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© 2011 www.mastermathmentor.com - 24 - Demystifying the MC AB Calculus Exam
I. Horizontal and Vertical Tangent Lines
How to find them: You need to work with
!
" f x( ) , the derivative of function f. Express
!
" f x( ) as a fraction. Horizontal tangent lines: set
!
" f x( ) = 0 and solve for values of x in the domain of f. Vertical tangent lines: find values of x where
!
" f x( ) is undefined (the denominator of
!
" f x( ) = 0 ). In both cases, to find the point of tangency, plug in the x values you found back into the function f. However, if both the numerator and denominator of
!
" f x( ) are simultaneously zero, no conclusion can be made about tangent lines. These types of problems go well with implicit differentiation.
47. The graph of
!
" f , the derivative of f is shown to the right for -2 ≤ x < 6. At what values of x does f have a horizontal tangent line? A. x = 0 only B. x = -1, x = 1, x = 4 C. x = 2 only D. x = -2, x = 2, x = 6 E. x = -2, x = 0, x = 2, x = 6
!
E. Horizontal tangent lines occur when " f x( ) = 0.
This occurs at x = #2, x = 0, x = 2,x = 6
48. At which points is the tangent line to the curve
!
8x2
+ 2y2
= 6xy +14 vertical? I. (-2, -3) II (3, 8) III. (4, 6) A. I only B. II only C. III only D. I and II only E. I and III only
!
A. 8x 2 + 2y 2 = 6xy +14
16x + 4ydy
dx= 6x
dy
dx+ 6y"
dy
dx=
6y #16x
4y # 6x=
3y # 8x
2y # 3x
Tangent line vertical when dy
dx is undefined so 2y = 3x.
#2,#3( ) is on the curve but 4,6( ) is not.
49. How many horizontal (H) and vertical (V) tangent lines does the graph of
!
x2
=4
3y3" y
4 have?
A. 2 H, 2V B. 1 H, 2V C. 1 H, 3V D. 0 H, 1V E. 0 H, 2V
!
C. 2x = 4y 2" 4y 3( )
dy
dx#
dy
dx=
2x
4y 2 1" y( )
There is a horizontal tangent at x =1 and vertical tangent lines at y = 0 and y =1.
But when y =1, there are 2 possible values for x ± 13( ) so there are 3 vertical tangents.
© 2011 www.mastermathmentor.com - 25 - Demystifying the MC AB Calculus Exam
J. Related Rates This is a difficult topic to represent. There are many types of questions that could be interpreted as related rates problems. It can be argued that straight-line motion problems fall under the category of related rates. So do the types of questions that interpret a derivative as a rate of change. In this section, we will only focus on questions that use geometric shapes with different quantities changing values. What you are finding: In related rates problems, any quantities, given or asked to calculate, that are changing are derivatives with respect to time. If you are asked to find how a volume is changing, you are
being asked for
!
dV
dt. Typically in related rates problems, you need to write the formula for a quantity in
terms of a single variable and differentiate that equation with respect to time, remembering that you are actually performing implicit differentiation. If the right side of the formula contains several variables, typically you need to link them somehow. Constants may be plugged in before the differentiation process but variables that are changing may only be plugged in after the differentiation process.
50. A tube is in the shape of a right circular cylinder. The height is increasing at the rate of 1.5 inches/sec while the radius is decreasing that the rate of 0.5 inches/sec. At the time that the radius is 4 inches and the length is 14 inches, how is the lateral surface area of the tube changing? (The lateral surface area of a right circular cylinder with radius r and height h is 2πrh).
A. Decreasing at
!
1 in2
sec B. Increasing at
!
1 in2
sec C. Decreasing at
!
2! in2
sec D. Increasing at
!
26! in2
sec E. Increasing at
!
38! in2
sec
!
C. dA
dt= 2" r
dh
dt+ h
dh
dt
#
$ %
&
' ( = 2" 4 1.5( ) +14 )0.5( )[ ] = 2" )1( ) = )2" in
2sec
51. A sphere with radius r has volume
!
4
3"r
2 and surface area
!
4"r2 . At a certain instant the rate of
decrease of its volume is twice the rate of decrease of its surface area. What is the radius of the sphere at that point in time?
A.
!
1
4 B.
!
1
2 C. 1 D. 2 E. 4
!
E. 4"r2 dr
dt= 2 8"r( )
dr
dt# 4"r =16" # r = 4.
52. A large block of ice in the shape of a cube is melting. All sides of the cube melt at the same rate. At the
time that the block is s feet on each side, its surface area is decreasing at the rate of
!
24 ft2
hr . At what rate is the volume of the block decreasing at that time?
A.
!
12s ft3
hr B.
!
6s ft3
hr C.
!
4s ft3
hr D.
!
2s ft3
hr E.
!
s ft3
hr
!
