AP' CALCULUS AB 2002 SCORING GUIDELINES
APR5 - Solutions
Question 1
Let f and g be the functions given by f(x) = eZ and g(x) = lnx.
(a) Find the area of the region enclosed by the graphs of f and g between x
1
=-
2
and x = 1.
(b) Find the volume of the solid generated when the region enclosed by the graphs of f and g between
1
x = - and x = 1 is revolved about the line y = 4.
2
(c) Let h be the function given by h(x) = f(x) - g(x). Find the absolute minimum value of h(x) on the
1
closed interval - < x 5 1, and find the absolute maximum value of h(x) on the closed interval
2 1
- < x 5 1. Show the analysis that leads to your answers.
2 -
(b) Volume =
TJ'( ( 4 - lnx)' - ( 4 - eZ)')dx
?4
I I
I
1 : limits and constant
2 : integrand
< -1 > each error
Note: 0 / 2 if not of the form
- 1 : considers h'(x) = 0
1 : identifies critical point
3 .
C Q S ED
~~JTF-K~AL
wt7H 0 D
and endpoints as candidates
1 : answers
Absolute minimum value and absolute
maximum value occur a t the critical point or
at the endpoints.
Note: Errors in computation come off
the third point.
APR5
Page 1
The absolute minimum is 2.330.
The absolute maximum is 2.718.
Copyright O 2002 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
2
AP@CALCULUS A 6 2002 SCORING GUIDELINES
APR5 - Solutions
Question 2
The rate a t which people enter a n amusement park on a given day is modeled by the function E defined by
15600
E(t) =
(t2 - 24t 160)'
+
The rate a t which people leave the same amusement park on the same day is modeled by the function L defined by
Both E ( t ) and L ( t ) are measured in people per hour and time t is measured in hours after midnight. These
functions are valid for 9 5 t 5 23, the hours during which the park is open. At time t = 9, there are no people in
the park.
(a) How many people have entered the park by 5:00 P.M. ( t = 17 )? Round answer t o the nearest whole number.
(b) The price of admission t o the park is $15 until 5:00 P.M. ( t = 17 ). After 5:00 P.M., the price of admission to
the park is $11. How many dollars are collected from admissions to the park on the given day? Round your
answer t o the nearest whole number.
(c) Let ~ ( t =) J ' ( E ( ~ )
-
9
L(x))dx for 9 5 t 5 23. The value of H ( 1 7 ) to the nearest whole number is 3725.
Find the value of H1(17) and explain the meaning of H (17) and H1(17) in the context of the park.
(d) At what time t, for 9 5 t 5 23, does the model predict that the number of people in the park is a maximum?
(a)
J1'E ( t )dt = 6004.270
3
6004 people entered the park by 5 pm.
t )
(b) 15JL7~ ( dt
9
+ 11JZ3~ ( tdt) = 104048.165
I
1 : limits
1 : integrand
1 : answer
1 : setup
17
The amount collected was $104,048.
or
l:
~ ( tdt) = 1271.283
1271 people entered the park between 5 pm and
11 pm, so the amount collected was
$15. (6004)
+ $11
(1271) = $104,041.
- 1 : value of H1(17)
(c) H1(17)= E(17) - L(17) = -380.281
2 : meanings
There were 3725 people in the park at t = 17.
The number of people in the park was decreasing
3 .
1 : meaning of H(17)
at the rate of approximately 380 people/hr at
1 : meaning of H1(17)
time t = 17.
< -1 > if
no reference to t =APR5
17
1 : E ( t ) - L(t) = 0
( d ) H1(t)= E ( t ) - L(t) = 0
t = 15.794 or 15.795
1 : answer
Copyright O 2002 by College Entrance Examination Board. All rights reserved.
Advanced Placement Proglam and AP are registered trademarks of the College Entrance Examination Board.
3
Page 2
AP@CALCULUS AB 2002 SCORING GUIDELINES
Question 3
APR5 - Solutions
An object moves along the s a x i s with initial position x(0) = 2. The velocity of the object a t time t 2 0 is given
by v ( t ) = sin
(a) What is the acceleration of the object a t time t = 4?
(b) Consider the following two statements.
< t < 4.5, the velocity of the object is decreasing.
For 3 < t < 4.5, the speed of the object is increasing.
For 3
Statement I:
Statement 11:
Are either or both of these statements correct? For each statement provide a reason why it is correct or not
correct.
(c) What is the total distance traveled by the object over the time interval 0 5 t 5 4 ?
(d) What is the position of the object at time t = 4 ?
(a) a(4) = v1(4)=
1 : answer
3
(b) On 3 < t < 4.5:
7r
7r
a(t) = v l ( t ) = -cos
t <o
3
3
Statement 1 is correct since a ( t ) < 0 .
Statement I1 is correct since v ( t ) < 0 and
a(t) < 0 .
1 : I correct, with reason
(- )
1 : I1 correct
1 : reason for I1
k4I
(c) Distance =
v(t)ldt = 2.387
OR
1:
I
limits of 0 and 4 on an integral
of v ( t ) or I v ( t )(
or
uses x(0) and x(4) to compute
distance
1 : handles change of direction a t
v ( t ) = 0 when t = 3
student's turning point
1 : answer
011 if incorrect turning point or
,
(d) x(4) = x(0)
+
4
no turning point
1 : integral
v ( t ) d t = 3.432
1 : answer
OR
1
2
{
I I
APR5
Page 3
1 : answer
011 if no constant of integration
Copyright O 2002 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.