AP Calculus BC 8 Scoring Guidelines 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 9, the association is composed of more than 5, schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents,, high schools, and,5 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 ecellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns. 8 The College Board. All rights reserved. College Board, AP Central, Advanced Placement Program, AP, 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. 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. AP Central is the online home for AP teachers: apcentral.collegeboard.com.
8 SCORING GUIDELINES Question Let R be the region bounded by the graphs of y = sin( π ) and y =, as shown in the figure above. (a) Find the area of R. (b) The horizontal line y = splits the region R into two parts. Write, but do not evaluate, an integral epression for the area of the part of R that is below this horizontal line. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the -ais is a square. Find the volume of this solid. (d) The region R models the surface of a small pond. At all points in R at a distance from the y-ais, the depth of the water is given by h ( ) =. Find the volume of water in the pond. (a) sin ( π ) = at = and = Area ( ( ) ( )) = sin π d = : : limits : integrand (b) = at r =.59889 and s =.6759 s The area of the stated region is ( ( )) r d : { : limits : integrand (c) Volume = ( sin ( π ) ( )) d = 9.978 : { : integrand (d) Volume = ( )( sin ( π ) ( )) d = 8.69 or 8.7 : { : integrand 8 The College Board. All rights reserved.
8 SCORING GUIDELINES Question t (hours) 7 8 9 Lt ()(people) 56 76 6 5 8 Concert tickets went on sale at noon ( t = ) and were sold out within 9 hours. The number of people waiting in line to purchase tickets at time t is modeled by a twice-differentiable function L for t 9. Values of Lt () at various times t are shown in the table above. (a) Use the data in the table to estimate the rate at which the number of people waiting in line was changing at 5: P.M. ( t = 5.5 ). Show the computations that lead to your answer. Indicate units of measure. (b) Use a trapezoidal sum with three subintervals to estimate the average number of people waiting in line during the first hours that tickets were on sale. (c) For t 9, what is the fewest number of times at which L () t must equal? Give a reason for your answer. (d) The rate at which tickets were sold for t 9 is modeled by rt () = 55te t tickets per hour. Based on the model, how many tickets were sold by P.M. ( t =, ) to the nearest whole number? L( 7) L( ) 5 6 (a) L ( 5.5) = = 8 people per hour 7 (b) The average number of people waiting in line during the first hours is approimately L( ) + L( ) L() ( ) ( ) ( ) ( ) + L L ( ) + L ( ) + + = 55.5 people (c) L is differentiable on [, 9 ] so the Mean Value Theorem implies L () t > for some t in (, ) and some t in (, 7 ). Similarly, L () t < for some t in (, ) and some t in ( 7, 8 ). Then, since L is continuous on [, 9 ], the Intermediate Value Theorem implies that L () t = for at least three values of t in [, 9 ]. OR The continuity of L on [, ] implies that L attains a maimum value there. Since L( ) > L( ) and L( ) > L( ), this maimum occurs on (, ). Similarly, L attains a minimum on (, 7 ) and a maimum on (, 8 ). L is differentiable, so L () t = at each relative etreme point on (, 9 ). Therefore L () t = for at least three values of t in [, 9 ]. [Note: There is a function L that satisfies the given conditions with L () t = for eactly three values of t.] (d) rt () dt= 97.78 There were approimately 97 tickets sold by P.M. : { : estimate : units : trapezoidal sum : { : : : considers change in sign of L : analysis : conclusion OR : considers relative etrema of L on (, 9) : analysis : conclusion : { : integrand : limits and answer 8 The College Board. All rights reserved.
8 SCORING GUIDELINES Question h ( ) h ( ) h ( ) h ( ) ( h ) ( ) 99 8 8 8 88 8 58 9 7 75 8 8 6 5 6 Let h be a function having derivatives of all orders for >. Selected values of h and its first four derivatives are indicated in the table above. The function h and these four derivatives are increasing on the interval. (a) Write the first-degree Taylor polynomial for h about = and use it to approimate h (.9 ). Is this approimation greater than or less than h (.9 )? Eplain your reasoning. (b) Write the third-degree Taylor polynomial for h about = and use it to approimate h (.9 ). (c) Use the Lagrange error bound to show that the third-degree Taylor polynomial for h about = approimates h (.9) with error less than. (a) P ( ) = 8 + 8( ), so h(.9) P (.9) = 67. P (.9) < h (.9) since h is increasing on the interval. : P ( ) : : P (.9) : P(.9) < h(.9 ) with reason 88 8 (b) P ( ) = 8 + 8( ) + ( ) + ( ) 6 8 h(.9) P (.9) = 67.988 : : P ( ) : P (.9) (c) The fourth derivative of h is increasing on the interval (, so ) 58 ma h ( ) =..9 9 58.9 Therefore, h(.9) P (.9) 9! =.77 < : form of Lagrange error estimate : { : reasoning 8 The College Board. All rights reserved.
