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26: (a) Parabolic segment. (b) Tangent sweep of parabolic segment cut oﬀ by the x axis. (c) Region obtained by doubling the lengths of the tangent segments in (b). of the rectangle. Archimedes made the stunning discovery that the area is exactly one-third that of the rectangle, or R/3. We will deduce this by a simple geometric approach using tangent sweeps. The parabola has equation y = x2 , but we shall not need this. 26b. 26b is obtained by drawing all tangent lines to the parabola between 0 and x and cutting them oﬀ at the x axis.
Power functions have linear subtangents. In fact, for a nonzero constant b we have f (x) = Kx1/b for a constant K = 0 if and only if s(x) = bx. In particular, the parabola f (x) = x2 has subtangent s(x) = x/2, and the hyperbola f (x) = 1/x has subtangent s(x) = −x. 21 (left) shows how tangent lines to the parabola f (x) = x2 can be easily constructed by joining (x/2, 0) to (x, x2 ). 21 (right) illustrates the tangent construction for the hyperbola f (x) = 1/x. Here x − s(x) = 2x, so the tangent line passes through the points (2x, 0) and (x, 1/x).
Because the angle T P C is a right angle, the chord P T is perpendicular to the normal P C and hence is tangent to the cycloid. Thus, the cycloidal cap is a tangent sweep. 4. Then the other extremity P moves to point P , with P T equal in length and parallel to P T . Obviously, the segments P T are chords of a circular disk congruent to the rolling disk. By Mamikon’s theorem, the area of the tangent sweep P ODT is equal to that of the tangent cluster T C P . 1. 1) Throughout this chapter we employ square brackets to designate areas of regions.