Unformatted text preview: AP® Calculus AB
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2003 SCORING GUIDELINES
Question 1
Let R be the shaded region bounded by the graphs of y = x and y = e 3x and the vertical line x = 1, 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 horizontal
line y = 1.
(c) The region R is the base of a solid. For this solid, each cross section
perpendicular to the xaxis is a rectangle whose height is 5 times the length of its
base in region R. Find the volume of this solid. Point of intersection
e 3x = x 1: Correct limits in an integral in at (T, S) = (0.238734, 0.488604)
1 (
T (a) Area = x e 3x (a), (b), or (c) )dx 2: = 0.442 or 0.443 1 ((1
T (b) Volume = e 3x 2 (1 ) x 2 ) 1 : integrand
1 : answer 2 : integrand )dx < 3: 1 > reversal < 1 > error with constant < 1 > omits 1 in one radius < or 1.423 or 1.424 = 0.453 2 > other errors 1 : answer (c) Length = x e Height = 5 ( x 2 : integrand 3x e 3x ) < 1 > incorrect but has
x 3:
Volume = 1
T 5( x e 3x 2 ) e 3x as a factor dx = 1.554 1 : answer Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 2 AP® CALCULUS AB
2003 SCORING GUIDELINES
Question 2
A particle moves along the xaxis so that its velocity at time t is given by v (t ) = ( t + 1 ) sin t2
.
2 At time t = 0, the particle is at position x = 1.
(a) Find the acceleration of the particle at time t = 2. Is the speed of the particle increasing at t = 2 ?
Why or why not?
(b) Find all times t in the open interval 0 < t < 3 when the particle changes direction. Justify your answer.
(c) Find the total distance traveled by the particle from time t = 0 until time t = 3. (d) During the time interval 0 3, what is the greatest distance between the particle and the t origin? Show the work that leads to your answer. 1 : a(2) (a) a(2) = v (2) = 1.587 or 1.588
v(2) = 3 sin(2) < 0 2: with reason Speed is decreasing since a(2) > 0 and v(2) < 0 .
t2
=
2
or 2.506 or 2.507 (b) v(t ) = 0 when
t= 2 Since v(t ) < 0 for 0 < t < 2:
2 1 : speed decreasing 1: t = 2 only 1 : justification and v(t ) > 0 for 2 < t < 3 , the particle changes directions at
t= 2. (c) Distance = 3
0 1 : limits v(t ) dt = 4.333 or 4.334 3: 1 : integrand
1 : answer (d) 2
0 v(t ) dt = 1 : ± (distance particle travels 3.265 x ( 2 ) = x (0) + 2
0 2:
v(t )dt = 2.265 while velocity is negative)
1 : answer Since the total distance from t = 0 to t = 3 is
4.334, the particle is still to the left of the origin
at t = 3. Hence the greatest distance from the
origin is 2.265. Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 3 AP® CALCULUS AB
2003 SCORING GUIDELINES
Question 3
The rate of fuel consumption, in gallons per minute, recorded during an airplane flight is given by a
twicedifferentiable and strictly increasing function R
of time t. The graph of R and a table of selected
values of R ( t ), for the time interval 0 t 90 minutes, are shown above.
(a) Use data from the table to find an approximation
for R ( 45 ) . Show the computations that lead to your answer. Indicate units of measure.
(b) The rate of fuel consumption is increasing fastest at time t = 45 minutes. What is the value of R ( 45 ) ? Explain your reasoning.
(c) Approximate the value of 90
0 R(t )dt using a left Riemann sum with the five subintervals indicated by the data in the table. Is this numerical approximation less than the value of 90
0 R(t )dt ? Explain your reasoning.
(d) For 0 < b 90 minutes, explain the meaning of 1
b
units of measure in both answers. plane. Explain the meaning of (a) R (45) b
0 b
0 R ( t )dt in terms of fuel consumption for the R ( t )dt in terms of fuel consumption for the plane. Indicate R(50) R(40)
55 40
=
50 40
10
2
= 1.5 gal/min 1 : a difference quotient using
numbers from table and 2: interval that contains 45
1 : 1.5 gal/min2 (b) R (45) = 0 since R (t ) has a maximum at 2: t = 45 . (c) 90
0 R(t ) dt 1 : R (45) = 0
1 : reason
1 : value of left Riemann sum (30)(20) + (10)(30) + (10)(40) 2: +(20)(55) + (20)(65) = 3700 1 : “less” with reason Yes, this approximation is less because the
graph of R is increasing on the interval.
(d) b
0 2 : meanings R(t ) dt is the total amount of fuel in gallons consumed for the first b minutes.
1b
R(t ) dt is the average value of the rate of
b0
fuel consumption in gallons/min during the 1 : meaning of 3: b
0 R(t ) dt 1b
R(t ) dt
b0
1 > if no reference to time b 1 : meaning of
< 1 : units in both answers first b minutes. Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 4 AP® CALCULUS AB
2003 SCORING GUIDELINES
Question 4
Let f be a function defined on the closed interval 3 x 4 with f ( 0 ) = 3. The graph of f , the derivative of f, consists of one line segment and a semicircle, as shown above.