8 SCORING GUIDELINES Question A particle moves along the -ais so that its velocity at time t, for t 6, is given by a differentiable function v whose graph is shown above. The velocity is at t =, t =, and t = 5, and the graph has horizontal tangents at t = and t =. The areas of the regions bounded by the t-ais and the graph of v on 5, 6 are 8,, and, respectively. At time t =, the particle is at =. the intervals [, ], [, 5 ], and [ ] (a) For t 6, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer. (b) For how many values of t, where t 6, is the particle at = 8? Eplain your reasoning. (c) On the interval < t <, is the speed of the particle increasing or decreasing? Give a reason for your answer. (d) During what time intervals, if any, is the acceleration of the particle negative? Justify your answer. (a) Since vt () < for < t < and 5 < t < 6, and vt () > for < t < 5, we consider t = and t = 6. ( ) = + v( t) dt = 8 = 6 ( 6) = + v( t) dt = 8 + = 9 Therefore, the particle is farthest left at time t = when its position is ( ) =. (b) The particle moves continuously and monotonically from ( ) = to ( ) =. Similarly, the particle moves continuously and monotonically from ( ) = to ( 5) = 7 and also from ( 5) = 7 to ( 6) = 9. By the Intermediate Value Theorem, there are three values of t for which the particle is at t () = 8. (c) The speed is decreasing on the interval < t < since on this interval v < and v is increasing. (d) The acceleration is negative on the intervals < t < and < t < 6 since velocity is decreasing on these intervals. : identifies t = as a candidate 6 : : considers vt () dt : conclusion : : positions at t =, t = 5, and t = 6 : description of motion : conclusion with reason : { : justification 8 The College Board. All rights reserved.
8 SCORING GUIDELINES Question 5 The derivative of a function f is given by f ( ) = ( ) e for >, and f () = 7. (a) The function f has a critical point at =. At this point, does f have a relative minimum, a relative maimum, or neither? Justify your answer. (b) On what intervals, if any, is the graph of f both decreasing and concave up? Eplain your reasoning. (c) Find the value of f (. ) (a) f ( ) < for < < and f ( ) > for > Therefore, f has a relative minimum at =. : minimum at = : : justification (b) f ( ) = e + ( ) e = ( ) e f ( ) > for > : f ( ) : with reason f ( ) < for < < Therefore, the graph of f is both decreasing and concave up on the interval < <. (c) f ( ) = f( ) + f ( ) d = 7 + ( ) e d u = dv = e d du = d v = e f ( ) = 7 + ( ) e e d = 7 + (( ) e e ) = 7 + e e : : uses initial condition : integration by parts 8 The College Board. All rights reserved.
8 SCORING GUIDELINES Question 6 dy y Consider the logistic differential equation = ( 6 y). Let y = f() t be the particular solution to the dt 8 differential equation with f ( ) = 8. (a) A slope field for this differential equation is given below. Sketch possible solution curves through the points (, ) and (, 8 ). (Note: Use the aes provided in the eam booklet.) (b) (c) (d) (a) Use Euler s method, starting at t = with two steps of equal size, to approimate f (). Write the second-degree Taylor polynomial for f about t =, and use it to approimate f (). What is the range of f for t? : : solution curve through (,8) : solution curve through (,) (b) ( ) ( )( ) () 7 ( )( ) f 8 + = 7 7 5 f + = 8 6 (c) d y = dy ( 6 y) + y ( dy ) dt 8 dt 8 dt dy 8 f( ) = 8; f ( ) = = ( 6 8) = ; and dt t= 8 d y 8 5 f ( ) = = ( )( ) + ( ) = dt 8 8 t= The second-degree Taylor polynomial for f about 5 t = is P () t = 8 t + t. 9 f() P () = (d) The range of f for t is 6 < y 8. : : Euler s method with two steps : approimation of f () d y : : dt : second-degree Taylor polynomial : approimation of f () 8 The College Board. All rights reserved.