B. A = 6s2 "
dA
dt=12s
ds
dt"#24 =12s
ds
dt"
ds
dt=#2
s ft hr
V = s3 "
dV
dt= 3s
2 ds
dt= 3s
2 #2
s
$
% &
'
( ) = #6s ft
3hr
© 2011 www.mastermathmentor.com - 26 - Demystifying the MC AB Calculus Exam
53. Matthew is visiting Gregory at his home on North Street. Shortly after Matthew leaves, Gregory realizes that Matthew left his wallet and begins to chase him. When Gregory is 3 miles from the
!
90° intersection along North Street traveling at 40 mph towards the intersection, Matthew is 4 miles along East street traveling away from the intersection at 25 mph. At that time, how fast is the distance between the two men changing?
A. getting closer at 4 mph B. getting further away at 44 mph C. getting closer at 44 mph D. Getting closer at 5 mph E. getting closer at 7 mph
!
A. Let s be the distance between them. s2 = g2 + m2 by Pythag Thm, s = 5
2sds
dt= 2g
dg
dt+ 2m
dm
dt" 5
ds
dt= 3 #40( ) + 4 25( )
ds
dt= #4 mph
54. A right triangle has legs a and b and hypotenuse c. The lengths of leg a and leg b are changing but at a
certain instant, the area of the triangle is not changing. Which statement must be true?
A. a = b B.
!
da
dt=db
dt C.
!
da
dt= "
db
dt D.
!
ada
dt= "b
db
dt E.
!
adb
dt= "b
da
dt
!
E. A =1
2ab"
dA
dt=
1
2adb
dt+ b
da
dt
#
$ %
&
' ( = 0
adb
dt= )b
da
dt
55. A lighthouse is 2 km from a point P along a straight shoreline and its
light makes 1 revolution per minute. How fast, in km/min, is the beam of light moving along the shoreline when it is 1 km from point P?
A. 2.5 B.
!
10" C.
!
20" D.
!
5" E. 10
!
D. tan" =x
2
sec2"d"
dt=
1
2
dx
dt#
dx
dt= 2
5
2
$
% &
'
( )
2
2*( ) = 5* km min
© 2011 www.mastermathmentor.com - 27 - Demystifying the MC AB Calculus Exam
56. (Calc) An isosceles triangle with its two equal sides of 4 inches has the angle between them changing at the rate of
!
10° min . How fast is the 3rd side of the triangle changing, in inches/min, when the angle between the two given sides is
!
60°?
A. 0.605 B. 17.321 C. 34.641 D. 1.209 E. 8.660
!
D. Draw an angle bisector creating a 60° angle and a right triangle.
!!!!!!Call the opposite side x. If the angle is changing at 10° min, half the
!!!!!!angle is changing at the rate of 5° min =10"
180
sin# =x
4$ cos#
d#
dt=
1
4
dx
dt or cos30°
10"
180
%
& '
(
) * =
1
4
dx
dt$
dx
dt= .605
!!!!!But that represents half of the entire base of the triangle. So the base is changing at 1.209 in min.
!!!!!!Note : the problem can be attacked as a law of cosines problem:
!!!!!!!x2 = 42 + 42 + 2 4( ) 4( )cos# $ 2x
dx
dt= 32sin#
d#
dt Now x = 4, # = 60° and
d#
dt=
20"
180
%
& '
(
) *
2 4( )dx
dt= 32
3
2
%
& '
(
) *
20"
180=1.209 in min.
57. (Calc) A conical tank is leaking water at the rate of
!
75in3min. At the
same time, water is being pumped into the tank at a constant rate. The tank’s height is 60 in while its top diameter is 20 inches. If the water level is rising at the rate of
!
5in min when the height of the water is 10 inches high, find the rate in which water is being pumped into the tank
to the nearest
!
in3min . (The volume of a cone is given by
!
V =1
3"r
2h )
A.
!
44 in3
min B.
!
175 in3
min C.
!
250 in3
min D.
!
119 in3
min E.
!
31 in3
min
!
D. V =1
3"r2
h r
h=
10
60# r =
h
6
V ="
3
h2
36
$
% &
'
( ) h =
"h3
108
dV
dt* 75 =
"h2
36
dh
dt
$
% &
'
( ) =
100"
365( ) = 43.633
dV
dt= 75 + 43.633 +119 in
3min.
© 2011 www.mastermathmentor.com - 28 - Demystifying the MC AB Calculus Exam
K. Straight-Line Motion – Derivatives
What you are finding: Typically in these problems, you work your way from the position function
!
x t( ) to the velocity function
!
v t( ) to the acceleration function
!
a t( ) by the derivative process. Finding when a particle is stopped involves setting the velocity function
!
v t( ) = 0. Speed is the absolute value of velocity. Determining whether a particle is speeding up or slowing down involves finding both the velocity and acceleration at a given time. If
!
v t( ) and a t( ) have the same sign, the particle is speeding up at a specific value of t. If their signs are different, then the particle is slowing down.
