AP CALCULUS AB 2001 SCORING GUIDELINES. Question 3

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1 AP CALCULUS AB SCORING GUIDELINES Question A car is traveling on a straight road with velocity ft/sec at time t =. For > t > 8 seconds, the car s acceleration at (), in ft/sec, is the piecewise linear function defined by the graph above. (a) Is the velocity of the car increasing at t = seconds? Why or why not? (b) At what time in the interval > t > 8, other than t =, is the velocity of the car ft/sec? Why? (c) On the time interval > t > 8, what is the car s absolute maximum velocity, in ft/sec, and at what time does it occur? Justify your answer. (d) At what times in the interval > t > 8, if any, is the car s velocity equal to zero? Justify your answer. (a) Since v= () a() and a(), the velocity is increasing at t =. : answer and reason (b) At time t = because v() Гv() a( t) dt. : t : reason (c) The absolute maximum velocity is ft/sec at t =. The absolute maximum must occur at t = or at an endpoint. v() a( t) dt 8 () ()() v() at () dt so v(8) v() : : t : absolute maximum velocity : identifies t and t 8 as candidates or indicates that v increases, decreases, then increases : eliminates t 8 (d) The car s velocity is never equal to. The absolute minimum occurs at t = where v() a( t) dt Г. : answer : reason Copyright by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 78

2 SCORING GUIDELINES (Form B) Question EJAHL=Г! # EIIDMEJDABECKHA=>LADACH=FDB BD=I=DHEJ=J=CAJEA=JN$AJ CN # BJ@J BH Г! > N > # $ N =.E@C$ C= $ =@C== $ > MD=JEJAHL=IEIC@A?HA=IECKIJEBOOKH=IMAH? MD=JEJAHL=IEIJDACH=FDBC??=LA@MKIJEBOOKH=IMAH BJ@J KIECIENIK>EJAHL=IBACJD J! Г! $ = C$ # # C$ $! C C = $ $ B$! C == $ C$ B$ > CEI@A?HA=IECГ! =@ # IE?A Г! C= N BN BHN =@N! # KIJEBE?=JE? DACH=FDBCEI??=LA@M$# IE?A EJAHL= C B EI@A?HA=IECJDEIEJAHL= Г! Г JH=FAE@=AJD@ Copyright by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 79

3 SCORING GUIDELINES Question DACH=FDBJDABK?JE B IDM=>LA?IEIJIBJMEAIACAJIAJ C >AJDA BK?JECELA>O CN B =.E@ C Г C = Г =@ C == Г >.HMD=JL=KAIB N EJDAFAEJAHL= HA=IEC?.HMD=JL=KAIB N N Г Г EI C E?HA=IEC-NF=EOKH EIJDACH=FDB JDA=NAIFHLE@A@IAJ?DJDACH=FDB C JDA?IA@EJAHL= Г = Г! CГ BJ@J Г BJ@J Г C= Г BГ C== Г B= Г! Г > CEIE?HA=IECГ N >A?=KIA C= N BN JDEIEJAHL=? DACH=FDBCEI??=LA@M N >A?=KIAC== N B= N JDEIEJAHL= H >A?=KIAC= N BN EJAHL= CГ! C= Г C== Г EJAHL= HA=I EJAHL= HA=I C Г C C =FFHFHE=JAE?HA=IEC@A?HA=IEC =@??=LEJO>AD=LEH Г LAHJE?==IOFJJA Copyright by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. 8

4 SCORING GUIDELINES (Form B) Question Let f be a function defined on the closed interval [,7]. The graph of f, consisting of four line segments, is shown above. Let g be the function given by gx ( ) = ftdt ( ). (a) Find g (, ) g ( ), and g ( ). x (b) Find the average rate of change of g on the interval x. (c) For how many values c, where < c <, is g () c equal to the average rate found in part (b)? Explain your reasoning. (d) Find the x-coordinate of each point of inflection of the graph of g on the interval < x < 7. Justify your answer. (a) g() = f( t) dt = ( + ) = g () = f() = g() = f() = = : : g() : g() : g() (b) g() g() = () ftdt 7 ()() + ( + ) = = ( ) : g() g() = f( t) dt : answer (c) There are two values of c. We need 7 = g( c) = f( c) The graph of f intersects the line places between and. 7 y = at two : answer of : reason Note: / if answer is by MVT (d) x = and x = because g = f changes from increasing to decreasing at x =, and from decreasing to increasing at x =. : x = and x = only : justification (ignore discussion at x = ) Copyright by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 8

