AP Calculus AB Free Response Notebook

Transcription

1 AP Calculus AB Free Response Notebook

2 Table of Contents Area and Volume Charts with Riemann Sums, MVT, Ave. Rates/Values Analyzing Graph of f Slope Fields with differential Equations Related Rates Accumulation Functions Implicit Differentiation Particle Motion Charts of f, f, f Functions/Misc

3 998 AP Calculus AB Scoring Guidelines. Let R be the region bounded by the ais, the graph of y =, and the line = 4. (a) Find the area of the region R. (b) Find the value of h such that the vertical line = h divides the region R into two regions of equal area. (c) Find the volume of the solid generated when R is revolved about the ais. (d) The vertical line = k divides the region R into two regions such that when these two regions are revolved about the ais, they generate solids with equal volumes. Find the value of k. (a) y y = : A = : answer 4 d R O 4 5 A = 4 d = 4 / = 6 or 5. (b) h d = 8 h/ = 8 or h h = 6 or.5 or.59 4 d = d h/ = 6 h/ h { : equation in h : answer 4 (c) V = π ( ) d = π 4 = 8π or 5. or 5. : limits and constant : integrand : answer k (d) π ( ) d = 4π π k = 4π or k π ( 4 ) d = π ( ) d k π k = 8π π k { : equation in k : answer k = 8 or.88 Copyright 998 College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board.

4 AB{ / BC{ 999. The shaded region, R, is bounded by the graph of y = and the line y =4,asshown in the gure above. (a) Find the area of R. (b) Find the volume of the solid generated by revolving R about the {ais. (c) There eists a number k, k>4, such that when R is revolved about the line y = k, the resulting solid has the same volume as the solid in part (b). Write, but do not solve, an equation involving an integral epression that can be used to nd the value of k. O y y y 4 (a) Area = Z ; (4 ; ) d Z = (4 ; ) d ( : integral : answer = = 4 ; =:666 or :667 (b) Volume = = Z 4 ; ( ) d ; Z (6 ; 4 ) d 8 : limits and constant >< : integrand >: : answer = 6 ; 5 5 = 56 5 = 6:849 or 6:85 (c) Z ; (k ; ) ; (k ; 4) d = : limits and constant >< : integrand 4 <; > each error >: : equation

5 !Å#ALCULUSÅ!"nÅÅ"#n,ETÅÅBEÅTHEÅSHADEDÅREGIONÅINÅTHEÅFIRSTÅQUADRANTÅENCLOSEDÅBYÅTHEÅGRAPHSÅOF Y E X ÅY COSX ÅANDÅTHEÅYAXISÅASÅSHOWNÅINÅTHEÅFIGUREÅABOVE A &INDÅTHEÅAREAÅOFÅTHEÅREGIONÅ B &INDÅTHEÅVOLUMEÅOFÅTHEÅSOLIDÅGENERATEDÅWHENÅTHEÅREGIONÅÅISÅREVOLVED ABOUTÅTHEÅXAXIS C 4HEÅREGIONÅÅISÅTHEÅBASEÅOFÅAÅSOLIDÅ&ORÅTHISÅSOLIDÅEACHÅCROSSÅSECTION PERPENDICULARÅTOÅTHEÅXAXISÅISÅAÅSQUAREÅ&INDÅTHEÅVOLUMEÅOFÅTHISÅSOLID.5.5 y R O.5.5 y = e y = cos EGIONÅ E X COSX ÅATÅXÅÅÅÅ! Å #ORRECTÅLIMITSÅINÅANÅINTEGRALÅINÅA ÅB ORÅC A!REA! X E COS X DX ÅÅORÅ Å ÅINTEGRAND ÅANSWER! X B 6OLUME Å Q COS E X DX Å Q ÅÅÅORÅ ÅINTEGRANDÅANDÅCONSTANT Å ÅÅÅÅÅ ÅEACHÅERROR ÅANSWER! C 6OLUME COS X E X DX Å ÅINTEGRAND ÅÅÅÅÅ ÅEACHÅERROR ÅÅÅÅÅÅ.OTEÅÅIFÅNOTÅOFÅTHEÅFORM Å D ÅÅÅÅÅÅÅÅÅÅÅÅÅK F X G X DX C ÅANSWER Copyright by College Entrance Eamination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Eamination Board.

6 AP CALCULUS AB SCORING GUIDELINES Question Let R and S be the regions in the first quadrant shown in the figure above. The region R is bounded by the -ais and the graphs of y Г and y tan. The region S is bounded by the y-ais and the graphs of y Г and y tan. (a) Find the area of R. (b) Find the area of S. (c) Find the volume of the solid generated when S is revolved about the -ais. Point of intersection Г tan at ( AB, ) (.955,.6575) (a) Area R = or Area R = or A d Г d =.79 tan A B / Г ( Гy) Г tan y dy =.79 : : limits : integrand : answer A d d =.79 Area R = Г Г Г Г tan (b) Area S = or Area S = or Г Г tand =.6 or.6 A B Г / tan ydy ( Гy) dy =.6 or.6 B : : limits : integrand : answer Area S B / / Г = ( Гy) dy Г ( Гy) Г tan ydy =.6 or.6 A (c) Volume = Г Г tan =.65 or 8. or 8. d : : limits and constant : integrand : answer Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 4

7 SCORING GUIDELINES (Form B) Question N O =@ O " Г N =IIDMEJDABECKHA=>LA N =.E@JDA=HA=B4 >.E@JDALKABJDAIE@CAAH=JA@MDA4EIHALLA@ =>KJJDAN=NEI? 6DAHACE4EIJDA>=IAB=IE@.HJDEIIE@A=?D?HII IA?JEFAHFA@E?K=HJJDAN=NEIEI=IGK=HA.E@JDA LKABJDEIIE@ 4ACE4! N N " Г = )HA= N =JN"&%$$") )! N " Г N N! "H! # > 8KA )! N " N Г N!&&"H!&&#H"'? 8KA &''% )! N " Г N N Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 5 +HHA?JEEJIE=EJACH=E=> H? EJACH=@ =IMAH EJACH=@=@?IJ=J! Г A=?DAHHH =IMAH EJACH=@ Г A=?DAHHH! JA BN Г =IMAH

8 SCORING GUIDELINES Question AJ B =@ C >AJDABK?JEICELA>O BN A =@CN N N =.E@JDA=HA=BJDAHACEA?IA@>OJDACH=FDIB B =@ C >AJMAA N =@N >.E@JDALKABJDAIE@CAAH=JA@MDAJDAHACEA?IA@>OJDACH=FDIB B =@ C >AJMAA N =@ N EIHALLA@=>KJJDAEA O "? AJ D >AJDABK?JECELA>ODN BN ГCN.E@JDA=>IKJAEEKL=KAB DN JDA?IA@EJAHL= > N > =@BE@JDA=>IKJA=NEKL=KABDN JDA?IA@EJAHL= > N > 5DMJDA==OIEIJD=JA=@IJOKH=IMAHI N = )HA= A Г H! N > 8KA " Г N Г " %## H!$' N? D= N B= N ГC= N A Г N N #$%"! )>IKJAEEKL=KA=@=>IKJA =NEKL=KA??KH=JJDA?HEJE?=FEJH =JJDAA@FEJI D#$%"!!! D#!"& D %& 6DA=>IKJAEEKEI!! 6DA=>IKJA=NEKEI %& EJACH= =IMAH EEJI=@?IJ=J EJACH=@ Г A=?DAHHH " JA EBJBJDABH > 4N Г = =IMAH?IE@AHID= N E@AJEBEAI?HEJE?=FEJ! =@A@FEJI=I?=@E@=JAI =IMAHI JA-HHHIE?FKJ=JE?ABB JDAJDEH@FEJ Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 6

9 SCORING GUIDELINES Question Let R be the shaded region bounded by the graphs of y the vertical line =, as shown in the figure above. (a) Find the area of R. = and y = e and (b) Find the volume of the solid generated when R is revolved about the horizontal (c) line y =. The region R is the base of a solid. For this solid, each cross section perpendicular to the -ais is a rectangle whose height is 5 times the length of its base in region R. Find the volume of this solid. Point of intersection e = at (T, S) = (.874,.48864) : Correct limits in an integral in (a), (b), or (c) (a) Area = ( ) T e d =.44 or.44 : : integrand : answer ( ) (b) Volume = ( ) ( ) T e d =.45 or.4 or.44 : : integrand < > reversal < > error with constant < > omits in one radius < > other errors : answer (c) Length = e Height = 5( e ) e d =.554 Volume = 5( ) T : : integrand < > incorrect but has e as a factor : answer Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com. 7

10 4 SCORING GUIDELINES (Form B) Question Let R be the region enclosed by the graph of y =, the vertical line =, and the -ais. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the horizontal line y =. (c) Find the volume of the solid generated when R is revolved about the vertical line =. (a) Area = d = 8 : : limits : integrand : answer ( ) (b) Volume = π 9 ( ) =.57 or.58 d : : limits and constant : integrand : answer (c) Volume = π ( ( y + )) = 47.5 dy : : limits and constant : integrand : answer Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 8

11 4 SCORING GUIDELINES Question Let f and g be the functions given by f ( ) = ( ) and g( ) = ( ) for. The graphs of f and g are shown in the figure above. (a) Find the area of the shaded region enclosed by the graphs of f and g. (b) Find the volume of the solid generated when the shaded region enclosed by the graphs of f and g is revolved about the horizontal line y =. (c) Let h be the function given by h ( ) = k( ) for. For each k >, the region (not shown) enclosed by the graphs of h and g is the base of a solid with square cross sections perpendicular to the -ais. There is a value of k for which the volume of this solid is equal to 5. Write, but do not solve, an equation involving an integral epression that could be used to find the value of k. (a) Area = ( f( ) g( ) ) d = ( ( ) ( ) ) d =. : { : integral : answer ( ) (b) Volume = π ( g( ) ) ( f( ) ) d = π (( ( ) ) ( ( ) ) ) d 4 : = 6.79 : limits and constant : integrand each error Note: if integral not of form b c ( R ( ) r ( ) ) d a : answer (c) Volume = ( h ( ) g ( )) d ( k( ) ( ) ) d = 5 : { : integrand : answer Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 9

