Be Prepared
for the
AP
Calculus
Exam
Mark Howell
Gonzaga High School, Washington, D.C.
Martha Montgomery
Fremont City Schools, Fremont, Ohio
Practice exam contributors:
Benita Albert
Oak Ridge High School, Oak Ridge, Tennessee
Thomas Dick
Oregon State University
Joe Milliet
St. Mark's School of Texas, Dallas, Texas
Skylight Publishing
Andover, Massachusetts
Copyright © 2005 by Skylight Publishing
Chapter 10. Annotated Solutions to Past Free-Response Questions
This material is provided to you as a supplement to the book
Be Prepared for the AP Calculus Exam (ISBN 0-9727055-5-4).
You are not authorized to publish or distribute it in any form without
our permission. However, you may print out one copy of this chapter
for personal use and for face-to-face teaching for each copy of the Be
Prepared book that you own or receive from your school.
Skylight Publishing
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2003 AB
AP Calculus Free-Response
Solutions and Notes
Question 1
(a)
1
y = x and y = e −3x intersect at a point (a, y) where­ a ≈ 0.238734 .
Area of R =
(b)
1
a
)
x − e −3x dx ­ ≈ 0.443 .
R = 1 − e −3 x ; r = 1 − x ;
Volume =
(c)
∫(
∫
1
a
1
a
b = x − e −3 x ; h = 5
Volume =
∫
1
a
(
(
π ( R 2 − r 2 ) dx = ∫ π (1 − e −3 x ) − 1 − x
(
2
) ) dx ­ ≈ 1.424 .
2
)
x − e −3 x .
1
bh dx = ∫ 5
a
(
x − e −3 x
)
2
dx ­ ≈ 1.554 .
Notes:
1.
Store a in a named variable in your calculator (see Appendix: Calculator Skills).
3
4
FREE-RESPONSE SOLUTIONS ~ 2003 AB
Question 2
(a)
a (2) = v′(2) ­ ≈ 1.587 . Since v(2) = −3sin(2) < 0 and a (2) > 0 , the velocity and
acceleration have different signs1, so the speed is decreasing.
(b)
t2
= kπ ⇒ t = ± 2kπ . Of these values only
2
t = 2π falls into the interval 0 < t < 3. The velocity indeed changes its sign at that
t2
point in time because the sign of (t + 1) does not change and the sign of sin
2
1
changes when t = 2π .
We must have v(t ) = 0, so t = −1 or
∫
3
2
v(t ) dt ­ ≈ 4.333 .
(c)
Distance traveled =
(d)
By the Candidate Test, the maximum distance over 0 ≤ t ≤ 3 can be reached at
t = 2π or t = 0 or t = 3 . The distance from the origin 3 at t = 2π is
x(0) + ∫
2π
0
v(t ) dt
0
­ ≈ 1 − 3.265 = 2.265 .
3
and at t = 3 is x(0) + ∫ v(t ) dt
0
The distance from the origin at t = 0 is 1
­ ≈ 1 − 2.196 = 1.196 .4
The maximum is 2.265.
Notes:
1.
Justification is required for full credit.
2.
This is shorter than
3.
From the origin, not from x(0)!
4.
Another line of reasoning may be used. Since the total distance traveled over
0 ≤ t ≤ 3 is 4.333 and the particle got to the position −2.265 , it will be still to the left
of the origin at t = 3 and therefore the maximum distance from the origin cannot be
reached at t = 3.
∫
2π
0
v(t ) dt +
∫
3
2π
v(t ) dt .
FREE-RESPONSE SOLUTIONS ~ 2003 AB
Question 3
R(50) − R (40) 55 − 40
=
= 1.5 gal/min 2 .
10
10
(a)
R′(45) ≈
(b)
R′′(45) = 0 since R′(t ) has a maximum at t = 45 .
(c)
∫
1
90
0
2
R(t ) dt ≈ 30 ⋅ 20 + 10 ⋅ 30 + 10 ⋅ 40 + 20 ⋅ 55 + 20 ⋅ 65 = 3700 gal .
This
approximation is less than the actual value because, for an increasing function, a
left-hand Riemann sum underestimates the integral.
(d)
∫
b
0
R(t ) dt (gal 3) is the total fuel consumed over the time interval 0 ≤ t ≤ b .
1 b
R(t ) dt (gal/min 3) is the average rate of fuel consumption over the time
b ∫0
interval 0 ≤ t ≤ b .
Notes:
1.
You have to give a correct reason for full credit.
2.
An additional check: lengths of the subintervals, 30, 10, 10, 20, 20, must add up
to 90.
3.
Units must be specified for full credit.
5
6
FREE-RESPONSE SOLUTIONS ~ 2003 AB
Question 4
(a)
f is increasing on [–3, –2] since f ′( x) is positive for –3 < x < –2 (and nowhere
else).
(b)
f ′ changes from decreasing to increasing at x = 0 and from increasing to
decreasing at x = 2. Therefore, there are two points of inflection: at x = 0 and at
x = 2.
(c)
y = f (0) + f ′(0) ⋅ x = 3 − 2 x .
(d)
f (−3) = f (0) + ∫
0
1
 f ′( x) dx = 3 − ∫ f ′( x) dx = 3 −  − 2  . 1
0
−3
2

