페이지 이미지
PDF

bers of this inequality approach a common value, their limit; that is, ,

^ ^ \x0, when t' == t.

Hence also

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[merged small][ocr errors]
[ocr errors]
[ocr errors]

Q. E. D. (Cf. Art. 43.)

53. Area of a Hyperbolic Sector. Let the perpendicular p be dropped from any point P, of the equilateral hyperbola x* y2 = a2, upon its asymptote, meeting the

latter in S, and let OS = The following properties of this hyperbola are well known and are proved in elementary works on conic sections:

sp = a" / 2,

x-y=PV2, _

x=(s+p)/v/2,

y = (s-p) /V 2.

[graphic]
[ocr errors]
[ocr errors]

(4) . Deduce these formulae also geometrically from the constructions of Arts. 48, 53, assuming for the definitions of sinh u and cosh u the ratios NP / a and ON I a. [Burnside: Messenger of Mathematics, vol. xx, pp. 145-148.]

(5) . In the figure of Art. 53 show that the trapezoids SaAPsS,, S,P^P%S,, etc., are equal in area to the corresponding triangles OAP,, OP^P,, etc., and consequently to each other.

(6) . Show that when the hyperbolic sector OAP (Art. 53) increases uniformly, the corresponding segment OS, laid off on the asymptote, increases proportionately to its own length.

(7) . Assuming a=i in the equilateral hyperbola of Art. 48, and that the area of any sector is \ u, prove that

j)TM1^ [(smn «)/*]= 1. (Use the method of Art. 55.)

55. An Approximate Value of Natural Base.

We may determine between what integers the numerical value of e must lie, by substituting their equivalents in x and y for the terms of the inequality:

Triangle OAP > sector OA VP > triangle ORS, * as represented in Fig. 28. For p our present purpose it will involve no loss of generality and it will simplify the computation to assume OA 1, so that the equation of the hyperbola is x' —f 1. The sectorial area OA VP, as 0 R A N previously found in Art. 53, is

28 then \u, the area of OAP is

obviously \ y, and for that of ORS we may write

\ OR X (ordinate of S).

[graphic]

To determine OR and this ordinate, write the equations to the tangent RS and the line OP, and find the ordinate of their intersection, and the intercept of the former on the ^r-axis. The results are:

, V = x~^Tr $ ~ V 2/(*-l),

the equation to the tangent RS, f and rj being the current co-ordinates of the line;

the equation to OP;

0R==y/2{x-i)
y

the required intercept on the x-axis; and

[ocr errors]

* Had the assumption a — 1 not been made, this inequality would have been

The equations for RSax\<\ OP and the expressions for OR and V would have been correspondingly changed, but the final results would have been the same as those given above.

« 이전계속 »