12 5 SCORING GUIDELINES (Form B) Question Let f and g be the functions given by f ( ) = + sin( ) and g( ) = e. Let R be the shaded region in the first quadrant enclosed by the graphs of f and g 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 -ais. (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the -ais are semicircles with diameters etending from y = f( ) to y = g( ). Find the volume of this solid. The graphs of f and g intersect in the first quadrant at ( S, T ) = (.569, ). : correct limits in an integral in (a), (b), or (c) (a) Area = ( f( ) g( ) ) d S = ( + sin( ) e ) d S =.49 : integrand : : answer S ( ) (b) Volume = π ( f( ) ) ( g( ) ) d S = π (( + sin( ) ) ( e ) ) d = 4.66 or 4.67 : integrand each error Note: if integral not of form : b c ( R ( ) r ( ) ) d a : answer (c) Volume S π f( ) g( ) = S π + sin( ) e = =.77 or.78 d d : integrand : : answer Copyright 5 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

13 5 SCORING GUIDELINES Question Let f and g be the functions given by f ( ) = + sin ( π ) and g ( ) = 4. Let 4 R be the shaded region in the first quadrant enclosed by the y-ais and the graphs of f and g, and let S be the shaded region in the first quadrant enclosed by the graphs of f and g, as shown in the figure above. (a) Find the area of R. (b) Find the area of S. (c) Find the volume of the solid generated when S is revolved about the horizontal line y =. f ( ) = g( ) when + sin ( π ) = 4. 4 f and g intersect when =.788 and when =. Let a =.788. a (a) ( g ( ) f ( )) d =.64 or.65 : : limits : integrand : answer (b) ( f ( ) g ( )) d =.4 a : : limits : integrand : answer ( ) (c) π ( f( ) + ) ( g( ) + ) d = or a : { : integrand : limits, constant, and answer Copyright 5 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

14 6 SCORING GUIDELINES (Form B) Question Let f be the function given by f ( ) = + cos. Let R 4 be the shaded region in the second quadrant bounded by the graph of f, and let S be the shaded region bounded by the graph of f and line l, the line tangent to the graph of f at =, as shown above. (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the horizontal line y =. (c) Write, but do not evaluate, an integral epression that can be used to find the area of S. For <, f( ) = when =.7. Let P =.7. (a) Area of R = f( ) d =.9 P : { : integral : answer ( ) (b) Volume = π ( f( ) + ) 4 d = 59.6 P 4 : : limits and constant : integrand : answer (c) The equation of the tangent line l is y =. The graph of f and line l intersect at A = A Area of = (( ) ( )) S f d : : tangent line : integrand : limits 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

15 6 SCORING GUIDELINES Question Let R be the shaded region bounded by the graph of y = ln and the line y =, as shown above. (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the horizontal line y =. (c) Write, but do not evaluate, an integral epression that can be used to find the volume of the solid generated when R is rotated about the y-ais. ln ( ) = when =.5859 and.469. Let S =.5859 and T =.469 T (a) Area of R = ( ln ( ) ( ) ) d =.949 S : : integrand : limits : answer T ( ) (b) Volume = π ( ln ( ) + ) ( + ) d S = 4.98 or 4.99 : { : integrand : limits, constant, and answer ( ) T (c) Volume = y π ( + ) ( ) S y e dy : { : integrand : limits and constant 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

16 7 SCORING GUIDELINES (Form B) Question Let R be the region bounded by the graph of y = e and the horizontal line y =, and let S be the region bounded by the graph of y = e and the horizontal lines y = and y =, as shown above. (a) Find the area of R. (b) Find the area of S. (c) Write, but do not evaluate, an integral epression that gives the volume of the solid generated when R is rotated about the horizontal line y =. e = when =.44657,.5594 Let P = and Q =.5594 R = d =.54 e : P Q (a) Area of ( ) : integrand : limits : answer (b) e = when =, ( ) Area of S = e d Area of R =.66 Area of R =.546 : : integrand : limits : answer P OR ( ) ( ) + + ( ) e d Q P e d = =.546 Q Q = d : { : integrand P (c) Volume π ( e ) ( ) : constant and limits 7 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). 4

17 7 SCORING GUIDELINES Question Let R be the region in the first and second quadrants bounded above by the graph of below by the horizontal line y =. (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the -ais. (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the -ais are semicircles. Find the volume of this solid. y = and + + = when =± : correct limits in an integral in (a), (b), or (c) (a) Area = d = 7.96 or : { : integrand : answer (b) Volume = π d = : { : integrand : answer (c) Volume π = + π = d = + d : { : integrand : answer 7 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). 5

18 8 SCORING GUIDELINES (Form B) Question Let R be the region in the first quadrant bounded by the graphs of y = and y =. (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the vertical line =. (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the y-ais are squares. Find the volume of this solid. The graphs of y (, ) and ( 9, ). = and y = intersect at the points 9 (a) ( ) d = 4.5 OR ( ) y y dy = 4.5 : : limits : integrand : answer ( ) y y dy 4 : (b) π ( + ) ( + ) 7π = =.6 or.6 5 : constant and limits : integrand : answer (c) ( y y ) dy = : { : integrand 8. : limits and answer 8 The College Board. All rights reserved. Visit the College Board on the Web: 6

19 8 SCORING GUIDELINES Question Let R be the region bounded by the graphs of y = sin( π ) and y = 4, as shown in the figure above. (a) Find the area of R. (b) The horizontal line y = splits the region R into two parts. Write, but do not evaluate, an integral epression for the area of the part of R that is below this horizontal line. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the -ais is a square. Find the volume of this solid. (d) The region R models the surface of a small pond. At all points in R at a distance from the y-ais, the depth of the water is given by h ( ) =. Find the volume of water in the pond. (a) sin ( π ) = 4 at = and = Area ( ( ) ( )) = sin π 4 d = 4 : : limits : integrand : answer (b) 4 = at r = and s =.6759 s The area of the stated region is ( ( 4 )) r d : { : limits : integrand (c) Volume = ( sin ( π ) ( 4 )) d = : { : integrand : answer (d) Volume = ( )( sin ( π ) ( 4 )) d = 8.69 or 8.7 : { : integrand : answer 8 The College Board. All rights reserved. Visit the College Board on the Web: 7

20 99 AB4/BC Consider the curve defined by the equation y + cosy = + for y π. (a) Find dy d in terms of y. (b) Write an equation for each vertical tangent to the curve. (c) Find d y in terms of y. d Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com 8

21 99 AB4/BC Solution (a) dy dy sin y = d d dy ( sin y) = d dy = d sin y (b) dy undefined when sin y = d π y = π + = + π = (c) d y d d sin y = d dy cos y d = ( sin y) cos y sin y = = ( sin y) cos y ( sin y) Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com 9

22 99 AB Let R be the region between the graphs of y = + sin(π) and y = from = to =. (a) Find the area of R. (b) (c) Set up, but do not integrate an integral epression in terms of a single variable for the volume of the solid generated when R is revolved about the -ais. Set up, but do not integrate an integral epression in terms of a single variable for the volume of the solid generated when R is revolved about the y-ais. Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com

23 99 AB Solution (a) sin( π ) = + A d = cos( π ) π = ( ) π π = + π (b) V π ( sin( π )) or 4 = + d / π π arcsin π y dy+ y ( y ) dy ( ) (c) V π sin( π ) or π = + d ydy+ π arcsin ( y ) arcsin ( y ) dy π π Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com

24 99 AB Let R be the region enclosed by the graphs of y = e, y =( ), and the line =. (a) Find the area of R. (b) (c) Find the volume of the solid generated when R is revolved about the -ais. Set up, but do not integrate, an integral epression in terms of a single variable for the volume of the solid generated when R is revolved about the y-ais.

25 99 AB Solution (a) = ( ) A e d e d = + = e ( ) 4 = ( e ) = e 4 (b) V = π e ( ) e 5 d ( ) = π π 5 e e 7 = π = π 5 or e ( y) dy π y( y) V = π + ln y dy 5/ = π y + π y y ln y y e 7 = π + π e π 5 = 4 4 e ( ) (c) V = π e d or e ( ) dy π ( y) V = π y + ln dy

26 989 AB Let R be the region in the first quadrant enclosed by the graph of y = y =, and the y-ais , the line (a) Find the area of R. (b) Set up, but do not integrate, an integral epression in terms of a single variable for the volume of the solid generated when R is revolved about the -ais. (c) Set up, but do not integrate, an integral epression in terms of a single variable for the volume of the solid generated when R is revolved about the y-ais. Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com 4

27 989 AB Solution (a) Area = d = ( ) 6 / 64 8 = 4 = (b) Volume about -ais V = ( ) or π V = ( 6 4) d π π + d (c) Volume about y-ais V = π ( 6+ 4 ) or V = π d 4 4 y y 4 dy π 6 dy Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com 5

28 SCORING GUIDELINES Question - J #$ J Г " J $ J '&' J Г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Г% Г!& & 6DAHAMAHA!% #FAFAEJDAF=H=JJ% 6DAK>AHBFAFAEJDAF=HM=I@A?HA=IEC =JJDAH=JAB=FFHNE=JAO!&FAFADH=J = J -J ГJ J#%'"H#%'# Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 6 EEJI! EJACH=@ =IMAH IAJKF L=KAB = % A=ECI! A=ECB % A=ECB = % Г EBHABAHA?AJJ % -J ГJ =IMAH

29 SCORING GUIDELINES (Form B) Question A tank contains 5 gallons of heating oil at time t =. During the time interval t hours, heating oil is pumped into the tank at the rate H ( t) = + ( + ln( t + ) ) gallons per hour. During the same time interval, heating oil is removed from the tank at the rate t R( t) = sin gallons per hour. 47 (a) How many gallons of heating oil are pumped into the tank during the time interval t hours? (b) Is the level of heating oil in the tank rising or falling at time t = 6 hours? Give a reason for your answer. (c) How many gallons of heating oil are in the tank at time t = hours? (d) At what time t, for t, is the volume of heating oil in the tank the least? Show the analysis that leads to your conclusion. (a) Htdt () = 7.57 or 7.57 : : integral : answer (b) H(6)(6) R =.94, so the level of heating oil is falling at t = 6. : answer with reason (c) 5 + ( Ht ( )( Rt) dt =.5 or.6 : : limits : integrand : answer (d) The absolute minimum occurs at a critical point or an endpoint. Ht ()() Rt = when t = 4.79 and t =.8. The volume increases until t = 4.79, then decreases until t =.8, then increases, so the absolute minimum will be at t = or at t = ( Ht ()() Rt) dt =.78 : : sets Ht ( )( Rt) = : volume is least at t =.8 : analysis for absolute minimum Since the volume is 5 at t =, the volume is least at t =.8. Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com. 7