5 SCORING GUIDELINES Question Let f be a function defined on the closed interval x with f ( ) =. The graph of f, the derivative of f, 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 f on the open interval < x <. Justify your answer. (c) Find an equation for the line tangent to the graph of f at the point (, ). (d) Find f ( ) and f ( ). Show the work that leads to your answers. (a) The function f is increasing on [, ] since f > for x <. : interval : reason (b) x = and x = f changes from decreasing to increasing at x = and from increasing to decreasing at x = : x = and x = only : justification (c) f () = Tangent line is y = x +. : equation (d) f() f( ) 9 f( ) = f() + = f() f() = f() t dt = ()() ()() = = f() t dt ( ) = 8 () = 8 + f() = f() 8 + = + : ( ) : ± (difference of areas of triangles) : answer for f ( ) using FTC : ± ( 8 () ) (area of rectangle area of semicircle) : answer for f () using FTC Copyright by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 8

6 SCORING GUIDELINES Question The graph of the function f shown above consists of a semicircle and three line segments. Let g be the function x given by g( x) = f( t) dt. (a) Find g ( ) and g (. ) (b) Find all values of x in the open interval (, ) at which g attains a relative maximum. Justify your answer. (c) Find the absolute minimum value of g on the closed interval [, ]. Justify your answer. (d) Find all values of x in the open interval (, ) at which the graph of g has a point of inflection. 9 (a) g( ) = f( t) dt = ( )( + ) = g ( ) = f( ) = : g( ) : g ( ) (b) g has a relative maximum at x =. This is the only x-value where g = f changes from positive to negative. : x = : justification (c) The only x-value where f changes from negative to positive is x =. The other candidates for the location of the absolute minimum value are the endpoints. : identifies x = as a candidate : : g( ) = : justification and answer g( ) = g( ) = f( t) dt = g π ( ) ( ) 9 π = + = So the absolute minimum value of g is. (d) x =,, correct values each missing or extra value Copyright by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 8

7 SCORING GUIDELINES (Form B) Question The graph of the function f above consists of three line segments. (a) Let g be the function given by g( x) = f( t) dt. For each of g(, ) g (, ) and g (, ) find the value or state that it does not exist. (b) For the function g defined in part (a), find the x-coordinate of each point of inflection of the graph of g on the open interval < x <. Explain your reasoning. x x (c) Let h be the function given by hx ( ) = f( tdt ). Find all values of x in the closed interval x for which hx ( ) =. (d) For the function h defined in part (c), find all intervals on which h is decreasing. Explain your reasoning. (a) g( ) = f( t) dt = ( )( ) = g ( ) = f( ) = g ( ) does not exist because f is not differentiable at x =. : : g( ) : g ( ) : g ( ) (b) x = g = f changes from increasing to decreasing at x =. : x = (only) : reason (c) x =,, correct values each missing or extra value (d) h is decreasing on [, ] h = f < when f > : interval : reason Copyright by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 8

8 SCORING GUIDELINES Question A car is traveling on a straight road. For t seconds, the car s velocity vt (), in meters per second, is modeled by the piecewise-linear function defined by the graph above. (a) Find vt () dt. Using correct units, explain the meaning of () vt dt. (b) For each of v ( ) and v ( ), find the value or explain why it does not exist. Indicate units of measure. (c) Let at () be the car s acceleration at time t, in meters per second per second. For < t <, write a piecewise-defined function for at (). (d) Find the average rate of change of v over the interval 8 t. Does the Mean Value Theorem guarantee a value of c, for 8 < c <, such that v ( c) is equal to this average rate of change? Why or why not? (a) vt () dt= ( )( ) + ( )( ) + ( 8)( ) = The car travels meters in these seconds. { : value : meaning with units (b) v ( ) does not exist because vt () v( ) vt () v( ) lim = = lim. t t + t t v ( ) = = m sec : v ( ) does not exist, with explanation : : v ( ) : units (c) if < t < at () = if < t < if < t < at () does not exist at t = and t =. : finds the values,, : identifies constants with correct intervals (d) The average rate of change of v on [ 8, ] is v( ) v( ) 8 = m sec. 8 No, the Mean Value Theorem does not apply to v on [ 8, ] because v is not differentiable at t =. : average rate of change of v on [ 8, ] : answer with explanation Copyright by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and 8 (for AP students and parents).

