ap03_sg_calculus_ab_26472

ap03_sg_calculus_ab_26472 - AP® Calculus AB 2003 Scoring...

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Unformatted text preview: AP® Calculus AB 2003 Scoring Guidelines 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 Question 1 Let R be the shaded region bounded by the graphs of y = x and y = e 3x and the vertical line x = 1, 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 horizontal line y = 1. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis 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 3x = x 1: Correct limits in an integral in at (T, S) = (0.238734, 0.488604) 1 ( T (a) Area = x e 3x (a), (b), or (c) )dx 2: = 0.442 or 0.443 1 ((1 T (b) Volume = e 3x 2 (1 ) x 2 ) 1 : integrand 1 : answer 2 : integrand )dx < 3: 1 > reversal < 1 > error with constant < 1 > omits 1 in one radius < or 1.423 or 1.424 = 0.453 2 > other errors 1 : answer (c) Length = x e Height = 5 ( x 2 : integrand 3x e 3x ) < 1 > incorrect but has x 3: Volume = 1 T 5( x e 3x 2 ) e 3x as a factor dx = 1.554 1 : answer Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 2 AP® CALCULUS AB 2003 SCORING GUIDELINES Question 2 A particle moves along the x-axis so that its velocity at time t is given by v (t ) = ( t + 1 ) sin t2 . 2 At time t = 0, the particle is at position x = 1. (a) Find the acceleration of the particle at time t = 2. Is the speed of the particle increasing at t = 2 ? Why or why not? (b) Find all times t in the open interval 0 < t < 3 when the particle changes direction. Justify your answer. (c) Find the total distance traveled by the particle from time t = 0 until time t = 3. (d) During the time interval 0 3, what is the greatest distance between the particle and the t origin? Show the work that leads to your answer. 1 : a(2) (a) a(2) = v (2) = 1.587 or 1.588 v(2) = 3 sin(2) < 0 2: with reason Speed is decreasing since a(2) > 0 and v(2) < 0 . t2 = 2 or 2.506 or 2.507 (b) v(t ) = 0 when t= 2 Since v(t ) < 0 for 0 < t < 2: 2 1 : speed decreasing 1: t = 2 only 1 : justification and v(t ) > 0 for 2 < t < 3 , the particle changes directions at t= 2. (c) Distance = 3 0 1 : limits v(t ) dt = 4.333 or 4.334 3: 1 : integrand 1 : answer (d) 2 0 v(t ) dt = 1 : ± (distance particle travels 3.265 x ( 2 ) = x (0) + 2 0 2: v(t )dt = 2.265 while velocity is negative) 1 : answer Since the total distance from t = 0 to t = 3 is 4.334, the particle is still to the left of the origin at t = 3. Hence the greatest distance from the origin is 2.265. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 3 AP® CALCULUS AB 2003 SCORING GUIDELINES Question 3 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 0 t 90 minutes, are shown above. (a) Use data from the table to find an approximation 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 ) ? Explain your reasoning. (c) Approximate the value of 90 0 R(t )dt using a left Riemann sum with the five subintervals indicated by the data in the table. Is this numerical approximation less than the value of 90 0 R(t )dt ? Explain your reasoning. (d) For 0 < b 90 minutes, explain the meaning of 1 b units of measure in both answers. plane. Explain the meaning of (a) R (45) b 0 b 0 R ( t )dt in terms of fuel consumption for the R ( t )dt in terms of fuel consumption for the plane. Indicate R(50) R(40) 55 40 = 50 40 10 2 = 1.5 gal/min 1 : a difference quotient using numbers from table and 2: interval that contains 45 1 : 1.5 gal/min2 (b) R (45) = 0 since R (t ) has a maximum at 2: t = 45 . (c) 90 0 R(t ) dt 1 : R (45) = 0 1 : reason 1 : value of left Riemann sum (30)(20) + (10)(30) + (10)(40) 2: +(20)(55) + (20)(65) = 3700 1 : “less” with reason Yes, this approximation is less because the graph of R is increasing on the interval. (d) b 0 2 : meanings R(t ) dt is the total amount of fuel in gallons consumed for the first b minutes. 