30 4 SCORING GUIDELINES (Form B) Question For t, the rate of change of the number of mosquitoes on Tropical Island at time t days is t Rt = 5 tcos mosquitoes per day. There are mosquitoes on Tropical Island at 5 modeled by () ( ) time t =. (a) Show that the number of mosquitoes is increasing at time t = 6. (b) At time t = 6, is the number of mosquitoes increasing at an increasing rate, or is the number of mosquitoes increasing at a decreasing rate? Give a reason for your answer. (c) According to the model, how many mosquitoes will be on the island at time t =? Round your answer to the nearest whole number. (d) To the nearest whole number, what is the maimum number of mosquitoes for t? Show the analysis that leads to your conclusion. (a) Since R ( 6) = 4.48 >, the number of mosquitoes is increasing at t = 6. : shows that R ( 6) > (b) R ( 6) =.9 Since R ( 6) <, the number of mosquitoes is increasing at a decreasing rate at t = 6. : : considers R ( 6) : answer with reason (c) + Rt () dt= To the nearest whole number, there are 964 mosquitoes. : integral : : answer (d) Rt () = when t =, t =.5π, or t = 7.5π Rt () > on < t <.5π Rt () < on.5π < t < 7.5π Rt () > on 7.5π < t < The absolute maimum number of mosquitoes occurs at t =.5π or at t =..5π + Rt () dt= 9.57, There are 964 mosquitoes at t =, so the maimum number of mosquitoes is 9, to the nearest whole number. 4 : : absolute maimum value : integral : answer : analysis : computes interior critical points : completes analysis Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 8

31 4 SCORING GUIDELINES Question Traffic flow is defined as the rate at which cars pass through an intersection, measured in cars per minute. The traffic flow at a particular intersection is modeled by the function F defined by t () 8 4sin ( ) Ft = + for t, where F() t is measured in cars per minute and t is measured in minutes. (a) To the nearest whole number, how many cars pass through the intersection over the -minute period? (b) Is the traffic flow increasing or decreasing at t = 7? Give a reason for your answer. (c) What is the average value of the traffic flow over the time interval t 5? Indicate units of measure. (d) What is the average rate of change of the traffic flow over the time interval t 5? Indicate units of measure. (a) Ft () dt= 474 cars : : limits : integrand : answer (b) F ( 7) =.87 or.87 Since F ( 7) <, the traffic flow is decreasing at t = 7. : answer with reason 5 (c) () cars min 5 Ft dt= : : limits : integrand : answer (d) F( 5) F( ) 5 =.57 or.58 cars min : answer Units of cars min in (c) and cars min in (d) : units in (c) and (d) Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 9

32 5 SCORING GUIDELINES (Form B) Question A water tank at Camp Newton holds gallons of water at time t =. During the time interval t 8 hours, water is pumped into the tank at the rate t () 95 tsin ( ) W t = gallons per hour. 6 During the same time interval, water is removed from the tank at the rate t () 75sin ( ) Rt = gallons per hour. (a) Is the amount of water in the tank increasing at time t = 5? Why or why not? (b) To the nearest whole number, how many gallons of water are in the tank at time t = 8? (c) At what time t, for t 8, is the amount of water in the tank at an absolute minimum? Show the work that leads to your conclusion. (d) For t > 8, no water is pumped into the tank, but water continues to be removed at the rate R() t until the tank becomes empty. Let k be the time at which the tank becomes empty. Write, but do not solve, an equation involving an integral epression that can be used to find the value of k. (a) No; the amount of water is not increasing at t = 5 since W( 5) R( 5) =.9 <. : answer with reason 8 (b) + ( W() t R() t ) dt = gallons : : limits : integrand : answer (c) W() t R() t = t =, ,.9748 t (hours) gallons of water : interior critical points : amount of water is least at : t = or : analysis for absolute minimum The values at the endpoints and the critical points show that the absolute minimum occurs when t = or k (d) Rt () dt= 8 : limits : : equation Copyright 5 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

33 5 SCORING GUIDELINES Question The tide removes sand from Sandy Point Beach at a rate modeled by the function R, given by 4πt () = + ( ) Rt 5sin. 5 A pumping station adds sand to the beach at a rate modeled by the function S, given by 5 t St () =. + t Both R() t and St () have units of cubic yards per hour and t is measured in hours for t 6. At time t =, the beach contains 5 cubic yards of sand. (a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure. (b) Write an epression for Yt (), the total number of cubic yards of sand on the beach at time t. (c) Find the rate at which the total amount of sand on the beach is changing at time t = 4. (d) For t 6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers. 6 (a) Rt () dt=.85 or.86 yd : { : integral : answer with units (b) Y( t) = 5 + ( S( ) R( ) ) d t : : integrand : limits : answer (c) Y () t = S() t R() t Y ( 4) = S( 4) R( 4) =.98 or.99 yd hr : answer (d) Y () t = when St () Rt () =. The only value in [, 6 ] to satisfy St () = Rt () is a = : : sets Y () t = : critical t-value : answer with justification t Y() t 5 a The amount of sand is a minimum when t = 5.7 or 5.8 hours. The minimum value is cubic yards. Copyright 5 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

34 6 SCORING GUIDELINES Question At an intersection in Thomasville, Oregon, cars turn t left at the rate Lt () = 6 tsin ( ) cars per hour over the time interval t 8 hours. The graph of y = L() t is shown above. (a) To the nearest whole number, find the total number of cars turning left at the intersection over the time interval t 8 hours. (b) Traffic engineers will consider turn restrictions when Lt () 5 cars per hour. Find all values of t for which Lt () 5 and compute the average value of L over this time interval. Indicate units of measure. (c) Traffic engineers will install a signal if there is any two-hour time interval during which the product of the total number of cars turning left and the total number of oncoming cars traveling straight through the intersection is greater than,. In every two-hour time interval, 5 oncoming cars travel straight through the intersection. Does this intersection require a traffic signal? Eplain the reasoning that leads to your conclusion. 8 (a) Lt () dt 658 cars (b) Lt () = 5 when t =.48, 6.66 Let R =.48 and S = 6.66 Lt () 5 for t in the interval [ R, S ] S Lt () dt= S R cars per hour R (c) For the product to eceed,, the number of cars turning left in a two-hour interval must be greater than 4. 5 OR Lt () dt= 4.9 > 4 The number of cars turning left will be greater than 4 on a two-hour interval if Lt () on that interval. Lt () on any two-hour subinterval of [.54, 5.86 ]. : { : setup : answer : : t-interval when L() t 5 : average value integral : answer with units : considers 4 cars : valid interval [ h, h ] + 4 : h+ : value of Lt () dt h : answer and eplanation 4 : OR : considers cars per hour : solves Lt () : discusses hour interval : answer and eplanation Yes, a traffic signal is required. 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents).

35 7 SCORING GUIDELINES Question The amount of water in a storage tank, in gallons, is modeled by a continuous function on the time interval t 7, where t is measured in hours. In this model, rates are given as follows: (i) The rate at which water enters the tank is f () t = t sin ( t) gallons per hour for t 7. (ii) The rate at which water leaves the tank is 5 for t < gt () = gallons per hour. for < t 7 The graphs of f and g, which intersect at t =.67 and t = 5.76, are shown in the figure above. At time t =, the amount of water in the tank is 5 gallons. (a) How many gallons of water enter the tank during the time interval t 7? Round your answer to the nearest gallon. (b) For t 7, find the time intervals during which the amount of water in the tank is decreasing. Give a reason for each answer. (c) For t 7, at what time t is the amount of water in the tank greatest? To the nearest gallon, compute the amount of water at this time. Justify your answer. 7 gallons : { : integral (a) f() t dt 864 : answer (b) The amount of water in the tank is decreasing on the intervals t.67 and t 5.76 because f () t < g() t for t <.67 and < t < (c) Since f () t g() t changes sign from positive to negative only at t =, the candidates for the absolute maimum are at t =,, and 7. t (hours) gallons of water f() t dt 5( ) = f() t dt ( 4) = : { : intervals : reason : identifies t = as a candidate : integrand 5 : : amount of water at t = : amount of water at t = 7 : conclusion The amount of water in the tank is greatest at hours. At that time, the amount of water in the tank, rounded to the nearest gallon, is 57 gallons. 7 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

36 For time AP CALCULUS AB 8 SCORING GUIDELINES (Form B) t t hours, let rt () ( e ) Question = represent the speed, in kilometers per hour, at which a car travels along a straight road. The number of liters of gasoline used by the car to travel kilometers is g =.5 e. modeled by ( ) ( ) (a) How many kilometers does the car travel during the first hours? (b) Find the rate of change with respect to time of the number of liters of gasoline used by the car when t = hours. Indicate units of measure. (c) How many liters of gasoline have been used by the car when it reaches a speed of 8 kilometers per hour? kilometers : { : integral (a) r() t dt = 6.7 : answer dg dg d d (b) = ; = rt () dt d dt dt dg dg r( ) dt = d t= = 6.7 = (.5)( ) = 6 liters hour : uses chain rule : { : answer with units (c) Let T be the time at which the car s speed reaches 8 kilometers per hour. Then, rt ( ) = 8 or T =.45 hours. 4 : : equation rt () = 8 : distance integral : answer At time T, the car has gone T T ( ) = r( t) dt= kilometers and has consumed g( ( T )) =.57 liters of gasoline. 8 The College Board. All rights reserved. Visit the College Board on the Web: 4