9 SCORING GUIDELINES (Form B) Question The rate, in calories per minute, at which a person using an exercise machine burns calories is modeled by the function f. In the figure above, f () t = t + t + for t and f is piecewise linear for t. (a) Find f ( ). Indicate units of measure. (b) For the time interval t, at what time t is f increasing at its greatest rate? Show the reasoning that supports your answer. (c) Find the total number of calories burned over the time interval t 8 minutes. (d) The setting on the machine is now changed so that the person burns f () t + c calories per minute. For this setting, find c so that an average of calories per minute is burned during the time interval t 8. (a) f ( ) = = calories/min/min : f ( ) and units (b) f is increasing on [, ] and on [, ]. 9 On (, ), f () t = = since f has constant slope on this interval. On (, ), f () t = t + t and f () t = t + = when t =. This is where f has a maximum on [, ] since f > on (, ) and f < on (, ). On [, ], f is increasing at its greatest rate when t = because f ( ) = >. 8 (c) f() t dt = 9 ( ) + ( )( 9+ ) + ( ) = calories 8 (d) We want ( () ). f t + c dt = This means + c = (). So, c =. OR Currently, the average is = calories/min. Adding c to f () t will shift the average by c. So c = to get an average of calories/min. : f on (, ) : shows f has a max at t = on (, ) : : shows for < t <, f () t < f ( ) : answer { : method : answer { : setup : value of c The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 8

10 The graph of the function f shown above consists of six line segments. Let g be the function given by x g( x) = f( t) dt. (a) Find g (, ) g (, ) and g (. ) (b) Does g have a relative minimum, a relative maximum, or neither at x =? Justify your answer. SCORING GUIDELINES Question (c) Suppose that f is defined for all real numbers x and is periodic with a period of length. The graph above shows two periods of f. Given that g ( ) =, find g ( ) and write an equation for the line tangent to the graph of g at x = 8. (a) g( ) = f( t) dt = g ( ) = f( ) = g ( ) = f ( ) = : : g( ) : g ( ) : g ( ) (b) g has a relative minimum at x = because g = f changes from negative to positive at x =. { : answer : reason (c) g ( ) = and the function values of g increase by for every increase of in x. g( ) = g( ) = 8 g( 8) = f() t dt + f() t dt = g( ) + g( ) = : : g( ) : g( 8) : : g ( 8) : equation of tangent line g ( 8) = f( 8) = f( ) = An equation for the line tangent to the graph of g at x = 8 is y = ( x 8 ). The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 87

11 7 SCORING GUIDELINES (Form B) Question Let f be a function defined on the closed interval x with f ( ) =. The graph of f, the derivative of f, consists of two semicircles and two line segments, as shown above. (a) For < x <, find all values x at which f has a relative maximum. Justify your answer. (b) For < x <, find all values x at which the graph of f has a point of inflection. Justify your answer. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Explain your reasoning. (d) Find the absolute minimum value of f ( x ) over the closed interval x. Explain your reasoning. (a) f ( x) = at x =,, f changes from positive to negative at and. Thus, f has a relative maximum at x = and at x =. { : x -values : justification (b) f changes from increasing to decreasing, or vice versa, at x =,, and. Thus, the graph of f has points of inflection when x =,, and. { : x -values : justification (c) The graph of f is concave up with positive slope where f is increasing and positive: < x < and < x <. { : intervals : explanation (d) Candidates for the absolute minimum are where f changes from negative to positive (at x = ) and at the endpoints ( x =, ). π f( ) = + f ( x) dx = + π > f ( ) = f( ) = + f ( x) dx = + > The absolute minimum value of f on [, ] is f ( ) =. : : identifies x = as a candidate : considers endpoints : value and explanation 7 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). 88