4
1


f (4) = f (0) + ∫ f ′( x) dx = 3 −  2 ⋅ 4 − π ⋅ 22  1, 2
0
2


−3
Notes:
1.
2.
To save time and avoid mistakes, do not simplify further.
1
2 ⋅ 4 − π ⋅ 22 represents the area of the rectangle minus the area of the semicircle.
2
Since the region is below the x-axis, its contribution to the integral is negative.
FREE-RESPONSE SOLUTIONS ~ 2003 AB
7
Question 5
(a)
(b)
2
dV d (π r h )
dh
dh
dh −5π h
h
=
= 25π
⇒ − 5π h = 25π
. Therefore,
.
=
=−
dt
25π
5
dt
dt
dt
dt
dh
h
dh
dt
=−
⇒
=−
⇒
dt
5
5
h
∫
dh
dt
1
= −∫
⇒ 2 h = − t +C .
5
5
h
2
 1

2
 − 5 t + 2 17   1

=
−
+
t
17
h(0) = 17 ⇒ C = 2 17 , h = 
 
 .
2
10






(c)
h(t ) = 0 ⇒ −
1
t + 17 = 0 ⇒ t = 10 17 .
10
8
FREE-RESPONSE SOLUTIONS ~ 2003 AB
Question 6
(a)
Yes. lim− f ( x) = x + 1
x →3
x =3
= 2; lim+ f ( x) = ( 5 − x ) x =3 = 2 . Therefore,
x →3
lim f ( x) = 2 = f (3) .
x →3
(b)
∫
5
0
f ( x)dx = ∫
3
3
0
5
x + 1 dx + ∫ ( 5 − x ) dx =
3
5
3

2
x2 
2
25
9
20
( x + 1) 2 +  5 x −  = ( 8 − 1) +  25 − − 15 +  = .
3
2 3 3
2
2 3


0
1 5
4
The average is ∫ f ( x)dx = .
5 0
3
(c)
g(x) is continuous at x = 3, so k 3 + 1 = m ⋅ 3 + 2 ⇒ 2k = 3m + 2 .
 k
for 0 < x < 3

g ′( x) =  2 x + 1
 m for 3 < x < 5

k
g ′( x) exists at x = 3, so = m ⇒ k = 4m . Substituting this expression for k in the
4
2
8
first equation, we get 8m = 3m + 2 ⇒ m = . k = 4m = .
5
5
2003 BC
AP Calculus Free-Response
Solutions and Notes
Question 1
See AB Question 1.
Question 2
(a)
Both x and y are decreasing along the segment BD of the curve, so both
dx
dy
and
dt
dt
are negative or zero at point C.
(b)
At point B,
πt
=
π
 π t +1 
dx
 πt 
= 0 ⇒ cos   = 0 or sin 
 = 0 . For t > 0,
dt
 6 
 2 
+ kπ (k = 0,1, 2,...) or
π t +1
6 2
2
1
solution for 0 < t < 9 is t = 3.
(c)
x′(8) = −9 cos
= kπ (k = 1, 2,...) . The only possible
8π
π 8 +1
9 y′(8) 5
5
sin
=− .
= ⇒ y′(8) = − . The velocity is
6
2
2 x′(8) 9
2
9 5
x′(8), y′(8) = − , − . The speed is
2 2
(d)
Distance =
∫
9
0
x′(t ) dt
2
2
9 5 x′(8) 2 + y′(8) 2 =   +   . 2
2 2
­ ≈ 39.255 .
Notes:
1.
From the first equality, t = 3 + 6k . 0 < t < 9 only when k = 0, t = 3. From the
second equality, t = 4k 2 − 1 . 0 < t < 9 only when k = 1, t = 3. It is a “coincidence”
that t = 3 satisfies both equations.
2.
Leave it at that to save time and avoid arithmetic mistakes.
9
10
FREE-RESPONSE SOLUTIONS ~ 2003 BC
Question 3
(a)
5
5
5
3
 25 
y = 1 + y 2 ⇒  − 1 y 2 = 1 ⇒ y = . x = y ⇒ x = .
3
4
3
4
 9

