_ap06_calcab_formb_sg

_ap06_calcab_formb_sg - AP® Calculus AB 2006 Scoring...

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Unformatted text preview: AP® Calculus AB 2006 Scoring Guidelines Form B The College Board: Connecting Students to College Success 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 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 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. © 2006 The College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Board. Admitted Class Evaluation Service, CollegeEd, connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All other products and services may be trademarks of their respective owners. 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 2006 SCORING GUIDELINES (Form B) Question 1 x3 x 2 x − − + 3cos x. Let R 4 3 2 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 x = 0, as shown above. Let f be the function given by f ( x ) = (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the horizontal line y = −2. (c) Write, but do not evaluate, an integral expression that can be used to find the area of S. For x < 0, f ( x ) = 0 when x = −1.37312. Let P = −1.37312. (a) Area of R = 0 ∫ P f ( x ) dx = 2.903 (b) Volume = π ∫ 0 P 2: ( ( f ( x ) + 2)2 − 4) dx = 59.361 (c) The equation of the tangent line l is y = 3 − A ⌡0 1 : integral 1 : answer ⎧ 1 : limits and constant ⎪ 4 : ⎨ 2 : integrand ⎪ 1 : answer ⎩ 1 x. 2 The graph of f and line l intersect at A = 3.38987. ⎮ Area of S = ⌠ { ⎧ 1 : tangent line ⎪ 3 : ⎨ 1 : integrand ⎪ 1 : limits ⎩ ((3 − 1 x ) − f ( x )) dx 2 © 2006 The College 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 2006 SCORING GUIDELINES (Form B) Question 2 Let f be the function defined for x ≥ 0 with f ( 0 ) = 5 and f ′, the () first derivative of f, given by f ′( x ) = e( − x 4 ) sin x 2 . The graph of y = f ′( x ) 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 1.7 < x < 1.9. Explain your reasoning. (b) On the interval 0 ≤ x ≤ 3, find the value of x at which f has an absolute maximum. Justify your answer. (c) Write an equation for the line tangent to the graph of f at x = 2. (a) On the interval 1.7 < x < 1.9, f ′ is decreasing and thus f is concave down on this interval. (b) f ′( x ) = 0 when x = 0, π , 2π , 3π , K On [0, 3] f ′ changes from positive to negative only at π . The absolute maximum must occur at x = π or at an endpoint. f ( 0) = 5 f ( π ) = f (0) + f ( 3) = f ( 0 ) + ∫0 π 2: { 1 : answer 1 : reason ⎧ 1 : identifies π and 3 as candidates ⎪ - or ⎪ indicates that the graph of f ⎪ 3: ⎨ increases, decreases, then increases ⎪ ⎪ 1 : justifies f ( π ) > f ( 3) ⎪ ⎩ 1 : answer f ′( x ) dx = 5.67911 3 ∫ 0 f ′( x ) dx = 5.57893 This shows that f has an absolute maximum at x = π . (c) f ( 2) = f ( 0) + 2 ∫ 0 f ′( x ) dx = 5.62342 ⎧ 2 : f ( 2 ) expression ⎪ 1 : integral ⎪ 1 : including f ( 0 ) term 4: ⎨ ⎪ 1 : f ′( 2 ) ⎪ ⎩ 1 : equation f ′( 2 ) = e−0.5 sin ( 4 ) = −0.45902 y − 5.623 = ( −0.459 )( x − 2 ) © 2006 The College 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 2006 SCORING GUIDELINES (Form B) Question 3 The figure above is the graph of a function of x, which models the height of a skateboard