Solutions to the
2007 AP Calculus AB Exam
Free Response Questions
Louis A. Talman
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver
The limits of the integral that gives the area are the solutions of the equation :
The required area is therefore
The required area is .
Using the method of washers, we find that the required volume is
The required volume is .
The diameter of the semicircle at is , so the radius is . Hence the area of the cross section at is . The required volume is therefore
The required volume is .
The amount of water that enters the tank during the time interval is :
That's about 8263.806 gallons.
From the graph and what we are given about the intersection points of the curves, we see that the rate at which water leaves the tank exceeds that at which it leaves the tank on the intervals and . It follows that the amount of water in the tank is decreasing on each of the intervals and .
The rate at which the amount of water in the tank increases is, as we have seen in part b), negative on the interval . It is positive on the interval . By the First Derivative Test, the amount of water in the tank has a local maximum at . We must also consider the amount of water in the tank when and the amount when . Over the interval , gallons of water have left the tank, while, according to part a), 8263.806 gallons have entered. When , the amount of water in the tank is therefore gallons. The amount of water in the tank at is :
The maximum occurs when and, to the nearest gallon, is gallons.
The functions and are differentiable for all real numbers, and therefore are continuous everywhere. As a composition of continuous differentiable functions, must be everywhere continuous and everywhere differentiable. If , then , while . But lies between and . By the Intermediate Value Property of continuous functions, there must be where .
As noted above, the function is everywhere differentiable. By the Mean Value Theorem, there must be a where , or .
If is the function given by , then by the Fundamental Theorem of Calculus and the Chain Rule, . Thus, .
Putting , we have . Thus, . The equation of the tangent line is .
The particle is farthest to the left when the function assumes its absolute minimum value on the interval . Because when and when , this minimum must lie somewhere in , and must therefore be at a critical point of . We have , and never vanishes. Hence the critical point of that we seek must lie at , which is the only zero of in the interval . We conclude that the particle is farthest left when .
Because and , the equation reduces to . This latter equation is true for only if , or if .
The linearization of at is the linear function . An approximate value for is feet. The curve is given to be concave downward, so the tangent line at each point lies (locally) above the curve. Thus, our estimate of ft is an overestimate.
Because , we have . Thus, .
The right Riemann sum corresponding to the data given is feet. By the Fundamental Theorem of Calculus, is the change, in feet, in the radius from its value when to its value when .
The function is given concave down, so is a decreasing function. Consequently, when lies anywhere in an interval . It follows that each term of the right Riemann sum is less than the area under the curve in the corresponding interval. The right Riemann sum therefore underestimates the integral.
If , then , and .
We must have , or if is to be a critical point. Then , so by the Second Derivative Test the critical point gives a relative minimum.
An inflection point lies where changes sign. Because is continuous on the interval in question, we must have at such a point. Thus, , or . But we want our inflection point to lie on the -axis, and this means we must have , or . This is equivalent to = 2, whence .
Created by Mathematica (May 12, 2007)