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AP Calculus Featured Question: July 2005
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by Ben Klein
Davidson College
Davidson, North Carolina
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Polar Coordinates and Conic Sections
The May 2005 Featured Question extended BC3 from the 2003 AP Calculus Examination, a problem that dealt in part with polar coordinates. The extension did not do much with polar coordinates, but we promised in May that we would revisit BC3 and do more with polar coordinates in a subsequent Featured Question. Prompted in part by the fact that one of the free-response problems on this year's BC examination involved polar coordinates, we decided to revisit BC3 from 2003 sooner rather than later.
We begin this Featured Question with a slight variation on the problem posed in BC3 from 2003, in which students were asked to use polar coordinates to find the area of a region bounded by a hyperbola and two line segments. We then explore similar situations involving first a parabola and then a straight line, rather than a hyperbola. The last part of this Featured Question uses the results from previous parts to verify a special case of a result of Archimedes relating the area of a parabolic segment to the area of a triangle inscribed in the segment.
Even though all of the computations called for in this Featured Question can be done in rectangular coordinates, we hope that you will use polar coordinates instead. We also hope that you will do all of the necessary calculations by hand, rather than with a computer algebra system. If you use the suggestions we have provided, you will find that the calculations are surprisingly simple.
(A) Consider the region in the first quadrant that is bounded by the graph of the hyperbola
x2 - y2 = 1, the line y = mx where 0 < m < 1 , and the x-axis.
(1) Show that if
, 
, then the line and the hyperbola do not meet.
(2) Express the hyperbola in polar coordinates and then show that the area of the region is given by 
.
(3) Show that 
and use this result to show that the area of the region is 
. [You can, of course, check this result by differentiating both sides. However, it is more challenging and interesting to evaluate the indefinite integral directly. If you do, you will find the trigonometric identity 
helpful.]
(B) Consider the region in the first quadrant that is bounded by the graph of the parabola
y = x2 and the line y = mx, where m is a positive parameter. This region is a segment of the parabola, i.e., a region bounded by an arc of the parabola and the line segment that joins the ends of the arc.
(1) Express the parabola in polar coordinates and then show that the area of the region is given by 
.
(2) Evaluate the integral in (1). [Hint: Let 
.]
(C) Let a, m, and k be positive parameters with mk < 1.
(1) Show that the line y = mx and the line with x-intercept a and slope 1/k meet at a point in the first quadrant.
Consider the triangle with vertices (0,0), (a,0), and the point of intersection found in (1).
(2) Express the line ky = x - a (the second line in (1)) in polar coordinates and then show that the area of the triangle is given by 
.
You could use a computer algebra system to evaluate the integral in (2), but the hints below provide a way to evaluate the integral by hand.
(3) Let 
and show that 
. Use this identity to evaluate the integral in (2).
(4) Verify your answer to (3) by noting that the y-coordinate of the point of intersection found in (1) is the height of the triangle if the base of the triangle is taken to be the segment [0,a] on the x-axis.
(D) Let m = 1 and consider the parabolic segment of part (B). Let x0 be in the interval (0,1) and consider the triangle whose vertices are (0,0), (1,1), and 
. This triangle is inscribed in the parabolic segment.
(1) Use the result from part (C) twice to show that the area of the inscribed triangle is 
. [Hint: If L is the line through (1,1) and 
, you need to consider first the two lines L and y = x then consider the two lines L and y = x0(x).]
(2) Show that the area of the inscribed triangle is maximized when 
and then show that slope of the tangent to the parabola y = x2 at
is 1. Use part (B) to find the area of the parabolic segment and show that this area is four-thirds of the maximum area of the inscribed triangle.
Note: Part (2) of (D) is a special case of a theorem of Archimedes on parabolic segments. Archimedes showed that the area of any parabolic segment is four-thirds of the area of the triangle of largest area that can be inscribed in the segment.
Complete the question before viewing the answers and explanation!
Ben Klein is currently the Beverly F. Dolan Professor of Mathematics at Davidson College in Davidson, North Carolina, where he has taught since 1971. Ben's relationship with AP Calculus began in 1990 when he served as a Reader at Clemson University. He has attended every reading since then and served as a Question Leader at the 2005 Reading. In 2003, he completed a four-year term on the AP Calculus Development Committee.
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