Solutions to the
2004 AP Calculus BC Exam (Form B)
Free Response Questions
Louis A. Talman
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver
Problem 1. (Thanks to Bill Kehlenbeck for pointing out a careless error that rendered the first three parts of this one wrong.)
The tangent vector is
At time ,
The slope of a line parallel to this vector is , so an equation of the line tangent to the curve at , the point corresponding to , is .
Total distance travelled over the interval is . Integrating numerically, we obtain
The -coordinate of the particle at time is . Numeric integration gives
We are given as the third-degree Taylor polynomial for a certain function about .
In any Taylor polynomial, the coefficient of is . Hence and .
Reasoning as in part a.), we find that , so that has a critical point at . Because , the Second Derivative Test allows us to conclude that has a local maximum at .
. There is not enough information to determine whether or not has a critical point at . This is because the third-degree Taylor polynomial carries no information about derivatives at any point other than the point about which the expansion has been done; it is determined solely by the values of the function and its first three derivatives at that point.
The Lagrange Remainder for the third-degree Taylor polynomial at has the form , where is some unknown number in the interval between and . Thus, for a certain . Because for all , this means that . But, from part c.) above, we know that . Hence, , whence .
We are given
(This arcane syntax assigns the correct values to v, v, etc.)
The Mid-Point Rule with four subintervals of equal length gives for the approximate value
is given in miles per minute, so the integral gives miles travelled during the time interval over which the integral is taken.
By Rolle's Theorem, acceleration--which is --must be zero at least once in the interval , because . Similarly, must be zero at least once in the interval , because . Thus, acceleration must vanish at least twice in the interval .
If the function models velocity, then acceleration is . Thus
At time , acceleration is .
Average velocity over is :
Inflection points are to be found where has relative extrema. Consequently, the function whose derivative is pictured has inflection points at and at .
b. Thanks to Tammy Brown, Becky Myers, and Sue Wall, who all pointed out that I'd blown this one completely.
The function is decreasing on the interval and increasing on the interval because is non-positive on the first of these intervals, and non-negative on the second. Consequently, takes on its absolute minimum value for the interval at .
The function has its absolute maximum at one of the points where or . (There can be no absolute maximum for at any point interior to because has in that interval no zeroes at which it undergoes a sign change from positive to negative as increases.) The area bounded by and the -axis on the interval is clearly larger than the area bounded by and the -axis on the interval , so . This means that , so that the absolute maximum value taken on by in the interval is .
We are given , so . Because , this means that an equation for the line tangent to the graph of at is .
The average value of on the interval is .
The volume of the solid generated when the region bounded by the graph of , the vertical lines and , and the -axis is revolved about the -axis is .
The average value of the areas of the cross sections perpendicular to the -axis is .
The improper integral is , so the improper integral diverges. However, .
We have .
If , then , so that the equation of the line ℓ is . This line meets the -axis when , so that the base of the triangle has length . Because the altitude of the triangle is 1, the area of is .
From what we have seen in parts a.) and b.), the area of the region is . Then , which is zero for only when (by the Quadratic Formula). Noting that for , while for , we conclude that the maximal area occurs when .
Created by Mathematica (May 13, 2004)