = 1. (a) Find all values of in he inerval 4 for which he speed of he paricle is. (b) Wrie an expression involving an inegral ha gives he posiion s ( ). Use his expression o find he posiion of he paricle a ime = 5. (c) Find all imes in he inerval 5 a which he paricle changes direcion. Jusify your answer. (d) Is he speed of he paricle increasing or decreasing a ime = 4? Give a reason for your answer. (a) Solve v ( ) = on 4. =.1 (or.17) and =.47 1 : considers : { 1 : answer v ( ) = (b) s ( ) = 1 + vx ( ) 5 s ( 5) = 1 + v( x) = 9.7 : 1 : s ( ) 1 : s( 5) (c) v ( ) = when =.5,.1775 v ( ) changes sign from negaive o posiive a ime =.5. v ( ) changes sign from posiive o negaive a ime = : { 1 : considers v ( ) = : answers wih jusificaion Therefore, he paricle changes direcion a ime =.5 and ime =.1 (or.17). (d) v ( 4) = <, a( 4) = v ( 4) = < : conclusion wih reason The speed is increasing a ime = 4 because velociy and acceleraion have he same sign. 1 The College Board.
4 1 SCORING GUIDELINES Quesion (minues) C ( ) (ounces) Ho waer is dripping hrough a coffeemaker, filling a large cup wih coffee. The amoun of coffee in he cup a ime,, is given by a differeniable funcion C, where is measured in minues. Seleced values of C ( ), measured in ounces, are given in he able above. (a) Use he daa in he able o approximae C (.5 ). Show he compuaions ha lead o your answer, and indicae unis of measure. (b) Is here a ime, 4, a which C ( ) =? Jusify your answer. (c) Use a midpoin sum wih hree subinervals of equal lengh indicaed by he daa in he able o approximae 1 1 he value of C( ) d. Using correc unis, explain he meaning of ( ) C d in he conex of he problem..4 (d) The amoun of coffee in he cup, in ounces, is modeled by B ( ) = 1 1 e. Using his model, find he rae a which he amoun of coffee in he cup is changing when = 5. C( 4) C( ) (a) C (.5) = 4 1 = 1. ounces min : { 1 : approximaion 1 : unis (b) C is differeniable C is coninuous (on he closed inerval) C( 4) C( ) 1.. = = 4 Therefore, by he Mean Value Theorem, here is a leas one ime, < < 4, for which C ( ) =. C( 4) C( ) 1 : : 4 1 : conclusion, using MVT (c) 1 1 ( ) [ ( 1) ( ) ( 5) ] C d C + C + C 1 = ( ) 1 = (. ) = 1.1 ou n ce s 1 ( ) C d is he average amoun of coffee in he cup, in ounces, over he ime inerval minues. (d) B ( ) = 1(.4) e =.4e.4.4.4( 5).4 B ( 5) =.4 e = ounces min e : : 1 : midpoin sum 1 : approximaion 1 : inerpreaion 1 : B ( ) 1 : B ( 5) 1 The College Board.
5 1 SCORING GUIDELINES Quesion 4 The figure above shows he graph of f, he derivaive of a wicediffereniable funcion f, on he closed inerval x. The graph of f has horizonal angen lines a x = 1, x =, and x = 5. The areas of he regions beween he graph of f and he xaxis are labeled in he figure. The funcion f is defined for all real numbers and saisfies f ( ) = 4. (a) Find all values of x on he open inerval < x < for which he funcion f has a local minimum. Jusify your answer. (b) Deermine he absolue minimum value of f on he closed inerval x. Jusify your answer. (c) On wha open inervals conained in < x < is he graph of f boh concave down and increasing? Explain your reasoning. (d) The funcion g is defined by ( ) ( ( )) 5 g x = f x. If f ( ) =, find he slope of he line angen o he graph of g a x =. (a) x = is he only criical poin a which f changes sign from negaive o posiive. Therefore, f has a local minimum a x =. (b) From par (a), he absolue minimum occurs eiher a x = or a an endpoin. f( ) = f( ) + f ( x) = f ( ) f ( x) = 4 1 = f( ) = f( ) + f ( x) f ( ) = 4 = f ( ) f ( x) = 4 7 = The absolue minimum value of f on he closed inerval [, ] is. (c) The graph of f is concave down and increasing on < x < 1 and < x < 4, because f is decreasing and posiive on hese inervals. (d) g ( x ) = [ f( x) ] f ( x) 5 [ ] ( ) ( ) g ( ) = f ( ) f = 4 = 75 1 : answer wih jusificaion : 1 : considers x = and x = 1 : answer 1 : jusificaion : { 1 : answer 1 : explanaion : g ( x) : 1 : answer 1 The College Board.
6 1 = + and g( x) ( x) AP CALCULUS AB 1 SCORING GUIDELINES Quesion 5 Le f( x) x x 4 = 4cos. Le R be he region 4 bounded by he graphs of f and g, as shown in he figure above. (a) Find he area of R. (b) Wrie, bu do no evaluae, an inegral expression ha gives he volume of he solid generaed when R is roaed abou he horizonal line y = 4. (c) The region R is he base of a solid. For his solid, each cross secion perpendicular o he xaxis is a square. Wrie, bu do no evaluae, an inegral expression ha gives he volume of he solid. (a) Area = [ g( x) f ( x) ] = ( ) ( + 4 ) x ( 4 ) ( 1 ) = 4cos x x x 4 4 = 4 sin x x 4x 1 1 = + 4 : 1 : inegrand : aniderivaive 1 : answer (b) Volume = ( 4 f ( x) ) ( 4 g( x) ) ( ) = x x 4c s x 4 ( 4 ( + 4) ) 4 o ( ) : { : inegrand 1 : limis and consan (c) Volume = [ g( x) f ( x) ] ( ) ( ) = 4cos x x x : { 1 : inegrand 1 : limis and consan 1 The College Board.
7 dy y Consider he differenial equaion e ( x x) AP CALCULUS AB 1 SCORING GUIDELINES Quesion. = Le y = f( x) be he paricular soluion o he differenial equaion ha passes hrough ( 1, ). (a) Wrie an equaion for he line angen o he graph of f a he poin ( 1, ). Use he angen line o approximae f ( 1. ). (b) Find y = f( x), he paricular soluion o he differenial equaion ha passes hrough ( 1, ). (a) ( ) dy = e 1 1 = ( x, y) = ( 1, ) An equaion for he angen line is y = ( x 1. ) f ( 1. ) ( 1. 1 ) =. dy 1 : a he poin ( x, y) = ( 1, ) : 1 : angen line equaion 1 : approximaion dy (b) = ( x ) x y e dy = ( x x y ) e y e = x x + C e = C C = 1 y e = x x + 1 y e = x + x 1 ( x x ) ( x x ) y = ln + 1 y = ln + 1 : 1 : separaion of variables : aniderivaives 1 : consan of inegraion 1 : uses iniial condiion 1 : solves for y Noe: max [1] if no consan of inegraion Noe: if no separaion of variables Noe: This soluion is valid on an inerval conaining x = 1 for which x + x 1 >. 1 The College Board.
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