ap04_sg_calculus_ab_36998

ap04_sg_calculus_ab_36998 - AP® Calculus AB 2004 Scoring...

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Unformatted text preview: AP® Calculus AB 2004 Scoring Guidelines The materials included in these files are intended for noncommercial use by AP teachers for course and exam preparation; permission for any other use ® must be sought from the Advanced Placement Program . Teachers may reproduce them, in whole or in part, in limited quantities, for face-to-face teaching purposes but may not mass distribute the materials, electronically or otherwise. This permission does not apply to any third-party copyrights contained herein. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here. The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 4,500 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns. For further information, visit www.collegeboard.com Copyright © 2004 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. PSAT/NMSQT is a registered trademark of the College Entrance Examination Board and National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service. Other products and services may be trademarks of their respective owners. For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. AP® CALCULUS AB 2004 SCORING GUIDELINES Question 1 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 F ( t ) = 82 + 4sin ( 2t ) for 0 ≤ t ≤ 30, 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 30-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 10 ≤ t ≤ 15 ? Indicate units of measure. (d) What is the average rate of change of the traffic flow over the time interval 10 ≤ t ≤ 15 ? Indicate units of measure. 30 1 : limits 3 : 1 : integrand 1 : answer (a) ∫0 (b) F ′ ( 7 ) = −1.872 or −1.873 Since F ′ ( 7 ) < 0, the traffic flow is decreasing at t = 7. 1 : answer with reason (c) 1 15 F ( t ) dt = 81.899 cars min 5 ∫10 1 : limits 3 : 1 : integrand 1 : answer (d) F (15 ) − F (10 ) = 1.517 or 1.518 cars min 2 15 − 10 1 : answer F ( t ) dt = 2474 cars 1 : units in (c) and (d) Units of cars min in (c) and cars min 2 in (d) Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 2 AP® CALCULUS AB 2004 SCORING GUIDELINES Question 2 Let f and g be the functions given by f ( x ) = 2 x (1 − x ) and g ( x ) = 3 ( x − 1) x for 0 ≤ x ≤ 1. 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 = 2. (c) Let h be the function given by h( x ) = k x (1 − x ) for 0 ≤ x ≤ 1. For each k > 0, 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 x-axis. There is a value of k for which the volume of this solid is equal to 15. Write, but do not solve, an equation involving an integral expression that could be used to find the value of k. (a) Area = = 1 ∫0 ( f ( x ) − g ( x ) ) dx 1 ∫0 ( 2 x (1 − x ) − 3( x − 1) x ) dx = 1.133 (b) Volume = π = π∫ 1 0 2: ∫0 ( ( 2 − g ( x ) ) 1 (( 2 − 3( x − 1) 2 ) − ( 2 − f ( x ) )2 dx ) 2 x ) − ( 2 − 2 x (1 − x ) )2 dx = 16.179 (c) Volume = { 1 : integral 1 : answer 1 : limits and constant 2 : integrand −1 each error 4: Note: 0 2 if integral not of form b c ∫ R 2 ( x ) − r 2 ( x ) dx a 1 : answer ( 1 ∫ 0 ( h( x ) − g ( x ) ) 2 dx 3: 1 2 ∫0 ( k x (1 − x ) − 3( x − 1) x ) dx = 15 { ) 2 : integrand 1 : answer Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 3 AP® CALCULUS AB 2004 SCORING GUIDELINES Question 3 () A particle moves along the y-axis so that its velocity v at time t ≥ 0 is given by v( t ) = 1 − tan −1 et . At time t = 0, the particle is at y = −1. (Note: tan −1 x = arctan x ) (a) Find the acceleration of the particle at time t = 2. (b) Is the speed of the particle increasing or decreasing at time t = 2 ? Give a reason for your answer. (c) Find the time t ≥ 0 at which the particle reaches its highest point. Justify your answer. (d) Find the position of the particle at time t = 2. Is the particle moving toward the origin or away from the origin at time t = 2 ? Justify your answer. (a) a( 2 ) = v′( 2 ) = −0.132 or −0.133 1 : answer (b) v( 2 ) = −0.436 Speed is increasing since a( 2 ) < 0 and v( 2 ) < 0. 