FLUXIONS. 447 
c for the constant quantity, which is the circular are of which the radius is a and the sine _ Inverse 
Inverse 
z, and this arc, supposing AD a quadrant, is PD ; let Method. 
alues, and 
pr to obrlete the foest, and we have 
== 
Fig. 10. 
ig. 9 
dz 2 d?z 23 
frteaetee- iF at dz 23 —~ *- 
This is Bernoulli’s Theorem. It may be otherwise 
readily deduced from Taylor's theorem, If in the de- 
velopement we make z = 0, all the terms, with the ex- 
ception of the constant quantity, vanish, therefore c is 
the value of the fluent when z = 0. 
d : 
Examp.e. To find the fiuent [== (a + x) by 
ate@ 
this method: When 2=0, then 1. (a + x) becomes 
1, (a), therefore c= 1. (a): And because z = 
dz —1 © dtz 2 
therefore dx (a+z2)?’ dx? (a+ 
1 
atx’ 
js? Oe ad 
z = x3 
Liepal=} (9) +iye +3(apayt Hapa 
&e. 
This series differs from that found in last article, but 
the one may be transformed into the other. 
Quadrature of Curves. 
150. Let «= AQ, (Fig. 14.) be the abscissa of a 
curve, y= PQ Phaaped ‘abe 2 Ards the ane CEOP, 
com ded between the indefinite ordinate PQ, and 
CE an ordinate having a given position. We have 
found, (art. 71.) that ds =yd-z, so that s=fyd «, 
We are now to apply this formula to particular exam- 
Exampre 1. Let the curve be a parabola, (Fig. 10.) 
in this case, a being put for the parameter, y* =a 2, 
and y=a* x4, andds= dz=a* x4ds, therefore, 
Coie tes deans 118) 
s=fyde=ia a* hemsyepe. 
Let us suppose the fluent to begin at the vertex A; 
then, when x= 0 we to have s=0. The ge- 
neral equation of the in this case becomes 
0=0-+6¢, therefore c=0; and sos =}y-x, which 
with what was found in Conic Sections, (Sect. 
IL. Prop.1.) If s were supposed to begin when z had 
some given value 5, we would have O=3a?4* 40; 
then c= —30%6*, ands=}a4 (2*¥—53), 
Ex. 2. Let the curve be a cizcle, (Fig. 9.) and sup- 
ely of the co-ordinates z = and y= 
to be at O the centre; put a for the radius. © By 
the nature of the sant hak An mprate therefore, 
ydz=dx/ (a*—x*): Apply to this fluxion the for- 
mula fudt=ut— ftdu, (making 4/ (a?— 2*) =u, 
and «== 1?) and we have 
edz 
da (at—x*)=2/ (a — x*) + 7 amar 14 
fe eda ST ray 
: io. natde 
Again, Waa) = 
bringing 
i teal 
es to a common de- 
nominator ; therefore, substivuting, and putting zy for 
ad : 
a? — xt) 
Now, (Art. 114, Rule V.) the fluent of a 
it be denoted by z, and we have 
ae af His daz+e. 
This is the complete fluent, which holds true, what- 
ever be its origin. Let us now suppose that the fluent 
begins when «= 0; in this case z= 0, and as then 
$= 0, it follows that c= 0; hence 
s (=area ODPQ)=42cy+ haz. 
Now if we join OP, the triangle POQ> }.ry, there- 
fore the sector POD = jaz. 
Ex. 3. Let the curve be an ellipse, (Fig. 11.) of Fig. 11. 
which the semitransverse and semiconjugate axes are a 
and 4; suppose the origin of the co-ordinates to be at 
A, one extremity of the transverse. The equation of 
the curve in? = (202 —24); hence 
safyds a dz @a2r—at), 
If a circle be described on the transverse axis as a dia« 
meter, and y’ be put for pQ, the ordinate of the circle, 
corresponding to the common ‘abscissa x, and s’ for the 
circular area AQp, we have y’? = 4/(2 ax —x*), and 
& = fy de=fde/ (2ax—2°). 
Hence, the variable parts of these two fluents have to 
each other the constant ratio of to 1, ot of 6 to a, and 
as they begin together, the fluents themselves must have 
the same ratio, that is s: s':: 6: a, as was shewn in 
Conic Sections, (Sect. VII.) 
Ex. 4. Let the curve be an 
25.) and AX, AY its asym ; and let it be re- 
uired to find the area included by the hyperbolic arc 
P, the straight lines CE, PQ, which are parallel to 
one asymptote, and EQ, the nt of the other 
acymptote, intercepted between Let CE, one 
the Is, have a given position. Put AE =a, 
EC = 6, AQ= 2, QP = y. By the nature of the curve, 
b 
zy = ab, hence y = “~~, and 
tafyds=ab f= 0b1(2) +c, 
By the nature of the problem, when z= a, then s= 0," 
therefore in this case,0 =abl. (a) +c, hence c= 
—abl.(a), and 
s=ab fl (z)—1(@)} =.a51. (=). 
If we suppose a = 5 = 1, then s =], (=), hence 
the area s is expressed by the Napierean logarithm of 
the abscissa AQ, and these areas serve as a geometrical 
representation of Napier’s logarithms. On this account 
they were called by the early writers hyperbolic loga- 
rithms, but improperly, as any logarithms whatever 
may be represented by hyperbolic-areas: See LoGa- 
RITHMS, 
Ex. 5. Let the hyperbolic area PAQ be required; sup: 
ing PQ to be an ordinate to CA, the Aa cra’ am 
(Fig. 26.) Let C, the centre, be the origin of the 
ordinates ; put CQ=., PQ= y, 
= a, the semiconjugate = d, 
the semitransverse axis 
e equation of the curve 
(Conic Sections, Sect. VIII.) is y= by (t— a); 
hence 
3 
equilateral hyperbola, (Fig. Fig, 25 
CO- Fig. 26, 
