AP Calculus AB 2004 Scoring Guidelines

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2 4 SCORING GUIDELINES Question 1 Traffic flow is defined as the rate at which cars pass through an intersection, measured in cars per minute. The traffic flow at a particular intersection is modeled by the function F defined by t () 8 4sin ( ) Ft = + for t, where F() t is measured in cars per minute and t is measured in minutes. (a) To the nearest whole number, how many cars pass through the intersection over the -minute period? (b) Is the traffic flow increasing or decreasing at t = 7? Give a reason for your answer. (c) What is the average value of the traffic flow over the time interval 1 t 15? Indicate units of measure. (d) What is the average rate of change of the traffic flow over the time interval 1 t 15? Indicate units of measure. (a) Ft () dt= 474 cars : 1 : limits 1 : integrand (b) F ( 7) = 1.87 or 1.87 Since F ( 7) <, the traffic flow is decreasing at t = 7. with reason 1 15 (c) () cars min 5 Ft dt= 1 : 1 : limits 1 : integrand (d) F( 15) F( 1) 15 1 = or cars min Units of cars min in (c) and cars min in (d) 1 : units in (c) and (d) Copyright 4 by College Entrance Eamination Board. All rights reserved.

3 4 SCORING GUIDELINES Question Let f and g be the functions given by f ( ) = ( 1 ) and g( ) = ( 1) for 1. The graphs of f and g are shown in the figure above. (a) Find the area of the shaded region enclosed by the graphs of f and g. (b) Find the volume of the solid generated when the shaded region enclosed by the graphs of f and g is revolved about the horizontal line y =. (c) Let h be the function given by h ( ) = k( 1 ) for 1. For each k >, the region (not shown) enclosed by the graphs of h and g is the base of a solid with square cross sections perpendicular to the -ais. There is a value of k for which the volume of this solid is equal to 15. Write, but do not solve, an equation involving an integral epression that could be used to find the value of k. 1 (a) Area = ( f( ) g( ) ) d 1 = ( ( 1 ) ( 1 ) ) d = 1.1 : { 1 : integral ( ) 1 1 (b) Volume = π ( g( ) ) ( f( ) ) d = π (( ( 1) ) ( ( 1 ) ) ) d 4 : = : limits and constant : integrand 1 each error Note: if integral not of form b c ( R ( ) r ( ) ) d a 1 (c) Volume = ( h ( ) g ( )) d 1 ( k( 1 ) ( 1) ) d = 15 : { : integrand Copyright 4 by College Entrance Eamination Board. All rights reserved.

4 4 SCORING GUIDELINES Question A particle moves along the y-ais so that its velocity v at time At time t =, the particle is at y = 1. (Note: tan 1 = arctan ) 1 t t is given by vt () = ( e) 1 tan. (a) Find the acceleration of the particle at time t =. (b) Is the speed of the particle increasing or decreasing at time t =? Give a reason for your answer. (c) Find the time t at which the particle reaches its highest point. Justify your answer. (d) Find the position of the particle at time t =. Is the particle moving toward the origin or away from the origin at time t =? Justify your answer. (a) a( ) = v ( ) =.1 or.1 (b) v ( ) =.46 Speed is increasing since a ( ) < and v ( ) <. with reason 1 t vt = when ( e ) (c) () tan = 1 t = ln ( tan() 1 ) =.44 is the only critical value for y. vt () > for < t < ln( tan() 1 ) vt () < for t > ln ( tan () 1 ) 1 : sets vt () = : 1 : identifies t =.44 as a candidate 1 : justifies absolute maimum yt () has an absolute maimum at t =.44. (d) y( ) = 1 + v( t) dt = 1.6 or 1.61 The particle is moving away from the origin since v ( ) < and y ( ) <. 4 : 1 : vt () dt 1 : handles initial condition 1 : value of y( ) with reason Copyright 4 by College Entrance Eamination Board. All rights reserved. 4

5 4 SCORING GUIDELINES Question 4 Consider the curve given by + 4y = 7 + y. dy y (a) Show that =. d 8y (b) Show that there is a point P with -coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P. d y (c) Find the value of at the point P found in part (b). Does the curve have a local maimum, a d local minimum, or neither at the point P? Justify your answer. (a) + 8yy = y + y ( 8y ) y = y y y = 8y 1 : implicit differentiation : 1 : solves for y (b) y 8y = ; y = When =, y = 6 y = + 4 = 5 and 7 + = 5 Therefore, P = (, ) is on the curve and the slope is at this point. dy 1 : = d : 1 : shows slope is at (, ) 1 : shows (, ) lies on curve (c) d y ( 8y )( y ) ( y )( 8y ) = d ( 8y ) d y ( 16 9)( ) At P = (, ), = =. d ( 16 9) 7 Since y = and y < at P, the curve has a local maimum at P. 4 : d y : d d y 1 : value of at (, ) d 1 : conclusion with justification Copyright 4 by College Entrance Eamination Board. All rights reserved. 5

6 4 SCORING GUIDELINES Question 5 The graph of the function f shown above consists of a semicircle and three line segments. Let g be the function given by g( ) = f( t) dt. (a) Find g ( ) and g (. ) (b) Find all values of in the open interval ( 5, 4) at which g attains a relative maimum. Justify your answer. (c) Find the absolute minimum value of g on the closed interval [ 5, 4 ]. Justify your answer. (d) Find all values of in the open interval ( 5, 4) at which the graph of g has a point of inflection. 1 9 (a) g( ) = f( t) dt = ( )( + 1) = g ( ) = f( ) = 1 : 1 : g( ) 1 : g ( ) (b) g has a relative maimum at =. This is the only -value where g = f changes from positive to negative. 1 : = : 1 : justification (c) The only -value where f changes from negative to positive is = 4. The other candidates for the location of the absolute minimum value are the endpoints. 1 : identifies = 4 as a candidate : 1 : g( 4) = 1 1 : justification and answer g( 5) = 4 g( 4) = f( t) dt = 1 g π ( 4) ( ) 9 1 π = + = So the absolute minimum value of g is 1. (d) =, 1, : correct values 1 each missing or etra value Copyright 4 by College Entrance Eamination Board. All rights reserved. 6

7 4 SCORING GUIDELINES Question 6 dy Consider the differential equation ( y 1. ) d = (a) On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the pink test booklet.) (b) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the y-plane. Describe all points in the y-plane for which the slopes are positive. (c) Find the particular solution y = f( ) to the given differential equation with the initial condition f ( ) =. (a) : 1 : zero slope at each point (, y) where = or y = 1 positive slope at each point (, y) where and y > 1 1 : negative slope at each point (, y) where and y < 1 (b) Slopes are positive at points (, y ) where and y > 1. 1 : description (c) 1 dy = d y 1 1 ln y 1 = + C y 1 = e e C y 1 = Ke, K = ± e = Ke = K y = 1+ e C 6 : 1 : separates variables : antiderivatives 1 : constant of integration 1 : uses initial condition 1 : solves for y 1 if y is not eponential Note: ma 6 [1----] if no constant of integration Note: 6 if no separation of variables Copyright 4 by College Entrance Eamination Board. All rights reserved. 7

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