37 8 SCORING GUIDELINES Question Oil is leaking from a pipeline on the surface of a lake and forms an oil slick whose volume increases at a constant rate of cubic centimeters per minute. The oil slick takes the form of a right circular cylinder with both its radius and height changing with time. (Note: The volume V of a right circular cylinder with radius r and height h is given by V = π r h. ) (a) At the instant when the radius of the oil slick is centimeters and the height is.5 centimeter, the radius is increasing at the rate of.5 centimeters per minute. At this instant, what is the rate of change of the height of the oil slick with respect to time, in centimeters per minute? (b) A recovery device arrives on the scene and begins removing oil. The rate at which oil is removed is R() t = 4 t cubic centimeters per minute, where t is the time in minutes since the device began working. Oil continues to leak at the rate of cubic centimeters per minute. Find the time t when the oil slick reaches its maimum volume. Justify your answer. (c) By the time the recovery device began removing oil, 6, cubic centimeters of oil had already leaked. Write, but do not evaluate, an epression involving an integral that gives the volume of oil at the time found in part (b). (a) When r = cm and h =.5 cm, and dr dt =.5 cm min. dv dr dh = πr h + πr dt dt dt = π( )(.5)(.5) + π( ) dh dt =.8 or.9 cm min dv dt dh dt = cm min dv dr : = and =.5 dt dt 4 : dv : epression for dt : answer dv dv (b) = R() t, so = when Rt () =. dt dt This occurs when t = 5 minutes. dv dv Since > for < t < 5 and < for t > 5, dt dt the oil slick reaches its maimum volume 5 minutes after the device begins working. : : Rt () = : answer : justification (c) The volume of oil, in cm, in the slick at time t = 5 minutes 5 is given by 6, + ( R() t ) dt. : limits and initial condition : { : integrand 8 The College Board. All rights reserved. Visit the College Board on the Web: 5

38 989 AB6 Oil is being pumped continuously from a certain oil well at a rate proportional to the amount of oil left in the well; that is, dy = ky, where y is the amount of oil left in the dt well at any time t. Initially there were,, gallons of oil in the well, and 6 years later there were 5, gallons remaining. It will no longer be profitable to pump oil when there are fewer than 5, gallons remaining. (a) Write an equation for y, the amount of oil remaining in the well at any time t. (b) At what rate is the amount of oil in the well decreasing when there are 6, gallons of oil remaining? (c) In order not to lose money, at what time t should oil no longer be pumped from the well? Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com 6

39 989 AB6 Solution (a) dy = ky dt kt y = Ce or dy = kdt y ln y = kt+ C kt C y = e + t = C =, C = ln 6 y = e kt t = 6 = e ln k = 6 6k 6 6 t t ln y = e = (b) dy ln ky 6 dt = = 6 5 = ln 5 Decreasing at ln gal/year (c) 5 = e kt = ln kt ln t = ln 6 ln = 6 = 6log ln ln 6 years after starting ln Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com 7

40 998 Calculus AB Scoring Guidelines 5. The temperature outside a house during a 4-hour period is given by ( ) πt F (t) = 8 cos, t 4, where F (t) is measured in degrees Fahrenheit and t is measured in hours. (a) Sketch the graph of F on the grid below. (b) Find the average temperature, to the nearest degree Fahrenheit, between t = 6 and t = 4. (c) An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of t was the air conditioner cooling the house? (d) The cost of cooling the house accumulates at the rate of $.5 per hour for each degree the outside temperature eceeds 78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 4 hour period? (a) Degrees Fahrenheit : bell shaped graph minimum 7 at t =, t = 4 only maimum 9 at t = only 6 (b) Avg. = 4 6 (c) = 8 ( ) 6 Time in Hours = 87.6 or F [ 8 cos cos 5. or 5. ( πt ) ( πt [ ( )] πt 8 cos dt )] t or 8.77 : integral : limits and /(4 6) : integrand : answer / if integral not of the form b a b a F (t) dt { : inequality or equation : solutions with interval (d) C = or or 5. ([ ( )] ) πt 8 cos 78 dt =.5(.974) = 5.96 $5. : integral : limits and.5 : integrand : answer / if integral not of the form k b a (F (t) 78) dt Copyright 998 College Entrance Eamination Board. All rights reserved. 8 Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board.

41 !Å#ALCULUSÅ!"n 7ATERÅISÅPUMPEDÅINTOÅANÅUNDERGROUNDÅTANKÅATÅAÅCONSTANTÅRATEÅOFÅÅGALLONSÅPERÅMINUTEÅ7ATERÅLEAKSÅOUT OFÅTHEÅTANKÅATÅTHEÅRATEÅOFÅ T ÅGALLONSÅPERÅMINUTEÅFORÅÅbÅTÅbÅÅMINUTESÅ!TÅTIMEÅTÅÅÅTHEÅTANK CONTAINSÅÅGALLONSÅOFÅWATER A (OWÅMANYÅGALLONSÅOFÅWATERÅLEAKÅOUTÅOFÅTHEÅTANKÅFROMÅTIMEÅTÅÅÅTOÅTÅÅÅMINUTES B (OWÅMANYÅGALLONSÅOFÅWATERÅAREÅINÅTHEÅTANKÅATÅTIMEÅTÅÅÅMINUTES C 7RITEÅANÅEXPRESSIONÅFORÅ!T ÅTHEÅTOTALÅNUMBERÅOFÅGALLONSÅOFÅWATERÅINÅTHEÅTANKÅATÅTIMEÅT D!TÅWHATÅTIMEÅTÅFORÅÅbÅTÅbÅÅISÅTHEÅAMOUNTÅOFÅWATERÅINÅTHEÅTANKÅAÅMAXIMUMÅ*USTIFYÅYOUR ANSWER -ETHODÅ A ÅÅ-ETHODÅÅ T DT T DEFINITEÅINTEGRAL ÅÅÅÅÅÅÅLIMITS ÅÅÅÅÅÅnÅORÅn Å ÅÅÅÅÅÅÅINTEGRAND ÅÅÅÅÅÅ-ETHODÅÅ,T ÅÅGALLONSÅLEAKEDÅINÅFIRSTÅTÅMINUTES ÅÅANSWER D, T, T T # nåorån DT -ETHODÅ, # ÅÅANTIDERIVATIVEÅWITHÅ,T T, Å # ÅÅSOLVESÅFORÅ# ÅUSINGÅÅ, ÅÅANSWER B ÅÅ ÅÅANSWER C -ETHODÅ T!T X DX T X DX nåorån ÅÅÅÅÅÅ-ETHODÅ D! T DT!T T T # ÅÅÅÅÅÅ # #!T T T T -ETHODÅ T Å T X DX nåorån -ETHODÅ ÅÅANTIDERIVATIVEÅWITHÅ# Å ÅÅANSWER D! a T T ÅWHENÅTÅÅ! a T ÅISÅPOSITIVEÅFORÅÅÅTÅÅÅANDÅNEGATIVEÅFOR ÅÅTÅÅÅ4HEREFOREÅTHEREÅISÅAÅMAXIMUM ATÅTÅÅ ÅSETSÅ! a T Å ÅSOLVESÅFORÅT ÅJUSTIFICATION Copyright by College Entrance Eamination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Eamination Board. 9

42 SCORING GUIDELINES (Form B) Question J 6DAK>AHBC=IJ B=FKJ=JE==A?D=CAI=JJDAH=JA= J Г! A Г C=IFAH@=OMDAHAJEIA=IKHA@E@=OI6DAHA=HA#C=IBJDAFKJ=JEJDA=A=J JEAJ6DA=AEI?IE@AHA@J>AI=BAMDAEJ?J=EI"C=IHAIIBFKJ=J = IJDA=KJBFKJ=JE?HA=IEC=JJEAJ'9DOHMDOJ >.HMD=JL=KABJMEJDAK>AHBC=IBFKJ=J>A=JEJIEEKKIJEBOOKH =IMAH? IJDA=AI=BAMDAJDAK>AHBC=IBFKJ=JEI=JEJIEEKKIJEBOOKH )ELAIJEC=JHKIAIJDAJ=CAJEA=FFHNE=JEJJ =JJ=I=@ABHJDA =KJBFKJ=JEJDA=A)JMD=JJEAJ@AIJDEI@AFHA@E?JJD=JJDA=A >A?AII=BA =IMAHMEJDHA=I $ = = ' Г! A Г Г$"$ IJDA=KJEIJE?HA=IEC=JJDEIJEA > = J Г! A Г J IAJI= J J #!!%"! ILAIBHJ = J EIAC=JELABHJ #! =@FIEJELA KIJEBE?=JE BHJ #! 6DAHABHAJDAHAEI=EEK=J J #!!%" Г J?!%" # Г! EJACH=@ EEJI!#""IJDA=AEII=BA!??KIEMEJDHA=I >=IA@EJACH=B= = Г! Г 6DA=AME>A?AI=BA IFABJ=CAJEA MDAJDA=KJ@A?HA=IAI>O)EA=H@A =IMAH FHA@E?JIJDEIMED=FFAMDAJ# Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 4

43 AB{ / BC{ 999. The rate at which water ows out of a pipe, in gallons per hour, is given by a dierentiable function R of time t. The table above shows the rate as measured every hours for a 4{hour period. (a) Use a midpoint Riemann sum with 4 subdivisions of equal length to approimate Z 4 R(t) dt. Using correct units, eplain the meaning of your answer in terms of water ow. (b) Is there some time t, <t<4, such that R (t) =? Justify your answer. (c) The rate of water ow R(t) can be approimated by Q(t) = ; t ; t. Use Q(t) toapproimate the 79 average rate of water ow during the 4{hour time period. Indicate units of measure. t R(t) (hours) (gallons per hour) (a) Z 4 R(t) dt 6[R() + R(9) + R(5) + R()] =6[:4+:+:+:] = 58.6 gallons This is an approimation to the total ow in gallons of water from the pipe in the 4{hour period. 8 : R() + R(9) + R(5) + R() >< : answer >: : eplanation (b) Yes Since R() = R(4) = 9:6, the Mean Value Theorem guarantees that there is a t, <t<4, such thatr (t) =. ( : answer : MVT or equivalent (c) Average rate of ow average value of Q(t) = Z (768 + t ; t ) dt = :785 gal/hr or.784 gal/hr 8 : limits and average value constant >< : Q(t) asintegrand >: : answer (units) Gallons in part (a) and gallons/hr in part (c), or equivalent. : units 4

44 Velocity (feet per second) v(t) O 998 Calculus AB Scoring Guidelines Time (seconds) t t v(t) (seconds) (feet per second) The graph of the velocity v(t), in ft/sec, of a car traveling on a straight road, for t 5, is shown above. A table of values for v(t), at 5 second intervals of time t, is shown to the right of the graph. (a) During what intervals of time is the acceleration of the car positive? Give a reason for your answer. (b) Find the average acceleration of the car, in ft/sec, over the interval t 5. (c) Find one approimation for the acceleration of the car, in ft/sec, at t = 4. Show the computations you used to arrive at your answer. (d) Approimate 5 v(t) dt with a Riemann sum, using the midpoints of five subintervals of equal length. Using correct units, eplain the meaning of this integral. (a) (b) (c) (d) Acceleration is positive on (, 5) and (45, 5) because the velocity v(t) is increasing on [, 5] and [45, 5] Avg. Acc. = v(5) v() 5 or.44 ft/sec Difference quotient; e.g. v(45) v(4) 5 v(4) v(5) 5 = = v(45) v(5) = or Slope of tangent line, e.g. 5 = 7 5 through (5, 9) and (4, 75): v(t) dt = 7 5 = ft/sec or = 6 5 ft/sec or = ft/sec 9 75 = ft/sec 5 4 [v(5) + v(5) + v(5) + v(5) + v(45)] = ( ) = 5 feet This integral is the total distance traveled in feet over the time to 5 seconds. : (, 5) : (45, 5) : reason Note: ignore inclusion of endpoints : answer { : method : answer Note: / if first point not earned : midpoint Riemann sum : answer : meaning of integral Copyright 998 College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks 4 of the College Entrance Eamination Board.

