Solutions to the
2006 AP Calculus AB Exam
Free Response Questions
Louis A. Talman
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver
The curves intersect where .
Answer: Either of the integrals or will do. Evaluation was not required; however
and, of course,
The rate at which left turns happen is:
The number of turns when is therefore
Answer: To the nearest whole number, this is 1658.
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.
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.
Answer: By the Fundamental Theorem of Calculus, . Thus, , while for , and for . By the First Derivative Test, has a relative minumum at .
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 .
Answer: Average acceleration is feet per second per second.
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.
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.
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 .
Answer: , when .
If , then , and .
If , then . Thus, , while .
Answer: . The equation of the line tangent to the graph of at is , or .
Created by Mathematica (May 8, 2006)