Solutions to the
2006 AP Calculus AB Exam
Free Response Questions

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

Part A

Problem 1.

a)

The curves intersect where .

b)

c)

Answer:  Either of the integrals or will do.  Evaluation was not required; however

and, of course,

Problem 2:

a)

The rate at which left turns happen is:

The number of turns when is therefore

Answer:  To the nearest whole number, this is 1658.

b)

From the graph, we see that   on where and .  Solving numerically for and , we find

Answer:  There are 150 or more left turns per hour approximately when , where is measured in hours.  The average during this interval is left turns per hour.

c)

During the two-hour interval ,

cars make left turns.  500 oncoming cars pass straight through the intersection in this two-hour period.  The product of these two numbers is , and this exceeds the threshhold of .

Answer:  The intersection requires a traffic signal. The reasoning is given in the preceding paragraph.

Problem 3:

a)

;
.

b)

Answer:  By the Fundamental Theorem of Calculus, .  Thus, , while for , and for .  By the First Derivative Test, has a relative minumum at .

c)

By periodicity, g(10) - g(5) = g(5) - g(0) = g(5) = 2.Hence, g(10) = g(10) - g(0) = [g(10) - g(5)] + [g(5) - g(0)] = 2 + 2 = 4.  Reasoning as in the previous sentence, we find that for any positive integer , we must have , so that .  Also, , by inspection of the graph.  Hence .  Also, the first of the following equalities being a consequence of the Fundamental Theorem of Calculus, and the second being a consequence of periodicity, .  So the equation of the line tangent to at is .

Answer:   , and the equation of the tangent line at is .

Part B

Problem 4:

a)

Answer:  Average acceleration is feet per second per second.

b)

Answer:   measures the distance in feet between the rocket's position at time to its position at time .  The midpoint Riemann sum with three subdivisions of equal length is feet.

c)

Answer:  For rocket B, we have, by the Fundamental Theorem of Calculus, ft/sec.  Thus Rocket A is traveling at 49 ft/sec when , so rocket B is traveling faster.

Problem 5:

a)

b)

If , then .  Thus, , as long as and .  Consequently, , or .  Now we have assumed that , so the latter equation is equivalent to .  When is near , must be near , so for such .  Hence, , or .   We may not have either or , so the domain of this solution is .

Problem 6:

a)

If , then , and .