INFINITESIMAL CALCULUS where r represents the area of the corresponding hyperbolic sector, represented by APl l j. Also, by analogy, we write sinh v c" - C v c 2 " - 1 . tana v= , &c. cosh v c v + c " c- v -(- 1 Again, we obviously have cos x = cosh ix , i sin x = sinh ix . Between hyperbolic functions many relations exist analogous to those between ordinary trigonometrical functions. For example, it is easily seen that we have cosh 2 ic-sinh*a;==l , cosh (x + y) = cosh x cosh y + sinh x sinh y , sinh (a: + 1/) = sinh x cosh y + cosli x sinh y, cosh 3x= 4 cosh 3 a: - 3 cosh x , sinh 3a:=4 siuh 3 ^ + 3 sinh a; . -y- sinh x= cosh x , -r- cosh a; = einh x , ax dx d I d 1 -j- tanh x= j-g- , r- coth x*= . . / cosh xdx = sinh a; , / sinh x dx = cosh x , /* (7 t /"" d~c I ^75- =tanh a; , / -r-4-r- = ~ c th a; . ./ cosn-,c J suih 2 ^ Hence The analogy between hyperbolic and trigonometrical functions will also appear as follows. If we make 2: = sec in the equation of the equilateral hyperbola x- - y- = l, we get y = ta.n <f>. Conse quently sec = cosh v. tan </> = sinh v. Moreover the equation v log(x + y}, gives 111 this case <p is called the hyperbolic amplitude of v ; and, by analogy, we write = amh v. Also when x = cosh v, we have v = lo" (x + Va: 2 - 1 ). Again, when y = sinh v, we have v = log (y + V?/ 2 + 1 ). Moreover, since -y is the sector whose corresponding hyperbolic cosine is .r, the connexion between v and x may be represented by the relation 7;=- sect cosh x. Hence we have sect cosh a; = log (x + Va; 2 - 1 ) . In like manner we get sect sinh x = log (a; + Va: 2 + 1) , sect tanh x= log l-x This notation exhibits the analogy between the elementary integrals in a remarkable manner, and even more strikingly when we intro duce the Continental notation, arc sin x, instead of sin- ^x, &c. Thus i~ dx , I- = sin - l x arc sin x ; J Vl-se 2 I J 1 /" A - = tan ~ l x = arc tan x ;
- 2
1 + a? - - = sect tanh a; . -Efc. 3. To find the area included between the cissoid of Diocles and its asymptote. The equation of this curve is if-(1a-x} = x and that of its asymptote is a: = 2rt. Hence the area in question is represented by Let x=2a sin- 6, and the integral becomes Ex. 4. To find the whole area of the curve Hence the whole area is represented by 2 sin"+ ] cos"- 1 ^^. yo The method of determining this integral has been exhibited in 164. In the preceding examples the area of any portion of a plane may be conceived as divided into a system of infinitesimal rectangles, dxdy, by lines drawn parallel to the axes of coordinates. Accord ingly any plane area may be represented by ffdxdif, taken be tween limits determined by the boundary of the area. Again, as in polar coordinates, the plane may be divided by a system of circles having the origin as centre, and also by a system of radii vectores drawn through the origin. In such coordinates the clement of area bounded by two circles of radii r and r + dr and by the radii vectores corresponding to the angles 6 and + d6 is plainly represented by rdrdO. Accordingly, any plane area may be represented by ffrdrde taken between the limits determined by its boundary. Hence, if the equation of a curve be given in polar coordinate*, the sectorial area S bounded by two radii, and the curve is repre sented by where o and are the values of corresponding to the limiting radii. (1) For example, in the class of spirals represented by r = aO m , we have Q a? 2 +i S -a- - -f const. 2 2m + 1 If the area be bounded by the prime vector = 0, this gives f/ ? 02m+l
- 2 2m + 1
Thus for the spiral of Archimedes, whose equation is r=a0, f ft & S = 3 =-- 6 " 6a In the spiral, r 2 = 0, we have S = - In the reciprocal to this, spiral, viz., r 2 = 2 , we have in which the sector is reckoned from = 1 (2) To find the area of a loop of the curve 7-2 = ft 2 cos H Q_ Here r = when nO = -^- , and r = a when 7(0 = 0. Consequently the area of a loop is represented by / a" I "" cos M0 c^. o ^ and, accordingly, is . It is easily seen that when n is a posi tive integer, the curve consists of n loops ; accordingly the entire area of the curve is a 2 . (3)_To find the area of the loop of the folium of Descartes, the equation of the curve being a,- 3 + y 3 = 3axi/. Transforming to polar coordinates, we get sin 2 _9o 2 /*2 sin 2 2y (sir? S Let tan = u, and this becomes ^L r _ 165. If from any point a perpendicular be drawn to any tangent to a curve, the locus of the foot of the perpendicular is called the pedal of the curve with respect to the assumed origin. If p and a> be the polar coordinates of the foot of the perpendi cular, the sectorial area of the pedal curve is plainly represented by taken between proper limits. The following remarkable connexion between the pedal areas with respect to the same closed curve, for different internal origins, is due to Steiner. Let A be the area of the pedal with respect to the origin 0, A the area for origin , and p, !> the corresponding perpendiculars, then we have
