2005 AP® Calculus BC Free-Response Questions
Workshop, Tuesday, Aug. 2, 2005
1. Let f and g be the functions given by f (x) = 14 + sin(πx) and g(x) = 4−x . Let R
be the shaded region in the first quadrant enclosed by the y-axis and the graphs
of f and g, and let S 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.
(solution, (a)): This area is the integral
Z a
1
−x
4 −
+ sin(πx)
dx,
4
0
where a is the smallest non-negative solution of f (x) = g(x). This is evaluated numerically with the result a ≈ 0.178. Therefore,
0.178
1
−x
Area of R ≈
4 −
+ sin(πx)
dx
4
0
0.178
4−x 1
1
= −
− x + cos(πx)
ln 4 4
π
0
≈ 0.065.
Z
(b) Find the area of S.
(solution, (b)): This area is the integral
Z b
a
1
+ sin(πx) − 4−x
4
dx,
where b > a is the next positive solution of f (x) = g(x). This is clearly
given by b = 1; therefore,
1
1
Area of S ≈
+ sin(πx) − 4−x dx
4
0.178
1
4−x 1
1
x − cos(πx) +
−
=
4
π
ln 4
0.178
≈ 0.410.
Z
(c) Find the volume of the solid generated when S is revolved about the horizontal line y = −1
(solution, (c)): This is easily evaluated by the method of washers:
Z
Volume = π
a
1
(f (x) + 1)2 − (g(x) + 1)2 dx ≈ 4.559
2. The curve above is drawn in the xy-plane and is described by the equation in
polar coordinates r = θ + sin(2θ), 0 ≤ θ ≤ π, where r is measured in meters
and θ is measured in radians. The derivative of r with respect to θ is given by
dr
= 1 + 2 cos(2θ).
dθ
(a) Find the area bounded by the curve and the x-axis.
(solution, (a)): This is given by the integral
Area
=
=
=
=
=
=
Z
1 π 2
r dθ
2 0
Z
1 π
(θ + sin 2θ)2 dθ
2 0
Z
1 π 2
(θ + 2θ sin 2θ + sin2 2θ) dθ
2 0
Z 1 π 2
1 − cos 4θ
θ + 2θ sin 2θ +
dθ
2 0
2
1 θ3
1
θ sin 4θ π
− θ cos 2θ + sin 2θ + −
2 3
2
2
8
0
3
3
π
1 θ
θ π
π
− θ cos 2θ +
− ≈ 4.382 meters2
=
2 3
2 0
6
4
(b) Find the angle θ that corresponds to the point on the curve with x-coordinate
−2.
(solution, (b)): Since x = r cos θ, and since r = θ + sin 2θ, we are to solve the
equation
(θ + sin 2θ) cos θ = −2
for θ, where, again, 0 ≤ θπ. The relevant solution is found numerically to
be θ ≈ 2.786 ≈ 0.887π .
2π dr
π
(c) For < θ <
,
is negative. What does this fact say about r? What
3
3 dθ
does this fact say about the curve?
π
2π
(solution, (c)): This fact says that on the interval < θ <
, r is decreasing;
3
3
in turn, since r > 0 on this interval, this means that over this interval the
curve is approaching the origin.
π
(d) Find the value of θ in the interval 0 ≤ θ ≤ that corresponds to the point
2
on the curve in the first quadrant with greatest distance from the origin.
Justify your answer.
(solution, (d)): The problem is, then to maximize the function r = θ + sin 2θ
on the given interval. Since r is differentiable on its domain of definition,
dr
the critical values of θ are precisely those values for which
= 0. Since we
dθ
dr
are given that
= 1 + 2 cos 2θ, we are to solve the equation
dθ
1 + 2 cos 2θ = 0,
for θ. This reduces to cos 2θ = − 12 ; the only relevant solution for θ is then
=
θ = √π3 . Finally, we have r(0) = 0, r( π2 ) = π2 , and r( π3 ) = π3 + sin 2π
3
3
π
π
π
+ 2 > 2 , and so we conclude that θ = 2 affords the maximum value of r
3
on the interval 0 ≤ θ ≤ π2 .
Distance
x (cm)
0
Temperature
T (x) (◦ C)
1
5
6
8
100 93 79 62 55
3. A metal wire of length 8 centimeters (cm) is heated at one end. The table above
give selected values of the temperature T (x), in degrees Celsius (◦ C), of the wire
x cm from the heated end. The function T is decreasing and twice differentiable.
(a) Estimate T 0 (7). Show the work that leads to your answer. Indicated units
of measure.
