Solutions to the
2005 AP Calculus BC Exam
Free Response Questions

Louis A. Talman
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver

Problem 1.

a)

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So the required area is

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b)

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c)

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Problem 2.

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a)

Area is

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b)

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c)

When , decreases as increases.  This means that as θ ranges upward from to , the corresponding point on the curve gets closer to the pole.

d)

We seek to find the absolute maximum value of on the interval .  This will be found where or or .  Now and .  This eliminates from contention.   when , or when .  We have .  We conclude that the maximum distance from the pole to the curve for occurs when .

Problem 3.

a)

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Thus is approximately degrees C per centimeter, or -7/2 degrees C. per centimeter.

b)

Average temperature of the wire is .  This is

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c)

We are given that is twice differentiable (although we are not told where).  This means that is continuous--on the interval [0,8], we hope, so that the problem is meaningful.  By the Fundamental Theorem of Calculus, degrees Celsius.  The integrand is the (instantaneous) rate at which changes per unit length at each point of the interval , and the integral gives net temperature change over the interval .

d)

By the Mean Value Theorem, there is a point where .  By the Mean Value Theorem again, there is a point where .  Note that, necessarily, .  A third application of the Mean Value Theorem---this time to ---assures us that there is a point such that .  Consequently, these data are not consistent with the assertion that for every in

Problem 4.

a)

b)

At a local minimum, we must have , and because we are on a solution curve this tells us that .  Hence, the -coordinate of this minimum is .

c)

Because , we have .  Consequently, .  Consequently, .  So .

d)

If , then .  This latter quantity is positive when and --as it surely is for our solution.  Hence the solution must be concave upward in the region where we are examining it.  Consequently tangent lines to the solution lie below the curve throughout the interval .  Because Euler's Method proceeds by replacing the curve with tangent lines, the method underestimates in this region.

Problem 5.

a)

gives the distance, in meters, traveled by the car during the time period .

b)

is the definition of .  We note that for , we have , so that when but is small.  Thus for such values of .  But when , we have , so that when and is small.  For such we therefore have .  Consequently, , but .  The two one-sided limits have different values, so the two-sided limit does not exist.  this means that does not exist.

Now when .    Hence for all for which , by the standard rule for differentiating polynomials.  It follows that .

c)

Acceleration is given by when , by when , and by when

d)

The average rate of change of over is .  The hypotheses of the Mean Value Theorem require that exist at every point of the interior of an interval on which we wish to apply the theorem.  The Mean Value Theorem is therefore inapplicable to on because does not exist.

Problem 6.

a)

The 6-th degree Taylor polynomial about for this function is

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b)

The coefficient of in the Taylor expansion of this function about is .

c)

We have .  Consequently, the Ratio Test guarantees that the series converges when , or when .  When , the general term of the series becomes .  This is a constant multiple of the harmonic series, and so diverges.  The interval of convergence is therefore .

Created by Mathematica  (May 6, 2005)