58. A particle moving along the x-axis such that at any time t ≥ 0 its position is given by
!
x t( ) = 2 " 30t + 81t 2 " 20t 3. For what values of t is the particle moving to the left?
A.
!
0 " t <27
20 B.
!
t >27
20 C.
!
0 " t <1
5,t >
5
2
D.
!
1
5< t <
5
2 E.
!
t >1
5
!
C. v t( ) = "60t2 +162t " 30 = "6 10t
2 " 27t + 5( ) = "6 2t " 5( ) 5t "1( )
v t( ) = 0 at t =5
2,t =
1
5 v t( ) < 0 on 0,
1
5
#
$ % &
' ( and
5
2,)
*
+ ,
&
' (
59. A particle moving along the x-axis such that at any time t ≥ 0 its velocity is given by
!
v t( ) =t3
ln t. What
is the acceleration of the particle at t = e ?
A.
!
6e2" 2e B.
!
6e3 C.
!
2e3 D.
!
4e2 E.
!
6e2" 4e
5 2
!
D. a t( ) = " v t( ) =3t
2ln t # t
3 1
t( ) 1
2 t( )
ln t( )2
a e( ) =3e
2ln e #
e2
2
ln e( )2
=
3e3 1
2
$
% & '
( ) #
e2
2
1
2
$
% & '
( )
2= 6e
2 # 2e2 = 4e
2
Better : v t( ) =t
3
ln t=
t3
1
2ln t
=2t
3
ln t
a t( ) = " v t( ) =6t
2ln t( ) # 2t
3 1
t( )ln t( )
2=
6t2
ln t( ) # 2t2
ln t( )2
a t( ) =6e
2 # 2e2
lne( )2
= 4e2
60. (Calc) A particle moves along a straight line with velocity given by
!
v t( ) = t " 2.5cos2t . What is the acceleration of the particle at t = 2 ?
A. 0.168 B. 0.238 C. 0.451 D. 0.584 E. 1.450
!
B. a t( ) =1" 2.5cos2t
ln2.5( ) "sin2t( ) 2( )# a t( ) =1" 2.5cos4
ln2.5( ) "sin4( ) 2( ) = 0.238
© 2011 www.mastermathmentor.com - 29 - Demystifying the MC AB Calculus Exam
61. A squirrel climbs a telephone pole and starts to walk along the telephone wire. Its velocity v of the squirrel at time t, 0 ≤ t ≤ 6 is given by the function whose graph is to the right. At what value of t does the squirrel change direction?
A. 1 and 5 only B. 2 only C. 2 and 4 only D. 1, 3, and 5 only E. 1, 3, 4 and 5 only
!
B. Direction changes ocurs when velocity switches from positive to negative
or from negative to positive. This occurs at time t = 2 only.
62. A particle moving along a straight line that at any time t ≥ 0 its velocity is given by
!
v t( ) =sin t
cos2t. For
which values of t is the particle speeding up?
I.
!
t ="
6 II.
!
t ="
3 III.
!
t =2"
3
A. I only B. II only C. III only D. I and II only E. I and III only
!
E. a t( ) =cos2t cos t " sin t "2sin2t( )
cos2t( )2
=cos2t cos t + 2sin t sin2t( )
cos2t( )2
v#
6
$
% & '
( ) =
12
12
> 0,a#
6
$
% & '
( ) =
12( ) 3
2( ) + 2 12( ) 3
2( )12( )
2> 0 v
#
3
$
% & '
( ) =
3
2
" 12
< 0,a#
3
$
% & '
( ) =
" 12( ) 1
2( ) + 2 3
2( ) 3
2( )" 1
2( )2
> 0
v2#
3
$
% &
'
( ) =
3
2
" 12
< 0,a2#
3
$
% &
'
( ) =
" 12( ) " 1
2( ) + 2 3
2( ) " 3
2( )" 1
2( )2
< 0 Speed increasing if v t( ) and a t( ) have the same signs.
63. Frank is driving along a straight road. At time t = 0, he starts up from a stop sign towards a traffic light
500 feet away. Frank’s distance from the stop sign is a differentiable function
!
F t( ) . At time t = 30 sec, his distance from the stop sign is 200 feet and his velocity is 40 ft/sec. His distance
!
D t( ) from the
traffic light is given by the equation
!
D t( )[ ]2
= 2500 " F t( )[ ]2
. At what rate is the distance between Frank and the traffic light decreasing at time t = 30 sec?
A.
!
30 ft sec B.
!
40 ft sec C.
!
80
3 ft sec D.
!
20
3 ft sec E.
!
550
3 ft sec
!