45 AP CALCULUS AB SCORING GUIDELINES Question The temperature, in degrees Celsius ( C), of the water in a pond is a t differentiable function W of time t. The table above shows the water (days) temperature as recorded every days over a 5-day period. (a) Use data from the table to find an approimation for W = (). Show the 6 9 computations that lead to your answer. Indicate units of measure. (b) Approimate the average temperature, in degrees Celsius, of the water 5 over the time interval > t > 5 days by using a trapezoidal approimation with subintervals of length t days. ( t /) (c) A student proposes the function P, given by Pt ( ) te Г, as a model for the Wt () ( C) temperature of the water in the pond at time t, where t is measured in days and Pt () is measured in degrees Celsius. Find P= (). Using appropriate units, eplain the meaning of your answer in terms of water temperature. (d) Use the function P defined in part (c) to find the average value, in degrees Celsius, of Pt () over the time interval > t > 5 days. 8 4 (a) Difference quotient; e.g. W(5) ГW() W = () N Г 5 Г C/day or W() ГW(9) W = () N Г Г 9 C/day or : : difference quotient : answer (with units) W(5) ГW(9) W = () N Г 5 Г 9 C/day (b) () (8) (4) () 76.5 Average temperature N (76.5) 5. C 5 : : trapezoidal method : answer (c) P= () e Г te Гt/ Гt/ t 4 e Г Г Г.549 C/day : : P = () (with or without units) : interpretation This means that the temperature is decreasing at the rate of.549 C/day when t = days. (d) 5 5 Гt / te dt C : : integrand : limits and average value constant : answer Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 4

46 SCORING GUIDELINES (Form B) Question A blood vessel is 6 millimeters (mm) long Distance with circular cross sections of varying diameter. (mm) Diameter The table above gives the measurements of the B() (mm) diameter of the blood vessel at selected points along the length of the blood vessel, where represents the distance from one end of the blood vessel and B () is a twice-differentiable function that represents the diameter at that point. (a) Write an integral epression in terms of B () that represents the average radius, in mm, of the blood vessel between = and = 6. (b) Approimate the value of your answer from part (a) using the data from the table and a midpoint Riemann sum with three subintervals of equal length. Show the computations that lead to your answer. 75 B () (c) Using correct units, eplain the meaning of d 5 in terms of the blood vessel. (d) Eplain why there must be at least one value, for < < 6, such that B ( ) =. (a) 6 B () d : 6 : limits and constant : integrand (b) B(6) B(8) B() + + = 6 [ 6( )] = 4 6 : : B(6) + B(8) + B() : answer (c) B ( ) B ( ) is the radius, so is the area of the cross section at. The epression is the volume in mm of the blood vessel between 5 : : volume in mm : between = 5 and = 75 mm and 75 mm from the end of the vessel. (d) By the MVT, B ( c) = for some c in (6, 8) and B ( c) = for some c in (4, 6). The MVT applied to B ( ) shows that B () = for some in the interval ( c c ),. : : eplains why there are two values of where B( ) has the same value : eplains why that means B ( ) = for < < 6 Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com Note: ma / if only eplains why B ( ) = at some in (, 6).

47 SCORING GUIDELINES Question 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 time t. The graph of R and a table of selected values of R( t ), for the time interval t 9 minutes, are shown above. (a) Use data from the table to find an approimation 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 = 45 minutes. What is the value of R ( 45 )? Eplain your reasoning. (c) Approimate the value of 9 Rt () dt using a left Riemann sum with the five subintervals indicated by the data in the table. Is this numerical approimation less than the value of 9 Rt () dt? Eplain your reasoning. b (d) For < b 9 minutes, eplain the meaning of ( ) R t dt in terms of fuel consumption for the b plane. Eplain the meaning of R ( t ) dt b in terms of fuel consumption for the plane. Indicate units of measure in both answers. (a) R(5) R(4) 55 4 R(45) = 5 4 =.5 gal/min : (b) R (45) = since R () t has a maimum at (c) t = Rt ( ) dt ()() + ()() + ()(4) + ()(55) + ()(65) = 7 Yes, this approimation is less because the graph of R is increasing on the interval. : : : a difference quotient using numbers from table and interval that contains 45 :.5 gal/min : R(45) = : reason : value of left Riemann sum : less with reason (d) b Rt () dt is the total amount of fuel in gallons consumed for the first b minutes. b Rt () dt b is the average value of the rate of fuel consumption in gallons/min during the first b minutes. Copyright by College Entrance Eamination Board. All rights reserved. 45 Available at apcentral.collegeboard.com. 4 : : meanings b : meaning of Rt ( ) dt b : meaning of Rt ( ) dt b < > if no reference to time b : units in both answers

48 4 SCORING GUIDELINES (Form B) Question A test plane flies in a straight line with t (min) positive velocity vt (), in miles per vt ()(mpm) minute at time t minutes, where v is a differentiable function of t. Selected values of vt () for t 4 are shown in the table above. (a) Use a midpoint Riemann sum with four subintervals of equal length and values from the table to 4 approimate vt () dt. Show the computations that lead to your answer. Using correct units, 4 eplain the meaning of vt () dtin terms of the plane s flight. (b) Based on the values in the table, what is the smallest number of instances at which the acceleration of the plane could equal zero on the open interval < t < 4? Justify your answer. t 7t (c) The function f, defined by f() t = 6 + cos( ) + sin ( ), is used to model the velocity of the 4 plane, in miles per minute, for t 4. According to this model, what is the acceleration of the plane at t =? Indicates units of measure. (d) According to the model f, given in part (c), what is the average velocity of the plane, in miles per minute, over the time interval t 4? (a) Midpoint Riemann sum is [ v( 5) + v( 5) + v( 5) + v( 5) ] = [ ] = 9 The integral gives the total distance in miles that the plane flies during the 4 minutes. : : v( 5) + v( 5) + v( 5) + v( 5) : answer : meaning with units (b) By the Mean Value Theorem, v () t = somewhere in the interval (, 5 ) and somewhere in the interval ( 5, ). Therefore the acceleration will equal for at least two values of t. : two instances : : justification (c) f ( ) =.47 or.48 miles per minute : answer with units 4 (d) Average velocity = () 4 f t dt = 5.96 miles per minute : : limits : integrand : answer Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 46 4

49 5 SCORING GUIDELINES Question Distance (cm) Temperature T( ) ( C) A metal wire of length 8 centimeters (cm) is heated at one end. The table above gives selected values of the temperature T( ), in degrees Celsius ( C, ) of the wire cm from the heated end. The function T is decreasing and twice differentiable. (a) Estimate T ( 7. ) Show the work that leads to your answer. Indicate units of measure. (b) Write an integral epression in terms of T( ) for the average temperature of the wire. Estimate the average temperature of the wire using a trapezoidal sum with the four subintervals indicated by the data in the table. Indicate units of measure. 8 8 (c) Find T ( ) d, and indicate units of measure. Eplain the meaning of ( ) T d in terms of the temperature of the wire. (d) Are the data in the table consistent with the assertion that T ( ) > for every in the interval < < 8? Eplain your answer. (a) T( 8) T( 6) = = Ccm 8 6 : answer 8 (b) ( ) 8 T d 8 Trapezoidal approimation for T( ) d: A = Average temperature C 8 A = 8 (c) T ( ) d = T( 8) T( ) = 55 = 45 C The temperature drops 45 C from the heated end of the wire to the other end of the wire., 5 is 7 9 = , 6 is 6 7 = T c = 5.75 for some c in the interval (, 5 ) T c = 8 for some c in the interval ( 5, 6 ). It follows that c, c. Therefore T (d) Average rate of change of temperature on [ ] Average rate of change of temperature on [ ] No. By the MVT, ( ) and ( ) T must decrease somewhere in the interval ( ) is not positive for every in [, 8 ]. : 8 : T( ) d 8 : trapezoidal sum : answer : { : value : meaning : two slopes of secant lines : { : answer with eplanation Units of Ccmin (a), and C in (b) and (c) : units in (a), (b), and (c) Copyright 5 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and 47 (for AP students and parents). 4

50 6 SCORING GUIDELINES (Form B) Question 6 t (sec) vt () ( ft sec ) at () ( ft sec ) A car travels on a straight track. During the time interval t 6 seconds, the car s velocity v, measured in feet per second, and acceleration a, measured in feet per second per second, are continuous functions. The table above shows selected values of these functions. 6 (a) Using appropriate units, eplain the meaning of vt () dtin terms of the car s motion. Approimate 6 vt () dtusing a trapezoidal approimation with the three subintervals determined by the table. (b) Using appropriate units, eplain the meaning of at () dtin terms of the car s motion. Find the eact value of at () dt. (c) For < t < 6, must there be a time t when vt () = 5? Justify your answer. (d) For < t < 6, must there be a time t when at () =? Justify your answer. 6 (a) vt () dtis the distance in feet that the car travels from t = sec to t = 6 sec. Trapezoidal approimation for 6 vt () dt: A = ( 4 + ) 5 + ( )( 5) + ( )( ) = 85 ft (b) at () dtis the car s change in velocity in ft/sec from t = sec to t = sec. a() t dt = v () t dt = v( ) v( ) = 4 ( ) = 6 ft/sec (c) Yes. Since v( 5) = < 5 < = v( 5 ), the IVT guarantees a t in ( 5, 5 ) so that vt () = 5. (d) Yes. Since v( ) = v( 5 ), the MVT guarantees a t in (, 5 ) so that at () = v () t =. Units of ft in (a) and ft/sec in (b) : { : eplanation : value : { : eplanation : value : v( 5) < 5 < v( 5) : : Yes; refers to IVT or hypotheses : v( ) = v( 5) : : Yes; refers to MVT or hypotheses : units in (a) and (b) 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 7 48