3
dx
2y
3
4
=
=
= .1
2
dy 2 1 + y
9 5
1+
16
∫
3
4
0
5 

2
 1 + y − y  dy ­ ≈ 0.346 .
3 

(b)
Area of S =
(c)
x = r cos θ ; y = r sin θ ⇒
x 2 − y 2 = 1 ⇔ r 2 cos 2 θ − r 2 sin 2 θ = 1 ⇔ r 2 =
(d)
Area of S =
∫
( 5) r2
tan −1 3
0
2
dθ = ∫
( 5)
tan −1 3
1
dθ .2
2
2 ( cos θ − sin θ )
2
0
Notes:
1.
You might notice that
2.
Or
∫
( 3)
cot −1 5
0
2y
2 1+ y
... (same as above).
2
=
1
.
cos θ − sin 2 θ
2
y 3
= .
x 5
FREE-RESPONSE SOLUTIONS ~ 2003 BC
Question 4
See AB Question 4.
Question 5
See AB Question 5.
11
12
FREE-RESPONSE SOLUTIONS ~ 2003 BC
Question 6
(a)
Since the coefficient of the x term in this Maclaurin series is 0, f ′(0) = 0 . Since the
1
coefficient of the x2 term in this Maclaurin series is − ,
3!
f ′′(0)
1
1
= − ⇒ f ′′(0) = − . The first derivative of f is 0 and the second derivative
2
3!
3
is negative, so f has a local maximum at x = 0.
(b)
This is a converging alternating series with decreasing terms, so the absolute value
1
 1 1
of the error is less than the first omitted term: f (1) − 1 −  < <
.
 3!  5! 100
(c)
y′ = −2
x
x3
x5
x 2 n −1
n
+ 4 − 6 + ... + ( −1) 2n
+ ... 1
3!
5!
7!
( 2n + 1)!
 x2

x4
x6
x2n
n
xy′ + y =  −2 + 4 − 6 + ... + ( −1) 2n
+ ... 
5!
7!
( 2n + 1)! 
 3!
 x2 x 4 x6