C. at t = 30, D 30( ) = 300 (either by formula or 500" 200 = 300.
Realize that v t( ) = # F t( ) = 40
2D t( ) $ # D t( ) = "2F t( ) $ # F t( )% 30 # D t( ) = "200 40( )
# D t( ) = "200 40( )
30= "
80
3 ft sec
© 2011 www.mastermathmentor.com - 30 - Demystifying the MC AB Calculus Exam
L. Function Analysis What you are finding: You have a function
!
f x( ) . You want to find intervals where
!
f x( ) is increasing and decreasing, concave up and concave down. You also want to find values of x where there is a relative minimum, a relative maximum, and points of inflection. How to find them: Find critical values – values
!
x = c where " f c( ) = 0 or " f c( ) is undefined.
!
f x( ) is increasing for values of k such that
!
" f k( ) > 0
!
f x( ) is decreasing for values of k such that
!
" f k( ) < 0. If
!
f x( ) switches from increasing to decreasing at c, there is a relative maximum at
!
c, f c( )( ) . If
!
f x( ) switches from decreasing to increasing at c, there is a relative minimum at
!
c, f c( )( ) . This is commonly called the first derivative test.
!
f x( ) is concave up for values of k such that
!
" " f k( ) > 0 .
!
f x( ) is concave down for values of k such that
!
" " f k( ) < 0 . If
!
f x( ) switches concavity at c, there is a point of inflection at
!
c, f c( )( ) . If you want to find the actual maximum or minimum value or find the value of the function at specific points, you need to use accumulated area and the FTC.
64. The graph of a twice-differentiable function f is shown in the figure to
the right. Which of the following is true? A.
!
f "2( ) < # f "2( ) < # # f "2( ) B.
!
f "2( ) < # # f "2( ) < # f "2( ) C.
!
" " f #2( ) < " f #2( ) < f #2( ) D.
!
" f #2( ) < f #2( ) < " " f #2( ) E.
!
" " f #2( ) < f #2( ) < " f #2( )
!
D. Since f is decreasing, " f #2( ) < 0 and since f is concave up, " " f #2( ) > 0.
Since f #2( ) = 0, the largest is " " f #2( ) and the smallest is " f #2( ).
65. If the graph of
!
" f x( ) is shown to the right, which of the following could be the graph of
!
y = f x( )?
I. II. III.
A. I only B. II only C. III only D. I and III only E. II and III only
!
A. Since " f x( ) > 0,x # 0, f x( ) is increasing. But since " f 0( ) = 0
f x( ) has a horizontal tangent at x = 0. For III, " f 0( ) does not exist.
© 2011 www.mastermathmentor.com - 31 - Demystifying the MC AB Calculus Exam
66. (Calc) The first derivative of a function f is given by
!
" f x( ) = x # 2( )esin 2x . How many points of inflection does the graph of f have on the interval 0 < x < 2π?
A. Two B. Three C. Four D. Five E. Six
!
B. " " f x( ) = 2cos2x x # 2( )esin 2x + esin 2x
As shown on the graph of " " f x( ), there are
3 values of x where it changes sign.
67. The graph of
!
" " f , the second derivative of f , is shown to the right. On which of the following intervals is
!
" f x( ) decreasing? A. [0, 1] B. [0, 1] and [2, 4] C. [0, 1] and [2, 5] D. [4, 5] E. [4, 6]
!
E. " f x( ) decreasing means that f is concave down. This occurs when " " f x( ) < 0 or on 4,6[ ]
Note : Being given a graph of " " f x( ) and asking questions about " f x( ) is exactly
the same as being given a graph of " f x( ) and asking questions about f x( ).
The testers do not get into the argument whether to use parentheses or brackets. Either is OK.
68. The graph of
!
" f x( ), the derivative of f , is shown to the right. Which of the following statements is not true?
A. f is increasing on 2 ≤ x ≤ 3. B. f has a local minimum at x = 1. C. f has a local maximum at x = 0. D. f is differentiable at x = 3. E. f is concave down on -2 ≤ x ≤ 1.
!
B. Local minima occur when " f x( ) switches from negative to positive. This occurs at x = 2.
A. is true as f is increasing when " f x( ) > 0
C. is true as local maxima occur when " f x( ) switches from postitive to negative.
D. is true as " f 3( ) exists. " " f 3( ) does not exist.
E. is true as f is concave down when " f x( )is decreasing.
© 2011 www.mastermathmentor.com - 32 - Demystifying the MC AB Calculus Exam
69. If
!
" f x( ) = 2x #1( )2
3x +1( )3
x #1( ), then f has which of the following relative extrema?
I. A relative maximum at
!
x ="1
3
II. A relative minimum at
!
x =1
2
III. A relative minimum at
!
x =1
A. I only B. III only C. I and II only D. I and III only E. I, II and III
70. (Calc) Let
!
" f x( ) = x #1( )2
cos x2 +1( ) . If c represents the largest value on the x-axis on [0, 2] where
there is a point of relative extrema of
!
f x( ) , and let d represents the smallest value on the x-axis on [0, 2] where there is a point of relative extrema of
!
f x( ) , find the value of c – d.
A. 0.244 B. 0.927 C. 1.171 D. -0.689 E. there are no relative extrema
!