1b R(t ) dt is the average value of the rate of b0 fuel consumption in gallons/min during the 1 : meaning of 3: b 0 R(t ) dt 1b R(t ) dt b0 1 > if no reference to time b 1 : meaning of < 1 : units in both answers first b minutes. Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 4 AP® CALCULUS AB 2003 SCORING GUIDELINES Question 4 Let f be a function defined on the closed interval 3 x 4 with f ( 0 ) = 3. The graph of f , the derivative of f, consists of one line segment and a semicircle, as shown above. (a) On what intervals, if any, is f increasing? Justify your answer. (b) Find the x-coordinate of each point of inflection of the graph of f 3 < x < 4. Justify your answer. on the open interval (c) Find an equation for the line tangent to the graph of f at the point ( 0, 3 ) . (d) Find f ( 3 ) and f ( 4 ) . Show the work that leads to your answers. (a) The function f is increasing on [ 3, 2] since f > 0 for x< 3 1 : interval 2: 2. 1 : reason (b) x = 0 and x = 2 f changes from decreasing to increasing at x = 0 and from increasing to decreasing at 2: 1 : x = 0 and x = 2 only 1 : justification x =2 (c) f (0) = 1 : equation 2 Tangent line is y = (d) f (0) 0 f ( 3) = 3 2x + 3. 1 = (1)(1) 2 f ( 3) = f (0) + 1: ± f (t )dt 1 (2)(2) = 2 3 2 1 (2 2) (difference of areas of triangles) 3 9 = 2 2 1 : answer for f ( 3) using FTC 4: f (4) f (0) = = f (4) = f (0) 4 0 ( (8 1: ± 8 12 (2) 2 8+2 = )= ) 12 (2) 2 (area of rectangle f (t )dt 8+2 area of semicircle) 5+2 1 : answer for f (4) using FTC Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 5 AP® CALCULUS AB 2003 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 h cubic inches per second. (The volume V of a cylinder 5 changing at the rate of with radius r and height h is V = r 2h. ) dh = dt (a) Show that h . 5 (b) Given that h = 17 at time t = 0 , solve the differential equation dh = dt h for 5 h as a function of t. (c) At what time t is the coffeepot empty? (a) V = 25 h 1: dV dh = 25 = dt dt dh = dt (b) 5h = 25 dh = dt h 5 ( h dV dt 1 : shows result 1 : computes 1 : antiderivatives 1 : constant of integration 1 dt 5 5: 1 : uses initial condition h = 17 when t = 0 1 t +C 5 1 : solves for h 2 17 = 0 + C h= 5 1 : separates variables h 5 1 dh = h 2 h= 3: h 5 dV = dt Note: max 2/5 [1-1-0-0-0] if no constant 1 t + 17 10 of integration ) 2 Note: 0/5 if no separation of variables (c) ( 1 t + 17 10 ) 2 =0 1 : answer t = 10 17 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 6 AP® CALCULUS AB 2003 SCORING GUIDELINES Question 6 Let f be the function defined by x + 1 for 0 ≤ x ≤ 3 f (x ) = 5 − x for 3 < x ≤ 5. (a) Is f continuous at x = 3 ? Explain why or why not. (b) Find the average value of f (x ) on the closed interval 0 ≤ x ≤ 5. (c) Suppose the function g is defined by k x + 1 for 0 ≤ x ≤ 3 g(x ) = mx + 2 for 3 < x ≤ 5, where k and m are constants. If g is differentiable at x = 3, what are the values of k and m ? 1 : answers “yes” and equates the values of the left- and right-hand 2: limits 1 : explanation involving limits (a) f is continuous at x = 3 because lim f (x ) = lim+ f (x ) = 2. x → 3− x →3 Therefore, lim f (x ) = 2 = f (3). x →3 (b) 5 ∫0 f (x ) dx = 3 ∫0 f (x ) dx + 3 5 ∫3 4: f (x ) dx ( ) 5 3 2 1 (x + 1) 2 + 5x − x 2 3 2 0 3 16 2 25 21 20 = − + − = 3 3 2 2 3 = ( Average value: )( ) 3 5 1 : k ∫ f (x ) dx + k ∫ f (x ) dx 0 3 (where k ≠ 0) 1 : antiderivative of 1 : antiderivative of 5 − x 1 : evaluation and answer 15 4 ∫0 f (x )dx = 3 5 (c) Since g is continuous at x = 3, 2k = 3m + 2. k 2 x + 1 for 0 < x < 3 g ′(x ) = m for 3 < x < 5 k lim g ′(x ) = and lim g ′(x ) = m − 4 x →3 x → 3+ Since these two limits exist and g is 1 : 2k = 3m + 2 k 3: 1: =m 4 1 : values for k and m differentiable at x = 3, the two limits are k equal. Thus = m. 4 8m = 3m + 2 ; m = 2 8 and k = 5 5 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 7 x +1 ...
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