ramp. The function meets the following requirements. (i) At x = 0, the value of the function is 0, and the slope of the graph of the function is 0. (ii) At x = 4, the value of the function is 1, and the slope of the graph of the function is 1. (iii) Between x = 0 and x = 4, the function is increasing. (a) Let f ( x ) = ax 2 , where a is a nonzero constant. Show that it is not possible to find a value for a so that f meets requirement (ii) above. x2 , where c is a nonzero constant. Find the value of c so that g meets requirement (ii) 16 above. Show the work that leads to your answer. (c) Using the function g and your value of c from part (b), show that g does not meet requirement (iii) above. (b) Let g ( x ) = cx3 − xn , where k is a nonzero constant and n is a positive integer. Find the values of k and n so that k h meets requirement (ii) above. Show that h also meets requirements (i) and (iii) above. (d) Let h( x ) = (a) f ( 4 ) = 1 implies that a = implies that a = (b) 1 and f ′( 4 ) = 2a ( 4 ) = 1 16 1 . Thus, f cannot satisfy (ii). 8 1 . 32 2 ( 4) 1 1 1 When c = , g ′( 4 ) = 3c ( 4 )2 − =3 (16 ) − = 1 16 32 2 32 g ( 4 ) = 64c − 1 = 1 implies that c = ⎧ 1 : a = 1 or a = 1 ⎪ 2: ⎨ 16 8 ⎪ 1 : shows a does not work ⎩ 1 : value of c () (c) (d) g ′( x ) = 32 x 1 x− = x ( 3x − 4 ) 32 8 32 4 g ′( x ) < 0 for 0 < x < , so g does not satisfy (iii). 3 ⎧ 1 : g ′( x ) 2: ⎨ ⎩ 1 : explanation 4n = 1 implies that 4n = k . k n 4n −1 n4n −1 n h′ ( 4 ) = = = = 1 gives n = 4 and k = 44 = 256. n k 4 4 ⎧ 4n ⎪1: k =1 ⎪ n -1 ⎪ 4 : ⎨ 1 : n4 =1 k ⎪ ⎪ 1 : values for k and n ⎪ 1 : verifications ⎩ h ( 4) = x4 ⇒ h ( 0 ) = 0. 256 4 x3 ⇒ h′ ( 0 ) = 0 and h′ ( x ) > 0 for 0 < x < 4. h′ ( x ) = 256 h( x) = © 2006 The College 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 2006 SCORING GUIDELINES (Form B) Question 4 The rate, in calories per minute, at which a person using an exercise machine burns calories is modeled by the function 1 3 f. In the figure above, f ( t ) = − t 3 + t 2 + 1 for 4 2 0 ≤ t ≤ 4 and f is piecewise linear for 4 ≤ t ≤ 24. (a) Find f ′ ( 22) . Indicate units of measure. (b) For the time interval 0 ≤ t ≤ 24, at what time t is f increasing at its greatest rate? Show the reasoning that supports your answer. (c) Find the total number of calories burned over the time interval 6 ≤ t ≤ 18 minutes. (d) The setting on the machine is now changed so that the person burns f ( t ) + c calories per minute. For this setting, find c so that an average of 15 calories per minute is burned during the time interval 6 ≤ t ≤ 18. (a) f ′( 22 ) = 15 − 3 = −3 calories/min/min 20 − 24 1 : f ′( 22 ) and units (b) f is increasing on [ 0, 4] and on [12, 16]. 15 − 9 3 = since f has On (12, 16 ) , f ′( t ) = 16 − 12 2 constant slope on this interval. 3 On ( 0, 4 ) , f ′( t ) = − t 2 + 3t and 4 3 f ′′( t ) = − t + 3 = 0 when t = 2. This is where f ′ 2 has a maximum on [ 0, 4] since f ′′ > 0 on ( 0, 2 ) and f ′′ < 0 on ( 2, 4 ) . ⎧ 1 : f ′ on ( 0, 4 ) ⎪ 1 : shows f ′ has a max at t = 2 on ( 0, 4 ) ⎪ 4: ⎨ ⎪ 1 : shows for 12 < t < 16, f ′( t ) < f ′( 2 ) ⎪ 1 : answer ⎩ On [ 0, 24] , f is increasing at its greatest rate when 3 t = 2 because f ′( 2 ) = 3 > . 