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Unformatted text preview: AP® Calculus AB
2004 Scoring Guidelines The materials included in these files are intended for noncommercial use by
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2004 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 F ( t ) = 82 + 4sin ( 2t ) for 0 ≤ t ≤ 30, 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 30minute
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 10 ≤ t ≤ 15 ? Indicate units of
measure.
(d) What is the average rate of change of the traffic flow over the time interval 10 ≤ t ≤ 15 ? Indicate
units of measure.
30 1 : limits 3 : 1 : integrand 1 : answer (a) ∫0 (b) F ′ ( 7 ) = −1.872 or −1.873
Since F ′ ( 7 ) < 0, the traffic flow is decreasing
at t = 7. 1 : answer with reason (c) 1 15
F ( t ) dt = 81.899 cars min
5 ∫10 1 : limits 3 : 1 : integrand 1 : answer (d) F (15 ) − F (10 )
= 1.517 or 1.518 cars min 2
15 − 10 1 : answer F ( t ) dt = 2474 cars 1 : units in (c) and (d) Units of cars min in (c) and cars min 2 in (d) Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 2
Let f and g be the functions given by f ( x ) = 2 x (1 − x ) and g ( x ) = 3 ( x − 1) x for 0 ≤ x ≤ 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 = 2.
(c) Let h be the function given by h( x ) = k x (1 − x ) for 0 ≤ x ≤ 1. For each
k > 0, 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 xaxis.
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 expression that
could be used to find the value of k. (a) Area = = 1 ∫0 ( f ( x ) − g ( x ) ) dx
1 ∫0 ( 2 x (1 − x ) − 3( x − 1) x ) dx = 1.133 (b) Volume = π
= π∫ 1
0 2: ∫0 ( ( 2 − g ( x ) )
1 (( 2 − 3( x − 1) 2 ) − ( 2 − f ( x ) )2 dx ) 2 x ) − ( 2 − 2 x (1 − x ) )2 dx = 16.179 (c) Volume = { 1 : integral
1 : answer 1 : limits and constant 2 : integrand −1 each error 4: Note: 0 2 if integral not of form b c ∫ R 2 ( x ) − r 2 ( x ) dx
a 1 : answer ( 1 ∫ 0 ( h( x ) − g ( x ) ) 2 dx 3: 1 2
∫0 ( k x (1 − x ) − 3( x − 1) x ) dx = 15 { ) 2 : integrand
1 : answer Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 3 () A particle moves along the yaxis so that its velocity v at time t ≥ 0 is given by v( t ) = 1 − tan −1 et . At time t = 0, the particle is at y = −1. (Note: tan −1 x = arctan x )
(a) Find the acceleration of the particle at time t = 2.
(b) Is the speed of the particle increasing or decreasing at time t = 2 ? Give a reason for your answer.
(c) Find the time t ≥ 0 at which the particle reaches its highest point. Justify your answer.
(d) Find the position of the particle at time t = 2. Is the particle moving toward the origin or away from
the origin at time t = 2 ? Justify your answer. (a) a( 2 ) = v′( 2 ) = −0.132 or −0.133 1 : answer (b) v( 2 ) = −0.436
Speed is increasing since a( 2 ) < 0 and v( 2 ) < 0. 1 : answer with reason (c) v( t ) = 0 when tan −1 et = 1 () t = ln ( tan (1) ) = 0.443 is the only critical value for y. 1 : sets v ( t ) = 0 3 : 1 : identifies t = 0.443 as a candidate 1 : justifies absolute maximum v( t ) > 0 for 0 < t < ln ( tan (1) )
v( t ) < 0 for t > ln ( tan (1) )
y ( t ) has an absolute maximum at t = 0.443. (d) y ( 2 ) = −1 + 2 ∫0 v( t ) dt = −1.360 or −1.361 The particle is moving away from the origin since
v( 2 ) < 0 and y ( 2 ) < 0. 1 : 2 v( t ) dt
∫0 1 : handles initial condition
4: 1 : value of y ( 2 ) 1 : answer with reason Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 4
Consider the curve given by x 2 + 4 y 2 = 7 + 3x y. (a) Show that dy 3 y − 2 x
=
.
