Unformatted text preview: AP® Calculus AB
2003 Scoring Guidelines
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For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. AP® CALCULUS AB
2003 SCORING GUIDELINES (Form B)
Question 1
Let f be the function given by f (x ) = 4x 2
be the line y = 18 3x , where x 3 , and let is tangent to the graph of f. Let R be the region bounded by the graph of f and the xaxis, and let S be the region bounded by the
graph of f, the line , and the xaxis, as shown above.
is tangent to the graph of y = f (x ) at (a) Show that the point x = 3.
(b) Find the area of S. (c) Find the volume of the solid generated when R is revolved about the xaxis. (a) f (x ) = 8x 3x 2 ; f (3) = 24 f (3) = 36 27 = 1 : finds f (3) and f (3) 3 27 = 9 finds equation of tangent line y= 3) + 9 = 3(x 1: 3x + 18, (b) f (x ) = 0 at x = 4 2 : integral for nontriangular region The line intersects the xaxis at x = 6.
4
1
Area = (3)(9)
( 4x 2 x 3 )dx
3
2
= 7.916 or 7.917 1 : limits 4: 3 + ( ( 18 1 : answer ( 4x 2 3x ) 1
(2)(18
2 1 : integrand
1 : area of triangular region OR
4 shows (3,9) is on both the
graph of f and line which is the equation of line . Area = or 2: Tangent line at x = 3 is x 3 ) )dx 12) = 7.916 or 7.917 (c) Volume = 4
0 ( 4x 2 = 156.038 x 3 ) dx
2 1 : limits and constant 3: or 490.208 1 : integrand
1 : answer Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 2 AP® CALCULUS AB
2003 SCORING GUIDELINES (Form B)
Question 2
A tank contains 125 gallons of heating oil at time t = 0. During the time interval 0 t 12 hours, heating oil is pumped into the tank at the rate H (t ) = 2 + 10
( 1 + ln ( t + 1 ) ) gallons per hour. During the same time interval, heating oil is removed from the tank at the rate
R ( t ) = 12 sin t2
47 gallons per hour. (a) How many gallons of heating oil are pumped into the tank during the time interval 0 t 12 hours? (b) Is the level of heating oil in the tank rising or falling at time t = 6 hours? Give a reason for your
answer.
(c) How many gallons of heating oil are in the tank at time t = 12 hours? (d) At what time t, for 0 t 12, is the volume of heating oil in the tank the least? Show the analysis that leads to your conclusion. (a) 12
0 1 : integral H (t ) dt = 70.570 or 70.571 (b) H (6) R(6) = 2: 1 : answer 1 : answer with reason 2.924, so the level of heating oil is falling at t = 6.
(c) 125 + 12
0 ( H (t ) 1 : limits R(t ) ) dt = 122.025 or 122.026 3: 1 : integrand
1 : answer (d) The absolute minimum occurs at a critical point 1 : sets H (t ) or an endpoint.
H (t ) 1 : volume is least at R(t ) = 0 when t = 4.790 and t = 11.318. The volume increases until t = 4.790, then
decreases until t = 11.318, then increases, so the 3: t = 11.318
1 : analysis for absolute
minimum absolute minimum will be at t = 0 or at
t = 11.318.
125 + 11.318
0 ( H (t ) R(t ) = 0 R(t ) ) dt = 120.738 Since the volume is 125 at t = 0, the volume is
least at t = 11.318.
Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 3 AP® CALCULUS AB
2003 SCORING GUIDELINES (Form B)
Question 3
A blood vessel is 360 millimeters (mm) long Distance
x (mm)
Diameter
B(x) (mm) with circular cross sections of varying diameter.
The table above gives the measurements of the 0 60 120 180 240 300 360 24 30 28 30 26 24 26 diameter of the blood vessel at selected points
along the length of the blood vessel, where x represents the distance from one end of the blood vessel and B(x ) is a twicedifferentiable function that represents the diameter at that point.
(a) Write an integral expression in terms of B(x ) that represents the average radius, in mm, of the
blood vessel between x = 0 and x = 360.
(b) Approximate the value of your answer from part (a) using the data from the table and a midpoint
Riemann sum with three subintervals of equal length. Show the computations that lead to your answer.
275 (c) Using correct units, explain the meaning of
125 B(x ) 2
dx in terms of the blood vessel.
2 (d) Explain why there must be at least one value x, for 0 < x < 360, such that B (x ) = 0. (a) (b) (c) 1
360 360 B(x )
0 2 1 : limits and constant dx 2: 1
B(60) B(180) B(300)
120
+
+
360
2
2
2
1
[ 60 ( 30 + 30 + 24 ) ] = 14
360
B(x )
is the radius, so
2 B(x )
2 = 2: 2 1 : integrand 1 : B(60) + B(180) + B(300)
1 : answer 1 : volume in mm 3 is the area of
2: the cross section at x. The expression is the 1 : between x = 125 and
x = 275 volume in mm3 of the blood vessel between 125
mm and 275 mm from the end of the vessel.
