ON THE TRIGO>fOME'rRY OF THE PARABOLA. 



83 



The corresponding arc of the parabola will be given by the following series: — 



n(?n . e)=2/«ri + — + — ^ + &C.1, 



^ ^ L 123 ^ 12:345 ^ 1234.567 J 



123 ' 12345 1234567 



since the subtangent in this case is equal to m sec e tan e= — (^^ — e~^)- 



4 



XXIII. If we now extend this inquiry, and ask what is the magnitude of the 

 amplitude of the arc of the parabola which shall render the difference between 

 this parabolic arc and its subtangent equal to n times the distance between 

 the focus and the vertex, we shall have, as before, by the terms of the question, 



n(m .6) — m sec tan d=nm. 

 But, in general, n(m. 6) -m see 6 tan d=m^ sec Odd; 

 hence we must have 



n=: rsec6rf0 = log(sec9+tan6), or sec0+ tan 0=e". 



Now we may solve this equation in two ways; either by making n a given 

 number, and then determine the value of sec + tan d, which may be called 

 the base; or we may assign an arbitrary value to sec 0+ tan 9, and then 

 derive the value of fi. Taking the latter course, let, for example, 



secO + tan0=lO, then«=loglO; 



or putting S for this angle, secS+tang=10 (27) 



Hence as every number whose logarithm is to be exhibited must be put 

 under the form sec 0+ tan 0, which is of the form l+x, since the limiting 

 value of sec is 1, we discover the reason why in developing the logarithm 

 of a number, the number itself must be put under the form 1+x, and not 

 simply under that of x. 



XXIV. Given a number to find its logarithm, may be exhibited by the fol- 



Fig. 5. 



lowing geometrical construction 



Let SVP be a parabola. Through 

 the focus S draw the perpendicular SQ 

 to the axis VS. Through V let a tan- 

 gent of indefinite length be drawn, 

 which may be called the scalar. On this 

 tangent take the line VN to represent 

 the given number. Join NS, and make 

 the angle NST always equal to the 

 angle NSQ. Draw TP at right angles 

 to TS. This line will touch the para- 

 bola in the point P, and the arc of the 

 parabola VP diminished by the sub- 

 tangent PT, or the tangential difference 

 for the arc VP, will be the logarithm 

 of VN. 



The line SN makes the angle (— + -] 



with the axis of the parabola. 



When SN'=VS= the unit m, the angle N'SQ is equal to half a right 

 angle. Hence the point T in this case will coincide with V. The parabolic 

 arc therefore vanishes, or the logarithm of I is 0. When sec 0+ tan = 1, 

 0=0. 



When the number is less than 1, the point N will fall below N' in the 

 position n. Hence wSQ is greater than half a right angle. Therefore T 

 will fall hehw the axis in the point T' ; and if we draw through T' a tangent 



g2 



