.
Solutions to the
2011 AP Calculus BC, Form B,
Free Response Questions
Louis A. Talman, Ph.D.
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver
Problem 1
(a)
According to the model, the height of the water in the can at the end of the 60-day period is .
(b)
The average rate of change in the height of water in the can over the 60-day period is given by 
, where we have inserted the value of the integral obtained in part (a).
(c)
The volume V of water in the can at time t is given by V(t) = 100 π S(t), so V'(t) = 100 π S'(t). Consequently, 
cubic millimeters per second.
(d)
We have 
. Using S'(t) as given, we find that 
, while 
. Because D is a continuous function on [0,60], it follows from the Intermediate Value Theorem that there is a time 
such that 
, which is to say that 
, or the two rates are the same.
Problem 2
(a)
The area of the polar curve r = r(θ) corresponding to the interval α≤θ≤β is given by 
, so we must evaluate 
. The exact area is
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This is about
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Numeric integration yields substantially the same result:
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(b)
We are to solve the equation r(θ) cos θ = (3 θ + sin θ ) cos θ== -3, for θ between π/2 and π. To this end, let f(θ) = (3 θ + sin θ) cos θ + 3. We apply Newton's method to solve the equation f(θ) = 0. We take 
as our initial approximation.
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To the nearest thousandth, the solution we seek is θ = 2.017. The y-coordinate of the point P is then
(3 θ + sin θ) sin θ, or
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To the nearest thousandth, y = 6.272.
(c)
We have y(θ) = r(θ) sin θ, so that y'(θ)= r'(θ) sin θ + r(θ) cos θ.
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When θ = 2 π/3, this gives
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But 
, and 
is given as 2. Thus, when θ = 2π/3, we have 
. This is the y-component of the velocity of the particle at the instant in question; it is negative, so the particle is moving downward.
Problem 3
(a)
The area of the pictured region R is 
.
(b)
A cross-section of this solid perpendicular to the y-axis at y = t is a rectangle whose height is 2 t and whose base extends from the curve 
to the curve x = 6 - y. The area of such a cross-section is therefore 
, so the required integral is 
.
Note: It is not required to evaluate the integral. However, the curious may wish to know that the volume is 
.
(c)
The slope of the line y = 6 - x is -1, so we seek a point on the curve 
where y' = 1. But 
when 
, or 
. The point P therefore has coordinates 
.
Problem 4
(a)
The average value of f over [0,5] is 
.
(b)
We have 
(c)
If 
, then g'(x) = f(x), by the Fundamental Theorem of Calculus. But f(x) < 0 for all x ∈ (0, 5), and for all so g is increasing on [0, 5]. The derivative f' can change sign only at the critical points x = 3 and x = 8, so from the fact, evident from the graph and what is given, that f has a minimum at x = 3 and a maximum at x = 8, we see that f', which is g'', must be positive on (3, 8). It follows that g is concave upward on (3, 8), or, depending upon how upward concavity is defined, possibly on [3, 8]. It follows that g is both increasing and concave upward on (3, 5], or, for some definitions of upward concavity, on [3, 5].
(d)
The required arc-length is given by 
. Taking 
, we find that 2 dx = dt, that x = 0 when t = 0, and that x = 10 when t = 20. Consequently, we may replace the integral in the first sentence of this paragraph by
the values of the last two of these integrals having been given in the statement of the problem.
Problem 5
(a)
Ben's acceleration at time t = 5 is approximately [v(10) - v(0)]/(10 - 0) = (2.3 - 2.0)/10 = 0.03 meters per second per second.
(b)
The integral 
is the integral of Ben's speed. It measures the total distance he has traveled when 0 ≤ t ≤60. We have 
, so the total distance Ben traveled during this minute is about 139 meters.
(c)
We have [B(60) - B(40)]/(60 - 40) = (49 - 9)/(60 - 40) = 40/20 = 2. By the Mean Value Theorem, there must be a time 
when 
. [Note: We may apply the Mean Value Theorem here because we are given that B is a twice differentiable function. Consequently, B is continuous on [40, 60] and differentiable on (40, 60), and these are precisely the conditions that the hypotheses of the Mean Value Theorem require.]
(d)
From 
, we find that 2 L L' = 2 B B'=2 B v. Thus, when t = 40, we have 
. However, when t = 40, we also have 
, or L = 15. Thus, at t = 40, 45 = 2 L L' =2·15·L', or 
meters per second.
Problem 6
(a)
We may substitute 
for x in the Maclaurin series for ln(1 + x) to obtain that for f. The Maclaurin series for 
is therefore
(b)
The series for f must converge in the open interval (-1,1) because x=0 is the center of the expansion and the radius of convergence is given to be 1.
The only real issue is whether or not the series for f converges at the endpoints. When x = 1, the series becomes 
which is the convergent alternating harmonic series. When x = -1, the series becomes 
, which is the negative of the divergent harmonic series, and therefore diverges.
The interval of convergence for the Maclaurin series for f is therefore (-1, 1].
(c)
The Maclaurin series for f' is
+… .
Consequently, the first four nonzero terms of the Maclaurin series for 
are
.
Replacing 
with the first two terms of this series in 
gives
.
(d)
The Maclaurin series for g begins with the terms 
,
and we have been given that the series meets the hypotheses of the Alternating Series Test. We used the first two terms above to approximate g(1) in part (c) above. Hence the error in our approximation is bounded by the magnitude of the third term, which is 
when x = 1.
[Note: In fact, it can be shown that 
,
so the approximation of part (c) is a pretty miserable one.]
