ap03_sg_calculus_ab_b_26473

ap03_sg_calculus_ab_b_26473 - AP® Calculus AB 2003 Scoring...

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Unformatted text preview: AP® Calculus AB 2003 Scoring Guidelines Form B The materials included in these files are intended for 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 noncommercial, face-to-face teaching purposes. This permission does not apply to any third-party copyrights contained herein. This material may not be mass distributed, electronically or otherwise. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here. These materials were produced by Educational Testing Service® (ETS®), which develops and administers the examinations of the Advanced Placement Program for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, and their programs, services, and employment policies are guided by that principle. The College Board is a national nonprofit membership association whose mission is to prepare, inspire, and connect students to college and opportunity. Founded in 1900, the association is composed of more than 4,300 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 22,000 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 equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. For further information, visit www.collegeboard.com Copyright © 2003 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. AP Central is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark jointly owned by the College Entrance Examination Board and the 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 2003 SCORING GUIDELINES (Form B) Question 1 Let f be the function given by f (x ) = 4x 2 be the line y = 18 3x , where x 3 , and let is tangent to the graph of f. Let R be the region bounded by the graph of f and the x-axis, and let S be the region bounded by the graph of f, the line , and the x-axis, as shown above. is tangent to the graph of y = f (x ) at (a) Show that the point x = 3. (b) Find the area of S. (c) Find the volume of the solid generated when R is revolved about the x-axis. (a) f (x ) = 8x 3x 2 ; f (3) = 24 f (3) = 36 27 = 1 : finds f (3) and f (3) 3 27 = 9 finds equation of tangent line y= 3) + 9 = 3(x 1: 3x + 18, (b) f (x ) = 0 at x = 4 2 : integral for non-triangular region The line intersects the x-axis at x = 6. 4 1 Area = (3)(9) ( 4x 2 x 3 )dx 3 2 = 7.916 or 7.917 1 : limits 4: 3 + ( ( 18 1 : answer ( 4x 2 3x ) 1 (2)(18 2 1 : integrand 1 : area of triangular region OR 4 shows (3,9) is on both the graph of f and line which is the equation of line . Area = or 2: Tangent line at x = 3 is x 3 ) )dx 12) = 7.916 or 7.917 (c) Volume = 4 0 ( 4x 2 = 156.038 x 3 ) dx 2 1 : limits and constant 3: or 490.208 1 : integrand 1 : answer Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 2 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B) Question 2 A tank contains 125 gallons of heating oil at time t = 0. During the time interval 0 t 12 hours, heating oil is pumped into the tank at the rate H (t ) = 2 + 10 ( 1 + ln ( t + 1 ) ) gallons per hour. During the same time interval, heating oil is removed from the tank at the rate R ( t ) = 12 sin t2 47 gallons per hour. (a) How many gallons of heating oil are pumped into the tank during the time interval 0 t 12 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 = 12 hours? (d) At what time t, for 0 t 12, is the volume of heating oil in the tank the least? Show the analysis that leads to your conclusion. (a) 12 0 1 : integral H (t ) dt = 70.570 or 70.571 (b) H (6) R(6) = 2: 1 : answer 1 : answer with reason 2.924, so the level of heating oil is falling at t = 6. (c) 125 + 12 0 ( H (t ) 1 : limits R(t ) ) dt = 122.025 or 122.026 3: 1 : integrand 1 : answer (d) The absolute minimum occurs at a critical point 1 : sets H (t ) or an endpoint. H (t ) 1 : volume is least at R(t ) = 0 when t = 4.790 and t = 11.318. The volume increases until t = 4.790, then decreases until t = 11.318, then increases, so the 3: t = 11.318 1 : analysis for absolute minimum absolute minimum will be at t = 0 or at t = 11.318. 125 + 11.318 0 ( H (t ) R(t ) = 0 R(t ) ) dt = 120.738 Since the volume is 125 at t = 0, the volume is least at t = 11.318. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 3 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B) Question 3 A blood vessel is 360 millimeters (mm) long Distance x (mm) Diameter B(x) (mm) with circular cross sections of varying diameter. The table above gives the measurements of the 0 60 120 180 240 300 360 24 30 28 30 26 24 26 diameter of the blood vessel at selected points along the length of the blood vessel, where x represents the distance from one end of the blood vessel and B(x ) is a twice-differentiable function that represents the diameter at that point. (a) Write an integral expression in terms of B(x ) that represents the average radius, in mm, of the blood vessel between x = 0 and x = 360. (b) Approximate 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. 