Unformatted text preview: AP® Calculus AB 2007 Scoring Guidelines The College Board: Connecting Students to College Success
The College Board is a notforprofit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven 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 bestknown 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. © 2007 The College Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Central, SAT, and the acorn logo are registered trademarks of the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. Permission to use copyrighted College Board materials may be requested online at: www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program: apcentral.collegeboard.com. AP® CALCULUS AB 2007 SCORING GUIDELINES
Question 1
Let R be the region in the first and second quadrants bounded above by the graph of y = below by the horizontal line y = 2. 20 and 1 + x2 (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the xaxis. (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the xaxis are semicircles. Find the volume of this solid. 20 = 2 when x = ±3 1 + x2 1 : correct limits in an integral in (a), (b), or (c) 20 − 2 ⎞ dx = 37.961 or 37.962 (a) Area = ⌠ ⎛ ⎮⎜ ⎟ 2 ⌡−3 ⎝ 1 + x ⎠ 3 2: { 1 : integrand 1 : answer ⎞ ⌠ ⎛ ⎛ 20 ⎞ 2 (b) Volume = π ⎮ ⎜ ⎜ − 22 ⎟ dx = 1871.190 2⎟ ⌡−3 ⎝ ⎝ 1 + x ⎠ ⎠ 3 3: { 2 : integrand 1 : answer π ⌠ 1 20 (c) Volume = ⎮ ⎛ ⎛ − 2 ⎞ ⎞ dx ⎜ 2⎜ ⎟⎟ 2 2 ⌡−3 ⎝ ⎝ 1 + x ⎠⎠
2 3 3: { 2 : integrand 1 : answer = π ⌠ ⎛ 20 − 2 ⎞ dx = 174.268 ⎟ 8 ⎮−3 ⎜ 1 + x 2 ⎝ ⎠ ⌡
2 3 © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CALCULUS AB 2007 SCORING GUIDELINES
Question 2
The amount of water in a storage tank, in gallons, is modeled by a continuous function on the time interval 0 ≤ 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 ) = 100t 2 sin ( t ) gallons per hour for 0 ≤ t ≤ 7. (ii) The rate at which water leaves the tank is ⎧ 250 for 0 ≤ t < 3 gallons per hour. g (t ) = ⎨ ⎩2000 for 3 < t ≤ 7 The graphs of f and g, which intersect at t = 1.617 and t = 5.076, are shown in the figure above. At time t = 0, the amount of water in the tank is 5000 gallons.
(a) How many gallons of water enter the tank during the time interval 0 ≤ t ≤ 7 ? Round your answer to the nearest gallon. (b) For 0 ≤ 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 0 ≤ 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. (a) ∫ 0 f ( t ) dt ≈ 8264 gallons 7 2: { { 1 : integral 1 : answer 1 : intervals 1 : reason (b) The amount of water in the tank is decreasing on the intervals 0 ≤ t ≤ 1.617 and 3 ≤ t ≤ 5.076 because f ( t ) < g ( t ) for 0 ≤ t < 1.617 and 3 < t < 5.076. (c) Since f ( t ) − g ( t ) changes sign from positive to negative only at t = 3, the candidates for the absolute maximum are at t = 0, 3, and 7. t (hours) gallons of water 0 3 5000 2: ⎧ 1 : identifies t = 3 as a candidate ⎪ 1 : integrand ⎪ 5 : ⎨ 1 : amount of water at t = 3 ⎪ 1 : amount of water at t = 7 ⎪ ⎩ 1 : conclusion 5000 + ∫ 0 f ( t ) dt − 250 ( 3) = 5126.591 ∫3