(d) By the MVT, B (c 1 ) = 0 for some c1 in 2 : explains why there are two (60, 180) and B (c 2 ) = 0 for some c2 in values of x where B (x ) has (240, 360). The MVT applied to B (x ) shows
that B (x ) = 0 for some x in the interval 3: the same value
1 : explains why that means (c 1, c 2 ) . B (x ) = 0 for 0 < x < 360 Note: max 1/3 if only explains why
B (x ) = 0 at some x in (0, 360).
Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 4 AP® CALCULUS AB
2003 SCORING GUIDELINES (Form B)
Question 4
A particle moves along the xaxis with velocity at time t 1 + e1 t . 0 given by v ( t ) = (a) Find the acceleration of the particle at time t = 3.
(b) Is the speed of the particle increasing at time t = 3 ? Give a reason for your answer.
(c) Find all values of t at which the particle changes direction. Justify your answer. (d) Find the total distance traveled by the particle over the time interval 0 (a) a(t ) = v (t ) = e1
a(3) = e t 2: 2 (b) a(3) < 0
v(3) = t 3. 1 : v (t )
1 : a(3) 1 : answer with reason
1 +e <0 2 Speed is increasing since v(3) < 0 and a(3) < 0 .
1 : solves v(t ) = 0 to (c) v(t ) = 0 when 1 = e 1 t , so t = 1.
2: v(t ) > 0 for t < 1 and v(t ) < 0 for t > 1 . 3
0
1 = ( = 1 : limits v(t ) dt ( 1 + e1 t 0 e1 t1
0 t 3 )dt + 1 ) + ( t + e1 =( 1 1 + e ) + (3 + e = e +e 2 1 : integrand (1
t3
1 2 e1 t )dt 4: )
1 e1 1) 1 OR
1 : any antiderivative t x (0) = e
x (1) = 1 : evaluates x (t ) when 2 x (3) = e 1 : antidifferentiation
1 : evaluation OR
x (t ) = t 1 : justifies change in
direction at t = 1 Therefore, the particle changes direction at t = 1. (d) Distance = get t = 1 2 4: 3
x (0) ) + ( x (1) Distance = ( x (1) = ( 2 + e) + ( 1 + e
= e +e 2 2 x (3) ) t = 0, 1, 3
1 : evaluates distance
between points ) 1 : evaluates total distance 1 Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 5 AP® CALCULUS AB
2003 SCORING GUIDELINES (Form B)
Question 5
Let f be a function defined on the closed interval [0, 7]. The graph of
f, consisting of four line segments, is shown above. Let g be the function given by g (x ) = x
2 f (t )dt. (a) Find g ( 3 ), g ( 3 ), and g ( 3 ) .
(b) Find the average rate of change of g on the interval 0
(c) x 3. For how many values c, where 0 < c < 3, is g (c ) equal to the
average rate found in part (b)? Explain your reasoning. (d) Find the xcoordinate of each point of inflection of the graph of
g on the interval 0 < x < 7. Justify your answer. (a) g(3) = 3
2 f (t ) dt = 1
(4 + 2) = 3
2 1 : g (3) 3: g (3) = f (3) = 2
g (3) = f (3) = (b) g (3) g (0)
3 0
4 4
=
2 1 : g (3) 2 13
f (t ) dt
30
11
1
7
(2)(4) + (4 + 2) =
=
32
2
3 = ( 1 : g (3) 2: ) g(0) = 3
0 f (t ) dt 1 : answer 1 : answer of 2 (c) There are two values of c.
7
We need = g (c) = f (c)
3
The graph of f intersects the line y = 1 : g(3) 2:
7
at two
3 1 : reason Note: 1/2 if answer is 1 by MVT places between 0 and 3.
1 : x = 2 and x = 5 only (d) x = 2 and x = 5
because g = f changes from increasing to 2: decreasing at x = 2, and from decreasing to 1 : justification
(ignore discussion at x = 4) increasing at x = 5. Copyright © 2003 by College Entrance Examination Board. All rights reserved.
Available at apcentral.collegeboard.com. 6 AP® CALCULUS AB
2003 SCORING GUIDELINES (Form B)
Question 6
Let f be the function satisfying f (x ) = x f (x ) for all real numbers x, where f ( 3 ) = 25. (a) Find f ( 3 ) . dy
= x y with the initial
dx (b) Write an expression for y = f (x ) by solving the differential equation condition f ( 3 ) = 25. (a) f (x ) = f (x ) + x f (x )
=
2 f (x ) f (x ) + x2
2 2 : f (x ) 3:
9 19
25 + =
2
2 f (3) = (b) < 2 > product or chain rule error
1 : value at x = 3 1 : separates variables 1
dy = x dx
y 1 : antiderivative of dy term
1 : antiderivative of dx term 1
2 y = x2 + C
2 6: 1 : constant of integration
1 : uses initial condition f (3) = 25 2 25 = y= 12
11
(3) + C ; C =
2
2 1 : solves for y Note: max 3/6 [111000] if no 1 2 11
x+
4
4 constant of integration
Note: 0/6 if no separation of variables y= (1x
4 2 + 11
4 ) 2 = 12
2
( x + 11 )
16 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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