AP Calculus ABScoring Guidelines© 2018 The College Board. College Board, Advanced Placement Program, AP, AP Central, and the acorn logo are registered trademarks of the College Board. Visit the College Board on the Web: www.collegeboard.org.AP Central is the official online home for the AP Program: apcentral.collegeboard.org2018www.mymathscloud.com
AP® CALCULUS AB/CALCULUS BC2018 SCORING GUIDELINES© 2018 The College Board. Visit the College Board on the Web: www.collegeboard.org. Question 1(a)( )3000270r t dt=∫According to the model, 270 people enter the line for the escalator during the time interval 0300.t≤≤ 1 : integral2 : 1 : answer(b)( )()( )30030000200.7200.7 30080r tdtr t dt+− =+−⋅ =∫∫According to the model, 80 people are in line at time 300.t= 1 : considers rate out2 : 1 : answer(c)Based on part (b), the number of people in line at time 300t=is 80.The first timet that there are no people in line is 80300414.2860.7+= (or 414.285) seconds.1 : answer(d)The total number of people in line at timet, 0300,t≤≤is modeled by ( )0200.7 .tr x dxt+−∫( )120.7033.013298,166.574719rttt− =⇒==tPeople in line for escalator0201t3.8032t158.07030080The number of people in line is a minimum at time 33.013t= seconds, when there are 4 people in line.( ) 1 : considers 0.70 1 : identifies 33.0134 : 1 : answers 1 : justificationrtt−==www.mymathscloud.com
AP® CALCULUS AB2018 SCORING GUIDELINES© 2018 The College Board. Visit the College Board on the Web: www.collegeboard.org. Question 2 (a)( )32.118v′= −The acceleration of the particle at time 3t= is 2.118.−1 : answer(b)( ) ( )( )( )33003051.760213xxvtdtvtdt=+=−+=−∫∫The position of the particle at time 3t= is 1.760.−( )30 1 : 3 : 1 : uses initial condition 1 : answervtdt∫(c) ( )3.502.844vtdt=∫ (or 2.843)( )3.503.737vtdt=∫The integral ( )3.50vtdt∫ is the displacement of the particle over the time interval 03.5.t≤≤The integral ( )3.50vtdt∫ is the total distance traveled by the particle over the time interval 03.5.t≤≤( )()3.503.50 1 : answers 2 : interpretations of 3 : and vtdtvtdt∫∫(d) ( )()2vtxt′=()211.57054vttt= −⇒=The two particles are moving with the same velocity at time 1.571t= (or 1.570).( )()2 1 : sets 2 : 1 : answervtxt′=www.mymathscloud.com
AP® CALCULUS AB/CALCULUS BC2018 SCORING GUIDELINES© 2018 The College Board. Visit the College Board on the Web: www.collegeboard.org. Question 3 (a)()( )( )( )()()()5115511319253 913222ffgxdxfgxdx−−−= += −= −−− + = −− =∫∫{ 1 : integral2 : 1 : answer(b)( )( )()()()()6 361 1336213633224216244410333xxgxdxgxdxgxdxdxxdxx===+= +−= + −= + −− =∫∫∫∫∫()2 1 : split at 33 : 1 : antiderivative of 24 1 : answerxx=−(c)The graph offis increasing and concave up on 01x<<and 46x<< because ()()0fxgx′=> and ( )( )fxgx′= is increasing on those intervals.{ 1 : intervals2 : 1 : reason(d)The graph offhas a point of inflection at 4x=because ( )( )fxgx′= changes from decreasing to increasing at 4.x={ 1 : answer2 : 1 : reasonwww.mymathscloud.com
AP® CALCULUS AB/CALCULUS BC2018 SCORING GUIDELINES© 2018 The College Board. Visit the College Board on the Web: www.collegeboard.org. Question 4 (a)( )( )( )75116567522HHH−−′≈==−( )6H′ is the rate at which the height of the tree is changing, in meters per year, at time 6t= years.{ 1 : estimate2 : 1 : interpretation with units(b)( )( )53622532HH−−==−BecauseH is differentiable on 35,t≤≤H is continuous on 35.t≤≤By the Mean Value Theorem, there exists a value c, 35,c<< such that ( )2.Hc′=()( )53 1 : 532 : 1 : conclusion using Mean Value TheoremHH−−(c)The average height of the tree over the time interval 210t≤≤is given by ( )1021.102H t dt−∫( )()()1021.52266211.11211 152221238812636583257H t dt+++ +++ +≈ ⋅⋅ ⋅ ⋅==∫The average height of the tree over the time interval 210t≤≤ is 26332 meters.{ 1 : trapezoidal sum2 : 1 : approximation(d)( )501Gxx= ⇒=( )()( )()()()()221100100110011xxdddxdxdxGxGxdtdxdtdtdtxx+ −⋅=⋅=⋅=⋅++( )()()2110030.03411xdGxdt== ⋅=+According to the model, the rate of change of the height of the tree with respect to time when the tree is 50 meters tall is 34 meter per year. ( )() 2 : 3 : 1 : answerdGxdtNote: max 13 [1-0] ifno chain rulewww.mymathscloud.com
AP® CALCULUS AB2018 SCORING GUIDELINES© 2018 The College Board. Visit the College Board on the Web: www.collegeboard.org. Question 5 (a)The average rate ofchange offon the interval 0xπ≤≤is( ) ( )01.0ffeππππ−−−=−1 : answer(b)( )cossinxxfx ex ex′=−( )( )( )3232323 33cossin2 22feeeππππ ππ′=−=The slope of the line tangent to the graph off at 32xπ= is 32.eπ( ) 1 : 2 : 1 : slopefx′(c)( )50cossin0,44fxxxxxππ′=⇒ − =⇒= =x( )fx014π412eπ54π5412eπ−2π2eπThe absolute minimum value off on 02xπ≤≤ is 541.2eπ−( ) 1 : sets 05 1 : identifies ,3 : 44as candidates 1 : answer with justificationfxxxππ′===(d)()2lim0xfxπ→=Because g is differentiable, g is continuous.( )( )2lim02xgxgππ→==By L’Hospital’s Rule, ( )( )( )( )222limlim.2xxfxf xegxg xπππ→→′−==′ 1 : is continuous at 2and limits equal 03 : 1 : applies L’Hospital’s Rule 1 : answergxπ=Note: max 13 [1-0-0] if no limit notation attached to a ratio of derivatives www.mymathscloud.com