7 3 7 5126.591 + f ( t ) dt − 2000 ( 4 ) = 4513.807 The amount of water in the tank is greatest at 3 hours. At that time, the amount of water in the tank, rounded to the nearest gallon, is 5127 gallons. © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CALCULUS AB 2007 SCORING GUIDELINES
Question 3
x 1 2 3 4 f ( x)
6 9 10 –1 f ′( x )
4 2 –4 3 g( x)
2 3 4 6 g ′( x )
5 1 2 7 The functions f and g are differentiable for all real numbers, and g is strictly increasing. The table above gives values of the functions and their first derivatives at selected values of x. The function h is given by h( x ) = f ( g ( x ) ) − 6. (a) Explain why there must be a value r for 1 < r < 3 such that h( r ) = −5. (b) Explain why there must be a value c for 1 < c < 3 such that h′( c ) = −5. (c) Let w be the function given by w( x ) =
g ( x) ∫1 f ( t ) dt. Find the value of w′( 3) . (d) If g −1 is the inverse function of g, write an equation for the line tangent to the graph of y = g −1 ( x ) at x = 2. (a) h(1) = f ( g (1) ) − 6 = f ( 2 ) − 6 = 9 − 6 = 3 h( 3) = f ( g ( 3) ) − 6 = f ( 4 ) − 6 = −1 − 6 = −7 Since h ( 3) < −5 < h (1) and h is continuous, by the Intermediate Value Theorem, there exists a value r, 1 < r < 3, such that h ( r ) = −5. (b) h( 3) − h(1) −7 − 3 = = −5 3 −1 3 −1 Since h is continuous and differentiable, by the Mean Value Theorem, there exists a value c, 1 < c < 3, such that h′ ( c ) = −5.
⎧ 1 : h(1) and h( 3) 2: ⎨ ⎩ 1 : conclusion, using IVT h ( 3) − h (1) ⎧ ⎪1: 2: ⎨ 3 −1 ⎪ 1 : conclusion, using MVT ⎩ (c) w′ ( 3) = f ( g ( 3) ) ⋅ g ′ ( 3) = f ( 4 ) ⋅ 2 = −2 2: { 1 : apply chain rule 1 : answer (d) g (1) = 2, so g −1 ( 2 ) = 1. ( g −1 )′ ( 2 ) = 1 1 1 = = g ′ (1) 5 ′ g −1 ( 2 ) g ( ) An equation of the tangent line is y − 1 = 1 ( x − 2). 5 ⎧ 1 : g −1 ( 2 ) ⎪ ⎪ 3 : ⎨ 1 : g −1 ′ ( 2 ) ⎪ ⎪ 1 : tangent line equation ⎩ () © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CALCULUS AB 2007 SCORING GUIDELINES
Question 4
A particle moves along the xaxis with position at time t given by x( t ) = e−t sin t for 0 ≤ t ≤ 2π . (a) Find the time t at which the particle is farthest to the left. Justify your answer. (b) Find the value of the constant A for which x( t ) satisfies the equation Ax′′( t ) + x′( t ) + x( t ) = 0 for 0 < t < 2π . (a) x′( t ) = −e−t sin t + e−t cos t = e−t ( cos t − sin t ) x′( t ) = 0 when cos t = sin t. Therefore, x′( t ) = 0 on π 5π and t = . 0 ≤ t ≤ 2π for t = 4 4 The candidates for the absolute minimum are at π 5π t = 0, , , and 2π . 44
t 0 x( t ) e0 sin ( 0 ) = 0
e
− ⎧ 2 : x′( t ) ⎪ 1 : sets x′( t ) = 0 ⎪ 5: ⎨ ⎪ 1 : answer ⎪ 1 : justification ⎩ π
4 5π 4 π
4
5π 4 sin e − ( π4 ) > 0 5π sin ( ) < 0 4
5π . 4
⎧ 2 : x′′( t ) ⎪ 1 : substitutes x′′( t ) , x′( t ) , and x( t ) ⎪ 4: ⎨ into Ax′′( t ) + x′( t ) + x( t ) ⎪ ⎪ 1 : answer ⎩ 2π e−2π sin ( 2π ) = 0 The particle is farthest to the left when t = (b) x′′( t ) = −e −t ( cos t − sin t ) + e−t ( − sin t − cos t ) = −2e −t cos t
Ax′′( t ) + x′( t ) + x( t ) = A −2e −t cos t + e −t ( cos t − sin t ) + e −t sin t = ( −2 A + 1) e −t cos t =0 1 Therefore, A = . 2 ( ) © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CALCULUS AB 2007 SCORING GUIDELINES
Question 5