(solution, (a)): Since T is known to be differentiable, we may apply the symmetric difference quotient to compute the derivative:
T (a + h) − T (a − h)
.
h→0
2h
T 0 (a) = lim
This gives a convenient approximation for T 0 (7) :
T (8) − T (6)
55 − 62
7
=
= − ◦ C per cm of length.
2
2
2
(b) Write an integral expression in terms of T (x) for the average temperature of
the wire. Estimate the average temperature of the wire using a trapezoidal
sum with the four subintervals indicated by the data in the table. Indicate
units of measure.
(solution, (b)): First of all, the average temperature—expressed as an integral—
is given by
T 0 (7) ≈
1
Avg Temp =
8
Z
8
T (x) dx.
0
We may approximate this via four trapezoids constructed from the given
data; the result is then
Z
8
T (x) dx ≈
0
1
[(100 + 93) + 4(93 + 70) + (70 + 62) + 2(62 + 55)] ≈ 605.5,
2
and so the average temperature over the length of wire is given by
Avg temperature ≈
Z
1
× 605.5 = 75.6875.
8
8
(c) Find
T 0 (d) dx, and indicate units of measure. Explain the meaning of
0
Z 8
T 0 (d) dx in terms of the temperature of the wire.
0
Z
(solution, (c)): This is easy:
8
T (x) dx = T (8) − T (0) = 55 − 100 = −45 ◦ C,
0
which represents the total temperature drop over the length of wire.
(d) Are the data in the table consistent with the assertion that T 00 (x) > 0 for
every x in the interval 0 < x < 8? Explain your answer.
(solution, (d)): Under the assumption that T 00 (x) > 0 on the interval 0 < x < 8
we would observe that every finite difference quotient computed from the
data would represent a the slope of a chord lying entirely below the graph.
As such, the slopes of these chords would increase for increasing values of x.
However, note that
T (1) − T (0)
T (6) − T (5)
>
,
1
1
|
{z
} |
{z
}
−7
−8
which contradicts the assumption that T 00 (x) > 0 over the interval of definition.
4. Consider the differential equation
dy
= 2x − y.
dx
(a) On the axes provided, sketch a slope field for the given differential equation
at the twelve points indicated, and sketch the solution curve that passes
through the point (0, 1).
(solution, (a)):
6
C
•C
C
2 •A
•B
1@•@
•
•
•
1
A
•
•
A
B
B
A
•A
A
−1
•
• 2
(b) The solution curve that passes through the point (0, 1) has a local minimum
at x = ln 23 . what is the y-coordinate of this local minimum?
(solution, (b)): Since the solution function y = y(x) is a differentiable function
dy
of x, at the local minimum we must have dx
= 0. Since we are told that at
dy this point x = ln 32 , this condition becomes 0 = dx
= 2 ln 23 − y, which
3
x=ln
implies that y = 2 ln
3
.
2
2
(c) Let y = f (x) be the particular solution to the given differential equation
with the initial condition f (0) = 1. Use Euler’s method, starting at x = 0
with two steps of equal size, to approximate f (−0.4). Show the work that
leads to your answer.
(solution, (c)): We shall be using step size of 0.2; since we are to approximate
f (−0.4), we shall have ∆x = −0.2 in each case. Euler’s method is based on
ynew − yold = ∆y ≈
dy
∆x,
dx
i.e., that
ynew
dy ≈ yold + ∆x.
dx (xold , yold )
We have, then, that
f (−0.2) ≈ 1 + (2x − y)
(−0.2)
(0,1)
= 1 + (−1)(−0.2) = 1.2.
Therefore,
f (−0.4) ≈ 1.2 + (2x − y)
(−0.2)
(−0.2,1.2)
= 1.2 + (−1.6)(−0.2) = 1.52.
d2 y
in terms of x and y. Determine whether the approximation found
dx2
in part (c) is less than or greater than f (−0.4). Explain your reasoning.
(d) Find
(solution, (d)): First of all, we have
d2 y
d
dy
=
(2x − y) = 2 −
= 2 − (2x − y) = 2 + y − 2x.
2
dx
dx
dx
2
d x
This says that in the second quadrant, we have dx
2 > 0, i.e., that the solution
curve over this interval is concave up. Since our approximations are based
on local linear approximations, we see that the estimates will, at each
stage underestimate the true values of the solution y = f (x).
v(t)
(4, 20)
20 6 •
Velocity
(m/s) 10 5 15
(16, 20)
•
BB
B
B
B
B
B
B
B
B
B
B
-t
4 8 12 16 20 24
Time (seconds)
5. A car is traveling on a straight road. For 0 ≤ t ≤ 24 seconds, the car’s velocity
v(t), in meters per second, is modeled by the piecewise-linear function defined
by the graph above.