dx 8 y − 3 x (b) Show that there is a point P with xcoordinate 3 at which the line tangent to the curve at P is
horizontal. Find the ycoordinate of P.
d2y
at the point P found in part (b). Does the curve have a local maximum, a
dx 2
local minimum, or neither at the point P ? Justify your answer. (c) Find the value of (a) (b) 2 x + 8 y y′ = 3 y + 3x y′
(8 y − 3x ) y′ = 3 y − 2 x
3y − 2x
y′ =
8 y − 3x 1 : implicit differentiation
2: 1 : solves for y ′ 3y − 2x
= 0; 3 y − 2 x = 0
8 y − 3x 1 : dy = 0 dx 3: 1 : shows slope is 0 at ( 3, 2 ) 1 : shows ( 3, 2 ) lies on curve When x = 3, 3 y = 6
y=2
32 + 4⋅22 = 25 and 7 + 3⋅3⋅2 = 25
Therefore, P = ( 3, 2 ) is on the curve and the slope
is 0 at this point. (c) d 2 y ( 8 y − 3x )( 3 y ′ − 2 ) − ( 3 y − 2 x )( 8 y ′ − 3)
=
dx 2
( 8 y − 3 x )2
d 2 y (16 − 9 )( −2 )
2
=
=− .
2
2
7
dx
(16 − 9 )
Since y ′ = 0 and y ′′ < 0 at P, the curve has a local
maximum at P. At P = ( 3, 2 ) , d2y
2:
dx 2 4: d2y
at ( 3, 2 ) 1 : value of
dx 2 1 : conclusion with justification Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 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 ( x ) = x ∫−3 f ( t ) dt. (a) Find g ( 0 ) and g ′( 0 ) . (b) Find all values of x in the open interval ( −5, 4 ) at which
g attains a relative maximum. 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 x in the open interval ( −5, 4 ) at which
the graph of g has a point of inflection. (a) g ( 0) = 1 0 1 : g ( 0)
2: 1 : g ′( 0 ) 9 ∫−3 f ( t ) dt = 2 ( 3)( 2 + 1) = 2 g ′( 0 ) = f ( 0 ) = 1 (b) g has a relative maximum at x = 3.
This is the only xvalue where g ′ = f changes from
positive to negative. 1:x = 3
2: 1 : justification (c) The only xvalue where f changes from negative to
positive is x = − 4. The other candidates for the
location of the absolute minimum value are the
endpoints. 1 : identifies x = − 4 as a candidate 3 : 1 : g ( −4 ) = −1 1 : justification and answer g ( −5 ) = 0
g ( −4) =
g ( 4) = −4 ∫−3 ( f ( t ) dt = −1 ) 9
π
13 − π
+ 2−
=
2
2
2 So the absolute minimum value of g is −1. (d) x = −3, 1, 2 2 : correct values
−1 each missing or extra value Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 6
Consider the differential equation dy
= x 2 ( y − 1) .
dx (a) On the axes provided, sketch a slope field for the given
differential equation at the twelve points indicated.
(Note: Use the axes 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 xyplane. Describe all points
in the xyplane for which the slopes are positive.
(c) Find the particular solution y = f ( x ) to the given differential
equation with the initial condition f ( 0 ) = 3. (a) 1 : zero slope at each point ( x, y ) where x = 0 or y = 1 positive slope at each point ( x, y )
2: where x ≠ 0 and y > 1 1: negative slope at each point ( x, y ) where x ≠ 0 and y < 1 (b) Slopes are positive at points ( x, y )
where x ≠ 0 and y > 1. (c) 1 : description 1
dy = x 2 dx
y −1
1
ln y − 1 = x3 + C
3
y −1 = 1 : separates variables 2 : antiderivatives 1 : constant of integration
6: 1 : uses initial condition 1 : solves for y 0 1 if y is not exponential 13
C 3x
ee
13
x y − 1 = Ke 3 , K = ± eC
2 = Ke0 = K
y =1+ Note: max 3 6 [12000] if no constant of
integration
Note: 0 6 if no separation of variables 13
x
2e 3 Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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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
 Dq

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