C. It is necessary to look at " f very close to the x - axis to see that " f
switches from positive to negative at x = 0.756 and that " f switches
from negative to positive at x =1.927. 1.927 # 0.756 =1.171.
71. Let g be a strictly decreasing function such that
!
g x( ) < 0 for all real numbers x. If
!
f x( ) = x "1( )2
g x( ) , which of the following is true?
A. f has a relative minimum at x = 1 B. f has a relative maximum at x = 1 C. f will be a strictly decreasing function D. f will be a strictly increasing function E. It cannot be determined if f has a relative extrema
!
E. g x( ) < 0 and " g x( ) < 0
" f x( ) = x #1( )2" g x( ) + 2 x #1( )g x( ) = x #1( ) x #1( ) " g x( ) + 2g x( )[ ]
There is a critical point at x =1. If x >1, " f x( ) = + + #( ) + #( )[ ] = #( )
If x <1, " f x( ) = # # #( ) + #( )[ ] which may be positive or negative.
So f is increasing to the right of x =1 but not enough information to the left of x =1
!
D. Critical values at x ="1
3,x =
1
2,x = 1
© 2011 www.mastermathmentor.com - 33 - Demystifying the MC AB Calculus Exam
72. If
!
" f x( ) =x2# 5x # 5
e#x
, for what values of x is
!
f x( ) concave down?
A. All values of x B. No values of x C.
!
"2 < x < 5
D.
!
x > 5 or x < "2 E.
!
1" 53
2< x <
1+ 53
2
!
C. " " f x( ) =e#x
2x # 5( ) # #e#x( ) x
2# 5x # 5( )
e#2x
=e#x
2x # 5 + x2# 5x # 5( )
e#2x
=x
2# 3x #10
e#x
" " f x( ) =x # 5( ) x + 2( )
e#x
< 0 when # 2 < x < 5
73. The function g is defined by
!
g x( ) = x2" f x( ) where the graph of
!
" f x( ) is pictured to the right. Find and classify all points of relative extrema of
!
g x( ) .
A. Relative maximum at x = 2 only B. Relative minimum at x = 2 only C. Relative maximum at x = 2 and relative minimum at x = 0 D. Relative maximum at x = 0 and relative minimum at x = 2 E. Relative minimum at x = 0 and relative maximum at x = 4
!
D. " g x( ) = 2x # " f x( ) = 0
!!!!!There are two values of x where 2x = " f x( ) : x = 0 and x = 2.
74. The derivative of a function f is defined by
!
" f x( ) =g x( ), #1$ x $1
1# 2ex +1
,#3 $ x < #1
% & '
.
The graph of the continuous function
!
" f is shown to the right with the graph of g being a semicircle. Which of the following statements are true?
I. f has a relative maximum at x =
!
"1" ln2 II. f has an inflection point at x = 1 III. f is increasing on (1,3)
A. I only B. II only C. I and II only D. II and III only E. I, II and III
!
C. Relative Max occurs when " f switches from positive to negative. This occurs when 1# 2ex +1 = 0
2ex +1 =1$ e
x +1 =1
2$ x +1= ln
1
2
%
& ' (
) * = #ln2$ x = #1# ln2
Inflection points occur when " f switches from decreasing to increasing. This occurs at x =1.
Since " f < 0 on 1,3( ), f is decreasing on 1,3( ).
© 2011 www.mastermathmentor.com - 34 - Demystifying the MC AB Calculus Exam
M. Second Derivative Test
What you are finding: Students normally find points of relative maximum and relative minimum of
!
f x( ) by using the first derivative test, finding points where a function switches from increasing to decreasing or vice versa. But students may be forced to use the 2nd derivative test, as seen below. However, this type of problem is rare on the multiple choice section. How to use it: Find critical values c where
!
" f c( ) = 0. a) If
!
" " f c( ) > 0, f is concave up at c and there is a relative minimum at x = c. b) If
!
" " f c( ) < 0, f is concave down at c and there is a relative maximum at x = c. c) If
!
" " f c( ) = 0, the 2nd derivative test is inconclusive. ex. f x( ) = x4 at x = 0( )
75. If
!
f x( ) = k sin x + ln x where k is a constant has a critical point at x = π, determine which statement below is true.
A. f has a relative minimum at x = π B. f has a relative maximum at x = π C. f has a point of inflection at x = π D. f has no concavity at x = π E. None of the above
!
B. f x( ) = k sin x + ln x
" f x( ) = k cos x +1
x# " f $( ) = k cos$ +
1
$= 0# k =
1
$
" " f x( ) =%1
$sin x %
1
x2# " " f $( ) =
%1
$sin$ %
1
$2
= %1
$2
!!!Since " f $( ) = 0 and " " f $( ) < 0, there is a relative maximum at x = $.
76. Consider the differential equation
!
dy
dx= x
2" 2y + 3. If
!
y = f x( ) is the solution to the differential
equation, at what point does f have a relative minimum?
