. Differential and integral calculus, an introductory course for colleges and engineering schools. er, be derived from the power series, and from this relation andfrom the equation of the circle, x2 + y2 = a2, we could derive byinspection the equations x = a cos 8, y = a sin 6, which are parametric equations of the circle. The geometricmeaning of the parameter 6 would remain to be ascertained. Now in just this way we may connect the hyperbolic func-tions with the equilateral hyperbola. Being given the relationch2 0 — sh2 0=1, and the equation of the curve, x2 — y2 = a2, wecan write down at onc
. Differential and integral calculus, an introductory course for colleges and engineering schools. er, be derived from the power series, and from this relation andfrom the equation of the circle, x2 + y2 = a2, we could derive byinspection the equations x = a cos 8, y = a sin 6, which are parametric equations of the circle. The geometricmeaning of the parameter 6 would remain to be ascertained. Now in just this way we may connect the hyperbolic func-tions with the equilateral hyperbola. Being given the relationch2 0 — sh2 0=1, and the equation of the curve, x2 — y2 = a2, wecan write down at once the equations x = a ch 6, y = a sh 0, and in these we have a pair of parametric equations of the geometric meaning of the parameter 0 will be determinedpresently. Consider next the unit circle, x2 + y2 = 1, and let u be thearea of the shaded sector, whose arc is 0. Then u = | 0, 0 = 2 u, * In Peirces Short Table of Integrals may be found a collection of for-mulae relating to hyperbolic functions. 396 and CALCULUS x = cos 6 = cos 2 u, y = sin 6 = sin 2 w, - = tan 2 w, 255 P(x,y). and here we have x, y, and - ex- x pressed as trigonometric functions of the double area 2 u. Following the method presently to be employed with the hyperbolic functions, we shall derive these very simple relations again by means of definite integrals. We have u = area OAPB — ^ OAPB = H x dy = / Vl - y2 dy Jy=0 JO = - (yVl - y2 + sin1^) = -xy + ^sin-^. Hence 1 . sin-1 y and y = sm2u. Further, = VI x = v i — y2 = Vl — sin2 2 u = cos 2 u, and we have the same expressions for x and y as before, but wehave obtained them here without making any use of the repre-sentation of 6 as an angle. Consider now the equilateral hyperbola x2 — y2 = 1, and let ube the area of the shaded sector. Then u = area OAPB — ~ OAPB = f^x dy = f Vl + y2 dy Jy=0 JO = \ [2/Vl + y2 + log (y + VF+y2)]= ^ xy + ^ log (y + vT+l/2). §255Therefore THE HYPERBOLIC FUNCTIONS u = g log (y + Vl + i/2) =
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Keywords: ., bookcentury1900, bookdecade1910, booksubjectcalculu, bookyear1912