275 (c) Using correct units, explain the meaning of 125 B(x ) 2 dx in terms of the blood vessel. 2 (d) Explain why there must be at least one value x, for 0 < x < 360, such that B (x ) = 0. (a) (b) (c) 1 360 360 B(x ) 0 2 1 : limits and constant dx 2: 1 B(60) B(180) B(300) 120 + + 360 2 2 2 1 [ 60 ( 30 + 30 + 24 ) ] = 14 360 B(x ) is the radius, so 2 B(x ) 2 = 2: 2 1 : integrand 1 : B(60) + B(180) + B(300) 1 : answer 1 : volume in mm 3 is the area of 2: the cross section at x. The expression is the 1 : between x = 125 and x = 275 volume in mm3 of the blood vessel between 125 mm and 275 mm from the end of the vessel. (d) By the MVT, B (c 1 ) = 0 for some c1 in 2 : explains why there are two (60, 180) and B (c 2 ) = 0 for some c2 in values of x where B (x ) has (240, 360). The MVT applied to B (x ) shows that B (x ) = 0 for some x in the interval 3: the same value 1 : explains why that means (c 1, c 2 ) . B (x ) = 0 for 0 < x < 360 Note: max 1/3 if only explains why B (x ) = 0 at some x in (0, 360). Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 4 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B) Question 4 A particle moves along the x-axis with velocity at time t 1 + e1 t . 0 given by v ( t ) = (a) Find the acceleration of the particle at time t = 3. (b) Is the speed of the particle increasing at time t = 3 ? Give a reason for your answer. (c) Find all values of t at which the particle changes direction. Justify your answer. (d) Find the total distance traveled by the particle over the time interval 0 (a) a(t ) = v (t ) = e1 a(3) = e t 2: 2 (b) a(3) < 0 v(3) = t 3. 1 : v (t ) 1 : a(3) 1 : answer with reason 1 +e <0 2 Speed is increasing since v(3) < 0 and a(3) < 0 . 1 : solves v(t ) = 0 to (c) v(t ) = 0 when 1 = e 1 t , so t = 1. 2: v(t ) > 0 for t < 1 and v(t ) < 0 for t > 1 . 3 0 1 = ( = 1 : limits v(t ) dt ( 1 + e1 t 0 e1 t1 0 t 3 )dt + 1 ) + ( t + e1 =( 1 1 + e ) + (3 + e = e +e 2 1 : integrand (1 t3 1 2 e1 t )dt 4: ) 1 e1 1) 1 OR 1 : any antiderivative t x (0) = e x (1) = 1 : evaluates x (t ) when 2 x (3) = e 1 : antidifferentiation 1 : evaluation OR x (t ) = t 1 : justifies change in direction at t = 1 Therefore, the particle changes direction at t = 1. (d) Distance = get t = 1 2 4: 3 x (0) ) + ( x (1) Distance = ( x (1) = ( 2 + e) + ( 1 + e = e +e 2 2 x (3) ) t = 0, 1, 3 1 : evaluates distance between points ) 1 : evaluates total distance 1 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 5 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B) Question 5 Let f be a function defined on the closed interval [0, 7]. The graph of f, consisting of four line segments, is shown above. Let g be the function given by g (x ) = x 2 f (t )dt. (a) Find g ( 3 ), g ( 3 ), and g ( 3 ) . (b) Find the average rate of change of g on the interval 0 (c) x 3. For how many values c, where 0 < c < 3, is g (c ) equal to the average rate found in part (b)? Explain your reasoning. (d) Find the x-coordinate of each point of inflection of the graph of g on the interval 0 < x < 7. Justify your answer. (a) g(3) = 3 2 f (t ) dt = 1 (4 + 2) = 3 2 1 : g (3) 3: g (3) = f (3) = 2 g (3) = f (3) = (b) g (3) g (0) 3 0 4 4 = 2 1 : g (3) 2 13 f (t ) dt 30 11 1 7 (2)(4) + (4 + 2) = = 32 2 3 = ( 1 : g (3) 2: ) g(0) = 3 0 f (t ) dt 1 : answer 1 : answer of 2 (c) There are two values of c. 7 We need = g (c) = f (c) 3 The graph of f intersects the line y = 1 : g(3) 2: 7 at two 3 1 : reason Note: 1/2 if answer is 1 by MVT places between 0 and 3. 1 : x = 2 and x = 5 only (d) x = 2 and x = 5 because g = f changes from increasing to 2: decreasing at x = 2, and from decreasing to 1 : justification (ignore discussion at x = 4) increasing at x = 5. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 6 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B) Question 6 Let f be the function satisfying f (x ) = x f (x ) for all real numbers x, where f ( 3 ) = 25. (a) Find f ( 3 ) . dy = x y with the initial dx (b) Write an expression for y = f (x ) by solving the differential equation condition f ( 3 ) = 25. (a) f (x ) = f (x ) + x f (x ) = 2 f (x ) f (x ) + x2 2 2 : f (x ) 3: 9 19 25 + = 2 2 f (3) = (b) < 2 > product or chain rule error 1 : value at x = 3 1 : separates variables 1 dy = x dx y 1 : antiderivative of dy term 1 : antiderivative of dx term 1 2 y = x2 + C 2 6: 1 : constant of integration 1 : uses initial condition f (3) = 25 2 25 = y= 12 11 (3) + C ; C = 2 2 1 : solves for y Note: max 3/6 [1-1-1-0-0-0] if no 1 2 11 x+ 4 4 constant of integration Note: 0/6 if no separation of variables y= (1x 4 2 + 11 4 ) 2 = 12 2 ( x + 11 ) 16 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 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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