a aA 
~— 
“TOF DE t D+ 2 
va c = a A- Id 
ead 7a 
A sel 
series, but it is incomplete, because it contains only 
one arbitrary constant ity. Let us next consider 
the case of » = 1, and putting A’, B’, C’, &e. for the 
ing values of the coefficients A, B, C, &e. of 
which the SEE ET: eae il 
—aA’ ot 
/_ a See 
P= Gea) (2) “ ba: 
c re ham Al = 
(1+2)* 
—_—a@ 
1.2. $(n+8)(2n-+5) (847) (242)5 
“This 'eivéel's aeodnid Hcbanglati : aie, 
liy adding theltwo valties of 4, we get aa 
yr=At Bo cet Dee t 
ae Certs petty yas 
i 191. 
a (art. 19 gr ; 
D’= 
+ &e. 
But then we may assume y= az, and the equation 
owill become a(a—1) + a@=0, which gives two values 
of a. Let them be a and 2’. The two incomplete 
fluents are y= “x and y= «'z", and their sum y=«2" 
64 
+-«'x is the complete primitive equation. See Euler 
Instit. Calc. Integ. vol. ii. cap. 7 ; Lacroix Traite du Calc. 
Diff. &c. Part i. chap. 6. This last work, as it is the 
latest and most copious, may. i i 
ing the most complete ‘view 
of any extant, 
Miscellaneous Problems. 
193. Having now given as full a view of the prin- 
ciples of the Reseed erlesig as we conceive to be 
compatible with the nature of our work, we shall con- 
clude with a few more examples of its application to 
the resolution of parti lems. 
__ Pros. 1. To find the length of the enlarged meri- 
dian in Mercator’ s, or rather Wright's projection of the 
FLUXIONS. 
of e and z, then, by the, principles of the projection, —Miscelle 
be considered as exhibit- ; 
: 468 
rad 
jet se Reale therefore, putting the ratio of the fluxions Problems. 
v cos.0 i 
. > gr wa? ee , z 1 
Pane Oe peat Peep eee odes eae 
apd dt oe Therefore, taking the fluent (art. 
146), 2=1. { tan. (45°44 of. Pui no correction 
is wanted, because when v= 0, then z=]. (tan. 45°) 
=I. (1) = 0, as it should be, 
If we compare the fluxion of z with that of the dif- 
ference of a parabolic arc and its tangent, (art. 158,) it 
will appear that they; are identical. Hence we have 
this elegant theorem... If, from the focus of a parabo- 
la, a dicular. be ‘drawn to any tangent to the 
curve, and a circle be Pn sem ee are as a cen- 
tre to throngh vertex ; the meridional 8 
SE ee to the are of the circle between Bote 
tex, and the perpendicular to the tangent, is equal to 
the excess of the parabolic arc between the vertex and 
point of contact, above that portion of the tangent 
which is intercepted between the same point and per- 
Henry Bond, in the 1650, discovered, by chance, 
that the eniarged meridian might be expressed by the 
logarithmic tangents of half the complements of the la- 
titudes, a rule easily found from the preceding solution; 
but the difficulty of proving this was then considered so 
great, that Mercator off to wager a sum of money 
against any person that should undertake to prove it, 
either true or false. James Gregory, however, proved 
it in his Exercitationes Geometric, published in 1668, 
and afterwards Barrow, in his Geometrical Lectures ; 
their demonstrations, however, were intricate. After- 
wards Dr Wallis and Dr Halley gave demonstrations, 
which were sufficiently simple saidalagent 
Pros. 2. A body T proceeds uniformly 
straight line BC, (Fig. 43), and a body S in pursuit of T, 
moves always direct! 'y towards it, with a velocity which 
is to that of T in the given ratio of 1 to n ;“wHat is the 
nature of the curve described by S? 
Let the tangent AB, which makes right angles with 
BC, be put=a, the abscissa BR=z, the ordinate SR=y, 
dz 
the are AS = =, then the subtangent BASVa, 
: - 
(art 67,) and BT =e — 9 Now BT and AS being 
described in the same time, they are to each other as 
the velocities n and 1, therefore BT = n x AS, that is 
c— vs =n z, and hence, taking the fluxions, making 
dy constant, — y = ndz; butdz=,/(d2x* + dy?), 
therefore 
ndy _ ae 
yy ode+dyy 
Put dz=pdy, then d?x=dp dy; hence, by sub- 
stitution in the second member, 
ady___ ap" 
y  A(+py 
Bid Ging eat tila i 
cml (el fr+ vit pt. 
along a Fig. 43. 