51 6 SCORING GUIDELINES Question 4 t (seconds) vt () (feet per second) Rocket A has positive velocity vt () after being launched upward from an initial height of feet at time t = seconds. The velocity of the rocket is recorded for selected values of t over the interval t 8 seconds, as shown in the table above. (a) Find the average acceleration of rocket A over the time interval t 8 seconds. Indicate units of measure. 7 (b) Using correct units, eplain the meaning of vt () dt in terms of the rocket s flight. Use a midpoint Riemann sum with subintervals of equal length to approimate vt () dt. (c) Rocket B is launched upward with an acceleration of at () = feet per second per second. At time t + t = seconds, the initial height of the rocket is feet, and the initial velocity is feet per second. Which of the two rockets is traveling faster at time t = 8 seconds? Eplain your answer. 7 (a) Average acceleration of rocket A is : answer v( 8) v( ) 49 5 ft sec = = 8 8 (b) Since the velocity is positive, vt () dtrepresents the distance, in feet, traveled by rocket A from t = seconds to t = 7 seconds. 7 : eplanation : : uses v( ), v( 4 ), v( 6) : value A midpoint Riemann sum is [ v( ) + v( 4) + v( 6) ] = [ ] = ft (c) Let vb () t be the velocity of rocket B at time t. vb () t = dt = 6 t + + C t + = vb ( ) = 6 + C vb () t = 6 t + 4 v ( 8) = 5 > 49 = v( 8) B 4 : : 6 t + : constant of integration : uses initial condition : finds vb ( 8 ), compares to v( 8 ), and draws a conclusion Rocket B is traveling faster at time t = 8 seconds. Units of ft sec in (a) and ft in (b) : units in (a) and (b) 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 5 49

52 7 SCORING GUIDELINES Question 5 t (minutes) r () t (feet per minute) The volume of a spherical hot air balloon epands as the air inside the balloon is heated. The radius of the balloon, in feet, is modeled by a twice-differentiable function r of time t, where t is measured in minutes. For < t <, the graph of r is concave down. The table above gives selected values of the rate of change, r (), t of the radius of the balloon over the time interval t. The radius of the balloon is feet when 4 t = 5. (Note: The volume of a sphere of radius r is given by V = π r. ) (a) Estimate the radius of the balloon when t = 5.4 using the tangent line approimation at t = 5. Is your estimate greater than or less than the true value? Give a reason for your answer. (b) Find the rate of change of the volume of the balloon with respect to time when t = 5. Indicate units of measure. (c) Use a right Riemann sum with the five subintervals indicated by the data in the table to approimate r () t dt. Using correct units, eplain the meaning of () r t dt in terms of the radius of the balloon. (d) Is your approimation in part (c) greater than or less than r () t dt? Give a reason for your answer. (a) r( 5.4) r( 5) + r ( 5) Δ t = + (.4) =.8 ft : Since the graph of r is concave down on the interval { : estimate : conclusion with reason 5 < t < 5.4, this estimate is greater than r ( 5.4 ). dv dt dv dt 4 π r (b) = ( ) t = 5 dr dt = 4π( ) = 7π ft min (c) r ( t) dt ( 4.) + (.) + (.) + 4(.6) + (.5) = 9. ft r () t dt is the change in the radius, in feet, from t = to t = minutes. (d) Since r is concave down, r is decreasing on < t <. Therefore, this approimation, 9. ft, is less than r () t dt. : dv : dt : answer : { : approimation : eplanation : conclusion with reason Units of ft min in part (b) and ft in part (c) : units in (b) and (c) 7 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). 5

53 8 SCORING GUIDELINES (Form B) Distance from the river s edge (feet) Question Depth of the water (feet) 7 8 A scientist measures the depth of the Doe River at Picnic Point. The river is 4 feet wide at this location. The measurements are taken in a straight line perpendicular to the edge of the river. The data are shown in the table above. The velocity of the water at Picnic Point, in feet per minute, is modeled by vt = 6 + sin t+ for t minutes. () ( ) (a) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approimate the area of the cross section of the river at Picnic Point, in square feet. Show the computations that lead to your answer. (b) The volumetric flow at a location along the river is the product of the cross-sectional area and the velocity of the water at that location. Use your approimation from part (a) to estimate the average value of the volumetric flow at Picnic Point, in cubic feet per minute, from t = to t = minutes. π (c) The scientist proposes the function f, given by f( ) ( ) = 8sin, as a model for the depth of the 4 water, in feet, at Picnic Point feet from the river s edge. Find the area of the cross section of the river at Picnic Point based on this model. (d) Recall that the volumetric flow is the product of the cross-sectional area and the velocity of the water at a location. To prevent flooding, water must be diverted if the average value of the volumetric flow at Picnic Point eceeds cubic feet per minute for a -minute period. Using your answer from part (c), find the average value of the volumetric flow during the time interval 4 t 6 minutes. Does this value indicate that the water must be diverted? (a) ( + 7) ( 7 + 8) ( 8 + ) ( + ) = 5 ft : trapezoidal approimation (b) 5 () vt dt = or 87.7 ft min π 8sin d =. or. ft : { : integra 4 (c) ( ) 4 (d) Let C be the cross-sectional area approimation from part (c). The average volumetric flow is 6 () 8.9 or 8.9 ft min. C v t dt = 4 Yes, water must be diverted since the average volumetric flow for this -minute period eceeds ft min. : : : limits and average value constant : integrand : answer : answer : volumetric flow integral : average volumetric flow : answer with reason 8 The College Board. All rights reserved. Visit the College Board on the Web: 5

54 8 SCORING GUIDELINES Question t (hours) Lt ()(people) Concert tickets went on sale at noon ( t = ) and were sold out within 9 hours. The number of people waiting in line to purchase tickets at time t is modeled by a twice-differentiable function L for t 9. Values of Lt () at various times t are shown in the table above. (a) Use the data in the table to estimate the rate at which the number of people waiting in line was changing at 5: P.M. ( t = 5.5 ). Show the computations that lead to your answer. Indicate units of measure. (b) Use a trapezoidal sum with three subintervals to estimate the average number of people waiting in line during the first 4 hours that tickets were on sale. (c) For t 9, what is the fewest number of times at which L () t must equal? Give a reason for your answer. (d) The rate at which tickets were sold for t 9 is modeled by rt () = 55te t tickets per hour. Based on the model, how many tickets were sold by P.M. ( t =, ) to the nearest whole number? L( 7) L( 4) 5 6 (a) L ( 5.5) = = 8 people per hour 7 4 (b) The average number of people waiting in line during the first 4 hours is approimately L( ) + L( ) L() ( ) ( ) ( ) ( ) + L L ( ) + L 4 ( 4 ) = 55.5 people (c) L is differentiable on [, 9 ] so the Mean Value Theorem implies L () t > for some t in (, ) and some t in ( 4, 7 ). Similarly, L () t < for some t in (, 4 ) and some t in ( 7, 8 ). Then, since L is continuous on [, 9 ], the Intermediate Value Theorem implies that L () t = for at least three values of t in [, 9 ]. OR The continuity of L on [, 4 ] implies that L attains a maimum value there. Since L( ) > L( ) and L( ) > L( 4 ), this maimum occurs on (, 4 ). Similarly, L attains a minimum on (, 7 ) and a maimum on ( 4, 8 ). L is differentiable, so L () t = at each relative etreme point on (, 9 ). Therefore L () t = for at least three values of t in [, 9 ]. [Note: There is a function L that satisfies the given conditions with L () t = for eactly three values of t.] (d) rt () dt= There were approimately 97 tickets sold by P.M. : { : estimate : units : trapezoidal sum : { : answer : : : considers change in sign of L : analysis : conclusion OR : considers relative etrema of L on (, 9) : analysis : conclusion : { : integrand : limits and answer 8 The College Board. All rights reserved. Visit the College Board on the Web: 5

55 !Å#ALCULUSÅ!"n 4HEÅFIGUREÅABOVEÅSHOWSÅTHEÅGRAPHÅOFÅ F a ÅTHEÅDERIVATIVEÅOF THEÅFUNCTIONÅFÅFORÅnÅbÅXÅbÅÅ4HEÅGRAPHÅOFÅ F a ÅHAS HORIZONTALÅTANGENTÅLINESÅATÅXÅÅnÅXÅÅÅANDÅXÅÅÅAND AÅVERTICALÅTANGENTÅLINEÅATÅXÅÅ A &INDÅALLÅVALUESÅOFÅXÅFORÅnÅÅXÅÅÅATÅWHICHÅF ATTAINSÅAÅRELATIVEÅMINIMUMÅ*USTIFYÅYOURÅANSWER y y = f'() O B &INDÅALLÅVALUESÅOFÅXÅFORÅnÅÅXÅÅÅATÅWHICHÅFÅATTAINSÅAÅRELATIVEÅMAXIMUMÅ*USTIFYÅYOURÅANSWER C &INDÅALLÅVALUESÅOFÅXÅFORÅnÅÅXÅÅÅATÅWHICHÅ Faa X D!TÅWHATÅVALUEÅOFÅXÅFORÅnÅbÅXÅbÅÅDOESÅFÅATTAINÅITSÅABSOLUTEÅMAXIMUMÅ*USTIFYÅYOURÅANSWER ÅA ÅXÅÅn Fa X ÅCHANGESÅFROMÅNEGATIVEÅTOÅPOSITIVEÅATÅXÅÅn Å ÅANSWER ÅJUSTIFICATION ÅB XÅÅn Fa X ÅCHANGESÅFROMÅPOSITIVEÅTOÅNEGATIVEÅATÅXÅÅn Å ÅANSWER ÅJUSTIFICATION C F aa X EXISTSÅANDÅ F a ISÅDECREASINGÅONÅTHEÅINTERVALS ÅAND Å Å Å ƒ D XÅÅ 4HEÅABSOLUTEÅMAXIMUMÅMUSTÅOCCURÅATÅXÅÅnÅORÅATÅAN ENDPOINT F F ÅBECAUSEÅFÅISÅINCREASINGÅONÅnn 4HEÅGRAPHÅOFÅ F a SHOWSÅTHATÅTHEÅMAGNITUDEÅOFÅTHE NEGATIVEÅCHANGEÅINÅFÅFROMÅX ÅTOÅX ÅISÅSMALLER THANÅTHEÅPOSITIVEÅCHANGEÅINÅFÅFROMÅX ÅTOÅX 4HEREFOREÅTHEÅNETÅCHANGEÅINÅFÅISÅPOSITIVEÅFROMÅX ÅTO X ÅAND F F ÅÅOÅ F ÅISÅTHEÅABSOLUTE MAXIMUM ÅANSWER ÅIDENTIFIESÅX ÅANDÅX Å ÅÅÅÅÅÅASÅCANDIDATES ÅÅÅÅÅ ÅORÅ ÅÅÅÅÅÅINDICATESÅTHATÅTHEÅGRAPHÅOFÅF ÅÅÅÅÅÅINCREASESÅDECREASESÅTHENÅINCREASES ÅJUSTIFIESÅF F Copyright by College Entrance Eamination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Eamination Board. 5

