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Unformatted text preview: AP® Calculus AB
2005 Scoring Guidelines
Form B The College Board: Connecting Students to College Success
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2005 SCORING GUIDELINES (Form B)
Question 1
Let f and g be the functions given by f ( x ) = 1 + sin ( 2 x ) and g ( x ) = e x 2 . Let R be the shaded region in the first quadrant enclosed by
the graphs of f and g as shown in the figure above.
(a) Find the area of R.
(b) Find the volume of the solid generated when R is revolved about the
xaxis.
(c) The region R is the base of a solid. For this solid, the cross sections
perpendicular to the xaxis are semicircles with diameters extending
from y = f ( x ) to y = g ( x ) . Find the volume of this solid. The graphs of f and g intersect in the first quadrant at
( S , T ) = (1.13569, 1.76446 ) . S ∫0 ( f ( x ) − g ( x ) ) dx
S
= ∫ (1 + sin ( 2 x ) − e x 2 ) dx
0 (a) Area = 1 : correct limits in an integral in (a), (b),
or (c) ⎧ 1 : integrand
2: ⎨
⎩ 1 : answer = 0.429 (b) Volume = π ∫0 ( ( f ( x ) )
S ⎮
= π⌠ S ⌡0 2 ) − ( g ( x ) )2 dx ((1 + sin ( 2x )) 2 ( − ex ) 22 ⎧ 2 : integrand
⎪
−1 each error
⎪
⎪
Note: 0 2 if integral not of form
3: ⎨
b
⎪
c ∫ R 2 ( x ) − r 2 ( x ) dx
a
⎪
⎪ 1 : answer
⎩ ) dx ( = 4.266 or 4.267 ) S 2
⌠ π ⎛ f ( x) − g( x) ⎞
(c) Volume = ⎮
⎜
⎟ dx
2
⎠
⌡0 2 ⎝
S ⎧ 2 : integrand
3: ⎨
⎩ 1 : answer 2 ⌠ π ⎛ 1 + sin ( 2 x ) − e x 2 ⎞
=⎮
⎟ dx
2⎜
2
⎠
⌡0 ⎝
= 0.077 or 0.078 Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 2 A water tank at Camp Newton holds 1200 gallons of water at time t = 0. During the time interval
0 ≤ t ≤ 18 hours, water is pumped into the tank at the rate W ( t ) = 95 t sin 2 ( 6t ) gallons per hour. During the same time interval, water is removed from the tank at the rate
R( t ) = 275sin 2 ( 3t ) gallons per hour. (a) Is the amount of water in the tank increasing at time t = 15 ? Why or why not?
(b) To the nearest whole number, how many gallons of water are in the tank at time t = 18 ?
(c) At what time t, for 0 ≤ t ≤ 18, is the amount of water in the tank at an absolute minimum? Show the
work that leads to your conclusion.
(d) For t > 18, no water is pumped into the tank, but water continues to be removed at the rate R( t )
until the tank becomes empty. Let k be the time at which the tank becomes empty. Write, but do not
solve, an equation involving an integral expression that can be used to find the value of k.
(a) No; the amount of water is not increasing at t = 15
since W (15 ) − R (15 ) = −121.09 < 0. (b) 1200 + 18 ∫0 (W ( t ) − R( t ) ) dt = 1309.788 ⎧ 1 : limits
⎪
3 : ⎨ 1 : integrand
⎪ 1 : answer
⎩ 1310 gallons (c) W ( t ) − R( t ) = 0
t = 0, 6.4948, 12.9748 ⎧ 1 : interior critical points
⎪ 1 : amount of water is least at
⎪
3: ⎨
t = 6.494 or 6.495
⎪
⎪
⎩ 1 : analysis for absolute minimum t (hours) gallons of water 0
6.495
12.975
18 1 : answer with reason 1200
525
1697
1310 The values at the endpoints and the critical points
show that the absolute minimum occurs when
t = 6.494 or 6.495. (d) k ∫18 R( t ) dt = 1310 ⎧ 1 : limits
2: ⎨
⎩ 1 : equation Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 3 A particle moves along the xaxis so that its velocity v at time t, for 0 ≤ t ≤ 5, is given by ( ) v( t ) = ln t 2 − 3t + 3 . The particle is at position x = 8 at time t = 0.
(a) Find the acceleration of the particle at time t = 4.
(b) Find all times t in the open interval 0 < t < 5 at which the particle changes direction. During which
time intervals, for 0 ≤ t ≤ 5, does the particle travel to the left?
(c) Find the position of the particle at time t = 2.
(d) Find the average speed of the particle over the interval 0 ≤ t ≤ 2. (a) 5
a( 4 ) = v′( 4 ) =
7 (b) v( t ) = 0 1 : answer ⎧ 1 : sets v( t ) = 0
⎪
3 : ⎨ 1 : direction change at t = 1, 2
⎪ 1 : interval with reason
⎩ 2 t − 3t + 3 = 1
t 2 − 3t + 2 = 0
( t − 2 ) ( t −1) = 0
t = 1, 2
v( t ) > 0 for 0 < t < 1
v( t ) < 0 for 1 < t < 2
v( t ) > 0 for 2 < t < 5
The particle changes direction when t = 1 and t = 2.
