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AP Calculus Question of the Month: December
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by Lin McMullin
Educational Consultant and Writer
Niantic, Connecticut
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Fun Investigations with a CAS: Third-Degree Polynomials, Part 2
Recall that last month we noticed a nice feature of cubic polynomials; namely, that the tangent at the average of two roots had an x-intercept at the third root. Now we investigate areas bounded by two tangents to a cubic polynomial. This month's question uses techniques from both differential and integral calculus -- finding slopes, writing the equations of tangent lines, and finding the areas of regions between two graphs.
This month's investigation begins with any cubic function. Draw a tangent line at any point, other than the point of inflection of the cubic. This tangent will intersect the cubic at a second point; draw a tangent line at this second point. The second tangent will intersect the cubic at a third point. Let A1 be the area of the region between the first tangent line and the cubic, and let A2 be the area of the region between the cubic and the second tangent line. A general graph is given below. The interesting result is that the ratio A2 : A1 is constant.
The investigation requires three steps, and a computer algebra system (CAS) is helpful.
(A) Find the ratio A2 : A1.
(B) Prove that the ratio is constant.
At this point, before you start to work on the problem, you may wish to consider the following.
1. Since the ratio is given as constant, you may use any cubic to find it. Pick a simple one, and pick "nice" points. A nice way to use this idea in class is to divide the class into groups and give each group the equation of a different (simple) cubic, and possibly a convenient initial point, to work with. When they all come up with the same ratio, you have an open invitation to prove the general result.
2. To find the second point in which the tangent line at x = a (say) meets the cubic you have chosen, you will need to find the roots of a (new) third degree equation, equating the cubic and the equation of the tangent line. You can do this without using a computer algebra system (a CAS) or even a numeric solver. Note that x = a is a double root of the new third degree equation. So, if you divide the third degree expression by (x - a)2 , the quotient will be a linear function whose x-intercept is the x-coordinate of the second point of intersection.
3. Proving that the ratio is constant in general is more difficult; this is a problem for a CAS.
Complete the question before viewing the answers and explanation!
Lin McMullin, an educational consultant and writer with extensive experience teaching AP Calculus, lives in Niantic, Connecticut. As a College Board consultant he has presented AP Calculus institutes and workshops in the United States and Europe, and is a Table Leader at the Reading. His work as a writer includes co-authorship of the popular D&S Marketing Systems review books for the AP Calculus Exam and Teaching AP®
Calculus, a book especially for AP Calculus teachers.
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