56 4 SCORING GUIDELINES (Form B) Question 4 The figure above shows the graph of f, the derivative of the function f, on the closed interval 5. The graph of f has horizontal tangent lines at = and =. The function f is twice differentiable with f ( ) = 6. (a) Find the -coordinate of each of the points of inflection of the graph of f. Give a reason for your answer. (b) At what value of does f attain its absolute minimum value on the closed interval 5? At what value of does f attain its absolute maimum value on the closed interval 5? Show the analysis that leads to your answers. (c) Let g be the function defined by g( ) = f( ). Find an equation for the line tangent to the graph of g at =. (a) = and = because the graph of f changes from increasing to decreasing at =, and changes from decreasing to increasing at =. : =, = : : reason (b) The function f decreases from = to = 4, then increases from = 4 to = 5. Therefore, the absolute minimum value for f is at = 4. The absolute maimum value must occur at = or at = 5. f( 5 ) f( ) = f ( t) dt < 5 Since f( 5) < f( ), the absolute maimum value occurs at =. 4 : : indicates f decreases then increases : eliminates = 5 for maimum : absolute minimum at = 4 : absolute maimum at = (c) g ( ) = f( ) + f ( ) g ( ) = f( ) + f ( ) = 6 + ( ) = 4 g( ) = f( ) = : : g ( ) : tangent line Tangent line is y = 4( ) + Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 54 5

57 6 SCORING GUIDELINES (Form B) Question Let f be the function defined for with f ( ) = 5 and f, the ( 4) first derivative of f, given by f ( ) = e sin ( ). The graph of y = f ( ) is shown above. (a) Use the graph of f to determine whether the graph of f is concave up, concave down, or neither on the interval.7 < <.9. Eplain your reasoning. (b) On the interval, find the value of at which f has an absolute maimum. Justify your answer. (c) Write an equation for the line tangent to the graph of f at =. (a) On the interval.7 < <.9, f is decreasing and thus f is concave down on this interval. : { : answer : reason (b) f ( ) = when =, π, π, π, K On [, ] f changes from positive to negative only at π. The absolute maimum must occur at = π or at an endpoint. f ( ) = 5 ( π ) ( ) ( ) f = f + f d = f( ) = f( ) + f ( ) d = π : identifies π and as candidates - or - indicates that the graph of f : increases, decreases, then increases : justifies f( π ) > f( ) : answer This shows that f has an absolute maimum at = π. (c) f( ) = f( ) + f ( ) d = 5.64 f ( ) = e.5 sin( 4) =.459 y 5.6 = (.459)( ) : f ( ) epression : integral 4 : : including f ( ) term : f ( ) : equation 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 55

58 8 SCORING GUIDELINES Question 4 A particle moves along the -ais so that its velocity at time t, for t 6, is given by a differentiable function v whose graph is shown above. The velocity is at t =, t =, and t = 5, and the graph has horizontal tangents at t = and t = 4. The areas of the regions bounded by the t-ais and the graph of v on 5, 6 are 8,, and, respectively. At time t =, the particle is at =. the intervals [, ], [, 5 ], and [ ] (a) For t 6, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer. (b) For how many values of t, where t 6, is the particle at = 8? Eplain your reasoning. (c) On the interval < t <, is the speed of the particle increasing or decreasing? Give a reason for your answer. (d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer. (a) Since vt () < for < t < and 5 < t < 6, and vt () > for < t < 5, we consider t = and t = 6. ( ) = + v( t) dt = 8 = 6 ( 6) = + v( t) dt = 8 + = 9 Therefore, the particle is farthest left at time t = when its position is ( ) =. (b) The particle moves continuously and monotonically from ( ) = to ( ) =. Similarly, the particle moves continuously and monotonically from ( ) = to ( 5) = 7 and also from ( 5) = 7 to ( 6) = 9. By the Intermediate Value Theorem, there are three values of t for which the particle is at t () = 8. (c) The speed is decreasing on the interval < t < since on this interval v < and v is increasing. (d) The acceleration is negative on the intervals < t < and 4 < t < 6 since velocity is decreasing on these intervals. : identifies t = as a candidate 6 : : considers vt () dt : conclusion : : positions at t =, t = 5, and t = 6 : description of motion : conclusion : answer with reason : { : answer : justification 8 The College Board. All rights reserved. Visit the College Board on the Web: 56

59 989 AB5 The figure above shows the graph of f, the derivative of a function f. The domain of f is the set of all real numbers such that. (a) For what values of does the graph of f have a horizontal tangent? (b) For what values of in the interval Justify your answer. (,) does f have a relative maimum? (c) For value of is the graph of f concave downward? Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com 57

60 989 AB5 Solution (a) f ( ) horizontal tangent = = 7,, 4, 8 (b) Relative maima at these points =, 8 because f changes from positive to negative at (c) f concave downward f decreasing (, ), ( 6,) Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com 58

61 4 SCORING GUIDELINES (Form B) Question 5 dy 4 Consider the differential equation ( y. ) d = (a) On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the test booklet.) (b) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the y-plane. Describe all points in the y-plane for which the slopes are negative. (c) Find the particular solution y = f( ) to the given differential equation with the initial condition f ( ) =. (a) : : zero slope at each point (, y) where = or y = positive slope at each point (, y) where and y > : negative slope at each point (, y) where and y < (b) Slopes are negative at points (, y ) where and y <. : description (c) 4 dy = d y 5 ln y = + C 5 y = e e y = Ke, K = ± e = Ke = K y = e C C 6 : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y if y is not eponential Note: ma 6 [----] if no constant of integration Note: 6 if no separation of variables Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 59 6

62 4 SCORING GUIDELINES Question 6 dy Consider the differential equation ( y. ) d = (a) On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the pink test booklet.) (b) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the y-plane. Describe all points in the y-plane for which the slopes are positive. (c) Find the particular solution y = f( ) to the given differential equation with the initial condition f ( ) =. (a) : : zero slope at each point (, y) where = or y = positive slope at each point (, y) where and y > : negative slope at each point (, y) where and y < (b) Slopes are positive at points (, y ) where and y >. : description (c) dy = d y ln y = + C y = e e C y = Ke, K = ± e = Ke = K y = + e C 6 : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y if y is not eponential Note: ma 6 [----] if no constant of integration Note: 6 if no separation of variables Copyright 4 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6 7

63 5 SCORING GUIDELINES (Form B) Question 6 dy y Consider the differential equation =. Let d y = f( ) be the particular solution to this differential equation with the initial condition f ( ) =. (a) On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the test booklet.) (b) Write an equation for the line tangent to the graph of f at =. (c) Find the solution y = f( ) to the given differential equation with the initial condition f ( ) =. (a) : zero slopes : : nonzero slopes ( 4 ) (b) Slope = = y = ( + ) : equation (c) y dy = d = + C y 4 = + C; C = y = = : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y Note: ma 6 [----] if no constant of integration Note: 6 if no separation of variables Copyright 5 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6 7

64 Consider the differential equation dy d AP CALCULUS AB 5 SCORING GUIDELINES =. y Question 6 (a) On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the pink test booklet.) (b) Let y = f ( ) be the particular solution to the differential equation with the initial condition (). f = Write an equation for the line tangent to the graph of f at (, ) and use it to approimate f (. ). (c) Find the particular solution y = f ( ) to the given differential equation with the initial condition f () =. (a) : zero slopes : { : nonzero slopes (b) The line tangent to f at (, ) is y + = ( ). Thus, f (.) is approimately.8. : : equation of the tangent line : approimation for f (.) (c) dy = d y ydy= d y = + C = + C; C = y = + Since the particular solution goes through (, ), y must be negative. Thus the particular solution is y =. 5 : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y Note: ma 5 [----] if no constant of integration Note: 5 if no separation of variables Copyright 5 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and 6 (for AP students and parents). 7

65 6 SCORING GUIDELINES (Form B) Question 5 dy Consider the differential equation = ( y ) cos ( π ). d (a) On the aes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the aes provided in the eam booklet.) (b) There is a horizontal line with equation y = c that satisfies this differential equation. Find the value of c. (c) Find the particular solution y = f ( ) to the differential equation with the initial condition f () =. (a) : zero slopes : { : all other slopes (b) The line y = satisfies the differential equation, so c =. : c = (c) dy = cos( π ) d ( y ) ( y ) = sin( π ) + C π = sin ( π ) + C y π = sin( π ) + C = C π = sin ( π ) + y π π y = sin( π) + π π y = for sin ( π) + π < < 6 : : separates variables : antiderivatives : constant of integration : uses initial condition : answer Note: ma 6 [----] if no constant of integration Note: 6 if no separation of variables 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6 6

66 6 SCORING GUIDELINES Question 5 dy + y Consider the differential equation =, where. d (a) On the aes provided, sketch a slope field for the given differential equation at the eight points indicated. (Note: Use the aes provided in the pink eam booklet.) (b) Find the particular solution y = f ( ) to the differential equation with the initial condition f ( ) = and state its domain. (a) : sign of slope at each point and relative steepness of slope lines in rows and columns (b) dy = d + y ln + y = ln + K ln + y = e + y = C = C + y = + K y = and < or y = and < 7 : : separates variables : antiderivatives 6 : : constant of integration : uses initial condition : solves for y Note: ma 6 [ ---- ] if no constant of integration Note: 6 if no separation of variables : domain 6 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6 64

