Arc Length, Area, and the Arcsine Function Andrew M

Arc Length, Area, and the Arcsine Function Andrew M. Rockett Long Island University Greenvale, NY 11548 Mathematics Magazine, March 1983, Volume 56, Number 2, pp. 104–110. eometric interpretations of algebraic quantities provide an essential motivation for elementary calculus. The “calculus with analytic geometry” titles of many Gintroductory texts allude to the reliance on pictorial images in the development of the derivative and the definite integral. While geometric considerations dominate most presentation of the trigonometric functions and their derivatives, there is a near universal switch to algebraic methods for the introduction and study of the inverse trigonometric functions (see [6], [3], and so on). This note shows how the geometric ideas used in the definitions of the definite integral, the trigonometric functions, and their derivatives may be continued in the discussion of the corresponding inverse functions. Reference will be made to the relation of these geometric ideas to the development of the theory of elliptic functions and to Euler’s method of finding algebraic addition theorems for circular, hyperbolic, and lemniscate sines. “Typical calculus” versus the arcsine as arc length Taking [6] and [3] as our models, we note that in the typical modern textbook, after the definite integral has been defined, the applications include the area between two curves and the arc length formula. Since few integration techniques are available, the arc length 5 s d eb! 1 9s d2 problems are restricted to “nice” curves y f x such that the integral a 1 f x dx is particularly simple and sometimes an author’s apology is offered for the lack of interesting applications (see [3], p. 429). The introduction of the trigonometric functions follows a review of radian measure as arc length measured from the point s1, 0d on the unit circle x2 1 y2 5 1. The sine and cosine of a real number u are defined as the coordinates of the point sx, yd on the unit circle u radians from s1, 0d (see FIGURE 1). Then the properties of sin u and cos u are derived from the symmetries of the circle and the other trigonometric functions are defined in terms of the sine and cosine. The derivatives of the sine and the cosine are found as consequences of limssin udyu 5 1. This limit is established by equating arc u→0 length along the edge of the unit circle with the area of the sector determined by the arc sin Figure 2,u 5 2 ? area AOBd and then squeezing this area between two triangular regions. After studying the calculus of the six trigonometric functions s"fsxd"d, the corresponding inverse functions s"f 21sxd"d are sought by reversing the graphs s"y 5 fsxd" becomes "x 5 f21syd"d, making arbitrary choices for the “principal values” (see [6], pp. 295–6), s 21s dd s 21s dd 5 and then calculating Dx f x from the identity f f x x. FIGURE 1 FIGURE 2 FIGURE 3 FIGURE 4 In contrast to exploiting the definitions and properties of inverse functions, the arcsine function can be approached in a more geometric way. Since sin u was defined as the y- coordinate of the point on the unit circle an arc length of u away from s1, 0d, a geometric attempt to invert this function would ask “given that the second coordinate of the point is y, from what arc length did it come?” There are two “small” answers to this question (see Figure 3) and then infinitely more separated from each other by multiples of 2p. The “principal value” of the inverse sine function may be introduced naturally as the smallest distance from the stating point, s1, 0d, that will work, and this arc length may be called the “arc sine of y” and written “arcsin (y)” or the “inverse sine of y” and written "sin21syd." We will use the “arcsine” notation for the remainder of our discussion. Thus the arcsine function has arcsin syd 5 u where 2py2 ≤ u ≤ py2 and sinu 5 y. eb! 1 9s d2 Since arcsin (y) is an arc length, the arc length formula a 1 f t dt can be applied to fstd 5 !1 2 t2 from t 5 0 to t 5 y (see FIGURE 4) to find that y arcsinsyd 5 E !1 1 f9std2 dt 0 y t2 5 E 1 1 dt ! 