1 : answer with reason (c) v( t ) = 0 when tan −1 et = 1 () t = ln ( tan (1) ) = 0.443 is the only critical value for y. 1 : sets v ( t ) = 0 3 : 1 : identifies t = 0.443 as a candidate 1 : justifies absolute maximum v( t ) > 0 for 0 < t < ln ( tan (1) ) v( t ) < 0 for t > ln ( tan (1) ) y ( t ) has an absolute maximum at t = 0.443. (d) y ( 2 ) = −1 + 2 ∫0 v( t ) dt = −1.360 or −1.361 The particle is moving away from the origin since v( 2 ) < 0 and y ( 2 ) < 0. 1 : 2 v( t ) dt ∫0 1 : handles initial condition 4: 1 : value of y ( 2 ) 1 : answer with reason Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 4 AP® CALCULUS AB 2004 SCORING GUIDELINES Question 4 Consider the curve given by x 2 + 4 y 2 = 7 + 3x y. (a) Show that dy 3 y − 2 x = . dx 8 y − 3 x (b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P. d2y at the point P found in part (b). Does the curve have a local maximum, a dx 2 local minimum, or neither at the point P ? Justify your answer. (c) Find the value of (a) (b) 2 x + 8 y y′ = 3 y + 3x y′ (8 y − 3x ) y′ = 3 y − 2 x 3y − 2x y′ = 8 y − 3x 1 : implicit differentiation 2: 1 : solves for y ′ 3y − 2x = 0; 3 y − 2 x = 0 8 y − 3x 1 : dy = 0 dx 3: 1 : shows slope is 0 at ( 3, 2 ) 1 : shows ( 3, 2 ) lies on curve When x = 3, 3 y = 6 y=2 32 + 4⋅22 = 25 and 7 + 3⋅3⋅2 = 25 Therefore, P = ( 3, 2 ) is on the curve and the slope is 0 at this point. (c) d 2 y ( 8 y − 3x )( 3 y ′ − 2 ) − ( 3 y − 2 x )( 8 y ′ − 3) = dx 2 ( 8 y − 3 x )2 d 2 y (16 − 9 )( −2 ) 2 = =− . 2 2 7 dx (16 − 9 ) Since y ′ = 0 and y ′′ < 0 at P, the curve has a local maximum at P. At P = ( 3, 2 ) , d2y 2: dx 2 4: d2y at ( 3, 2 ) 1 : value of dx 2 1 : conclusion with justification Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 5 AP® CALCULUS AB 2004 SCORING GUIDELINES Question 5 The graph of the function f shown above consists of a semicircle and three line segments. Let g be the function given by g ( x ) = x ∫−3 f ( t ) dt. (a) Find g ( 0 ) and g ′( 0 ) . (b) Find all values of x in the open interval ( −5, 4 ) at which g attains a relative maximum. Justify your answer. (c) Find the absolute minimum value of g on the closed interval [ −5, 4]. Justify your answer. (d) Find all values of x in the open interval ( −5, 4 ) at which the graph of g has a point of inflection. (a) g ( 0) = 1 0 1 : g ( 0) 2: 1 : g ′( 0 ) 9 ∫−3 f ( t ) dt = 2 ( 3)( 2 + 1) = 2 g ′( 0 ) = f ( 0 ) = 1 (b) g has a relative maximum at x = 3. This is the only x-value where g ′ = f changes from positive to negative. 1:x = 3 2: 1 : justification (c) The only x-value where f changes from negative to positive is x = − 4. The other candidates for the location of the absolute minimum value are the endpoints. 1 : identifies x = − 4 as a candidate 3 : 1 : g ( −4 ) = −1 1 : justification and answer g ( −5 ) = 0 g ( −4) = g ( 4) = −4 ∫−3 ( f ( t ) dt = −1 ) 9 π 13 − π + 2− = 2 2 2 So the absolute minimum value of g is −1. (d) x = −3, 1, 2 2 : correct values −1 each missing or extra value Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 6 AP® CALCULUS AB 2004 SCORING GUIDELINES Question 6 Consider the differential equation dy = x 2 ( y − 1) . dx (a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the axes 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 xy-plane. Describe all points in the xy-plane for which the slopes are positive. (c) Find the particular solution y = f ( x ) to the given differential equation with the initial condition f ( 0 ) = 3. (a) 1 : zero slope at each point ( x, y ) where x = 0 or y = 1 positive slope at each point ( x, y ) 2: where x ≠ 0 and y > 1 1: negative slope at each point ( x, y ) where x ≠ 0 and y < 1 (b) Slopes are positive at points ( x, y ) where x ≠ 0 and y > 1. (c) 1 : description 1 dy = x 2 dx y −1 1 ln y − 1 = x3 + C 3 y −1 = 1 : separates variables 2 : antiderivatives 1 : constant of integration 6: 1 : uses initial condition 1 : solves for y 0 1 if y is not exponential 13 C 3x ee 13 x y − 1 = Ke 3 , K = ± eC 2 = Ke0 = K y =1+ Note: max 3 6 [1-2-0-0-0] if no constant of integration Note: 0 6 if no separation of variables 13 x 2e 3 Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). 7 ...
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This note was uploaded on 10/01/2009 for the course OC 9876 taught by Professor Dq during the Spring '09 term at UC Merced.

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