A. (0, 0) B. (-1, 2) C. (4, 9) D.
!
0,3
2
"
# $
%
& ' E. (5, 14)
!
E. d
2y
dx2
= 2x " 2dy
dx
To have a relative minimum, dy
dx= 0 and
d2y
dx2
> 0
A. dy
dx= 3 B.
dy
dx= 0,
d2y
dx2
= "2 C. dy
dx=1
D. dy
dx= 0,
d2y
dx2
= 0 E. dy
dx= 0,
d2y
dx2
=10
© 2011 www.mastermathmentor.com - 35 - Demystifying the MC AB Calculus Exam
N. Absolute Extrema What you are finding: the highest value or lowest value of a function, typically in an interval. Another way of asking the question is to ask for the range of the function. Although these types of problems use
differential calculus, students can be asked to utilize the fact that
!
f x( ) = f 0( ) + " f t( )0
x
# dt to find an extreme
value of f on an interval when either given an expression for
!
" f x( ) or a graphical representation for it. How to find it: a) find critical points (x-values where the derivative = 0 or is not defined). b) use 1st derivative test to determine x-values when the function is increasing/decreasing. c) evaluate function at critical points and at the endpoints. This may involve using the FTC to find accumulated change.
77. The absolute maximum value of
!
y = x2" x "12 on -4 ≤ x ≤ 4 is
A. 0.5 B. 4 C. 8 D. 11.75 E. 12.25
!
E. y = x2 " x "12 =
x2 " x "12 if x > 4 or x < "3
"x2 + x +12 if " 3 # x # 4
$ % &
' y =2x "1 if x > 4 or x < "3
"2x +1 if " 3 # x # 4
$ % &
So max. occurs either at x =1
2, x = "4, x = "3,x = 4
y .5( ) =12.25, y "4( ) = 8, y 3( ) = 0, y 4( ) = 0
78. The difference in maximum acceleration and minimum acceleration attained on the interval 0 ≤ x ≤ 3
by the particle whose velocity is given by
!
v t( ) = 2t 3 "12t 2 +18t "1 is A. 4 B. 6 C. 18 D. 21 E. 24
!
E. a t( ) = 6t2" 24t +18# $ a t( ) =12t " 24 = 0# t = 2
a 0( ) =18, a 2( ) = "6, a 3( ) = 0
!!!!!Maximum acceleration =18 Minimum acceleration = "6 Range = 24
79. (Calc) The rate at which the gasoline is changing in the tank of a hybrid car is modeled by
!
f t( ) = t + .5sin t " 2.5 gallons per hour, t hours after a 6-hour trip starts. At what time during the 6-hour trip was the gasoline in the tank going down most rapidly?
A. 0 B. 2.292 C. 3.228 D. 4.203 E. 6
!
A. " f t( ) =1
2 t+ .5cos t = 0. # t = 2.292 or t = 4.203.
Since the change is negative, we want the value of t when f t( ) is a minimum.
f 0( ) = $2.5 f 2.292( ) = $0.611 f 4.203( ) = $0.886 f 6( ) = $0.190
© 2011 www.mastermathmentor.com - 36 - Demystifying the MC AB Calculus Exam
80. The graph of
!
" f , the derivative of f , is shown to the right for 0 ≤ x ≤ 8. The areas of the regions between the graph of
!
" f and the axis are 8, 2 and 2 respectively. If
!
f 0( ) =10, what is the minimum value of f on the interval 0 ≤ x ≤ 8?
A. -12 B. -10 C. -8 D. 0 E. 10
!
D. Since f x( ) = f 0( ) + " f t( )0
x
# dt, there is a value c where " f t( )0
c
# dt = $10
so at that value, f c( ) = f 0( ) + " f t( )0
c
# dt =10 $10 = 0.
81. (Calc) Let f be the function with the first derivative defined as
!
" f x( ) = sin x3# 4x + 2( ) for 0 ≤ x ≤ 2.
Let c be the value of x where f attains its minimum value in the interval 0 ≤ x ≤ 2 and let d be the value of x where f attains its maximum value in the interval 0 ≤ x ≤ 2. Find
!
c " d .
A. 0.539 B. 1.136 C. 1.461 D. 1.675 E. 2
!
B. Since f x( ) = f 0( ) + " f t( )0
x
# dt, c and d are at the locations shown in the graphs.
82. (Calc) Cellophane-wrapped Tastycakes come off the assembly line at the rate of
!
C t( ) = 5 + 4sin"t
6
#
$ %
&
' ( .
They are boxed at the rate of
!
B t( ) =20t
1+ t 2.
!
C t( ) and B t( ) are measured in hundreds of cakes per hour,
and t is measured in hours for 0 ≤ t ≤ 5. There are 500 cakes waiting to be boxed at the start of the 5-hour shift. How long (in hours) does it take the number of cakes waiting to be boxed to go from a minimum to a maximum during the 5-hour shift?