Z 24
Z 24
(a) Find
v(t) dt. Using correct units, explain the meaning of
v(t) dt.
0
0
(solution, (a)): Since v(t) is non-negative over the domain of definition, the
given integral is simply the area under the velocity curve and above the
t-axis. Geometrically, this is the area of a trapezoid of height 20 and bases
24 and 12. The relevant area is then 20 × 18 = 360. Since the above integral
is (again, since v is non-negative) the total distance traveled in the car over
the time interval, 0 ≤ t ≤ 24, we conclude that over the first 24 seconds, the
car has traveled a total distance of 360 meters.
(b) For each of v 0 (4) and v 0 (20), find the value or explain why it does not exist.
Indicate units of measure.
(solution, (b)): v 0 (4) does not exist as the graph of v is not smooth at t = 4.
At t = 20, the graph of v is smooth, linear, and has slope − 20
= −2.5.
8
Therefore, v 0 (20) = −2.5 meters per second per second.
(c) Let a(t) be the car’s acceleration at time t, in meters per second per second.
For 0 < t < 24, write a piecewise-defined function for a(t).
(solution, (c)): The acceleration a(t) = v 0 (t), and hence is not defined at the
values t = 4, 16 (as the velocity graph is not smooth at these points). On
the interval 0 < t < 4 the derivative of v(t) is the slope of the linear segment
defining v(t) over this interval, viz., 20
= 5. On the interval 4 < t < 16, the
4
derivative of v(t) is clearly 0, and on the interval 16 < t < 24 the derivative
of v(t) is (as observed above) −2.5. That is to say, the acceleration function
has the piecemeal definition


if 0 < t < 4,
 5
a(t) =
0
if 4 < t < 16,

−2.5 if 16 < t < 24.
(d) Find the average rate of change of v over the interval 8 ≤ t ≤ 20. Does the
Mean value Theorem guarantee a value of c, for 8 < c < 20, such that v 0 (c)
is equal to this average rate of change? Why or why not?
(solution, (d)): The average rate of change of v over the interval 8 ≤ t ≤ 20 is
given by
Avg rate of change of v =
v(20) − v(8)
10 − 20
5
=
= − m/sec2 .
20 − 8
12
6
To apply the Mean Value Theorem to v would require that v be differentiable over the interval 8 < t < 20, which it is not, as the graph of v is
not smooth at t = 16. Therefore, nothing further can be inferred here.
6. Let f be a function with derivatives of all orders and for which f (2) = 7. When
n is odd, the nth derivative of f at x = 2 is 0. When n is even and n ≥ 2, the
(n − 1)!
nth derivative of f at x = 2 is given by f (n) (2) =
.
3n
(a) Write the sixth-degree Taylor polynomial for f about x = 2.
(solution, (a)): The sixth-degree Taylor polynomial for f about x = 2 is given
by
T6 (x, 2) =
6
X
f (j) (2)
j=0
= 7+
j!
(x − 2)j
1
1
1
(x − 2)2 +
(x − 2)4 +
(x − 2)6 .
2
4
2·3
4·3
6 · 36
(b) In the Taylor series for f about x = 2, what is the coefficient of (x − 2)2n
for n ≥ 1?
(solution, (b)): The coefficient of (x − 2)2n is
f 2n (2)
(2n − 1)!
1
=
= 2n
.
(2n)!
3 (2n)!
2n · 32n
(c) Find the interval of convergence of the Taylor series for f about x = 2.
Show the work that leads to your answer.
(solution, (c): We first find the radius of convergence by using the comparison
test and requiring that
f 2(n+1) (2)|x − 2|2(n+1)
n→∞
f 2n (2)|x − 2|2n
32n 2n|x − 2|2
= lim 2(n+1)
n→∞ 3
2(n + 1)
2
|x − 2|
2n
=
lim
n→∞
9
2n + 2
|x − 2|2
=
.
9
1 >
lim
This implies that the radius of convergence is 3, i.e., that the Taylor series
certainly converges on the interval −1 < x < 5.
It remains to check the behavior at the endpoints x = −1 and x = 5. Since
only even powers are involved, in both cases the infinite series reduces to
7+
∞
X
n=1
∞
X
32n
1
1
=
7
+
= 7 + × Harmonic Series,
2n
3 · 2n
2n
2
n=1
which of course diverges. Therefore, the interval of convergence is −1 < x < 5 .