The particle travels to the left when 1 < t < 2. (c) ( ∫0 ln ( u − 3u + 3) du
2
s ( 2 ) = 8 + ∫ ln ( u 2 − 3u + 3) du
0
s( t ) = s( 0 ) + t = 8.368 or 8.369 (d) ) ⎧ 1 : 2 ln u 2 − 3u + 3 du
∫0
⎪
⎪
3: ⎨
1 : handles initial condition
⎪
⎪
⎩ 1 : answer 2 12
v( t ) dt = 0.370 or 0.371
2 ∫0 ⎧ 1 : integral
2: ⎨
⎩ 1 : answer Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 4 The graph of the function f above consists of three line
segments.
(a) Let g be the function given by g ( x ) = x ∫− 4 f ( t ) dt. For each of g ( −1) , g ′( −1) , and g ′′( −1) , find the
value or state that it does not exist.
(b) For the function g defined in part (a), find the
xcoordinate of each point of inflection of the graph
of g on the open interval − 4 < x < 3. Explain
your reasoning.
(c) Let h be the function given by h( x ) = 3 ∫ x f ( t ) dt. Find all values of x in the closed interval − 4 ≤ x ≤ 3 for which h( x ) = 0.
(d) For the function h defined in part (c), find all intervals on which h is decreasing. Explain your
reasoning. (a) g ( −1) = −1 1 15 ∫− 4 f ( t ) dt = − 2 ( 3)( 5) = − 2 g ′( −1) = f ( −1) = −2
g ′′( −1) does not exist because f is not differentiable
at x = −1. ⎧ 1 : g ( −1)
⎪
3 : ⎨ 1 : g ′( −1)
⎪ 1 : g ′′( −1)
⎩ (b) x =1
g ′ = f changes from increasing to decreasing
at x = 1. ⎧ 1 : x = 1 (only)
2: ⎨
⎩ 1 : reason (c) x = −1, 1, 3 2 : correct values
−1 each missing or extra value (d) h is decreasing on [ 0, 2]
h′ = − f < 0 when f > 0 ⎧ 1 : interval
2: ⎨
⎩ 1 : reason Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 5 Consider the curve given by y 2 = 2 + xy.
(a) Show that dy
y
=
.
dx 2 y − x (b) Find all points ( x, y ) on the curve where the line tangent to the curve has slope 1
.
2 (c) Show that there are no points ( x, y ) on the curve where the line tangent to the curve is horizontal.
(d) Let x and y be functions of time t that are related by the equation y 2 = 2 + xy. At time t = 5, the
dy
dx
= 6. Find the value of
at time t = 5.
value of y is 3 and
dt
dt (a) (b) 2 y y′ = y + x y′
( 2 y − x ) y′ = y
y
y′ =
2y − x ⎧ 1 : implicit differentiation
2: ⎨
⎩ 1 : solves for y ′ y
1
=
2y − x 2
2y = 2y − x
x=0
y=± 2 y
1
⎧1:
=
⎪
2y − x 2
2: ⎨
⎪ 1 : answer
⎩ y
=0
2y − x
y=0
The curve has no horizontal tangent since ⎧1: y = 0
2: ⎨
⎩ 1 : explanation ( 0,
(c) 2 ) , ( 0, − 2 ) 02 ≠ 2 + x ⋅ 0 for any x. (d) When y = 3, 32 = 2 + 3 x so x = 7
.
3 ⎧ 1 : solves for x
⎪
3 : ⎨ 1 : chain rule
⎪ 1 : answer
⎩ dy dy dx
y
dx
=
⋅
=
⋅
2 y − x dt
dt
dx dt
3
9 dx
dx
At t = 5, 6 =
⋅
=
⋅
7 dt 11 dt
6−
3
dx
22
=
dt t = 5
3 Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 6 dy − xy 2
. Let
=
dx
2
y = f ( x ) be the particular solution to this differential
equation with the initial condition f ( −1) = 2. Consider the differential equation (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 test booklet.)
(b) Write an equation for the line tangent to the graph of
f at x = −1.
(c) Find the solution y = f ( x ) to the given differential equation with the initial condition f ( −1) = 2.
(a) ⎧ 1 : zero slopes
2: ⎨
⎩ 1 : nonzero slopes − ( −1) 4
=2
2
y − 2 = 2 ( x + 1) (b) Slope = (c) 1 : equation 1
x
dy = − dx
2
2
y ⎧ 1 : separates variables
⎪ 2 : antiderivatives
⎪
6 : ⎨ 1 : constant of integration
⎪ 1 : uses initial condition
⎪
⎪
⎩ 1 : solves for y 1
x2
=−
+C
4
y
1
1
1
− = − + C; C = −
2
4
4
1
4
y= 2
=2
1
x
x +1
+
4
4
− Note: max 3 6 [12000] if no
constant of integration
Note: 0 6 if no separation of variables Copyright © 2005 by College 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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