2 2 0 1 t 2 y 1 5 E dt (1) ! 2 2 0 1 t and then, by the Fundamental Theorem of Calculus, it follows that 1 D sarcsinsydd 5 . y !1 2 y2 Since arcsins1d 5 py2, we also have a simple example of an improper integral: 1 1 y 1 p E dt 5 limE dt 5 . ! 2 2 → ! 2 2 0 1 t y 1 0 1 t 2 For the inverse tangent, a similar argument yields an arctangent function arctanswd 5 u with principal value 2py2 < u < py2 such that arctan(w) is the distance along the unit circle from s1, 0d to the point sx, yd with yyx 5 w. Since x 5 !1 2 y2, we can solve yyx 5 w to find y 5 wy!1 1 w2. From the arc length formula (1) we have y! 1 2 w 1 w 1 arctanswd 5 E dt. ! 2 2 0 1 t Making the change of variable t 5 uy!1 1 u2, so that t 5 0 when u 5 0 and t 5 wy!1 1 w2 when u 5 w, we find that w 1 arctanswd 5 E !1 1 u2 du ! 1 2s 1 2d 0 1 u 1 u w 1 5 E du, 1 2 0 1 u and so 1 D sarctanswdd 5 . w 1 1 w2 For the inverse secant of w (where |w| ≥ 1d, if we seek the smallest positive arc length from s1, 0d to the point with first coordinate 1/w, we obtain an arcsecant function arcsecswd 5 u with principal value 0 ≤ u < py2 and py2 < u ≤ p. For w > 1, the arc length u is py2 2 f (see FIGURE 5) so we have p arcsecswd 5 2 arcsins1ywd 2 and so 1 21 D sarcsecswdd 5 2 ? w !1 2 s1ywd2 w2 1 5 for w > 1. w!w2 2 1 For w < 21, we have by symmetry that arcsecswd 5 p 2 arcsecs2wd 3 and so s s dd 5 2 s s2 dd Dw arcsec w Dw arcsec w and thus 1 D sarcsecswdd 5 for |w| > 1. w |w|!w2 2 1 (We omit discussion of the formula ! 22 y w 1 w 1 arcsecswd 5 E dt for w ≥ 1 ! 2 2 0 1 t FIGURE 5 FIGURE 6 as it results in a subtle improper integral calculation.) The formulas for the inverse cosine, cotangent, and cosecant can be obtained by similar arguments. Area and the arcsine Let us now study the arcsine as an area. Let 0 < y < 1 be given and let A denote the area bounded by the line xstd 5 s!1 2 y2yyd ? t, the circle xstd 5 !1 2 t2, and the x-axis t 5 0 (see FIGURE 6). Since arc length along the edge of the unit circle equals twice the area of the sector determined by the arc, we have that arcsinsyd 5 2 ? A y !1 2 y2 5 2 ? E1!1 2 t2 2 ? t2 dt 0 y y 5 ? E ! 2 2 2 ! 2 2 2 1 t dt y 1 y . (2) 0 Differentiating (2), we have y2 D sarcsinsydd 5 !1 2 y2 1 y !1 2 y2 1 5 , !1 2 y2 4 as before. Alternatively, our integral expression (2) for arcsin(y) provides a motivating example for the integration by parts formula eu ? dv 5 u ? v 2 ev ? du. Letting u 5 !1 2 t2 and dv 5 dt, we have 2t E!1 2 t2 dt 5 t!1 2 t2 2 Et dt !1 2 t2 s!1 2 t2d2 2 1 5 t!1 2 t2 2 E dt !1 2 t2 1 5 t!1 2 t2 2 E!1 2 t2 dt 1 E dt !1 2 t2 so 1 2E!1 2 t2 dt 5 t!1 2 t2 1 E dt. !1 2 t2 But then y s d 5 ? E ! 2 2 2 ! 2 2 arcsin y 2 1 t dt y 1 y 0 y 1 5 y!1 2 y2 1 E dt 2 y!1 2 y2 ! 2 2 0 1 t y 1 5 E dt, ! 2 2 0 1 t as before. This development of the arcsine as the area bounded by the x-axis, the circle xstd 5 !1 2 t2 from t 5 0 to t 5 y, and the line from the origin to the point s!1 2 y2, yd may be adapted to the hyperbola xstd 5 !1 1 t2 to find the “inverse hyperbolic sine” sinh21syd 5 ln|!1 1 y2 1 y|. However, unlike the circle, there is no simple relation between arc length and area for the hyperbola (see[4], §6 and 7; for a representation of arc length on the hyperbola, see [1], §60, and the corresponding geometric construction is §61). On the other hand, a similar procedure to develop a “lemniscate sine” using the lemniscate instead of the circle must be carried out in terms of arc length (see [4], §8). Rationalization of the arcsine We now turn to the problem of expressing the integrand in (1) as a rational function; that is, we wish to express 1 2 t2 as the square of a rational function.

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