x 2n
n
+ 1 − + − + ... + ( −1)
+ ...  =
( 2n + 1) ! 
 3! 5! 7!
x2
x4
x6
x2n
n
+ ... =
1 − 3 + 5 − 7 + ... + ( −1) ( 2n + 1)
3!
5!
7!
( 2n + 1)!
= 1−
2n
x2 x4 x6
n x
+ − + ... + ( −1)
+ ... = cos x .
2! 4! 6!
( 2n ) !
Notes:
1.
This problem does not ask you to write formulas for the general term of each series
involved, and it may be easier, in fact, to work without general terms. However,
when this question was graded, there was a considerable debate as to how many
terms had to be shown for full credit. For those who worked with general terms,
there was no such question.
2.
If you notice that the given series multiplied by x is the series for sin x and that
xy′ + y = ( xy )′ , your proof can be greatly simplified:
xy = x −
x3 x5 x 7
+ − + ... = sin x ⇒ ( xy )′ = cos x .
3! 5! 7!
2003 AB (Form B)
AP Calculus Free-Response
Solutions and Notes
Question 1
(a)
f ′(3) = 8 x − 3 x 2
= −3 . The tangent line to the graph of f ( x) at x = 3 is
x =3
y = f (3) + f ′(3) ⋅ ( x − 3) = 9 + (−3) ⋅ ( x − 3) = −3 x + 18 , the same as l.
(b)
y = f ( x) intersects the x-axis at x = 4 and l
intersects the x-axis at x = 6. Area of
1
27
+ MPQ = (6 − 3) ⋅ f (3) =
. Area of
2
2
y
P
4
S = area of +MPQ − ∫ f ( x) dx =
3
4
27
− ∫ ( 4 x 2 − x 3 ) dx ­ ≈ 7.917 .
3
2
(c)
Volume = π ∫
4
0
( 4x
2
M
3
Q
4
6
x
− x 3 ) dx ­ ≈ 490.208 .
2
13
14
FREE-RESPONSE SOLUTIONS ~ 2003 AB (FORM B)
Question 2
∫
12
H (t ) dt ­ ≈ 70.571 gallons.
(a)
Amount of oil pumped in =
(b)
Falling. The rate of change of the volume is [ H (t ) − R(t ) ] t =6 ­ ≈ −2.924 < 0 .
(c)
Volume = 125 + ∫
(d)
Applying the Candidate Test. V(0) = 125. V(125) = 122.026 (from Part (c)). Local
dV
= 0 ⇒ H (t ) = R(t ) ⇒ ­ t ≈ 4.790 or t ≈ 11.318 .
minima may occur when
dt
12
0
V (4.790) = 125 + ∫
[ H (t ) − R(t )] dt ­ ≈ 122.026
4.790
0
V (11.318) = 125 + ∫
0
[ H (t ) − R(t )] dt ­ ≈ 149.4 .1
11.318
0
is at t = 11.318 hours.
gallons.
[ H (t ) − R(t )] dt ­ ≈ 120.7 .
The absolute minimum for V(t)
2
Notes:
1.
Instead of calculating V(4.790) you could state that H (t ) − R (t ) is positive for
t < 4.790, so V(t) can’t have a local minimum at t = 4.790.
2.
Simply graphing V (t ) = ∫ [ H (τ ) − R (τ ) ] dτ and finding its minimum will not get
t
0
you full credit because reading a maximum or minimum from a graph is not a
permitted calculator operation.
FREE-RESPONSE SOLUTIONS ~ 2003 AB (FORM B)
15
Question 3
1 360 B( x)
dx . 1
∫
0
360
2
(a)
Average radius =
(b)
1 360 B( x)
1
1
1
dx ≈
⋅120 ⋅ [ B(60) + B(180) + B(300) ] = ( 30 + 30 + 24 ) .
∫
0
360
2
360
2
6
(c) This integral represents the total volume (in mm3)2 of the segment of the blood
vessel between x = 125 and x = 275.
(d) By the MVT, B′( x) > 0 for some 0 ≤ x ≤ 60 and B′( x) < 0 for some 60 ≤ x ≤ 120 .
Therefore (by the IVT), B′( x1 ) = 0 for some 0 ≤ x1 ≤ 120 . Similarly, B′( x) < 0 for
some 180 ≤ x ≤ 300 and B′( x) > 0 for some 300 ≤ x ≤ 360 , so B′( x2 ) = 0 for some
180 ≤ x2 ≤ 360 . Applying the MVT (or, its special case, Rolle’s theorem) to B′( x)
we conclude that B′′( x) = 0 for some x1 < x < x2 .3
Notes:
1.
Average radius, not diameter!
2.
You must state the units to get full credit.
3.
You may notice that B(60) = B(180), so by the MVT (or Rolle’s) B′( x1 ) = 0 for
some x1 from [60, 180]. The same for some x2 from [240, 360]. From the MVT (or
Rolle’s) applied to B′( x) , we conclude that B′′( x) must be equal to 0 somewhere
between x1 and x2.
16
FREE-RESPONSE SOLUTIONS ~ 2003 AB (FORM B)
Question 4
(a)
a(t ) = v′(t ) = −e1−t . a (3) = v′(3) = −e −2 .
(b)
Since v(3) = −1 + e −2 < 0 and a (3) < 0 , the velocity and acceleration have the same
sign 1, so the speed is increasing.
(c)
We must have v(t ) = 0, so t = 1 . The velocity indeed changes sign at t = 1:
2
v(t ) > 0 for t < 1 and v(t ) < 0 for t > 1 .
(d)
Distance traveled =
∫
3
0
1
3
v(t ) dt = ∫ −1 + e1−t dt + ∫ −1 + e1−t dt =
0
1
( −t − e ) − ( − t − e ) = ( − 2 + e ) − ( − 1 − e ) = −1 + e + e
1−t
1
0
1−t
3
−2
−2
1
Notes:
1.
Reason is required for full credit.
2.
Justification is required for full credit.
3.
Or: x(t ) = −t − e1−t ⇒ x(0) = −e; x(1) = −2; x(3) = −3 − e −2 .
Distance traveled = ( x(1) − x(0) ) + ( x(1) − x(3) ) = −1 + e + e −2 .
.3
FREE-RESPONSE SOLUTIONS ~ 2003 AB (FORM B)
17
Question 5
3
2−4
= −2 .
3− 2
(a)
g (3) = ∫ f (t ) dt = 3 . g ′(3) = f (3) = 2 . g ′′(3) = f ′(3) =
(b)
The rate of change of g is f. The average of f(x) on 0 ≤ x ≤ 3 is
3
1 3
1 2
1
7
f (t ) dt = ∫ f (t ) dt + ∫ f (t ) dt = (4 + 3) = .
∫
2
3 0
3 0
3
3
(c)
Again, g ′(c) = f (c) . The line y =
2
(
)
7
intersects the graph of f in two points with
3
x-coordinates in the interval (0, 3).
(d)
g ′ (which is f) changes from increasing to decreasing at x = 2 and from decreasing
to increasing at x = 5. Therefore, there are two inflection points: at x = 2 and at
x = 5.
18
FREE-RESPONSE SOLUTIONS ~ 2003 AB (FORM B)
Question 6
(a)
f ′′( x) =
d
d
f ′( x) =
x f ( x) =
dx
dx
f ( x) + x
f ′( x)
=
2 f ( x)
f ( x) +
x2
.
2
9
f ′′(3) = 25 + .
2
(b)
dy
=x y ⇒
dx
∫
dy
x2
= ∫ x dx ⇒ 2 y = + C .
2
y
2
 1  x 2 11  
32
9 11
2 25 = + C ⇒ C = 10 − = . y =   +   .1
2
2 2
 2  2 2 
Notes:
1.
You have to solve for y but stop here, do not simplify any further.
2003 BC (Form B)
AP Calculus Free-Response
Solutions and Notes
Question 1
See AB Question 1.
Question 2
(a)
Area of R =
(b)
Area of R =
(c)
Area of R =
∫
∫
∫
1
1 − ( x − 1) dx + ∫
2
0
1
0
2
1
2 − x 2 dx .
)
(
 2 − y 2 − 1 − 1 − y 2  dy .1