A. 1.560 B. 1.869 C. 3.131 D. 4.691 E. 5
!
C. The first screen is of C t( ) and B t( ). Since C t( ) and B t( ) are rates of change,
we are interested in where B t( ) " C t( ) = 0 (2nd screen). The area between the
curve and the x" axis represents the number of cakes still to be boxed
plus the original 500( ). At the 1st intersection, some additional cakes have gathered.
At the 2nd intersection, more cakes have been boxed than are coming off the line,
but by 5 hours, many more cakes are coming off the line than are being boxed. So
the minimum occurs at t = 1.869 and the maximum occurs at t = 5. 5"1.869 = 3.131.
© 2011 www.mastermathmentor.com - 37 - Demystifying the MC AB Calculus Exam
O. Computation of Riemann Sums What you are finding: Riemann sums are approximations for definite integrals, which we know represent areas under curves. There are numerous real-life models for areas under curves so this is an important concept. Typically these types of problems show up when we are given data points as opposed to algebraic functions. How to find them: Given data points:
!
x x0
x1
x2
... xn"2 xn"1 xn
f x( ) f x0( ) f x
1( ) f x2( ) ... f xn"2( ) f xn"1( ) f xn( )
Assuming equally spaced x-values:
!
xi+1 " xi = b
Left Riemann Sums:
!
S = b f x0( ) + f x
1( ) + f x2( ) + ...+ f xn"1( )[ ]
Right Riemann Sums:
!
S = b f x1( ) + f x
2( ) + f x3( ) + ...+ f xn( )[ ]
Trapezoids:
!
S =b
2f x
0( ) + 2 f x1( ) + 2 f x
2( ) + ...+ 2 f xn"2( ) + 2 f xn"1( ) + f xn( )[ ]
If bases are not the same (typical in AP questions), you have to compute the area of each
trapezoid:
!
1
2xi+1 " xi( ) f xi( ) + f xi+1( )[ ]
Midpoints: This is commonly misunderstood. For example, you cannot draw a rectangle halfway between
!
x = 2 and x = 3 because you may not know
!
f 2.5( ) . You can’t make up data. So in the table above the first midpoint rectangle would be drawn halfway between
!
x0 and x
2 which is x
1. So
assuming that the x-values are equally spaced, the midpoint sum is
!
S = 2b f x1( ) + f x
3( ) + f x5( ) + ...[ ] .
83. The graph of the function f is shown to the right for -2 ≤ x ≤ 2. Four
calculations are made:
LS – Left Riemann sum approximation of
!
f x( )"2
2
# dx with 4
subintervals of equal length
RS – Right Riemann sum approximation of
!
f x( )"2
2
# dx with 4
subintervals of equal length
TS – Trapezoidal sum approximation of
!
f x( )"2
2
# dx with 4 subintervals of equal length
DI –
!
f x( )"2
2
# dx
Arrange the results of the calculations from highest to lowest.
A. DI – RS – TS - LS B. RS – TS – DI - LS C. TS – DI – LS - RS D. LS – DI – TS - RS E. RS – DI – TS - LS
!
B. LS - the smallest RS - the largest TS - slight over - approx DI - Exact
© 2011 www.mastermathmentor.com - 38 - Demystifying the MC AB Calculus Exam
84. The function f is continuous on the closed interval [0, 8] and has the values given in the table below.
The trapezoidal approximation for
!
f x( ) dx0
8
" found with 3 subintervals is 20k. What is the value of k?
!
x 0 3 5 8
f x( ) 5 k27 10
A. 4 B.
!
±4 C. 8 D. -8 E. No values of k
!
A. 3
25 + k 2( ) +
2
2k
2 + 7( ) +3
27 +10( ) = 20k
15 + 3k2 + 2k
2 +14 + 51= 40k" 5k2# 40k # 80 = 0
5 k2# 8k #16( ) = 0" k # 4( )
2
= 0" k = 4
85. The expression
!
1
75ln76
75+ ln
77
75+ ln
78
75+ ...+ ln2
"
# $
%
& ' is a Riemann sum approximation for
A.
!
lnx
75
"
# $
%
& ' dx
1
2
( B.
!
lnx
75
"
# $
%
& ' dx
76
150
( C.
!
1
75ln x dx
76
150
" D.
!
ln x dx
1
2
" E.
!
1
75ln x dx
1
2
"
!
D. This is a right Riemann sum. The base is 1
75 and the heights are ln 1+
1
75
"
# $
%
& ' ,ln 1+
2
75
"
# $
%
& ' ...ln 1+
75
75
"
# $
%
& '
So this is ln x dx1
2
( .
86. An oil slick forms in a lake. The oil is 4 feet deep. The slick is 18 feet
wide and is divided into 6 sections of equal width having measurements as shown in the figure to the right. Tim uses a trapezoidal sum with six trapezoids to approximate the volume while Doug uses a midpoint sum using three equally spaced rectangles. What is the difference between their approximations of the volume of the slick in cubic feet?