(a) On what intervals, if any, is f increasing? Justify your answer.
(b) Find the xcoordinate of each point of inflection of the graph of f
3 < x < 4. Justify your answer. on the open interval
(c) Find an equation for the line tangent to the graph of f at the
point ( 0, 3 ) . (d) Find f ( 3 ) and f ( 4 ) . Show the work that leads to your answers. (a) The function f is increasing on [ 3, 2] since
f > 0 for x< 3 1 : interval 2: 2. 1 : reason (b) x = 0 and x = 2
f changes from decreasing to increasing at x = 0 and from increasing to decreasing at 2: 1 : x = 0 and x = 2 only
1 : justification x =2 (c) f (0) = 1 : equation 2 Tangent line is y = (d) f (0) 0 f ( 3) = 3 2x + 3. 1
= (1)(1)
2
f ( 3) = f (0) + 1: ± f (t )dt
1
(2)(2) =
2 3
2 1
(2 2) (difference of areas
of triangles) 3
9
=
2
2 1 : answer for f ( 3) using FTC 4:
f (4) f (0) =
= f (4) = f (0) 4
0 ( (8 1: ± 8 12
(2)
2 8+2 = )= ) 12
(2)
2
(area of rectangle f (t )dt
8+2 area of semicircle) 5+2 1 : answer for f (4) using FTC Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 5 AP® CALCULUS AB
2003 SCORING GUIDELINES
Question 5
A coffeepot has the shape of a cylinder with radius 5 inches, as shown in the figure
above. Let h be the depth of the coffee in the pot, measured in inches, where h is a
function of time t, measured in seconds. The volume V of coffee in the pot is h cubic inches per second. (The volume V of a cylinder 5 changing at the rate of with radius r and height h is V = r 2h. ) dh
=
dt (a) Show that h
.
5 (b) Given that h = 17 at time t = 0 , solve the differential equation dh
=
dt h
for
5 h as a function of t.
(c) At what time t is the coffeepot empty? (a) V = 25 h 1: dV
dh
= 25
=
dt
dt dh
=
dt (b) 5h
=
25 dh
=
dt h
5 ( h dV
dt
1 : shows result 1 : computes 1 : antiderivatives
1 : constant of integration 1
dt
5 5: 1 : uses initial condition h = 17
when t = 0 1
t +C
5 1 : solves for h 2 17 = 0 + C
h= 5 1 : separates variables h
5 1
dh =
h
2 h= 3: h 5 dV
=
dt Note: max 2/5 [11000] if no constant 1
t + 17
10 of integration ) 2 Note: 0/5 if no separation of variables (c) ( 1
t + 17
10 ) 2 =0 1 : answer t = 10 17
Copyright © 2003 by College Entrance Examination Board. All rights reserved.
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2003 SCORING GUIDELINES
Question 6
Let f be the function defined by x + 1 for 0 ≤ x ≤ 3 f (x ) = 5 − x
for 3 < x ≤ 5. (a) Is f continuous at x = 3 ? Explain why or why not.
(b) Find the average value of f (x ) on the closed interval 0 ≤ x ≤ 5.
(c) Suppose the function g is defined by k x + 1 for 0 ≤ x ≤ 3 g(x ) = mx + 2 for 3 < x ≤ 5, where k and m are constants. If g is differentiable at x = 3, what are the values of k and m ? 1 : answers “yes” and equates the values of the left and righthand 2: limits 1 : explanation involving limits (a) f is continuous at x = 3 because lim f (x ) = lim+ f (x ) = 2. x → 3− x →3 Therefore, lim f (x ) = 2 = f (3).
x →3 (b) 5 ∫0 f (x ) dx = 3 ∫0 f (x ) dx +
3 5 ∫3 4: f (x ) dx ( ) 5 3
2
1
(x + 1) 2 + 5x − x 2
3
2
0
3
16 2
25 21
20
=
−
+
−
=
3
3
2
2
3 = ( Average value: )( ) 3 5 1 : k ∫ f (x ) dx + k ∫ f (x ) dx
0 3 (where k ≠ 0)
1 : antiderivative of 1 : antiderivative of 5 − x
1 : evaluation and answer 15
4
∫0 f (x )dx = 3
5 (c) Since g is continuous at x = 3, 2k = 3m + 2.
k 2 x + 1 for 0 < x < 3
g ′(x ) = m for 3 < x < 5 k
lim g ′(x ) = and lim g ′(x ) = m
−
4
x →3
x → 3+
Since these two limits exist and g is 1 : 2k = 3m + 2 k
3: 1: =m 4 1 : values for k and m differentiable at x = 3, the two limits are
k
equal. Thus = m.
4
8m = 3m + 2 ; m = 2
8
and k =
5
5 Copyright © 2003 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
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