12 !Å#ALCULUSÅ!"nÅÅ"#n WOÅRUNNERSÅ!ÅANDÅ"ÅRUNÅONÅAÅSTRAIGHTÅRACETRACKÅFOR ÅbÅTÅbÅÅSECONDSÅHEÅGRAPHÅABOVEÅWHICHÅCONSISTSÅOFÅTWOÅLINE SEGMENTSÅSHOWSÅTHEÅVELOCITYÅINÅMETERSÅPERÅSECONDÅOFÅUNNER!ÅHEÅVELOCITYÅINÅMETERSÅPERÅSECONDÅOFÅUNNERÅ"ÅISÅGIVENÅBY T THEÅFUNCTIONÅVÅDEFINEDÅBYÅVT T A &INDÅTHEÅVELOCITYÅOFÅUNNERÅ!ÅANDÅTHEÅVELOCITYÅOFÅUNNER Time (seconds) "ÅATÅTIMEÅTÅÅÅSECONDSÅ)NDICATEÅUNITSÅOFÅMEASURE B &INDÅTHEÅACCELERATIONÅOFÅUNNERÅ!ÅANDÅTHEÅACCELERATIONÅOFÅUNNERÅ"ÅATÅTIMEÅTÅÅÅSECONDS )NDICATEÅUNITSÅOFÅMEASURE C &INDÅTHEÅTOTALÅDISTANCEÅRUNÅBYÅUNNERÅ!ÅANDÅTHEÅTOTALÅDISTANCEÅRUNÅBYÅUNNERÅ"ÅOVERÅTHEÅTIME INTERVALÅÅbÅTÅbÅÅSECONDSÅ)NDICATEÅUNITSÅOFÅMEASURE Velocity of Runner A (meters per second) (, ) (, ) t ÅA UNNERÅ!ÅVELOCITY Å ÅÅORÅÅMETERSSEC UNNERÅ"ÅV ÅÅÅMETERSSEC Å ÅVELOCITYÅFORÅUNNERÅ! ÅVELOCITYÅFORÅUNNERÅ" ÅB UNNERÅ!ÅACCELERATIONÅÅ ÅÅÅ METERSSEC UNNERÅ"ÅA Va T T Å ÅMETERSSEC Å ÅACCELERATIONÅFORÅUNNERÅ! ÅACCELERATIONÅFORÅUNNERÅ" ÅC UNNERÅ!ÅDISTANCE Å ÅÅÅMETERS UNNERÅ"ÅDISTANCE Å T DT ÅÅÅMETERS T ÅDISTANCEÅFORÅUNNERÅ! ÅÅÅÅÅÅ ÅMETHOD ÅÅÅÅÅÅ ÅANSWER Å ÅDISTANCEÅFORÅUNNERÅ" ÅÅÅÅÅÅ ÅINTEGRAL ÅÅÅÅÅÅ ÅANSWER UNITS METERSSECÅINÅPARTÅA Å METERSSEC ÅINÅPARTÅB ÅAND METERSÅINÅPARTÅC ÅORÅEQUIVALENT ÅUNITS Copyright by College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Examination Board. 89

13 8 SCORING GUIDELINES (Form B) Question Let g be a continuous function with g ( ) =. The graph of the piecewise-linear function g, the derivative of g, is shown above for x 7. (a) Find the x-coordinate of all points of inflection of the graph of y = g( x) for < x < 7. Justify your answer. (b) Find the absolute maximum value of g on the interval x 7. Justify your answer. (c) Find the average rate of change of g( x ) on the interval x 7. (d) Find the average rate of change of g ( x) on the interval x 7. Does the Mean Value Theorem applied on the interval x 7 guarantee a value of c, for < c < 7, such that g ( c) is equal to this average rate of change? Why or why not? (a) g changes from increasing to decreasing at x = ; g changes from decreasing to increasing at x =. { : x -values : justification Points of inflection for the graph of y = g( x) occur at x = and x =. (b) The only sign change of g from positive to negative in the interval is at x =. g = + = + + = g( ) = 7 g( 7) = + g ( x) dx = + ( ) + = ( ) g ( x) dx ( ) : : identifies x = as a candidate : considers endpoints : maximum value and justification The maximum value of g for x 7 is. (c) g( 7) g( ) = = 7 ( ) : difference quotient { : answer (d) g ( 7) g ( ) ( ) = = 7 ( ) No, the MVT does not guarantee the existence of a value c with the stated properties because g is not differentiable for at least one point in < x < 7. : average value of g ( x) : answer No with reason 8 The College Board. All rights reserved. Visit the College Board on the Web: 9

AP CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES

AP CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES The figure above shows the graph of the piecewise-linear function f. For 4, the function g is defined by g( ) = f ( t) (a) Does g have a relative minimum, a relative maimum, or neither at =? Justify your