67 Consider the differential equation AP CALCULUS AB 7 SCORING GUIDELINES (Form B) dy y. d = + Question 5 (a) On the aes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the aes provided in the eam booklet.) d y (b) Find in terms of and y. Describe the region in the y-plane in d which all solution curves to the differential equation are concave up. (c) Let y = f( ) be a particular solution to the differential equation with the initial condition f ( ) =. Does f have a relative minimum, a relative maimum, or neither at =? Justify your answer. (d) Find the values of the constants m and b, for which y = m + b is a solution to the differential equation. (a) : Sign of slope at each point and relative steepness of slope lines in rows and columns. (b) (c) d y dy = + = + y d d Solution curves will be concave up on the half-plane above the line y = +. dy d (, ) d y = + = and d (, ) Thus, f has a relative minimum at (, ). = + > : d y : d : description : answer : { : justification (d) Substituting y = m + b into the differential equation: m = + ( m + b) = ( m + ) + ( b ) Then = m + and m = b : m = and b =. : value for m : { : value for b 7 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents). 65

68 8 SCORING GUIDELINES Question 5 dy y Consider the differential equation =, where. d (a) On the aes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the aes provided in the eam booklet.) (b) Find the particular solution y = f( ) to the differential equation with the initial condition f ( ) =. (c) For the particular solution y = f( ) described in part (b), find lim f ( ). (a) : zero slopes : { : all other slopes (b) dy = d y ln y = + C y = e + C y = e e y = ke, where k = ± e = ke k = e C ( ) ( ) f = e, > C 6 : : separates variables : antidifferentiates : includes constant of integration : uses initial condition : solves for y Note: ma 6 [----] if no constant of integration Note: 6 if no separation of variables ( ) (c) lim e e = : limit 8 The College Board. All rights reserved. Visit the College Board on the Web: 66

69 998 AP Calculus AB Scoring Guidelines 4. Let f be a function with f() = 4 such that for all points (, y) on the graph of f the slope is given by +. y (a) Find the slope of the graph of f at the point where =. (b) Write an equation for the line tangent to the graph of f at = and use it to approimate f(.). (c) Find f() by solving the separable differential equation dy d = + with the initial y condition f() = 4. (d) Use your solution from part (c) to find f(.). (a) dy d = + y dy d = y = 4 = + 4 = 4 8 = : answer (b) y 4 = ( ) f(.) 4 (. ) { : equation of tangent line : uses equation to approimate f(.) (c) f(.). + 4 = 4. y dy = ( + ) d y dy = ( + ) d y = + + C 4 = + + C 4 = C y = y = is branch with point (, 4) f() = : separates variables : antiderivative of dy term : antiderivative of d term 5 : uses y = 4 when = to pick one function out of a family of functions : solves for y / if solving a linear equation in y / if no constant of integration Note: ma /5 if no separation of variables Note: ma /5 [----] if substitutes value(s) for, y, or dy/d before antidifferentiation (d) f(.) = : answer, from student s solution to the given differential equation in (c) Copyright 998 College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 67

70 !Å#ALCULUSÅ!"n DY X #ONSIDERÅTHEÅDIFFERENTIALÅEQUATIONÅ Y DX E A &INDÅAÅSOLUTIONÅY F X ÅTOÅTHEÅDIFFERENTIALÅEQUATIONÅSATISFYINGÅ F B &INDÅTHEÅDOMAINÅANDÅRANGEÅOFÅTHEÅFUNCTIONÅFÅFOUNDÅINÅPARTÅA Y ÅA E DY X DX Y E X # Y E X # LN Y X # LN # ÅÅ # E LN Y X E ÅSEPARATESÅVARIABLES ÅANTIDERIVATIVEÅOFÅDYÅTERM ÅANTIDERIVATIVEÅOFÅDX ÅTERM ÅCONSTANTÅOFÅINTEGRATION Å ÅUSESÅINITIALÅCONDITIONÅF ÅSOLVESÅFORÅY Å.OTEÅÅIFÅY ÅISÅNOTÅAÅLOGARITHMICÅFUNCTIONÅOFÅX.OTEÅMAXÅÅ;=ÅIFÅNOÅCONSTANTÅOF INTEGRATION.OTEÅÅIFÅNOÅSEPARATIONÅOFÅVARIABLES B $OMAINÅX E X E X E E ANGEÅd Y d ÅX E ÅDOMAIN Å ÅÅÅÅÅÅ.OTEÅÅIFÅÅISÅNOTÅINÅTHEÅDOMAIN ÅRANGE.OTEÅÅIFÅYÅISÅNOTÅAÅLOGARITHMICÅFUNCTIONÅOFÅX Copyright by College Entrance Eamination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Eamination Board. 68

71 SCORING GUIDELINES (Form B) Question Г N O = AJO BN >AJDAF=HJE?K=HIKJEJJDACELA@EBBAHAJE=AGK=JEBH N # IK?DJD=JJDAEA O Г EIJ=CAJJJDACH=FDBB.E@JDAN?H@E=JABJDAFEJB J=CA?O=@@AJAHEAMDAJDAHBD=I=?==NEK?=EEKHAEJDAH=JJDEI FEJKIJEBOOKH=IMAH > AJO CN >AJDAF=HJE?K=HIKJEJJDACELA@EBBAHAJE=AGK=JEBH Г N & MEJDJDAEEJE=?@EJE C$ Г".E@O ГO ГO=! Г O! Г IBD=I=?=EEK=JJDEIFEJ H *A?=KIABEI?JEKKIBH N # JDAHA EIFIEJELAJJDA HECDJBN!6DAHABHABD=I=?= EEK=JN! > O@O! O! N Г N + & & Г& + +& O $ N ГN $ O Г $ N ГN $ N!!?=EEK KIJEBE?=JE IAF=H=JAIL=HE=>AI =JE@AHEL=JELAB@OJAH =JE@AHEL=JELAB@N JAH $?IJ=JBEJACH=JE KIAIEEJE=?@EJEC$ Г" ILAIBHO JA=N!$EB?IJ=J BEJACH=JE JA$EBIAF=H=JEBL=HE=>AI Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 69 6

72 SCORING GUIDELINES (Form B) Question 6 5DEF)EIJH=LAEC@KAMAIJJM=H@ECDJDKIA4?=J= DHAJN>AJDA@EIJ=?A>AJMAA5DEF)=@ECDJDKIA 4?=JJEAJ=@AJO>AJDA@EIJ=?A>AJMAA5DEF*=@ ECDJDKIA4?=JJEAJ=IIDMEJDABECKHA=>LA =.E@JDA@EIJ=?AEEAJAHI>AJMAA5DEF)=@ 5DEF*MDAN"=@O! >.E@JDAH=JAB?D=CAEDHBJDA@EIJ=?A>AJMAAJDAJMIDEFIMDAN" =@O!? AJ 4 >AJDA=CAIDMEJDABECKHA.E@JDAH=JAB?D=CAB 4 EH=@E=IFAHDKH MDAN"=@O! =,EIJ=?A! " # =IMAH > H @O ANFHAIIEBH@EIJ=?A H " HANFE?EJO Г?D=EHKAAHHH H N O @J N O "Г#! Г$ # O? J= 4 4 N Г O N # " JH=I?A@AJ=BK?JEAHHH )JN"=@O! IA? 4 " JA $ " #! 4 Г # $ BK?JE &# % AL=K=JE H=@E=IDH # # Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 7 7

73 SCORING GUIDELINES Question 5 )?J=EAHD=IJDAID=FAB=FAHECDJ?EH?K=H?A=IIDMEJDABECKHA =>LA6DADAECDJBJDA?J=EAHEI?=@JDA@E=AJAHBJDAFAECEI?9=JAHEJDA?J=EAHEIAL=FH=JECIJD=JEJI@AFJD D EI?D=CEC=JJDA Г!?IJ=JH=JAB?DH 6DALKAB=?ABDAECDJ D =@H=@EKI H EICELA>O 8 H D! =.E@JDALKA 8 BM=JAHEJDA?J=EAHMDA D #?@E?=JAKEJI BA=IKHA >.E@JDAH=JAB?D=CABJDALKABM=JAHEJDA?J=EAHMEJDHAIFA?JJJEAMDA D #?@E?=JA KEJIBA=IKHA? 5DMJD=JJDAH=JAB?D=CABJDALKABM=JAHEJDA?J=EAH@KAJAL=FH=JEEI@EHA?JOFHFHJE= JJDAANFIA@IKHB=?A=HA=BJDAM=JAH9D=JEIJDA?IJ=JBFHFHJE=EJO # # #! = 9DAD#H 8# #?! H # > D I 8 D D D D! " #! # Г? " & DH D @D #! #! Г # # D#H! " #! Г? D Г " Г! H Г! H Г! =HA= "! 6DA?IJ=JBFHFHJE=EJOEIГ! KEJIB? E==@? DH E>! 8 MDAD# H DE=H> 8 =I=BK?JEBAL=HE=>A E=H> Г?D=EHKAHFH@K?JHKAAHHH AL=K=JE=J D IDMI E@AJEBEAI?IJ=JB FHFHJE=EJO?HHA?JKEJIE==@> Copyright by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 7 6

74 AP CALCULUS AB SCORING GUIDELINES Question 5 A coffeepot has the shape of a cylinder with radius 5 inches, as shown in the figure above. Let h be 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 5 h cubic inches per second. (The volume V of a cylinder with radius r and height h is V = r h. ) dh h (a) Show that. dt = 5 dh h (b) Given that h = 7 at time t =, solve the differential equation = for dt 5 h as a function of t. (c) At what time t is the coffeepot empty? (a) V = 5h dv dh = 5 = 5 h dt dt dh 5 h h = = dt 5 5 : dv : = 5 h dt dv : computes dt : shows result (b) dh dt = h 5 dh = dt h 5 h = t + C 5 7 = + C ( ) t 7 h = + : separates variables : antiderivatives : constant of integration 5 : : uses initial condition h = 7 when t = : solves for h Note: ma /5 [----] if no constant of integration Note: /5 if no separation of variables (c) ( ) 7 t + = : answer t = 7 Copyright by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com. 7 6