π
4
0
( 2)
2
2
( 2 cos θ ) dθ .2
dθ + ∫ π
2
π
2
2
4
Notes:
1.
In x = 1 ± 1 − y 2 we choose 1 − 1 − y 2 because we need x < 1.
2.
Just set up, do not simplify anything.
Question 3
See AB Question 3.
19
20
FREE-RESPONSE SOLUTIONS ~ 2003 BC (FORM B)
Question 4
(a)
Velocity vector = 6e3t − 7e −7 t , 9e3t + 2e −2t .
The speed =
[ x′(0)] + [ y′(0)]
2
2
=
( −1)
2
+ 112 .
(b)
dy
dy 9e3t + 2e−2t 9
dy dt 9e3t + 2e −2t
=
= 3t
. lim =
= .
dx dx 6e − 7e −7 t t →∞ dx 6e3t − 7e −7 t 6
dt
(c)
The tangent line is horizontal when
9e 3 t + 2e − 2 t
(d)
dy
= 0 ⇔ 9e3t + 2e −2t = 0 . Since
dt
> 0 for all t, no such t exists.
The tangent line is vertical when
dx
7
1 7
= 0 ⇔ 6e3t − 7e−7 t = 0 ⇔ e3t + 7 t = ⇔ t = ln   .
dt
6
10  6 
Question 5
See AB Question 3.
FREE-RESPONSE SOLUTIONS ~ 2003 BC (FORM B)
Question 6
(a)
(b)
f ′′(2)
f ′′′(2)
( x − 2) 2 +
( x − 2)3 + ... +
2!
3!
n
f ( ) (2)
( x − 2) n + ... =
n!
( n + 1) ! ( x − 2)n + ... =
2!
3!
4!
1 + ( x − 2) +
( x − 2) 2 +
( x − 2)3 + ... +
2
3
3
2!3
3!3
n !3n
( n + 1) ( x − 2)n + ...
2
3
4
1 + ( x − 2) + 2 ( x − 2) 2 + 3 ( x − 2)3 + ... +
3
3
3
3n
f ( x) = f (2) + f ′(2)( x − 2) +
By the Ratio Test, the series converges when
n+2
( x − 2) n +1
n +1
x−2
n+2 x−2
lim 3
lim
=
=
< 1 . The series diverges when
n →∞ n + 1
n →∞ n + 1
3
3
n
( x − 2)
3n
x−2
> 1 . The radius of convergence is 3.
3
(c)
1
1
1
1
g ( x) = 3 + ( x − 2) + ( x − 2) 2 + 2 ( x − 2)3 + 3 ( x − 2) 4 + ... + n ( x − 2) n +1 + ...
3
3
3
3
(d)
The series for g diverges at x = −2 because the n-th term does not approach 0:
( −4 )
lim
n →∞
3n
n +1
n
4
 4
= −4 ⋅ lim  −  does not exist because − > 1 .
n →∞
3
 3
21