t (minutes) r ′( t ) (feet per minute) 0 5.7 2 4.0 5 2.0 7 1.2 11 0.6 12 0.5 The volume of a spherical hot air balloon expands as the air inside the balloon is heated. The radius of the balloon, in feet, is modeled by a twicedifferentiable function r of time t, where t is measured in minutes. For 0 < t < 12, 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 0 ≤ t ≤ 12. The radius of the balloon is 30 feet when 4 t = 5. (Note: The volume of a sphere of radius r is given by V = π r 3 . ) 3 (a) Estimate the radius of the balloon when t = 5.4 using the tangent line approximation 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 approximate ∫0 12 r ′( t ) dt. Using correct units, explain the meaning of ∫0 12 r ′( t ) dt in terms of the radius of the balloon. (d) Is your approximation in part (c) greater than or less than (a) r ( 5.4 ) ≈ r ( 5 ) + r ′( 5 ) Δt = 30 + 2 ( 0.4 ) = 30.8 ft Since the graph of r is concave down on the interval 5 < t < 5.4, this estimate is greater than r ( 5.4 ) . (b) dV 4 dr =3 π r2 dt 3 dt dV = 4π ( 30 )2 2 = 7200π ft 3 min dt t = 5 ∫0 12 r ′( t ) dt ? Give a reason for your answer. 2: { 1 : estimate 1 : conclusion with reason () ⎧ 2 : dV ⎪ 3: ⎨ dt ⎪ 1 : answer ⎩ (c) ∫0
∫0 12 r ′( t ) dt ≈ 2 ( 4.0 ) + 3 ( 2.0 ) + 2 (1.2 ) + 4 ( 0.6 ) + 1( 0.5 ) r ′( t ) dt is the change in the radius, in feet, from 2: = 19.3 ft
12 { 1 : approximation 1 : explanation t = 0 to t = 12 minutes.
(d) Since r is concave down, r ′ is decreasing on 0 < t < 12. Therefore, this approximation, 19.3 ft, is less than 1 : conclusion with reason ∫0 12 r ′( t ) dt.
1 : units in (b) and (c) Units of ft 3 min in part (b) and ft in part (c) © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). AP® CALCULUS AB 2007 SCORING GUIDELINES
Question 6
Let f be the function defined by f ( x ) = k x − ln x for x > 0, where k is a positive constant.
(a) Find f ′( x ) and f ′′( x ) . (b) For what value of the constant k does f have a critical point at x = 1 ? For this value of k, determine whether f has a relative minimum, relative maximum, or neither at x = 1. Justify your answer. (c) For a certain value of the constant k, the graph of f has a point of inflection on the xaxis. Find this value of k. (a) f ′( x ) = k 1 − 2x x ⎧ 1 : f ′( x ) 2: ⎨ ⎩ 1 : f ′′( x )
2 1 f ′′( x ) = − kx −3 4 + x −2 (b) f ′(1) = 1 k −1 = 0 ⇒ k = 2 2 1 When k = 2, f ′(1) = 0 and f ′′(1) = − + 1 > 0. 2 f has a relative minimum value at x = 1 by the Second Derivative Test. ⎧ 1 : sets f ′(1) = 0 or f ′( x ) = 0 ⎪ 1 : solves for k ⎪ 4: ⎨ ⎪ 1 : answer ⎪ 1 : justification ⎩ (c) At this inflection point, f ′′( x ) = 0 and f ( x ) = 0. f ′′( x ) = 0 ⇒ 1 4 −k + 2 =0⇒k = 32 x 4x x ln x f ( x ) = 0 ⇒ k x − ln x = 0 ⇒ k = x ⎧ 1 : f ′′( x ) = 0 or f ( x ) = 0 ⎪ 3 : ⎨ 1 : equation in one variable ⎪ 1 : answer ⎩ Therefore, ln x 4 = x x ⇒ 4 = ln x ⇒ x = e4 4 ⇒k= 2 e © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). ...
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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.
 Spring '09
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