2 1 ( 4 )( 9 + 15 ) + 2 (15 ) 2 = 132 calories 2: { 1 : method 1 : answer 1 18 ( f ( t ) + c ) dt = 15. 12 6 This means 132 + 12c = 15(12). So, c = 4. (c) 2: { 1 : setup 1 : value of c 18 ∫6 f ( t ) dt = 6 ( 9 ) + (d) We want ∫ OR 132 = 11 calories/min. 12 Adding c to f ( t ) will shift the average by c. So c = 4 to get an average of 15 calories/min. Currently, the average is © 2006 The College 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 2006 SCORING GUIDELINES (Form B) Question 5 Consider the differential equation dy = ( y − 1) 2 cos (π x ) . dx (a) On the axes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the axes provided in the exam 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 ( x ) to the differential equation with the initial condition f (1) = 0. (a) 2: (b) The line y = 1 satisfies the differential equation, so c = 1. (c) 1 dy = cos (π x ) dx ( y − 1)2 − ( y − 1)−1 = 1 π sin (π x ) + C 1: c =1 Note: max 3 6 [1-2-0-0-0] if no constant of integration Note: 0 6 if no separation of variables π 1 1 = sin (π x ) + 1 1− y π π 1 : zero slopes 1 : all other slopes ⎧ 1 : separates variables ⎪ 2 : antiderivatives ⎪ 6 : ⎨ 1 : constant of integration ⎪ 1 : uses initial condition ⎪ ⎩ 1 : answer 1 1 = sin (π x ) + C 1− y π 1 1 = sin (π ) + C = C 1− y { = sin (π x ) + π y =1− π for −∞ < x < ∞ sin (π x ) + π © 2006 The College 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 2006 SCORING GUIDELINES (Form B) Question 6 t (sec) v( t ) ( ft sec) ( ft a( t ) sec2 0 15 25 30 35 50 60 –20 –30 –20 –14 –10 0 10 1 5 2 1 2 4 2 ) A car travels on a straight track. During the time interval 0 ≤ t ≤ 60 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. (a) Using appropriate units, explain the meaning of 60 ∫30 v( t ) dt in terms of the car’s motion. Approximate v( t ) dt using a trapezoidal approximation with the three subintervals determined by the table. (b) Using appropriate units, explain the meaning of of 60 ∫ 30 30 ∫0 30 ∫0 a( t ) dt in terms of the car’s motion. Find the exact value a( t ) dt. (c) For 0 < t < 60, must there be a time t when v( t ) = −5 ? Justify your answer. (d) For 0 < t < 60, must there be a time t when a( t ) = 0 ? Justify your answer. (a) 60 ∫ 30 v( t ) dt is the distance in feet that the car travels 2: { 1 : explanation 1 : value 2: { 1 : explanation 1 : value from t = 30 sec to t = 60 sec. Trapezoidal approximation for A= (b) 30 ∫0 60 ∫30 v( t ) dt : 1 1 1 (14 + 10 ) 5 + (10 )(15) + (10 )(10 ) = 185 ft 2 2 2 a( t ) dt is the car’s change in velocity in ft/sec from t = 0 sec to t = 30 sec. 30 ∫0 a( t ) dt = 30 ∫0 v′( t ) dt = v ( 30 ) − v ( 0 ) = −14 − ( −20 ) = 6 ft/sec (c) Yes. Since v( 35 ) = −10 < −5 < 0 = v( 50 ) , the IVT guarantees a t in ( 35, 50 ) so that v( t ) = −5. ⎧ 1 : v( 35 ) < −5 < v( 50 ) 2: ⎨ ⎩ 1 : Yes; refers to IVT or hypotheses (d) Yes. Since v( 0 ) = v( 25 ) , the MVT guarantees a t in ( 0, 25 ) so that a( t ) = v′( t ) = 0. ⎧ 1 : v( 0 ) = v( 25 ) 2: ⎨ ⎩ 1 : Yes; refers to MVT or hypotheses Units of ft in (a) and ft/sec in (b) 1 : units in (a) and (b) © 2006 The College 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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