A. 9 B. 36 C. 12 D. 3 E. 0
!
B. Tim : Area =1
23( ) 3 + 30 + 48 + 60 + 56 + 28 + 5( ) = 345" Volume = 4 345( ) =1380
Doug : Area = 6 15 + 30 +14( ) = 354 " Volume = 4 354( ) =1416
Difference : 36
© 2011 www.mastermathmentor.com - 39 - Demystifying the MC AB Calculus Exam
P. Integration Techniques
What you are finding: Integration is anti-differentiation. If
!
d
dxf x( ) dx = g x( ) then g x( )" dx = f x( ) + C .
While a derivative can be taken for any expression, not all expressions can be integrated. Just as differentiation of complicated expressions involve u-substitution, substitution can be used to integrate some expressions. It is important to add on the constant of integration when performing anti-differentiation. How to find it: These are usually straightforward. Students need to know these integration formulas.
!
undu =
un+1
n +1+ C
du
u="" ln u + C e
u= e
u+ C" a
u=
1
lnaau
+ C"
sinu du = #cosu + C" cosu du = sinu + C" sec2u du = tanu + C"
cscu du = #cot u + C" secutanu = secu + C" cscucot u du = #cscu + C"du
a2 # u2
" = sin#1 u
a+ C
du
a2
+ u2" =
1
atan
#1 u
a+ C
du
u u2 # a2
" =1
asec
#1 u
a+ C
87.
!
x4 " x( )
2
# dx =
A.
!
x9
9+x3
3+ C B.
!
x9
9"x6
6+x3
3+ C C.
!
x9
9"x6
3+x3
3+ C
D.
!
x4" x
2( )3
3+ C E.
!
x4" x
2( )3
3 4x3" 2x( )
+ C
!
C. x8 " 2x
5 + x2( )# dx =
x9
9"x
6
3+x
3
3+ C
88.
!
ex+2
8x +16 " dx =
A.
!
2ex+2
+ C B.
!
ex+2
2+ C C.
!
ex+2
8+ C
D.
!
2ex+2
+ C E.
!
2ex+2
2+ C
!
E. e
x+2
8x +16 " dx =
ex+2
8 x + 2( ) " dx =
ex+2( )
1 2
2 2 x + 2( )1 2
" dx
u = x + 2( )1 2
du =1
2 x + 2( )1 2dx
ex+2( )
1 2
2 2 x + 2( )1 2
" dx =1
2eu " du =
1
2eu =
2ex+2
2+ C
© 2011 www.mastermathmentor.com - 40 - Demystifying the MC AB Calculus Exam
89.
!
x + 2
x2
+ 4 " dx =
A.
!
ln x2 + 4( ) + tan"1 x( )
2+ C B.
!
tan"1 x
2
#
$ % &
' ( + C C.
!
ln x + 2( ) + C
D.
!
1
2ln x
2 + 4( ) + tan"1x
2
#
$ % &
' ( + C E.
!
1
2ln x
2 + 4( ) + tan"1x
2
#
$ % &
' (
)
* +
,
- . + C
!
D. x + 2
x2 + 4
" dx =x
x2 + 4
+2
x2 + 4
#
$ %
&
' ( " dx
x
x2 + 4
" dx : u = x2 + 4,du = 2x dx)
x
x2 + 4
" dx =1
2
du
u=
1
2ln x
2 + 4( ) "
2
x2 + 4
" dx : u = x,a = 2)2
x2 + 4
" = 2du
a2 + u2
= 21
2
#
$ % &
' ( " tan
*1 x
2
#
$ % &
' (
x
x2 + 4
+2
x2 + 4
#
$ %
&
' ( " dx =
1
2ln x
2 + 4( ) + tan*1 x
2
#
$ % &
' ( + C
90.
!
sin2x
1" sin22x
# dx =
A.
!
2ln cos2x( ) + C B.
!
1
2ln cos2x( ) + C C.
!
1
2sec2x tan2x + C
D.
!
1
2sec2x + C E.
!
2tan2x + C
!
D. sin2x
1" sin22x
# dx =sin2x
cos22x
# dx = tan2x sec2x dx =#1
2sec2x + C
91. (Calc) Let
!
F x( ) be an antiderivative of
!
ln x " 3( )2
x. If
!
F 1( ) = 2, then F 2( ) =
A. -11.092 B. -8.769 C. 2.092 D. 6.908 E. 16.724
!
D. F x( ) =ln x " 3( )
2
x# dx u = ln x " 3,du =
1
xdx
F x( ) =ln x " 3( )
3
3+ C$ F 1( ) =
0 " 3( )3
3+ C = 2$"9 + C = 2$ C =11
F x( ) =ln x " 3( )
3
3+11$ F 2( ) =
ln2 " 3( )3
3+11= 6.908